Mechanics of Materials in SI Units. Solutions Manual [10 ed.] 9781292178202

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1–1. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A.

0.3 m A 30� 0.1 m

80 N

45�

Solution Equations of Equilibrium: +

Q©Fx¿ = 0;

NA - 80 cos 15° = 0 Ans.

NA = 77.3 N a+ ©Fy¿ = 0;

VA - 80 sin 15° = 0 Ans.

VA = 20.7 N a+

©MA = 0;

MA + 80 cos 45°(0.3 cos 30°) - 80 sin 45°(0.1 + 0.3 sin 30°) = 0 MA = - 0.555 N # m

Ans.

or a+

©MA = 0;

MA + 80 sin 15°(0.3 + 0.1 sin 30°) -80 cos 15°(0.1 cos 30°) = 0 MA = - 0.555 N # m

Ans.

Negative sign indicates that MA acts in the opposite direction to that shown on FBD.

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished. Ans:

NA = 77.3 N, VA = 20.7 N, MA = -0.555 N # m

1

CH 01.indd 2

11/29/10 10:18:17 AM

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1–2. Determine the resultant internal loadings on the cross section at point D.

C

1m

F

2m

1.25 kN/m

Solution

A

Support Reactions: Member BC is the two force member. 4 a+ ΣMA = 0;   FBC (1.5) - 1.875(0.75) = 0 5

D

E

0.5 m 0.5 m 0.5 m

B

1.5 m

FBC = 1.1719 kN + c ΣFy = 0;  Ay +

4 (1.1719) - 1.875 = 0 5

             Ay = 0.9375 kN + ΣFx = 0; S

3 (1.1719) - Ax = 0 5              Ax = 0.7031 kN

Equations of Equilibrium: For point D + ΣFx = 0;  ND - 0.7031 = 0 S            ND = 0.703 kN

Ans.

+ c ΣFy = 0;  0.9375 - 0.625 - VD = 0             VD = 0.3125 kN

Ans.

a+ ΣMD = 0;  MD + 0.625(0.25) - 0.9375(0.5) = 0 MD = 0.3125 kN # m

Ans.

Ans: ND = 0.703 kN, VD = 0.3125 kN, MD = 0.3125 kN # m 2

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1–3. 

C

1m

F

2m

1.25 kN/m

Solution

A

Support Reactions: Member BC is the two-force member. 4 a+ ΣMA = 0;   FBC (1.5) - 1.875(0.75) = 0 5

D

E

0.5 m 0.5 m 0.5 m

B

1.5 m

FBC = 1.1719 kN + c ΣFy = 0;  Ay +

4 (1.1719) - 1.875 = 0 5

Ay = 0.9375 kN 3 + ΣFx = 0;   (1.1719) - Ax = 0 S 5 Ax = 0.7031 kN Equations of Equilibrium: For point F + bΣFx′ = 0;  NF - 1.1719 = 0 Ans.

NF = 1.17 kN a + ΣFy′ = 0;    VF = 0

Ans.

a+ ΣMF = 0;      MF = 0

Ans.

Equations of Equilibrium: For point E + ΣFx = 0;  NE - 3 (1.1719) = 0 d 5 Ans.

NE = 0.703 kN + c ΣFy = 0;  VE - 0.625 +

4 (1.1719) = 0 5 Ans.

VE = - 0.3125 kN a+ ΣME = 0;  - ME - 0.625(0.25) +

4 (1.1719)(0.5) = 0 5

ME = 0.3125 kN # m

Ans.

Negative sign indicates that VE acts in the opposite direction to that shown on FBD.

Ans: NF = 1.17 kN, VF = 0, MF = 0, NE = 0.703 kN, VE = - 0.3125 kN, ME = 0.3125 kN # m 3

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*1–4. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Determine the resultant internal loadings acting on the cross section at C.

600 N/m A

B

D

C

1m

1m

1m

1.5 m

1.5 m 900 N

Solution Support Reactions: We will only need to compute By by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a. a+ ΣMA = 0;

By(4.5) - 600(2)(2) - 900(6) = 0

By = 1733.33 N

Internal Loadings: Using the result of By, section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b, + ΣFx = 0; S

Ans.

NC = 0

+ c ΣFy = 0;

VC - 600(1) + 1733.33 - 900 = 0

a+ ΣMC = 0;

1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0

VC = -233 N MC = 433 N # m

Ans.

Ans.

The negative sign indicates that VC acts in the opposite sense to that shown on the free-body diagram.

Ans: NC = 0, VC = - 233 N, MC = 433 N # m 4

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1–5. Determine the resultant internal loadings in the beam at cross sections through points D and E. Point E is just to the right of the 3-kip 15-kNload. load.

315kip kN 1.5 kip/ ft 25 kN/m

A

D 62 ft m

26m ft

B

E

4 ftm 1.5

4 ftm 1.5

C

Solution Support Reactions: For member AB a + ©MB = 0;

50(4/3) = 0= 0 9.00(4) –-AA y(4) y(12)

+ : ©Fx = 0; + c ©Fy = 0;

kN Ayy ==16.67 3.00 kip

Bx = 0 By ++ 16.67 3.00 –- 50 9.00 = 0= 0

Byy == 33.33 6.00 kip kN

Equations of Equilibrium: For point D + : ©Fx = 0; + c ©Fy = 0;

Ans.

ND = 0 3.00 -– 12.5 2.25 –-VV 16.67 DD==0 0

Ans.

V kN 0.750 kip VDD == 4.17 a + ©MD = 0;

2 - 3.00(6) MD ++ 12.25( 2.25(2) 0 3 ) – 16.67(2) = 0

kN #· ftm MD == 25.17 13.5 kip

Ans.

Equations of Equilibrium: For point E + : ©Fx = 0; + c ©Fy = 0;

Ans.

NE = 0 –33.33 6.00 -– 15 3 -– V VE = 00 E=

Ans.

- 9.00 kN kip VEE == –48.33 a + ©ME = 0;

ME ++33.33(1.5) 6.00(4) ==00 M kN #· ft m ME = –50.00 - 24.0 kip

Ans.

Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD. 1 2 (25)(4)

8 3  m

15 kN

= 50 kN

4 3  m

2m

1 2 (12.5)(2)

2m

4 3  m

= 12.5 kN

33.33 kN

15 kN

1.5 m

2 3  m

16.67 kN

Ans: ND = 0, VD = 4.17 kN,

MD = 25.0 kN # m, NE = 0, VE = -48.3 kN, ME = -50.0 kN # m 5

CH 01.indd 3

11/30/10 9:39:58 AM

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1–6. The shaft is supported by a smooth thrust bearing at B and a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E.

B

A

1m

C

E

1m

1m

D

1m

1800 N 3600 N

Solution Support Reactions: We will only need to compute Cy by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a. a + ©MB = 0;

Cy(2) + 1800(1) - 3600(3) = 0

Cy = 4500 N

Internal Loadings: Using the result for Cy, section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b, + : ©Fx = 0;

NE = 0

+ c ©Fy = 0;

VE + 4500 - 3600 = 0

Ans. VE = - 900 N

Ans.

a + ©ME = 0; 4500(1) - 3600(2) - ME = 0 ME = -2700 N . m = -2.70 kN . m

Ans.

The negative signs indicates that VE and ME act in the opposite sense to that shown on the free-body diagram.

Ans: NE = 0, VE = -900 N, ME = -2.7 kN .m 6

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1–7. Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 2000-N load is applied along the centroidal axis of the member.

a

b 30⬚

2000 N

2000 N

b

A a

(a) + : ©Fx = 0;

Na - 2000 = 0 Na = 2000 N

Ans.

+ T©Fy = 0;

Va = 0

Ans.

R+ ©Fx = 0;

Nb - 2000 cos 30° = 0

(b)

Nb = 1732 N +Q©F = 0; y

Ans.

Vb - 2000 sin 30° = 0 Vb = 1000 N

Ans.

Ans: Na = 2000 N, Va = 0, Nb = 1732 N, Vb = 1000 N 7

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0.2 m 0.2 m

*1–8. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at G.

0.4 m

E B

0.6 m

G

F 0.3 m

H

C

D

0.5 m

A

75�

Solution Support Reactions: We will only need to compute FEF by writing the moment equation of equilibrium about D with reference to the free-body diagram of the hook, Fig. a. a + ©MD = 0;

FEF(0.3) - 600(9.81)(0.5) = 0

FEF = 9810 N

Internal Loadings: Using the result for FEF, section FG of member EF will be considered. Referring to the free-body diagram, Fig. b, + : ©Fx = 0; + c ©Fy = 0; a + ©MG = 0;

9810 - NG = 0

NG = 9810 N = 9.81 kN

Ans. Ans.

VG = 0

Ans.

MG = 0

Ans: NG = 9.81 kN, VG = 0, MG = 0 8

CH 01.indd 13

11/29/10 10:18:28 AM

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1–9. The floor crane is used to lift a 600-kg concrete pipe. Determine the resultant internal loadings acting on the cross section at H.

0.2 m 0.2 m

0.4 m G

E B

0.6 m F 0.3 m

H

C

D

0.5 m

A

75�

Solution Support Reactions: Referring to the free-body diagram of the hook, Fig. a. a + ©MF = 0;

Dx(0.3) - 600(9.81)(0.5) = 0

Dx = 9810 N

+ c ©Fy = 0;

Dy - 600(9.81) = 0

Dy = 5886 N

Subsequently, referring to the free-body diagram of member BCD, Fig. b, a + ©MB = 0;

FAC sin 75°(0.4) - 5886(1.8) = 0

+ : ©Fx = 0;

FAC = 27 421.36 N

Bx + 27 421.36 cos 75° - 9810 = 0 Bx = 2712.83 N

+ c ©Fy = 0;

27 421.36 sin 75° - 5886 - By = 0

By = 20 601 N

Internal Loadings: Using the results of Bx and By, section BH of member BCD will be considered. Referring to the free-body diagram of this part shown in Fig. c, + : ©Fx = 0;

NH + 2712.83 = 0

NH = -2712.83 N = - 2.71 kN

Ans.

+ c ©Fy = 0;

- VH - 2060 = 0

VH = -20601 N = - 20.6 kN

Ans.

a + ©MD = 0;

MH + 20601(0.2) = 0

MH = - 4120.2 N # m = -4.12 kN # m Ans.

The negative signs indicates that NH, VH, and MH act in the opposite sense to that shown on the free-body diagram.

Ans: NH = -2.71 kN, VH = -20.6 kN, MH = -4.12 kN # m 9

CH 01.indd 14

11/29/10 10:18:29 AM

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1–10. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point C. Assume the reactions at the supports A and B are vertical.

4 kN/m

A

B D

C 1.5 m

3m

1.5 m

Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0;  By(6) -

1 (4)(6)(2) = 0 2

By = 4.00 kN

Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through C, Fig. b, + ΣFx = 0;     NC = 0 S + c ΣFy = 0;    VC + 4.00 a+ ΣMC = 0;     

4.00(4.5) -

Ans. 1 (3)(4.5) = 0 2

Ans.

VC = 2.75 kN

1 (3)(4.5)(1.5) - MC = 0   2

MC = 7.875 kN # m

Ans.

Ans: NC = 0, VC = 2.75 kN, MC = 7.875 kN # m 10

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1–11. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point D. Assume the reactions at the supports A and B are vertical.

4 kN/m

A

B D

C 1.5 m

3m

1.5 m

Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0;  By(6) -

1 (4)(6)(2) = 0 2

By = 4.00 kN

Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through D, Fig. b, + ΣFx = 0;     ND = 0 S

Ans.

+ c ΣFy = 0;    VD + 4.00 a+ ΣMD = 0;      4.00(1.5) -

1 (1.00)(1.5) = 0 2

VD = -3.25 kN

Ans.

1 (1.00)(1.5)(0.5) - MD = 0   2

MD = 5.625 kN # m

Ans.

The negative sign indicates that VD acts in the sense opposite to that shown on the FBD.

Ans: ND = 0, VD = - 3.25 kN, MD = 5.625 kN # m 11

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*1–12. a

The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section a–a that passes through point D.

225 mm 30 b B

A D b F E

Solution

a

150 mm F C

Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, + ΣFx = 0; d

Na - a + 100 = 0

+ c ΣFy = 0;

Va - a = 0

a+ ΣMD = 0; - Ma - a - 100(0.15) = 0

Na - a = -100 N

Ans. Ans.

Ma - a = -15 N # m

Ans.

The negative sign indicates that Na - a and Ma - a act in the opposite sense to that shown on the free-body diagram.

Ans: Na - a = -100 N, Va - a = 0, Ma - a = - 15 N # m 12

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1–13. a

The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section b–b that passes through point D.

225 mm 30 b B

A D b F E

Solution

a

150 mm F C

Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, ΣFx′ = 0;

Nb - b + 100 cos 30° = 0   Nb - b = - 86.6 N

Ans.

ΣFy′ = 0;

Vb - b = 50 N Vb - b - 100 sin 30° = 0    

Ans.

a+ ΣMD = 0;

Mb - b = -15 N # m - Mb - b - 100(0.15) = 0    

Ans.

The negative sign indicates that Nb–b and Mb–b act in the opposite sense to that shown on the free-body diagram.

Ans: Nb - b = - 86.6 N, Vb - b = 50 N, Mb - b = -15 N # m 13

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1–14. The boom DF of the jib crane and the column DE uniformweight wieghtofof the and hoistload andweigh load have aauniform 50750 lb/ft.N/m. If theIfhoist weigh N, determine the resultant internal in loadings in 300 lb, 1500 determine the resultant internal loadings the crane thecross cranesections on crossthrough sectionspoints through points A, B, and C. on A, B, and C.

D

2.4 8 ftm

2 ft 1.5 5 ftm

F

A

B

0.9 3 ftm

0.6 m C 1500 N 300 lb

2.1 7 ftm

Solution

E

Equations of Equilibrium: For point A + ; © Fx = 0; + c © Fy = 0;

0.750(0.9) = 0.675 kN

NA = 0

Ans.

150 300 == 00 VA –-0.675 – 1.500

0.45 m 0.45 m

VA == 2.175 450 lbkN a + ©MA = 0;

Ans.

1.500 kN

1.500(0.9) –M MAA –-0.675(0.45) 150(1.5) -– 300(3) = 0= 0 0.750(3.3) = 2.475 kN

MA = -1.65 kN . m

Ans.

Negative sign indicates that MA acts in the opposite direction to that shown on FBD. 1.65 m

Equations of Equilibrium: For point B

1.500 kN

+ ; © Fx = 0;

NB = 0

+ c © Fy = 0;

550 -– 1.5 300==0 0 VBB –- 2.475

Ans. 0.750(3.9) = 2.925 kN

VB = 3.975 kN a + © MB = 0;

1.65 m

Ans.

–M 1.500(3.3) -MBB–-2.475(1.65) 550(5.5) -– 300(11) = =0 0

0.750(1.5) = 1.125 kN

MB = -9.03 kN . m

Ans.

1.95 m

1.95 m 1.500 kN

Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ; © Fx = 0; + c © Fy = 0;

VC = 0

Ans.

–N -NCC –-1.125 250 –- 2.925 650 -– 1.500 300 ==00 kNlb = - 1.20 kip - 1200 NCC == –5.55

a + ©MC = 0;

Ans.

–M 1.500(3.9) MCC –-2.925(1.95) 650(6.5) -– 300(13) = =0 0 MC = -11.6 kN . m

Ans.

Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD.

Ans: NA  0, VA  2.175 kN, MA  1.65 kN · m, NB  0, VB  3.975 kN, MB  9.03 kN · m, VC  0, NC  5.55 kN, MC  11.6 kN · m 14

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1–15. The metal stud punch is subjected to a force of 120 N on the handle. Determine the magnitude of the reactive force at the pin A and in the short link BC. Also, determine the resultant internal loadings acting on the cross section at point D.

120 N

60

50 mm 100 mm E

B

30

50 mm D

100 mm 300 mm

A C

Solution

200 mm

Member: FBC cos 30°(50) - 120(500) = 0

a+ΣMA = 0;

Ans.

FBC = 1385.6 N = 1.39 kN + c ΣFy = 0;

Ay - 1385.6 - 120 cos 30° = 0          Ay = 1489.56 N

+ ΣFx = 0; d

         Ax - 120 sin 30° = 0;    Ax = 60 N

FA = 21489.562 + 602

 

Ans.

= 1491 N = 1.49 kN

Segment: a+ ΣFx′ = 0;

ND - 120 = 0 Ans.

                  ND = 120 N +Q ΣFy′ = 0; a+ ΣMD = 0;

Ans.

VD = 0            MD - 120(0.3) = 0                   MD = 36.0 N # m

Ans.

Ans: FBC = 1.39 kN, FA = 1.49 kN, ND = 120 N, VD = 0, MD = 36.0 N # m 15

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*1–16. Determine the resultant internal loadings acting on the cross section at point E of the handle arm, and on the cross section of the short link BC.

120 N

60

50 mm 100 mm E

B

30

50 mm D

100 mm 300 mm

A C

Solution

200 mm

Member: a+ ΣMA = 0;     FBC cos 30°(50) - 120(500) = 0 FBC = 1385.6 N = 1.3856 kN Segment: +bΣF = 0; x′

Ans.

        NE = 0

a + ΣFy′ = 0;       VE - 120 = 0;       VE = 120 N

Ans.

a+ ΣME = 0;      ME - 120(0.4) = 0;  ME = 48.0 N # m

Ans.

Short link: + ΣFx = 0;       V = 0 d

Ans.

+ c ΣFy = 0;        1.3856 - N = 0;    N = 1.39 kN

Ans.

a+ ΣMH = 0;   M = 0

Ans.

Ans: NE = 0, VE = 120 N, ME = 48.0 N # m, Short link: V = 0, N = 1.39 kN, M = 0 16

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1–17. The forged steel clamp exerts a force of F = 900 N on the wooden block. Determine the resultant internal loadings acting on section a–a passing through point A.

200 mm F � 900 N

a

30�

F � 900 N

A

a

Solution Internal Loadings: Referring to the free-body diagram of the section of the clamp shown in Fig. a, ©Fy¿ = 0;

900 cos 30° - Na - a = 0

Na - a = 779 N

Ans.

©Fx¿ = 0;

Va - a - 900 sin 30° = 0

Va - a = 450 N

Ans.

a + ©MA = 0;

900(0.2) - Ma - a = 0

Ma - a =

180 N # m

Ans.

Ans: Na-a = 779 N, Va-a = 450 N,

900(0.2) - Ma-a = 0, Ma-a = 180 N # m 17

CH 01.indd 12

11/29/10 10:18:26 AM

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1–18. Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is >ft22 acts N/m fixed to to the the ground groundand andaauniform uniformpressure pressureofof500 7 lb perpendicular to the face of the sign.

z ft 3m ft 2m

ft 3m 22

7 lb/ft 500 N/m

ft 6m B

Solution

A

©Fx = 0;

(VB)x –- 7500 105 = 0; 0;

7500lbN = 7.5 kN (VB)x = 105

©Fy = 0;

(VB)y = 0

Ans.

©Fz = 0;

(NB)z = 0

Ans.

ft 4m

Ans.

©Mx = 0;

(MB)x = 0

Ans.

©My = 0;

# ft= 56.25 kN · m ·m (MB)y –- 7500(7.5) 105(7.5) = 0;  0; (M(M = 788Nlb B)y B=)y56250

Ans.

©Mz = 0;

(TB)z 105(0.5) = 0;  0; (TB(T )z 3750 = 52.5 ft 3.75 kN · m – 7500(0.5) )zB= N ·lbm# =

Ans.

x

y

0.5 m

500(5)(3) = 7500 N

7.5 m

Ans: (VB)x = 7.5 kN, (VB)y = 0, (NB)z = 0, (MB)x = 0, (MB)y = 56.25 kN # m,

(TB)z = 3.75 kN # m 18

CH 01.indd 16

11/30/10 9:39:59 AM

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1–19. Determine the resultant internal loadings acting on the cross section at point C in the beam. The load D has a mass of 300 kg and is being hoisted by the motor M with constant velocity.

2m

2m

2m 0.1 m

0.1 m E

C

A

B 1m

1.5 m D

M

Solution + ΣFx = 0;   NC + 2.943 = 0;    NC = - 2.94 kN d

Ans.

+ c ΣFy = 0;   VC - 2.943 = 0;      VC = 2.94 kN

Ans.

a+ ΣMC = 0;   - MC - 2.943(0.6) + 2.943(0.1) = 0             MC = - 1.47 kN # m

Ans.

Ans: NC = -2.94 kN, VC = 2.94 kN, MC = -1.47 kN # m 19

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*1–20. Determine the resultant internal loadings acting on the cross section at point E. The load D has a mass of 300 kg and is being hoisted by the motor M with constant velocity.

2m

2m

2m 0.1 m

0.1 m E

C

A

B 1m

1.5 m D

M

Solution + ΣFx = 0;   NE + 2943 = 0 S +T

Ans.

NE = - 2.94 kN ΣFy = 0;   - 2943 - VE = 0

Ans.

VE = - 2.94 kN a+ ΣME = 0;    ME + 2943(1) = 0

ME = - 2.94 kN # m

Ans.

Ans: NE = - 2.94 kN, VE = - 2.94 kN, ME = - 2.94 kN # m 20

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1–21. Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A, and the jaws at B are smooth.

20 N

40 mm

120 mm

15 mm C A

B

D 80 mm 20 N

30�

Solution + c ©Fy = 0; + : ©Fx = 0; +d ©MC = 0;

- VC + 60 = 0;

Ans.

VC = 60 N

Ans.

NC = 0 - MC + 60(0.015) = 0;

Ans.

MC = 0.9 N.m

Ans: VC = 60 N, NC = 0, MC = 0.9 N # m 21

CH 01.indd 9

11/29/10 10:18:23 AM

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1–22. Determine the resultant internal loading on the cross section through point D of the pliers.

20 N

40 mm

120 mm

15 mm C A

B

D 80 mm 20 N

30�

Solution R+ ©Fy = 0;

VD - 20 cos 30° = 0;

VD = 17.3 N

Ans.

+ b©Fx = 0;

ND - 20 sin 30° = 0;

ND = 10 N

Ans.

+d ©MD = 0;

MD - 20(0.08) = 0;

MD = 1.60 N.m

Ans.

Ans: VD = 17.3 N, ND = 10 N, MD = 1.60 N # m 22

CH 01.indd 9

11/29/10 10:18:23 AM

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1–23. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section at point C. The 400-N forces act in the –z direction and the 200-N and 80-N forces act in the +y direction. The journal bearings at A and B exert only y and z components of force on the shaft.

z

400 mm 150 mm 150 mm 200 mm 200 mm 300 mm

80 N 80 N 200 N

Support Reactions:

A

160(0.4) + 400(0.7) - Ay(1.4) = 0

x

Ay = 245.71 N ΣFy = 0;

y

D

C

Solution ΣMz = 0;

B

200 N 400 N 400 N

- 245.71 - By + 400 + 160 = 0 By = 314.29 N

ΣMy = 0;

800(1.1) - Az(1.4) = 0

Az = 628.57 N

ΣFz = 0;

Bz + 628.57 - 800 = 0

Bz = 171.43 N

Equations of Equilibrium: For point C ΣFx = 0; ΣFy = 0;

Ans.

(NC)x = 0 - 245.71 + (VC)y = 0

Ans.

(VC)y = - 246 N ΣFz = 0;

ΣMx = 0; ΣMy = 0;

ΣMz = 0;

628.57 - 800 + (VC)z = 0 (VC)z = - 171 N

Ans.

(TC)x = 0

Ans.

(MC)y - 628.57(0.5) + 800(0.2) = 0 (MC)y = - 154 N # m

Ans.

(MC)z - 245.71(0.5) = 0

(MC)z = - 123 N # m

Ans.

Ans: (NC)x = 0, (VC)y = -246 N, (VC)z = -171 N, (TC)x = 0, (MC)y = -154 N # m, (MC)z = -123 N # m 23

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* 1.24.

FF  N � 400 80 lb

The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a–a.

a

30 30�

5.75 0.23mm in. A 4 mm 0.16 in.

45 45� a

Solution Equations of Equilibrium: For section a–a +

Q©Fx¿ = 0;

VA –-400 80 cos 15° = 0 Ans.

VA = 386.37 77.3 lb N a+ ©Fy¿ = 0;

NA –-400 80 sin 15° = 0 Ans.

103.53 NA = 20.7 lb N a + ©MA = 0;

MAA –-400 80 sin sin 15°(0.004) 15°(0.16) ++ 400 80 cos = 0= 0 –M cos15°(0.23) 15°(0.00575) 1.808lbN# in. ·m MA = 14.5

Ans.

400 N

0.00575 m

0.004 m

Ans: VA = 386.37 N, NA = 103.53 N, MA = 1.808 N . m 24

CH 01.indd 7

11/29/10 10:18:20 AM

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1–25. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section at point D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y direction. The journal bearings at A and B exert only y and z components of force on the shaft.

z

400 mm 150 mm 150 mm 200 mm 200 mm 300 mm

80 N 80 N 200 N

Support Reactions:

A

160(0.4) + 400(0.7) - Ay (1.4) = 0

x

Ay = 245.71 N ΣFy = 0;

y

D

C

Solution ΣMz = 0;

B

200 N 400 N 400 N

- 245.71 - By + 400 + 160 = 0 By = 314.29 N

ΣMy = 0;

800(1.1) - Az(1.4) = 0

Az = 628.57 N

ΣFz = 0;

Bz + 628.57 - 800 = 0

Bz = 171.43 N

Equations of Equilibrium: For point D ΣFx = 0; ΣFy = 0;

Ans.

(ND)x = 0 (VD)y - 314.29 + 160 = 0

Ans.

(VD)y = 154 N ΣFz = 0;

ΣMx = 0; ΣMy = 0;

ΣMz = 0;

171.43 + (VD)z = 0 (VD)z = - 171 N

Ans.

(TD)x = 0

Ans.

171.43(0.55) + (MD)y = 0

(MD)y = - 94.3 N # m

Ans.

314.29(0.55) - 160(0.15) + (MD)z = 0 (MD)z = - 149 N # m

Ans.

Ans: (ND)x = 0, (VD)y = 154 N, (VD)z = - 171 N, (TD)x = 0, (MD)y = - 94.3 N # m, (MD)z = - 149 N # m 25

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1–26. The serving tray T used on an airplane is supported on each side by an arm. The tray is pin connected to the arm at A, and at B there is a smooth pin. (The pin can move within the slot in the arms to permit folding the tray against the front passenger seat when not in use.) Determine the resultant internal loadings acting on the cross section of the arm through point C when the tray arm supports the loads shown.

12 N 9N 15 mm B 60⬚

100 mm A

150 mm

T

500 mm

VC

C

MC NC

b+ ©Fx = 0;

NC + 9 cos 30° + 12 cos 30° = 0;

NC = - 18.2 N

Ans.

a+ ©Fy = 0;

VC - 9 sin 30° - 12 sin 30° = 0;

VC = 10.5 N

Ans.

a + ©MC = 0; - MC - 9(0.5 cos 60° + 0.115) - 12(0.5 cos 60° + 0.265) = 0 MC = - 9.46 N

#

m

Ans.

Ans: NC = - 18.2 N VC = 10.5 N MC = - 9.46 N 26

#

m

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1–27. z

The pipe has a mass of 12 kg>m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B.

300 N

400 N A C x

Solution

ΣFy = 0; ΣFz = 0;

2m

2m

Internal Loadings: Referring to the FBD of the right segment of the pipe assembly sectioned through B, Fig. a, ΣFx = 0;

1m

B

(VB)x + 300 = 0 4 (NB)y + 400 + 500 a b = 0 5

(VB)x = -300 N

Ans.

(NB)y = -800 N

Ans.

5

3 4

y 500 N

3 (VB)z - 2 3 12(2)(9.81) 4 - 500 a b = 0 5

(VB)z = 770.88 N = 771 N  Ans.

ΣMx = 0;

3 (MB)x - 12(2)(9.81)(1) - 12(2)(9.81)(2) - 500 a b(2) 5 (MB)x = 2106.32 N # m = 2.11 kN # m

ΣMy = 0;

(TB)y + 300(2) = 0

ΣMz = 0;

(MB)z - 300(2) = 0

- 400(2) = 0 Ans.

(TB)y = -600 N # m (MB)z = 600 N # m

Ans. Ans.

The negative signs indicates that (VB)x, (NB)y, and (TB)y act in the sense opposite to those shown in the FBD.

Ans: (VB)x = - 300 N, (NB)y = - 800 N, (VB)z = 771 N, (MB)x = 2.11 kN # m, (TB)y = - 600 N # m, (MB)z = 600 N # m 27

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*1–28. The brace and drill bit is used to drill a hole at O. If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A.

z

O x

�150 30 lb FFxx N

A

225 mm 9 in. 3 mm in. 225 75 9 in. mm 150 mm 6 in. 150 mm 6 in. 150 mm 6 in.

� 50 10 N lb Fz  �250 50 lb FFyy N y

Solution Internal Loading: Referring to the free-body diagram of the section of the drill and brace shown in Fig. a, ©Fx = 0; ©Fy = 0; ©Fz = 0; ©Mx = 0; ©My = 0; ©Mz = 0;

30 =500 A VA B x - 150

250=500 A NA B y - 50

30 lb N A VA B x 5= 150

Ans.

= 50 10 N lb A VA B z 5

Ans.

50 lb N A NA B y 5= 250

50 5 = 00 A VA B z - 10

Ans.

22.5 lb N #· ft m A MA B x5= 33.75

50(0.675)=500 A MA B x - 10(2.25)

Ans.

= 33.75 22.5 lbN# ·ftm A TA B y 5

30(0.75) = 5 00 A TA B y - 150(0.225)

Ans.

- 37.5 lb N# ·ftm A MA B z 5= 256.25

30(1.25) = 5 00 A MA B z + 150(0.375)

Ans.

The negative sign indicates that (MA)Z acts in the opposite sense to that shown on the free-body diagram.

0.15 m 150 N 0.225 m

0.15 m 0.225 m

0.225 m

50 N

0.15 m

250 N

Ans: (VA)x = 150 N, (NA)y = 250 N, (VA)z = 50 N, (MA)x = 33.75 N # m, (TA)y = 33.75 N # m, (MA)z = −56.25 N # m

28

CH 01.indd 18

11/29/10 10:18:34 AM

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1–29. The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determine the resultant internal loadings acting on the cross section at point B. Hint: The distance from the centroid C of segment AB to point O is CO = 0.9745r.

D

B

r C

45 22.5

O 90

A

Solution p rw = 0; 4

ΣFz = 0;

VB -

ΣFx = 0;

NB = 0

ΣMx = 0;

TB -

ΣMy = 0;

MB +

p rw(0.09968r) = 0; 4 p rw(0.3729 r) = 0; 4

Ans.

VB = 0.785 w r

Ans. TB = 0.0783 w r 2

Ans.

MB = -0.293 w r 2

Ans.

Ans: VB = 0.785wr, NB = 0, TB = 0.0783wr 2, MB = - 0.293wr 2 29

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1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = -N, dM>du = - T, and dT>du = M.

M  dM V  dV

T  dT N  dN

M V N

du

T

Solution ΣFx = 0; N cos

du du du du + V sin - (N + dN) cos + (V + dV) sin = 0 2 2 2 2

(1)

ΣFy = 0; N sin

du du du du - V cos + (N + dN) sin + (V + dV) cos = 0 2 2 2 2

(2)

ΣMx = 0; T cos

du du du du + M sin - (T + dT) cos + (M + dM) sin = 0 2 2 2 2

(3)

ΣMy = 0; T sin

du du du du - M cos + (T + dT) sin + (M + dM) cos = 0 2 2 2 2

Since

du du du du is can add, then sin = , cos = 1 2 2 2 2

Eq. (1) becomes Vdu - dN +

(4)

dVdu = 0 2

Neglecting the second order term, Vdu - dN = 0 dN = V QED du Eq. (2) becomes Ndu + dV +

dNdu = 0 2

Neglecting the second order term, Ndu + dV = 0 dV = - N QED du Eq. (3) becomes Mdu - dT +

dMdu = 0 2

Neglecting the second order term, Mdu - dT = 0 dT = M QED du Eq. (4) becomes Tdu + dM +

dTdu = 0 2

Neglecting the second order term, Tdu + dM = 0 dM = - T QED du

Ans: N/A 30

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  The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90° 2 .

P

P u A

Solution Equations of Equilibrium: R + ΣFx = 0;

V - P cos u = 0

V = P cos u

Q + ΣFy = 0;

N - P sin u = 0

N = P sin u

Average Normal Stress and Shear Stress: Area at u plane, A′ =  savg =

N P sin u P = = sin2 u A A′ A sin u

tavg =

V P cos u = A A′ sin u =

A . sin u Ans.

P P sin u cos u = sin 2u A 2A

Ans.

Ans: savg = 31

P P sin2 u, tavg = sin 2u A 2A

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*1–32. 4 kN

The built-up shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points.

B

A

6 kN

C

8 kN

6 kN E

D

Solution At D: sD =

P = A

4(103) p 2 4 (0.028

- 0.02 2)

Ans.

= 13.3 MPa (C)

At E: sE =

P = A

8(103) p 4

(0.012 2)

Ans.

= 70.7 MPa (T)

Ans: sD = 13.3 MPa, sE = 70.7 MPa 32

CH 01.indd 22

11/29/10 10:18:36 AM

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1–33. The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the glue can withstand a maximum average shear stress of 800 kPa, determine the maximum allowable clamping force F.

50 mm 45�

F glue 25 mm

F

Solution Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;

F cos 45° - V = 0

V =

2 F 2

Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain

tavg

V ; = A

2 F 2 800(10 ) = 1.25(10 - 3) 3

Ans.

F = 1414 N = 1.41 kN

Ans: F = 1.41 kN 33

CH 01.indd 45

11/29/10 10:19:24 AM

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1–34. The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the clamping force is F = 900 N, determine the average shear stress developed in the glued shear plane.

50 mm 45�

F glue 25 mm

F

Solution Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. + : ©Fx = 0;

900 cos 45° - V = 0

V = 636.40 N

Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3)m2. We obtain tavg =

V 636.40 = 509 kPa = A 1.25(10 - 3)

Ans.

Ans. V = 636.40 N, tavg = 509 kPa 34

CH 01.indd 45

11/29/10 10:19:24 AM

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1–35. Determine the largest intensity w of the uniform loading that can be applied to the frame without causing either the average normal stress or the average shear stress at section b–b to exceed s = 15 MPa and t = 16 MPa, respectively. Member CB has a square cross section of 30 mm on each side.

B

b

w

b 3m

Solution

C

A

Support Reactions: FBD(a) 4m

4 a+ ΣMA = 0;    FBC (3) - 3w(1.5) = 0   FBC = 1.875w 5 Equations of Equilibrium: For section b–b, FBD(b) + ΣFx = 0;    4 (1.875w) - Vb - b = 0   Vb - b = 1.50w S 5 3 + c ΣFy = 0;    (1.875w) - Nb - b = 0   Nb - b = 1.125w 5 Average Normal Stress and Shear Stress: The cross-sectional area of section b–b, 5A ; where A = (0.03)(0.03) = 0.90 ( 10-3 ) m2. A′ = 3 5 Then A′ = (0.90) ( 10-3 ) = 1.50 ( 10-3 ) m2. 3 Assume failure due to normal stress. (sb - b)Allow =

Nb - b ′

A

;

15 ( 106 ) =

1.125w 1.50 ( 10-3 )

w = 20000 N>m = 20.0 kN>m Assume failure due to shear stress. (tb - b)Allow =

Vb - b A′

;

16(106) =

1.50w 1.50(10-3)

w = 16000 N>m = 16.0 kN>m (Controls !)

Ans.

Ans: w = 16.0 kN>m (Controls !) 35

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*1–36. The supporting wheel on a scaffold is held in place on the leg using a 4-mm-diameter pin. If the wheel is subjected to a normal force of 3 kN, determine the average shear stress in the pin. Assume the pin only supports the vertical 3-kN load.

Solution

+ c ΣFy = 0;     3 kN # 2V = 0;     V = 1.5 kN

tavg

3 kN

1.5 ( 103 ) V = = p = 119 MPa 2 A 4 (0.004)

Ans.

Ans: tavg = 119 MPa 36

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1–37. If P = 5 kN, determine the average shear stress in the pins at A, B, and C. All pins are in double shear, and each has a diameter of 18 mm.

0.5 m

6P

3P

P 1.5 m

P 2m

1.5 m

0.5 m

B 5

C

3

A

4

Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, 3 5(0.5) + 30(2) + 15(4) + 5(5.5) - FBC a b(6) = 0 5

a+ ΣMA = 0;

FBC = 41.67 kN

a+ ΣMB = 0;

Ay(6) - 5(0.5) - 15(2) - 30(4) - 5(5.5) = 0  Ay = 30.0 kN

+ ΣFx = 0;     S

4 41.67 a b - Ax = 0 5

Thus,

             Ax = 33.33 kN

FA = 2 Ax2 + Ay2 = 233.332 + 30.02 = 44.85 kN

Average Shear Stress: Since all the pins are subjected to double shear, FBC 41.67 VB = VC = = kN = 20.83 kN (Fig. b) and VA = 22.42 kN (Fig. c) 2 2 For pins B and C 20.83 ( 103 ) VC = = 81.87 MPa = 81.9 MPa p A ( 0.0182 ) 4 ( 22.42 103 ) VA = tA = = 88.12 MPa = 88.1 MPa p A ( 0.0182 ) 4 tB = tC =

Ans.

Ans.

Ans: tB = tC = 81.9 MPa, tA = 88.1 MPa 37

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1–38. Determine the maximum magnitude P of the loads the beam can support if the average shear stress in each pin is not to exceed 80 MPa. All pins are in double shear, and each has a diameter of 18 mm.

0.5 m

6P

3P

P 1.5 m

P 2m

1.5 m

0.5 m

B 5

C

3

A

4

Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0;

3 P(0.5) + 6P(2) + 3P(4) + P(5.5) - FBC a b(6) = 0 5 FBC = 8.3333P

a+ ΣMB = 0;

Ay(6) - P(0.5) - 3P(2) - 6P(4) - P(5.5) = 0  Ay = 6.00P

+ ΣFx = 0;      S

4 8.3333P a b - Ax = 0 5

Thus,

                     Ax = 6.6667P

FA = 2 Ax2 + Ay2 = 2(6.6667P)2 + (6.00P)2 = 8.9691P

Average Shear Stress: Since all the pins are subjected to double shear, FBC 8.3333P VB = VC = = = 4.1667P (Fig. b) and VA = 4.4845P (Fig. c). Since 2 2 pin A is subjected to a larger shear force, it is critical. Thus tallow =

VA ; A

80 ( 106 ) =

4.4845P p ( 0.0182 ) 4

P = 4.539 ( 103 ) N = 4.54 kN

Ans.

Ans: P = 4.54 kN 38

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1–39. Determine the average normal stress in each of the 20-mm-diameter bars of the truss. Set P = 40 kN.

C

P

1.5 m

B

A

Solution Internal Loadings: The force developed in each member of the truss can be determined by using the method of joints. First, consider the equilibrium of joint C, Fig. a,

2m

4 + ΣFx = 0;           40 - FBC a b = 0           FBC = 50 kN (C) S 5

3 + c ΣFy = 0;           50 a b - FAC = 0 5

           FAC = 30 kN (T)

Subsequently, the equilibrium of joint B, Fig. b, is considered 4 + ΣFx = 0;          50 a b - FAB = 0           FAB = 40 kN (T) S 5

Average Normal Stress: The cross-sectional area of each of the bars is A =

p (0.022) = 0.3142(10 - 3) m2. We obtain, 4 50(103) FBC (savg)BC = = = 159 MPa A 0.3142(10 - 3)

Ans.

(savg)AC =

30(103) FAC = = 95.5 MPa A 0.3142(10 - 3)

Ans.

(savg)AB =

40(103) FAB = = 127 MPa A 0.3142(10 - 3)

Ans.

Ans: (savg)BC = 159 MPa, (savg)AC = 95.5 MPa, (savg)AB = 127 MPa 39

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*1–40. If the average normal stress in each of the 20-mm-diameter bars is not allowed to exceed 150 MPa, determine the maximum force P that can be applied to joint C.

C

P

1.5 m

B

A

Solution Internal Loadings: The force developed in each member of the truss can be determined by using the method of joints. First, consider the equilibrium of joint C, Fig. a. 4 + ΣFx = 0;          P - FBC a b = 0            FBC = 1.25P(C) S 5 + c ΣFy = 0;

3          1.25P a b - FAC = 0 5

2m

        FAC = 0.75P(T)

Subsequently, the equilibrium of joint B, Fig. b, is considered. + ΣFx = 0; S

4            1.25P a b - FAB = 0 5

       FAB = P(T)

Average Normal Stress: Since the cross-sectional area and the allowable normal stress of each bar are the same, member BC, which is subjected to the maximum normal force, is the critical member. The cross-sectional area of each of the bars is p A = (0.022) = 0.3142(10 - 3) m2. We have, 4 (savg)allow =

FBC 1.25P ;   150(106) = A 0.3142(10 - 3) P = 37 699 N = 37.7 kN

Ans.

Ans: P = 37.7 kN 40

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  Determine the maximum average shear stress in pin A of the truss. A horizontal force of P = 40 kN is applied to joint C. Each pin has a diameter of 25 mm and is subjected to double shear.

C

P

1.5 m

B

A

Solution Internal Loadings: The forces acting on pins A and B are equal to the support reactions at A and B. Referring to the free-body diagram of the entire truss, Fig. a, ΣMA = 0;

By(2) - 40(1.5) = 0            

            By = 30 kN

+ ΣFx = 0; S

40 - Ax = 0

                 Ax = 40 kN

2m

+ c ΣFy = 0;        30 - Ay = 0                  Ay = 30 kN Thus, FA = 2Ax2 + Ay 2 = 2402 + 302 = 50 kN

Since pin A is in double shear, Fig. b, the shear forces developed on the shear planes of pin A are VA =

FA 50 = = 25 kN 2 2

p Average Shear Stress: The area of the shear plane for pin A is AA = (0.0252) = 4 0.4909(10 - 3) m2. We have (tavg)A =

25(103) VA = = 50.9 MPa AA 0.4909(10 - 3)

Ans.

Ans: (tavg)A = 50.9 MPa 41

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1–42. The pedestal has a triangular cross section as shown. If it is subjected to a compressive force of 2250 N, specify the x and y coordinates for the location of point P(x, y), where the load must be applied on the cross section, so that the average normal stress is uniform. Compute the stress and sketch its distribution acting on the cross section at a location removed from the point of load application.

2250 N 300 mm

P(x,y) y

x

75 mm 150 mm

Solution x

 2 (75)  1 (75)(300) +  75 + 1 (150)  1 (150)(300) 3 3    100 mm = 2 2 1 (75)(300) + 1 (150)(300) 2 2

Ans.

y=

1 (300 3

Ans.

σ=

P = A

mm) = 100 mm

2250

Ans.

1 (0.225)(0.3) 2 3

= 66.67(10 ) N/m 2 = 66.7 kPa 75 mm + 1 (150 mm ) 3

2 (75 3

mm )

1 (300 3

mm )

Ans: x = 100 mm, y = 100 mm, s = 66.7 kPa 42

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  The plate has a width of 0.5 m. If the stress distribution at the support varies as shown, determine the force P applied to the plate and the distance d to where it is applied.

4m d

P

x

s  (15x1/2) MPa

30 MPa

Solution The resultant force dF of the bearing pressure acting on the plate of area dA = b dx = 0.5 dx, Fig. a, 1

1

dF = sb dA = (15x2)(106)(0.5dx) = 7.5(106)x2 dx L

+ c ΣFy = 0;

dF - P = 0 L0

4m

1

7.5(106)x 2 dx - P = 0

P = 40(106) N = 40 MN

Ans.

Equilibrium requires L

a+ ΣMO = 0; L0

4m

xdF - Pd = 0 1

x[7.5(106)x2 dx] - 40(106) d = 0 Ans.

d = 2.40 m

Ans: P = 40 MN, d = 2.40 m 43

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*1–44.

27 kN

The joint is subjected to the axial member force of 27 kN. Determine the average normal stress acting on sections AB and BC. Assume the member is smooth and is 40 mm thick. 60⬚ A

C

40 mm

20⬚ 112 mm

B

Solution + c ©Fy = 0;

- 27 sin 60° + NBC cos 20° = 0 NBC = 24.88 kN

+ : ©Fx = 0;

NAB - 27 cos 60° - 24.88 sin 20° = 0 NAB = 22.01 kN

sAB =

NAB 22.01 (103) = = 13.76(106) N/m2 = 13.8 MPa AAB (0.04)(0.04)

Ans.

sBC =

NBC 24.88 (103) = = 5.554(106) N/m2 = 5.55 MPa (0.04)(0.112) ABC

Ans.

Ans: sAB = 13.8 MPa, sBC = 5.55 MPa 44

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1–45. The plastic block is subjected to an axial compressive force of 600 N. Assuming that the caps at the top and bottom distribute the load uniformly throughout the block, determine the average normal and average shear stress acting along section a–a.

600 N

a 150 mm 30 a 50 mm

Solution

50 mm 50 mm

Along a–a:

600 N

+bΣFx = 0;  V - 600 sin 30° = 0 V = 300 N +R ΣFy = 0;     - N + 600 cos 30° = 0 N = 519.6 N sa - a = ta - a =

519.6 (0.05) 1 cos0.130° 2 300

(0.05) 1 cos0.130° 2

= 90.0 kPa

Ans.

= 52.0 kPa

Ans.

Ans: sa - a = 90.0 kPa, ta - a = 52.0 kPa 45

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1–46. The column is made of concrete having a density of 2.30 Mg>m3. At its top B it is subjected to an axial compressive force of 15 kN. Determine the average normal stress in the column as a function of the distance z measured from its base.

z

15 kN B

180 mm

Solution

4m

+ c ΣFy = 0     P - 15 - 9.187 + 2.297z = 0

z

P = 24.187 - 2.297z s =

P 24.187 - 2.297z = = (238 - 22.6z) kPa A p(0.18)2

Ans.

x

y

Ans: s = (238 - 22.6z) kPa 46

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1–47. If P = 15 kN, determine the average shear stress in the pins at A, B, and C. All pins are in double shear, and each has a diameter of 18 mm.

0.5 m C 30 B

P

4P 1m

4P 1.5 m

1.5 m

2P 0.5 m

A

Solution For pins B and C: tB = tC = For pin A:

82.5 ( 103 ) V = p 18 2 = 324 MPa A 4 1 1000 2

Ans.

FA = 2(82.5)2 + (142.9)2 = 165 kN tA =

82.5 ( 103 ) V = p 18 2 = 324 MPa A 4 1 1000 2

Ans.

Ans: tB = 324 MPa, tA = 324 MPa 47

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* 1–48. The driver of the sports car applies his rear brakes and causes the tires to slip. If the normal force on each rear tire is 1800 N and the coefficient of kinetic friction between the tires and the pavement is µ k = 0.5, determine the average shear stress developed by the friction force on the tires. Assume the rubber of the tires is flexible and each tire is filled with an air pressure of 225 K P 1800 N

Solution From the tire pressure, p =

N ; A

225(103) =

1800 A

A = 0.008 m2

The friction is F = mkN = 0.5(1800) = 900 N Then the shear stress is tavg =

F 900 = = 112.5(103) N/m2 = 112.5 KPa 0.008 A

Ans.

Ans: tavg = 112.5 K Pa 48

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1–49. The beam is supported by two rods AB and CD that have cross-sectional areas of 12 mm2 and 8 mm2, respectively. If d = 1 m, determine the average normal stress in each rod.

B

D 6 kN d

A

3m

C

Solution a+ ΣMA = 0;  FCD(3) - 6(1) = 0 FCD = 2 kN + c ΣFy =     F 0; AB - 6 + 2 = 0 FAB = 4 kN sAB =

4 ( 103 ) FAB = = 333 MPa AAB 12 ( 10-6 )

Ans.

sCD =

2 ( 103 ) FCD = = 250 MPa ACD 8 ( 10-6 )

Ans.

Ans: sAB = 333 MPa, sCD = 250 MPa 49

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1–50. The beam is supported by two rods AB and CD that have cross-sectional areas of 12 mm2 and 8 mm2, respectively. Determine the position d of the 6-kN load so that the average normal stress in each rod is the same.

B

D 6 kN d

A

3m

C

Solution a+ ΣMO = 0;  FCD(3 - d) - FAB(d) = 0 s =

(1)

FCD FAB = 12 8 (2)

FAB = 1.5 FCD From Eqs. (1) and (2), FCD(3 - d) - 1.5 FCD(d) = 0 FCD(3 - d - 1.5 d) = 0 3 - 2.5 d = 0

Ans.

d = 1.20 m

Ans: d = 1.20 m 50

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1–51. The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at its center. If it is rotating in the horizontal plane at a constant angular rate of v, determine the average normal stress in the bar as a function of x.

L 2

L 2 V x

Solution Equation of Motion: + ΣFx = maN; d

1 1 N = m c (L - 2x) d v2 c (L + 2x) d 2 4 =

mv2 2 ( L - 4x2 ) 8

Average Normal Stress: s =

N mv2 2 ( L - 4x2 ) = A 8A

Ans.

Ans: s = 51

mv2 2 ( L - 4x2 ) 8A

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37.5 1.5 mm in.

*1–52. The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 2 kN.

4 kN 800 lb

30 30�

251 mm in.

4 kN 800 lb

251 mm in.

30 30�

Solution a+ ©Fy = 0;

N sin 30° = 0; = 0; N –-2 400 sin 30°

kN lb N ==1200

+Q©Fx = 0;

2400 coscos 30°30° – V-=V0; = 0;

kNlb V ==1.732 346.41

= A′

(0.0375)(0.025) = 1.875(10 −3 ) m 2 sin 30° 2 kN

N 1(10 3 ) = σ = = 533 kPa A′ 1.875(10 −3 ) m 2

= t

Ans.

2 kN

Ans.

V 1.732(10 3 ) = = 924 kPa A′ 1.875(10 −3 ) m 2

Ans:

σ= 533 kPa t 924 kPa = 52

CH 01.indd 26

11/29/10 10:18:39 AM

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1–53. P

The pier is made of material having a specific weight g. If it has a square cross section, determine its width w as a function of z so that the average normal stress in the pier remains constant. The pier supports a constant load P at its top where its width is w1.

w1 w1 z w

w L

Solution Assume constant stress s1, then at the top, P s1 = w1 2

(1)

For an increase in z the area must increase, dA =

g A dz g dW dA =   or   = dz s1 s1 s1 A

For the top section: A

g z dA = dz s1 L0 LA1 A In

g A = z s1 A1 A = A1e ( s ) z g

1

A = w2 A1 = w 21 w = w1 e ( 2 s ) z g

1

From Eq. (1), 2

w = w1 e

cw 1g d z 2P

Ans.

Ans: 2

w = w1 e 53

cw 1g d z 2P

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  The 2-Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the average normal stress in the cables. The diameters of AB and AC are 12 mm and 10 mm, respectively.

A

30

45 C

Solution

B

G

Internal Loadings: The normal force developed in cables AB and AC can be determined by considering the equilibrium of the hook for which the free-body diagram is shown in Fig. a. ΣFx′ = 0;  2000(9.81) cos 45° - FAB cos 15° = 0  FAB = 14 362.83 N (T) ΣFy′ = 0;  2000(9.81) sin 45° - 14 362.83 sin 15° - FAC = 0  FAC = 10 156.06 N (T) Average Normal Stress: The cross-sectional areas of cables AB and AC are p p AAB = (0.0122) = 0.1131(10 - 3) m2  and  AAC = (0.012) = 78.540(10 - 6) m2 . 4 4 We have sAB =

FAB 14 362.83 = = 127 MPa AAB 0.1131(10 - 3)

Ans.

sAC =

FAC 10 156.06 = = 129 MPa AAC 78.540(10 - 6)

Ans.

Ans: sAB = 127 MPa, sAC = 129 MPa 54

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1–55. The 2-Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the diameter of cable AB so that the average normal stress in this cable is the same as in the 10-mm-diameter cable AC.

A

30

45 C

Solution

B

G

Internal Loadings: The normal force in cables AB and AC can be determined by considering the equilibrium of the hook for which the free-body diagram is shown in Fig. a. ΣFx′ = 0; 2000(9.81) cos 45° - FAB cos 15° = 0

FAB = 14 362.83 N (T)

ΣFy′ = 0; 2000(9.81) sin 45° - 14 362.83 sin 15° - FAC = 0 FAC = 10 156.06 N (T) Average Normal Stress: The cross-sectional areas of cables AB and AC are p p AAB = d AB2 and AAC = (0.012) = 78.540(10 - 6) m2. 4 4 Here, we require sAB = sAC FAC FAB = AAB AAC 14 362.83 10 156.06 = p 2 -6 d 78.540(10 ) 4 AB Ans.

d AB = 0.01189 m = 11.9 mm

Ans: d AB = 11.9 mm 55

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*1–56. Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the 3 kN force is applied to the ring at  B, determine the angle u so that the average normal stress in each rod is equivalent. What is this stress?

C

5 4

3

B

A u 3 kN

Solution Method of Joints: Referring to the FBD of joint B, Fig. a, 3 + c ΣFy = 0;  FBC a b - 3 cos u = 0 FBC = 5 cos u kN 5 + ΣFx = 0; S

4 (5 cos u)a b - 3 sin u - FAB = 0  FAB = (4 cos u - 3 sin u) kN 5

Average Normal Stress: sAB =

sBC =

[4 cos u - 3 sin u] ( 103 ) 250 ( 106 ) FAB = = (4 cos u - 3 sin u) p p AAB (0.004)2 4 (5 cos u) ( 103 ) 555.56 ( 106 ) FBC = = c d cos u p p ABC (0.006)2 4

It is required that sAB = sBC 250 ( 10 p

6

)

(4 cos u - 3 sin u) = c

1.7778 cos u - 3 sin u = 0 tan u =

555.56 ( 106 ) d cos u p

1.7778 3 Ans.

u = 30.65° = 30.7° Then s = sBC = c

555.56 ( 106 ) d cos 30.65° = 152.13 ( 106 ) Pa = 152 MPa p

Ans.

Ans: u = 30.7°, s = 152 MPa 56

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1–57. The bar has a cross-sectional area of 400(10−6) m2. If it is subjected to a triangular axial distributed loading along its length which is 0 at x = 0 and 9 kN>m at x = 1.5 m, and to two concentrated loads as shown, determine the average normal stress in the bar as a function of x for 0 … x 6 0.6 m.

8 kN 4 kN x 0.6 m

0.9 m

Solution Internal Loading: Referring to the FBD of the right segment of the bar sectioned at x, Fig. a, + ΣFx = 0; S

8 + 4 + N =

1 (6x + 9)(1.5 - x) = 0 2

5 18.75

Average Normal Stress: s =

N = A =

- 3x2 6 kN

( 18.75 - 3x2 )( 103 ) 400 ( 10-6 )

5 46.9

- 7.50x2 6 MPa

Ans.

57

Ans: s = 546.9 - 7.50x2 6 MPa

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  The bar has a cross-sectional area of 400(10−6) m2. If it is subjected to a uniform axial distributed loading along its length of 9 kN>m, and to two concentrated loads as shown, determine the average normal stress in the bar as a function of x for 0.6 m 6 x … 1.5 m.

8 kN 4 kN x 0.6 m

0.9 m

Solution Internal Loading: Referring to a FBD of the right segment of the bar sectioned at x, + ΣFx = 0; S

4 + 9 (1.5 - x) - N = 0 N =

5 17.5

Average Normal Stress: s =

N = A =

- 9x 6 kN

( 17.5 - 9x )( 103 ) 400 ( 10-6 )

5 43.75

- 22.5x 6 MPa

Ans.

58

Ans: s = 5 43.75 - 22.5x 6 MPa

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  The two steel members are joined together using a 30° scarf weld. Determine the average normal and average shear stress resisted in the plane of the weld.

15 kN

Internal Loadings: Referring to the FBD of the upper segment of the member sectioned through the scarf weld, Fig. a, ΣFx = 0;

N - 15 sin 30° = 0

N = 7.50 kN

ΣFy = 0;

V - 15 cos 30° = 0

V = 12.99 kN

A = 0.02 a

40 mm

15 kN

Average Normal and Shear Stress: The area of the scarf weld is

Thus,

20 mm

30

Solution

0.04 b = 1.6 ( 10-3 ) m2 sin 30°

s =

7.50 ( 103 ) N = = 4.6875 ( 106 ) Pa = 4.69 MPa An 1.6 ( 10-3 )

Ans.

t =

12.99 ( 103 ) V = = 8.119 ( 106 ) Pa = 8.12 MPa Av 1.6 ( 10-3 )

Ans.

Ans: s = 4.69 MPa, t = 8.12 MPa 59

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*1–60. The bar has a cross-sectional area of 400(10−6) m2. If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads, determine the average normal stress in the bar as a function of x for 0 6 x … 0.5 m.

w  8 kN/m

6 kN 3 kN

x 0.5 m

0.75 m

Solution Equation of Equilibrium: + ΣFx = 0;   - N + 3 + 6 + 8(1.25 - x) = 0 S N = (19.0 - 8.00x) kN Average Normal Stress: s =

(19.0 - 8.00x)(103) N = A 400(10-6) Ans.

= (47.5 - 20.0x) MPa

Ans: s = (47.5 - 20.0x) MPa 60

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1–61. The bar has a cross-sectional area of 400(10−6) m2. If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads, determine the average normal stress in the bar as a function of x for 0.5 m 6 x … 1.25 m.

w  8 kN/m

6 kN 3 kN

x 0.5 m

0.75 m

Solution Equation of Equilibrium: + ΣFx = 0;   - N + 3 + 8(1.25 - x) = 0 S N = (13.0 - 8.00x) kN Average Normal Stress: s =

(13.0 - 8.00x) ( 103 ) N = A 400 ( 10-6 ) Ans.

= (32.5 - 20.0x) MPa

Ans: s = (32.5 - 20.0x) MPa 61

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1–62. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for 0 … x 6 a.

w0

x a

a

Solution Equation of Equilibrium: w + ΣFx = 0;   - N + 1 a 0 x + w0 b (a - x) + 1 w0 a = 0 S 2 a 2 N =

w0 ( 2a2 - x2 ) 2a

Average Normal Stress: s =

N = A

w0 2a

( 2a2 - x2 ) A

=

w0 ( 2a2 - x2 ) 2aA

Ans.

Ans: s = 62

w0 ( 2a2 - x2 ) 2aA

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1–63. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for a 6 x … 2a.

w0

x a

a

Solution Equation of Equilibrium: w + ΣFx = 0;   - N + 1 c 0 (2a - x) d (2a - x) = 0 S 2 a N =

w0 (2a - x)2 2a

Average Normal Stress: s =

N = A

w0 2a

(2a - x)2 A

=

w0 (2a - x)2 2aA

Ans.

Ans: s = 63

w0 (2a - x)2 2aA

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*1–64. Determine the greatest constant angular velocity v of the flywheel so that the average normal stress in its rim does not exceed s = 15 MPa. Assume the rim is a thin ring having a thickness of 3 mm, width of 20 mm, and a mass of 30 kg>m. Rotation occurs in the horizontal plane. Neglect the effect of the spokes in the analysis. Hint: Consider a free-body diagram of a semicircular segment of the ring. The center of mass for this segment is located at nr = 2r>π from the center.

v

0.8 mm

Solution + T ΣFn = m(aG)n;

2T = m( r )v2 2sA = ma

2r 2 bv p

2 ( 15 ( 106 ) ) (0.003)(0.020) = p(0.8)(30)a v = 6.85 rad>s

2(0.8) 2 bv p

Ans.

Ans: v = 6.85 rad>s 64

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1–65. Determine the largest load P that can be applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa, respectively. Member CB has a square cross section of 25 mm on each side.

B

2m

a a A

Solution

C 1.5 m

Analyze the equilibrium of joint C using the FBD shown in Fig. a,

P

4 + c ΣFy = 0; FBC a b - P = 0 FBC = 1.25P 5

Referring to the FBD of the cut segment of member BC Fig. b. + ΣFx = 0; S + c ΣFy = 0;

3 Na - a - 1.25P a b = 0 5

4 1.25P a b - Va - a = 0 5

The cross-sectional area of section 1.0417(10 - 3) m2. For Normal stress, sallow =

Na - a = 0.75P Va - a = P a–a

is

Aa - a = (0.025)a

0.025 b = 3>5

Na - a 0.75P ; 150(106) = Aa - a 1.0417(10 - 3)

P = 208.33(103) N = 208.33 kN For Shear Stress tallow =

Va - a P ; 60(106) = Aa - a 1.0417(10 - 3)

P = 62.5(103) N = 62.5 kN (Controls!)

Ans.

Ans: P = 62.5 kN 65

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1–66. The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normal stress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.

B

C

3 ftm 0.9

A

4 ftm 1.2 P

E

D

4 ft 1.2 m

0.75 P

Solution Joint A: FAB 13.33 66.67(10 ) = = 85.5 MPa A AB 780(10−6)

(T)

FAE 53.33(103) = 68.4 MPa = = A AE 780(10−6)

(C)

sAB = sAE

FAB = 66.67 kN

3

FAE = 53.33 kN

Ans. 40 kN

Ans.

Joint E:

FEB =30 kN FED = 53.33 kN

53.33 kN

sED =

FED 53.33(103) = 68.4 MPa = A ED 780(10−6)

(C)

Ans.

sEB =

FEB 6.0 30(103) = = 38.5 MPa A EB 780(10−6)

(T)

Ans.

30 kN

FBC =146.67 kN 66.67 kN

Joint B: sBC =

FBC 29.33 146.67(103) = 188 MPa = A BC 780(10−6)

(T)

Ans.

sBD =

3 FBD 23.33 116.67(10 ) = = 150 MPa −6 A BD 780(10 )

(C)

Ans.

30 kN

FBD =116.67 kN

Ans: Joint A: sAB sAE Joint E: sED sEB Joint B: sBC

= = = = =

85.5 MPa (T), 68.4 MPa (C) 68.4 MPa (C), 38.5 MPa (T) 188 MPa (T),

66

CH 01.indd 23

11/29/10 10:18:37 AM

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1–67. The bars of the truss each have a cross-sectional the maximum maximum average average normal normal stress in area of 780 1.25mm in2.2.IfIfthe determine the the maximum any bar bar isisnot nottotoexceed exceed140 20MPa, ksi, determine magnitude P of the loads that can be applied to the truss.

B

0.9 3 ftm

A

1.2 4 ftm

Solution P

Joint A: + c ©Fy = 0;

-P +

E

D

1.2 m 4 ft

0.75 P

3 F = 0 5 AB

FAB = 1.6667P + : ©Fx = 0;

C

4 -FAE + (1.6667P)a b = 0 5 FAE = 1.3333P

(T)

FEB

(C) FED

1.3333 P

Joint E: + c ©Fy = 0;

0.75 P

FEB - 0.75P = 0 FEB = 0.75P

(T) FBC

+ : ©Fx = 0;

1.3333P - FED = 0 FED = 1.3333P

1.6667 P

0.75 P

FBD

(C)

Joint B: + c ©Fy = 0;

3 3 FBD - 0.75P - (1.6667P)a b = 0 5 5 FBD = 2.9167P

+ : ©Fx = 0;

(C)

4 4 FBC – (2.9167P)a b - (1.6667P)a b = 0 5 5

FBC = 3.6667P

(T)

The highest stressed member is BC:

3.6667 P = σ BC σ= 140(106 ) max ; 780(10 −6 ) P = 29.78(10 3 ) N = 29.8 kN

Ans.

Ans: P = 29.8 kN 67

CH 01.indd 24

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*1–68. 2

The radius of the pedestal is defined by r = (0.5e−0.08 y ) m, where y is in meters. If the material has a density of 2.5 Mg>m3, determine the average normal stress at the support. 2

3m

r  0.5e0.08y

y

Solution

r

2

2

A = p(0.5) = 0.7854 m dV = p ( r

2

) dy = p(0.5) ( e

V = 10 p(0.5) ( e 3

0.5 m

2

2

2

- 0.08y

)

- 0.08y2 2

) 2 dy = 0.7854 103 ( e - 0.08y ) 2dy 2

W = rg V = (2500)(9.81)(0.7854) 10 ( e 3

W = 19.262 ( 103 ) 10 ( e 3

s =

2

- 0.08y

- 0.08y2 2

) dy

) 2dy = 38.849 kN

W 38.849 = = 49.5 kPa A 0.7854

Ans.

Ans: s = 49.5 kPa 68

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1–69. The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt Determine thethe maximum load P has a diameter of 0.3 7.5 in. mm. Determine maximum load that can P that canbebeapplied appliedtotothe themember member ifif the the allowable shear for the the bolts bolts isistallow ksi and tallow==8412MPa stress for and the allowable 20 MPa. ksi. average normal stress is sallow ==140

60� P

P

Solution a + ©Fy = 0;

N - P sin 60° = 0 N = 0.8660 P

b+ ©Fx = 0;

V - P cos 60° = 0 V = 0.5P

Assume failure due to shear:

V 0.5P ; 84(106 ) tallow = = 2Ab 2  π4 (0.00752 )   P = 14.84(10 3 ) N = 14.84 kN Assume failure due to normal force:

N 0.8660 P ; 140(106 ) sallow = = 2Ab 2  π4 (0.00752 )   P = 14.28(10 3 ) N = 14.3 kN (contπols)

Ans.

Ans: P = 14.3 kN (controls) 69

CH 01.indd 54

11/29/10 10:19:35 AM

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1–70. Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are 10 mm thick, determine to the nearest multiples of 5 mm the smallest dimension h of the horizontal segment so that it does not fail in shear. The average shear stress for the segment is tallow = 2.1 MPa.

B 13

800 4 kNlb

5

12

h A

Solution V 1.538(10 3 ) ; 2.1(106 ) tallow = = A (0.01)h

4 kN = 3.692 kN

h = 0.07326 m = 73.26 mm

V = 1.538 kN

Use h = 75 mm

Ans.

Ans: h = 0.07326 m, use h = 75 mm 70

CH 01.indd 50

11/29/10 10:19:31 AM

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1–71. The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is tallow = 35 MPa.

a A

d a 20 mm

500 mm 200 N

Solution a+ ΣMA = 0;

Fa - a (20) - 200(500) = 0 Fa - a = 5000 N

tallow =

Fa - a ; Aa - a

35(106) =

5000 d(0.025) Ans.

d = 0.00571 m = 5.71 mm

Ans: d = 5.71 mm 71

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*1–72. The lapbelt assembly is to be subjected to a force of 800 N. Determine (a) the required thickness t of the belt if the allowable tensile stress for the material is (st)allow = 10 MPa, (b) the required lap length dl if the glue can sustain an allowable shear stress of (tallow)g = 0.75 MPa, and (c) the required diameter dr of the pin if the allowable shear stress for the pin is (tallow)p = 30 MPa.

800 N 45 mm

t dl

dr 800 N

Solution Allowable Normal Stress: Design of belt thickness. (st)allow =

P ; A

10 A 106 B =

800 (0.045)t Ans.

t = 0.001778 m = 1.78 mm Allowable Shear Stress: Design of lap length. VA 400 ; 0.750 A 106 B = (tallow)g = A (0.045) dt

Ans.

Allowable Shear Stress: Design of pin size. VB 400 ; 30 A 106 B = p 2 (tallow)P = A 4 dr

Ans.

dt = 0.01185 m = 11.9 mm

dr = 0.004120 m = 4.12 mm

Ans: t = 1.78 mm, dt = 11.9 mm, dr = 4.12 mm 72

CH 01.indd 51

11/29/10 10:19:31 AM

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1–73. The cotter is used to hold the two rods together. Determine the smallest thickness t of the cotter and the smallest diameter d of the rods. All parts are made of steel for which the failure normal stress is sfail = 500 MPa and the failure shear stress is t fail = 375 MPa. Use a factor of safety of (F.S.)t = 2.50 in tension and (F.S.)s = 1.75 in shear.

30 kN d

t

40 mm

10 mm

Solution Allowable Normal Stress: Design of rod size sallow =

sfail P = ; F.S A

500 ( 10

6

)

2.5

d

30 ( 10

3

=

)

30 kN

p 2 4d

Ans.

d = 0.01382 m = 13.8 mm Allowable Shear Stress: Design of cotter size. tallow =

tfail V = ; F.S A

375 ( 106 ) 1.75

=

15.0 ( 103 ) (0.01)t Ans.

t = 0.0070 m = 7.00 mm

Ans: d = 13.8 mm, t = 7.00 mm 73

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1–74. The truss is used to support the loading shown Determine the required cross-sectional area of member BC if the allowable normal stress is sallow = 165 MPa.

4000 N

2000 N

2m

2m

F

B 30⬚ A

Solution

E

60⬚

2m

2m

C D

45⬚

For the entire truss, a + © MA = 0;

-2000(2) - 4000(2.8284) + 2(2.8284)(Dy) = 0 Dy = 2707.11 N

For method of section, a + © MF = 0;

2707.11(2.8284) - FBC (1.7932) = 0 F

FBC = 4270.06 N

FEF

Using the given allowable stress, sallow =

FBC ; A

165(106) =

2.0705sin 60° = 1.7932 m

4270.06 A

2 A = 25.88(10−6) m2 = 25.9 mm

Ans.

FCF

2 = 2.0705 m cos 15°

FBC 2.8284 m

Ans: A = 25.9 mm2 74

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1–75. If the allowable tensile stress for wires AB and AC is sallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 6 kN.

C B

5

45

4 3

A

Solution Normal Forces: Analyzing the equilibrium of joint A, Fig. a, + ΣFx = 0; S + c ΣFy = 0;

3 FAC a b - FAB sin 45° = 0 5

(1)

4 FAC a b + FAB cos 45° - 6 = 0 5

P

(2)

Solving Eqs. (1) and (2)

FAC = 4.2857 kN  FAB = 3.6365 kN Average Normal Stress: For wire AB, sallow =

FAB ; AAB

200 ( 106 ) =

3.6365 ( 103 ) p 2 d 4 AB

d AB = 0.004812 m = 4.81 mm

Ans.

For wire AC, sallow =

FAC ; AAC

200 ( 106 ) =

4.2857 ( 103 ) p 2 d 4 AC

d AC = 0.005223 m = 5.22 mm

Ans.

Ans: d AB = 4.81 mm, d AC = 5.22 mm 75

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*1–76. If the allowable tensile stress for wires AB and AC is sallow = 180 MPa, and wire AB has a diameter of 5 mm and AC has a diameter of 6 mm, determine the greatest force P that can be applied to the chain.

C B

5

45

4 3

A

Solution Normal Forces: Analyzing the equilibrium of joint A, Fig. a, + ΣFx = 0; S + c ΣFy = 0;

3 FAC a b - FAB sin 45° = 0 5

(1)

4 FAC a b + FAB cos 45° - P = 0 5

P

(2)

Solving Eqs. (1) and (2)

FAC = 0.7143P  FAB = 0.6061P Average Normal Stress: Assuming failure of wire AB, sallow =

FAB ; AAB

180 ( 106 ) =

0.6061P p ( 0.0052 ) 4

P = 5.831 ( 103 ) N = 5.83 kN Assume the failure of wire AC, sallow =

FAC ; AAC

180 ( 106 ) =

0.7143P p ( 0.0062 ) 4

P = 7.125 ( 103 ) N = 7.13 kN Choose the smaller of the two values of P, Ans.

P = 5.83 kN

Ans: P = 5.83 kN 76

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1–77. The spring mechanism is used as a shock absorber for a load applied to the drawbar AB. Determine the force in each spring when the 50-kN force is applied. Each spring is originally unstretched and the drawbar slides along the smooth guide posts CG and EF. The ends of all springs are attached to their respective members. Also, what is the required diameter of the shank of bolts CG and EF if the allowable stress for the bolts is sallow = 150 MPa?

k  80 kN/m C H

A

B

k¿  60 kN/m

Solution

k¿  60 kN/m

G

F 200 mm

Equations of Equilibrium: a+ ΣMH = 0;

E

200 mm D

- FBF (200) + FAG(200) = 0 FBF = FAG = F

+ c ΣFy = 0;

(1)

2F + FH - 50 = 0

50 kN

Required, ∆ H = ∆ B ;

FH F = 80 60 (2)

F = 0.75 FH  Solving Eqs. (1) and (2) yields, FH = 20.0 kN

Ans.

FBF = FAG = F = 15.0 kN

Ans.

Allowable Normal Stress: Design of bolt shank size. s allow =

P ; A

150 ( 106 ) =

15.0 ( 103 ) p 2 4d

d = 0.01128 m = 11.3 mm Ans.

d EF = d CG = 11.3 mm

Ans: FH = 20.0 kN, FBF = FAG = 15.0 kN, d EF = d CG = 11.3 mm 77

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1–78. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load P = 1500 N, determine the required minimum diameter of pins B and C. Use a factor of safety of 2 against failure. The pins are made of material having a failure shear stress of tfail = 150 MPa, and each pin is subjected to double shear.

P A

300 mm

30 mm

100 mm

B 60�

C

D

Solution Internal Loadings: The forces acting on pins B and C can be determined by considering the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. a + ©MC = 0;

1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0 FBD = 5905.36 N

+ : ©Fx = 0;

Cx - 5905.36 cos 60° = 0

+ c ©Fy = 0;

5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N

Cx = 2952.68 N

Thus, FB = FBD = 5905.36 N

FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N

Since both pins are in double shear, FB 5905.36 VB = = = 2952.68 N 2 2

VC =

FC 4666.98 = = 2333.49 N 2 2

Allowable Shear Stress: tfail 150 tallow = = = 75 MPa F.S. 2 Using this result, VB tallow = ; AB

75(106) =

2952.68 p 2 d 4 B Ans.

dB = 0.007080 m = 7.08 mm tallow =

VC ; AC

75(106) =

2333.49 p 2 d 4 C Ans.

dC = 0.006294 m = 6.29 mm

Ans: dB = 7.08 mm, dC = 6.29 mm 78

CH 01.indd 61

11/29/10 10:19:42 AM

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1–79. The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load of P = 1500 N, determine the factor of safety of pins B and C against failure if they are made of a material having a shear failure stress of tfail = 150 MPa. Pin B has a diameter of 7.5 mm, and pin C has a diameter of 6.5 mm. Both pins are subjected to double shear.

P A

300 mm

30 mm

100 mm

B 60�

C

D

Solution Internal Loadings: The forces acting on pins B and C can be determined by considerning the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig. a. + ©MC = 0;

1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = 0 FBD = 5905.36 N

+ : ©Fx = 0;

Cx - 5905.36 cos 60° = 0

+ c ©Fy = 0;

5905.36 sin 60° - 1500 - Cy = 0 Cy = 3614.20 N

Cx = 2952.68 N

Thus, FB = FBD = 5905.36 N

FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N

Since both pins are in double shear, VB =

FB 5905.36 = = 2952.68N 2 2

VC =

FC 4666.98 = = 2333.49 N 2 2

Allowable Shear Stress: The areas of the shear plane for pins B and C are p p A B = (0.00752) = 44.179(10 - 6)m2 and A C = (0.00652) = 33.183(10 - 6)m2. 4 4 We obtain

A tavg B B = A tavg B C =

VB 2952.68 = 66.84 MPa = AB 44.179(10 - 6) VC 2333.49 = 70.32 MPa = AC 33.183(10 - 6)

Using these results, tfail = (F.S.)B = t A avg B B tfail (F.S.)C = = A tavg B C

150 = 2.24 66.84

Ans.

150 = 2.13 70.32

Ans.

Ans: (F.S.)B = 2.24, (F.S.)C = 2.13 79

CH 01.indd 62

11/29/10 10:19:43 AM

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*1–80. Determine the required diameter of the pins at A and B if the allowable shear stress for the material is tallow = 100 MPa. Both pins are subjected to double shear.

2 kN/m B A

3m

Solution Support Reactions: Member BC is a two force member. a+ ΣMA = 0;

FBC sin 45°(3) - 6(1.5) = 0

C

FBC = 4.243 kN + c ΣFy = 0;

Ay + 4.243 sin 45° - 6 = 0 Ay = 3.00 kN

+ ΣFx = 0; d

Ax - 4.243 cos 45° = 0 Ax = 3.00 kN

Allowable Shear Stress: Pin A and pin B are subjected to double shear. FA = 23.002 + 3.002 = 4.243 kN and FB = FBC = 4.243 kN. Therefore, VA = VB = t allow =

4.243 = 2.1215 kN 2 V ; A

100 ( 106 ) =

2.1215 ( 103 ) p 2 4d

d = 0.005197 m = 5.20 mm Ans.

d A = d B = d = 5.20 mm

Ans: d A = d B = 5.20 mm 80

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1–81. The steel pipe is supported on the circular base plate and concrete pedestal. If the thickness of the pipe is t  = 5  mm and the base plate has a radius of 150 mm, determine the factors of safety against failure of the steel and concrete. The applied force is 500 kN, and the normal failure stresses for steel and concrete are (sfail)st = 350 MPa and (sfail)con = 25 MPa, respectively.

t

500 kN 100 mm r

Solution Average Normal and Bearing Stress: The cross-sectional area of the steel pipe and the bearing area of the concrete pedestal are Ast = p(0.12 - 0.0952) = 0.975(10 - 3)p m2 and (Acon)b = p(0.152) = 0.0225p m2. We have (savg)st =

500(103) P = = 163.24 MPa Ast 0.975(10 - 3)p

(savg)con =

500(103) P = = 7.074 MPa (Acon)b 0.0225p

Thus, the factor of safety against failure of the steel pipe and concrete pedestal are (F.S.)st =

(sfail)st (savg)st

(F.S.)con =

=

(sfail)con (savg)con

350 = 2.14 163.24 =

Ans.

25 = 3.53 7.074

Ans.

Ans: (F.S.)st = 2.14, (F.S.)con = 3.53 81

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1–82. The steel swivel bushing in the elevator control of an airplane is held in place using a nut and washer as shown in Fig. (a). Failure of the washer A can cause the push rod to separate as shown in Fig. (b). If the average shear stress is tavg = 145 MPa, determine the force F that must be applied to the bushing that will cause this to happen. The washer is 1.5 mm thick.

20 mm F

F A (a)

(b)

Solution From the given average shear stress, tavg =

V ; A

145(106) =

F 2p(0.01)(0.0015)

F = 13.67(103) = 13.7 kN

Ans.

Ans: F = 13.7 kN 82

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1–83. Determine the required minimum thickness t of member AB and edge distance b of the frame if P = 40 kN and the factor of safety against failure is 2. The wood has a normal failure stress of s fail = 42 MPa, a nd shear failure stress of tfail = 10.5 MPa.

P 75 mm B

75 7 mm t

A

30⬚ b

30⬚

C

Solution Internal Loadings: The normal force developed in member AB can be determined by considering the equilibrium of joint A. Fig. a. + : ©Fx = 0;

FAB cos 30° - FAC cos 30° = 0

FAC = FAB

+ c ©Fy = 0;

2FAB sin 30° - 40 = 0

FAB = 40 kN

Subsequently, the horizontal component of the force acting on joint B can be determined by analyzing the equilibrium of member AB, Fig. b. + : ©Fx = 0;

(FB)x - 40 cos 30° = 0

(FB)x = 34.64 kN

Referring to the free-body diagram shown in Fig. c, the shear force developed on the shear plane a–a is + : ©Fx = 0;

Va - a - 34.64 = 0

Va - a = 34.64 kN

Allowable Normal Stress: sfail 42 = = 21 M Pa sallow = F.S. 2 tallow =

tfail 10.5 = = 5.25 M Pa F.S. 2

Using these results, sallow =

tallow =

FAB ; AAB Va - a ; Aa - a

40(103) (0.075)t t = 0.02540 m = 25.4 mm 21(106) =

5.25(106) =

Ans.

34.64(103)

(0.075)b b = 0.08798 m = 88.0 mm

Ans.

Ans: t = 25.4 mm, b = 88.0 mm 83

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*1–84. Determine the maximum allowable load P that can be safely supported by the frame if t = 30 mm and b = 90 mm. The wood has a normal failure stress of sfail = 42 MPa, and shear failure stress of tfail = 10.5 MPa. Use a factor of safety against failure of 2.

P 75 mm B

75 mm t

A

30⬚ b

30⬚

C

Solution Internal Loadings: The normal force developed in member AB can be determined by considering the equilibrium of joint A. Fig. a. + : ©Fx = 0; FAB cos 30° - FAC cos 30° = 0 FAC = FAB 2FAB sin 30° - P = 0

+ c ©Fy = 0;

FAB = P

Subsequently, the horizontal component of the force acting on joint B can be determined by analyzing the equilibrium of member AB, Fig. b. + (FB)x = 0.8660P : ©Fx = 0; (FB)x - P cos 30° = 0 Referring to the free-body diagram shown in Fig. c, the shear force developed on the shear plane a–a is + Va - a = 0.8660P : ©Fx = 0; Va - a - 0.8660P = 0 Allowable Normal and Shear Stress: sallow =

sfail 42 = = 21 MPa F.S. 2

tallow =

tfail 10.5 = = 5.25 MPa F.S. 2

Using these results, sallow =

FAB ; AAB

21(106) =

P 0.075(0.03)

P = 47.25(103) N = 47.25 kN tallow =

Va - a ; Aa - a

5.25(106) =

0.8660P 0.075(0.09)

P = 40.92(103) N = 40.9 kN (controls)

Ans.

Ans: P = 40.9 kN 84

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  The hanger is supported using the rectangular pin. Determine the magnitude of the allowable suspended load P if the allowable bearing stress is (sb)allow = 220 MPa, the allowable tensile stress is (st)allow = 150 MPa, and the allowable shear stress is tallow = 130 MPa. Take t = 6 mm, a = 5 mm and b = 25 mm.

20 mm 75 mm 10 mm a

37.5 mm

Allowable Normal Stress: For the hanger P ; A

150 1 106 2 =

b

37.5 mm

t

Solution

(st)allow =

a

P

P (0.075)(0.006)

P = 67.5 kN

Allowable Shear Stress: The pin is subjected to double shear. Therefore, V = tallow =

V ; A

130 1 106 2 =

P 2

P>2 (0.01)(0.025)

P = 65.0 kN

Allowable Bearing Stress: For the bearing area (sb)allow =

P ; A

220 1 106 2 =

P>2 (0.005)(0.025) Ans.

P = 55.0 kN (Controls!)

Ans: P = 55.0 kN 85

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1–86. The hanger is supported using the rectangular pin. Determine the required thickness t of the hanger, and dimensions a and b if the suspended load is P = 60 kN. The allowable tensile stress is (st)allow = 150 MPa, the allowable bearing stress is (sb)allow = 290 MPa, and the allowable shear stress is tallow = 125 MPa.

20 mm 75 mm 10 mm a

37.5 mm

Allowable Normal Stress: For the hanger P ; A

150 1 106 2 =

b

37.5 mm

t

Solution

(st)allow =

a

P

60(103) (0.075)t Ans.

t = 0.005333 m = 5.33 mm

Allowable Shear Stress: For the pin tallow =

V ; A

125 1 106 2 =

30(103) (0.01)b Ans.

b = 0.0240 m = 24.0 mm

Allowable Bearing Stress: For the bearing area (sb)allow =

P ; A

290 1 106 2 =

30(103) (0.0240) a Ans.

a = 0.00431 m = 4.31 mm

Ans: t = 5.33 mm, b = 24.0 mm, a = 4.31 mm 86

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1-87. The assembly is used to support the distributed loading ofww==500 10 lb>ft kN/m. Determine factor of safety of safety with loading of . Determine thethe factor with respect to yielding for the steel rod BC and the pins at A and B if the yield stress for the steel in tension is s 250 MPa shear 125aMPa. The of rod hasin., a y =in y = has and shear ty and . Thetrod diameter 0.40 = 18inksi diameter of 13 mm, and the pins each have a diameter of and the pins each have a diameter of 0.30 in. 10 mm.

C

1.2 4 ftm A

Solution For rod BC: s =

P = A

B 3

10(10 ) π

4

(0.0132 )

= 75.34 MPa

0.9 3 ftm w

sy

250 36 ==3.32 F. S. = = 2.71 s 13.26 75.34

0.3 1 ftm

Ans. 0.9 m

For pin B: tB =

VB 5(10 ) = π = 63.66 MPa A (0.012 ) 4 tY

F. S. =

tB

= =

FBC = 10 kN

Az = 6 kN

3

VB = 5 kN Ay = 4 kN

125 18 ==1.96 1.53 63.66 11.79

Ans.

VB = 5 kN 10(1.2) = 12 kN 10 kN

0.6 m

Pin B

For pin A:

FA =

4 2 + 6 2 = 7.211 kN

VA = 3.606 kN VA = 3.606 kN

VA = tA =

7.211 kN = 3.606 kN 2

7.211 kN Pin A

VA 3.606(10 3 ) = = 45.91 MPa π (0.012 ) A 4

F. S. =

tY tA

=

125 = 2.72 45.91

Ans.

Ans: (F.S.)rod = 3.32, (F.S.)pinB = 1.96 (F.S.)pinA = 2.72 87

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* 1–88. If the allowable allowableshear shearstress stressfor foreach each 0.30of of thethe 10-mm12.5 and ksi, in.-diameter steel at B, A,and B, and is tallow diameter steel pinspins at A, C is Ctallow = 90=MPa, and the allowable normal for13-mm-diameter the 0.40-in.-diameter the allowable normal stressstress for the rod is rod , determine largest intensity of = 22 ksi sallowis =sallow 150 MPa, determine the the largest intensity w ofwthe the uniform distributed be suspended uniform distributed loadload that that can can be suspended fromfrom the the beam. beam.

C

4 ftm 1.2 A

B 3 ftm 0.9 w 1 ftm 0.3

Solution Assume failure of pins B and C: 3 1.667w 0.5w(10 ) tallow == 90 12.5 = =p p 2 2 ) ) 4 (10 4 (0.3

0.5w

0.3605w 0.3605w

0.5w

1.0w

w == 14.14 0.530kN/m   kip>ft

0.721w

Ans.

(controls) (controls)

Assume Assumefailure failureof ofpins pinsA: A:

FBC = 1.0w

Ax = 0.6w

FFAA = 2(0(2w) = 2.404 w .6 w)2 + ((1.333w) 0.4 w)2 =2 0.721w

Ay = 0.4w 1.2w

1.202w 3) 0.3605w(10 12.5 ttallow = = pp 22 allow = 90 (10 )) 44(0.3 ww = 19.61 0.735 kN/m kip>ft Assumefailure failureof ofrod rodBC: BC: Assume ssallow 22 == allow = 150

3.333w 3) 1.0w(10 p p 22 4 (13 )) 4 (0.4

ww = 19.91 0.829 kN/m kip>ft

Ans: w = 14.14 kN/m (controls) 88

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  The compound wooden beam is connected together by a bolt at B. Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is 1st 2 allow = 150 MPa and the allowable bearing stress for the wood is 1sb 2 allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt.

2m

3 kN

2m

2 kN 1.5 kN 1.5 m 1.5 m 1.5 m

1.5 m C

A

D B

Solution From FBD (a): a + ΣMD = 0;

FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 0 (1)

4.5 FB - 6 FC = - 7.5 From FBD (b): a + ΣMA = 0;

FB(5.5) - FC(4) - 3(2) = 0 (2)

5.5 FB - 4 FC = 6 Solving Eqs. (1) and (2) yields FB = 4.40 kN;

FC = 4.55 kN

For bolt: sallow = 150(106) =

4.40(103) p 2 4 (d B)

d B = 0.00611 m Ans.

= 6.11 mm For washer: sallow = 28 (104) =

4.40(103) p 2 4 (d w

- 0.006112) Ans.

d w = 0.0154 m = 15.4 mm

Ans: dB = 6.11 mm, dw = 15.4 mm 89

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  The two aluminum rods support the vertical force of P = 20 kN. Determine their required diameters if the allowable tensile stress for the aluminum is sallow = 150 MPa.

B

C

A

45

P

Solution + c ΣFy = 0;   FAB sin 45° - 20 = 0;      FAB = 28.284 kN + ΣFx = 0;   28.284 cos 45° - FAC = 0;   FAC = 20.0 kN S For rod AB: sallow =

FAB ; AAB

       150(106) =

28.284(103) p 2 4 dAB

Ans.

dAB = 0.0155 m = 15.5 mm For rod AC: sallow =

FAC ; AAC

       150(106) =

20.0(103) p 4

2 dAC

Ans.

dAC = 0.0130 m = 13.0 mm

Ans: dAB = 15.5 mm, dAC = 13.0 mm 90

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  The two aluminum rods AB and AC have diameters of 10 mm and 8 mm, respectively. Determine the largest vertical force P that can be supported. The allowable tensile stress for the aluminum is sallow = 150 MPa.

B

C

A

45

P

Solution + c ΣFy = 0;    FAB sin 45° - P = 0;     P = FAB sin 45°

(1)

+ ΣFx = 0;    FAB cos 45° - FAC = 0 S

(2)

Assume failure of rod AB: sallow =

FAB FAB ;  150(106) = p 2 AAB (0.01) 4

FAB = 11.78 kN From Eq. (1), P = 8.33 kN Assume failure of rod AC: sallow =

FAC FAC ;  150(106) = p 2 AAC 4 (0.008)

FAC = 7.540 kN Solving Eqs. (1) and (2) yields: FAB = 10.66 kN;  P = 7.54 kN Choose the smallest value Ans.

P = 7.54 kN

Ans: P = 7.54 kN 91

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*1–92. The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d1 of the top disk, the largest diameter d2 of the opening, and the largest diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (sb)allow = 350 MPa and allowable shear stress is tallow = 125 MPa.

140 kN d1 B

20 mm A

10 mm

C d3 d2

Solution Allowable Shear Stress: Assume shear failure for disk C. tallow =

140 ( 103 ) V ; 125 ( 106 ) = A pd 2(0.01) Ans.

d 2 = 0.03565 m = 35.7 mm Allowable Bearing Stress: Assume bearing failure for disk C. (sb)allow =

P ; 350 ( 106 ) = A

140 ( 103 ) p 4

1 0.035652 - d 23 2

Ans.

d 3 = 0.02760 m = 27.6 mm

Allowable Bearing Stress: Assume bearing failure for disk B. (sb)allow =

140 ( 103 ) P ; 350 ( 106 ) = p 2 A 4 d1 d 1 = 0.02257 m = 22.6 mm

Since d 3 = 27.6 mm 7 d 1 = 22.6 mm, disk B might fail due to shear. t =

140 ( 103 ) V = = 98.7 MPa 6 tallow = 125 MPa (O.K!) A p(0.02257)(0.02)

Therefore,

Ans.

d 1 = 22.6 mm

Ans: d 2 = 35.7 mm, d 3 = 27.6 mm, d 1 = 22.6 mm 92

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1–93. The aluminium bracket A is used to support the centrally applied load of If 40itkN. it has a thickness constant centrally applied load of 8 kip. has If a constant thickness 12 mm, determine theheight smallest height in of 0.5 in.,ofdetermine the smallest h in orderh to order to prevent failure. failure shearstress stress is prevent a sheara shear failure. The The failure shear tfail == 160 MPa. Use factorofofsafety safetyfor forshear shearofofF.S. F.S. == 2.5. Use a afactor 23 ksi.

A

h

Solution Equation of Equilibrium: + c ©Fy = 0;

V V –-408 ==00

840kip kN

V –=408.00 kNkip

Allowable Shear Stress: Design of the support size tallow =

tfail V = ; F.S A

40 kN

160(106 ) 40(10 3 ) = 2.5 h(0.012)

V = 40 kN

Ans.

h = 0.05208 m = 52.1 mm

Ans: h = 52.1 mm 93

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  The rods AB and CD are made of steel. Determine their smallest diameter so that they can support the dead loads shown. The beam is assumed to be pin connected at A and C. Use the LRFD method, where the resistance factor for steel in tension is f = 0.9, and the dead load factor is gD = 1.4. The failure stress is sfail = 345 MPa.

B

D 6 kN

5 kN

4 kN

A

Solution

C 2m

2m

3m

3m

Support Reactions: a+ ΣMA = 0;

FCD(10) - 5(7) - 6(4) - 4(2) = 0 FCD = 6.70 kN

a+ ΣMC = 0;

4(8) + 6(6) + 5(3) - FAB(10) = 0 FAB = 8.30 kN

Factored Loads: FCD = 1.4(6.70) = 9.38 kN FAB = 1.4(8.30) = 11.62 kN For rod AB 0.9[345(106)] p a

d AB 2 b = 11.62(103) 2

Ans.

d AB = 0.00690 m = 6.90 mm

For rod CD 0.9[345(106)] p a

d CD 2 b = 9.38(103) 2

Ans.

d CD = 0.00620 m = 6.20 mm

Ans: d AB = 6.90 mm, d CD = 6.20 mm 94

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1–95. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the size of square bearing plates A¿ and B¿ required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN.

40 kN/m

A 1.5 m

P

A¿

B¿ 3m

B 1.5 m

Solution Referring to the FBD of the bean, Fig. a a + ©MA = 0;

NB(3) + 40(1.5)(0.75) - 100(4.5) = 0

NB = 135 kN

a + ©MB = 0;

40(1.5)(3.75) - 100(1.5) - NA(3) = 0

NA = 25.0 kN

For plate A¿ , NA (sb)allow = ; A A¿

1.5(106) =

25.0(103) a2A¿ Ans.

aA¿ = 0.1291 m = 130 mm For plate B¿ , sallow =

NB ; A B¿

1.5(106) =

135(103) a2B¿ Ans.

aB¿ = 0.300 m = 300 mm

Ans: aA′ = 130 mm, aB′ = 300 mm 95

CH 01.indd 64

11/29/10 10:19:46 AM

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*1–96. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A¿ and B¿ have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively.

40 kN/m

A 1.5 m

P

A¿

B¿ 3m

B 1.5 m

Solution Referring to the FBD of the beam, Fig. a, a + ©MA = 0;

NB(3) + 40(1.5)(0.75) - P(4.5) = 0

NB = 1.5P - 15

a + ©MB = 0;

40(1.5)(3.75) - P(1.5) - NA(3) = 0

NA = 75 - 0.5P

For plate A¿ , NA (sb)allow = ; A A¿

1.5(106) =

(75 - 0.5P)(103) 0.15(0.15)

P = 82.5 kN For plate B¿ , NB ; (sb)allow = A B¿

1.5(106) =

(1.5P - 15)(103) 0.25(0.25) Ans.

P = 72.5 kN (Controls!)

Ans: P = 72.5 kN (Controls!) 96

CH 01.indd 65

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R1–1. The circular punch B exerts a force of 2 kN on the top of the plate A. Determine the average shear stress in the plate due to this loading.

2 kN B 4 mm

A

2 mm

Solution Average Shear Stress: The shear area A = p(0.004)(0.002) = 8.00(10 - 6)p m2 tavg =

2(103) V = 79.6 MPa = A 8.00(10 - 6)p

Ans.

Ans: tavg = 79.6 MPa 97

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R1–2. C

Determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is sallow = 200 MPa and the allowable shear stress for the pins is ta llow = 70 MPa.

1.5mm in. 40

Solution

B

Referring to the FBD of member AB, Fig. a, a + ©MA = 0;

2(8)(4) - F sinsin 60°60° (8)(2.4) = 0= 0 FBC F=BC9.238 kipkN 30(2.4)(1.2) –BC FBC = 41.57

+ : ©Fx = 0;

9.238 41.57 cos 60° –-AAx x= = 0 0

+ c ©Fy = 0;

9.238 2(8) + +AAyy = 0 0 41.57 sin 60° –- 30(2.4)

60� 60

8 ftm 2.4

A

2 kip/ft 30 kN/m

A x = 4.619 Ax kip = 20.785 kN AA kip 36.00 kN y y==8.00

Thus, the force acting on pin A is 2 2 2 2 FA == 2Ax2A+2x A+y2 A 4.619 +.008.00 = 9.238 F = 41.57 kN kip = 2y =202 .785 + 36

Pin A is subjected to single shear, Fig. c, while pin B is subjected to double shear, Fig. b. 9.238 FBCFBC 41.57 kN kip V 41.57 kN   = = = 20.785 VA ==FF 9.238 kip VBV= = 4.619 AA= = B = 2 2 2 2 For member BC FBC 41.57(10 3 ) ;= sallow = 200(106 ) = t 0.005196 m = 5.196 mm A BC (0.04)t 1 6 mm UseUse t = t =in. 4 For pin A, VA ; tallow = AA

For pin B, VB ; tallow = AB

= 70(106 )

= 70(106 )

Ans.

41.57(10 3 ) = dA 0.02750 = m 27.50 mm π d2 4 A 1 UseUse dA d=A = 1 28inmm 8 20.785(10 3 ) = dB 0.01944 = m 19.44 mm π d2 4 B 13 UseUse dB d=B = 20 inmm 16

Ans.

Ans. FBC = 41.57 kN

1.2 m

1.2 m

FA = 41.57 kN

30(2.4) kN

Ans: t = 6 mm, dA = 28 mm, dB = 20 mm 98

CH 01.indd 69

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R1–3. The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a–a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b–b.

8 mm

a 7 mm

18 mm

b

8 kN

b a 30 mm

Solution P = A

8 (103)

= 208 MPa

Ans.

(tavg)a =

8 (103) V = = 4.72 MPa A p (0.018)(0.030)

Ans.

(tavg)b =

8 (103) V = = 45.5 MPa A p (0.007)(0.008)

Ans.

ss =

p 4

(0.007)2

Ans: ss = 208 MPa, (tavg)a = 4.72 MPa, (tavg)b = 45.5 MPa 99

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*R1–4.

The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 2.0 kN/m and the column FC has a weight of 3.0 kN/m, determine the resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of both the beam and column in the calculation. kN Given: wb := 2.0 L := 3.6m d := 1.8m m kN wc := 3.0 m a := 3.6m

H := 4.8m e := 1.2m b := 3.6m

c := 1.2m

Solution Beam AB: L c :=

+ ΣΜA=0;

+ ΣF y=0;

2

2

L +c

vb :=

c

hb :=

L

Lc Lc By⋅ ( L) − wAB⋅ ( L ) ( 0.5 ⋅ L ) = 0 L By := wb⋅ ( L) ⋅ ⎛⎜ 0.5 ⎞ ⎝ L⎠ m By = 7936.64 lb 2 s Ay := −By + wb⋅ ( L ) −Ay − By + wb⋅ ( L) = 0 By By = FBC⋅ vb FBC := vb −FBC⋅ ( h) + Ax = 0 Ax := FBC⋅ hb

( )

+

ΣF x=0;

( )

Ay = 3.6 kN FBC = 11.38 kN Ax = 10.8 kN

Segment AD:

+

ΣF x=0;

+

ΣF y=0;

+ ΣΜD=0;

ND + Ax = 0 −Ay + wb⋅ ( d) + VD = 0

ND := −Ax

ND = −10.8 kN

Ans.

VD := Ay − wb⋅ ( d)

VD = 0 kN

Ans.

MD + ⎡wb⋅ ( d)⎤ ⋅ ( 0.5 ⋅ d) − Ay⋅ ( d) = 0 ⎣ ⎦ MD := −⎡wb⋅ ( d)⎤ ⋅ ( 0.5 ⋅ d) + Ay⋅ ( d) ⎣ ⎦

100

MD = 3.24 kN⋅ m

Ans.

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* R1–4. Continued

Member CG:

Hb :=

2

H +b

2

Column FC:

+ ΣΜC=0;

H

vc :=

Fx⋅ ( H) − Ax⋅ c = 0

hc :=

Hb

b Hb

c Fx := Ax⋅ ⎛⎜ ⎞ H

⎝ ⎠

Fx = 2.7 kN

( )

( )

+

ΣF x=0;

FBC⋅ hb − Ax + Fx − FCG⋅ hc = 0

+

ΣF y=0;

FCG = 4.5 kN hc −Fy + By + wc⋅ ( H) + FBC⋅ vb + FCG⋅ vc = 0 FCG :=

( )

FBC⋅ hb − Ax + Fx

( ) ( ) Fy := By + wc⋅ ( H) + FBC⋅ ( vb) + FCG⋅ ( vc) Fy = 25.2 kN

Segment FE:

+ +

ΣF x=0;

VE − Fx = 0

VE := Fx

VE = 2.7 kN

Ans.

ΣF y=0;

NE + wc⋅ ( e) − Fy = 0

NE := −wc⋅ ( e) + Fy

NE = 21.6 kN

Ans.

−ME + Fy⋅ ( e) = 0

ME := Fy⋅ ( e)

+ ΣΜE=0;

ME = 30.24 kN⋅m

Ans.

Ans: ND = −10.8 kN, VD = 0 kN, MD = 3.24 kN ⋅ m NE = 21.6 kN, VE = 2.7 kN, ME = 30.24 kN ⋅ m 101

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R1–5. Determine the average punching shear stress the circular shaft creates in the metal plate through section AC and BD. Also, what is the average bearing stress developed on the surface of the plate under the shaft?

40 kN

50 mm A

B 10 mm

C 60 mm

Solution Average Shear and Bearing Stress: The area of the shear plane and the bearing area on  the punch are AV = p(0.05)(0.01) = 0.5(10 - 3)p m2 and Ab = p ( 0.122 - 0.062 ) = 2.7(10 - 3)p m2. We obtain 4 tavg = sb =

D

40(103) P = = 25.5 MPa AV 0.5(10 - 3)p

120 mm

Ans.

40(103) P = 4.72 MPa = Ab 2.7(10 - 3)p

Ans.

Ans: tavg = 25.5 MPa, sb = 4.72 MPa 102

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R1–6.  The 150 mm by 150 mm block of aluminum supports a compressive load of 6  kN. Determine the average normal and shear stress acting on the plane through section a–a. Show the results on a differential volume element located on the plane.

6 kN

a

30

Solution a

Equation of Equilibrium: + QΣFx = 0;

Va - a - 6 cos 60° = 0

Va - a = 3.00 kN

a + ΣFy = 0;

Na - a - 6 sin 60° = 0

Na - a = 5.196 kN

150 mm

Average Normal Stress and Shear Stress: The cross sectional Area at section a–a is A = a

0.15 b(0.15) = 0.02598 m2. sin 60°

sa - a =

5.196(103) Na - a = = 200 kPa A 0.02598

Ans.

ta - a =

3.00(103) Va - a = = 115 kPa A 0.02598

Ans.

Ans: sa - a = 200 kPa, ta - a = 115 kPa 103

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R1–7.  The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members.

5 kN

40 mm

30 mm A 25 mm

Solution

5 kN

For the 40 - mm - dia. rod: s40 =

5 (103) P = p = 3.98 MPa 2 A 4 (0.04)

Ans.

For the 30 - mm - dia. rod: s30 =

5 (103) V = p = 7.07 MPa 2 A 4 (0.03)

Ans.

Average shear stress for pin A: tavg =

2.5 (103) P = p = 5.09 MPa 2 A 4 (0.025)

Ans.

Ans: s40 = 3.98 MPa, s30 = 7.07 MPa, tavg = 5.09 MPa 104

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*R1–8.  The cable has a specific weight g (weight/volume) and crosssectional area A. Assuming the sag s is small, so that the cable’s length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C.

A

B s C L/2

L/2

Solution Equation of Equilibrium: a+ ΣMA = 0;

Ts -

gAL L a b = 0 2 4 T =

gAL2 8s

Average Normal Stress: gAL2

gL2 T 8s s = = = A A 8s

Ans.

Ans: s = 105

gL2 8s

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2–1. An air-filled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until the ball’s diameter becomes 175 mm, determine the average normal strain in the rubber.

Solution d0 = 150 mm d = 175 mm e =

pd - pd 0 175 - 150 = = 0.167 mm>mm 150 pd 0

Ans.

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

Ans: e = 0.167 mm>mm

106

CH 02.indd 73

11/29/10 10:20:54 AM

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2–2. A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having an outer diameter of 125 mm, determine the average normal strain in the strip.

Solution L0 = 375 mm L = p(125 mm) e =

L - L0 125p - 375 = = 0.0472 mm/mm 375 L0

Ans.

Ans. e = 0.0472 mm/mm 107

CH 02.indd 73

11/29/10 10:20:54 AM

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2–3.  If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain in wires CE and BD.

D

E

4m

P A

Solution ∆LCE ∆LBD = 3 7 ∆LBD =

3 (10) 7

B

3m

C

2m

2m

= 4.286 mm

PCE =

∆LCE 10 = = 0.00250 mm>mm L 4000

Ans.

PBD =

∆LBD 4.286 = = 0.00107 mm>mm L 4000

Ans.

Ans: PCE = 0.00250 mm>mm, PBD = 0.00107 mm>mm 108

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*2–4. The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin B through an angle of 2°. Determine the average normal strain in each wire. The wires are unstretched when the lever is in the horizontal position.

G 200 mm

F 200 mm 300 mm

300 mm

E

B

A

C

200 mm

D

H

Solution

2° bp rad = 0.03491 rad. 180 Since u is small, the displacements of points A, C, and D can be approximated by Geometry: The lever arm rotates through an angle of u = a dA = 200(0.03491) = 6.9813 mm dC = 300(0.03491) = 10.4720 mm dD = 500(0.03491) = 17.4533 mm Average Normal Strain:  The unstretched length of wires AH, CG, and DF are LAH = 200 mm, LCG = 300 mm, and LDF = 300 mm. We obtain (Pavg)AH = (Pavg)CG = (Pavg)DF =

dA 6.9813 = = 0.0349 mm>mm LAH 200

Ans.

dC 10.4720 = = 0.0349 mm > mm LCG 300

Ans.

dD 17.4533 = = 0.0582 mm >mm LDF 300

Ans.

109

Ans: (Pavg)AH = 0.0349 mm>mm (Pavg)CG = 0.0349 mm > mm (Pavg)DF = 0.0582 mm >mm

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  P

The pin-connected rigid rods AB and BC are inclined at u = 30° when they are unloaded. When the force P is applied u becomes 30.2°. Determine the average normal strain in wire AC.

B

u

Solution

A

Geometry: Referring to Fig. a, the unstretched and stretched lengths of wire AD are

u

600 mm

C

LAC = 2(600 sin 30°) = 600 mm LAC ′ = 2(600 sin 30.2°) = 603.6239 mm Average Normal Strain: (Pavg)AC =

LAC ′ - LAC 603.6239 - 600 = = 6.04(10 - 3) mm>mm LAC 600

Ans.

Ans: (Pavg)AC = 6.04(10 - 3) mm>mm 110

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2–6. The wire AB is unstretched when u = 45°. If a load is applied to the bar AC, which causes u to become 47°, determine the normal strain in the wire.

B

u

L

Solution 2

2

L = L +

2 L′AB

C

- 2LL′AB cos 43°

A L

L′AB = 2L cos 43° PAB = =

L′AB - LAB LAB 2L cos 43° - 22L

= 0.0343

22L

Ans.

Ans: PAB = 0.0343 111

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  If a horizontal load applied to the bar AC causes point A to be displaced to the right by an amount ∆L, determine the normal strain in the wire AB. Originally, u = 45°.

B

u

L

Solution

C

L′AB = 4( 22L ) + ∆L - 2 ( 22L ) (∆L) cos 135° 2

2

A L

= 22L2 + ∆L2 + 2L∆L

PAB = =

=

L′AB - LAB LAB

22L2 + ∆L2 + 2L∆L - 22L 22L

C

1 +

∆L2 ∆L + - 1 L 2L2

Neglecting the higher - order terms, 1

PAB

∆L 2 = a1 + b - 1 L = 1 + =

1 ∆L + ..... - 1 2 L

(binomial theorem)

0.5∆L L

Ans.

Also, PAB =

∆L sin 45° 22L

=

0.5 ∆L L

Ans.

Ans: PAB = 112

0.5∆L L

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*  y

The rectangular plate is subjected to the deformation shown by the dashed line. Determine the average shear strain gxy in the plate.

150 mm B

3 mm

200 mm

Solution x

A

Geometry: u′ = tan-1 u = a

3 mm

3 = 0.0200 rad 150

p + 0.0200b rad 2

Shear Strain:

gxy =

p p p - u = - a + 0.0200b 2 2 2

Ans.

= - 0.0200 rad

Ans: gxy = - 0.0200 rad 113

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  y

The square deforms into the position shown by the dashed lines. Determine the shear strain at each of its corners, A, B, C, and D, relative to the x, y axes. Side D′B′ remains horizontal.

D¿

B¿

3 mm B

D

53 mm

50 mm 91.5

Solution

C A

Geometry:

x

50 mm

B′C′ = 2(8 + 3)2 + (53 sin 88.5°)2 = 54.1117 mm 2

C¿ 8 mm

2

C′D′ = 253 + 58 - 2(53)(58) cos 91.5° = 79.5860 mm

B′D′ = 50 + 53 sin 1.5° - 3 = 48.3874 mm cos u =

(B′D′)2 + (B′C′)2 - (C′D′)2 2(B′D′)(B′C′) 2

=

48.3874 + 54.11172 - 79.58602 = - 0.20328 2(48.3874)(54.1117)

u = 101.73° b = 180° - u = 78.27° Shear Strain: (gA)xy = (gB)xy =

p 91.5° b = - 0.0262 rad - pa 2 180°

Ans.

p p 101.73° b = - 0.205 rad - u = - pa 2 2 180°

(gC)xy = b (gD)xy = p a

Ans.

p 78.27° p b = pa = - 0.205 rad 2 180° 2

Ans.

88.5° p b = - 0.0262 rad 180° 2

Ans.

Ans: (gA)xy (gB)xy (gC)xy (gD)xy 114

= = = =

-0.0262 rad -0.205 rad -0.205 rad -0.0262 rad

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  Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB. If a force is applied to the end B of the member and causes it to rotate by u = 0.5°, determine the normal strain in the cable. Originally the cable is unstretched.

u P

B

800 mm

Solution Geometry: Referring to the geometry shown in Fig. a, the unstretched and stretched lengths of cable AB are LAB = 26002 + 8002 = 1000 mm

A

C

600 mm

LAB′ = 26002 + 8002 - 2(600)(800) cos 90.5° = 1004.18 mm

Average Normal Strain: PAB =

LAB′ - LAB 1004.18 - 1000 = = 0.00418 mm>mm LAB 1000

Ans.

Ans: PAB = 0.00418 mm>mm 115

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  Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB. If a force is applied to the end B of the member and causes a normal strain in the cable of 0.004 mm>mm, determine the displacement of point B. Originally the cable is unstretched.

u P

B

800 mm

Solution Geometry: Referring to the geometry shown in Fig. a, the unstretched and stretched lengths of cable AB are LAB = 26002 + 8002 = 1000 mm

A

C

600 mm

LAB′ = 26002 + 8002 - 2(600)(800) cos (90° + u) LAB′ = 21(106) - 0.960(106) cos (90° + u)

Average Normal Strain: PAB =

Thus,

21(106) - 0.960(106) cos (90° + u) - 1000 LAB′ - LAB ;  0.004 = LAB 1000 u = 0.4784° a

p b = 0.008350 rad 180°

∆ B = uLBC = 0.008350(800) = 6.68 mm

Ans.

Ans: ∆ B = 6.68 mm 116

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*2–12. y

Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines.

12 mm 4 mm 3 mm

B

8 mm

C 300 mm

Solution

D

Geometry: Referring to the geometry shown in Fig. a, the small-angle analysis gives 7 a = c = = 0.022876 rad 306 5 b = = 0.012255 rad 408 2 u = = 0.0049383 rad 405

400 mm

A

2 mm

x

5 mm

Shear Strain: By definition, (gA)xy = u + c = 0.02781 rad = 27.8(10-3) rad

Ans.

(gB)xy = a + b = 0.03513 rad = 35.1(10-3) rad

Ans.

Ans: (gA)xy = 27.8(10-3) rad (gB)xy = 35.1(10-3) rad 117

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2–13. y

Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines.

12 mm 4 mm 3 mm

B

8 mm

C 300 mm

Solution

D

Geometry: Referring to the geometry shown in Fig. a, the small-angle analysis gives 4 a = c = = 0.013201 rad 303 2 u = = 0.0049383 rad 405 5 b = = 0.012255 rad 408

400 mm

A

2 mm

x

5 mm

Shear Strain: By definition (gxy)C = a + b = 0.02546 rad = 25.5(10-3) rad

Ans.

(gxy)D = u + c = 0.01814 rad = 18.1(10-3) rad

Ans.

Ans: (gxy)C = 25.5(10-3) rad (gxy)D = 18.1(10-3) rad 118

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2–14. y

The material distorts into the dashed position shown. Determine the average normal strains Px, Py and the shear strain gxy at A, and the average normal strain along line BE.

15 mm C

30 mm D

50 mm B 200 mm E

Solution

A

150 mm

50 mm F

x

Geometry: Referring to the geometry shown in Fig. a, tan u =

15 p ;  u = (3.4336°) a rad b = 0.05993 rad 250 180°

LAC ′ = 2152 + 1502 = 262725 mm

EE′ 50 BB′ 200 = ;  BB′ = 12 mm   = ;  EE′ = 6 mm 15 250 30 250

x′ = 150 + EE′ - BB′ = 150 + 6 - 12 = 144 mm LBE = 21502 + 1502 = 15022 mm  LB ′E ′ = 21442 + 1502 = 243236 mm

Average Normal and Shear Strain: Since no deformation occurs along x axis,

Ans.

(Px)A = 0 (Py)A =

LAC′ - LAC 262725 - 250 = = 1.80(10-3) mm>mm LAC 250

Ans.

By definition, Ans.

(gxy)A = u = 0.0599 rad PBE =

LB′E′ - LBE 243236 - 15022 = = - 0.0198 mm>mm LBE 15022

Ans.

Ans: (Px)A = 0 (Py)A = 1.80(10-3) mm>mm (gxy)A = 0.0599 rad PBE = -0.0198 mm>mm 119

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  y

The material distorts into the dashed position shown. Determine the average normal strains along the diagonals AD and CF.

15 mm C

30 mm D

50 mm B 200 mm E

Solution

A

150 mm

50 mm F

x

Geometry: Referring to the geometry shown in Fig. a, LAD = LCF = 21502 + 2502 = 285000 mm

LAD′ = 2(150 + 30)2 + 2502 = 294900 mm LC′F = 2(150 - 15)2 + 2502 = 280725 mm

Average Normal Strain: PAD = PCF =

LAD′ - LAD 294900 - 285000 = = 0.0566 mm>mm LAD 285000

LC′F - LCF 280725 - 285000 = = -0.0255 mm>mm LCF 285000

Ans. Ans.

Ans: PAD = 0.0566 mm>mm PCF = -0.0255 mm>mm 120

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*2–16. The nylon cord has an original length L and is tied to a bolt at A and a roller at B. If a force P is applied  to the roller, determine the normal strain in the cord when the roller is at C, and at D. If the cord is originally unstrained when it is at C, determine the normal strain P′D when the roller moves to D. Show that if the displacements ∆ C and ∆ D are small, then P′D = PD − PC .

D C P

B

C

L

Solution LC = 2L2 + ∆ 2C

=

A

2L2 + ∆ 2C - L L

PC =

L 41 +

For small ∆ C, PC = 1 +

2 C 2

1 ∆L 2

- L

L

=

C

1 + a

∆ 2C L2

b - 1

1 ∆ 2C 1 ∆ 2C a 2b - 1 = 2 L 2 L2

Ans.

In the same manner, PD =

1 ∆ 2D 2 L2

PD ′ =

2L2 + ∆ 2D - 2L2 + ∆ 2C

Ans.

2L2 + ∆ 2C

=

41 +

For small ∆ C and ∆ D, PD ′ =

PD ′ =

11

+

2 1 ∆C 2 L2

11

∆ 2C - ∆ 2D 2

2L -

2

∆ 2C

-

11 ∆ 2C 2

+

2

2 1 ∆D 2 L2

2

=

1 2L2 1 2L2

∆ 2D L2

- 41 +

41 +

∆ 2C L2

∆ 2C L2

( ∆ 2C - ∆ 2D ) ( 2L2 + ∆ 2C )

+

1 2 L

=

1 ( ∆ 2C - ∆ 2D ) = PC - PD 2L2

QED

Also this problem can be solved as follows: AC = L sec uC ;  AD = L sec uD PC =

L sec uC - L = sec uC - 1 L

PD =

L sec uD - L = sec uD - 1 L

Expanding sec u sec u = 1 +

u2 5 u4 + ...... 2! 4!

121

D

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*2–16. Continued

For small u neglect the higher order terms sec u = 1 +

u2 2

Hence, PC = 1 +

u C2 u C2 - 1 = 2 2

PD = 1 +

u D2 u D2 - 1 = 2 2

PD ′ =

L sec uD - L sec uC sec uD = - 1 = sec uD cos uC - 1 L sec uC sec uC

Since cos u = 1 -

u2 u4 + ...... 2! 4!

sec uD cos uC = a1 + = 1 -

u C2 u D2 ......ba1 ......b 2 2

u 2C u 2C u 2D u 2D + 2 2 4

Neglecting the higher order terms sec uD cos uC = 1 + PD ′ = c 1 +

u 2C u 2D 2 2

u C2 u 22 u 12 u D2 d - 1 = 2 2 2 2

QED

= PD - PC

Ans: 1 2 1 = 2

PC = PD 122

∆ 2C L2 ∆ 2D L2

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2–17. P

A thin wire, lying along the x axis, is strained such that each point on the wire is displaced ∆x = kx2 along the x axis. If k is constant, what is the normal strain at any point P along the wire?

x

x

Solution P =

d(∆x) dx

Ans.

= 2 k x

Ans: P = 2kx 123

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2–18. y

The block is deformed into the position shown by the dashed lines. Determine the average normal strain along line AB.

15 mm

30 mm 70 mm 30 mm 55 mm B¿ B

100 mm

110 mm

Solution Geometry:

x

A

AB = 21002 + (70 - 30)2 = 107.7033 mm

30 mm

70 mm

AB′ = 3(70 - 30 - 15)2 + ( 1102 - 152 ) = 111.8034 mm

Average Normal Strain: PAB = =

AB′ - AB AB 111.8034 - 107.7033 107.7033

= 0.0381 mm>mm = 38.1 ( 10-3 ) mm

Ans.

Ans: PAB = 38.1 ( 10-3 ) mm 124

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2–19. Nylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated.

y 2 mm P

3 mm 5 mm 3 mm 5 mm 3 mm

x

Solution 1 mm

From the geometry shown in Fig. a 1 θ = tan- 1 a b = 11.31° = 0.1974 rad 5 The shear strain is

π

π g =−  + θ  = −0.1974 πad = − 0.197 πad 2 2 

Ans.

5 mm

θ

(a)

Ans: g = −0.197 rad 125

CH 02.indd 75

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*2–20.

u � 2�

The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. Assume the columns are rigid and rotate about their lower supports.

u � 2�

B

3m

1m

A

4m

Solution Geometry: The vertical displacement is negligible xA = (1) ¢

2° ≤ p = 0.03491 m 180°

xB = (4) ¢

2° ≤ p = 0.13963 m 180°

x = 4 + xB - xA = 4.10472 m Ans.

A¿B¿ = 232 + 4.104722 = 5.08416 m AB = 232 + 42 = 5.00 m Average Normal Strain: eAB = =

A¿B¿ - AB AB 5.08416 - 5 = 16.8 A 10 - 3 B m>m 5

Ans: A¿B¿ = 5.08416 m 126

CH 02.indd 86

11/29/10 10:21:06 AM

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2–21. y

The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal AC, and the average shear strain at corner A relative to the x, y axes.

6 mm 400 mm 2 mm

2 mm

6 mm C

D

300 mm

Solution

2 mm

Geometry: The unstretched length of diagonal AC is

A

LAC = 23002 + 4002 = 500 mm

400 mm

x

B 3 mm

Referring to Fig. a, the stretched length of diagonal AC is LAC′ = 2(400 + 6)2 + (300 + 6)2 = 508.4014 mm

Referring to Fig. a and using small angle analysis, f =

2 = 0.006623 rad 300 + 2

a =

2 = 0.004963 rad 400 + 3

Average Normal Strain: Applying Eq. 2, (Pavg)AC =

LAC′ - LAC 508.4014 - 500 = = 0.0168 mm>mm LAC 500

Ans.

Shear Strain: Referring to Fig. a, Ans.

(gA)xy = f + a = 0.006623 + 0.004963 = 0.0116 rad

Ans: (Pavg)AC = 0.0168 mm>mm, (gA)xy = 0.0116 rad 127

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2–22.

y

The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B.

A

16 mm D

B

x

3 mm 3 mm 16 mm

16 mm

Solution

C

16 mm

Applying trigonometry to Fig. a f = tan - 1 a a = tan - 1 a

13 p rad b = 39.09° a b = 0.6823 rad 16 180°

16 p rad b = 50.91° a b = 0.8885 rad 13 180°

By the definition of shear strain,

A gxy B A = A gxy B B =

p p - 2f = - 2(0.6823) = 0.206 rad 2 2

Ans.

p p - 2a = - 2(0.8885) = - 0.206 rad 2 2

Ans.

Ans:

Agxy BA = 0.206 rad A gxy B B = -0.206 rad 128

CH 02.indd 77

11/29/10 10:20:57 AM

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2–23. y

Determine the shear strain gxy at corners A and B if the plate distorts as shown by the dashed lines.

5 mm

2 mm 2 mm

B

C

4 mm

300 mm

D

Solution

400 mm

A

2 mm

x

3 mm

Geometry: For small angles, a = c =

2 = 0.00662252 rad 302

 b = u =

2 = 0.00496278 rad 403

Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 1 10 - 3 2 rad

Ans.

(gA)xy = u + c

= 0.0116 rad = 11.6 1 10 - 3 2 rad

Ans.

Ans: (gB)xy = 11.6(10 - 3) rad, (gA)xy = 11.6(10 - 3) rad 129

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*2–24. y

Determine the shear strain gxy at corners D and C if the plate distorts as shown by the dashed lines.

5 mm

2 mm 2 mm

B

C

4 mm

300 mm

D

Solution

400 mm

A

2 mm

x

3 mm

Geometry: For small angles, a = c =

2 = 0.00496278 rad 403

b = u =

2 = 0.00662252 rad 302

Shear Strain: (gC)xy = a + b = 0.0116 rad = 11.6 1 10 - 3 2 rad

Ans.

(gD)xy = u + c

= 0.0116 rad = 11.6 1 10 - 3 2 rad

Ans.

Ans: (gC)xy = 11.6(10 - 3) rad, (gD)xy = 11.6(10 - 3) rad 130

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2–25. y

Determine the average normal strain that occurs along the diagonals AC and DB.

5 mm

2 mm 2 mm

B

C

4 mm

300 mm

D

Solution

400 mm

A

2 mm

x

3 mm

Geometry: AC = DB = 24002 + 3002 = 500 mm

DB′ = 24052 + 3042 = 506.4 mm

A′C′ = 24012 + 3002 = 500.8 mm

Average Normal Strain: PAC =

A′C′ - AC 500.8 - 500 = AC 500

= 0.00160 mm>mm = 1.60 1 10 - 3 2 mm>mm

Ans.

= 0.0128 mm>mm = 12.8 1 10 - 3 2 mm>mm

Ans.

PDB =

DB′ - DB 506.4 - 500 = DB 500

Ans: PAC = 1.60 1 10 - 3 2 mm>mm 131

PDB = 12.8 1 10 - 3 2 mm>mm

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2–26. If the unstreched length of the bowstring is 887.5 mm, determine the average normal strain in the string when it is streched to the position shown.

18 mm in. 450

6 in. 150 mm 18 mm in. 450

Geometry: Referring to Fig. a, the stretched length of the string is 4502 + 150 948.68in. mm 62 2= =37.947 L = 2L¿ = 2 218 Average Normal Strain: eavg =

L - L0 37.947 –-887.5 35.5 948.68 == 0.0689 0.0689mm/mm in.>in. = L0 35.5 887.5

Ans.

450 mm

450 mm

150 mm

Ans: eavg = 0.0689 mm/mm 132

CH 02.indd 75

11/29/10 10:20:56 AM

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2–27. y

The triangular plate is fixed at its base, and its apex  A is given a horizontal displacement of 5 mm. Determine the shear strain, gxy, at A.

45

45

x¿

A¿ 5 mm

800 mm 2

L = 2800 + 5 - 2(800)(5) cos 135° = 803.54 mm sin 135° sin u = ; 803.54 800

gxy =

A

45

Solution 2

800 mm

x

u = 44.75° = 0.7810 rad

p p - 2u = - 2(0.7810) 2 2 Ans.

= 0.00880 rad

Ans: gxy = 0.00880 rad 133

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*2–28. y

The triangular plate is fixed at its base, and its apex  A is given a horizontal displacement of 5 mm. Determine the average normal strain Px along the x axis.

45

45

x¿

A¿ 5 mm

800 mm 2

L = 2800 + 5 - 2(800)(5) cos 135° = 803.54 mm Px =

A

45

Solution 2

800 mm

803.54 - 800 = 0.00443 mm>mm 800

Ans.

x

Ans: Px = 0.00443 mm>mm 134

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2–29. y

The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px′ along the x′ axis.

45

800 mm

45

x¿

A

45

Solution

A¿ 5 mm

800 mm

L = 800 cos 45° = 565.69 mm Px′ =

5 = 0.00884 mm>mm 565.69

Ans.

x

Ans: Px′ = 0.00884 mm>mm 135

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2–30.

y y�  x2

The rubber band AB has an unstretched length of 1 m. If it is fixed at B and attached to the surface at point A¿, determine the average normal strain in the band.The surface is defined by the function y = (x2) m, where x is in meters.

A¿ ft 1m B

Solution

ft 1m

x

A

Geometry: L =

L0

1m ft

A

1 + a

dy 2 b dx dx

However y = x2 then L = =

L0

dy = 2x dx

1m ft

21 + 4 x2 dx

1 1m ft C 2x21 + 4 x2 + ln A 2x + 21 + 4x2 B D �0 4

= 1.47894 ft m

Average Normal Strain: L - L0 1.47894 - 1 = = 0.479 m/m ft>ft eavg = L0 1

Ans.

Ans: eavg = 0.479 m/m 136

CH 02.indd 89

11/29/10 10:21:08 AM

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2–31. y

The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal BD, and the average shear strain at corner B relative to the x, y axes.

6 mm 400 mm 2 mm

2 mm

6 mm C

D

300 mm

Solution

2 mm

Geometry: The unstretched length of diagonal BD is

A

LBD = 23002 + 4002 = 500 mm

400 mm

x

B 3 mm

Referring to Fig. a, the stretched length of diagonal BD is LB′D′ = 2(300 + 2 - 2)2 + (400 + 3 - 2)2 = 500.8004 mm

Referring to Fig. a and using small angle analysis, f =

2 = 0.004963 rad 403

a =

3 = 0.009868 rad 300 + 6 - 2

Average Normal Strain: Applying Eq. 2, (Pavg)BD =

LB′D′ - LBD 500.8004 - 500 = = 1.60(10 - 3) mm>mm Ans. LBD 500

Shear Strain: Referring to Fig. a, (gB)xy = f + a = 0.004963 + 0.009868 = 0.0148 rad

Ans.

Ans: (Pavg)BD = 1.60(10 - 3) mm>mm, (gB)xy = 0.0148 rad 137

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*2–32. The nonuniform loading causes a normal strain in the shaft p that can be expressed as Px = k sin a xb , where  k is a L constant. Determine the displacement of the center C and the average normal strain in the entire rod.

C A

B L — 2

L — 2

Solution Px = k sin a (∆x)C =

L0

p xb L L>2

Px dx =

L0

L>2

k sin a

p xb dx L

L>2 p L L p = -k a b cos a xb ` = - k a bacos - cos 0b p p L 2 0

= (∆x)B =

kL p L0

Ans.

L

k sin a

p xbdx L

L L p L 2kL = - k a b cos a xb ` = - k a b(cos p - cos 0) = p p p L 0

Pavg =

(∆x)B L

=

2k p

Ans.

Ans:

kL p 2k = p

(∆x)C = Pavg 138

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2–33. y

The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements u A  and vB respectively, determine the normal strain in the fiber when it is in position A′ B′.

B¿ vB B L A

uA A¿

u x

Solution Geometry: LA′B′ = 2(L cos u - u A)2 + (L sin u + vB)2

= 2L2 + u 2A + v2B + 2L(vB sin u - u A cos u)

Average Normal Strain: PAB =

=

LA′B′ - L L

C

1 +

u 2A + v2B 2

L

+

2(vB sin u - u A cos u) L

- 1

Neglecting higher terms u 2A and v2B PAB = J1 +

2(vB sin u - u A cos u) L

1 2

R - 1

Using the binomial theorem: PAB = 1 + =

2u A cos u 1 2vB sin u a b + c -1 2 L L

vB sin u u Acos u L L

Ans.

Ans: PAB = 139

vB sin u u Acos u L L

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2–34. If the normal strain is defined in reference to the final length ∆s′, that is, P= = lim a ∆s′ S 0

∆s′ - ∆s b ∆s′

instead of in reference to the original length, Eq. 2–2, show that the difference in these strains is represented as a second-order term, namely, P - P= = P P′.

Solution P =

∆s′ - ∆s ∆s

P - P= =

∆s′ - ∆s ∆s′ - ∆s ∆s ∆s′

=

∆s′2 - ∆s∆s′ - ∆s′∆s + ∆s2 ∆s∆s′

=

∆s′2 + ∆s2 - 2∆s′∆s ∆s∆s′

=

(∆s′ - ∆s)2 ∆s∆s′

= ¢

∆s′ - ∆s ∆s′ - ∆s ≤¢ ≤ ∆s ∆s′

= P P=

(Q.E.D)

Ans: N/A 140

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3–1. A tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use a scale of 25 mm = 140 MPa and 25 mm= 0.05 mm >mm. Redraw the elastic region, using the same stress scale but a strain scale of 25 mm = 0.001 mm >mm.

= A

Load (kN) Elongation (mm) 0 7.0 21.0 36.0 50.0 53.0 53.0 54.0 75.0 90.0 97.0 87.8 83.3

1 = π (0.01252 ) 122.72(10 −6 ) m 2 4

L = 50 mm. s(M Pa)

e(mm> mm)

0

0

57.07

0.00025

171.21

0.00075

293.51

0.00125

407.66

0.00175

432.12

0.0025

432.12

0.0040

440.27

0.010

611.49

0.020

733.79

0.050

790.86

0.140

715.85

0.200

679.16 Eapprox =

0 0.0125 0.0375 0.0625 0.0875 0.125 0.2 0.5 1.0 2.5 7.0 10.0 11.5

0.230 6

336(10 ) 9 ) N/m 2 224 GPa = 224(10= 0.0015

Ans.

n sio ver is y r ina M). It ry lim S a I e ( n r i l p nua elim t a me sen ns Ma his pr e so s. e r r p t o a i e t r a ut tion e ons s' Sol olu ll be hat her i s t t t u r y wi sol SM cto kel raft se tru ven li the I he d anual The e Ins t e m g s th and parin rs in thi of nd re ro ible r a p s e s ed po of and rrect ge co sta sions e b is om e will . s d e e Th blish u p e r

Ans: (sult)approx = 770 MPa, (sR)approx = 652 MPa, (sY)approx = 385 MPa, Eapprox = 224 GPa 141

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3–2. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience.

Solution

00 232.4 33.2 45.5 318.5 49.4 345.8 51.5 360.5 53.4 373.8

0 0 0.0006 0.0006 0.0010 0.0010 0.0014 0.0014 0.0018 0.0018 0.0022

0.0022

420

232.4(106 ) − 0 9 = ) N/m 2 387 GPa = 387.33(10 0.0006 − 0

Ans. 350

Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress–strain diagram (shown shaded). = (U i )r

eP (mm/mm) (in./in.)

σ (MPa)

Modulus of Elasticity: From the stress–strain diagram E =

S(MPa) (ksi)

1 N  m 3 N⋅m 232.4(106 ) = ) 69.7 kJ/m 3 =  0.0006  69.72(10 2  m m 2   m3

Ans.

280 232.4 210

140

70

e (mm/mm)

Ans: E = 387 GPa , ur = 69.7 kJ/m 3 142

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3–3. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is sr = 53.4 373.8ksi. MPa

(MPa) S (ksi) 00 232.4 33.2 45.5 318.5 49.4 345.8 51.5 360.5 53.4 373.8

Solution Modulus of Toughness: The modulus of toughness is equal to the area under the stress–strain diagram (shown shaded).

= (U i )t

1 N  m 232.4(106 ) 2  (0.0004 + 0.0010)  2  m m  

s (MPa) 420

350 318.5 280

N  m 1 55.3(106 ) 2   0.0012  2  m m  +

0 0 0.0006 0.0006 0.0010 0.0010 0.0014 0.0014 0.0018 0.0018 0.0022 0.0022

373.8

m N   +  318.5(106 ) 2   0.0012  m m   +

e P(mm/mm) (in./in.)

232.4 210

N  m 1 86.1(106 ) 2   0.0004   2 m m 

140

3 N⋅m = 595.28(10 = ) 595 kJ/m 3 m3

Ans.

70

e (mm/mm)

Ans: (ui)t  595 kJ/m3 143

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*3–4. The stress–strain diagram for a metal alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.

Solution From the stress–strain diagram, Fig. a, = E

283(106 ) 9 = 283(10= ) N/m 2 283 GPa 0.001

Ans.

= σ Y 283 = MPa σ σ /t 547 GPa

Thus, 6  π 2   = PY σ= = ) 32.01(10 3= ) N 32.0 kN Y A  283(10 )  4 (0.012 

Ans.

6  π 2  3  = = ) 61.86(10= ) N 61.9 kN Pσ /t σ= σ /t A  547(10 )  4 (0.012 

Ans.

Ans: E  283 GPa, PY  32.0 kN, Pult  61.9 kN 144

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3–5. The stress–strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. If the specimen is loaded until it is stressed to 500 MPa, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded.

Solution From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is = E

283(106 ) 9 = 283(10= ) N/m 2 283 GPa 0.001

when the specimen is unloaded, its normal strain recovered along line AB, Fig. a, which has a gradient of E. Thus

Elasticity Recovery =

500(106 ) = = 0.001767 mm/mm E 283(109 )

σ

Amount of Elastic Recovery = (0.001767 mm/mm)(50 mm) = 0.08834 mm = 0.0883 mm

Ans.

PP = 0.05 Thus, the permanent set0.0 is eP = 0.08 – 0.001767 = 0.07823 mm/mm = 0.047(2) = PPL Then, the ¢L increase in gauge length= is0

Ans.

Ans: Elastic Recovery  0.0883 mm 욼L  3.91 mm 145

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3-6. The stress–strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material.

Solution The Modulus of resilience is equal to the area under the stress–strain diagram up to the proportional limit. sPL = 283 MPa

PPL = 0.001 m/m

Thus, (Ui)r =

1 N  m 1 3 N⋅m sPLPPL =  283(106 ) =  0.001  141.5(10 )  2 2 m 2 m  m3 = 141.5 kJ/m 3

Ans.

The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number is 32. Thus, N  m  6 N⋅m (U i )  = 32 100(106 ) = ) 128 MJ/m 3 =  0.04  128(10 ult  aprox  m m 2   m3 

Ans.

Ans: (ui)r = 141.5 kJ>m3, [(Ui)ult]approx = 128 MJ>m3 146

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3–7. A specimen isisoriginally long, hashas a diameter of originally300 1 ftmm long, a diameter 0.5 12 in.,mm, and isand subjected to a force 500 lb. theWhen force of is subjected to aofforce ofWhen 2.5 kN. is increased from 500 lb to 1800 lb, the the force is increased from 2.5 kN to 9specimen kN,, the elongates specimen 0.009 in. 0.225 Determine the modulus of elasticity for the elongates mm. Determine the modulus of elasticity for material if it if remains linear elastic. the material it remains linear elastic.

Solution Normal Stress and Strain: Applying s =

= σ1 = σ2 = ∆ε

dL P and e = . A L

2.5(10 3 ) 6 = 22.10(10 = ) N/m 2 22.10 MPa π (0.012 2 ) 4 9(10 3 ) 6 = 79.58(10 = ) N/m 2 79.58 MPa 2 (0.012 ) 4

π

0.225 = 0.000750 mm/mm 300

Modulus of Elasticity: E =

∆σ (79.58 − 22.10)(106 ) Pa 76.6 GPa = = 76.63(109 ) = 0.000750 ∆ε

Ans.

Ans: E  76.6 GPa 147

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*3–8. The strut is supported by a pin at C and an A-36 AB. If the the wire wire has has aa diameter diameter of of 0.2 5 mm, steel guy wire AB. in., determine how much it stretches when the distributed load acts on the strut.

A

60� 60 200 lb/ft 3.4 kN/m

Solution C

Here, we are only interested in determining the force in wire AB. a + ©MC = 0;

11 FAB cos 60°(2.7) 60°(9) -–  (3.4)(2.7)(0.9) 600kN lb (200)(9)(3) = 0= 0  F FAB AB cos AB ==3.06 22

B 9 ftm 2.7

The normal stress the wire is FAB 3.06(10 3 ) 6 155.84(10 = ) N/m 2 155.84 MPa = π sAB = = 2 AAB (0.005 ) 4 250ksi MPa, Hooke’s Lawcan canbe beapplied applied to determine , Hooke’s Law determine the thestrain strain Since sAB 6 sY = 36 in wire. sAB = EPAB;

155.84(106) = 200(109)PAB 0.7792(10–3 - 3) mm/mm AB = PeAB = 0.6586(10 ) in>in

9(12)3) 2.7(10 The unstretched length of the wire is LAB = = 3117.69 mm. Thus, the wire sin sin 60° 60° stretches -3 0.7792(10–3 )(3117.69) )(124.71) dAB = PAB LAB = 0.6586(10

= 2.429 mm = 2.43 mm

Ans.

1  (3.4)(2.7) kN 2

0.9 m

1.8 m

Ans: dAB = 2.43 mm 148

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3–9. The s -P diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.

s (MPa) 0.385

0.077 1

0.385

0.077 = 0.0385 MPa 2

(U i )r =

Ans.

1 0.077(106 ) N/m 2  (2 m/m)  2 3

3

= 77.0(10 ) N ⋅ m/= m 77.0 kJ/m

(U i )t = +

P (mm/mm)

s (MPa)

Solution E =

2 2.25

0.077 3

1

Ans.

2 2.25

P (mm/mm)

1 0.077(106 ) N/m 2  (2 m/m)  2

{

}

1  0.077(106 ) + 0.385(106 ) N/m 2 [(2.25 − 2) m/m ]  2 

3 = 134.75 N ⋅ m/m= 135 kJ/m 3

Ans.

Ans: E = 0.0385 MPa, (ui)r = 77.0 kJ>m3, (ui)t = 135 kJ>m3 149

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3–10. A structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 20 kN is to be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 1 m long and its elongation is 0.5 mm? Ezr = 100 GPa, sY = 400 MPa. The material has elastic behavior.

Solution Allowable Normal Stress: F.S. = 3 =

sy sallow 57.5 400 sallow

sallow = 19.17 133.33ksi MPa sallow = 19.17 5 = 133.33

P A 3 4 20(10 ) AA

2 2 A == 150 0.2087 = 0.209 in2 mmin

Ans.

Stress–Strain Relationship: Applying Hooke’s law with e =

0.02 d 0.5 == 0.0005 mm/mm = 0.000555 in.>in. 3 L 3 (12) 1(10 ) 33 )(0.0005) = 50 MPa s == Ee Ee==100(10 14 A 10 = 7.778 ksi B (0.000555)

Normal Force: Applying equation s =

P . A

P ==sA sA= =50(150) 7.778 (0.2087) kip P = 7500 N== 1.62 7.5 kN

Ans.

Ans: A  150 mm 2, P  7.5 kN 150

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3–11.

s (ksi) (MPa)

A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the proportional limit,(b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method.

70 490 60 420 50 350 40 280 30 210 20 140 10 70 00

Solution

0.02 0.02 0.002 0.002

0.04 0.04 0.004 0.004

0.06 0.06 0.006 0.006

0.08 0.08 0.008 0.008

0.10 0.10 0.010 0.010

(in./in.) PP (mm/mm)

Proportional Limit and Yield Strength: From the stress–strain diagram, Fig. a, 44 ksi spl ==308 MPa

Ans.

60 ksi sYY ==420 MPa

Ans.

Modulus of Elasticity: From the stress–strain diagram, the corresponding strain for is is Thus, sPL == 308 44 ksi eplepl= =0.004 MPa 0.004in.>in. mm/mm. Thus, E5 =

44 –- 00 308 33 5=77.0(10 ) )MPa 11.0(10 ksi = 77.0 GPa 0.004 –-00 0.004

Ans.

Ans: spl = 308 MPa sY = 420 MPa E = 77.0 GPa 1 51

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*3–12.

(ksi) ss(MPa)

A tension test was performed on an aluminum 2014-T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the modulus of resilience; and (b) modulus of toughness.

70 490 60 420 50 350 40 280 30 210 20 140 10 70 0

Solution

0.02 0.002

0.04 0.004

0.06 0.006

0.08 0.008

0.10 0.010

(in./in.) PP(mm/mm)

Modulus of Resilence The modulus of resilence is equal to the area under the stress–strain diagram up to the propotional limit. From the stress–strain diagram, spl ==308 44 ksi MPa

e pl == 0.004 0.004mm/mm in.>in.

Thus, 1 2

1 2

A Ui B r = splepl = [308(106)](0.004) = 0.616 MJ/m3

Ans.

Modulus of Toughness: The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number of squares is 65. Thus,

C A Ui Bt Dapprox = 65[70(106)](0.01) = 45.5 MJ/m3

Ans.

s (MPa)

490 420 350 280 210 140 70 e (mm/mm)

Ans:

AUi Br = 0.616 MJ/m3

152

CH 03.indd 106

C A Ui Bt Dapprox = 45.5 MJ/m3

1/18/11 2:08:24 PM

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3.13. 8000 40 kNlb

A bar having a length of 125 mm and cross-sectional area of 437.5 mm2 is subjected to an axial force of 40 kN. If the bar streches 0.05 mm, determine the modulus of elasticity of the material. The material has linear-elastic behavior.

8000 40 kNlb

5 in. 125 mm

Solution Normal Stress and Strain:

s5

40(103) P 5 5 91.43(106) N/m2 5 91.43 MPa 437.5(10−6) A

e5

dL 0.05 5 5 0.000400 mm/mm L 125

Modulus of Elasticity: E5

91.43(106) s 5 5 228.57(109) N/m2 5 229 GPa e 0.000400

Ans.

Ans: E 5 229 GPa 1 53

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3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 6.5 mm, determine how much it streches when a load of P ==3600 kN acts onon thethe pipe. pipe. lb acts

B

1.2 4 ftm

P A

D C 0.9 3 ftm

0.9 3 ftm

Solution Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a

FBD

a + ©MA = 0;

= ( 45 ) (0.9) − 3(1.8)

0

= 7.50 kN FBD

The normal stress developed in the wire is

σ= BD

FBD = ABD

7.50(10 3 ) 2 = 226.02(106 ) N/m = 226.02 MPa 2 (0.0065 ) 4

π

Since sBD 6 sy = 36 , Hooke’s Law cancan bebe applied totodetermine 250ksi MPa, Hooke’s Law applied determinethe thestrain straininin the wire. sBD = EPBD;

226.02(106 ) = 200(109 ) PAB PBD = 1.1301(10 −3 ) mm/mm

1.5=m.60 Thus, the the in. Thus, The unstretched length of the wire is LBD = 23 0.922 ++ 14.222==5ft wire stretches dBD = PBD LBD = 1.1301(10 −3 )[1.5(10 3 )]

= mm 1.70 mm = 1.6951

Ans.

3 kN

0.9 m

0.9 m

Ans: dBD = 1.70 mm 1 54

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3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. BD. If If the the wire wire has hasaadiameter diameterofof0.25 6 mm, in., determine the load load PP ifif the the end end CCisisdisplaced displaced1.875 0.075mm in. downward.

B

4 ftm 1.2

P A

D C 3 ftm 0.9

Solution

3 ftm 0.9

Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a FBD A 45 B (0.9) (3) -–P(6) P(1.8)= =0 0

a + ©MA = 0;

FBD = 2.50 P

The unstretched length for wire BD is LBD = 23 ft m. = 60 in. From From the the 0.922 ++ 14.222==51.5 geometry shown in Fig. b, the stretched length of wire BD is LBD¿ =

15002 + 1.6752 - 2(1500)(1.675) cos 143.13° = 1501.3403 mm

Thus, the normal strain is PBD =

LBD¿ - LBD 1501.3403 – 1500 5 0.8936(10−3) mm/mm = 1500 LBD

Then, the normal stress can be obtain by applying Hooke’s Law. 9 6 = )[0.8936(10 −3 )] 178.71(10 = ) N−m 2 178.71 MPa sBD = EPBD 200(10=

MPa, result is valid. 36 ksi Since sBD 6 sy = 250 , thethe result is valid. sBD =

FBD ; ABD

178.71(106 ) =

2.50 P π

(0.00652 ) 4

P = 2.3721(10 3 ) N = 2.37 kN

Ans.

LBD = 1500 mm

0.9 m

0.9 m

1.675 mm

Ans:

P = 2.37 kN

155

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*3–16. Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used for connections. If a nut on the bolt is tightened so that the six 3-mm high heads of the indicator are strained 0.1 mm>mm, and leave a contact area on each head of 1.5 mm2, determine the tension in the bolt shank. The material has the stress–strain diagram shown.

s (MPa) 3 mm

600 450

0.0015

0.3

P (mm/ mm)

Solution Stress–Strain Relationship: From the stress–strain diagram with P = 0.1 mm>mm 7 0.0015 mm>mm s - 450 600 - 450 = 0.1 - 0.0015 0.3 - 0.0015 s = 499.497 MPa Axial Force: For each head P = sA = 499.4971 ( 106 ) (1.5) ( 10-6 ) = 749.24 N Thus, the tension in the bolt is Ans.

T = 6 P = 6(749.24) = 4495 N = 4.50 kN

Ans: T = 4.50 kN 156

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3–17.

B

The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the diameter of the post is 80 mm.

2m P

A

Solution

C 0.75 m 0.75 m

D

0.5 m

From the stress–strain diagram, E =

32.2(10)6 = 3.22(109) Pa 0.01

s (MPa) 100 95

Thus,

70 60

40(10 ) FAB = p = 31.83 MPa 2 AAB 4 (0.04)

sAB =

eAB

31.83(106) sAB = 0.009885 mm>mm = = E 3.22(109)

sCD

40(103) FCD = = p = 7.958 MPa 2 ACD 4 (0.08)

50 tension

40 32.2 20 0

6

eCD =

compression

80 3

7.958(10 ) sCD = 0.002471 mm>mm = E 3.22(109)

0

0.01 0.02 0.03 0.04

P (mm/mm)

dAB = eABLAB = 0.009885(2000) = 19.771 mm dCD = eCDLCD = 0.002471(500) = 1.236 mm Angle of tilt a: tan a =

18.535 ; 1500

Ans.

a = 0.708°

Ans. a = 0.708° 157

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3–18.

B

The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm.

2m P

A

C

Solution

0.75 m 0.75 m

D

0.5 m

Rupture of strut AB: sR =

FAB ; AAB

50(106) =

s (MPa)

P>2

; p 2 4 (0.012)

100 95

Ans.

P = 11.3 kN (controls)

70 60

Rupture of post CD: sR =

FCD ; ACD

compression

80

95(106) =

50

P>2

tension

40 32.2

p 2 4 (0.04)

20

P = 239 kN

0

0

0.01 0.02 0.03 0.04

P (mm/mm)

Ans. P = 11.3 kN (controls) 1 58

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3–19.  The stress–strain diagram for a bone is shown, and can be P = 0.45110-6 2 s + described by the equation -12 3 0.36110 2 s , where s is in kPa. Determine the yield strength assuming a 0.3% offset.

P

s

P  0.45(106)s + 0.36(1012)s3 P

P

Solution P = 0.45(10-6)s + 0.36(10-12)s3, dP = E =

1 0.45(10-6)

+ 1.08(10-12) s2 2 ds

ds 1 ` = = 2.22(106) kPa = 2.22 GPa dP 0.45(10 - 6) s=0

The equation for the recovery line is s = 2.22(106)(P - 0.003). This line intersects the stress–strain curve at sYS = 2027 kPa = 2.03 MPa

Ans.

Ans: sYS = 2.03 MPa 159

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*3–20. The stress–strain diagram for a bone is shown and can be described by the equation P = 0.45110-6 2 s + 0.36110-12 2 s3, where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200-mm-long region just before it fractures if failure occurs at P = 0.12 mm>mm.

P

s

P  0.45(106)s + 0.36(1012)s3 P

P

Solution When P = 0.12 120(10-3) = 0.45 s + 0.36(10-6)s3 Solving for the real root: s = 6873.52 kPa ut = ut =

LA L0

dA =

L0

6873.52

(0.12 - P)ds

6873.52

(0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds 6873.52

= 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4  0 = 613 kJ>m3

Ans. Ans.

d = PL = 0.12(200) = 24 mm

Ans: u t = 613 kJ>m3, d = 24 mm 160

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3–21. The two bars are made of polystyrene, which has the stress–strain diagram shown. If the cross-sectional area of bar AB is 975 mm2 and BC is 2600 mm2, determine the largest force P that can be supported before any member ruptures. Assume that buckling does not occur.

P 1.2 m

C

B

1m

A (MPa)

Solution

175

5 + c gFy = 0; FAB - P = 0; 61  61   6  + FBC = P   = ; ©F x = 0;  0  5   61  Assuming failure of bar BC:

61 P 5

(C)

FBC = 1.20 P

(T)

FAB =

140

(1)

105

(2)

FBC ; ABC

35(106 ) =

tension

35 0

From the stress–strain diagram (sR)t = 35 M Pa s =

compression

70

0

0.20

0.40

0.60

0.80

(mm/mm)

FBC 3 = FBC 91.0(10 = ) N 91.0 kN ; 2600(10 −6 )

From Eq. (2), P = 75.83 kN Assuming failure of bar AB: From stress–strain diagram (sR)c = 175 M Pa s =

FAB ; AAB

175(106 ) =

FAB 3 ; = FAB 170.625(10 = ) N 170.625 kN 975(10 −6 )

From Eq. (1), P ⫽ 109.23 kN Choose the smallest value P = 75.8 kN

Ans.

Ans: P = 75.8 kN 161

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3–22. The two bars are made of polystyrene, which has the stress–strain diagram shown. Determine the cross-sectional area of each bar so that the bars rupture simultaneously when the load P = 13.5 kN. Assume that buckling does not occur.

P 1.2 m

C

B

1m

A (MPa)

Solution

175

+ c ©Fy = 0;

 5  FAB=   − 13.5= 0  61 

FAB= 21.09 kN

+ : ©Fx = 0;

 6  − FBC + 21.06  =  0  61 

FBC 16.2 kN =

140 105

compression

70 tension

35

For member BC:

0

(smax)t =

FBC ; ABC

35(106 ) =

0

0.20

0.40

0.60

0.80

(mm/mm)

16.2(10 3 ) ; ABC

−3 = ) m 2 462.9 = mm 2 463 mm 2 ABC 0.4629(10 =

Ans.

For member BA: (smax)c =

FBA ; ABA

175(106 ) =

21.09(10 3 ) ; AAB

−3 = ) m 2 120.5 = mm 2 121 mm 2 AAB 0.1205(10 =

Ans.

Ans: ABC = 463 mm2, ABA = 121 mm2 162

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3–23. The stress–strain diagram for many metal alloys can be described analytically using the Ramberg-Osgood three parameter equation P = s>E + ksn, where E, k, and n are determined from measurements taken from the diagram. Using the stress–strain diagram shown in the 3 E == 30110 210 GPa figure, take E and determine determine the the other two 2 ksiand parameters k and n and thereby obtain an analytical expression for the curve.

ss(MPa) (ksi) 560 80 420 60 280 40 140 20 0.1 0.2 0.3 0.4 0.5

–6) P (10–6

Solution Choose, s == 280 MPa, e e= =0.1 0.1 40 ksi, s == 420 MPa, e e= =0.3 0.3 60 ksi, 0.1== 0.1

40 280 nn k(280) ++ k(40) 33 30(10 )) 210(10

0.3== 0.3

60 420 nn k(420) ++ k(60) 33 30(10 )) 210(10

0.098667 == k(280) 0.098667 k(40)nn 0.29800 == k(420) 0.29800 k(60)nn 0.3310962 = (0.6667)n ln (0.3310962) = n ln (0.6667) n = 2.73

Ans.

k = 21.0(10-9)

Ans.

Ans: n = 2.73 k = 21.0(10-9)

1 63

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*3–24. The s–P diagram for a collagen fiber bundle from which a human tendon is composed is shown. If a segment of the Achilles Achilles tendon tendon at atAAhas hasaalength lengthofof165 6.5mm in. and an approximate cross-sectional area of 145 mm 0.229 in2, determine its elongation if the foot supports a load of 125 625 lb, N, which causes a tension in the tendon of 1718.75 N. 343.75 lb.

ss(MPa) (ksi) 31.50 4.50

A

26.25 3.75 21.00 3.00 15.75 2.25 11.85 10.50 1.50

625 125 N lb

5.25 0.75 0.035 0.05

0.10

P (mm/mm) (in./in.)

Solution σ=

P 1718.75 = = 11.85 MPa A 145(10 −6 )

From the graph e == 0.035 0.035mm/mm in.>in. Ans.

0.035(165) = 5.775 d = eL = 0.035(6.5) 0.228 mm in.

Ans: d = 5.775 mm

164

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3–25. The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Ep = 2.70 GPa, np = 0.4.

300 N

300 N 200 mm

Solution s =

P = A

Plong =

300 p 2 4 (0.015)

= 1.678 MPa

1.678 ( 106 ) s = = 0.0006288 E 2.70 ( 109 ) Ans.

d = Plong L = 0.0006288(200) = 0.126 mm Plat = - nPlong = - 0.4(0.0006288) = -0.0002515 ∆d = Platd = -0.0002515(15) = -0.00377 mm

Ans.

Ans: d = 0.126 mm, ∆d = - 0.00377 mm 165

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3–26. The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve are 50 mm long. Determine the axial pressure p that must be applied to the top of the plug to cause it to contact the sides of the sleeve. Also, how far must the plug be compressed downward in order to do this? The plug is made from a material for which E = 5 MPa, n = 0.45.

p

Solution Plat =

d′ - d 32 - 30 = = 0.06667 mm>mm d 30

v = -

Plat Plat 0.06667 ;  Plong = = = - 0.1481 mm>mm Plong v 0.45

p = s = E Plong = 5 ( 106 ) (0.1481) = 741 kPa

Ans.

d =  Plong L  =  -0.1481(50)  = 7.41 mm

Ans.

Ans: p = 741 kPa, d = 7.41 mm 166

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3–27. The elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. When the applied load on the specimen is 50 kN, the diameter is 12.67494 mm. Determine Poisson’s ratio for the material.

s (MPa) 490

Solution

0.007

P (mm/ mm)

Average Normal Stress: 50 ( 10 ) N = = 394.71 ( 106 ) Pa = 394.71 MPa p A 2 ( 0.0127 ) 4 3

s =

Average Normal Strain: Referring to the stress–strain diagram, the modulus of 490 ( 106 ) elasticity is E = = 70.0 ( 109 ) Pa = 70.0 GPa. 0.007 Plong = Plat =

394.71 ( 106 ) s = = 0.0056386 mm>mm E 70.0 ( 109 )

d - d0 12.67494 - 12.7 = = - 0.0019732 d0 12.7

Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio, that is n = -

( - 0.019732) Plat = = 0.350 Plong 0.0056386

Ans.

Ans: n = 0.350 167

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*3–28. The elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. If a load of P = 60 kN is applied to the specimen, determine its new diameter and length. Take n = 0.35.

s (MPa) 490

Solution

0.007

P (mm/ mm)

Average Normal Stress: 60 ( 10 ) N = = 473.65 ( 106 ) Pa = 473.65 MPa p A 2 ( 0.0127 ) 4 3

s =

Average Normal Strain: Referring to the stress–strain diagram, the modulus of 490 ( 106 ) elasticity is E = = 70.0 ( 109 ) Pa = 70.0 GPa. 0.007 Plong =

473.65 ( 106 ) s = = 0.0067664 mm>mm E 70.0 ( 109 )

Thus,

dL = Plong L0 = 0.0067664(50.8) = 0.34373 mm

Then

Ans.

L = L0 + dL = 50.8 + 0.34373 = 51.1437 mm

Poisson’s Ratio: The lateral strain can be related to the longitudinal strain using Poisson’s ratio.

Plat = - nPlong = - 0.35(0.0067664) = - 0.0023682 mm>mm

Thus,

dd = Plat d = - 0.0023682(12.7) = -0.030077 mm

Then d = d 0 + dd = 12.7 + ( - 0.030077) = 12.66992 mm Ans.

= 12.67 mm

Ans: L = 51.1437 mm, d = 12.67 mm 168

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3–29. The brake pads for a bicycle tire are made of rubber. If a frictional force of 50 N is applied to each side of the tires, determine the average shear strain in the rubber. Each pad has cross-sectional dimensions of 20 mm and 50 mm. Gr = 0.20 MPa. 50 mm

Solution

10 mm

10 mm

Average Shear Stress: The shear force is V = 50 N. t =

V 50 = = 50.0 kPa A 0.02(0.05)

Shear-Stress – Strain Relationship: Applying Hooke’s law for shear t = Gg 50.0 ( 10

3

) = 0.2 ( 106 ) g Ans.

g = 0.250 rad

Ans: g = 0.250 rad 169

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3–30.

P 2

The lap joint is connected together using a 30 mm diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the shear strain developed in the shear plane of the bolt when P = 340 kN.

P 2

P

t (MPa) 525 350

Solution Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the free-body diagram shown in Fig. a. + : ©Fx = 0;

340 - 2V = 0

0.005

0.05

g (rad)

V = 170 kN

Shear Stress and Strain: V 170(10 3 ) 6 240.50(10 = ) N/m 2 240.50 MPa t = = = π (0.032 ) A 4 Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b. 240.50 350 = ; g 0.005

g = 3.4357(10-3) rad = 3.44(10-3) rad

Ans.

Ans: g = 3.44(10-3) rad 170

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3–31

P 2

The lap joint is connected together using a 30 mm diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the permanent shear strain in the shear plane of the bolt when the applied force P = 680 kN is removed.

P 2

P

t (MPa) 525

Solution

350

Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the free-body diagram shown in Fig. a. + : ©Fx = 0;

680 - 2V = 0

V = 340 kN 0.005

Shear Stress and Strain:

0.05

g (rad)

340(10 3 ) V 6 481.00(10 = ) N/m 2 481.00 MPa = π t= = 2 A (0.03 ) 4

Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b. 481.00 -350 = 525 - 350 ; g - 0.005 0.05 - 0.005

g = 0.03869 rad

When force P is removed, the shear strain recovers linearly along line BC, Fig. b, with a slope that is the same as line OA. This slope represents the shear modulus. = G

350(10 3 ) 9 = 700.0(10 = ) Pa 70.0 GPa 0.005

Thus, the elastic recovery of shear strain is t = Ggr;

481.00(106) = 70.0(109)gr

gr = 6.8715(10-3) rad

And the permanent shear strain is gP = g - gr = 0.03869 - 6.874(10-3) = 0.031815 rad = 0.0318 rad

Ans.

Ans: gP = 0.0318 rad 171

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*3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr = - tan g = - tan1P>12phGr22. For small angles we can write dy>dr = - P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = ro. From the result compute the deflection y = d of the plug.

P

h

ro

y

d

ri r y

Shear Stress–Strain Relationship: Applying Hooke’s law with tA

g =

P = . 2p r h

tA P = G 2p h G r

dy P = - tan g = - tan a b dr 2p h G r

(Q.E.D)

If g is small, then tan g = g. Therefore, dy P = dr 2p h G r

At r = ro,

y = -

dr P 2p h G L r

y = -

P ln r + C 2p h G

0 = -

P ln ro + C 2p h G

y = 0

C =

Then, y =

ro P ln r 2p h G

At r = ri,

y = d d =

P ln ro 2p h G

ro P ln ri 2p h G

Ans.

Ans. d =

172

ro P ln ri 2p h G

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3–33. The support consists of three rigid plates, which are connected together using two symmetrically placed rubber pads. If a vertical force of 5 N is applied to plate A, determine the approximate vertical displacement of this plate due to shear strains in the rubber. Each pad has cross-sectional dimensions of 30 mm and 20 mm. Gr = 0.20 MPa.

C

B

40 mm

40 mm

A

Solution tavg

V 2.5 = = = 4166.7 Pa A (0.03)(0.02)

g =

4166.7 t = 0.02083 rad = G 0.2(106)

5N

Ans.

d = 40(0.02083) = 0.833 mm

Ans. d = 0.833 mm 173

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3–34. A shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate A when the vertical load P is applied. Assume that the displacement is small so that d = a tan g ≈ ag.

P

d A

h

Solution P Average Shear Stress: The rubber block is subjected to a shear force of V = . 2

a

a

P

t =

V P 2 = = A bh 2bh

Shear Strain: Applying Hooke’s law for shear P

g =

t P 2bh = = G G 2bhG

Thus, d = ag =

Pa 2bhG

Ans.

Ans: d = 174

Pa 2bhG

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R3–1.

s (ksi) (MPa)

The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. When the applied load is 45 kN, the new diameter of the specimen is 12.4780 mm. Calculate the shear modulus Gal for the aluminum.

350

P (mm/mm) (in./in.)

0.00480

Solution From the stress–strain diagram,

σ 350(106 ) 2 72.92 GPa = = 72.92(109 ) N/m= ε 0.00480

Eal=

When specimen is loaded with a 45-kN load,

σ=

= ε lonε

ε lat = = V

= Gal

45(10 3 )

P = A

2 = 366.69(106 ) N/m= 366.69 MPa 2 (0.0125 ) 4

π

σ

= Eal

366.69(106 ) = 5.0289(10 −3 ) mm/mm 72.92(109 )

d′ − d 12.4780 − 12.5 = = −1.76(10 −3 ) mm/mm d 12.5

ε ε lonε

lat =

−1.76(10 −3 ) = 0.3500 5.0289(10 −3 )

72.92(109 ) Eat = = 27.00(109 )= N/m 2 27.0 GPa 2(1 + v) 2(1 + 0.3500)

Ans.

Ans:

Gal = 27.0 GPa 175

CH 03.indd 116

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R3–2.

s (ksi) (MPa)

The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. If the applied load is 40 kN, determine the new diameter of the specimen. The shear modulus is Gal = 27 GPa.

350

0.00480

P (mm/mm) (in./in.)

Solution σ=

40(10 3 )

P = A

2 = 325.95(106 ) N/m= 325.95 MPa 2 (0.0125 ) 4

π

From the stress–strain diagram, E =

350(106 ) 9 ) N/m 2 72.92 GPa = 72.92(10= 0.00480

ε lonε= = G

σ

= E

325.95(106 ) = 4.4702(10 −3 ) mm/mm 72.92(109 )

E 72.92(109 ) = ; 27(109 ) = ; v 0.3503 2(1 + v) 2(1 + v)

−vε lonε = −0.3503[4.4702(10 −3 )] = −1.5659(10 −3 ) mm/mm ε lat = ∆d =ε lat d =[−1.5659(10 −3 )](12.5) =−0.01957 mm d′= d + ∆d= 12.5 − 0.01957= 12.4804 mm

Ans.

Ans: d′ = 12.4804 mm 176

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R3–3. The rigid beam rests in the horizontal position on two 2014-T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm, determine the placement x of the applied 80-kN load so that the beam remains horizontal. What is the new diameter of cylinder A after the load is applied? nal = 0.35.

80 kN x

220 mm

A

B

210 mm

3m

Solution a+ΣMA = 0;  FB(3) - 80(x) = 0;     FB =

80x 3

a+ΣMB = 0;   - FA(3) + 80(3 - x) = 0;     FA =

(1) 80(3 - x)

(2)

3

Since the beam is held horizontally, dA = dB P

s =

s P A ;  P = = A E E P

PL A d = PL = a b L = E AE dA = dB;

80(3 - x) 3

(220)

AE

=

80x 3

(210) AE

80(3 - x)(220) = 80x(210) Ans.

x = 1.53 m From Eq. (2), FA = 39.07 kN sA =

39.07(103) FA = = 55.27 MPa p A (0.032) 4

Plong =

55.27(106) sA = = - 0.000756 E 73.1(109)

Plat = - nPlong = - 0.35( - 0.000756) = 0.0002646 d′A = d A + d Plat = 30 + 30(0.0002646) = 30.008 mm

Ans.

Ans: x = 1.53 m, d A ′ = 30.008 mm 177

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*R3–4.

C P

When the two forces are placed on the beam, the diameter of the A-36 steel rod BC decreases from 40 mm to 39.99 mm. Determine the magnitude of each force P.

1m

A

P 1m

1m

1m

B 0.75 m

Solution Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the free-body diagram of the beam shown in Fig. a. 4 FBC a b (3) - P(2) - P(1) = 0 5

a + ©MA = 0;

FBC = 1.25P

Normal Stress and Strain: The lateral strain of rod BC is Plat =

d - d0 39.99 - 40 = = - 0.25(10 - 3) mm>mm d0 40

Plat = - nPa;

- 0.25(10-3) = - (0.32)Pa Pa = 0.78125(10-3) mm>mm

Assuming that Hooke’s Law applies, sBC = EPa;

sBC = 200(109)(0.78125)(10-3) = 156.25 MPa

Since s 6 sY, the assumption is correct. sBC =

FBC ; ABC

156.25(106) =

1.25P p A 0.042 B 4

P = 157.08(103)N = 157 kN

Ans.

Ans: P = 157 kN 178

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R3–5.

C P

If P = 150 kN, determine the elastic elongation of rod BC and the decrease in its diameter. Rod BC is made of A-36 steel and has a diameter of 40 mm.

1m

P 1m

1m

1m

B

A

0.75 m

Solution Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram of the beam shown in Fig. a. a + ©MA = 0;

4 FBC a b (3) - 150(2) - 150(1) = 0 5

FBC = 187.5 kN

Normal Stress and Strain: The lateral strain of rod BC is sBC =

187.5(103) FBC = = 149.21 MPa p ABC A 0.042 B 4

Since s 6 sY, Hooke’s Law can be applied. Thus, sBC = EPBC;

149.21(106) = 200(109)PBC PBC = 0.7460(10-3) mm>mm

The unstretched length of rod BC is LBC = 27502 + 10002 = 1250 mm. Thus the elongation of this rod is given by dBC = PBCLBC = 0.7460(10-3)(1250) = 0.933 mm

Ans.

We obtain, Plat = - nPa ;

Plat = - (0.32)(0.7460)(10-3) = - 0.2387(10-3) mm>mm

Thus, dd = Plat dBC = - 0.2387(10-3)(40) = - 9.55(10-3) mm

Ans.

Ans. dBC = 0.933 mm dd = -9.55(10-3) mm 179

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R3–6. The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 4 kN, determine the normal strain in the bolts. Each bolt has a diameter of 5 mm. If σY = 280 MPa and Est = 200 GPa, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released?

C L

H

Solution Normal Stress:

σ=

P = A

4(10 3 )

2 = 203.72(106 ) N/m= 203.72 MPa < sg = 280 MPa 2 (0.005 ) 4

π

Normal Strain: Since s 6 sg, Hooke’s law is still valid.

= ε

σ

= E

203.72(106 ) = 1.0186(10 −3 ) mm/mm = 1.02(10 −3 ) mm/mm 200(109 )

If the nut is unscrewed, the load is zero. Also, the nut is not stressed beyond sg and so no permanent strain will be developed. Therefore, the strain e = 0

Ans.

Ans.

Ans: e = 0.0010186 mm>mm, eunscr = 0 180

CH 03.indd 118

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R3–7. The stress–strain diagram for polyethylene, which is used to sheath coaxial cables, is determined from testing a specimen that has a gauge length of 250 mm. If a load P on the specimen develops a strain of P = 0.024 mm/mm, determine the approximate length of the specimen, measured between the gauge points, when the load is removed. Assume the specimen recovers elastically.

Solution From the graph s = 14 Mpa, P = 0.004 mm/mm, E =

s 14(106) = 3.50(109) Pa = 3.50 GPa = P 0.004

P = 0.024 mm/mm, s = 25.67 Mpa L' = 250 mm + 0.024(250) = 256 mm Elastic strain recovery: Prec =

s 25.67(106) = = 7.3333(10−3) mm/mm E 3.50(109)

δ = PrecL = [7.3333(10−3)](250) mm/mm = 1.8333 mm L = L' − δ = 256 mm − 1.8333 mm = 254.167 mm

Ans.

Ans: 254.167 mm 181

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*R3–8. The pipe with two rigid caps attached to its ends is subjected to an axial force P. If the pipe is made from a material having a modulus of elasticity E and Poisson’s ratio  n, determine the change in volume of the material.

ri ro L P

Section a – a

a

a

Solution Normal Stress: The rod is subjected to uniaxial loading. Thus, slong =

P and slat = 0. A

P

dV = AdL + 2prLdr = APlong L + 2prLPlatr Using Poisson’s ratio and noting that AL = pr 2L = V, dV = PlongV - 2nPlongV = Plong (1 - 2n)V =

slong E

(1 - 2n)V

Since slong = P>A, dV = =

P (1 - 2n)AL AE PL (1 - 2n) E

Ans.

Ans: dV = 182

PL (1 - 2n) E

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R3–9. The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. Eal = 70 GPa, Emg = 45 GPa.

50 mm A

30 mm

Solution Normal Stress: 8(103)

sb =

P = Ab

p 2 4 (0.008 )

ss =

P = As

p 2 4 (0.02

= 159.15 MPa

8(103) - 0.0122)

= 39.79 MPa

Normal Strain: Applying Hooke’s Law Pb =

159.15(106) sb = = 0.00227 mm>mm Eal 70(109)

Ans.

Ps =

39.79(106) ss = = 0.000884 mm>mm Emg 45(109)

Ans.

Ans: Pb = 0.00227 mm>mm, Ps = 0.000884 mm>mm 183

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R3–10. An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces. If the top plate displaces 2  mm horizontally when it is subjected to a horizontal force P = 2 kN, determine the shear modulus of the polymer. The width of the block is 100 mm. Assume that the polymer is linearly elastic and use small angle analysis.

400 mm P  2 kN

200 mm

Solution Normal and Shear Stress: t =

2(103) V = = 50 kPa A 0.4(0.1)

Referring to the geometry of the undeformed and deformed shape of the block shown in Fig. a, g =

2 = 0.01 rad 200

Applying Hooke’s Law, t = Gg;

50(103) = G(0.01) Ans.

G = 5 MPa

Ans: G = 5 MPa 184

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4–1. The A992 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 60 mm2, determine the displacement of B and A, Neglect the size of the couplings at B, C, and D.

D 0.75 m C 60⬚

60⬚

1.50 m 3.30 kN

3.30 kN

Solution

B 5

3

3

3

4

3

10.4(10 )(1.50) 16.116(10 )(0.75) PL + = dB = © -6 9 AE 60(10 )(200)(10 ) 60(10 - 6)(200)(109)

2 kN

= 0.00231 m = 2.31 mm

Ans.

5 4

A

0.50 m 2 kN

8 kN

3

dA = dB +

8(10 )(0.5) 60(10 - 6)(200)(109)

= 0.00264 m = 2.64 mm

Ans.

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

Ans:

dB = 2.31 mm, dA = 2.64 mm 185

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4–2.  The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with respect to end D if the diameters of each segment are dAB = 20 mm, dBC = 25 mm, and dCD = 12 mm. Take Ecu = 126 GPa.

2m

3.75 m

22.5 kN

36 kN

2.5 m 9 kN

A 22.5 kN B

C 9 kN

27 kN D

Solution PL −36(10 3 )(2000) 9(10 3 )(3750) 27(10 3 )(2500) = + + AE  π (0.02 2 ) 126(109 )  π (0.0252 ) 126(109 )  π (0.012 2 ) 126(109 ) 4   4   4  

δ A −D = ∑

= 3.4635 = mm 3.46 mm

Ans.

The positive sign indicates that end A moves away from end D.

Ans: dA>D = 3.46 mm away from end D. 186

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4–3. The composite shaft, consisting of aluminum, copper, and steel sections, is subjected to the loading shown. Determine the displacement of end A with respect to end D and the normal stress in each section. The cross-sectional area and modulus of elasticity for each section are shown in the figure. Neglect the size of the collars at B and C.

Aluminum Eal = 70 GPa AAB = 58 mm2

Copper Ecu = 126 GPa ABC = 77 mm2

A

B 450 mm

16 kN

ACD = 39 mm2 8 kN

16 kN

9 kN

Steel Est = 200 GPa

C

300 mm

8 kN

7 kN D

400 mm

Solution σ= AB

9(10 3 ) PAB 2 = = 155.17(106 ) N−m = 155 MPa AAB 58(10 −6 )

(T)

Ans.

σ= BC

23(10 3 ) PBC 2 = = 298.70(106 ) N−m = 299 MPa ABC 77(10 −6 )

(C)

Ans.

σ = CD

7(10 3 ) PBC 2 = = 179.49(106 ) N−m = 179 MPa ABC 39(10 −6 )

(C)

Ans.

−23(10 3 )(300) 9(10 3 )(450) −7(10 3 )(400) PL = + + AE  58(10 −6 )  70(109 )  77(10 −6 ) 126(109 )  39(10 −6 )  200(109 )         

δ ND = ∑

= −0.07263 mm = −0.0726 mm

Ans.

The negative sign indicates end A moves towards end D.

Ans: sAB = 155 MPa 1T2, sBC = 299 MPa 1C2, sCD = 179 MPa 1C2, dA>D = 0.0726 mm towards end D. 187

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*4–4. Determine the displacement of B with respect to C of the composite shaft in Prob. 4–3.

Aluminum Eal = 70 GPa AAB = 58 mm2

Copper Ecu = 126 GPa ABC = 77 mm2

A

B 450 mm

dB>C =

−23(10 3 )(300) PL = −0.7112 mm = −0.711 mm = AE  77(10 −6 ) 126(109 )   

16 kN

ACD = 39 mm2 8 kN

16 kN

9 kN

Steel Est = 200 GPa

C

300 mm

8 kN

7 kN D

400 mm

Ans.

The negative sign indicates end B moves towards end C.

Ans.

dB>C = −0.711 mm 188

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4–5. The 2014-T6 aluminium rod has a diameter of 30 mm and supports the load shown. Determine the displacement of end A with respect to end E. Neglect the size of the couplings.

A

B

8 kN 4 kN 4m

C

D

6 kN 2m

E 2 kN

2m

2m

Solution dA>E = Σ =

PL 1 = [8(4) + 4(2) - 2(2) + 0(2)] ( 103 ) AE AE 36 ( 103 )

p (0.03)2(73.1) ( 109 ) 4

= 0.697 ( 10 - 3 ) = 0.697 mm

Ans.

Ans: dA>E = 0.697 mm 189

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4–6. The A992 steel drill shaft of an oil well extends 3600 m into the ground. Assuming that the pipe used to drill the well is suspended freely from the derrick at A, determine the maximum average normal stress in each pipe segment and the elongation of its end D with respect to the fixed end at A. The shaft consists of three different sizes of pipe, AB, BC, and CD, each having the length, weight per unit length, and cross-sectional area indicated.

A 2

AAB = 1600 mm wAB = 50 N/m

1500 m B

2

ABC = 1125 mm wBC = 40 N/m ACD = 800 mm2 wCD = 30 N/m

1500 m C D

600 m

Solution

σA =

P 50(1500) + 78000 2 = = 95.625(106 ) N−m= 95.6 MPa A 1600(10 −6 )

Ans.

σ= B

P 40(1500) + 18000 2 = = 69.33(106 ) N−m= 69.3 MPa A 1125(10 −6 )

Ans.

σ C=

30(600) P 2 = = 22.5(106 ) N−m= 22.5 MPa A 800(10 −6 )

Ans.

dD = ©

600 m 1500 m 1500 m P(x) dx (40 x + 18000)dx 30 x dx (50 x + 78000)dx + + = 1125(10 −6 )  200(109 ) 800(10 −6 )  200(109 ) 1600(10 −6 )  200(109 ) L0 L0 L A(x) E L0         

= 0.8952 = m 0.895 m

Ans.

Ans. sA = 95.6 MPa, sB = 69.3 MPa, sC = 22.5 MPa, dD = 0.895 m 190

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4–7. P

The truss is made of three A-36 steel members, each having a cross-sectional area of 400 mm2. Determine the horizontal displacement of the roller at C when P = 8 kN.

B

5 kN

0.8 m

A

C

Solution By observation the horizontal displacement of roller C is equal to the displacement of point C obtained from member AC.

0.8 m

0.6 m

FCA = 5.571 kN dC =

5.571 ( 103 ) (1.40) FCAL = = 0.0975 mm S AE (400) ( 10 - 6 ) (200) ( 106 )

Ans.

Ans: dC = 0.0975 mm S 191

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*4–8. The truss is made of three A-36 steel members, each having a cross-sectional area of 400 mm2. Determine the magnitude P required to displace the roller to the right 0.2 mm.

P B

5 kN

0.8 m

A

C

Solution 0.8 m

a+ MA = 0;   - P(0.8) - 5(0.8) + Cy(1.4) = 0

0.6 m

Cy = 0.5714 P + 2.857 4 + c ΣFy =     C 0; y - FBC a b = 0 5 FBC = 1.25 Cy

+ ΣFx = 0;   S

- FAC + 1.25 Cy (0.6) = 0 FAC = 0.75 Cy = 0.4286 P + 2.14286

Require, dCA = 0.0002 =

(0.4286 P + 2.14286) ( 103 ) (1.4) (400) ( 10-6 ) (200) ( 109 ) Ans.

P = 21.7 kN

Ans: P = 21.7 kN 192

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4–9. The assembly consists of two 10-mm diameter red brass C83400 copper rods AB and CD, a 15-mm diameter 304 stainless steel rod EF, and a rigid bar G. If P = 5 kN, determine the horizontal displacement of end F of rod EF.

300 mm A

450 mm B

P E

4P F

C

DG

P

Solution Internal Loading: The normal forces developed in rods EF, AB, and CD are shown on the free-body diagrams in Figs. a and b. p Displacement: The cross-sectional areas of rods EF and AB are AEF = (0.0152) = 4 p 56.25(10 - 6)p m2 and AAB = (0.012) = 25(10 - 6)p m2. 4 dF = Σ =

PEF LEF PAB LAB PL = + AE AEF Est AAB Ebr 20(103)(450) -6

9

56.25(10 )p(193)(10 )

+

5(103)(300) 25 ( 10 - 6)p(101)(109) Ans.

= 0.453 mm The positive sign indicates that end F moves away from the fixed end.

Ans: dF = 0.453 mm 193

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4–10. The assembly consists of two 10-mm diameter red brass C83400 copper rods AB and CD, a 15-mm diameter 304 stainless steel rod EF, and a rigid bar G. If the horizontal displacement of end F of rod EF is 0.45 mm, determine the magnitude of P.

300 mm A

450 mm B

P E

4P F

C

DG

P

Solution Internal Loading: The normal forces developed in rods EF, AB, and CD are shown on the free-body diagrams in Figs. a and b. Displacement: The cross-sectional areas of rods EF and AB are AEF = 56.25(10 - 6 )p m2 and AAB =

p (0.012 ) = 25(10 - 6 )p m2. 4

dF = Σ

PEF LEF PAB LAB PL = + AE AEF Est AAB Ebr

0.45 =

4P(450) -6

9

56.25(10 )p(193)(10 )

+

p (0.0152) = 4

P(300) -6

25(10 )p(101)(109) Ans.

P = 4967 N = 4.97 kN

Ans: P = 4.97 kN 194

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4–11. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the vertical displacement of of the the 2.5-kN 500-lb load if the members were originally horizontal when the load was 16 mmin2.2. applied. Each wire has a cross-sectional area of 0.025

E

F

0.9 3 ftm

Solution

0.54 1.8 ftm

C 0.6 2 ftm

1 ftm 0.3

I

Internal Forces in the wires:

A

0.9 3 ftm

FBD (b) FBG(1.2) – 2.5(0.9) = 0

+ c ©Fy = 0;

FAH + 1.875 – 2.5 = 0

1.5 5 ftm

H

D

a + ©MA = 0;

G

B 1 ft 0.3 m 2.5 500kN lb

FFBC 375.0 kN lb BC = 1.875 = 125.0 FF = 0.625 kN lb AH AH

FBD (a) a + ©MD = 0;

0.625(0.3)= =00 FF = 0.2083 FCF (3) -– 125.0(1) = 41.67kN lb CF CF(0.9) CF

+ c ©Fy = 0;

0.2083-– 125.0 0.625 = 0 0 FDE = 0.4167 kN lb FF FDE = 83.33 DE++ 41.67 DE

0.3 m

0.6 m

Displacement:

δD =

FDE LDE 0.4167(10 3 )(0.9)(10 3 ) = = 0.121438 mm ADE E [16(10 −6 )][193(109 )]

δC =

FCF LCF 0.2083(10 3 )(0.9)(10 3 ) = = 0.060719 mm ACF E [16(10 −6 )][193(109 )]

0.3 m

0.9 m

2.5 kN

0.060719 mm

′ δH

0.060719 ′ = = ; δH 0.040479 mm 0.6 0.9

0.3 m

0.6 m

δ H = 0.040479 + 0.060719 = 0.101198 mm δA H =

FAH LAH 0.625(10 3 )(0.54)(10 3 ) = = 0.109294 mm AAH E [16(10 −6 )][193(109 )]

0.101198 + 0.109294 = 0.210492 mm δA = δ H + δ A −H =

δB =

FBG LBG 1.875(10 3 )(1.5)(10 3 ) = = 0.910784 mm ABG E [16(10 −6 )][193(109 )]

δ I′ 0.700291 = = ; δ I′ 0.525219 mm 0.9 1.2 δ I = 0.210492 + 0.525219 = 0.7357 mm = 0.736 mm

0.9 m

0.3 m

Ans.

0.210492 mm

0.700291 mm

Ans: dl = 0.736 mm 195

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*4–12. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the 2.5-kN load load isis applied. applied. The The members were originally 500-lb horizontal, and each wire has a cross-sectional area of 16 mmin2.2. 0.025

E

F

G

3 ftm 0.9

Solution

1.8 ft 0.54 m

5 ftm 1.5

H

D

C 1 ft 0.3 m

2 ftm 0.6 I

Internal Forces in the wires:

A

B 3 ftm 0.9

FBD (b) a + ©MA = 0;

FBG 2.5(0.9)==00 F (4) -– 500(3) BG(1.2)

+ c ©Fy = 0;

+ 1.875 = 0 FF 375.0 –-2.5 500 = 0 AH + AH

1 ft 0.3 m 500kN lb 2.5

FFBG 375.0kN lb BG = 1.875 FF = 0.625 kN lb = 125.0 AH AH

FBD (a) a + ©MD = 0;

0.625(0.3)= =00 FF = 0.2083 FCF (3) -– 125.0(1) = 41.67kN lb CF CF(0.9) CF

+ c ©Fy = 0;

0.2083-– 125.0 0.625 = 0 0 FDE = 0.4167 kN lb FF FDE = 83.33 DE++ 41.67 DE

0.3 m

0.6 m

Displacement:

δD =

FDE LDE 0.4167(10 3 )(0.9)(10 3 ) = = 0.121438 mm ADE E [16(10 −6 )][193(109 )]

δC =

FCF LCF 0.2083(10 3 )(0.9)(10 3 ) = = 0.060719 mm ACF E [16(10 −6 )][193(109 )]

0.3 m

0.9 m

2.5 kN

0.060719 mm

′ δH

0.060719 ′ = = ; δH 0.040479 mm 0.6 0.9

0.3 m

0.6 m

′ + δC = 0.040479 + 0.060719 = 0.101198 mm δH = δH tan α =

δA H =

0.060719 ; α 0.003865 = = ° 0.00387° 0.9(10 3 )

Ans.

FAH LAH 0.625(10 3 )(0.54)(10 3 ) = = 0.109294 mm AAH E [16(10 −6 )][193(109 )]

δA = δH + δA H = 0.101198 + 0.109294 = 0.210492 mm δB =

FBG LBG 1.875(10 3 )(1.5)(10 3 ) = = 0.910784 mm ABG E [16(10 −6 )][193(109 )]

β = tan

0.700291 1.2(10 3 )

0.9 m

β 0.03344 ;= = ° 0.0334°

0.3 m

0.210492 mm

0.700291 mm

Ans. α = 0.00387° β = 0.0334° 196

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4–13. A spring-supported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries have a total weight of 4 kN, determine the displacement of the pipe when it is attached to the support.

F B

D

k G 0.75 m

Solution

0.75 m k H

E

A

C

Internal Force in the Rods: 0.25 m 0.25 m

FBD (a) a + ©MA = 0;

FCD (0.5) - 4(0.25) = 0

+ c ©Fy = 0;

FAB + 2.00 - 4 = 0

FCD = 2.00 kN

FAB = 2.00 kN

FBD (b) + c ©Fy = 0;

FEF - 2.00 - 2.00 = 0

FEF = 4.00 kN

Displacement: dD = dE =

FEFLEF = AEFE

dA>B = dC>D =

4.00(103)(750) p 4

(0.012)2(193)(109)

PCDLCD = ACDE

= 0.1374 mm

2(103)(750) p 4

(0.005)2(193)(109)

= 0.3958 mm

dC = dD + dC>D = 0.1374 + 0.3958 = 0.5332 mm Displacement of the spring dsp =

Fsp k

=

2.00 = 0.0333333 m = 33.3333 mm 60

dlat = dC + dsp Ans.

= 0.5332 + 33.3333 = 33.9 mm

Ans. dD = 0.1374 mm, dA>B = 0.3958 mm, dC = 0.5332 mm, dtot = 33.9 mm 1 97

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4–14. A spring-supported pipe hanger consists of two springs, which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe is displaced 82 mm when it is filled with fluid, determine the weight of the fluid.

F B

D

k G 0.75 m

Solution

0.75 m k H

E

A

C

Internal Force in the Rods: 0.25 m 0.25 m

FBD (a) a + ©MA = 0; + c ©Fy = 0;

FCD(0.5) - W(0.25) = 0

FCD =

W - W = 0 2

W 2

FAB +

FAB =

W 2

FBD (b) + c ©Fy = 0;

FEF -

W W = 0 2 2

FEF = W

Displacement: dD = dE =

FEFLEF = AEFE

W(750) p 2 9 4 (0.012) (193)(10 )

= 34.35988(10 - 6) W dA>B = dC>D =

FCDLCD = ACDE

W 2 (750) p 2 (0.005) (193)(109) 4

= 98.95644(10 - 6) W dC = dD + dC>D = 34.35988(10 - 6) W + 98.95644(10 - 6) W = 0.133316(10 - 3) W Displacement of the spring dsp =

Fsp k

=

W 2

60(103)

(1000) = 0.008333 W

dlat = dC + dsp 82 = 0.133316(10 - 3) W + 0.008333W Ans.

W = 9685 N = 9.69 kN

Ans. W = 9.69 kN 1 98

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4–15. The steel bar has the original dimensions shown in the figure. If it is subjected to an axial load of 50 kN, determine the change in its length and its new cross-sectional dimensions at section a–a. Est = 200 GPa, nst = 0.29.

50 kN

a

A

60 mm 20 mm 20 mm

B 200 mm 350 mm

a

C

20 mm 50 mm

D

50 kN

200 mm

Solution dA>D = Σ

2(50) ( 103 ) (200) 50 ( 103 ) (350) PL = + AE (0.02)(0.05)(200) ( 109 ) (0.06)(0.05)(200) ( 109 ) Ans.

= 0.129 mm  dB>C =

50 ( 10 ) (350) PL = = 0.02917 mm AE (0.06)(0.05)(200) ( 109 )

 PBC =

dB>C 0.02917 = 0.00008333 LBC 350

3

 Plat = - v Plong = - (0.29)(0.00008333) = - 0.00002417 h′ = 50 - 50(0.00002417) = 49.9988 mm

Ans.

w′ = 60 - 60(0.00002417) = 59.9986 mm

Ans.

Ans: dA>D = 0.129 mm, h′ = 49.9988 mm, w′ = 59.9986 mm 199

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*4–16. The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measured from the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings. A

B

C

D

5 kN

Solution

8m

Internal Force: As shown on FBD. Displacement: dA =

PL = AE

- 5.00 (103)(8) p 4

(0.42 - 0.32) 200(109)

= - 3.638(10 - 6) m = - 3.64 A 10 - 3 B mm

Ans.

Negative sign indicates that end A moves towards end D.

Ans. dA = -3.64(10- 3) mm 200

CH 04.indd 122

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4–17. The bar has a length L and cross-sectional area A. Determine its elongation due to the force P and its own weight.The material has a specific weight g (weight>volume) and a modulus of elasticity E. L

Solution d = =

L P(x) dx 1 = (gAx + P) dx AE L0 L A(x) E

gAL2 gL2 1 PL a + PL b = + AE 2 2E AE

P

Ans.

Ans. d =

L gL 2 1 PL (gAx + P) dx = + AE L0 2E AE

2 01

CH 04.indd 131

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4–18. The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area of each rod figure.If If aaforce forceof of30 6 kip is given in the figure. kN is applied to the ring F, determine the horizontal displacement of point F.

D

4 ftm 1.2

C in2 2

ACD 1 mm ACD � 600

20.6 ft m E

10.3ftm F

6 kip 30 kN 2

�2 1200 in mm EFEF 1 ftmAA 0.3

1.5mm in2 2 AA AB�900 AB

Solution

B

6 ftm 1.8

2

A

Internal Force in the Rods: a + ©MA = 0; + ©F = 0; : x

FCD (3) -– 6(1) = 0= 0 FCD = =2.00 kip F 30(0.3) FCD 10 kN CD(0.9) 6 30 - 2.00 – 10 –- FF == 0 0 AB AB

FABFAB == 4.00 20 kip kN

0.6 m

FEF = 30 kN

Displacement:

0.3 m

3

= δ F −E

0.6

0.1667 mm

FAB LAB 20(10 3 )(1.8)(10 3 ) = = 0.3333 mm AAB E [900(10 −6 )][120(109 )]

= δA

δ E′

0.3 m

3

FCDLCD 10(10 )(1.2)(10 ) = = 0.1667 mm ACDE [600(10 −6 )][120(109 )]

= δC

0.6 m

=

FEF LEF 30(10 3 )(0.3)(10 3 ) = = 0.0625 mm AEF E [1200(10 −6 )][120(109 )] 0.1667 , 0.9

δ E′ = 0.1111 mm

δ E = δC + δ E′ = 0.1667 + 0.1111 = 0.2778 mm δ= δ E + δ F −E F = 0.2778 + 0.0625 = 0.3403 mm = 0.340 mm

Ans.

Ans. = δ F 0.3403 mm 202

CH 04.indd 128

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4–19.

D

1.2 4 ftm

C 2 2 mm

ACD � 600 ACD 1 in

The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a forceof 30 kN is applied to the ring F, determine the angle of tilt of bar AC.

20.6 ft m E

F 10.3ftm

30 kN 6 kip 2

AA 1.5mm in2 2 AB AB�900 B

1.8 6 ftm

A

Solution Internal Force in the Rods: a + ©MA = 0; + ©F = 0; : x

F (3) -– 6(1) = 0= 0 FCD = =2.00 kip FCD 30(0.3) FCD 10 kN CD(0.9) 6 30 - –2.00 FAB = 0 10 –-FAB = 0

0.6 m

FAB kip FAB= =4.00 20 kN

δC =

10(10 3 )(1.2)(10 3 ) FCDLCD = = 0.1667 mm ACDE [600(10 −6 )][120(109 )]

δA =

20(10 3 )(1.8)(10 3 ) FAB LAB = = 0.3333 mm AAB E [900(10 −6 )][120(109 )]

0.6 m

FEF = 30 kN 0.3 m

Displacement:

0.3 m

0.16666667 mm

0.3333 − 0.1667 −1 δ A − δ C tan −1 θ tan = = LAC 0.9(10 3 ) = 0.01061 = ° 0.0106°

Ans.

Ans.

θ 0.0106° = 203

2

�2 1200 in mm 0.3 EFEF 1 ftmAA

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* 4–20. The pipe is stuck in the ground so that when it is pulled upward the frictional force along its length varies linearly from zero at B to fmax (force/length) at C. Determine the initial force P required to pull the pipe out and the pipe’s elongation just before it starts to slip. The pipe has a length L, cross-sectional area A, and the material from which it is made has a modulus of elasticity E.

P B

L

Solution From FBD (a) + c ΣFy =    P 0;

1 (F L) = 0 2 max P =

fmax

Fmax L 2

C

Ans.

From FBD (b) +T ΣFy =     0;

P(x) +

Fmax L 1 Fmax x a bx = 0 2 L 2 P(x) =

d = =

L P(x)

dx

L0 A(x)E

=

L0

L

Fmax L 2AE

Fmax L Fmax x2 2 2L L

dx -

Fmax x

2

L0 2AEL

dx

Fmax L2 3AE

Ans.

Ans:

204

P =

Fmax L , 2

d =

Fmax L2 3AE

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4–21. The post is made of Douglas fir and has a diameter of 100 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance distributed around the post that is triangular along its sides; that is, it varies from w = 0 at y = 0 to w = 12 kN>m at y = 2 m, determine the force F at its  bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.

20 kN A y w

2m

12 kN/m B

Solution

F

Equation of Equilibrium: Referring to the FBD of the entire post, Fig. a, + c ΣFy = 0;

F +

1 (12)(2) - 20 = 0 2

Ans.

F = 8.00 kN

Normal Force: Referring to the FBD of the upper segment of the post sectioned at arbitrary distance y, Fig. b + c ΣFy = 0;

1 (6y)(y) - 20 - P(y) = 0 2

Py = ( 3y2 - 20 ) kN

Displacement: For Douglas Fir, E = 13.1 GPa L

dA>B = =

N(y)dy

L0 A(y)E

=

1 AE L0

2 meters

( 3y2 - 20 ) dy

2 meters 1 ( y3 - 20y ) ` AE 0

= = -

32 kN # m AE

32 ( 103 ) p ( 0.12 ) [13.1 ( 109 ) ] 4

= - 0.3110 ( 10-3 ) m = - 0.311 mm

Ans.

The sign indicates that end A moves toward end B.

Ans: F = 8.00 kN, dA>B = - 0.311 mm 205

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4–22. The post is made of Douglas fir and has a diameter of 100 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w = 4 kN>m at y = 0 to  w = 12 kN>m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.

20 kN A y w

2m

12 kN/m B

Solution

F

Equation of Equilibrium: Referring to the FBD of the entire post, Fig. a, + c ΣFy = 0;

F +

1 (4 + 12)(2) - 20 = 0 2

Ans.

F = 4.00 kN

Normal Force: Referring to the FBD of the upper segment of the post sectioned at arbitrary distance y, Fig. b, + c ΣFy = 0;

(4 + 2y)y - 20 - P(y) = 0

P(y) = ( 2y2 + 4y - 20 ) kN

Displacement: For Douglas Fir, E = 13.1 GPa L

dA>B =

N(y)dy

L0 A(y)E

= =

1 AE L0

2m

( 2y2 + 4y - 20 ) dy

2m 1 2 3 a y + 2y2 - 20yb ` AE 3 0

= -

80 kN # m 3 AE

80(103)

= 3c

p ( 0.12 ) d [13.1 ( 109 ) ] 4

= - 0.2592 ( 10-3 ) m = - 0.259 mm

Ans.

Ans: F = 4.00 kN, dA>B = -0.259 mm 206

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4–23. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm2 and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied.

C 300 N/m 1.5 m A

2m

B

D 2m

Solution a+ ΣMA = 0; 1200(2) - TCB(0.6)(2) = 0 TCB = 2000 N dB>C =

(2000)(2.5) PL = = 0.0051835 AE 14(10 - 6)(68.9)(109)

(2.5051835)2 = (1.5)2 + (2)2 - 2(1.5)(2) cos u u = 90.248° u = 90.248° - 90° = 0.2478° = 0.004324 rad Ans.

dD = u r = 0.004324(4000) = 17.3 mm

Ans: dD = 17.3 mm 207

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*4–24. The weight of the kentledge exerts an axial force of P = 1500 kN on the 300-mm diameter high-strength concrete bore pile. If the distribution of the resisting skin friction developed from the interaction between the soil and the surface of the pile is approximated as shown, determine the resisting bearing force F for equilibrium. Take p0 = 180 kN>m. Also, find the corresponding elastic shortening of the pile. Neglect the weight of the pile.

P p0

12 m

Solution

F

Internal Loading: By considering the equilibrium of the pile with reference to its entire free-body diagram shown in Fig. a. We have + c ©Fy = 0;

F +

1 (180)(12) - 1500 = 0 2

F = 420 kN

Ans.

Also, p(y) =

180 y = 15y kN>m 12

The normal force developed in the pile as a function of y can be determined by considering the equilibrium of the sectional of the pile with reference to its free-body diagram shown in Fig. b. + c ©Fy = 0;

1 (15y)y + 420 - P(y) = 0 2

P(y) = (7.5y2 + 420) kN

p Displacement: The cross-sectional area of the pile is A = (0.32) = 0.0225p m2. 4 We have 12 m P(y)dy (7.5y2 + 420)(103)dy = 0.0225p(29.0)(109) L0 A(y)E L0 L

d =

12 m

=

L0

c 3.6587(10 - 6)y2 + 0.2049(10 - 3) d dy

= c 1.2196(10 - 6)y3 + 0.2049(10 - 3)y d

12 m 0

= 4.566(10 - 3) m = 4.57 mm

Ans.

Ans: F = 420 kN, d = 4.57 mm 208

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4–25. Determine the elongation of the aluminum strap when it is subjected to an axial force of 30 kN. Eal = 70 GPa.

15 mm 30 kN 250 mm

50 mm

6 mm 15 mm 800 mm

30 kN 250 mm

Solution d = (2) =

d2 Ph PL ln + Et(d 2 - d 1) d 1 AE 2(30) ( 103 ) (250)

(70) ( 10 ) (0.006)(0.05 - 0.015) 9

= 2.37 mm

aln

30 ( 103 ) (800) 50 b + 15 (0.006)(0.05)(70) ( 109 )

Ans.

Ans: d = 2.37 mm 209

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4–26. The ball is truncated at its ends and is used to support the bearing load P. If the modulus of elasticity for the material is E, determine the decrease in the ball’s height when the load is applied.

P

r 2

r r 2

Solution Displacement: Geometry: A(y) = px2 = p ( r 2 - y2 ) Displacement: When x = L

d = = =

r , 2

y = {

23 r 2

P(y) dy

L0 A(y) E 23

r 2 dy P 23 pE L- 2 r r 2 - y2 23

r + y 2r P 1 c ln d` r p E 2r r - y - 23 2

=

P [ ln 13.9282 - ln 0.07180 ] 2p r E

=

2.63 P prE

Ans.

Ans: d = 210

2.63 P prE

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4–27. The linkage is made of two pin-connected A-36 2 members,each each having a cross-sectional steel members, having a cross-sectional area ofarea 1.5 inof . 1000 mm2. If force a vertical P is = 250 kN istoapplied to If a vertical of Pforce = 50ofkip applied point A, point A, determine itsdisplacement vertical displacement at A. determine its vertical at A.

P A 0.6 2 ftm B

Solution

C 0.45 1.5 ftm

0.45 1.5 ftm

Analysing the equilibrium of Joint A by referring to its FBD, Fig. a, + ©F = 0 ; : x

3 3 FAC a b - FAB a b = 0 5 5

4 250== 00 -2F a b - 50 5

+ c ©Fy = 0

The

initial

FAC = FAB = F F - 31.25kN kip F ==–156.25

of members AB and AC is 12 in 0.45 ++0.26 == 0.75 m.ft)a The axialb deformation members AB and AC is deformation of members (2.50 = 30 in. Theofaxial L == 21.5 1 ft AB and AC is 22

δ=

length

22

FL (−156.25)(10 3 )(750) = = −0.58594 mm AE [1000(10 −6 )][200(109 )]

The negative sign indicates that end A moves toward B and C. From the geometry 1.5 0.45 shown in Fig. b, u = tan - 1 a Thus, bb = 36.87°. 36.87°.Thus, 2 0.6

(δ= A )ν

0.58594 δ = = 0.7324 = mm 0.732 mm ↓ coθθ coθ 36.87°

Ans.

d = 0.58594 mm

P = 250 kN

0.6 m

0.45 m

Ans. (dA)v = 0.732 mm T 211

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*4–28. The linkage is made of two pin-connected A-36 2 steel members, members,each each having a cross-sectional having a cross-sectional area ofarea 1.5 inof . 1000 mm2. Determine the magnitude forcetoPdisplace needed Determine the magnitude of the forceofP the needed to displace point A 0.625 mm downward. point A 0.025 in. downward.

P A

Solution

2 ftm 0.6 B

Analysing the equilibrium of joint A by referring to its FBD, Fig. a + ©F = 0; : x + c ©Fy = 0;

The

initial

3 3 FAC a b - FAB a b = 0 5 5 4 -2F a b - P = 0 5

C 1.5 ftm 0.45

FAC = FAB = F

1.5 ftm 0.45

F = - 0.625 P

length

of members AB and AC are 12 in deformation of members (2.50 = 30 in. Theofaxial L == 21.5 0.452 2++0.2622 ==0.75 m.ft)a The axialb deformation members AB and AC is 1 ft AB and AC is FL −0.625P(750) δ= = = −2.34375(10 −6 )P AE [1000(10 −6 )][200(109 )] The negative sign indicates that end A moves toward B and C. From the geometry 1.5 0.45 shown in Fig. b, we obtain u = tan - 1 a Thus bb = 36.87°. 36.87°.Thus, 2 0.6

(δ= A )ν

0.025 = 0.625 5

δ

coθθ 3 ))P P 0.4310(10 -–6 2.34375(10 cos 36.87°

3 = P 213.33(10 = ) N 213 kN

Ans.

(dA)v = 0.625 mm d = 2.34375(10–6)P

0.6 m

0.45 m

Ans. = P 213 kN

212

CH 04.indd 134

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4–29. The bar has a cross-sectional area of 1800 mm2, and E = 250 GPa. Determine the displacement of its end A when it is subjected to the distributed loading.

x

A

Solution P(x) =

L0

4 ftm 1.5

x

w dx = 500

P(x) dx = AE L0

L0

x

0

1

(

)

x3 dx = 375x 4/3 N

)1.5 m

L

dA =

� 500x1/3 N/m lb/in. w

x 4/3 ) dx ( 375=

[1800(10 −6 )][250(109 )]

1.5 m

3  −6 0.8333(10 −6= )  x 7/3  0.9199(10 = ) m 0.920 mm 7 0

–7 0.0128 in. dA ==9.199(10 )m

Ans.

= 0.00092 mm

Ans: dA = 2.990 mm 213

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4–30. Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P.

P d2 t

h

d1

Solution

P

d 1h + (d 2 - d 1)x d2 - d1 w = d1 + x = h h d =

P(x) dx

L A(x)E

=

P E L0

h

dx [d 1h + (d 2 - d 1)x]t h

h

= = = =

Ph dx E t L0 d 1h + (d 2 - d 1)x h h d1 h d2 - d1 Ph dx Ph = a b c ln a1 + xb d 0 E t d 1 h L0 1 + d 2 - d 1 x E t d1 h d2 - d1 d1 h 0 d1 h

d2 - d1 d1 + d2 - d1 Ph Ph c ln a1 + bd = c ln a bd E t(d 2 - d 1) d1 E t(d 2 - d 1) d1 d2 Ph c ln d  E t (d 2 - d 1) d1

Ans.

Ans: d =

214

d2 Ph c ln d E t (d 2 - d 1) d1

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4–31. The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected kN, determine determine the average normal to an axial force force of of 150 30 kip, stress in the concrete and in each rod. Each rod has a 20 mm. diameter of 0.75 in.

1004mm in. 150 kN 30 kip

Solution Equations of Equilibrium: + c ©Fy = 0;

[1]

150== 00 6Pst + Pcon - 30

1.2 3 ftm

Compatibility: dst = dcon Pst (1.2) Pcon (1.2) =  π (0.02 2 ) [200(109 )]  π (0.2 2 ) − 6 π (0.02 2 ) [29.0(109 )] 4  4 4 

( )

150 kN

[2]

Pstst ==0.073368 0.064065PP con con Solving Eqs. [1] and [2] yields: Pstst ==7.641 1.388kN kip

kN Pcon ==104.152 21.670 kip

Average Normal Stress:

σ= st

σ= con

Pcon = Acon

Pst 7.641(10 3 ) 2 = = 24.32(106 ) N−m = 24.3 MPa π (0.02 2 ) Ast 4 104.15(10 3 ) π

(0.2 2 ) − 6 4

= 3.527(106 ) N− = m 2 3.53 MPa

( π4 ) (0.022 )

Ans.

Ans.

Ans. sst = 24.3 MPa, scon = 3.53 MPa 215

CH 04.indd 145

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*4–32. The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force force of of 150 30 kip, kN,determine determine the required diameter of each rod so that one-fourth of the load is carried by the concrete and three-fourths by the steel.

1004mm in. 150 kN 30 kip

Solution kNisisrequired requiredto todistribute distributein insuch such aa manner manner that that 3/4 3/4 Equilibrium: The force of 150 30 kip of the force is carried by steel and 1/4 of the force is carried by concrete. Hence Pst =

3 (30) (150)== 22.5 112.5kip kN 4

Pcon =

1.2 3 ftm

1 (30) == 7.50 (150) 37.5 kip kN 4

Compatibility: dst = dcon PstL Pcon L = AstEst Acon Econ

Ast =

Pst Acon Econ Pcon Est

( )

(112.5)  π4 (0.2 2 ) − 6 π4 (0.02 2 )  29(109 ) π     6   d2 = 4 (37.5)  200(109 )   d = 0.05221 m = 52.2 mm

Ans.

Ans. d = 52.2 mm 216

CH 04.indd 145

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4–33.  The A-36 steel pipe has a 6061-T6 aluminum core. It is subjected to a tensile force of 200 kN. Determine the average normal stress in the aluminum and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm.

400 mm 200 kN

200 kN

Solution Equations of Equilibrium: + d ΣFx = 0;   Pal + Pst - 200 = 0

(1)

Compatibility: dal = dst Pal(400) p 4

( 0.072 ) (68.9) ( 109 )

=

Pst(400) p 4

( 0.082 - 0.072 ) (200) ( 109 ) (2)

Pal = 1.125367 Pst Solving Eqs. (1) and (2) yields:

Pst = 94.10 kN  Pal = 105.90 kN

Average Normal Stress: sal =

105.90 ( 103 ) Pal = p = 27.5 MPa 2 Aal 4 ( 0.07 )

sst =

Pst = Ast

94.10 ( 103 ) p 4

( 0.082 - 0.072 )

Ans. Ans.

= 79.9 MPa

Ans: sal = 27.5 MPa, sst = 79.9 MPa 217

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4–34. The 304 stainless steel post A has a diameter of mmand andisissurrounded surroundedby byaared red brass brass C83400 C83400 tube B. d == 50 2 in. kN is applied Both rest on the rigid rigid surface. surface. If If aa force forceof of25 5 kip to the rigid cap, determine the average normal stress developed in the post and the tube.

5 kip 25 kN

B in. 2008mm

A

B A

3 in. 75 mm

Solution d

Equations of Equilibrium: + c ©Fy = 0;

[1]

Pst + Pbr - 25 5 == 0

0.5mm in. 12 25 kN

Compatibility: dst = dbr

Pst (200) Pbπ(200) 5 2  9 2 π (0.05 ) [193(10 )]  4 (0.15 − 0.126 2 ) [101(109 )]   4  π

[2]

Pstst 5 P = 0.72120 0.69738 P Pbr br Solving Eqs. [1] and [2] yields: Pbr ==14.525 2.9457kN kip

= 2.0543 PP = 10.475 kNkip st st

Average Normal Stress:

14.525(10 3 ) 2 = 2.792(106 ) N−m = 2.79 MPa π (0.152 − 0.126 2 ) 4

σ= bπ

Pbπ = Abπ

σ= θt

Pθt 10.475(10 3 ) 2 = = 5.335(106 ) N−m = 5.34 MPa π (0.052 ) Aθt 4

Ans.

Ans.

Ans. sbr = 2.79 MPa, sst = 5.34 MPa 218

CH 04.indd 146

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4–35. The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a kN isis applied applied to to the the rigid rigid cap, cap, determine the force of 25 5 kip required diameter d of the steel post so that the load is shared equally between the post and tube.

25 kN 5 kip

B

B 200 8mm in.

A

A

75 mm 3 in.

Solution The force force of of525 is shared equally brass steel. Hence Equilibrium: The kipkN is shared equally by by thethe brass andand steel. Hence

d

12 0.5mm in.

Pst = Pbr = P = 12.5 2.50 kN kip Compatibility: dst = dbr PL PL = AstEst AbrEbr Ast = a

AbrEbr Est

 π (0.152 − 0.126 2 ) 101(109 ) p 2 4   bd = 4 193(109 )

Ans.

d 5 0.05888 m = 58.9 mm

Ans. d 5 58.9 mm 219

CH 04.indd 147

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*4–36. The A-36 steel pipe has an outer radius of 20 mm and an inner radius of 15 mm. If it fits snugly between the fixed walls before it is loaded, determine the reaction at the walls when it is subjected to the load shown.

B

A

300 mm

C

8 kN 8 kN 700 mm

Solution + ΣFx = 0;    FA + FC - 16 = 0 S

(1)

By superposition: + )    0 = - ∆ + d (S C C 0 =

- 16 (300) AE

+

FC (1000) AE Ans.

FC = 4.80 kN

From Eq. (1), Ans.

FA = 11.2 kN

Ans: FC = 4.80 kN, FA = 11.2 kN 220

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  The 10-mm-diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20 mm, and its inner diameter is 10 mm. If the yield stress for the steel is (sY)st = 640 MPa, and for the bronze (sY)br = 520 MPa, determine the magnitude of the largest elastic load P that can be applied to the assembly. Est = 200 GPa, Ebr = 100 GPa.

P

10 mm

20 mm

Solution + c ΣFy = 0;    Pst + Pbr - P = 0

(1)

Assume failure of bolt: p Pst = (sY)st(A) = 640 ( 106 ) a b ( 0.012 ) 4

P

= 50265.5 N

dst = dbr PstL p 4

( 0.012 ) (200) ( 109 )

=

PbrL p 4

( 0.022 - 0.012 ) (100) ( 109 )

Pst = 0.6667 Pbr 50 265.5 = 0.6667 Pbr Pbr = 75398.2 N From Eq. (1) P = 50265.5 + 75398.2 = 125663.7 N = 126 kN

Ans.

(controls)

Assume failure of sleeve: p Pbr = (sY)br(A) = 520 ( 106 ) a b ( 0.022 - 0.012 ) = 122 522.11 N 4 Pst = 0.6667 Pbr = 0.6667(122 522.11) = 81 681.4 N From Eq. (1), P = 122 522.11 + 81 681.4 = 204 203.52 N = 204 kN

Ans: P = 126 kN 221

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  The 10-mm-diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20 mm, and its inner diameter is 10 mm. If the bolt is subjected to a compressive force of P = 20 kN, determine the average normal stress in the steel and the bronze. Est = 200 GPa, Ebr = 100 GPa.

P

10 mm

20 mm

Solution + c ΣFy = 0;    Pst + Pbr - 20 = 0

(1)

dst = dbr PstL p 4

( 0.012 ) (200) ( 109 )

=

PbrL p 4

P

( 0.022 - 0.012 ) (100) ( 109 ) (2)

Pst = 0.6667 Pbr Solving Eqs. (1) and (2) yields Pst = 8 kN  Pbr = 12 kN sst =

8 ( 103 ) Pst = p = 102 MPa 2 Ast 4 ( 0.01 )

sbr =

Pbr = Abr

12 ( 103 ) p 4

( 0.022 - 0.012 )

Ans.

Ans.

= 50.9 MPa

Ans: sst = 102 MPa, sbr = 50.9 MPa 222

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4–39. If column AB is made from high strength pre-cast concrete and reinforced with four 20 mm diameter A-36 steel rods, determine the average normal stress developed in the concrete and in each rod. Set P = 350 kN.

P

P

A

a

225 mm

a 225 mm

3m

Section a-a

Solution Equation of Equilibrium: Referring to the free-body diagram of the cut part of the concrete column shown in Fig. a, + c ©Fy = 0;

Pcon + 4Pst - 2(350) = 0

B

(1)

Compatibility Equation: Since the steel bars and the concrete are firmly bonded, their deformation must be the same. Thus, dcon = dst Pcon (3) Pst (3) = 0.225(0.225) − 4 π (0.02 2 ) [29.0(109 )]  π (0.2 2 ) [200(109 )] 4 4   

( )

Pcon = 22.7859Pst

(2)

Solving Eqs. (1) and (2), Pst = 26.13 kN

Pcon = 595.47 kN

Normal Stress: Applying Eq. (1-6), scon =

Pcon 595.47(10 3 ) 6 = 12.06(10 = ) N m 2 12.1 Mpa = Acon 0.225(0.225) − 4 π (0.02 2 )

(4)

Pst 26.13(10 3 ) 6 83.18(10 = ) N m 2 83.2 Mpa = π sst = = 2 Ast (0.02 ) 4

( )

Ans.

Ans.

Ans: scon = 12.1 MPa , sst = 83.2 MPa 223

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*4–40. If column AB is made from high strength pre-cast concrete and reinforced with four 20 mm diameter A-36 steel rods, determine the maximum allowable floor loadings P. The allowable normal stress for the high strength concrete and the steel are (sallow)con = 18 MPa and (sallow)st = 170 MPa, respectively.

P

P

A

a

225 mm

a 225 mm

3m

Solution Equation of Equilibrium: Referring to the free-body diagram of the cut part of the concrete column shown in Fig. a, + c ©Fy = 0;

Pcon + 4Pst - 2P = 0

Section a-a

B

(1)

Compatibility Equation: Since the steel bars and the concrete are firmly bonded, their deformation must be the same. Thus, dcon = dst Pcon (3)

0.225(0.225) − 4 

( 4 ) (0.02 π

2

) [29.0(10 )]  9

=

Pst (3) (0.2 ) [200(109 )] 4 

π

2

Pcon = 22.7859Pst

(2)

Solving Eqs. (1) and (2), Pst = 0.07467P

Pcon = 1.7013P

Allowable Normal Stress: (scon)allow =

Pcon ; Acon

18(106) =

1.7013P p 0.225(0.225) - 4 a b (0.022) 4

P = 522.31(103) N = 522 kN (controls) (sst)allow =

Pst ; Ast

170(106) =

Ans.

0.07467P p (0.022) 4

P = 715.28(103) N = 715.28 kN

Ans: P = 522 kN 224

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4–41. Determine the support reactions at the rigid supports A and C. The material has a modulus of elasticity of E.

3 d 4

d P B

A 2a

C a

Solution Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, + ΣFx = 0; S

(1)

P - FA - FC = 0

Compatibility Equation: Using the method of superposition, Fig. b, + )    d = d - d (S P FC

0 =

FC =

P(2a) p a d 2 bE 4 9 P 17

- ≥

FCa 2 p 3 a db E 4 4

+

FC(2a) p a d 2 bE 4

¥ Ans.

Substituting this result into Eq. (1), FA =

8 P 17

Ans.

Ans:

9 P, 17 8 FA = P 17 FC =

225

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  If the supports at A and C are flexible and have a stiffness k, determine the support reactions at A and C. The material has a modulus of elasticity of E.

3 d 4

d P B

A 2a

C a

Solution Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, + ΣFx = 0;    P - FA - FC = 0 S

(1)

Compatibility Equation: Using the method of superposition, Fig. b, + )     d = d - d (S C P FC FC(2a) P(2a) FC FCa FC P + = ≥ ¥ + ¥ - ≥ + 2 k k k p 2 p 2 p 3 a d bE a d bE a db E 4 4 4 4 9(8ka + pd 2E)

136ka + 18pd 2E

FA = a

64ka + 9pd 2E bP 136ka + 18pd 2E

Substituting this result into Eq. (1),

Ans.

d P

FC = c

Ans.

Ans: FC = c FA = a 226

9(8ka + pd 2E) 136ka + 18pd 2E

d P,

64ka + 9pd 2E bP 136ka + 18pd 2E

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4–43. The tapered member is fixed connected at its ends A and B B and and isissubjected subjectedto toaaload loadPP= =357kN kipatatx x= 750 = 30mm in. Determine the reactions at the supports. The material is 50 mm thick is made froom 2014-T6 aluminum. 2 in. thick andand is made from 2014-T6 aluminum.

A

B P

1506 mm in.

3 in. 75 mm

x

Solution

1500 mm 60 in.

Equilibrium. Refering to the FBD of the member, Fig. a + ©F = 0; : x

FA + FB - 735== 00

(1)

Compatibility. From the geometry shown in Fig. b    1.5 − x  y= 2 0.0375 +  [0.05(3 − x)] m  (0.0375) = 1.5    

Thus, the cross-sectional area of the member as a function of x is A= (0.05)[ 0.05(3 − x)] = [ 0.0025(3 − x)] m 2

It is required that dA>B = 0 0.75 m

– -

0

dx FAA dx ++ 0.0025(3 – x)(E)

0.75 m

-

0

FA dx 3–x

1.5 m

+

0.75 m

FA ln(3 − x) 0

0.75 m

1.5 m

0.75 m

FB dx 3–x

F FBBdx == 00 0.0025(3 – x)(E)

=0 1.5 m

0 − FB ln(3 − x) 0.75 m =

 2.25   1.5  0 FA ln   − FB ln  =  3   2.25  FA = 1.4094 FB

(2)

Solving Eqs. (1) and (2) FA = 20.47 = 20.5 kN

Ans.

FB = 14.53 = 14.5 kN

Ans.

35 kN FA

FB (a)

y 0.0375 m 0.0375 m

x

1.5 − x (b)

Ans. y = 75 - 0.025x, FA = 20.47 kN, FB = 14.53 kN 227

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*4–44. The tapered member is fixed connected at its ends A and B and is subjected to a load P. Determine the location x of the load and its greatest magnitude so that the average normal stress in the bar does not exceed σallow = 28 MPa. The number is 50 mm thick.

A

B P

1506 mm in.

3 in. 75 mm

x

Solution

1500 mm 60 in.

Equilibrium. Refering to the FBD of the member, Fig. a + ©F = 0; : x

(1)

FA + FB - P = 0

Compatibility. From the geometry shown in Fig. b    1.5 − x  y= 2 0.0375 +  [0.05(3 − x)] m  (0.0375) = 1.5    

Thus, the cross-sectional area of the member as a function of x is A= (0.05)[ 0.05(3 − x)] = [ 0.0025(3 − x)] m 2

It is required that dA>B = 0 x

– -

0 x

-

dx FAA dx ++ 0.0025(3 – x)(E) FA dx

0

3–x

1.5 m

+

1.5 m x

FB dx 3–x

x

x

F FBBdx == 00 0.0025(3 – x)(E)

=0 1.5 m

FA ln(3 − x) 0 − FB ln(3 − x) x

0 =

 3 − x  1.5  0 FA ln   − FB ln  3 − x =  3   

(2)

The maximum normal stress in segment AC occurs at position x and in segment BC occurs at B. Then = σ allow

σ allow =

FA FA ; = 28(106 ) = FA [70(10 3 )(3 − x)] N A 0.0025(3 − x) FB FB ; 28(106 ) = = FB 105(10 3 ) N (0.05)(0.075) A

P

FB (a)

Substitute these results into Eq. (2)  1.5  3− x 3 70(10 3 )(3 − x)ln  0  − 105(10 )ln  =  3  3−x 3−x ln    3  3− x    3 

3− x

3− x

 1.5  = ln   3−x

 1.5  =  3−x

C

FA

y

0.0375 m

0.0375 m

1.5 1.5 − x

x (b)

1.5

Solving numerically, x = 0.721698 m = 722 m

Ans.

Then, FA =[70(10 3 )(3 − 0.7216980) =159.48(10 3 ) N

Substitute the results of FA and FB into Eq. (1) 159.48(10 3 ) + 105(10 3 ) − P = 0 3 P 264.48(10 = = ) 264 N

Ans.

Ans. x = 722 m, P = 264 N

228

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4–45. The rigid bar supports the uniform distributed load kN/m.Determine Determinethe theforce force in in each each cable cable if each cable of 90 6 kip>ft. 3 mm 200 GPa. has a cross-sectional cross-sectional area areaof of36 andEE= = 31110 0.05 in22, ,and 2 ksi.

C

ft 26 m

6 kip/ft 90 kN/m

Solution

A

Equilibrium: a + ©MA = 0;

TCB a

2 25

(1) –-270(1.5) b (3) 54(4.5) + TCD a

Geometry and compatibility: u = tan - 1

D

B ft 13 m

2 25

(3)==00 b9

ft 13 m

ft 13 m

(1) 270 kN

62 = 45° 62

1.5 m

1.5 m

2 (8.4853)22– -2(1)(2.8284) 2(3)(8.4853) cos cos u9 LB¿C = (1) (( 2 22++ (2.8284)

Also, 2.8284 m

(3)2 ++ (2.8284) cos u9 (8.4853)22– -2(3)(2.8284) 2(9)(8.4853) cos L2D¿C = (9)

(2)

Thus, eliminating cos u¿ . -L2B¿C (0.176778)

2m 5m

5m 1m

1m

+ 1.590978 =

-L2B¿C (0.176778) -L2B¿C

=

=

–L2D'C(0.058926)

2 0.058926LD'C

2 0.3333LD'C

+ 1.001735

+ 0.589243

+ 3.333

But, LB¿C = 5 + δBC,

LD¿C = 5 + δDC,

Neglect squares or d¿ B since small strain occurs. 2 2 + d+BCd)BC =) 5 =+ 45 2 5+ d2BC L2B¿C = ( 545 2 2 245 dDC L2D¿C = ( 245 =) 5=+ 45 2 5+ d2DC 5 + d+DCd)DC

5 + 2 5δ BC = 0.3333(5 + 2 5δ DC ) + 3.333 2 5δ BC = 0.3333(2 5δ DC )

dDC = 3dBC Thus, TCD 245 TCBTCB 5 5 245 5 3 = 3 AE AE AE AE TCD = 3 TCB From Eq. (1). TCD = 135.84 kN = 136 kN

Ans. Ans.

TCB = 45.28 kN = 45.3 kN

Ans. TCD = 136 kN, TCB = 45.3 kN

229

CH 04.indd 160

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4–46. The rigid bar is originally horizontal and is supported by two cables each having a cross-sectional area and EE ==31110 200 32GPa. of 36 0.05mm in2,2, and ksi. Determine the slight rotation of the bar when the uniform load is applied.

C

ft 26 m

Solution See solution of Prob. 4-51.

6 kip/ft 90 kN/m

TCD = 135.84 27.1682kN kip

A D

B

(135.84)(10 3 ) 5 TCDLCD δ DC = = = 0.04219 m AE [36(10 −6 )][200(109 )]

ft 13 m

ft 13 m

ft 13 m

Using Eq. (2) of Prob. 4-51, ( 5 + 0.04219)2 =+ (3)2 (2.8284)2 − 2(3)(2.8284)coθθ ′ = θ ′ 45.9023°

Thus, Ans.

¢u = 45.9023° – 45° = 0.9023° = 0.902°

Ans. ¢u = 0.902° 2 30

CH 04.indd 161

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4–47. The specimen represents a filament-reinforced matrix system made from plastic (matrix) and glass (fiber). If there are n fibers, each having a cross-sectional area of Af and modulus of Ef, embedded in a matrix having a cross-sectional area of Am and modulus of Em, determine the stress in the matrix and in each fiber when the force P is applied on the specimen.

P

P

Solution + c ΣFy = 0;

(1)

- P + Pm + Pf = 0

dm = df PfL PmL = ; AmEm nAfEf

Pm =

AmEm P nAfEf f

(2)

Solving Eqs. (1) and (2) yields Pm =

AmEm P; nAfEf + AmEm

Pf =

nAfEf nAfEf + AmEm

P

Normal stress:

sm =

sf =

Pm = Am Pf nAf

=

a a

AmEm - Pb nAfEf + AmEm Am

nAfEf nAfEf + AmEm nAf

Pb

=

=

Em P nAfEf + AmEm Ef

nAfEf + AmEm

Ans.

Ans.

P

Ans: sm = sf =

231

Em P, nAfEf + AmEm Ef nAfEf + AmEm

P

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*4–48. The rigid beam is supported by the three suspender bars. Bars AB and EF are made of aluminum and bar CD is made of steel. If each bar has a cross-sectional area of 450 mm2, determine the maximum value of P if the allowable stress is (sallow)st = 200 MPa for the steel and (sallow)al = 150 MPa for the aluminum. Est = 200 GPa, Eal = 70 GPa.

B al

D

F

st A

2m

al C

E

0.75 m 0.75 m 0.75 m 0.75 m P

Solution

2P

Equation of Equilibrium: Referring to the FBD of the rigid beam Fig. a, a+ ΣMA = 0;

FCD(1.5) + FEF(3) - P(0.75) - 2P(2.25) = 0 (1)

1.5 FCD + 3 FEF = 5.25 P a+ ΣME = 0;

2 P(0.75) + P(2.25) - FCD(1.5) - FAB(3) = 0 (2)

1.5 FCD + 3 FAB = 3.75 P Compatibility: Referring to the displacement diagram of the rigid beam, Fig. b, dCD - dAB dEF - dAB = 1.5 3 2dCD = dEF + dAB 2a

FCDL A 3 200 ( 10

9

)4

b =

FCD =

FEFL A 3 70 ( 10

9

)4

+

FABL A 3 70 ( 109 )4

10 (F + FAB) 7 EF

(3)

Solving Eqs. (1), (2) and (3), FEF = 0.8676 P

FAB = 0.3676 P

FCD = 1.7647 P

Assuming that bar EF fails. Then (sallow)al =

FEF ; A

150 ( 106 ) =

0.8676 P 450 ( 10-6 )

P = 77.80 ( 103 ) N = 77.80 kN Assuming that bar CD fails. Then (sallow)st =

FCD ; A

200 ( 106 ) =

1.7647 P 450 ( 10-6 )

P = 51.0 ( 103 ) N = 51.0 kN(control!) Ans.

Ans: P = 51.0 kN 232

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  If the gap between C and the rigid wall at D is initially 0.15 mm, determine the support reactions at A and D when the force P = 200 kN is applied. The assembly is made of solid A-36 steel cylinders.

600 mm

600 mm

0.15 mm

P A

50 mm

D B

25 mm

C

Solution Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, + ΣFx = 0;      200 ( 103 ) - FD - FA = 0 S

(1)

Compatibility Equation: Using the method of superposition, Fig. b, + )    d = dP - dFD (S 0.15 =

200(103)(600) p (0.052)(200)(109) 4

FD (600) FD (600) - C + S p p (0.052)(200)(109) (0.0252)(200)(109) 4 4 Ans.

FD = 20 365.05 N = 20.4 kN Substituting this result into Eq. (1),

Ans.

FA = 179 634.95 N = 180 kN

Ans: FD = 20.4 kN, FA = 180 kN 233

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  The support consists of a solid red brass C83400 copper post surrounded by a 304 stainless steel tube. Before the load is applied the gap between these two parts is 1 mm. Given the dimensions shown, determine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials.

P A

1 mm

0.25 m

Solution

60 mm 80 mm

Require,

10 mm

dst = dbr + 0.001 Fst(0.25) p[(0.05)2 - (0.04)2]193(109)

=

Fbr(0.25) p(0.03)2(101)(109)

+ 0.001

0.45813 Fst = 0.87544 Fbr + 106

(1)

+ c ΣFy = 0;

(2)

Fst + Fbr - P = 0

Assume brass yields, then (Fbr)max = sg Abr = 70(106)(p)(0.03)2 = 197 920.3 N (Pg)br = sg >E =

70.0(106) 101(109)

= 0.6931(10 - 3) mm>mm

dbr = (Pg)brL = 0.6931(10 - 3)(0.25) = 0.1733 mm 6 1 mm Thus only the brass is loaded. Ans.

P = Fbr = 198 kN

Ans: P = 198 kN 234

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  The assembly consists of two red brass C83400 copper rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rod EF of diameter 40 mm, and a rigid cap G. If the supports at A, C, and F are rigid, determine the average normal stress developed in the rods.

300 mm

450 mm 40 kN

A

30 mm

B

E

F 40 mm

C

30 mm

D

40 kN G

Solution Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the free-body diagram of the assembly shown in Fig. a, + ΣFx = 0; S

2F + FEF - 2 3 40 ( 103 )4 = 0

(1)

Compatibility Equation: Using the method of superposition, Fig. b, + ) 0 = - dP + dEF (S 0 = -



40(103)(300) p (0.032)(101)(109) 4

+ ≥

1 FEF >2 2 (300) FEF (450) ¥ + p p (0.042)(193)(109) (0.032)(101)(109) 4 4

FEF = 42 483.23 N

Substituting this result into Eq. (1), F = 18 758.38 N Normal Stress: We have, sAB = sCD =

sEF =

F 18 758.38 = = 26.5 MPa p ACD (0.032) 4

Ans.

FEF 42 483.23 = = 33.8 MPa p AEF (0.042) 4

Ans.

Ans: sAB = sCD = 26.5 MPa, sEF = 33.8 MPa 235

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*  The bolt AB has a diameter of 20 mm and passes through a sleeve that has an inner diameter of 40 mm and an outer diameter of 50 mm. The bolt and sleeve are made of A-36 steel and are secured to the rigid brackets as shown. If the bolt length is 220 mm and the sleeve length is 200 mm, determine the tension in the bolt when a force of 50 kN is applied to the brackets.

200 mm 25 kN

25 kN B

A

25 kN

25 kN 220 mm

Solution Equation of Equilibrium: + ΣFx = 0;     S

Pb + Ps - 25 - 25 = 0 (1)

Pb + Ps - 50 = 0 Compatibility: db = ds Pb(220) p 4

( 0.02 ) 200 ( 10 ) 2

9

=

Ps(200) p 4

( 0.05 - 0.042 ) (200) ( 109 ) 2

(2)

Pb = 0.40404 Ps Solving Eqs. (1) and (2) yields: Ps = 35.61 kN

Ans.

Pb = 14.4 kN

Ans: Pb = 14.4 kN 236

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4–53.  The 2014-T6 aluminum rod AC is reinforced with the firmly bonded A992 steel tube BC. If the assembly fits snugly between the rigid supports so that there is no gap at C,  determine the support reactions when the axial force of 400 kN is applied. The assembly is attached at D.

D A 400 mm

400 kN B

A992 steel

800 mm a

Solution

a

50 mm 25 mm

Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a,

C

FD + (FC)al + (FC)st - 400(103) = 0

+ c ΣFy = 0;

2014–T6 aluminum alloy

Section a–a

(1)

Compatibility Equation: Using the method of superposition, Fig. b, (+ T) 0 = +

0 = dp - dFC 400(103)(400) 2

9

p(0.025 )(73.1)(10 )

- £

(FC)al(800) 2

9

p(0.025 )(73.1)(10 )

+

[(FC)al + (FC)st](400) p(0.0252)(73.1)(109)

400(103) = 3(FC)al + (FC)st

§ (2)

Also, since the aluminum rod and steel tube of segment BC are firmly bonded, their deformation must be the same. Thus, (dBC)st = (dBC)al (FC)st(800) 2

2

9

p(0.05 - 0.025 )(200)(10 )

=

(FC)al(800) p(0.0252)(73.1)(109)

(FC)st = 8.2079(FC)al

(3)

Solving Eqs. (1) and (2), (FC)al = 35.689 kN

(FC)st = 292.93 kN

Substituting these results into Eq. (1), Ans.

FD = 71.4 kN Also, FC = (FC)st + (FC)al = 35.689 + 292.93

Ans.

= 329 kN

Ans: FD = 71.4 kN, FC = 329 kN 237

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4–54.  The 2014-T6 aluminum rod AC is reinforced with the firmly bonded A992 steel tube BC. When no load is applied to the assembly, the gap between end C and the rigid support is 0.5 mm. Determine the support reactions when the axial force of 400 kN is applied.

D A 400 mm

400 kN B

A992 steel

800 mm a

Solution

a

50 mm 25 mm

Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a,

C

FD + (FC)al + (FC)st - 400(103) = 0

+ c ΣFy = 0;

2014–T6 aluminum alloy

Section a–a

(1)

Compatibility Equation: Using the method of superposition, Fig. b, (+ T)

dC = dP - dFC

0.5 = +

400(103)(400) 2

9

p(0.025 )(73.1)(10 )

- ≥

(FC)al (800) 2

9

p(0.025 )(73.1)(10 )

+

[(FC)al + (FC)st](400) p(0.0252)(73.1)(109)

220.585(103) = 3(FC)al + (FC)st

¥ (2)

Also, since the aluminum rod and steel tube of segment BC are firmly bonded, their deformation must be the same. Thus, (dBC)st = (dBC)al (FC)st (800) 2

2

9

p(0.05 - 0.025 )(200)(10 )

=

(FC)al (800) p(0.0252)(73.1)(109) (3)

(FC)st = 8.2079(FC)al Solving Eqs. (2) and (3), (FC)al = 19.681 kN

(FC)st = 161.54 kN

Substituting these results into Eq. (1), Ans.

FD = 218.777 kN = 219 kN Also, FC = (FC)al + (FC)st = 19.681 + 161.54

Ans.

= 181 kN

Ans: FD = 219 kN, FC = 181 kN 238

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4–55.  The three suspender bars are made of A992 steel and have equal cross-sectional areas of 450 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown.

A 2m

C

B 80 kN

50 kN E

D 1m

1m

1m

F 1m

Solution Referring to the FBD of the rigid beam, Fig. a, + c ΣFy = 0; a+ ΣMD = 0;

FAD + FBE + FCF - 50(103) - 80(103) = 0

(1)

FBE(2) + FCF(4) - 50(103)(1) - 80(103)(3) = 0

(2)

Referring to the geometry shown in Fig. b, dBE = dAD + a dBE =

dCF - dAD b(2) 4

1 1 d + dCF 2 2 AD

FCF L FBE L 1 FADL = a + b AE 2 AE AE

(3)

FAD + FCF = 2 FBE

Solving Eqs. (1), (2), and (3) yields FBE = 43.33(103) N

FAD = 35.83(103) N

FCF = 50.83(103) N

Thus, sBE =

43.33(103) FBE = = 96.3 MPa A 0.45(10 - 3)

Ans.

sAD =

35.83(103) FAD = = 79.6 MPa A 0.45(10 - 3)

Ans. Ans.

sCF = 113 MPa

Ans: sBE = 96.3 MPa, sAD = 79.6 MPa, sCF = 113 MPa 239

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*4–56. The three A-36 steel wires each have a diameter of 2  mm and unloaded lengths of LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the force in each wire after the 150-kg mass is suspended from the ring at A.

B

C

D

5

5

4

4 3

3

A

Solution Equations of Equilibrium: + ΣFx = 0; S + c ΣFy = 0;

3 3 F - FAB = 0 5 AD 5

FAD = FAB = F

4 2a F b + FAC - 150(9.81) = 0 5

(1)

1.6F + FAC - 1471.5 = 0

Compatibility: dAD = dAC cos u Since the displacement is very small, cos u = dAD = F(2) AE

=

4 5

4 d 5 AC 4 FAC (1.6) c d 5 AE

(2)

F = 0.640 FAC Solving Eqs. (1) and (2) yields: FAC = 727 N

Ans.

FAB = FAD = F = 465 N

Ans.

Ans: FAC = 727 N, FAB = 465 N 240

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4–57. The A-36 steel wires AB and AD each have a diameter of 2 mm and the unloaded lengths of each wire are LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the required diameter of wire AC so that each wire is subjected to the same force when the 150-kg mass is suspended from the ring at A.

B

C

D

5

5

4

4 3

3

A

Solution Equations of Equilibrium: Each wire is required to carry the same amount of load. Hence FAB = FAC = FAD = F Compatibility: dAD = dAC cos u Since the displacement is very small, cos u = dAD = F(2) p 4

( 0.0022 ) E

=

4 5

4 d 5 AC F(1.6) p 2 4 dACE

Ans.

d AC = 0.001789 m = 1.79 mm

Ans: d AC = 1.79 mm 241

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4–58. The post is made from 606l-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A and B, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originally uncompressed, determine the reactions at A and B when the force P = 40 kN is applied to the collar.

A 0.25 m

P C

0.25 m

Solution

k  200 MN/m B

Equations of Equilibrium: FA + FB + Fsp - 40 ( 103 ) = 0

+ c ΣFy = 0;

(1)

Compatibility: 0 = dp - dB 0 =

40 ( 103 ) (0.25) p 4

( 0.052 ) 68.9 ( 109 ) ( FB + Fsp ) (0.25) + 2 9 4 ( 0.05 ) 68.9 ( 10 )

-£p

FB + p 2 ( ) 0.05 68.9 ( 109 ) 4 0.25

Fsp + 200 ( 106 )

§ (2)

FB + Fsp = 23119.45 Also, dsp = dBC Fsp 200 ( 10

6

)

=

FB + p 2 9 4 ( 0.05 ) 68.9 ( 10 ) 0.25

Fsp + 200 ( 106 ) (3)

FB = 2.7057 Fsp Solving Eq. (2) and (3) yields Fsp = 6238.9 N

Ans.

FB = 16880.6 N = 16.9 kN Substitute the results into Eq. (1)

Ans.

FA = 16880.6 N = 16.9 kN

Ans: FB = 16.9 kN, FA = 16.9 kN 242

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4–59. The post is made from 606l-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A and B, and at its center C there is a coiled spring attached to the rigid collar. If  the spring is originally uncompressed, determine the compression in the spring when the load of P = 50 kN is applied to the collar.

A 0.25 m

P C

0.25 m

Solution

k  200 MN/m B

Compatibility: 0 = dp - dB 0 =

50 ( 103 ) (0.25) p 4

( 0.052 ) 68.9 ( 109 ) ( FB + Fsp ) (0.25) + 2 9 4 ( 0.05 ) 68.9 ( 10 )

-£p

FB + p 2 ( 0.05 ) 68.9 ( 109 ) 4 0.25

Fsp + 200 ( 106 )

FB + Fsp = 28899.31

§

(1)

Also, dsp = dBC Fsp 200 ( 10

6

)

=

FB + p 2 9 4 ( 0.05 ) 68.9 ( 10 ) 0.25

Fsp + 200 ( 106 ) (2)

FB = 2.7057 Fsp Solving Eqs. (1) and (2) yield Fsp = 7798.6 N  FB = 21100.7 N Thus, dsp =

Fsp k

=

7798.6 200 ( 106 )

= 0.0390 ( 10 - 3 ) m = 0.0390 mm

Ans.

Ans: dsp = 0.0390 mm 243

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300 12 mm in.

50 mm 2 in.

250 10 mm in.

Solution Equilibrium: + ©F = 0; : x

2Fst - Fal = 0

Compatibility: dst = 0.125(10−3) - dal Fst (0.3) Fal (0.25) = 0.125(10 −3 ) −  π (0.012 2 ) [200(109 )]  π (0.052 ) [68.9(109 )] 4 4   = 13.2629 Fst 125(10 3 ) − 1.8480 Fal

Solving, Fst = 7.3708(103) N Fal = 14.7416(103) N Normal stress: srod =

7.3708(10 3 ) Fst 6 = 65.17(10 = ) N/m 2 65.2 MPa = π 2 (0.012 ) Ast 4

Ans.

scyl =

14.7416(10 3 ) Fal 6 = 7.508(10 = ) N/m 2 7.51 MPa = π 2 Aal (0.05 ) 4

Ans.

( )

( )

Ans. srod = 65.2 MPa, scyl = 7.51 MPa

0.125 (10−3) m

2 44

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300 12 mm in.

4–61. The press consists of two rigid heads that are held together by the two A-36 steel 12-mm-diameter rods. A 6061-T6-solid-aluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder. Determine the angle through which the screw can be turned before the rods or the specimen begin to yield. The single-threaded screw on the bolt has a lead of 0.25 mm. Note: The lead represents the distance the screw advances along its axis for one complete turn of the screw.

50 mm 2 in.

Solution

250 10 mm in.

Equilibrium: + ©F = 0; : x

2Fst - Fal = 0

Compatibility: dst = d - dal Fst (0.3) Fal (0.25) = d− 2  9 π  (0.052 ) [68.9(109 )] (0.012 ) [200(10 )] 4 4   π

= Fst (109 )d − 1.8480 Fal 13.2629

(1)

Assume steel yields first, 6 = σ Y 250(10 = )

Fst

; Fst 28.27(10 3 ) N =

( π4 ) (0.0122 )

Then Fal = 56.55(10 3 ) N = σ al

56.55(10 3 ) 6 = 28.80(10 = ) N/m 2 28.80 MPa π (0.052 ) 4

( )

4.50 6 37 ksi MPa, steel yields first asfirst assumed. From From Eq. (1), 28.80ksi MPa < 255 steel yields as assumed. Eq. (1), d = 0.4795(10 −3 ) m = 0.4795 mm Thus, u 0.47950 = 360° 0.01 0.25 Ans.

u = 690.48° = 690°

Ans. u = 690° 245

CH 04.indd 162

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4–62. The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the posts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has a stiffness of k = 2 MN >m and an unstretched length of 1.02 m, determine the force in each post after the load is applied to the bar.

50 kN/m

A

C

B 1m

k

Solution

1m

1m

Equations of Equilibrium: a+ ΣMC = 0; + c ΣFy = 0;

FB (1) - FA(1) = 0

FA = FB = F

2F + Fsp - 100 ( 10

3

) =0

(1)

Compatibility:

(+T)

dA + 0.02 = dsp F(1) 600 ( 10 - 6 ) 9.65 ( 109 )

+ 0.02 =

Fsp 2.0 ( 106 )

0.1727F + 20 ( 103 ) = 0.5 Fsp

(2)

Solving Eqs. (1) and (2) yields: Ans.

FA = FB = F = 25581.7 N = 25.6 kN Fsp = 48 836.5 N

Ans: FA = FB = 25.6 kN 246

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4–63. The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the posts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has a stiffness of k = 2 MN >m and an unstretched length of 1.02 m, determine the vertical displacement of A and B after the load is applied to the bar.

50 kN/m

A

C

B 1m

k

Solution

1m

1m

Equations of Equilibrium: a+ ΣMC = 0;

FB (1) - FA(1) = 0 2F + Fsp - 100 ( 10

3

+ c ΣFy = 0;

FA = FB = F

) =0

(1)

Compatibility:

(+T)

dA + 0.02 = dsp F(1)

600 ( 10 - 6 ) 9.65 ( 109 )

+ 0.02 =

Fsp 2.0 ( 106 )

0.1727F + 20 ( 103 ) = 0.5 Fsp

(2)

Solving Eqs. (1) and (2) yields: F = 25 581.7 N  Fsp = 48 836.5 N Displacement: dA = dB = =

FL AE

25 581.7(1000) 600 ( 10 - 6 ) (9.65) ( 109 )

Ans.

= 4.42 mm

Ans: dA = dB = 4.42 mm 247

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*4–64. P

The assembly consists of two posts AB and CD each made from material 1 having a modulus of elasticity of E1 and a cross-sectional area A1, and a central post made from material 2 having a modulus of elasticity E2 and crosssectional area A2. If a load P is applied to the rigid cap, determine the force in each material.

d

d

A

E 1

B

C 1

2 F

L D

Solution Equilibrium: + c ΣFy = 0;

(1)

2F1 + F2 - P = 0

Compatibility: d = d1 = d2 F1 L F2L = A1 E1 A2 E2 Solving Eq. (1) and (2) yields: F1 = a F2 = a

F1 = a

A1 E1 bF A2 E2 2

(2)

A1 E1 bP 2 A1 E1 + A2 E2

Ans.

A2 E2 bP 2A1 E1 + A2 E2

Ans.

Ans:

A1 E1 bP, 2 A1 E1 + A2 E2 A2 E2 F2 = a bP 2A1 E1 + A2 E2 F1 = a

248

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4–65. P

The assembly consists of two posts AB and CD each made from material 1 having a modulus of elasticity of E1 and a cross-sectional area A1, and a central post EF made from material 2 having a modulus of elasticity E2 and a crosssectional area A2. If posts AB and CD are to be replaced by those having a material 2, determine the required crosssectional area of these new posts so that both assemblies deform the same amount when loaded.

d

d

A

E 1

B

C 1

2 F

L D

Solution + c ΣFy = 0;

(1)

2F1 + F2 - P = 0

Compatibility: din = d1 = d2 F1 L F2L = A1 E1 A2 E2

F1 = a

Solving Eq. (1) and (2) yields: F1 = a din

A1 E1 bP 2 A1 E1 + A2 E2

F2 L = = A2 E2

F2 = a

1 2A EA +E A E 2 PL 2

1

2

1

2

2

A2 E2

=

A1 E1 bF A2 E2 2

(2)

A2 E2 bP 2A1 E1 + A2 E2 PL 2A1 E1 + A2 E2

Compatibility: When material 1 has been replaced by material 2 for two side posts, then dfinal = d1 = d2 F1 L F2 L = A′1 E2 A2 E2

F1 = a

A′1 bF A2 2

(3)

Solving for F2 from Eq. (1) and (3) F2 = a dfinal = Requires,

A2 bP 2A′1 + A2 F2 L = A2 E2

1 2A′A+ 2

1

A2

2 PL

A2 E2

=

PL E2 ( 2A′1 + A2 )

din = dfinal PL PL = 2A1 E1 + A2 E2 E2 ( 2A′1 + A2 ) A′1 = a

E1 bA  E2 1

Ans.

Ans: A′1 = a 249

E1 bA E2 1

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4–66. P

The assembly consists of two posts AB and CD each made from material 1 having a modulus of elasticity of E1 and a cross-sectional area A1, and a central post EF made from material 2 having a modulus of elasticity E2 and a crosssectional area A2. If post EF is to be replaced by one having a material 1, determine the required cross-sectional area of this new post so that both assemblies deform the same amount when loaded.

d

d

A

E 1

B

C 1

2 F

L D

Solution + c ΣFy = 0;

(1)

2F1 + F2 - P = 0

Compatibility: din = d1 = d2 F1 L F2L = A1 E1 A2 E2

F1 = a

Solving Eq. (1) and (2) yields: F1 = a din

A1 E1 bP 2 A1 E1 + A2 E2

F2 L = = A2 E2

F2 = a

1 2A EA +E A E 2 P 2

1

2

1

2

2

A2 E2

=

A1 E1 bF A2 E2 2

(2)

A2 E2 bP 2A1 E1 + A2 E2

PL 2A1 E1 + A2 E2

Compatibility: When material 2 has been replaced by material 1 for central posts, then dfinal = d1 = d2 F2 L F1 L = A1 E1 A′2E1

F2 = a

A′2 bF A2 1

(3)

Solving for F1 from Eq. (1) and (3) F1 = a dfinal Requires,

A1 bP 2A1 + A2′

F1 L = = A1E1

1 2A A+

1

1

A′2

2 PL

A1 E1

=

PL E1 ( 2A1 + A′2 )

din = dfinal PL PL = 2A1E1 + A2 E2 E1 ( 2A1 + A′2 ) A′2 = a

E2 bA  E1 2

Ans.

Ans: A′2 = a 250

E2 bA E1 2

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4–67. The wheel is subjected to a force of 18 kN from the axle. Determine the force in each of the three spokes. Assume the rim is rigid and the spokes are made of the same material, and each has the same cross-sectional area.

B

120

C

0.4 m

A

120

18 kN

D

Solution Equations of Equilibrium: + ΣFx = 0; S

FAC cos 30° - FAD cos 30° = 0   FAC = FAD = F

+ c   Σ Fy = 0;   FAB + 2F sin 30° - 18 = 0 (1)

  FAB + F = 18 Compatibility: dAC = dAB cos 60° F(0.4) AE

=

FAB (0.4)

AE F = 0.5FAB

cos 60° (2)

Solving Eq. (1) and (2) yields: FAB = 12.0 kN (T)

Ans.

FAC = FAD = F = 6.00 kN (C)

Ans.

Ans: FAB = 12.0 kN (T) FAC = FAD = 6.00 kN (C) 251

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*4–68.  The C83400-red-brass rod AB and 2014-T6aluminum rod BC are joined at the collar B and fixed connected at their ends. If there is no load in the members when T1 = 10°C, determine the average normal stress in each member when T2 = 45°C. Also, how far will the collar be displaced? The cross-sectional area of each member is 1130 mm2.

B

A

1m

C

0.6 m

Solution Equilibrium: ©Fx = 0;

Fbr = Fal = F

Compatibility:

dA>C = 0 -

FalLBC Fbr LAB + abr¢T LAB + aal ¢T LBC = 0 AAB Ebr ABCEal F(1)

-

[(1130)(10−6)][101(109)]

-

[(1130)(10−6)][73.1(109)]

+ 18(10- 6)(45 - 10)(1)

F(0.6)

+ 23(10 - 6)(45 - 10)(0.6) = 0

F = 69.45(103) N

sbr = sal =

69.45(103) = 61.46(106) N/m2 = 61.5 MPa 1130(10−6)

61.46 MPa < (sg)al and (sg)br

dB = -

[69.45(103)][(1)(103)] [1130(10−6)][101(109)]

Ans. OK

+ 18(10 - 6)(45 - 10)[1(103)]

dB = 0.02147 mm = 0.0215 mm :

Ans.

Ans. sbr = 61.5 MPa dB = 0.0215 mm 252

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4–69. Three bars each made of different materials are connected together and placed between two walls when the temperature is T1 = 12°C. Determine the force exerted on the (rigid) supports when the temperature becomes T2 = 18°C. The material properties and cross-sectional area of each bar are given in the figure.

Steel Copper Brass Est � 200 GPa Ebr � 100 GPa Ecu � 120 GPa ast � 12(10�6)/�C abr � 21(10�6)/°C acu � 17(10�6)/�C Ast � 200 mm2

300 mm

Solution + ) (;

Abr � 450 mm2

200 mm

Acu � 515 mm2

100 mm

0 = ¢T - d

0 = 12(10 - 6)(6)(0.3) + 21 (10 - 6)(6)(0.2) + 17 (10 - 6)(6)(0.1) -

F(0.3) -6

9

200(10 )(200)(10 )

-

F(0.2) -6

9

450(10 )(100)(10 )

-

F(0.1) 515(10 - 6)(120)(109) Ans.

F = 4203 N = 4.20 kN

Ans. 0 = ∆T - d, F = 4.20 kN 253

CH 04.indd 175

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4–70. The steel bolt has a diameter of 7 mm and fits through an aluminum sleeve as shown. The sleeve has an inner diameter of 8 mm and an outer diameter of 10 mm. The nut at A is adjusted so that it just presses up against the sleeve. If the assembly is originally at a temperature of T1 = 20°C and then is heated to a temperature of T2 = 100°C, determine the average normal stress in the bolt and the sleeve. Est = 200 GPa, Eal = 70 GPa, ast = 14(10-6)>°C, aal = 23(10-6)>°C.

A

Compatibility: (ds)T - (db)T = (ds)F + (db)F 23(10 - 6)(100 - 20)L - 14(10 - 6)(100 - 20)L =

p 2 4 (0.01

FL + - 0.0082)70(109)

FL p 2 9 4 (0.007 )200(10 )

F = 1133.54 N Average Normal Stress: 1133.54 = 40.1 MPa - 0.0082)

ss =

F = As

sb =

F 1133.54 = 29.5 MPa = p 2 Ab 4 (0.007 )

Ans.

p 2 4 (0.01

Ans.

Ans. ss = 40.1 MPa, sb = 29.5 MPa 254

CH 04.indd 186

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4–71. The AM1004-T61 magnesium alloy tube AB is capped with a rigid plate E. The gap between E and end C of the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the normal stress developed in the tube and the rod if the temperature rises to 80° C. Neglect the thickness of the rigid cap.

25 mm a

Section a-a

E B

A a

Solution

300 mm

20 mm

C

25 mm

D

0.2 mm 450 mm

Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(80 - 30)(300) = 0.39 mm and

A dT)CD = aal ¢TLCD = 24(10 - 6)(80 - 30)(450) = 0.54 mm. Referring deformation diagram of the tube and the rod shown in Fig. a, d =

C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D

0.2 = C 0.39 -

F(300)

p A 0.025 - 0.02 B (44.7)(10 ) 2

2

9

S + C 0.54 -

F = 32 017.60 N

to

the

F(450) p 4

A 0.0252 B (68.9)(109)

S

Normal Stress: sAB = sCD =

F 32 017.60 = = 45.3 MPa AAB p A 0.0252 - 0.022 B

Ans.

F 32 017.60 = = 65.2 MPa p 2 ACD 4 A 0.025 B

Ans.

F = 107 442.47 N

Ans. sAB = 45.3 MPa, sCD = 65.2 MPa, F = 107,442.47 N 255

CH 04.indd 184

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*4–72. The AM1004-T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the highest temperature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid cap.

25 mm a

Section a-a

E B

A a

Solution

300 mm

20 mm

C

25 mm

D

0.2 mm 450 mm

Then sCD =

F 107 442.47 = = 218.88MPa 6 (sY)al p 2 ACD 4 A 0.025 B

(O.K.!)

Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(T - 30)(300) = 7.8(10 - 6) (T - 30) and

A dT B CD = aal ¢TLCD = 24(10 - 6)(T - 30)(450) = 0.0108(T - 30).

Referring to the deformation diagram of the tube and the rod shown in Fig. a, d =

C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D

0.2 = C 7.8(10 - 3)(T - 30) -

+ C 0.0108(T - 30) T = 172° C

107 442.47(300) p A 0.0252 - 0.022 B (44.7)(109)

107 442.47(450) p 4

A 0.0252 B (68.9)(109)

S

S

Ans.

Ans. T = 172° C 256

CH 04.indd 185

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4–73.  The pipe is made of A992 steel and is connected to the collars at A and B. When the temperature is 15°C, there is no axial load in the pipe. If hot gas traveling through the pipe causes its temperature to vary along the pipe defined by T = (35 + 30x)C, where x is in meter, determine the average normal stress in the pipe. The inner diameter is 50 mm, the wall thickness is 4 mm.

A

B 2.4 m

Solution Compatibility: The cross-sectional area is A 5 0 = dT - dF

Where

p 2 2 –6 2 (8 –  )
= 216(10 ) p m . 4

dT =

L0

L

a ¢T dx and ¢T 


x





x C.

2.4m 8ft

10) - 6 B 0 == 12(10 6.60 A–6 0

F (2.4) + 15 (20(40 + 30x) dxx)– −6 L0 [216(10 )π ][200(109 )]

2 2 15(8) 30(2.4) F (2.4) -6 6.60–6 00==12(10 )c 20(2.4) + + d – A 10 B B 40(8) 2 2 [216(10 −6 )π ][200(109 )]

F = 91.20(10 3 ) N

Average Normal Stress:

91.20(10 3 ) 6 134.40(10 = ) N−m 2 134 MPa = s5 = 216(10 −6 )π

Ans.

Ans: s = 134 MPa 257

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4–74. 4–74.  The bronze C86100 pipe has an inner radius of 0.5 and a awall in. IfIf the 12.5in. mm and wallthickness thicknessofof0.2 5 mm. the gas gas flowing through it changes the temperature of the pipe uniformly from T at to A TtoB = TB15°C B, determine the TAA ==60°C 200°F = 60°F at A at B,atdetermine the axial axial it exerts onwalls. the walls. fitted between forceforce it exerts on the The The pipepipe was was fitted between the the walls when = 60°F. walls when T =T 15°C.

A

B

8 ftm 2.4

Solution Temperature Gradient:

45°C

2.48– -x x T(x) ==1560+ +a a b 140= 60 = 200 - 17.5x T(x) b (45) – 18.75x 2.4 8 Compatibility: 0 = dT - dF

Where

15°C

2.4 m

dT = 1 a¢Tdx

2.4 m

2.4m 2ft

10) - 6 B 0 == 17(10 9.60 A–6 0

F (2.4) [(60 – 18.75x) – 15]-dx – [(200 - 17.5x) 60]  π (0.0352 − 0.0252 ) [103(109 )] L0 4 

2.4m 2ft

F (2.4) 9.60–6 00==17(10 ) - 6 B (45 − 18.75 x) dx − A 10 L 0  π (0.0352 − 0.0252 ) [103(109 )]  4

= F 18.57(10 3 ) N = 18.6 kN

Ans.

Ans: F = 18.6 kN 258

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4–75. 40-ft-long A-36 4–75.  The 12-m-long A-36steel steelrails rails on on a train track are laid with a small gap between them to allow for thermal expansion. Determine the required gap d so that the rails just touch one another when the temperature is increased from Using thisgap, gap,what whatwould wouldbe be the T11 ==–30°C - 20°FtotoTT 90°F. T 30°C. Using this 2 2= = axial force in the rails if the temperature were to rise to T The cross-sectional area of of each railrail is 3200 mm The cross-sectional area each is 5.10 in2. T33 ==40°C? 110°F?

d

d

40m ft 12

Solution Thermal Expansion: Note that since adjacent rails expand, each rail will be d required to expand on each end, or d for the entine rail. 2

12 m

8.64 mm

3 -6 12(10–6)[30 – (–30)](12)(10 ) )[90 - ( - 20)](40)(12) d = a¢TL = 6.60(10

= 8.64 mm

Ans.

Compatibility: + B A:

8.64 mm

= dTT –- ddFF 80.34848 .64 mm =

= 8.64 12(10 −6 )[40 − (−30)]12(10 3 ) −

F (12)(10 3 ) [3200(10 −6 )][200(109 )]

3 F 76.8(10 = ) N 76.8 kN =

Ans.

Ans: d = 8.64 mm, F = 76.8 kN 259

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*4–76. The device is used to measure a change in temperature. Bars AB and CD are made of A-36 steel and 2014-T6 aluminum alloy respectively. When the temperature is at at 75°F, 40°C,ACE ACEis isin in horizontal position. Determine thethe horizontal position. Determine the the vertial displacement pointeratat EE when when the vertical displacement of ofthethepointer temperature rises rises to to 150°F. 80°C.

6 mm 0.25 in.

A

723 mm in.

E

C

36 1.5mm in.

Solution

B

D

Thermal Expansion: –3 -6 -3 23(10–6)(80 – 40)(36) = 33.12(10 ) mm )(150 - 75)(1.5) = 1.44(10 ) in. A dT B CD = aal ¢TLCD = 12.8(10

–3 -6 12(10–6)(80 – 40)(36) = 17.28(10 ) mm - 3) in. )(150 - 75)(1.5) = 0.7425(10 A dT B AB = ast ¢TLAB = 6.60(10

From the geometry of the deflected bar AE shown Fig. b, dE = A dT B AB + C

A dT B CD - A dT B AB 6 0.25

–3 - 3) 1.44(10

(78) S (3.25)

6 mm

72 mm

-3 –3 0.7425(10 )

) –3 - 3 32.12(10 ) – 17.28(10 = 17.28(10 ) +) c+ B d (78) R (3.25) = 0.7425(10 6 0.25 = 0.2102 0.00981mm in.

Ans.

Ans. dE = 0.2102 mm 260

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4–77. The bar has a cross-sectional area A, length L, modulus of elasticity E, and coefficient of thermal expansion a. The temperature of the bar changes uniformly along its length from TA at A to TB at B so that at any point x along the bar T = TA + x(TB - TA)>L. Determine the force the bar exerts on the rigid walls. Initially no axial force is in the bar and the bar has a temperature of TA.

x A

B TB

TA

Solution + S

0 = ∆ T - dF 

(1)

However, d∆ T = a∆ T dx = a aTA + L

    ∆ T = a

TB - TA x - TA bdx L

TB - TA TB - TA 2 L x dx = a c x d` L 2L 0 L0

= ac

TB - TA aL Ld = (TB - TA) 2 2

From Eq. (1). 0 =

aL FL (TB - TA) 2 AE

F =

a AE (TB - TA) 2

Ans.

Ans: F = 261

aAE (TB - TA) 2

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4–78. When the temperature is at 30°C, the A-36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, the temperatures at ends A and B rise to 130°C and 80°C, respectively. If the temperature drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume each tank provides a rigid support at A and B.

150 mm 10 mm Section a – a 6m x

A

Solution

a a

B

Temperature Gradient: Since the temperature varies linearly along the pipe, Fig. a, the temperature gradient can be expressed as a function of x as T(x) = 80 +

50 50 (6 - x) = a130 xb°C 6 6

Thus, the change in temperature as a function of x is ∆T = T(x) - 30° = a130 -

50 50 xb - 30 = a100 xb°C 6 6

Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of dT = a

L

∆Tdx = 12(10 - 6)

L0

6m

a100 -

50 xbdx = 0.0054 m = 5.40 mm 6

Using the method of superposition, Fig. b, +) (S

0 = dT - dF 0 = 5.40 -

F(6000) 2

p(0.16 - 0.152)(200)(109)

F = 1 753 008 N Normal Stress: s =

F 1 753 008 = = 180 MPa A p(0.162 - 0.152)

Ans.

Ans: s = 180 MPa 262

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4–79. When the temperature is at 30°C, the A-36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, the temperatures at ends A and B rise to 130°C and 80°C, respectively. If the temperature drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume the walls of each tank act as a spring, each having a stiffness of k = 900 MN>m.

150 mm 10 mm Section a – a 6m x

A

Solution

a a

B

Temperature Gradient: Since the temperature varies linearly along the pipe, Fig. a, the temperature gradient can be expressed as a function of x as T(x) = 80 +

50 50 (6 - x) = a130 xb°C 6 6

Thus, the change in temperature as a function of x is ∆T = T(x) - 30° = a130 -

50 50 xb - 30 = a100 xb°C 6 6

Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of dT = a

L

∆Tdx = 12(10 - 6)

L0

6m

a100 -

50 xbdx = 0.0054 m = 5.40 mm 6

Using the method of superposition, Fig. b, +) (S

d = dT - dF F 6

900(10 )

(1000) = 5.40 - C

F(6000) 2

p(0.16 - 0.152)(200)(109)

+

F 900(106)

(1000) S

F = 1 018 361 N

Normal Stress: s =

F 1 018 361 = = 105 MPa A p(0.162 - 0.152)

Ans.

Ans: s = 105 MPa 263

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*4–80. When the temperature is at 30°C, the A-36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, it causes the temperature to vary along the pipe as T = (53  x2 - 20x + 120)°C, where x is in meters. Determine the normal stress developed in the pipe. Assume each tank provides a rigid support at A and B.

150 mm 10 mm Section a – a 6m x

A

Solution

a a

B

Compatibility Equation: The change in temperature as a function of x is 5 5 ∆T = T - 30° = a x2 - 20x + 120b - 30 = a x2 - 20x + 90b°C. If the pipe 3 3 is unconstrained, it will have a free expansion of dT = a

L

∆Tdx = 12(10 - 6)

L0

6m

5 a x2 - 20x + 90bdx = 0.0036 m = 3.60 mm 3

Using the method of superposition, Fig. b, +) (S

0 = dT - dF 0 = 3.60 -

F(6000) 2

p(0.16 - 0.152)(200)(109)

F = 1 168 672.47 N Normal Stress: s =

F 1 168 672.47 = = 120 MPa A p ( 0.162 - 0.152 )

Ans.

Ans: s = 120 MPa 264

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4–81. The 50-mm-diameter cylinder is made from Am 1004-T61 magnesium and is placed in the clamp when the temperature is T1 = 20° C. If the 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the force in the cylinder when the temperature rises to T2 = 130°C.

100 mm

150 mm

Solution + c ΣFy = 0;     Fst = Fmg = F dmg = dst amg Lmg ∆T 26(10 - 6)(0.1)(110) -

FmgLmg EmgAmg

= astLst ∆T +

FstLst EstAst

F(0.1) F(0.150) = 17(10 - 6)(0.150)(110) + p p 44.7(109) (0.05)2 193(109)(2) (0.01)2 4 4 Ans.

F = 904 N

Ans: F = 904 N 265

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4–82. The 50-mm-diameter cylinder is made from Am 1004-T61 magnesium and is placed in the clamp when the temperature is T1 = 15°C. If the two 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the temperature at which the average normal stress in either the magnesium or the steel first becomes 12 MPa.

100 mm

150 mm

Solution + c ΣFy = 0;     Fst = Fmg = F dmg = dst amg Lmg ∆T 26(10 - 6)(0.1)(∆T) -

FmgLmg EmgAmg

= astLst ∆T +

FstLst EstAst

F(0.1) F(0.150) = 17(10 - 6)(0.150)(∆T) + p p 44.7(109) (0.05)2 193(109)(2) (0.01)2 4 4

The steel has the smallest cross-sectional area. p F = sA = 12(106)(2) a b (0.01)2 = 1885.0 N 4

Thus,

∆T = 229° Ans.

T2 = 229° + 15° = 244° C

Ans: T2 = 244° C 266

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4–83. T he rigid block has a weight of 400 kN and is to be supported by posts A and B, which are made of A-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original length before they are loaded, determine the average normal stress developed in each post when post C is heated so that its temperature is increased 20°F. Each in2. mm2. by 10°C. Each post post has has aa cross-sectional cross-sectional area area of of 85000

A

C

B

31 mft

31 mft

Solution Equations of Equilibrium: a + ©MC = 0;

400 kN

FB(3) (3) = 0 (1) - FA(1)

+ c ©Fy = 0;

FA = FB = F

2F + FC - 400(103) = 0

[1] 1m

Compatibility: (+ T )

1m

(dC)F - (dC)T = dF FC L FL − 18(10 −6 )(10)L =  5000(10 −6 ) [101(109 )]  5000(10 −6 ) [200(109 )]    

1.9802 FC − F = 180(10 3 )

[2]

Solving Eqs. [1] and [2] yields: F = 123.393(103) N

FC = 153.214(103) N

average Normal Sress: sA = sB = sC =

123.393(103) F = = 24.68(106) N/m2 = 24.7 MPa 5000(10−6) A

153.214(103) FC = 30.64(106) N/m2 = 30.6 MPa = 5000(10−6) A

Ans. Ans.

Ans. sA = sB = 24.7 MPa, sC = 30.6 MPa 267

CH 04.indd 180

1/18/11 2:10:25 PM

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*4–84. The cylinder CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectional area of 375 mm2. Est = 200 GPa, Eal = 70 GPa, and aal = 23(10-6)>°C.

0.7 mm B

F C

– 240 mm 300 mm

D A

+ E

Solution dst = (dg)al - dal - 0.0007 Fst(0.3) -6

9

(125)(10 )(200)(10 )

= 23(10 - 6)(150)(0.24) -

F(0.24) (375)(10 - 6)(70)(109)

- 0.0007

12Fst = 128 000 - 9.1428F

(1)

+ c ΣFy = 0;

(2)

F - 2Fst = 0

Solving Eqs. (1) and (2) yields, FAB = FEF = Fst = 4.23 kN Ans.

FCD = F = 8.45 kN

Ans: FCD = 8.45 kN 268

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4–85. The cylinder CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance. Also, the two end rods AB and EF are heated from T1 = 30°C to T2 = 50°C. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a crosssectional area of 375 mm2. Est = 200 GPa, Eal = 70  GPa, ast = 12(10-6)>°C, and aal = 23(10-6)>°C.

0.7 mm B

F C

– 240 mm 300 mm

D A

+ E

Solution dst + (dT)st = (dT)al - dal - 0.0007 Fst(0.3) (125)(10 - 6)(200)(109)

+ 12(10 - 6)(50 - 30)(0.3)

= 23(10 - 6)(180 - 30)(0.24) -

Fal(0.24) 375(10 - 6)(70)(109)

- 0.0007

12.0Fst + 9.14286Fal = 56000

(1)

+ c ΣFy = 0;

(2)

Fal - 2Fst = 0

Solving Eqs. (1) and (2) yields: Ans.

FAB = FEF = Fst = 1.85 kN FCD = Fal = 3.70 kN

Ans: FAB = FEF = 1.85 kN 269

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4–86. t

The metal strap has a thickness t and width w and is subjected to a temperature gradient T1 to T2 (T1 6 T2). This causes the modulus of elasticity for the material to vary linearly from E1 at the top to a smaller amount E2 at the bottom. As a result, for any vertical position y, measured from the top surface, E = [(E2 - E1)>w]y + E1. Determine the position d where the axial force P must be applied so that the bar stretches uniformly over its cross section.

T1 P

w d

P

T2

Solution P = constant = P0 P0 =

s = E

s = P0 a

s E2 - E1 aa by + E1 b w

E2 - E1 y + E1 b w

+ ΣFx = 0: S

P =

L0

w

s t dy =

P = P0t a

P0 t a a

L0

LA

s dA = 0

m

P0 a

E2 - E1 y + E1 b t dy w

E2 - E1 E2 + E1 + E1w b = P0 t a bw 2 2

a+ ΣM0 = 0: P0 t a

P -

P(d) -

LA

y sdA = 0

w E2 + E1 E2 - E1 2 bwd = P0 aa by + E1yb t dy 2 w L0

E1 2 E2 + E1 E2 - E1 2 bwd = P0 t a w + w b 2 3 2

E2 + E1 1 bd = (2E2 + E1)w 2 6

d = a

2 E2 + E1 bw 3(E2 + E1)

Ans.

Ans: d = a 270

2E2 + E1 bw 3(E2 + E1)

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4–87. 5 mm

Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN.

40 mm

20 mm P

P r  10 mm 20 mm

Solution For the fillet: w 40 = = 2 h 20 From Fig. 4–23,

r 10 = = 0.5 h 20

      K = 1.4 smax = Ksavg = 1.4 a

8 (103) 0.02 (0.005)

= 112 MPa

b

For the hole: 2r 20 = = 0.5 w 40 From Fig. 4–24,

        K = 2.1 smax = Ksavg = 2.1 a

8 (103) (0.04 - 0.02)(0.005)

= 168 MPa

b

Ans.

Ans: smax = 168 MPa 271

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*4–88. If the allowable normal stress for the bar is s allow = 120 MPa, determine the maximum axial force P that can be applied to the bar.

5 mm 40 mm

20 mm P

P r  10 mm 20 mm

Solution Assume failure of the fillet. w 40 = = 2; h 20 From Fig. 4–23.  

r 10 = = 0.5 h 20

K = 1.4 sallow = sm ax = Ksavg 120 (106) = 1.4 a

P b 0.02 (0.005)

P = 8.57 kN

Assume failure of the hole. 2r 20 = = 0.5 w 40 From Fig. 4–24.  

K = 2.1 sallow = smax = Ksavg 120 (104) = 2.1 a

P b (0.04 - 0.02) (0.005)

Ans.

P = 5.71 kN (controls)

Ans: P = 5.71 kN 272

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4–89. The steel bar has the dimensions shown. Determine the maximum axial force P that can be applied so as not to exceed an allowable tensile stress of sallow = 150 MPa.

20 mm 60 mm

30 30 mm mm P

P r  15 mm 24 mm

Solution Assume failure occurs at the fillet: w 60 = = 2 h 30 From the text,  

and

r 15 = = 0.5 h 30

K = 1.4 smax = sallow = Ksavg 150 (106) = 1.4 c

P d 0.03 (0.02)

P = 64.3 kN

Assume failure occurs at the hole: 2r 24 = = 0.4 w 60 From the text,  

K = 2.2 smax = sallow = Ksavg 150 (106) = 2.2 c

P d (0.06 - 0.024) (0.02)

P = 49.1 kN (controls!)

Ans.

Ans: P = 49.1 kN 273

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4–90. r = 30 mm

The A-36 steel plate has a thickness of 12 mm. If sallow = 150 MPa, determine the maximum axial load P that it can support. Calculate its elongation, neglecting the effect of the fillets.

r = 30 mm

120 mm

C

B 60 mm P A

800 mm

60 mm P D

200 mm

200 mm

Solution Maximum Normal Stress at fillet: r 30 = = 0.5 and h 60

w 120 = = 2 h 60

From the text, K = 1.4 smax = sallow = Ksavg 150(106) = 1.4J

P R 0.06(0.012) Ans.

P = 77142.86 N = 77.1 kN Displacement: d = Σ =

PL AE 77142.86(400) 9

(0.06)(0.012)(200)(10 )

+

77142.86(800) (0.12)(0.012)(200)(109) Ans.

= 0.429 mm

Ans: P = 77.1 kN, d = 0.429 mm 274

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4–91. Determine the maximum axial force P that can be applied to the bar. The bar is made from steel and has an allowable stress of sallow = 147 21 ksi. MPa.

40.125 mm in. 25 mm 1.25 in.

37.5 1.875mm in.

P

P

Solution 15 mm 0.75 in.

Assume failure of the fillet. r 0.25 5 = 0.2 = = 0.2 h 1.25 25

r mmin. � 50.25

37.5 w 1.875 ==1.51.5 = 25 h 1.25

From Fig. 4-24, K = 1.73 sallow = smax = Ksavg

  P 147(106 ) = 1.73   (0.025)(0.004)   P = 8.497(10 3 ) N Assume failure of the hole. r 0.375 7.5 ==0.20 = 0.20 w 1.875 37.5 From Fig. 4-25, K = 2.45 sallow = smax = Ksavg

  P 147(106 ) = 2.45   (0.0375 − 0.015)(0.004)   P = 5.400(10 3 ) N

(controls)

Ans. Ans.

= 5.40 kN

Ans. P = 5.40 kN 275

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*4–92.

0.125 4 mm in. 1.25 in. 25 mm

1.875 in. 37.5 mm

Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN. P

P

Solution 0.75 in. 15 mm

At fillet: r 0.25 5 = 0.2 = = 0.2 h 1.25 25

r�  0.25 5 mmin.

w 1.875 37.5 ==1.5 = 1.5 h 1.25 25

From Fig. 4-24, K = 1.73

  8(10 3 ) P 6 = = σ max K= N−m 2 138 MPa  138.4(10 ) =   1.73  (0.025)(0.004)  A   At hole: r 0.375 7.5 ==0.20 = 0.20 w 1.875 37.5 From Fig. 4-25, K = 2.45

  8(10 3 ) 6 = = σ max 2.45  = ) N−m 2 218 MPa (Controls)  217.78(10 (0.0375 − 0.015)(0.004)   Ans.

Ans. = σ max 2.45 MPa 276

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4–93. The member is to be made from a steel plate that is 60.25 mmin.thick. hole isis drilled drilled through through its center, thick.If Ifa a25-mm 1-in. hole determine the approximate width w of the plate so that it can support an axial force of 3350 The 16.75lb. kN. Theallowable allowablestress stressis s ksi.MPa. isallow sallow= =22150

60.25 mmin. w

3350kN lb 16.75

3350 lb 16.75 kN

1 in. 25 mm

Solution sallow = smax = Ksavg

  16.75(10 3 ) 150(106 ) = K    (w − 0.025)(0.006)  = w 0.025 + 0.01861K

By trial and error, from Fig. 4-25, choose

r = 0.2; 0.17; w

K ==2.50 2.45 K

0.025 + 0.01861(2.50) = 0.07153 m = 71.5 mm w=

Since

0.5 r 12.5 0.17 = == 0.2 w 2.49 71.53

Ans.

OK

Ans. K = 2.50, w = 71.5 mm 277

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4–94.

12.5 mm 0.5 in.

A

The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry?

15 0.6mm in. 20 0.8mm in.

5 mm 0.2 in.

P

15 0.6mm in. B

Solution

42 MPa 6 ksi

36 ksi 252 MPa

Number of squares = 28 3 = = = 73.5(10 ) N 73.5 kN P 28[42(106 )](0.005)(0.0125)

σ avg = = K

Ans.

73.5(10 3 ) P 2 = 196(106 ) N−m= = 196 MPa A 2(0.015)(0.0125)

σ max 252 = = 1.286 = 1.29 σ avg 196

Ans.

Ans. P = 73.5 kN, K = 1.29 278

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4–95. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress concentration factor?

10 mm A 20 mm 80 mm

P

B 5 MPa 30 MPa

Solution Number of squares = 19 P = 19(5)(106)(0.02)(0.01) = 19 kN savg = K =

Ans.

19(103) P = = 23.75 MPa A 0.08(0.01)

smax 30 MPa = = 1.26 savg 23.75 MPa

Ans.

Ans: P = 19 kN, K = 1.26 279

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*4–96.

B

The three bars are pinned together and subjected to the load P. If each bar has a cross-sectional area A, length L, and is made from an elastic perfectly plastic material, for which the yield stress is sY, determine the largest load (ultimate load) that can be supported by the bars, i.e., the load P that causes all the bars to yield. Also, what is the horizontal displacement of point A when the load reaches its ultimate value? The modulus of elasticity is E.

L C

u L D

A

P

u L

Solution When all bars yield, the force in each bar is, FY = sYA + ©F = 0; : x

P - 2sYA cos u - sYA = 0 Ans.

P = sYA(2 cos u + 1) Bar AC will yield first followed by bars AB and AD. dAB = dAD = dA =

FY(L) sYAL sYL = = AE AE E

dAB sYL = cos u E cos u

Ans.

Ans. dA =

dAB sYL = cos u E cos u

280

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4–98. The weight is suspended from steel and aluminum wires, each having the same initial length of 3 m and cross-sectional area of 4 mm2. If the materials can be assumed to be elastic perfectly plastic, with (sY)st = 120 MPa and (sY)al = 70 MPa, determine the force in each wire if the weight is (a) 600 N and (b) 720 N. Eal = 70 GPa, Est = 200 GPa.

Aluminum

Steel

Solution + c ΣFy = 0;

(1)

Fal + Fst - W = 0

Assume both wires behave elastically. dal = dst;

FalL FstL = A(70) A(200) (2)

Fal = 0.35 Fst (a) When W = 600 N, solving Eqs. (1) and (2) yields: Fst = 444.44 N = 444 N

Ans.

Fal = 155.55 N = 156 N

Ans.

sal =

Fal 155.55 = = 38.88 MPa 6 (sg)al = 70 MPa Ast 4(10 - 6)

OK

sst =

Fst 444.44 = = 111.11 MPa 6 (sg)st = 120 MPa Ast 4(10 - 6)

OK

The elastic analysis is valid for both wires. (b) When W = 720 N, solving Eqs. (1) and (2) yields: Fst = 533.33 N:

Fst = 186.67 N

sal =

Fal 186.67 = = 46.67 MPa 6 (sg)al = 70 MPa Aal 4(10 - 6)

sst =

Fst 533.33 = = 133.33 MPa 7 (sg)st = 120 MPa Ast 4(10 - 6)

OK

Therefore, the steel wire yields. Hence, Fst = (sg)stAst = 120 ( 106 ) (4) ( 10 - 6 ) = 480 N

Ans.

From Eq. (1), Fal = 240 N

Ans.

sal =

240 4(10 - 6)

OK

= 60 MPa 6 (sg)al

Ans: (a) Fst = 444 N, Fal = 156 N, (b) Fst = 480 N, Fal = 240 N 282

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4–99. The weight is suspended from steel and aluminum wires, each having the same initial length of 3 m and cross-sectional area of 4 mm2. If the materials can be assumed to be elastic perfectly plastic, with (sY)st = 120 MPa and (sY)al = 70 MPa, determine the force in each wire if the weight is (a) 600 N and (b) 720 N. Eal = 70 GPa, Est = 200 GPa.

Fal

Aluminum

Fst

Steel

Solution Equations of Equilibrium: + c ΣFy = 0;

(1)

Fal + Fst - w = 0

Elastic Analysis: Assume both wires behave elastically. dal = dst Fal L A(70) ( 10

9

)

=

Fst L A(200) ( 109 ) (2)

Fal = 0.350 Fst a) When W = 600 N, solving Eq. (1) and (2) yields: Fst = 444.44 N = 444 N

Ans.

Fal = 155.55 N = 156 N

Ans.

Average Normal Stress: sal =

Fal 155.55 = = 38.88 MPa < (sg)al = 70.0 MPa Asl 4.00 ( 10-6 )

(OK!)

 sst =

Fst 444.44 = = 111.11 MPa < (sg)st = 120 MPa Ast 4.00 ( 10-6 )

(OK!)

The average normal stress for both wires do not exceed their respective yield stress. Therefore, the elastic analysis is vaild for both wires b) When W = 720 N, solving Eq. (1) and (2) yields: Fst = 533.33 N

Fal = 186.67 N

Average Normal Stress: sal =

Fal 186.67 = = 46.67 MPa < (sg)al = 70.0 MPa Aal 4.00 ( 10-6 )

sst =

Fst 533.33 = = 133.33 MPa > (sg)st = 120 MPa Ast 4.00 ( 10-6 )

(OK!)

Therefore, the steel wire yields. Hence, Fst = (sg)stAst = 120 ( 106 )( 4.00 )( 10-6 ) = 480 N From Eq. (1),

Ans. Ans.

Fal = 240 N sal =

240 4.00 ( 10-6 )

(OK!)

= 60.00 MPa < (sg)al

Ans: Fst = Fal = Fst = Fal = 283

444 N, 156 N, 480 N, 240 N

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*4–100.

1.2 4 ftm

The distributed loading is applied to the rigid beam, (MPa) which is supported by the three bars. Each bar has a ss(ksi) cross-sectional area of 780 mm2 and is made from a 60 material having a stress–strain diagram that can be 420 approximated by the two line segments shown. If a load of w = 400 kN/m is applied to the beam, determine the 252 36 stress in each bar and the vertical displacement of the beam.

Solution

0.0012 0.0012

a + ©MB = 0;

A A

B B

420 252

0.2 0.2

C

A

s (MPa)

1.2 4 ftm 1.5 m

w B

∋∋(in./in.) (mm/mm)

1.5 5 ftm C

w

FC(1.2) – FA(1.2) = 0; FA = FC = F

+ c ©Fy = 0;

0.0012

0.2

∋ (mm/mm)

(1)

2F + FB - 960 200 = 0

Since the loading and geometry are symmetrical, the bar will remain horizontal. Therefore, the displacement of the bars is the same and hence, the force in each bar is the same. From Eq. (1). 320 kN F = FB = 66.67 kip Thus,

σ= A σ= B σ= C

320(10 3 ) 2 = 410.26(106 ) N−m= 410 MPa 780(10 −6 )

Ans.

From the stress-strain diagram:

410.26 − 252 420 − 252 = = ; ε 0.18847 mm−mm ε − 0.0012 0.2 − 0.0012 3 δ ε= L 0.18847(1.5)(10= = ) 282.71 mm = 283 mm

1.2 m

Ans.

1.2 m

2.4(400) = 960 kN

s (MPa) 420 252

e mm/mm)

Ans. δ = 283 mm

284

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4–101.

4 ft 1.2 m

The distributed loading is applied to the rigid beam, which is supported by the three bars. Each bar has a cross-sectional area of 468 mm2 and is made from a material having a stress–strain diagram that can be approximated by the two line segments shown. Determine the intensity of the distributed loading w needed to cause the beam to be displaced downard 37.5 mm.

36 252

+ c ©Fy = 0;

C

A A

s (MPa) 420

0.0012

FC(1.2) 0; (4) -– FAA(1.2) (4) == 0;

B

60 420

Solution a + ©MB = 0;

A

s s(ksi) (MPa)

4 ft 1.2 m 1.5 m

252

0.2

w B B

∋ (in./in.) (mm/mm)

5 ft 1.5 m C C

w w

FA = FC = F

2F + FB - 2.4w 8 w ==00

0.0012

(1)

0.2

∋ (mm/mm)

Since the system and the loading are symmetrical, the bar will remain horizontal. Hence the displacement of the bars is the same and the force supported by each bar is the same. From Eq. (1), (2)

FB = F = 0.8 2.6667 w w From the stress-strain diagram: e =

37.5 1.5 = 50.025 0.025in.>in. mm/mm 51.5(10 (12) 3)

ss–-252 420 60– -252 36 36 5= ;  ; s 5 272.11 MPa 0.025 –-0.0012 0.0012 0.2 0.2– -0.0012 0.0012 Hence F = sA = [272.11(106)][468(10−6)] = 127.35(103) N From Eq. (2), w = 159.19(103) N/m = 159 kN/m

Ans.

1.2 m

1.2 m

2.42v

s (MPa) 420 252

e mm/mm)

Ans. w = 159 kN/m 285

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4–102. The rigid lever arm is supported by two A-36 steel wires having the same diameter of 4 mm. Determine the smallest force P that will cause (a) only one of the wires to yield; (b) both wires to yield. Consider A-36 steel as an elastic perfectly plastic material.

P

450 mm 150 mm 150 mm 30 A

Solution

E

300 mm

Equation of Equilibrium: Referring to the free-body diagram of the lever arm shown in Fig. a, a+ ΣME = 0;

C

B

D

FAB (300) + FCD (150) - P(450) = 0 (1)

2FAB + FCD = 3P

(a) Elastic Analysis: The compatibility equation can be written by referring to the geometry of Fig. b. dAB = a

300 bd 150 CD

dAB = 2dCD

FCD L FAB L = 2a b AE AE FCD =

1 F  2 AB

(2)

Assuming that wire AB is about to yield first, FAB = (sY)st AAB = 250 1 106 2 c

From Eq. (2),

FCD =

p 1 0.0042 2 d = 3141.59 N 4

1 (3141.59) = 1570.80 N 2

Substituting the result of FAB and FCD into Eq. (1), Ans.

P = 2618.00 N = 2.62 kN (b) Plastic Analysis: Since both wires AB and CD are required to yield, FAB = FCD = (sY)st A = 250 1 106 2 c

Substituting this result into Eq. (1),

p 1 0.0042 2 d = 3141.59 N 4 Ans.

P = 3141.59 N = 3.14 kN

Ans: (a) P = 2.62 kN, (b) P = 3.14 kN 286

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4–103. The 1500-kN weight is slowly set on the top of a post made of 2014-T6 aluminum with an A-36 steel core. If both materials can be considered elastic perfectly plastic, determine the stress in each material. Aluminum 251mm in.

Solution

mm 250in.

Steel

Equations of Equilibrium: + c ©Fy = 0;

[1]

Pst + Pal - 1500 300 == 0

Elastic Analysis: Assume both materials still behave elastically under the load. dst = dal Pal L Pst L =  π (0.052 ) [200(109 )]  π (0.12 − 0.052 ) [73.1(109 )] 4 4  

Pst = 0.9120 Pal Solving Eqs. [1] and [2] yields: Pal = 784.52 kN

Pst = 715.48 kN

Average Normal Stress:

784.52(10 3 ) = 133.18 MPa < (σ γ= )al 414 MPa π (0.12 − 0.052 ) 4

σ= al

Pal = Aal

σ= st

Pst 715.48(10 3 ) = = 364.39 MPa > (σ γ= )st 250 MPa π (0.052 ) Ast 4

(OK!)

Therefore, the steel core yields and so the elastic analysis is invalid. The stress in the steel is sst = (sγ)st = 250 36.0MPa ksi

Ans.

6  π 3 ) 4 (0.052 ) 490.87(10 = ) N 490.87 kN Pst = (sg)stAst =  250(10 =  

From Eq. [1] Pal = 186.90 1009.13kip kN Pal 1009.13(10 3 ) 6 = = 171.31(10 ) N−m 2 = 171.31 MPa < (σ γ )al 414 MPa sal = = π 2 2 Aal (0.1 − 0.05 ) 4 Ans.

Then sal = 171.31 MPa = 171 MPa

Ans. sst = 250 MPa, sal = 171 MPa 287

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*4–104. The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is sY = 530 MPa, and Est = 200 GPa, determine (a) the intensity of the distributed load w that can be placed on the beam that will cause only one of the wires to start to yield and (b) the smallest intensity of the distributed load that will cause both wires to yield. For the calculation, assume that the steel is elastic perfectly plastic.

E

D

800 mm A

B

C G

Solution

400 mm

250 mm

w 150 mm

Equations of Equilibrium: a + ©MA = 0;

FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0 [1]

0.4 FBE + 0.65 FCD = 0.32w (a) By observation, wire CD will yield first. p Then FCD = sg A = 530 A 106 B a b A 0.0042 B = 6.660 kN. 4 From the geometry dCD dBE = ; 0.4 0.65

dCD = 1.625dBE FCDL FBEL = 1.625 AE AE [2]

FCD = 1.625 FBE Using FCD = 6.660 kN and solving Eqs. [1] and [2] yields: FBE = 4.099 kN

Ans.

w = 18.7 kN>m (b) When both wires yield FBE = FCD = (sg)A p = 530 A 106 B a b A 0.0042 B = 6.660 kN 4

Substituting the results into Eq. [1] yields:

Ans.

w = 21.9 kN>m

Ans. w = 21.9 kN>m 288

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4–105. The rigid beam is supported by three 25-mm diameter A-36 steel rods. If the beam supports the force of P = 230 kN, determine the force developed in each rod. Consider the steel to be an elastic perfectly plastic material.

D

F

E

600 mm P A

400 mm

Solution Equation of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, FAD + FBE + FCF - 230(103) = 0

+ c ΣFy = 0;

(1)

        FBE(400) + FCF(1200) - 230(103)(800) = 0

a+ ΣMA = 0;

FBE + 3FCF = 460 ( 103 ) 

(2)

Elastic Analysis: Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =

dCF - dAD b(400) 1200

2 1 d + dCF 3 AD 3

FBEL 2 FADL 1 FCF L = a b + a b AE 3 AE 3 AE FBE =

2 1 F + FCF  3 AD 3

(3)

Solving Eqs. (1), (2), and (3) FCF = 131 428.57 N

FBE = 65 714.29 N FAD = 32 857.14 N

Normal Stress: sCF =

sBE =

sAD =

FCF 131428.57 = = 267.74 MPa 7 (sY)st p ACF (0.0252) 4

(N.G.)

FBE 65714.29 = = 133.87 MPa 6 (sY)st p ABE (0.0252) 4 FAD 32857.14 = = 66.94 MPa 6 (sY)st p AAD (0.0252) 4

(O.K.)

(O.K.)

Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using FCF = (sY)st ACF = 250 (106) c

p (0.0252) d = 122 718.46 N = 123 kN 4

289

Ans.

B

400 mm

C

400 mm

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4–105.

Continued

Substituting this result into Eq. (2), Ans.

FBE = 91844.61 N = 91.8 kN Substituting the result for FCF and FBE into Eq. (1),

Ans.

FAD = 15436.93 N = 15.4 kN sBE =

sAD =

FBE 91844.61 = = 187.10 MPa 6 (sY)st p ABE ( 0.0252 ) 4 FAD 15436.93 = = 31.45 MPa 6 (sY)st p AAD ( 0.0252 ) 4

(O.K.)

(O.K.)

Ans: FCF = 123 kN, FBE = 91.8 kN, FAD = 15.4 kN 290

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4–106. The rigid beam is supported by three 25-mm diameter A-36 steel rods. If the force of P = 230 kN is applied on the beam and removed, determine the residual stresses in each rod. Consider the steel to be an elastic perfectly plastic material.

D 600 mm

P A

400 mm

Solution Equation of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, FAD + FBE + FCF - 230 ( 103 ) = 0

+ c ΣFy = 0;

(1)

         FBE(400) + FCF(1200) - 230 ( 103 ) (800) = 0

a+ ΣMA = 0;

FBE + 3FCF = 460(103)

(2)

Elastic Analysis: Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =

dCF - dAD b(400) 1200

2 1 d + dCF 3 AD 3

(3)

FBE L 2 FAD L 1 FCF L = a b + a b AE 3 AE 3 AE FBE =

2 1 F + FCF  3 AD 3

(4)

Solving Eqs. (1), (2), and (4)

FCF = 131428.57 N

FBE = 65714.29 N FAD = 32857.14 N

Normal Stress: sCF =

sBE =

sAD =

FCF 131428.57 = = 267.74 MPa (T) 7 (sY)st p ACF ( 0.0252 ) 4 FBE 65714.29 = = 133.87 MPa (T) 6 (sY)st p ABE ( 0.0252 ) 4 FAD 32857.14 = = 66.94 MPa (T) 6 (sY)st p AAD ( 0.0252 ) 4

(N.G.)

(O.K.)

(O.K.)

Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using sCF = (sY)st = 250 MPa (T) FCF = sCF ACF = 250 ( 106 ) c

F

E

p ( 0.0252 ) d = 122718.46 N 4

291

B

400 mm

C

400 mm

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4–106.

Continued

Substituting this result into Eq. (2), FBE = 91844.61 N Substituting the result for FCF and FBE into Eq. (1), FAD = 15436.93 N sBE =

sAD =

FBE 91844.61 = = 187.10 MPa (T) 6 (sY)st p ABE ( 0.0252 ) 4 FAD 15436.93 = = 31.45 MPa (T) 6 (sY)st p AAD ( 0.0252 ) 4

(O.K.)

(O.K.)

Residual Stresses: The process of removing P can be represented by applying the force P′, which has a magnitude equal to that of P but is opposite in sense. Since the process occurs in a linear manner, the corresponding normal stress must have the same magnitude but opposite sense to that obtained from the elastic analysis. Thus, = = = sCF = 267.74 MPa (C) sBE = 133.87 MPa (C) sAD = 66.94 MPa (C)

Considering the tensile stress as positive and the compressive stress as negative, = (sCF)r = sCF + sCF = 250 + ( - 267.74) = - 17.7 MPa = 17.7 MPa (C)

Ans.

= (sBE)r = sBE + sBE = 187.10 + ( - 133.87) = 53.2 MPa (T)

Ans.

(sAD)r = sAD +

= sAD

= 31.45 + ( - 66.94) = - 35.5 MPa = 35.5 MPa (C)

Ans.

Ans: (sCF)r = 17.7 MPa (C), (sBE)r = 53.2 MPa (T), (sAD)r = 35.5 MPa (C) 292

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4–107. CC

The wire BC has a diameter of 3.4 mm and the material has the stress–strain characteristics shown in the figure. Determine the vertical displacement of the handle at D if the pull at the grip is slowly increased and reaches a magnitude of (a) P = 2250 N, (b) P = 3000 N.

40 in.mm 1000 AA

DD

BB 50 in.mm 1250

30 in. 750 mm

Solution

PP

s (ksi) (MPa)

Equations of Equilibrium: a + ©MA = 0;

[1]

F (50) - –P(80) = 0= 0 FBC P(2000) BC(1250)

2250lb, N, (a) From Eq. [1] when P = 450

720 lbN FBC = 3600

80 560 70 490

Average Normal Stress and Strain: sBC =

FBC = ABC

3600 720 pp 2 44 (0.0034 )

5 396.51(106) N/m2 5 396.51 MPa

0.007 0.007

0.12 0.12

(in./in.) P (mm/mm)

From the Stress–Strain diagram 396.51 490 70 = 5 ;  ; 0.007 eBC 0.007

0.005867mm/mm in.>in. eeBC BC5=0.005664 1250 mm

Displacement:

750 mm

dBC = eBCLBC = 0.005664(1000) = 5.6644 mm dD dD dBCdBC = 5 ; ;   80 2000 501250

dD 5 (1.6)(5.664) = 9.0631 mm = 9.06 mm

600 lb, (b) From Eq. [1] when P == 3000 N,   

560 490

Ans.

(εBC, 396.51)

F 960 lb FBC == 4800 N

Average Normal Stress and Strain: sBC =

FBC = ABC

4800 960 p 2 44 (0.0034 )

(0.007, 490)

(0.12, 560) x (εBC, 528.68)

e(mm/mm)

5 528.68(106) N/m2 5 528.68 MPa

From Stress–Strain diagram 56080– 490 528.68 − 490 - 70 5    eBC − 0.007 0.12 0.12– 0.007 - 0.007

eBC = 0.06944 mm/mm

Displacement: dBC = eBCLBC = 0.06944(1000) = 69.44 mm ddDD dBCdBC 8 = 5 ; ;   dD dD 5 (1.6)(69.44) = 111.11 mm = 111 mm 80 501250 2000

1250 mm

Ans.

750 mm

Ans. (a) dD = 9.06 mm, (b) dD = 111 mm

293

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*4–108.

P

The bar having a diameter of 50 mm is fixed connected at its ends and supports the axial load P. If the material is elastic perfectly plastic as shown by the stress–strain diagram, determine the smallest load P needed to cause segment CB to yield. If this load is released, determine the permanent displacement of point C.

A

B

C

2 ftm 0.6

3 ftm 0.9

(ksi) ss(MPa)

20 140

Solution

(in./in.) P (mm/mm)

0.001

When P is increased, region AC will become plastic first, then CB will become plastic. Thus,

F= A 140(106 )  π4 (0.052= ) 274.89(10 3 ) = N 274.89 kN A F= B σ=    + ©F = 0; : x

(1)

FA + FB - P = 0 kN P ==2(274.89) 2(62.832)= =549.78 125.66 kip P = 550 kN

Ans.

The deflection of point C is, dC = eL = (0.001)(3)(12) in. ; (0.9)(103=) =0.036 0.9 mm  d Consider the reverse of P on the bar.

FB′ (0.6) F ′ (0.9) 5 = B AE AE FA ¿ = 1.5 FB ¿ So that from Eq. (1) FB ¿ = 0.4P FA ¿ = 0.6P 3 FB′L 0.4  549.78(10 ) 0.9(10 ) = = 0.720 mm S AE  π (0.052 ) 140(106 )−0.001 4   ¢d = 0.036 - 0.0288 = 0.00720 mm  din. ; 0.9 – 0.720 5 0.180 3

= δC′

Ans.

Ans.

P = 550 kN, ¢d = 0.180 mm. ; 294

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4–109. The rigid beam is supported by the three posts A, B, and C of equal length. Posts A and C have a diameter of 75 mm and are made of a material for which E = 70 GPa and sY = 20 MPa. Post B has a diameter of 20 mm and is made of a material for which E′ = 100 GPa and sY ′ = 590 MPa. Determine the smallest magnitude of P so that (a) only rods A and C yield and (b) all the posts yield.

P

A

P

B

2m

2m

C

2m

2m

Solution ΣMB = 0;

FA = FC = Fal

+ c ΣFy = 0;

(1)

Fat + 2Fat - 2P = 0

(a) Post A and C will yield, Fal = (st)alA p = 20(104)a b(0.075)2 a = 88.36 kN

(Eal)r =

(sr)al Eal

=

20(104) 70(104)

= 0.0002857

Compatibility condition: dbr = dal = 0.0002857(L) Fbr (L) = 0.0002857 L p (0.02)2(100) ( 104 ) 4 Fbr = 8.976 kN sbr =

8.976(103) = 28.6 MPa 6 sg p (0.023) 4

OK.

From Eq. (1), 8.976 + 2(88.36) - 2P = 0 Ans.

P = 92.8 kN (b) All the posts yield: Fbr = (sg)brA p = (590)(104)a b(0.022) 4 = 185.35 kN

Fal = 88.36 kN From Eq. (1); 185.35 + 2(88.36) - 2P = 0 Ans.

P = 181 kN

Ans: P = 92.8 kN, P = 181 kN 295

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4–110. The rigid beam is supported by the three posts A, B, and C. Posts A and C have a diameter of 60 mm and are made of a material for which E = 70 GPa and sY = 20 MPa. Post B is made of a material for which E′ = 100 GPa and sY ′ = 590 MPa. If P = 130 kN, determine the diameter of post B so that all three posts are about to yield.

P

A

P

B

2m

2m

C

2m

2m

Solution + c ΣFy = 0;

(1)

2(Fg)al + Fbr - 260 = 0 (Fal)g = (sg)al A

From Eq. (1),

p = 20(106)a b(0.06)2 = 56.55 kN 4 2(56.55) + Fbr - 260 = 0 Fbr = 146.9 kN (sg)br = 590 ( 106 ) =

146.9 ( 103 ) p (d )3 4 B Ans.

d B = 0.01779 m = 17.8 mm

Ans: d B = 17.8 mm 296

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R4–1. The assembly consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the temperature is at 30° C, the gap between the rod and rigid member AE is 0.1 mm. Determine the normal stress developed in the bolts and the rod if the temperature rises to 130° C. Assume BF is also rigid.

A

C

0.1 mm E 25 mm

25 mm 300 mm

400 mm 50 mm B

D

F

Solution Equation of Equilibrium: Referring to the free-body diagram of the rigid cap shown in Fig. a, + c ΣFy = 0;

(1)

Fr - 2Fb = 0

Compatibility Equation: If the bolts and the rod are unconstrained, they will have a free expansion of (dT)b = ast ∆TLb = 12(10 - 6)(130 - 30)(400) = 0.48 mm and (dg)r = aal ∆TLr = 24(10 - 6)(130 - 30)(300) = 0.72 mm. Referring to the initial and final position of the assembly shown in Fig. b, (dT)r - dFr - 0.1 = (dT)b + dFb 0.72 -

Fr (300) Fb(400) - 0.1 = 0.48 + p p (0.052)(68.9)(109) (0.0252)(200)(109) 4 4 (2)

Fb + 0.5443Fr = 34361.17 Solving Eqs. (1) and (2). Fb + 16 452.29 N

Fr = 32 904.58 N

Normal Stress: sb =

sr =

Fb 16 452.29 = = 33.5 MPa p Ab 2 (0.025 ) 4

Ans.

Fr 32 904.58 = = 16.8 MPa p Ar (0.052) 4

Ans.

Ans: sb = 33.5 MPa, sr = 16.8 MPa 297

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R4–2. The assembly shown consists of two A992 steel bolts AB and EF and an 6061-T6 aluminum rod CD. When the temperature is at 30° C, the gap between the rod and rigid member AE is 0.1 mm. Determine the highest temperature to which the assembly can be raised without causing yielding either in the rod or the bolts. Assume BF is also rigid.

A

C

0.1 mm E 25 mm

25 mm 300 mm

400 mm 50 mm B

D

F

Solution Equation of Equilibrium: Referring to the free-body diagram of the rigid cap shown in Fig. a, + c ΣFy = 0;

(1)

Fp - 2Fb = 0

Normal Stress: Assuming that the steel bolts yield first, then p Fb = (sg)stAb = 250(106) c (0.0252) d = 122 718.46 N 4

Substituting this result into Eq. (1), Fp = 245 436.93 N Then, sp =

Fp Ap

=

245 436.93 = 125 MPa 6 (sg)al p (0.052) 4

(O.K!)

Compatibility Equation: If the assembly is unconstrained, the bolts and the post will have free expansion of (dT)b = ast ∆TLb = 12(10 - 6)(T - 30)(400) = 4.8(10 - 3)(T - 30) and (dT)p = aal ∆TLp = 24(10 - 6)T - 30)(300) = 7.2(10 - 3)(T - 30). Referring to the initial and final position of the assembly shown in Fig. b, (dT)p - dFp - 0.1 = (dT)b + dFb 7.2(10 - 3)(T - 30) -

245 436.93(300) p (0.052)(68.9)(109) 4

- 0.1 = 4.8(10 - 3)(T - 30) +

122 718.46(400) p (0.0252)(200)(109) 4 Ans.

T = 506.78° C = 507°C

Ans: T = 507°C 298

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R4–3. The rods each have the same 25-mm diameter and 600-mm length. If they are made of A992 steel, determine the forces developed in each rod when the temperature increases by 50° C.

C

600 mm 60 B

Solution

A

60

600 mm

Equation of Equilibrium: Referring to the free-body diagram of joint A shown in Fig. a, + c ΣFx = 0;

FAD sin 60° - FAC sin 60° = 0

+> ΣFx = 0;

FAB - 2F cos 60° = 0

D

FAC = FAD = F

(1)

FAB = F

Compatibility Equation: If AB and AC are unconstrained, they will have a free expansion of 1 dT 2 AB = 1 dT 2 AC = ast ∆TL = 12(10 - 6)(50)(600) = 0.36 mm. Referring to the initial and final position of joint A, dFAB -

1 dT 2 AB

= adT ′ b

AC

- dFAC ′

dAC = 2dAC. cos 60° = 2(dT)AC and dFAC ′ = 2dFAC. Thus, this equation becomes

Due to symmetry, joint A will displace horizontally, and dAC ′ = Thus, adT ′ b dFAB -

AC

1 dT 2 AB

FAB (600)



= 2 1 dT 2 AC - 2dAC

p 1 0.0252 2 (200)(109) 4

- 0.36 = 2(0.36) - 2C

FAB + 2F = 176 714.59

F(600) p 1 0.0252 2 (200)(109) 4

S

(2)

Solving Eqs. (1) and (2), FAB = FAC = FAD = 58 904.86 N = 58.9 kN (C)

Ans.

Ans: FAB = FAC = FAD = 58.9 kN (C) 299

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*R4–4. B

A

Two A-36 steel pipes, each having a cross-sectional area of 200 mm2, are screwed together using a union at B as shown. Originally the assembly is adjusted so that no load is on the pipe. If the union is then tightened so that its screw, having a lead of 0.55 mm, undergoes two full turns, determine the average normal stress developed in the pipe. Assume that the union at B and couplings at A and C are rigid. Neglect the size of the union. Note: The lead would cause the pipe, when unloaded, to shorten shorten 0.55 mm when the union is rotated one revolution.

3 ftm 0.9

C 2 ftm 0.6

2(0.55) = 1.10 mm

The loads acting on both segments AB and BC are the same since no external load acts on the system. 1.10 = dB>A + dB>C 1.10 5

P(0.9)(1000) [200(10−6)][200(109)]

1

P(0.6)(1000) [200(10−6)][200(109)]

P 5 29.33(103) 5 29.33 kN sAB = sBC =

29.33(103) P 5 5 146.67(106) N/m2 5 147 MPa A 200(10−6)

Ans.

Ans. sAB = sBC = 147 MPa 300

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R4-5. The force P is applied to the bar, which is composed of an elastic perfectly plastic material. Construct a graph to show how the force in each section AB and BC (ordinate) varies as P (abscissa) is increased. The bar has cross-sectional areas of 625 mm2 in region AB and 2500 mm2 in region BC, and σY = 210 MPa. Given: L AB := 150mm AAB := 625mm2 2

L BC := 50mm

ABC := 2500mm σ Y := 210MPa

Solution: Equations of equilibrium:

+ ΣF x=0;

P − FA − FC = 0

[1]

Elastic behavior:

+

0 = ∆ C − δC 0=

( P) ⋅ L AB E ⋅ AAB

−

⎡ ( FC) ⋅ LBC ( FC) ⋅ LAB⎤ + ⎢ ⎥ E⋅ AAB ⎣ E⋅ ABC ⎦

( )

0 = 6P − FC ( 0.5 + 6) Substituting [2] into [1]:

FC =

12 P 13

[2]

FA =

1 P 13

[3]

By comparison, segment BC will yield first. Hence,

( )

FC := σ Y ⋅ ABC From [2]: From [3]:

13 ⋅F 12 C 1 FA := ⋅P 13 P :=

FC = 525 kN P = 568.75 kN FA = 43.75 kN

When segment AB yields,

( ) FC := ( σ Y) ⋅ ABC

FA := σ Y ⋅ AAB

From [1]:

P := FA + FC

FA = 131.25 kN FC = 525 kN P = 656.25 kN

301

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2014-T6aluminum aluminum a diameter of rodrod has has a diameter of 0.5 in. R4–6. The 2014-T6 12 mm and isattached lightly attached to the rigid supports and and is lightly to the rigid supports at A andatBA when B1 when T1 = If the temperature T2 = –20°, T2 = -10°F, If 25°C. the temperature becomesbecomes and T = 70°F. andaxial an axial force = 80 appliedtotothe therigid rigid collar collar as lb Nis isapplied an force of Pof=P 16 shown, determine the reactions at A and B.

A

B

P/2 P/2 5 in. 125 mm

8 in. 200 mm

40 N

Solution

40 N

Compatibility:

40 N

40 N

+ 0 = ¢ - ¢ + d : B T B = 0

FB (325) 80(125) − 23(10 −6 )[25° − (−20°)](325) +  π (0.012 2 ) [73.1(109 )]  π (0.012 2 ) [73.1(109 )]   4 4 3

= ) 8.53 kN = FB 8.526(10

+ ©F = 0; : x

Ans.

40 N

FA = 8.526 kN

40 N

2(0.008) 2.1251– F-A F 2(0.040) + 8.526 =A0 = 0

FA = 8.606 kN = 8.61 kN

Ans.

Ans. FB = 8.53 kN, FA = 8.61 kN 302

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R4–7. The 2014-T6 aluminum rod has a diameter of 0.5mm in. and isis lightly lightlyattached attachedtoto rigid supports A 12 thethe rigid supports at Aatand and B when Determine the Pforce thatbemust B when T1 = T 40°C. Determine the force that P must ap1 = 70°F. be applied to the so that, when = 0°F, atthe plied to the collar so collar that, when T = 0°C, the Treaction B reaction is zero. at B is zero.

A

B

P/2 P/2 125 mm 5 in.

200 mm 8 in.

Solution + : = 0

0 = ¢ B - ¢ T + dB P(125) − 23(10 −6 )(40)(325) + 0  π (0.012 2 ) [73.1(109 )] 4 

3 = = P 19.776(10 ) N 19.8 kN

Ans.

Ans. P = 19.8 kN 303

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*R4–8. The rigid link is supported by a pin at A and two A-36 steel steelwires, wires,each each having an unstretched length of having an unstretched length of 12 in. 300 mm and cross-sectional area in of2. 7.8 mm2. Determine and cross-sectional area of 0.0125 Determine the force the force developed in the wires the link supports the developed in the wires when thewhen link supports the vertical vertical loadlb.of 1.75 kN. load of 350

300 12 mm in. C 125 5mm in. B 100 4mm in.

A

Solution 150 mm 6 in.

Equations of Equilibrium: a + ©MA = 0;

[1]

- FCC(225) (9) -– F (4) + +350(6) = 0= 0 –F FBB(100) 1.75(150)

1.75 350 kN lb

Compatibility: ddCC ddBB = 5 4 9 100 225

125 mm

(L) F F FB(L) FC(L) = 5 4AE 9AE 100AE 225AE

125 mm 100 mm

2.25FB − FC = 0‚

[2]

100 mm 150 mm

Solving Eqs. [1] and [2] yields:

1.75 kN

FB = 0.4330 kN = 433 N

Ans.

FC = 0.9742 kN = 974 N

Ans.

Ans. FB = 433 N‚ FC = 974 N‚ 304

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R4–9. The joint is made from three A992 steel plates that are bonded together at their seams. Determine the displacement of end A with respect to end B when the joint is subjected to the axial loads. Each plate has a thickness of 5 mm.

100 mm 23 kN

46 kN

B 23 kN

A 600 mm 200 mm

800 mm

Solution dA>B = Σ

46(103)(600) 46(103)(200) 23(103)(800) PL = + + AE (0.005)(0.1)(200)(109) 3(0.005)(0.1)(200)(109) (0.005)(0.1)(200)(109) Ans.

= 0.491 mm

Ans: dA>B = 0.491 mm 305

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5–1.  The solid shaft of radius r is subjected to a torque T. Determine the radius r′ of the inner core of the shaft that resists one-half of the applied torque (T>2). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.

r¿ r

T

Solution Tc Tr 2T =p 4 = 3 J r pr 2

a) tmax =

t =

a T2 br′

Since t = r′ =

b)

L0 L0 L0

r 2

r 2

r 2

=

p 4 2 (r′)

r 1

24

T p(r′)3

r′ t ; r max

T r′ 2T = a b 3 r pr 3 p(r′)

Ans.

= 0.841r

dT = 2p

L0

r′

tr2 dr r′

dT = 2p

r tmax r2 dr L0 r r′

dT = 2p

r 2T a 3 b r2 dr L0 r pr r′

4T T = 4 r3 dr 2 r L0 r′ =

r′ 1

24

Ans.

= 0.841r

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

306

Ans: r′ = 0.841r

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5–2.  The solid shaft of radius r is subjected to a torque T. Determine the radius r′ of the inner core of the shaft that resists one-quarter of the applied torque (T>4). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution.

r¿ r

T

Solution T(r) Tc 2T = p 4 = 3 J (r ) pr 2

a) tmax =

Since t = t′ = r′ = b) t =

r′ 2Tr′ t = r max pr 4

(T4 )r′ T′c′ 2Tr′ ;   4 = p 4 J′ pr 2 (r′) r 1

44

r r 2T 2T t = a 3b = r; c max r pr pr 4

dT = rt dA = r c L0

T 4

Ans.

= 0.707 r

dT =

r L

4T 4

r′ T 4T r4 | ; = 4 4 r 4 0

r′

dA = 2pr dr

2T 4T r d (2pr dr) = 4 r3dr pr 4 r

r3dr (r′)4 1 = 4 r4 Ans.

r′ = 0.707 r

Ans: r′ = 0.707 r 307

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5–3.  A shaft is made of an aluminum alloy having an allowable shear stress of tallow = 100 MPa. If the diameter of the shaft is 100 mm, determine the maximum torque T that can be transmitted. What would be the maximum torque T′ if a 75-mm-diameter hole were bored through the shaft? Sketch the shear-stress distribution along a radial line in each case.

T T¿

Solution Allowable Shear Stress: Torsion formula can be applied. For the solid shaft, tmax = tallow =

T(0.05) Tc ; 100 ( 106 ) = p 4 J 2 (0.05 )

T = 19.63 ( 103 ) N # m = 19.6 kN # m

Ans.

For the hollow shaft, tmax = tallow =

Tc ; 100 ( 106 ) = J

T′(0.05) p 2

(0.054 - 0.03754)

 T′ = 13.42 ( 103 ) N # m = 13.4 kN # m

Ans.

The shear stress at the inner surface where r = 0.0375 m is tr = 0.0375 m =

T′r J

=

13.42 ( 103 ) (0.0375) p 4 2 (0.05

- 0.03754)

= 75.0 ( 106 ) Pa = 75.0 MPa

The shear stress distribution along the radius of the cross-section of the solid and hollow shafts are shown in Figs. a and b respectively.

Ans: T = 19.6 kN # m, T′ = 13.4 kN # m 308

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  The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm and a wall thickness of 5 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear-stress distribution over the cross section.

600 N 75 mm

75 mm

5 mm 25 mm

600 N

Solution

T = 600(0.15) = 90 N # m tmax =

ti =

Tc = J

Tr = J

90(0.0175) p 2

[(0.0175)4 - (0.0125)4] 90(0.0125)

p 2

[(0.0175)4 - (0.0125)4]

Ans.

= 14.5 MPa

= 10.3 MPa

Ans: tmax = 14.5 MPa 309

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  The solid shaft is fixed to the support at C and subjected to the torsional loadings. Determine the shear stress at points A and B on the surface, and sketch the shear stress on volume elements located at these points.

35 mm

C A

B 20 mm

35 mm

300 Nm 800 Nm

Solution tB =

800(0.02) TB r = p = 6.79 MPa 4 J 2 (0.035 )

Ans.

tA =

500(0.035) TA c = 7.42 MPa = p 4 J 2 (0.035 )

Ans.

Ans: tB = 6.79 MPa, tA = 7.42 MPa 310

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  The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress in the shaft.

300 Nm

500 Nm

A

200 Nm

C

400 Nm

300 mm

Solution

D B

400 mm

Internal Torque: As shown on torque diagram. Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying torsion formula, abs = t max

=

500 mm

Tmax c J 400(0.015) p 2

(0.0154)

Ans.

= 75.5 MPa

Ans: abs = 75.5 MPa t max 311

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  The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall and three torques are applied to it, determine the absolute maximum shear stress developed in the pipe. 30 Nm 20 Nm 80 Nm

Solution tmax =

Tmax c = J

90(0.02) p 4 2 (0.02

- 0.01854) Ans.

= 26.7 MPa

Ans: tmax = 26.7 MPa 312

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*5–8. The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.

A

30 Nm 20 Nm

Solution 80 Nm

tmax =

Tmax c = J

90(0.02) p 2

(0.024 - 0.01854)

= 26.7 MPa

Ans..

Ans: tmax = 26.7 MPa 313

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  The solid aluminum shaft has a diameter of 50 mm and an allowable shear stress of tallow = 60 MPa. Determine the largest torque T1 that can be applied to the shaft if it is also subjected to the other torsional loadings. It is required that T1 act in the direction shown. Also, determine the maximum shear stress within regions CD and DE.

E D C

300 Nm

B

600 Nm

A

T1 900 Nm

Solution Internal Torque: Assuming that failure occurs at region BC of the shaft, where the torque will be greatest. Referring to the FBD of the right segment of the shaft sectioned through region BC, Fig. a g M x = 0;  T1 - 900 - TBC = 0  TBC = T1 - 900

Maximum Shear Stress: Applying the torsion formula, tmax = tallow =

(T1 - 900)(0.025) TBC c ;  60 ( 106 ) = p 4 J 2 ( 0.025 )

(T1)max = T1 = 2372.62 N # m = 2.37 kN # m

Ans.

using this result the torque diagram shown in Fig. b can be plotted. This indicates that region BC indeed is subjected to maximum internal torque, thus, the critical region. From the torque diagram, the internal torques in regions CD and DE are TCD = 872.62 N # m and TDE = 572.62 N # m respectively. (tmax)CD =

872.62 (0.025) TCD C = p = 35.55 ( 106 ) Pa = 35.6 MPa 4 J 2 ( 0.025 )

Ans.

(tmax)DE =

572.62 (0.025) TDC C = p = 23.33 ( 106 ) Pa = 23.3 MPa 4 J ( ) 0.025 2

Ans.

Ans: (T1)max = 2.37 kN # m, (tmax)CD = 35.6 MPa, (tmax)DE = 23.3 MPa 314

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  The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stress is maximum. Set T1 = 2000 N # m.

E D C

300 Nm

B

600 Nm

A

T1 900 Nm

Solution Internal Torque: The torque diagram plotted in Fig. a indicates that region BC of the shaft is subjected to the greatest internal torque; thus, the critical region where the absolute maximum shear stress occurs . Here TBC = 1100 N # m. Applying the torsion formula, tmax abs =

1100 (0.025) TBC c = p = 44.82 ( 106 ) Pa = 44.8 MPa 4 J 2 ( 0.025 )

Ans.

The shear stress distribution along the radius of the cross-section of the shaft is shown in Fig. b.

Ans: tabs = 44.8 MPa max

315

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5–11. The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses on the shaft’s surface and specify their locations, measured from the free end.

400 Nm

C

4 kNm/m B

A

0.5 m 800 Nm

d

0.2 m 0.2 m

0.5 m

Solution Internal Torque: The torque diagram plotted in Fig. a indicates that region 1.0 m 6 x 6 1.2 m of the shaft is subjected to the greatest internal torque where Tmax = 1200 N # m, whereas the minimum internal torque, Tmin = 0, occurs at x = 0.7 m. Shear Stress: Applying the torsion formula, abs = tmax

1200 (0.03) Tmax C = p = 28.29 ( 106 ) Pa = 28.3 MPa 4 J ( ) 0.03 2

Ans.

occurs within the region 1.0 m < x < 1.2 m abs = tmin

Ans.

Tmin C   occurs at x = 0.700 m J

Ans.

Ans: abs = 28.3 MPa, tmax for 1.0 m 6 x 6 1.2 m, abs = 0, tmin at x = 0.700 m 316

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*5–12. The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft if the allowable shear stress for the material is tallow = 60 MPa.

C

400 Nm 4 kNm/m B

A

0.5 m 800 Nm

d

0.2 m 0.2 m

0.5 m

Solution Internal Torque: The torque diagram plotted in Fig. a indicates that region 1.0 m 6 x 6 1.2 m of the shaft is subjected to the greatest internal torque Tmax = 1200 N # m; thus, that is the critical region where the absolute maximum shear stress occurs. Allowable Shear Stress: Applying the torsion formula, abs = tallow = tmax

60 ( 106 ) =

Tmax C J

1 d2 2 p d 4 2 122

1200

Ans.

d = 0.04670 m = 46.7 mm

Ans: d = 46.7 mm 317

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5–13. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller smallerpipe pipehas has outer diameter of in. 18.75 and anan outer diameter of 0.75 andmm an inner an inner diameter ofwhereas 17 mm, the whereas larger has diameter of 0.68 in., largerthe pipe has pipe an outer an outer of diameter mmdiameter and an inner diameter 21.5 diameter 1 in. andofan25inner of 0.86 in. If theof pipe is mm. If secured the pipetoisthe tightly to the wall at C, determine tightly wall secured at C, determine the maximum shear the maximum shear stress developed each section of the stress developed in each section of the in pipe when the couple pipe when the couple applied to the handles of the shown is applied to theshown handlesisof the wrench. wrench.

C

B

A 75 N in. 15 lb1506 mm 2008mm in.

Solution tAB

15Nlb 75

26.25(0.009375) Tc 6 = 62.54(10 = ) N−m 2 62.5 MPa = = π 4 4 J (0.009375 − 0.0085 ) 2

Ans.

26.25(0.0125) Tc 6 = 18.89(10 = ) N−m 2 18.9 MPa = π 4 4 J (0.0125 − 0.01075 ) 2

Ans.

tBC =

26.25 N · m TAB = 26.25 N · m 26.25 N · m TBC = 26.25 N · m

Ans: tAB  62.5 MPa, tBC  18.9 MPa 318

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5–14. A steel tube having an outer diameter of 60 mm is used to transmit 6.75 kW when turning at 27 revNmin. Determine the inner diameter d of the tube to the nearest mm if the allowable shear stress is tallow = 70 MPa.

d 60 mm

Solution  

ω =  27

πev  2π πad  1 min     = 0.9π πad−s min  1 πev  60 s 

P = Tω 6.75(10 3 ) = T(0.9π ) = T 2.387(10 3 ) N ⋅ m tmax = tallow =

70(106 ) =

Tc J

2.387(10 3 )(0.03) π

2

( 0.034 − ci4 )

ci = 0.01996 m = = 0.03992 = d 2= ci 2(0.01996) m 39.92 mm Use d = 40 mm

Ans.

Ans: Use d = 40 mm 319

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5–15. The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the shear stress at points A and B, and sketch the shear stress on volume elements located at these points.

2 kNm/m

C 0.4 m

600 Nm

B

A

0.4 m 0.3 m 0.3 m

400 Nm

d

Solution Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula tA =

400(0.03) TA c = p = 9.43 MPa 4 J 2 ( 0.03 )

Ans.

tB =

600(0.03) TB c = p = 14.1 MPa 4 J 2 ( 0.03 )

Ans.

Ans: tA = 9.43 MPa, tB = 14.1 MPa 320

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*5–16. The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses on the shaft’s surface, and specify their locations, measured from the fixed end C.

2 kNm/m

C 0.4 m

600 Nm

B

A

0.4 m 0.3 m 0.3 m

400 Nm

d

Solution

Internal Torque: From the torque diagram, the maximum torque Tmax = 1400 N # m occurs at the fixed support and the minimum torque Tmin = 0 occurs at x = 0.700 m. Shear Stress: Applying the torsion formula tabs

min

=

tabs = max

Tmin c = 0  occurs at  x = 0.700 m J

Ans.

1400(0.03) Tmax c = p = 33.0 MPa 4 J 2 ( 0.03 )

Ans. Ans.

occurs at  x = 0

According to Saint-Venant’s principle, application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading. Therefore, tabs obtained is not valid. max

Ans: tabs max

= 0 occurs at x = 0.700 m,

min

= 33.0 MPa occurs at x = 0

tabs 321

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5–17. The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft if the allowable shear stress for the material is tallow = 1.6 MPa.

2 kNm/m

C 0.4 m

600 Nm

B

A

0.4 m 0.3 m 0.3 m

400 Nm

d

Solution

Internal Torque: From the torque diagram, the maximum torque Tmax = 1400 N # m occurs at the fixed support. Allowable Shear Stress: Applying the torsion formula tabs = tallow = Tmax c max J 175 ( 106 ) =

1400 1 d2 2

1 2

p d 4 2 2

Ans.

           d = 0.03441 m = 34.4 mm

According to Saint-Venant’s principle, application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading. Therefore, the above analysis is not valid.

Ans: d = 34.4 mm 322

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5–18. The motor delivers a torque of 50 N # m to the shaft AB. This torque is transmitted to shaft CD using the gears at E and F. Determine the equilibrium torque Tⴕ on shaft CD and the maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts.

A

50 mm 30 mm B

Solution

35 mm T¿

Equilibrium: a + ©ME = 0;

50 - F(0.05) = 0

a + ©MF = 0;

T¿ - 1000(0.125) = 0

C

E

125 mm D

F

F = 1000 N

T¿ = 125 N # m

Ans.

Internal Torque: As shown on FBD. Maximum Shear Stress: Applying torsion Formula. (tAB)max =

50.0(0.015) TAB c = 9.43 MPa = p 4 J 2 (0.015 )

Ans.

(tCD)max =

125(0.0175) TCDc = 14.8 MPa = p 4 J 2 (0.0175 )

Ans.

Ans. (tAB)max = 9.43 MPa (tCD)max = 14.8 MPa 223

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5–19. If the applied torque on shaft CD is T¿ = 75 N # m, determine the absolute maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating.

A

50 mm 30 mm B 35 mm T¿

Solution

C

E

125 mm D

F

Equilibrium: a + ©MF = 0;

75 - F(0.125) = 0;

a + ©ME = 0;

600(0.05) - TA = 0

F = 600 N

TA = 30.0 N # m Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula (tEA)max =

30.0(0.015) TEA c = 5.66 MPa = p 4 J 2 (0.015 )

Ans.

(tCD)max =

75.0(0.0175) TCDc = 8.91 MPa = p 4 J 2 (0.0175 )

Ans.

Ans. (tEA)max = 5.66 MPa (tCD)max = 8.91 MPa 224

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*  The shaft has an outer diameter of 100 mm and an inner diameter of 80 mm. If it is subjected to the three torques, determine the absolute maximum shear stress in the shaft. The smooth bearings A and B do not resist torque.

E

A

D

10 kNm

Maximum Shear Stress: Referring to the torque diagram shown in Fig. a, region ED is subjected to the largest internal torque, which is Tmax = 10 kN # m. Thus, the absolute maximum shear stress occurs in this region. Applying the torsion formula, Tmax C = J

C

15 kNm

Solution

tabs = max

B

10 ( 103 ) (0.05) p 2

( 0.054 - 0.044 )

= 86.26 ( 106 ) Pa = 86.3 MPa

5 kNm

Ans.

Ans: abs = 86.3 MPa tmax 325

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  The shaft has an outer diameter of 100 mm and an inner diameter of 80 mm. If it is subjected to the three torques, plot the shear stress distribution along a radial line for the cross section within region CD of the shaft. The smooth bearings at A and B do not resist torque.

E

A

D

10 kNm

C

15 kNm

Solution Shear Stress: Referring to the torque diagram shown in Fig. a, Region CD of the shaft is subjected to an internal torque of TCD = 5 kN # m. The torsion formula will be applied. The maximum shear stress is (tmax)CD =

B

TCD C = J

5 ( 103 ) (0.05) p 2

( 0.054 - 0.044 )

5 kNm

= 43.13 ( 106 ) Pa = 43.1 MPa

The shear stress at the inner surface of the hollow shaft is

(tp = 0.04 m)CD =

TCDr = J

5 ( 103 ) (0.04) p 2

( 0.054 - 0.044 )

= 34.51 ( 106 ) Pa = 34.5 MPa

Also, tmax C

=

tp r

r

;   (tp = 0.04 m)CD = a C btmax = a

0.04 b(43 # 13) 0.05

= 34.51 MPa = 34.5 MPa

The shear stress distribution along the radius of the cross-section of the shaft is shown in Fig. b.

Ans: N/A 326

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5–22. If the gears are subjected to the torques shown, determine the maximum shear stress in the segments AB and BC of the A-36 steel shaft. The shaft has a diameter of 40 mm.

300 Nm 100 Nm A 200 Nm B

Solution The internal torque developed in segments AB and BC are shown in their respective FBDs, Figs. a and b. The polar moment of inertia of the shaft is J =

C

p ( 0.024 ) = 80 ( 10-9 ) p m4. Thus, 2

   (tAB)max =

300(0.02) TAB c = = 23.87 ( 106 ) Pa = 23.9 MPa J 80 ( 10-9 ) p

Ans.

   (tBC)max =

200(0.02) TBC c = = 15.92 ( 106 ) Pa = 15.9 MPa J 80 ( 10-9 ) p

Ans.

Ans: (tAB)max = 23.9 MPa, (tBC)max = 15.9 MPa 327

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5–23. If the gears are subjected to the torques shown, determine the required diameter of the A-36 steel shaft to the nearest mm if tallow = 60 MPa.

300 Nm 100 Nm A 200 Nm B

Solution The internal torque developed in segments AB and BC are shown in their respective FBDs, Fig. a and b.

C

Here, segment AB is critical since its internal torque is the greatest. The polar p d 4 pd 4 moment of inertia of the shaft is J = a b = . Thus, 2 2 32 tallow =

TC ; J

60 ( 106 ) =

300 ( d>2 ) pd4 >32

Ans.

d = 0.02942 m = 30 mm

Ans: d = 30 mm 328

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*5–24. The rod has a diameter of 25 mm and a weight of 10 lb/ft. Determine thethe maximum torsional stress in the 150 N/m. Determine maximum torsional stress in rod the at a at section located at Aatdue to the weight. rod a section located A due to rod’s the rod’s weight.

1.35 m 4.5 ft B

A

0.45 1.5 ftm

0.45 m 1.5 ft

1.2 4 ftm

Solution Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;

TAA –-150(1.2)(0.6) 10(4)(2) = 0= 0

12in # ft· m 108 TATA==80 lb N a

0.6 m 150(1.2) N

The polar moment of inertia of the cross section at A is

= J

π = (0.01254 ) 38.35(10 −9 ) m 4 . Thus 2 tmax =

TA c 108(0.0125) 6 = 35.20(10 = ) N−m 2 35.2 MPa = J 38.35(10 −9 )

Ans.

Ans: tmax  35.2 MPa 329

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5–25. The rod has a diameter of 25 mm and a weight of 15 lb/ft. Determine thethe maximum torsional stress in the 225 N/m. Determine maximum torsional stress in rod the at a at section located at Bat due to the weight. rod a section located B due to rod’s the rod’s weight.

4.5 ft 1.35 m B

A

0.45 1.5 ftm

1.5 ft 0.45 m

4 ftm 1.2

Solution Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;

TBB –-225(1.2)(0.6) 162lbN# ft · ma 15(4)(2) = 0= 0 TBT= 120 B=

12 in

The polar moment of inertia of the cross-section at B is

= J

π = (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2 tmax =

162(0.0125) TB c 6 52.80(10 = ) N−m 2 52.8 MPa = = J 38.35(10 −9 )

Ans.

TB = 162 N · m

0.6 m

Tmax = 52.8 MPa

12.5 mm

225(1.2) N

Ans: tmax  52.8 MPa 330

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  The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F, which delivers 12 kW of power to the shaft while it is turning at 50 rev>s. If gears A, B, and C remove 3 kW, 4 kW, and 5 kW respectively, determine the maximum shear stress in the shaft within regions CF and BC. The shaft is free to turn in its support bearings D and E.

3 kW 4 kW D

A

5 kW 25 mm

B C

E

12 kW

F

Solution v = 50

rev 2p rad c d = 100 p rad>s s rev

TF =

12(103) P = = 38.20 N # m v 100 p

TA =

3(103) P = = 9.549 N # m v 100 p

TB =

4(103) P = = 12.73 N # m v 100 p

(tmax)CF =

38.20(0.0125) TCF c = p = 12.5 MPa 4 J 2 (0.0125 )

Ans.

(tmax)BC =

22.282(0.0125) TBC c = p = 7.26 MPa 4 J 2 (0.0125 )

Ans.

Ans: (tmax)CF = 12.5 MPa, (tmax)BC = 7.26 MPa 331

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5–27. The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F, which delivers 12 kW of power to the shaft while it is turning at 50 rev>s. If gears A, B, and C remove 3 kW, 4 kW, and 5 kW respectively, determine the absolute maximum shear stress in the shaft.

3 kW 4 kW D

A

5 kW 25 mm

B C

E

12 kW

F

Solution v = 50

rev 2p rad c d = 100 p rad>s s rev

TF =

12(103) P = = 38.20 N # m v 100p

TA =

3(103) P = = 9.549 N # m v 100p

TB =

4(103) P = = 12.73 N # m v 100p

From the torque diagram, Tmax = 38.2 N # m tabs = max

38.2(0.0125) Tc = 12.5 MPa = p 4 J 2 (0.0125 )

Ans.

Ans: tabs max

332

= 12.5 MPa

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**5–28. The drive shaft AB of an automobile is made of a having an an allowable allowableshear shearstress stressofoftallow tallow==856 MPa. If steel having If the ksi. the outer diameter ofshaft the shaft thedelivers engine outer diameter of the is 2.5isin.62.5 andmm the and engine delivers kW to the shaft is turning 1140rev>min, rev/min, 200 hp 165 to the shaft whenwhen it is itturning atat1140 determine the minimum minimum required required thickness thicknessof of the shaft’s wall. determine

B

A

Solution πev  2π πad  1 min   ω =  1140    = 38π πad−s min  1 πev  60 s   P = Tω 165(10 3 ) = T(38π ) = T 1.382(10 3 ) N ⋅ m tallow =

Tc J

56(106 ) =

[1.382(10 3 )](0.03125) π

2

(0.031254 − ri4 )

ri = 0.02608 m t = r0 − ri = 0.03125 − 0.02608 = 0.005170 m = 5.17 mm

Ans.

Ans. t = 5.17 mm 333

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5–29. The drive shaft AB of an automobile is to be thin-walled tube. tube.The Theengine enginedelivers delivers125 150kW hp designed as aa thin-walled when the shaft is turning at 1500 rev>min. rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is in.mm. The The material has an allowable shear stress is 2.5 62.5 material has an allowable shear of tallow 7 ksi. stress of = tallow = 50 MPa.

B

A

Solution  

ω =  1500

πev  2π πad  1 min    = 50π πad−s  min  1 πev  60 s 

P = Tω 125(10 3 ) = T(50π ) = T 795.77 N ⋅ m tallow =

Tc J

50(106 ) =

795.77(0.03125) π (0.031254 2

− ri4 )

ri = 0.02825 m t = r0 − ri = 0.03125 − 0.02825 = 0.002998 m = 3.00 mm

Ans.

Ans. t  3.00 mm 334

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5–30. A ship has a propeller drive shaft that is turning at 1500 rev>min rev/min while developing 1800 1500 hp. kW.IfIfit itis is8 ft 2.4long m long and and a diameter ofdetermine 100 mm, the determine theshear maximum has ahas diameter of 4 in., maximum stress shear in the shaft caused by torsion. in the stress shaft caused by torsion.

Solution Internal Torque:

 

ω =  1500

πev  2π πad  1 min     = 50.0π πad−s min  1 πev  60 s 

550 ft # lb>s P ==1500 1800kW hp a= 1500(103) N b ·=m/s 990 000 ft # lb>s 1 hp

T=

P 1500(10 3 ) = = 9549.30 N ⋅ m ω 50.0π

Maximum Shear Stress: Applying torsion formula tmax =

9549.30(0.05) Tc 3 = 48.63(10 = ) N−m 2 48.6 MPa = π (0.054 ) J

Ans.

2

Ans. tmax = 48.6 MPa 335

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5–31. The motor A develops a power of 300 W and turns its connected pulley at 90 rev>min. Determine the required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is tallow = 85 MPa.

60 mm 90 rev/min

A B

150 mm

Solution Internal Torque: For shafts A and B vA = 90

rev 2p rad 1 min a b = 3.00p rad>s min rev 60 s

P = 300 W = 300 N # m>s P 300 = = 31.83 N # m vA 3.00p

TA =

vB = vA a

rA 0.06 b = 3.00pa b = 1.20p rad>s rB 0.15

P = 300 W = 300 N # m>s TB =

P 300 = = 79.58 N # m vB 1.20p

Allowable Shear Stress: For shaft A tmax = tallow = 85 A 106 B =

TA c J 31.83 A d2A B

A B

p dA 4 2 2

dA = 0.01240 m = 12.4 mm

Ans.

For shaft B tmax = tallow = 85 A 106 B =

TB c J 79.58 A d2B B

A B

p dB 4 2 2

dB = 0.01683 m = 16.8 mm

Ans.

Ans. dA = 12.4 mm, dB = 16.8 mm 336

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*5–32. When drilling a well at constant angular velocity, the bottom end of the drill pipe encounters a torsional resistance TA. Also, soil along the sides of the pipe creates a distributed frictional torque along its length, varying uniformly from zero at the surface B to tA at A. Determine the minimum torque TB that must be supplied by the drive unit to overcome the resisting torques, and calculate the maximum shear stress in the pipe. The pipe has an outer radius ro and an inner radius ri.

TB B

L

Solution TA +

1 t L - TB = 0 2 A

TB =

2TA + t AL 2

tA A

TA

Ans.

Maximum shear stress: The maximum torque is within the region above the distributed torque. tmax =

Tc J

tmax =

c

(2TA + t AL) 2 p 4 2 (r o

-

d (ro)

r 4i )

=

(2TA + t AL)ro

Ans.

p(r 4o - r 4i )

Ans:

2TA + t AL , 2 (2TA + t AL)ro = p(r 4o - r 4i )

TB = tmax

337

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5–33. The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C, which delivers 3 kW of power to the shaft while it is turning at 50 rev>s. If gears A and B remove 1 kW and 2 kW, respectively, determine the maximum shear stress in the shaft within regions AB and BC. The shaft is free to turn in its support bearings D and E.

3 kW

2 kW 25 mm

1 kW A D

B

E

C

Solution TC =

3(103) P = 9.549 N # m = v 50(2p)

TA =

1 T = 3.183 N # m 3 C

(tAB)max =

3.183 (0.0125) TC = p = 1.04 MPa 4 J 2 (0.0125 )

Ans.

(tBC)max =

9.549 (0.0125) TC = 3.11 MPa = p 4 J 2 (0.0125 )

Ans.

Ans: (tAB)max = 1.04 MPa, (tBC)max = 3.11 MPa 338

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5–34. The shaft is subjected to a distributed torque along its length of t = (10x2) N # m>m, where x is in meters. If the maximum stress in the shaft is to remain constant at 80 MPa, determine the required variation of the radius c of the shaft for 0 … x … 3 m.

3m

x

c

t  (10x2) Nm/m

Solution T = t =

L

t dx =

L0

x

10 x2dx =

10 3 x 3

3 (10 Tc 3 )x c ; 80(106) = p 4 J 2 c

c 3 = 26.526(10-9) x3 Ans.

c = (2.98 x) mm

Ans: c = (2.98 x) mm 339

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5–35. The motor delivers 12 kW to the pulley A while turning turning at a constant rate of 1800 rpm. Determine to the 1 nearest of shaft of BC if BC the nearest multiples 5 mm thediameter smallest diameter shaft 8 in. theofsmallest allowable shearshear stressstress for steel is t=allow The does belt = 12The ksi. belt if the allowable is tallow 84 MPa. does noton slip onpulley. the pulley. not slip the

B

C 3 in. 75 mm

A

1.5 37.5in.mm

Solution The angular velocity of shaft BC can be determined using the pulley ratio that is vBC = a

rA 0.0375 rev 2p rad 1 min b vA = a b a 1800 ba ba b = 30p rad>s 0.075 rC min 1 rev 60 s

The power is 550 ft # n>s P ==12 (15 hp) b = 8250 ft # lb>s kW =a12(103) N · m/s 1 hp Thus, T =

P 12(10 8250 3) = = 127.32 N . m v 30p 30p

The polar moment of inertia of the shaft is J = tallow =

Tc ; J

84(106 ) =

pd4 p d 4 a b = . Thus, 2 2 32

127.32(d−2)

π d 4 −32

= d 0.01976 = m 19.76 mm d = 0.7639 in = Use d  20 mm

Ans.

Ans. d  20 mm 240

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*5–36. The gear motor can develop 1.6 kW when it turns at 450 rev/min. ofof 251mm, rev>min.IfIfthe theshaft shafthas hasa diameter a diameter in., determine the maximum shear stress developed in the shaft.

Solution The angular velocity of the shaft is v = ¢ 450

rev 2p rad 1 min ≤ ¢ ≤ ¢ ≤ = 15p rad>s min 1 rev 60 s

and the power is 550 ft # lb>s P ==1.6 2 hp 1100 ft # lb>s kW¢ = 1.6(103) N≤ · =m/s 1 hp Then

T=

P 1.6(10 3 ) = = 33.95 N ⋅ m ω 15π

The polar moment of inertia of the= shaft is J tmax =

π = (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2

33.95(0.0125) Tc 6 = 11.07(10 ) N−m 2 11.1 MPa == J 38.35(10 −9 )

Ans.

Ans. tmax = 11.1 MPa 341

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5–37. The gear motor can develop 2.4 kW when it turns 150150 rev>min. If Ifthe at rev/min. theallowable allowableshear shearstress stressfor for the the shaft shaft is 12MPa, ksi, determine tallow ==84 determine the the smallest smallest diameter diameter of the shaft 1 to the nearest multiples used. of 5be mm that can be used. 8 in. that can

Solution The angular velocity of the shaft is v = a 150

rev 2p rad 1 min ba ba b = 5p rad>s min 1 rev 60 s

and the power is 550 ft # lb>s P == 2.4 (3 kW hp) a= 2.4(103) N b· m/s = 1650 ft # lb>s 1 hp Then T =

P 1650 3) 12 in 2.4(10 # ft)a  152.79 ·m = = (105.04 lb N v 5p 5p

The polar moment of inertia of the shaft is J = tallow =

Tc ; J

84(106 ) =

pd4 p d 4 a b = . Thus, 2 2 32

152.79(d−2)

π d 4 −32 Ans.

= d 0.02100 = m 21.00 mm d = 0.8118 in. = use d  25 mm

Ans. d = 21.00 mm 342

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5–38. The 25-mm-diameter shaft on the motor is made of a material having an allowable shear stress of tallow = 75 MPa. If the motor is operating at its maximum power of 5 kW, determine the minimum allowable rotation of the shaft.

Solution Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.01254 B = 38.3495(10-9) m4. 2 tallow =

Tc ; J

75(106) =

T(0.0125) 38.3495(10-9)

T = 230.10 N # m Internal Loading: T =

P ; v

230.10 =

5(103) v

v = 21.7 rad>s

Ans.

Ans. v = 21.7 rad>s 343

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5–39. The drive shaft of the motor is made of a material having an allowable shear stress of tallow = 75 MPa . If the outer diameter of the tubular shaft is 20 mm and the wall thickness is 2.5 mm, determine the maximum allowable power that can be supplied to the motor when the shaft is operating at an angular velocity of 1500 rev>min.

Solution Internal Loading: The angular velocity of the shaft is v = a 1500

rev 2p rad 1 min ba ba b = 50p rad>s min 1 rev 60 s

We have T =

P P = v 50p

Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.014 - 0.00754 B = 10.7379(10-9) m4. 2

tallow =

Tc ; J

75(106) =

a

P b (0.01) 50p

10.7379(10-9)

P = 12 650.25 W = 12.7 kW

Ans.

Ans. P = 12.7 kW 344

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*5–40. The pump operates using the motor that has a power of 85 W. If the impeller at B is turning at 150 rev>min, determine the maximum shear stress in the 20-mm-diameter transmission shaft at A.

150 rev/min A

B

Solution Internal Torque: v = 150

rev 2p rad 1 min ¢ ≤ = 5.00p rad>s min rev 60 s

P = 85 W = 85 N # m>s T =

P 85 = = 5.411 N # m v 5.00p

Maximum Shear Stress: Applying torsion formula tmax = =

Tc J 5.411 (0.01) p 4 2 (0.01 )

Ans.

= 3.44 MPa

Ans: tmax = 3.44 MPa 345

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5–41. Two wrenches are used to tighten the pipe. If P = 300 N is applied to each wrench, determine the maximum torsional shear stress developed within regions AB and BC. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm. Sketch the shear stress distribution for both cases.

P 250 mm

C B A

Solution

250 mm

Internal Loadings: The internal torque developed in segments AB and BC of the pipe can be determined by writing the moment equation of equilibrium about the x axis by referring to their respective free - body diagrams shown in Figs. a and b. ©Mx = 0; TAB - 300(0.25) = 0

P

TAB = 75 N # m

And ©Mx = 0; TBC - 300(0.25) - 300(0.25) = 0

TBC = 150 N # m

Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254 - 0.014 B = 22.642(10 - 9)m4. 2

A tmax B AB =

75(0.0125) TAB c = 41.4 MPa = J 22.642(10 - 9)

A tAB B r = 0.01 m = A tmax B BC =

Ans.

TAB r 75(0.01) = 33.1 MPa = J 22.642(10 - 9)

150(0.0125) TBC c = 82.8 MPa = J 22.642(10 - 9)

A tBC B r = 0.01 m =

Ans.

TBC r 150(0.01) = 66.2 MPa = J 22.642(10 - 9)

The shear stress distribution along the radial line of segments AB and BC of the pipe is shown in Figs. c and d, respectively.

Ans: (tmax)AB = 41.4 MPa, (tmax)BC = 82.8 MPa 346

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5–42. Two wrenches are used to tighten the pipe. If the pipe is made from a material having an allowable shear stress of tallow = 85 MPa, determine the allowable maximum force P that can be applied to each wrench. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm.

P 250 mm

C B A

Solution

250 mm

Internal Loading: By observation, segment BC of the pipe is critical since it is subjected to a greater internal torque than segment AB. Writing the moment equation of equilibrium about the x axis by referring to the free-body diagram shown in Fig. a, we have ©Mx = 0; TBC - P(0.25) - P(0.25) = 0

P

TBC = 0.5P

Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254 - 0.014 B = 22.642(10 - 9)m4 2 tallow =

TBC c ; J

85(106) =

0.5P(0.0125) 22.642(10 - 9)

P = 307.93N = 308 N

Ans.

Ans. P = 308 N 347

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5–43.  The solid shaft has a linear taper from rA at one end to rB at the other. Derive an equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis.

T rB T B A

x

L

rA

Solution r = rB + =

rBL + (rA - rB)(L - x) rA - rB (L - x) = L L

rA(L - x) + rBx L

tmax =

Tc Tr 2T = p 4 = J pr3 2 r

= pc

2T rA(L - x) + rBx L

d

3

=

2TL3 p[rA(L - x) + rBx]3

Ans.

Ans: tmax =

348

2TL3 p[rA(L - x) + rBx]3

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*5–44. A motor delivers 375 kW to the shaft, which is tubular and has an outer diameter of 50 mm. If it is rotating at 200 radNs, determine its largest inner diameter to the nearest mm if the allowable shear stress for the material is tallow = 175 MPa.

A

B

150 mm

Solution P 375(10 3 ) N ⋅ m−s = T=

P 375(10 3 ) = = 1.875(10 3 ) N ⋅ m ω 200

tmax = tallow =

175(106 ) =

Tc J

1.875(10 3 )(0.025) π

2

( 0.0254 − ci4 )

ci = 0.02166 m = 0.04332 = m 43.32 mm = d 2= ci 2(0.02166) Use di = 43 mm

Ans.

Ans: Use di = 43 mm 349

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  The A-36 steel tubular shaft is 2 m long and has an outer diameter of 50 mm. When it is rotating at 40 rad>s, it transmits 25 kW of power from the motor M to the pump P. Determine the smallest thickness of the tube if the allowable shear stress is tallow = 80 MPa.

P

M

Solution The internal torque in the shaft is 25(103) P = = 625 N # m v 40 p The polar moment of inertia of the shaft is J = (0.0254 - ci 4). Thus, 2       T =

tallow =

Tc ;  80(106) = J

625(0.025) p 4 2 (0.025

- ci 4)

ci = 0.02272 m So that t = 0.025 - 0.02272 Ans.

= 0.002284 m = 2.284 mm = 2.28 mm

Ans: t = 2.28 mm 350

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  The A-36 solid steel shaft is 2 m long and has a diameter of 60 mm. It is required to transmit 60 kW of power from the moor M to the pump P. Determine the smallest angular velocity the shaft if the allowable shear stress is tallow = 80 MPa.

P

M

Solution The polar moment of inertia of the shaft is J = tallow =

Tc ; J

80(106) =

p (0.034) = 0.405(10-6)p m4. Thus, 2

T(0.03) 0.405(10-6)p

T = 3392.92 N # m P = Tv;

60(103) = 3392.92 v Ans.

v = 17.68 rad>s = 17.7 rad>s

Ans: v = 17.7 rad>s 351

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5–47.  The propellers of a ship are connected to an A-36 steel shaft that is 60 m long and has an outer diameter of 340 mm and inner diameter of 260 mm. If the power output is 4.5 MW when the shaft rotates at 20 rad>s, determine the maximum torsional stress in the shaft and its angle of twist.

Solution T =

4.5(106) P = = 225(103) N # m v 20

tmax =

f =

225(103)(0.170) Tc = = 44.3 MPa p J [(0.170)4 - (0.130)4] 2

Ans.

225 1 103 2 (60) TL = = 0.2085 rad = 11.9° p JG [(0.170)4 - (0.130)4)75(109) 2

Ans.

Ans: tmax = 44.3 MPa, f = 11.9° 352

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*5–48. The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain in the shaft is gmax = Tc>JG. What is the shear strain on an element located at point A, c>2 from the center of the shaft? Sketch the shear strain distortion of this element.

T

c/ 2

c

A

T

L

Solution From the geometry: gL = r f;  g = Since f = g =

rf L

TL , then JG

Tr JG

(1)

However the maximum shear strain occurs when r = c gmax =

Tc JG

QED

Shear strain when r = g =

T(c>2) JG

=

c is from Eq. (1), 2

Tc 2 JG

Ans.

Ans: g = 353

Tc 2 JG

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  The A-36 steel shaft has a diameter of 50 mm and is subjected to the distributed and concentrated loadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of the angle of twist of the shaft in radians versus x.

200 Nm/m

A

250 Nm x 0.5 m

C 0.5 m

B

Solution Internal Torque: As shown on FBD. Maximum Shear Stress: The maximum torque occurs at x = 0.5 m where Tmax = 150 + 200(0.5) = 250 N # m. tmax = ABS

250(0.025) Tmax c = 10.2 MPa = p 4 J 2 (0.025 )

Ans.

Angle of Twist: For 0 … x 6 0.5 m L0

L

=

L0

x

=

150x + 100x2 JG

f(x) =

= =

T(x) dx JG

(150 + 200x) dx JG

150x + 100x2

p 2

( 0.0254 ) 75.0 ( 109 )

3 3.26x

+ 2.17x2 4 (10-3) rad

At x = 0.5 m,  f = f C = 0.00217 rad

For 0.5 m 6 x 6 1 m  Since T(x) = 0, then f(x) = f C = 0.00217 rad

Ans: tmax = 10.2 MPa ABS

354

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  The 60-mm-diameter shaft is made of 6061-T6 aluminum having an allowable shear stress of tallow= 80 MPa. Determine the maximum allowable torque T. Also, find the corresponding angle of twist of disk A relative to disk C.

1.20 m A B 2T 3

Solution

T

Internal Loading: The internal torques developed in segments AB and BC of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: Segment AB is critical since it is subjected to a greater p internal torque. The polar moment of inertia of the shaft is J = (0.034) = 2 0.405 ( 10 - 6 ) p m4. We have tallow =

TAB c ; J

80(106) =

1.20 m

C

1 3T

(23 T)(0.03) 0.405(10 - 6)p

T = 5089.38 N # m = 5.09 kN # m

Ans.

Angle of Twist: The internal torques developed in segments AB and BC of the shaft 2 1 are TAB = (5089.38) = 3392.92 N # m and TBC = - (5089.38) = -1696.46 N # m. 3 3 We have f A>C = g f A>C =

TBCLBC TiLi TABLAB = + JiGi JGal JGal

3392.92(1.20) -6

9

0.405(10 )p(26)(10 )

+

- 1696.46(1.20) 0.405(10 - 6)p(26)(109) Ans.

= 0.06154 rad = 3.53°­

Ans: T = 5.09 kN # m, f A>C = 3.53° 355

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5–51. The 60-mm-diameter shaft is made of 6061-T6 aluminum. If the allowable shear stress is tallow = 80 MPa, and the angle of twist of disk A relative to disk C is limited so that it does not exceed 0.06 rad, determine the maximum allowable torque T.

1.20 m A B 2T 3

Solution

T

Internal Loading: The internal torques developed in segments AB and BC of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: Segment AB is critical since it is subjected to a greater p internal torque. The polar moment of inertia of the shaft is J = (0.034) = 2 -6 4 0.405(10 )p m . We have tallow =

TAB c ; J

80(103) =

1.20 m

C

1 3T

(23T)(0.03) 0.405(10 - 6)p

T = 5089.38 N # m = 5.089 kN # m Angle of Twist: It is required that f A>C = 0.06 rad. We have f A>C = Σ 0.06 =

TiLi TBC LBC TAB LAB = + JiGi JGal JGal (23T)(1.2)

0.405(10 - 6)p(26)(109)

+

(- 13T)(1.2) 0.405(10 - 6)p(26)(109)

T = 4962.14 N # m = 4.96 kN # m (controls)

Ans.

Ans: T = 4.96 kN # m 356

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  The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The shaft has a diameter of 40 mm.

1200 Nm

400 Nm B

600 Nm D C

600 mm A

Solution Internal Torque: The torque diagram shown in Fig. a can be plotted. From this diagram, TAC = 200 N # m, TCD = 800 N # m and TDB = 400 N # m.

400 mm

Angle of Twist: f B>A = g

500 mm

200 Nm

TL JG

1 (T L + TCD LCD + TDB LDB) JG AC AC 1 = [200(0.4) + 800(0.5) + ( - 400)(0.6)] JG 240 N # m2 = JG =

For A992 steel, G = 75 GPa. Then f B>A =

240 p 2

( 0.02 ) 3 75 ( 109 ) 4 4

= (0.01273 rad) a

180° b = 0.730° G p rad

Ans.

Ans: f B>A = 0.730° G 357

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5–53. The hydrofoil boat has an A-36 steel propeller shaft shaftisthat is long. 100 ftItlong. It is connected to an diesel in-lineengine diesel that 30 m is connected to an in-line engine that delivers a maximum power 2500 hp and that delivers a maximum power of 2000 kWofand causes the causes the shaft to rpm. rotate at outer 1700 diameter rpm. If ofthe shaft to rotate at 1700 If the theouter shaft diameter thethe shaft is 8thickness in. and the wall thickness is 38 the in., is 200 mmofand wall is 10 mm, determine determine shear the maximum shear in stress developed in the maximum stress developed the shaft. Also, what is shaft. Also, what is the of “wind orshaft angleatof twist in the the “wind up,” or angle twistup,” in the full power? shaft at full power?

30 mft 100

Solution Internal Torque:

 

ω =  1700

πev  2π πad  1 min     = 56.667π πad−s min  1 πev  60 s 

= = P 2000 kW 2000(10 3 ) N ⋅ m−s T=

P 2000(10 3 ) = = 11.23(10 3 ) N ⋅ m ω 56.667π

Maximum Shear Stress: Applying torsion Formula. tmax =

Tc J

11.23(10 3 )(0.1) 6 = 20.80(10 ) N−m 2 20.8 MPa == π (0.14 − 0.09 4 ) 2

Ans.

Angle of Twist: f =

11.23(10 3 )(30) TL = π JG [ 2 (0.14 − 0.09 4 )][75(109 )] = 0.08319 πad = 4.77°

Ans.

358

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5–54. The turbine develops 300 kW of power, which is transmitted to the gears such that both B and C receive an equal amount. If the rotation of the 100-mm-diameter A992 steel shaft is v = 600 rev>min., determine the absolute maximum shear stress in the shaft and the rotation of end D of the shaft relative to A. The journal bearing at D allows the shaft to turn freely about its axis.

A B

v

C

1.5 m

D

Solution

2m 1m

External Applied Torque: Gears B and C withdraw equal amount of power, 300 PB = PC = P = = 150 kw. Here, the angular velocity of the shaft is 2 v = a600

rev 2p rad 1 min ba ba b = 20p rad>s. Then, the torque exerted on gears min 1 rev 60 s

B and C can be determined from TB = TC

150 ( 103 ) p 7500 = = N#m p v 20p

Using this result, the torque diagram shown in Fig. a can be plotted. Maximum Shear Stress: From the torque diagram, we notice that the maximum 15000 torque Tmax = N # m occurs at region BA. Thus it is the critical region where p the absolute maximim shear stress occurs. Applying the torsion formula,   t abs = max

Tmax C = J

15000 p p 2

(0.05)

( 0.054 )

= 24.32 ( 106 ) Pa = 24.3 MPa

Angle of twist: From the torque diagram, TDC = 0, TCB = 15000 TBA = N#m p f D>A = g = =

Ans. 7500 N # m and p

TL 1 = (T L + TCB LCB + TBA LBA) JG JG DC DC

1 7500 15000 c0 + a b(2) + a b(1.5)d p p JG 37500 N # m2 p JG

For A992 steel, G = 75 GPa. Thus f D>A =

37500 p p c ( 0.054 ) d [75 ( 109 ) ] 2

= (0.016211 rad) a

= 0.9288°

180° b p rad

Ans.

= 0.929° Ans: t abs = 24.3 MPa max

f D>A = 0.929° 359

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5–55. The shaft is made of A992 steel. It has a diameter of  25 mm and is supported by bearings at A and D, which allow free rotation. Determine the angle of twist of B with respect to D.

A

B

90 Nm C

0.6 m 90 Nm

0.75 m

D 0.9 m

Solution The internal torques developed in segments BC and CD are shown in Figs. a and b. The polar moment of inertia of the= shaft is J

π

= (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2

TiLi TBC LBC TCD LCD = + FB>D = a JiGi J Gst J Gst

=

90(0.75) [38.35(10 −9 )]75(106 )

+0

= 0.02347 πad = 1.34°

Ans.

TCD = 0

(b)

TBC = 90 N ⭈ m

90 N ⭈ m

(a)

Ans: fB>D = 1.34° 360

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5–56. The shaft is made of A-36 steel. It has a diameter of 25 mm and supportedbybybearings bearingsatatAAand andD, D, which allow 1 in. and is is supported free rotation. Determine the angle of twist of gear C with respect to B.

A

B

90 60 Nm lbft C

0.6 2 ftm 90 lbft Nm 60

0.75 m 2.5 ft

D 0.9 3 ftm

Solution The internal torque developed in segment BC is shown in Fig. a The polar moment of inertia of the= shaft is J fC>B =

π

= (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2

TBC LBC 90(0.75) = J Gst [38.35(10 −9 )]75(106 )

= 0.02347 πad = 1.34°

Ans.

TBC = 90 N ⭈ m

90 N ⭈ m

(a)

Ans: fC>B = 1.34° 361

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5–57. The rotating flywheel-and-shaft, when brought to a sudden stop at D, begins to oscillate clockwise-counter-clockwise such that a point A on the outer edge of the fly-wheel is displaced through a 6-mm arc. Determine the maximum shear stress developed in the tubular A-36 steel shaft due to this oscillation. The shaft has an inner diameter of 24 mm and an outer diameter of 32 mm. The bearings at B and C allow the shaft to rotate freely, whereas the support at D holds the shaft fixed.

D 1.5 m

C

B A

Solution

75 mm

s = ru 6 = 75 f  f = 0.08 rad f = 0.08 =

TL JG T(1.5) J(75) ( 109 )

T = 4 ( 109 ) J tmax = =

TC J 4(109)(J)(0.016) J Ans.

= 64.0 MPa

Ans: tmax = 64.0 MPa 362

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5–58. The A992 steel shaft has a diameter of 50 mm and is subjected to the distributed loadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of the angle of twist of the shaft in radians versus x.

500 Nm

A

1000 Nm/m x 0.5 m

C 0.5 m

B

Solution Internal Torque: Referring to the FBD of the right segment of the shaft shown in Fig. a (0.5 m < x … 1 m) T(x) = {1000 - 1000x} N # m

ΣMx = 0;  T(x) - 1000 (1- x) = 0 And Fig. b (0 … x 6 0.5 m)

ΣMx = 0;  T(x) + 500 (0.5-x) - 1000 (0.5) = 0  TAc = {500x + 250} N # m

Maximum Shear Stress: The maximum torque at x = 0.5 m, Tmax = 1000 - 1000(0.5) = 500 N # m. Applying the torsion formula, abs = tmax

500 (0.025) TmaxC = p = 20.37 ( 106 ) Pa = 20.4 MPa 4 J 2 (0.025 )

where

Ans.

Angle of Twist: For A992 steel G = 75 GPa. For region 0 … x 6 0.5 m (between B and C).       f(x) = =

L

T(x)dx JG

=

3 ( 0.025 )4 3 75 ( 10 )4 Lo 1

4

p 2

5 0.005432 ( x2

9

x (500 x + 250) dx

+ x )6 rad

Ans.

At x = 0.5 m,  f = f C = 0.004074 rad. For region 0.5 m 6 x … 1 m       f (x) = f C +

L

T(x) dx

= 0.004074 +

JG x

3 ( 0.0254 ) 3 75 ( 109 ) 4 L0.5 m 1

p 2

(1000 - 1000x) dx

= { - 0.01086x2 + 0.02173 x - 0.004074} rad The maximum f occurs at where

Ans.

df = 0. dx

df = - 0.02173x + 0.02173 = 0 dx                x = 1m Thus, f max = f B = - 0.01086 ( 12 ) + 0.02173(1) - 0.004074 = 0.006791 rad The plot of f vs x is as shown in Fig. c.

Ans: abs = 20.4 MPa tmax For 0 … x 6 0.5 m, f(x) = 5 0.005432 ( x2 + x )6 rad For 0.5 m 6 x … 1 m, f(x) = { -0.01086x2 + 0.02173 x - 0.004074} rad 363

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  The shaft is made of A992 steel with the allowable shear stress of tallow = 75 MPa. If gear B supplies 15 kW of power, while gears A, C and D withdraw 6 kW, 4 kW and 5  kW, respectively, determine the required minimum diameter d of the shaft to the nearest millimeter. Also, find the corresponding angle of twist of gear A relative to gear D. The shaft is rotating at 600 rpm.

A B C

600 mm

D

600 mm 600 mm

Solution Internal Loading: The angular velocity of the shaft is v = a600

rev 1 min 2p rad ba ba b = 20p rad>s min 60 s 1 rev

Thus, the torque exerted on gears A, C, and D are TA =

6(103) PA = = 95.49 N # m v 20p

TC =

4(103) PC = = 63.66 N # m v 20p

TD =

5(103) PD = = 79.58 N # m v 20p

The internal torque developed in segments AB, CD, and BC of the shaft are shown in Figs. a, b, and c, respectively. Allowable Shear Stress: Segment BC of the shaft is critical since it is subjected to a greater internal torque.

tallow

TBC c = ;  75(106) = J

d 143.24 a b 2 p d 4 a b 2 2

d = 0.02135 m = 21.35 mm Ans. p Angle of Twist: The polar moment of inertia of the shaft is J = (0.0114) = 2 7.3205(10 - 9)p m4. We have Use

d = 22 mm

f A>D = Σ f A>D =

TiLi TBCLBC TCDLCD TABLAB = + + JiGi JGst JGst JGst

0.6 ( -95.49 + 143.24 + 79.58) 7.3205(10 )p(75)(109) -9

Ans.

= 0.04429 rad = 2.54°

Ans: Use d = 22 mm, f A>D = 2.54° F 364

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  Gear B supplies 15 kW of power, while gears A, C and D withdraw 6 kW, 4 kW and 5 kW, respectively. If the shaft is made of steel with the allowable shear stress of tallow = 75 MPa, and the relative angle of twist between any two gears cannot exceed 0.05 rad, determine the required minimum diameter d of the shaft to the nearest millimeter. The shaft is rotating at 600 rpm.

A B C

600 mm

D

600 mm 600 mm

Solution Internal Loading: The angular velocity of the shaft is v = a600

rev 1 min 2p rad ba ba b = 20p rad>s min 60 s 1 rev

Thus, the torque exerted on gears A, C, and D are TA =

6(103) PA = = 95.49 N # m v 20p

TC =

4(103) PC = = 63.66 N # m v 20p

TD =

5(103) PD = = 79.58 N # m v 20p

The internal torque developed in segments AB, CD, and BC of the shaft are shown in Figs. a, b, and c, respectively. Allowable Shear Stress: Segment BC of the shaft is critical since it is subjected to a greater internal torque.

tallow

TBC c = ; J

75(106) =

d 143.24 a b 2 p d 4 a b 2 2

d = 0.02135 m = 21.35 mm Angle of Twist: By observation, the relative angle of twist of gear D with respect to gear B is the greatest. Thus, the requirement is f D>B = 0.05 rad. ­­f D>B = Σ

TiLi TBC LBC TCD LCD = + = 0.05 JiGi JGst JGst

0.6 p d 4 9 2 ( 2 ) (75)(10 )

(143.24 + 79.58) = 0.05

d = 0.02455 m = 24.55 mm = 25 mm (controls!) Use

Ans.

d = 25 mm

Ans: Use d = 25 mm 365

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  The turbine develops 150 kW of power, which is transmitted to the gears such that C receives 70% and D receives 30%. If the rotation of the 100-mm-diameter A-36 steel shaft is v = 800 rev>min., determine the absolute maximum shear stress in the shaft and the angle of twist of end E of the shaft relative to B. The journal bearing at E allows the shaft to turn freely about its axis.

B

v

C

D

3m

E

4m

Solution

2m

rev 1 min 2p rad P = Tv;  150(10 ) W = T a800 ba ba b min 60 sec 1 rev 3

T = 1790.493 N # m

TC = 1790.493(0.7) = 1253.345 N # m TD = 1790.493(0.3) = 537.148 N # m Maximum torque is in region BC.    tmax =

1790.493(0.05) TC = = 9.12 MPa p 4 J 2 (0.05)

   f E>B = Σa =

Ans.

TL 1 b = [1790.493(3) + 537.148(4) + 0] JG JG 7520.171

p 4 9 2 (0.05) (75)(10 )

Ans.

= 0.0102 rad = 0.585°

Ans: tmax = 9.12 MPa, f E>B = 0.585° 366

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5–62. The turbine develops 150 kW of power, which is transmitted to the gears such that both C and D receive an equal amount. If the rotation of the 100-mm-diameter A-36 steel shaft is v = 500 rev>min., determine the absolute maximum shear stress in the shaft and the rotation of end B of the shaft relative to E. The journal bearing at E allows the shaft to turn freely about its axis.

B

v

C

D

3m

E

4m

Solution

2m

rev 1 min 2p rad P = Tv;  150(10 ) W = T a500 ba ba b min 60 sec 1 rev 3

T = 2864.789 N # m

TC = TD =

T = 1432.394 N # m 2

Maximum torque is in region BC. tmax =

2864.789(0.05) TC = = 14.6 MPa p 4 J 2 (0.05)

f B>E = Σa =

Ans.

TL 1 b = [2864.789(3) + 1432.394(4) + 0] JG JG 14323.945

p 4 9 2 (0.05) (75)(10 )

Ans.

= 0.0195 rad = 1.11°

Ans: tmax = 14.6 MPa, f B>E = 1.11° 367

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5–63. The 50-mm-diameter A992 steel shaft is subjected to the torques shown. Determine the angle of twist of the end A. 400 Nm B

200 Nm C

300 mm 600 mm

600 mm

Solution

800 Nm D A

Internal Torque: The torque diagram shown in Fig. a can be plotted. From this diagram, TAD = -800 N # m, TDC = - 1000 N # m and TCB = -600 N # m. Angle of Twist: f B>A = g

TL JG

=

1 (T L + TDC LDC + TCB LCB) JG AD AD

=

1 [( - 800)(0.6) + ( -1000)(0.6) + ( - 600)(0.3)] JG

= -

1260 N # m2 JG

For A992 steel, G = 75 GPa. Then fA =

1260 p 2

( 0.0254 ) 3 75 ( 109 ) 4

= ( -0.02738 rad) a

180° b = - 1.569° = 1.57° F p rad

Ans.

Ans: f A = 1.57° F 368

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*5–64. The 60-mm-diameter solid shaft is made of 2014-T6 aluminum and is subjected to the distributed and concentrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft.

1.5 kNm 2 kNm/m

B 0.4 m 0.6 m

A

Solution Internal Torque: Referring to the FBD of the right segments of the shaft shown in Fig. a and b gMx = 0;  TAC + 2000x = 0  TAC = ( - 2000x) N # m

And

gMx = 0;  TCB + 2000 (0.6) - 1500 = 0  TCB = 300 N # m

Angle of Twist: For 2014 - T6 Aluminum, G = 27 GPa.    f A = g

x

1 TL c T dx + TCB LCB d = JG JG L0 AC =

1

3

p 4 2 (0.03)

4 3 27 ( 10 ) 4

= ( - 0.006986 rad)a

9

c

L0

180° b p rad

0.6 m

( - 2000 x)dx + 300 (0.4) d

Ans .

= - 0.400° = 0.400° F

Ans: f A = 0.400° F 369

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5–65. The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end A when the torques are applied to the assembly as shown.

D

10 in. 250 mm

C

80 lbft 120 Nm A

30 in. 750 mm

40 60 lbft Nm

8 in. 200 mm 10 in. 250 mm

Solution

4 in. 100 mm 12 in. 300 mm

6 in.mm 150 B

Internal Torque: As shown on FBD. Angle of Twist: TL fE = a JG

60 F = 0.100 = 600 N

1 {−90(0.75) + 30(0.25)} = π  4 [ 2 (0.0125 )][75(109 )]

0.100 m 60 N · m

TAF = –60 N · m

- 0.01778 = rad = 0.01778 == –0.020861 0.020861 rad rad

 150   150  φF = = = 0.031291 πad  φE   (0.020861)  100   100  fA>F =

0.150 m

60 N · m

TCA = –600(0.150) = –90 N · m F = 600 N

TGF LGF JG

0.150 m TBA = 30 N · m

−60(0.25)  = π [ 2 (0.01254 )][75(109 )]

120 N · m

F = 600 N

 rad  rad = –0.0052152 - 0.004445 rad = 0.0052152 0.004445 rad

fF fE

fA = fF + fA>F

fA>F

 0.0052152 = 0.031291 0.02667 ++ 0.004445 = 0.036506 rad = 2.09°

Ans.

Ans. fA = 2.09° 370

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5–66. The A-36 steel bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t = (kx2) N # m>m, where x is in meters. If a torque of T = 50 N # m is applied to the bolt head, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume the shank has a constant radius of 4 mm. B

t

Solution

T = 50 Nm 50 mm

dT = t dx

T =

L0

x

A

0.05

0.05 m

x3 kx dx = k ` 3

= 41.667(10 - 6)k­­­­­­

2

0

50 - 41.6667(10 - 6) k = 0 k = 1.20(106) N>m2

In the general position, T =

f =

L

T(x)dx JG

=

1 JG L0

1.875 = JG

L0

x

1.20(106)x2dx = 0.4(106)x3

0.05 m

0.4(106)x4 1 = c 50x d JG 4 =

Ans.

[50 - 0.4(106)x3]dx 0.05 m

`

0

1.875 p 4 9 2 (0.004 )(75)(10 )

Ans.

= 0.06217 rad = 3.56°

Ans: k = 1.20(106) N>m2, f = 3.56° 371

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5–67. The A-36 steel bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t = (kx2>3) N # m>m, where x is in meters. If a torque of T = 50 N # m is applied to the bolt head, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume the shank has a constant radius of 4 mm. B

t

Solution

T = 50 Nm 50 mm

dT = t dx T =

L0

0.05

x

A

2 3 5 0.05 kx3 dx = k x3 ` = (4.0716)(10 - 3) k 5 0

50 - 4.0716 (10 - 3) k = 0

2

k = 12.3(103) N>m(3)

Ans.

In the general position, T =

L0

x

5

2

12.28(103)x3 dx = 7.368(103) x3

Angle of twist: f = = =

L

T(x) dx JG

=

1 JG L0

0.05 m

8

[50 - 7.3681(103)x3]dx

3 8 0.05 m 1 c 50x - 7.3681(103)a bx3 d  JG 8 0 1.5625

p 4 9 2 (0.004 )(75)(10 )

Ans.

= 0.0518 rad = 2.97°

Ans: k = 12.3(103) N>m2>3, f = 2.97° 372

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*5–68. The shaft of radius c is subjected to a distributed torque t, measured as torque>length of shaft. Determine the angle of twist at end A. The shear modulus is G.

B t0

x t = t0 1 + — L

( ( ) 2)

x L

Solution TB TB =

L

L

 f =

L

2 t0

t dx = 0

t dx = t 0

= t0 c x +

A

L

3

a1 +

x2 b dx L2

L x L 4 d ` = t 0 aL + b = t 0 L 2 3 3 3L 0

T (x) dx JG L

1 4 x3 c t 0 L - t 0 ax + b d dx JG L0 3 3L2

= =

L t0 4 7 t 0 L2 x2 x4 c Lx - a + bd ` = 2 JG 3 2 12 JG 12L 0

However J =

f =

p 4 c, 2

7 t 0 L2

Ans.

6 p c4 G

Ans: f = 373

7 t 0 L2 6 p c4 G

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5–69. The tubular drive shaft for the propeller of a hovercraft is 6 m long. If the motor delivers 4 MW of power to the shaft when the propellers rotate at 25 rad>s, determine the required inner diameter of the shaft if the outer diameter is 250 mm. What is the angle of twist of the shaft when it is operating? Take tallow = 90 MPa and G = 75 GPa.

6m

Solution Internal Torque: P = 4 ( 106 ) W = 4 ( 106 ) N # m>s T =

4 ( 106 ) P = = 160(103) N # m v 25

Maximum Shear Stress: Applying torsion formula. tmax = tallow =

Tc J 160 ( 103 ) ( 0.125 )

90 ( 106 ) = p 2

c 0.1254 - a

dt 4 b d 2

Ans.

d t = 0.2013 m = 201 mm    Angle of Twist: f =

TL = JG

160 ( 103 ) ( 6 ) p 2

( 0.1254 - 0.100654 ) 75 ( 109 ) Ans.

      = 0.0576 rad = 3.30°

Ans: d t = 201 mm, f = 3.30° 374

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5–70. The A-36 steel assembly consists of a tube having an outer outerradius radiusofof1 in. 25 and mma and wall thickness mm. wall athickness of 0.125of in.3Using Using rigidat plate it is connected the 1-in-diameter solid 25-mma rigidaplate B, itatisB, connected to theto solid diameter AB. Determine the rotation of the end tube’s shaft AB.shaft Determine the rotation of the tube’s C end if a C if a torque appliedtotothe thetube tubeat atthis this end. The torque of 200oflb25# N in.· m is is applied end A of the shaft is fixed supported.

B C 25 Nm 200 lbin.

100 mm 4 in.

A 150 mm 6 in.

Solution fB =

TABL 25(0.25) = = 0.0021730 πad  JG [ π2 (0.01254 )][75(109 )]

fC>B =

25 N · m TAB = 25 N · m

TCBL −25(0.1) =  JG [ π2 (0.0254 − 0.022 4 )][75(109 )]

TCB = −25 N · m 25 N · m

0.0001357 πad G = −0.0001357 πad = (+ G) fC = fB + fC>B = 0.0021730 + 0.0001357  rad== 0.113° 0.132° = 0.0023087 0.001964 rad

Ans.

Ans. fC = 0.113° 375

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5–71. The A-36 hollow steel shaft is 2 m long and has an outer diameter of 40 mm. When it is rotating at 80 rad>s, it transmits 32 kW of power from the engine E to the generator G. Determine the smallest thickness of the shaft if the allowable shear stress is tallow = 140 MPa and the shaft is restricted not to twist more than 0.05 rad.

E

G

Solution P = Tv 32(103) = T(80) T = 400 N # m Shear stress failure t =

Tc J

tallow = 140 (106) =

400 (0.02) p 4 2 (0.02

- ri 4 )

ri = 0.01875 m Angle of twist limitation: f = 0.05 =

TL JG 400(2) p 4 2 (0.02

- r i 4)(75)(109)

  ri = 0.01247 m

(controls)

     t = ro - ri = 0.02 - 0.01247 = 0.00753 m Ans.

= 7.53 mm

Ans: t = 7.53 mm 376

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*5–72. The A-36 solid steel shaft is 3 m long and has a diameter of 50 mm. It is required to transmit 35 kW of power from the engine E to the generator G. Determine the smallest angular velocity of the shaft if it is restricted not to twist more than 1°.

E

G

Solution f = 1°(p) 180°

=

TL JG T(3) p 4 9 2 (0.025 )(75)(10 )

T = 267.73 N # m P = Tv

35(103) = 267.73(v) Ans.

v = 131 rad>s

Ans: v = 131 rad>s 377

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5–73. The motor produces a torque of T = 20 N # m on gear A. If gear C is suddenly locked so it does not turn, yet B can freely turn, determine the angle of twist of F with respect to E and F with respect to D of the L2-steel shaft, which has an inner diameter of 30 mm and an outer diameter of 50 mm. Also, calculate the absolute maximum shear stress in the shaft. The shaft is supported on journal bearings at G at H.

A 100 mm 30 mm B

G

60 mm D

0.2 m

Solution

E C

0.8 m

H 40 mm

0.4 m

F

0.2 m

F (0.03) = 20 F = 666.67 N

T′ = (666.67) (0.1) = 66.67 N # m Since shaft is held fixed at C, the torque is only in region EF of the shaft. f F>E =

66.67(0.6) TL = = 0.999 ( 10 ) - 3 rad p JG 3 ( 0.025 ) 4 - ( 0.015 ) 4 4 75 ( 109 ) 2

Ans.

Since the torque in region ED is zero, f F>D = 0.999 (10) - 3 rad tmax =

Ans.

66.67(0.025) Tc = p J ( ( 0.025 ) 4 - ( 0.015 ) 4 ) 2 Ans.

= 3.12 MPa

Ans: f F>E = 0.999(10) -3 rad, f F>D = 0.999(10) -3 rad, tmax = 3.12 MPa 378

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5–74. The shaft has a radius c and is subjected to a torque per unit length of t0, which is distributed uniformly over the shaft’s entire length L. If it is fixed at its far end A, determine the angle of twist f of end B. The shear modulus is G.

A

f L

c B

t0

Solution f = 3 = =

T (x) dx JG

- t0 L x dx JG 30

- t 0 x2 L - t 0 L2 c d ` = JG 2 JG 2 0 - t 0 L2 2JG

However,  J = f =

=

- t 0 L2 4

pc G

=

p 4 c 2 t 0 L2

Ans.

p c4 G

Ans: f = 379

t 0L2 pc 4G

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5–75. The 60-mm-diameter solid shaft is made of A-36 steel and is subjected to the distributed and concentrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft due to these loadings.

2 kNm/m 600 N · m 400 Nm

B 0.8 m 0.6 m

A

Solution Internal Torque: As shown on FBD. Angle of Twist: fA = Σ =

TL JG - 400 (0.6)

p 2

( 0.03 ) 75.0 ( 10 ) 4

9

+ 3

0.8 m

0

(200 - 2000x)dx p 2

( 0.034 ) 75.0 ( 109 ) Ans.

= - 0.007545 rad = 0.432° F

Ans: f A = 0.432° F 380

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*5–76. The contour of the surface of the shaft is defined by the equation y = e ax, where a is a constant. If the shaft is subjected to a torque T at its ends, determine the angle of twist of end A with respect to end B. The shear modulus is G.

B

x

y = e ax T

y A L

T

Solution f =

T dx J(x)G L

where, J(x) =

p ax 4 (e ) 2

L

= = =

L 2T dx 2T 1 = ab ` pG 4 a e 4ax 0 pG L0 e 4ax

2T 1 1 T e 4aL - 1 a+ b = a b pG 4 a e 4aL 4a 2apG e 4aL T (1 - e - 4aL) 2apG

Ans.

Ans: f = 381

T (1 - e - 4aL) 2apG

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5–77. The steel shaft is made from two segments: AC has a diameter of 0.5 12 mm, andCB CBhas hasa adiameter diameterofof1 25 in, and in. mm. If it If is it is fixed at ends its ends and andsubjected subjectedtotoaa torque torque of fixed at its A A and B Band 300 N · m, determine the maximum shear stress in the shaft. Gst == 75 GPa. 32 ksi. 10.8110

A

0.5mm in. 12 C D 300 N  m

5 in. 125 mm

in. 251 mm

8 in. 200 mm

B 12 mm in. 300

Solution Equilibrium: TA + TB - 300 = 0

(1)

Compatibility condition: fD>A = fD>B

TA (0.125) π [ 2 (0.006 4 )]G

300 N · m

+

TA (0.2) π [ 2 (0.01254 )]G

TB (0.3) = π [ 2 (0.01254 )]G

TA = 0.11743TB

(2)

Solving Eqs. (1) and (2) yields

TA = 31.53 N ⋅ m

TB = 268.47 N ⋅ m

(τ AC )max =

TC 31.53(0.06) 2 = = 92.92(106 ) N−m= 92.9 MPa (Max) π (0.006 4 ) J 2

(τ DB )max =

TC 268.47(0.0125) 6 87.51(10 = ) N−m 2 87.5 MPa == π (0.01254 ) J 2

Ans.

Ans: (τ AC )max  92.9 MPa 382

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5–78. The steel shaft has a diameter of 40 mm and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB of the shaft. Gst  =  75 GPa.

3 kN

A 3 kN C 400 mm

50 mm 50 mm

B

600 mm

Solution Equilibrium: (1)

TA + TB - 3000(0.1) = 0 Compatibility condition: f C>A = f C>B TA (400) JG

=

TB(600) JG (2)

TA = 1.5 TB Solving Eqs (1) and (2) yields: TB = 120 N # m

TA = 180 N # m (tAC)max =

180(0.02) Tc = p = 14.3 MPa 4 J 2 (0.02 )

Ans.

(tCB)max =

120(0.02) Tc = p = 9.55 MPa 4 J 2 (0.02 )

Ans.

Ans: (tAC)max = 14.3 MPa, (tCB)max = 9.55 MPa 383

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5–79. The A992 steel shaft has a diameter of 60 mm and is fixed at its ends A and B. If it is subjected to the torques shown, determine the absolute maximum shear stress in the shaft.

200 Nm B 500 Nm

D

1m

1.5 m C A

1m

Solution Referring to the FBD of the shaft shown in Fig. a, ΣMx = 0; TA + TB - 500 - 200 = 0

(1)

Using the method of superposition, Fig. b f A = (f A)TA - (f A)T 0 =

TA (3.5) JG

- c

500 (1.5) JG

+

700 (1) JG

TA = 414.29 N # m

d

Substitute this result into Eq (1), TB = 285.71 N # m Referring to the torque diagram shown in Fig. c, segment AC is subjected to maximum internal torque. Thus, the absolute maximum shear stress occurs here. tabs = max

414.29 (0.03) TAC c = = 9.77 MPa p J 4 (0.03) 2

Ans.

Ans: tabs = 9.77 MPa max 384

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*5–80. The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the torque, determine the maximum shear stress in regions AC and CB.

B

2 kNm C

600 mm A 800 mm

Solution Equilibrium: Referring to the FBD of the shaft, Fig. a,    gMx = 0;  TA + TB - 2000 = 0

(1)

Compatibility: It is required that f CA = f CB TA LCA TB LCB = JG JG TA (0.8) JG

=

TB (0.6) JG (2)

TA = 0.75 TB Solving Eqs (1) and (2)

TB = 1142.86 N # m  TA = 857.14 N # m

Maximum Shear Stress: Applying the torsion formula, (tmax)AC =

857.14 (0.02) TA c = p = 68.21 ( 106 ) Pa = 68.2 MPa 4 J 2 ( 0.02 )

Ans.

(tmax)B C =

1142.86 (0.02) TB c = = 90.946 ( 106 ) Pa = 90.9 MPa p 4 J ( ) 0.02 2

Ans.

Ans: (tmax)AC = 68.2 MPa, (tmax)BC = 90.9 MPa 385

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5–81. The Am1004-T61 magnesium tube is bonded to the A-36 steel rod. If the allowable shear stresses for the magnesium and steel are (tallow)mg = 45 MPa and (tallow)st = 75 MPa, respectively, determine the maximum allowable torque that can be applied at A. Also, find the corresponding angle of twist of end A.

900 mm B

Solution

A

Equilibrium: Referring to the free-body diagram of the cut part of the assembly shown in Fig. a, ΣMx = 0; Tmg + Tst - T = 0

80 mm 40 mm

T

(1)

Compatibility Equation: Since the steel rod is bonded firmly to the magnesium tube, the angle of twist of the rod and the tube must be the same. Thus, (f st)A = (f mg)A TstL p (0.024)(75)(109) 2

=

TmgL p (0.044 - 0.024)(18)(109) 2 (2)

Tst = 0.2778Tmg Solving Eqs. (1) and (2), Tmg = 0.7826T

Tst = 0.2174T

Allowable Shear Stress: (tallow)mg =

Tmg c J

; 45(106) =

0.7826T(0.04) p (0.044 - 0.024) 2 T = 5419.25 N # m

(tallow) st =

Tstc ; J

75(106) =

0.2174T(0.02) p (0.024) 2 T = 4335.40 N # m = 4.34 kN # m (control!)

Ans.

Angle of Twist: Using the result of T, Tst = 942.48 N # m. We have fA =

942.48(0.9) TstL = = 0.045 rad = 2.58° p JstGst (0.024)(75)(109) 2

Ans.

Ans: T = 4.34 kN # m, f A = 2.58° 386

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5–82. The Am1004-T61 magnesium tube is bonded to the A-36 steel rod. If a torque of T = 5 kN # m is applied to end  A, determine the maximum shear stress in each material. Sketch the shear stress distribution.

900 mm B

Solution

A

Equilibrium: Referring to the free-body diagram of the cut part of the assembly shown in Fig. a, ΣMx = 0; Tmg + Tst - 5(103) = 0

80 mm 40 mm

T

(1)

Compatibility Equation: Since the steel rod is bonded firmly to the magnesium tube, the angle of twist of the rod and the tube must be the same. Thus, (f st)A = (f mg)A TstL p (0.024)(75)(109) 2

=

TmgL p (0.044 - 0.024)(18)(109) 2 (2)

Tst = 0.2778Tmg Solving Eqs. (1) and (2), Tmg = 3913.04 N # m

Tst = 1086.96 N # m

Maximum Shear Stress: (tst)max =

(tmg)max =

1086.96(0.02) Tstcst = = 86.5 MPa p Jst (0.024) 2 Tmgcmg Jmg

(tmg)  r = 0.02 m =

=

Tmg r Jmg

Ans.

3913.04(0.04) = 41.5 MPa p (0.044 - 0.024) 2 =

3913.04(0.02) = 20.8 MPa p (0.044 - 0.024) 2

Ans.

Ans.

Ans: (tst)max = 86.5 MPa, (tmg)max = 41.5 MPa, (tmg)  r = 0.02 m = 20.8 MPa 387

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5–83. A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N # m. If the steel portion has a diameter of 30 mm, determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa.

C

1.60 m

B 680 Nm A 0.75 m

Solution Compatibility Condition: f B> C = f B>A T(1.60) p 4 9 2 (c )(39)(10 )

=

T(0.75) p 4 9 2 (0.015 )(75)(10 )

c = 0.02134 m Ans.

d = 2c = 0.04269 m = 42.7 mm

Ans: d = 42.7 mm 388

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*5–84. Determine the absolute maximum shear stress in the shaft of Prob. 5–88.

C

1.60 m

B 680 Nm A 0.75 m

Solution Equilibrium, 2T = 680

T = 340 N # m

tabs occurs in the steel. See solution to Prob. 5–88. max

tabs = max

340(0.015) Tc = p 4 J 2 (0.015) Ans.

= 64.1 MPa

Ans: tabs = 64.1 MPa max 389

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5–85. The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T = 75 N · m is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in the brass and steel. Take Gst = 75 GPa, Gbr = 38 GPa.

0.9 3 ftm 0.6 2 ftm A A 12 0.5mm in.

Solution 25 mm C 1 in. C

Equilibrium:

B B T  50 75 lbft Nm T

(1)

Tbr + Tst - 50 75 = 0 Both the steel tube and brass core undergo the same angle of twist fC>B fC>B =

Tbr (0.6) Tst (0.6) TL = = π JG [ 2 (0.01254 )][38(109 )] [ π2 (0.0254 − 0.01254 )][75(109 )]

75 N · m

(2)

Tbr ==0.033778 0.032464TTstst Solving Eqs. (1) and (2) yields:

# ft; 48.428Nlb· m; Tstst ==72.549 fC = ©

1.572 TTbrbr==2.451 Nlb · m# ft

2.451(0.6) 75(0.9) TL +  = π 4 9 π JG [ 2 (0.0125 )][38(10 )] [ 2 (0.0254 )][75(109 )] = 0.002476 0.002019 rad ==0.142° 0.116° 

(tst)max AB =

Ans.

TABc 75(0.025) 6 = 3.056(10 = ) N−m 2 3.056 MPa = π J (0.0254 ) 2

(tst)max BC =

Tst c 72.549(0.025) = π J [ 2 (0.0254 − 0.01254 )]

6 = ) N−m 2 3.15 MPa (Max) = 3.153(10

Ans.

6

(gst)max =

(tst)max 3.153(10 ) = 42.0(10 −6 ) πad  = 9 G 75(10 )

Ans.

(tbr)max =

2.451(0.0125) Tbr c 6 = 0.7988(10 = ) N−m 2 0.799 MPa  = π 4 J (0.0125 ) 2

Ans.

(gbr)max =

(tbr)max 0.7988(106 ) = 21.0(10 −6 ) πad =  G 38(109 )

Ans.

390

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5–86. The shafts are made of A-36 steel and have the diameterof of100 4 in. If aIftorque ofof 1525kip same diameter mm. a torque kN# ·ftmisis applied to gear B, determine the absolute maximum shear stress developed in the shaft.

0.75 2.5mft 0.75 2.5mft

AA

Solution

BB

Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, we have (1)

F(0.15)-– 25 15 ==00 ©Mx = 0; TA + F(0.5)

1506mm in.

2515kN mft kip DD 30012 mm in.

CC EE

0.93mft

and (2)

©Mx = 0; F(1) = 00 F(0.3)-– T TEE =

Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: fCrC = fDrD

¢

TABLAB TDE LDE TBCLBC + ≤ rC = ¢ ≤ rD JGst JGst JGst

(0.9)(0.3) - TAA(0.75) (2.5) + F(0.15)(0.75)](0.15) F(0.5)(2.5) D (0.5) = =- –T TEE(3)(1) C[–T 0.5F==2.4T 2.4T TAA –-0.15F EE

(3)

Solving Eqs. (1), (2), and (3), we have F ==24.51 F 4.412kN kip

TE = 7.353 m TEkN= · 4.412 kip # ft

# ft TAA ==21.32 12.79kN kip· m

Maximum Shear Stress: By inspection, segment AB of shaft ABC is subjected to the greater torque.

A tmax B abs =

0.150 m

3

TAB c [21.32(10 )](0.05) = π (0.054 ) Jst 2 6 = ) N−m 2 109 MPa = 108.60(10

Ans.

TBC = F(0.150)

0.3 m

0.15 m

25 kN · m

Ans.

Atmax Babs = 109 MPa

391

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5–87. The shafts are made of A-36 steel and have the same diameter mm. a torque kN# ·ftmisisapplied applied to diameterof of100 4 in. If aIftorque ofof 1525kip gear B, determine the angle of twist of gear B.

2.5 ft 0.75 m

Solution

2.5 ft 0.75 m

Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, F(0.15) -– 25 ©Mx = 0; TA + F(0.5) 15 ==00

(1)

and

A B 15 kN kipm ft 25

6 in. 150 mm D

12mm in. 300

F(0.3)-– T TEE = ©Mx = 0; F(1) = 00

C E

(2) 3 ftm 0.9

Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: It is required that fCrC = fDrD

¢

TAB LAB TDE LDE TBC LBC + ≤ rC = ¢ ≤ rD JGst JGst JGst

(0.9)(0.3) - TAA(0.75) (2.5) + F(0.15)(0.75)](0.15) F(0.5)(2.5) D (0.5) = =- –T TEE(3)(1) C[–T TAA–-0.15F 0.5F==2.4T 2.4T T EE

(3)

Solving Eqs. (1), (2), and (3), TE =T7.353 kN · mkip # ft E = 4.412

F ==24.51 4.412kN kip F

# ft · m TA =TA12.79 kipkN = 21.32

- 12.79 kN kip·# m ft Angle of Twist: Here, TAB = - TA = –21.32 fB =

TAB LAB [−21.32(10 3 )](0.75) = π JGst [ 2 (0.054 )][75(109 )]

= –0.02172 - 0.01666rad rad==1.24° 0.955°  F

Ans.

0.150 m 0.15 m

0.3 m TBC = F(0.150)

25 kN · m

392

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*5–88. The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB.

2 kN A

2 kN

C

400 mm

50 mm

50 mm B

600 mm

Solution Equilibrium: (1)

TA + TB - 2(0.1) = 0 Compatibility: f C>A = f C>B TA(0.4) JG

=

TB(0.6) JG (2)

TA = 1.50TB Solving Eqs. (1) and (2) yields: TB = 0.080 kN # m

TA = 0.120 kN # m

Maximum Shear Stress:

( tAC ) max =

0.12 ( 103 ) (0.02) TA c = = 9.55 MPa p 4 J 2 ( 0.02 )

Ans.

( tCB ) max =

0.08 ( 103 ) (0.02) TB c = = 6.37 MPa p 4 J 2 ( 0.02 )

Ans.

Ans: (tAC)max = 9.55 MPa, (tCB)max = 6.37 MPa 393

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5–89. The two shafts are made of A-36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N # m is applied to the gear at E, determine the reactions at A and B.

B F

D

50 mm

0.75 m

100 mm

500 Nm E

C 1.5 m A

Solution Equilibrium: TA + F(0.1) - 500 = 0

(1)

TB - F(0.05) = 0

(2)

TA + 2TB - 500 = 0

(3)

From Eqs. (1) and (2)

Compatibility: 0.1f E = 0.05f F f E = 0.5f F TA(1.5) JG

= 0.5c

TB(0.75) JG

TA = 0.250TB

d

(4)

Solving Eqs. (3) and (4) yields: TB = 222 N # m

Ans.

TA = 55.6 N # m

Ans.

Ans: TB = 22.2 N # m, TA = 55.6 N # m 394

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5–90. The two shafts are made of A-36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N # m is applied to the gear at E, determine the rotation of this gear.

B F

D

50 mm

0.75 m

100 mm

500 Nm E

C 1.5 m A

Solution Equilibrium: TA + F(0.1) - 500 = 0

(1)

TB - F(0.05) = 0

(2)

From Eqs. (1) and (2) (3)

TA + 2TB - 500 = 0 Compatibility: 0.1f E = 0.05f F f E = 0.5f F TA(1.5) JG

= 0.5c

TB(0.75) JG

TA = 0.250TB

d

(4)

Solving Eqs. (3) and (4) yields: TB = 222.22 N # m

TA = 55.56 N # m

Angle of Twist: fE =

TAL = JG

55.56(1.5) p 2

(0.01254)(75.0)(109) Ans.

= 0.02897 rad = 1.66°

Ans: f E = 1.66° 395

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5–91. The two 1-m-long shafts are made of 2014-T6 aluminum. Each has a diameter of 30 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 900 N · m is applied to the top gear as shown, determine the maximum shear stress in each shaft.

A B C

900 600 Nm lbft E 80 4mm in. 40 2mm in.

13 m ft

D

F

Solution TA + F (0.08) − 900 = 0

(1)

TB − F (0.04) = 0

(2)

900 N · m

80 mm

40 mm

From Eqs. (1) and (2) TA + 2TB - 600 900 = 0 80(f 40(f 4(f ) )== 2(f F); EE F);

(3)

fE = 0.5fF

TAL TBL = 0.5 a b; JG JG

(4)

TA = 0.5TB

Solving Eqs. (3) and (4) yields:

# ft; 240Nlb· m; TBB ==360 (tBD)max =

120 TTAA= =180 N lb · m# ft

TB c 360(0.015) 6 = 67.91(10 = ) N−m 2 67.9 MPa = π J (0.0154 )

Ans.

TA c 180(0.015) 6 = 33.95(10 = ) N−m 2 34.0 MPa = π (0.0154 ) J

Ans.

2

(tAC)max =

2

Ans. (tBD)max = 67.9 MPa, (tAC)max = 34.0 MPa 396

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*5–92. If the shaft is subjected to a uniform distributed torque of t = 20 kN # m>m, determine the maximum shear stress developed in the shaft. The shaft is made of 2014-T6 aluminum alloy and is fixed at A and C.

400 mm

20 kNm/m 600 mm a A 80 mm 60 mm

Solution Equilibrium: Referring to the free-body diagram of the shaft shown in Fig. a, we have ΣMx = 0; TA + TC - 20(103)(0.4) = 0

B a C

Section a–a

(1)

Compatibility Equation: The resultant torque of the distributed torque within the region x of the shaft is TR = 20(103)x N # m. Thus, the internal torque developed in the shaft as a function of x when end C is free is T(x) = 20(103)x N # m, Fig. b. Using the method of superposition, Fig. c, fC = 0 = 0 =

1 fC 2 t

L0

0.4 m

L0

0.4 m

-

1 fC 2 T

T(x)dx JG

-

C

TCL JG

20(103)xdx JG

0 = 20(103) ¢

-

TC(1) JG

x2 0.4 m ≤` - TC 2 0

TC = 1600 N # m

Substituting this result into Eq. (1), TA = 6400 N # m Maximum Shear Stress: By inspection, the maximum internal torque occurs at support A. Thus,

1 tmax 2 abs

=

6400(0.04) TA c = = 93.1 MPa J p a0.044 - 0.034 b 2

Ans.

Ans: t abs = 93.1 MPa max

397

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5–93. The tapered shaft is confined by the fixed supports at A and B. If a torque T is applied at its mid-point, determine the reactions at the supports. A

T

2c

B c

L/2 L/ 2

Solution Equilibrium: (1)

TA + TB - T = 0 Section Properties: r(x) = c + J(x) = Angle of Twist: fT =

c c x = (L + x) L L

4 p c pc 4 c (L + x) d = (L + x)4 2 L 2L4

Tdx = L L J(x)G L2

L

Tdx pc 4 2L4

(L + x)4 G L

=

2TL4 dx pc 4 G LL2 (L + x)4 L

= = fB =

2TL4 1 c d ` 4 3pc G (L + x)3 L2

37TL 324 pc 4 G

Tdx = L J(x)G L0 =

TBdx pc 4 2L4

(L + x)4G

2TBL4

L

dx pc G L0 (L + x)4 4

= =

L

2TBL4

L 1 d ` 3 3pc G (L + x) 0 4

7TB L

c

12pc 4G

Compatibility: 0 = fT - fB 0 =

7TBL 37TL 324pc 4G 12pc 4G

TB =

37 T 189

Ans.

Substituting the result into Eq. (1) yields: TA =

152 T 189

Ans.

Ans: TB =

398

37 152 T, TA = T 189 189

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5–94. The shaft of radius c is subjected to a distributed torque t, measured as torque>length of shaft. Determine the reactions at the fixed supports A and B.

B t0

(

t  t0 1 

( Lx ) 2 )

x L

A

Solution

2t0

T(x) =

L0

By superposition:

x

t 0 a1 +

x2 x3 b dx = t ax + b 0 L2 3L2

(1)

0 = f - fB

0 =

L0

L

x3 b dx 2 7t 0L TB(L) 3L2 = - TB(L) JG JG 12

t 0 ax +

TB =

7t 0 L 12

Ans.

From Eq. (1), T = t 0 aL + TA +

4t 0 L L3 b = 3 3L2

7t 0 L 4t 0 L = 0 12 3 TA =

3t 0 L 4

Ans.

Ans: TB = 399

7t 0L 3t 0L , TA = 12 4

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5–95. If end B of the shaft, which has an equilateral triangle cross section, is subjected to a torque of T = 1200 N # m, determine the maximum shear stress developed in the shaft. Also, find the angle of twist of end B. The shaft is made from 6061-T1 aluminum.

0.6 m A

75 mm

B T

Solution Maximum Shear Stress: 20T 20(1200) 6 = 56.89(10 ) N−m 2 56.9 MPa tmax = = = 0.0753 a3

Ans.

Angle of Twist: f =

=

46TL a4G

46(1200)(0.6) (0.0754 )[26(109 )]

= 0.04026 πad = 2.31°

Ans.

Ans. tmax = 56.9 MPa f = 2.31° 400

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*5–96. If the shaft has an equilateral triangle cross section and is made from 606 |-T| aluminum alloy that has an allowable shear stress of tallow = 84 MPa, determine the maximum allowable torque T that can be applied to end B. Also, find the corresponding angle of twist of end B.

0.6 m A

75 mm

B

Solution

T

Allowable Shear Stress: tallow =

20T ; a3

84(106 ) =

20T 0.0753

= T 1.772(10 3 ) N= ⋅ m 1.77 kN ⋅ m

Ans.

Angle of Twist: f =

46TL a4G

46[1.772(10 3 )](0.6) =

(0.0754 )[26(109 )]

= 3.41° = 0.05945 πad

Ans.

Ans:

T = 1.77 kN # m, f = 3.41⬚

401

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5–97. The shaft is made of red brass C83400 and has an elliptical cross section. If it is subjected to the torsional loading, determine the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to end A.

A 20 Nm

50 Nm

30 Nm 2m C 1.5 m

50 mm 20 mm B

Solution Maximum Shear Stress: (tBC)max =

2TBC 2

pab

=

2(30.0) p(0.05)(0.022) Ans.

= 0.955 MPa (tAC)max =

2TAC 2

pab

=

2(50.0) p(0.05)(0.022) Ans.

= 1.59 MPa Angle of Twist:

f B>A = a =

(a2 + b2)T L p a3b3 G (0.052 + 0.022)

p(0.053)(0.023)(37.0)(109)

[( - 30.0)(1.5) + ( -50.0)(2)] Ans.

= -0.003618 rad = 0.207°

Ans: (tBC)max = 0.955 MPa, (tAC)max = 1.59 MPa, f B>A = 0.207° 402

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5–98. Solve Prob. 5–98 for the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to C.

A 20 Nm

50 Nm

30 Nm 2m C 1.5 m

50 mm 20 mm B

Solution Maximum Shear Stress: (tBC)max =

2TBC 2

pab

=

2(30.0) p(0.05)(0.022) Ans.

= 0.955 MPa (tAC)max =

2TAC 2

pab

=

2(50.0) p(0.05)(0.022) Ans.

= 1.59 MPa Angle of Twist: f B>C = =

(a2 + b2) TBC L p a3 b3 G (0.052 + 0.022)( - 30.0)(1.5) p(0.053)(0.023)(37.0)(109) Ans.

= - 0.001123 rad = 0.0643°

Ans: (tBC)max = 0.955 MPa, (tAC)max = 1.59 MPa, f B>C = 0.0643° 403

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5–99. If a = 25 mm and b = 15 mm, determine the maximum shear stress in the circular and elliptical shafts when the applied torque is T = 80 N # m. By what percentage is the shaft of circular cross section more efficient at withstanding the torque than the shaft of elliptical cross section?

b

a

a

Solution For the circular shaft: (tmax)c =

80(0.025) Tc = p = 3.26 MPa 4 J 2 (0.025 )

Ans.

For the elliptical shaft: (tmax)e =

2(80) 2T = = 9.05 MPa 2 pab p(0.025)(0.0152)

, more efficient = =

(tmax)e - (tmax)c (tmax)c

Ans.

(100,)

9.05 - 3.26 (100,) = 178, 3.26

Ans.

Ans: (tmax)c = 3.26 MPa, (tmax)e = 9.05 MPa, , more efficient = 178, 404

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*5–100. It is intended to manufacture a circular bar to resist torque; however, the bar is made elliptical in the process of manufacturing, with one dimension smaller than the other by a factor k as shown. Determine the factor by which the maximum shear stress is increased.

kd

d

d

Solution For the circular shaft: (tmax)c =

T 1 d2 2 Tc 16T = p d 4 = J p d3 2 122

For the elliptical shaft: (tmax)e =

2T 2T 16T = = 2 d kd 2 pab p k2 d3 p 1 2 21 2 2

Factor of increase in maximum shear stress = =

(tmax)e (tmax)c

=

16T p k 2d 3 16T p d3

1 k2

Ans.

Ans: Factor of increase = 405

1 k2

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5–101. The brass wire has a triangular cross section, 2 mm on a side. If the yield stress for brass is tY = 205 MPa, determine the maximum torque T to which it can be subjected so that the wire will not yield. If this torque is applied to the 4-m-long segment, determine the greatest angle of twist of one end of the wire relative to the other end that will not cause permanent damage to the wire. Gbr = 37 GPa.

T

4m T

Solution Allowable Shear Stress: tmax = tg = 205(106) =

20T a3 20T 0.0023

T = 0.0820 N # m

Ans.

Angle of Twist: f =

46(0.0820)(4) 46TL = a4G (0.0024)(37)(109) Ans.

= 25.5 rad

Ans: T = 0.0820 N # m, f = 25.5 rad 406

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5–102. If the solid shaft is made from red brass C83400 copper having an allowable shear stress of tallow = 28 MPa, determine the maximum allowable torque T that can be applied at B.

0.6 m

A

100 mm

50 50 mm 1.2 m mm

B T

C

Solution Equilibrium: Referring to the free-body diagram of the square bar shown in Fig. a, we have ©Mx = 0;

TA + TC - T = 0

(1)

Compatibility Equation: Here, it is required that fB>A = fB>C

7.10TA (0.6) 4

aG

=

7.10TC (1.2) a 4G

TA = 2TC

(2)

Solving Eqs. (1) and (2), TC =

1 T 3

TA =

2 T 3

Allowable Shear Stress: Segment AB is critical since it is subjected to the greater internal torque.

tallow =

4.81TA a3

;

28(106 ) =

( )

4.81 23 T 0.13

T 8.732(10 3 ) N= = ⋅ m 8.73 kN ⋅ m

Ans.

Ans: T = 8.73 kN ⋅ m 407

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5–103. If the solid shaft is made from red brass C83400 copper and it is subjected to a torque T = 8 kN · m at B, determine the maximum shear stress developed in segments AB and BC.

0.6 m

A

100 mm

Solution

50 50 mm 1.2 m mm

B T

C

Equilibrium: Referring to the free-body diagram of the square bar shown in Fig. a, we have ©Mx = 0;

(1)

TA + TC - 8 = 0

Compatibility Equation: Here, it is required that fB>A = fB>C

7.10TA (0.6) a 4G

=

7.10TC (1.2) a 4G

TA = 2TC

(2)

Solving Eqs. (1) and (2),

TC = 2.667 kN ⋅ m

TA = 5.333 kN ⋅ m

Maximum Shear Stress: (tmax)AB = (tmax)BC =

4.81TA 3

a

4.81TC a

3

4.81[5.333(10 3 )] 6 25.65(10 = ) N−m 2 25.7 MPa Ans. = = 0.13 4.81[2.667(10 3 )] 6 12.83(10 ) N−m 2 12.8 MPa Ans. = = = 0.13

Ans. (τ max ) AB 25.7 MPa, (τ max )BC 12.8 MPa = = 408

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*5–104. If the shaft is subjected to the torque of 3 kN # m, determine the maximum shear stress developed in the shaft. Also, find the angle of twist of end B. The shaft is made from A-36 steel. Set a = 50 mm.

600 mm

a

A

Solution

a

a

B 3 kNm

Maximum Shear Stress: 3

tmax =

2(3)(10 ) 2T = = 61.1 MPa 2 pab p(0.05)(0.0252)

Ans.

Angle of Twist: f = =

(a2 + b2)TL pa3b3G (0.052 + 0.0252)(3)(103)(0.6) p(0.053)(0.0253)(75)(109) Ans.

= 0.01222 rad = 0.700°

Ans: tmax = 61.1 MPa, f B = 0.700° 409

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  If the shaft is made from A-36 steel having an allowable shear stress of tallow = 75 MPa, determine the minimum dimension a for the cross section to the nearest millimeter. Also, find the corresponding angle of twist at end B.

600 mm

a

A

Solution

a

2T ; pab2

B 3 kNm

Allowable Shear Stress: tallow =

a

75(106) =

2(3)(103) p(a)(a2)2

      a = 0.04670 m Use

Ans.

a = 47 mm

Angle of Twist: f =

=

(a2 + b2)TL pa3b3G c 0.0472 + a

0.047 2 b d (3)(10 3)(0.6) 2

p(0.047 3)a

0.047 3 b (75)(10 9) 2

Ans.

= 0.01566 rad = 0.897°

Ans: Use a = 47 mm, f B = 0.897° 410

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  The plastic tube is subjected to a torque of 150 N # m. Determine the mean dimension a of its sides if the allowable shear stress is tallow = 60 MPa. Each side has a thickness of t = 3 mm.

a

a

150 Nm

Solution a2 1 a a Am = 4 c a ba b d = 2 2 2 2 tavg =

T ; 2 tAm

60(106) =

150 2(0.003) 12 a2 Ans.

a = 0.0289 m = 28.9 mm

Ans: a = 28.9 mm 411

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  The plastic tube is subjected to a torque of 150 N # m. Determine the average shear stress in the tube if the mean dimension a = 200 mm. Each side has a thickness of t = 3 mm.

a

a

150 Nm

Solution 1 0.2 0.2 Am = 4 c a ba b d = 0.02 m2 2 2 2 tavg =

T 150 = 2 t Am 2(0.003)(0.02)

Ans.

= 1.25 MPa

Ans: tavg = 1.25 MPa 412

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*5–108. For a given maximum shear stress, determine the factor by which the torque carrying capacity is increased if the half-circular section is reversed from the dashed-line position to the section shown. The tube is 0.1 2.5 in. mmthick. thick.

45 mm 1.80 in. 15 in. mm 0.6

30 mm 1.20 in. 12.5 0.5 mm in.

Solution A = m (0.0275)(0.04375) −

π (0.013752 ) = 0.9061(10 −3 ) m 2 2

′ (0.0275)(0.04375) + A = m

π (0.013752 ) = 1.5001(10 −3 ) m 2 2

tmax =

T 2t Am

T = 2 t Am tmax Factor = =

2t Am ¿ tmax 2t Am tmax Am ¿ 1.5001(10 −3 ) = 1.655 = 1.66 = Am 0.9061(10 −3 )

Ans.

Ans: Factor  1.66 413

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5–109. A torque of 200 N · m is applied to the tube. If the wall thickness is 2.5 mm, determine the average shear stress in the tube. 50 mm 50 mm

48.75 mm

Solution = Am

π (0.048752 ) = 1.8665(10 −3 ) m 2 4

tavg =

200 T 6 = 21.43(10 = ) N−m 2 21.4 MPa = 2t Am 2(0.0025)[1.8665(10 −3 )]

Ans.

Ans: tavg = 21.44 MPa 414

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5–110. The 6061-T6 aluminum bar has a square cross section of 25 mm by 25 mm. If it is 2 m long, determine the maximum shear stress in the bar and the rotation of one end relative to the other end.

C 1.5 m 20 Nm B

0.5 m A 80 Nm

60 N·m

Solution

25 mm

25 mm

Maximum Shear Stress: tmax =

4.81Tmax a

3

=

4.81(80.0) (0.0253)

Ans.

= 24.6 MPa

Angle of Twist: 7.10( - 20.0)(1.5) 7.10( -80.0)(0.5) 7.10TL f A>C = a 4 = + 4 9 aG (0.025 )(26.0)(10 ) (0.0254)(26.0)(109) Ans.

= - 0.04894 rad = 2.80°

Ans: tmax = 24.6 MPa, f A>C = 2.80° 415

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5–111. The aluminum strut is fixed between the two walls at A and B. If it has a 50 mm by 50 mm square cross section, and it is subjected to the torque of 120 N · m at C, determine the reactions at the fixed supports. Also, what is the angle of twist at C? Gal 5 27 GPa.

A C 0.6 2 ftm m 120 Nft 80 lb

B

0.9 3 ftm

Solution By superposition: 0 = f - fB 0 =

7.10(T 7.10(80)(2) 7.10(120)(0.6) 7.10(T B)(5) B)(1.5) -  4 4 4 G a4aG a Ga G

120 N · m

32 N lb ·# m ft TB = 48

Ans.

32 –-120 80 == 00 TA + 48 TA = 48 72 lb N #· ft m fC =

Ans.

7.10(48)(0.9) = 0.0018176 = πad 0.104° (0.054 )[27(109 )]

120 N · m 48 N · m

Ans.

Ans: T B  48 N · m, T A  72 N · m, fC  0.104° 416

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*5–112. Determine the constant thickness of the rectangular tube if average stress is not to exceed 84 MPa when a torque of T = 2 kN · m is applied to the tube. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown.

T 100 mm

Solution

50 mm

= Am (0.05)(0.1) = 0.005 m 2 tavg =

T 2t Am

84(106 ) =

2(10 3 ) 2t(0.005)

= t 0.002381 = m 2.38 mm

Ans.

Ans: t = 2.38 mm 417

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5–113. Determine the torque T that can be applied to the rectangular tube if the average shear stress is not to exceed 84 MPa. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown and the tube has a thickness of 3 mm.

T 100 mm

Solution 50 mm

= Am (0.05)(0.1) = 0.005 m 2 tavg =

T 2t Am ;

84(106 ) =

2(10 3 ) 2(0.003)(0.005)

= t 2.52(10 3 ) N= ⋅ m 2.52 kN ⋅ m

Ans.

Ans: T = 2.52 kN ⋅ m 418

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5–114. Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle. By what percentage is the torsional strength reduced when the eccentricity e is one-fourth of the difference in the radii?

ab 2 a

b e 2

e 2

Solution Average Shear Stress: For the aligned tube tavg =

T T = 2 t Am 2(a - b)(p) 1 a +2 b 2 2

T = tavg (2)(a - b)(p) a

For the eccentric tube

tavg =

a + b 2 b 2

T′ 2 t Am

t = a = a -

e e - a + bb = a - e - b 2 2 1 3 (a - b) - b = (a - b) 4 4

3 a + b 2 T′ = tavg (2) c (a - b) d (p) a b 4 2 Factor =

tavg (2) 3 34 (a - b) 4 (p) 1 a +2 b 2 2 T′ 3 = = T 4 tavg (2)(a- b)(p) 1 a +2 b 2 2

Percent reduction in strength = a1 -

3 b * 100, = 25, 4

Ans.

Ans: Percent reduction in strength = 25, 419

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5–115.  The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of t = 5 mm. If the allowable shear stress is tallow  =  56 MPa, and the tube is to resist a torque of T =   340 N · m, determine the necessary dimension b. The mean area for the ellipse is Am =  pb(0.5b).

b 0.5b

340 N.m

Solution tavg = tallow =

56(106 ) =

T 2tAm

340 2(0.005)[π b(0.5b)]

= b 0.01966 = m 19.7 mm

Ans.

Ans: b = 19.7 mm 420

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  The 304 stainless steel tube has a thickness of 10 mm. If the allowable shear stress is tallow = 80 MPa, determine the maximum torque T that it can transmit. Also, what is the angle of twist of one end of the tube with respect to the other if the tube is 4 m long? The mean dimensions are shown.

30 mm

70 mm

T

Solution Section Properties: Am = 0.07(0.03) = 0.00210 m2 L

ds = 2(0.07) + 2(0.03) = 0.200 m

Allowable Average Shear Stress: tavg = tallow = 80 ( 106 ) =

T 2 t Am T 2(0.01)(0.00210)

T = 3360 N # m = 3.36 kN # m

Ans.

Angle of Twist: f =

3360(4)(0.200) TL ds = 4 A2mG L t 4 ( 0.002102 ) (75.0) ( 109 ) (0.01) Ans.

= 0.2032 rad = 11.6°

Ans: T = 3.36 kN # m, f = 11.6° 421

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  The 304 stainless steel tube has a thickness of 10 mm. If the applied torque is T = 50 N # m, determine the average shear stress in the tube. The mean dimensions are shown. 30 mm

70 mm

T

Solution Section Properties: Am = 0.07(0.03) = 0.00210 m2 Average Shear Stress: tavg = =

T 2 t Am 50 2(0.01)(0.00210) Ans.

= 1.19 MPa

Ans: tavg = 1.19 MPa 422

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5–118. The plastic hexagonal tube is subjected to a torque of 150 N # m. Determine the mean dimension a of its sides if the allowable shear stress is tallow = 60 MPa. Each side has a thickness of t = 3 mm.

a

t = 3 mm

T = 150 Nm

Solution 1 Am = 4c (a cos 30°)(a sin 30°) d + (a)(2a)cos 30° = 2.5981 a2 2 tavg = tallow = 60 ( 106 ) =

T 2 t Am

150 (2)(0.003) ( 2.5981 a2 ) Ans.

a = 0.01266 m = 12.7 mm

Ans: a = 12.7 mm 423

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5–119.  The symmetric tube is made from a high-strength steel, having the mean dimensions shown and a thickness of 5 mm. If it is subjected to a torque of T = 40 N # m, determine the average shear stress developed at points A and B. Indicate the shear stress on volume elements located at these points.

20 mm

30 mm

60 mm A B

40 Nm

Solution Am = 4(0.04)(0.06) + (0.04)2 = 0.0112 m2 tavg =

T 2 t Am

(tavg)A = (tavg)B =

40 = 357 kPa 2(0.005)(0.0112)

Ans.

Ans: (tavg)A = (tavg)B = 357 kPa 424

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  The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diameter segment. Determine the radius of the elastic core produced in the smaller diameter segment. Neglect the stress concentration at the fillet.

T

60 mm 55 mm T

Solution Maximum Elastic Torque: For the larger diameter segment tY =

TY c J

TY =

tY J c

=

tY ( p2 )( 0.034 ) 0.03

= 13.5 ( 10-6 ) p tY Elastic - Plastic Torque: For the smaller diameter segment 2p 3 3 -6 -6 TP = 2p 3 tY c = 3 tY ( 0.0275 ) = 13.86 ( 10 ) ptY 7 13.5 ( 10 ) ptY . Applying Eq. 5–26 of the text, we have T = 13.5 ( 10-6 ) ptY =

p tY ( 4 c 3 - r3Y ) 6 p tY 3 4 ( 0.02753 ) - r3Y 4 6

Ans.

rY = 0.01298 m = 13.0 mm

Ans: rY = 13.0 mm 425

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5–121. The steel step shaft has an allowable shear stress of tallow = 8 MPa. If the transition between the cross sections has a radius r = 4 mm, determine the maximum torque T that can be applied.

50 mm

20 mm

T 2

T

20 mm

T 2

Solution Allowable Shear Stress: D 50 = = 2.5 d 20

and

r 4 = = 0.20 d 20

From the text, K = 1.25 tmax = tallow = K

Tc J t 2 (0.01) § 4 2 (0.01 )

8(106) = 1.25£ p T = 20.1 N # m

Ans.

Ans: T = 20.1 N # m 426

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5–122. The shaft is fixed to the wall at A and is subjected to the torques shown. Determine the maximum shear stress in the shaft. A fillet weld having a radius of 2.75 mm is used to connect the shafts at B.

900 Nm

A 100 Nm 50 mm

300 Nm

B 25 mm

Solution

C

Internal Torque: The internal torque in regions CD, BD and AE are indicated in their respective FBD, Fig. a, Maximum Shear Stress: For region CD,   (tmax)CD =

300(0.0125) TCDC = = 97.78 ( 106 ) Pa = 97.8 MPa (Max.) p J ( 0.01254 ) 2

Ans.

For Region AE,   (tmax)AE =

1100(0.025) TAE C = = 44.82 ( 106 ) Pa = 44.8 MPa p J ( 0.0254 ) 2

D 50 r 2.75 For the fillet, enter = = 2 and = = 0.11 into Fig. 5-32, we get d 25 d 25 K = 1.375.   (tmax)f = K a

200(0.0125) TBDC b = 1.375£ § = 89.64 ( 106 ) Pa = 89.6 MPa p J 4 ( 0.0125 ) 2

Ans: ( tmax ) = 97.8 MPa 427

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5–123. The steel shaft is made from two segments: AB and BC, which are connected using a fillet weld having a radius of 2.8 mm. Determine the maximum shear stress developed in the shaft.

C 50 mm

D

100 Nm

20 mm B A

60 Nm

Solution (tmax)CD =

40 Nm

100(0.025) TCDc = p 4 J 2 (0.025 )

= 4.07 MPa For the fillet: D 50 = = 2.5; d 20

r 2.8 = = 0.14 d 20

From Fig. 5-32, K = 1.325 (tmax)f = K

60(0.01) TABc d = 1.325 c p 4 J 2 (0.01 ) = 50.6 MPa (max)

Ans.

Ans. (tmax)f = 50.6 MPa (max) 428

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  The built-up shaft is to be designed to rotate at 450  rpm while transmitting 230 kW of power. Is this possible? The allowable shear stress is tallow = 150 MPa.

100 mm

60 mm

Solution Internal Torque: Here,    v = a450 Then,

   T =

rev 2p rad 1 min ba ba b = 15p rad>s min 1 rev 60 s

230 ( 103 ) P = = 4880.75 N # m v 15 p

Maximum Shear Stress: At the fillet,    tmax = tallow = K a          150 ( 106 ) = K£

TC b J 4880.75 (0.03) § p ( 0.034 ) 2

K = 1.30 Enter this value and r = 0.14 ; 60

D 100 r = = 1.67 into Fig. 5-32, we get = 0.14. Thus, d 60 d r = 8.4 mm

D-d 100- 60 = = 20 mm > r = 8.4 mm, it is possible to construct the 2 2 transition. Ans.

Since

Ans: It is possible. 429

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  The built-up shaft is designed to rotate at 450 rpm. If the radius of the fillet weld connecting the shafts is r = 13.2 mm, and the allowable shear stress for the material is tallow = 150 MPa, determine the maximum power the shaft can transmit. 100 mm

60 mm

Solution Internal Torque: Here,    v = a450 Then,

rev 2p rad 1 min ba ba b = 15p rad>s min 1 rev 60 s

T =

P P = v 15 p

D 100 r Maximum Shear Stress: Enter = = 1.67 and = 0.22 into Fig 5-32, we get d 60 d K = 1.2. Then tmax = tallow = K a

TC b J

150 ( 106 ) = 1.2£

P 15 p (0.03) § p 4 2 0.03

(

)

P = 249.82 ( 103 ) W = 250 kW

Ans.

Ans: P = 250 kW 430

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  A solid shaft has a diameter of 40 mm and length of 1 m. It is made from an elastic-plastic material having a yield stress of tY = 100 MPa. Determine the maximum elastic torque TY and the corresponding angle of twist. What is the angle of twist if the torque is increased to T = 1.2TY? G = 80 GPa.

Solution Maximum elastic torque TY, tY =

TY c J

TY =

100 ( 106 )( p2 )( 0.024 ) tY J = = 1256.64 N # m = 1.26 kN # m c 0.02

Ans.

Angle of twist: gY = f =

100 ( 106 ) tY = = 0.00125 rad G 80 ( 109 ) gY 0.00125 L = (1) = 0.0625 rad = 3.58° rY 0.02

Ans.

Also, f =

TY L = JG

1256.64(1) p 2

( 0.024 ) (80) ( 109 )

= 0.0625 rad = 3.58°

From Eq. 5–26 of the text, T =

p(100) ( 106 ) ptY 14 c 3 - r3Y 2 ;  1.2(1256.64) = 3 4 ( 0.023 ) - r3Y 4 6 6 rY = 0.01474 m

f′ =

gY 0.00125 L = (1) = 0.0848 rad = 4.86° rY 0.01474

Ans.

Ans: TY = 1.26 kN # m, f = 3.58°, f′ = 4.86° 431

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5–127.  Determine the torque needed to twist a short 2-mm-diameter steel wire through several revolutions if it is made from steel assumed to be elastic perfectly plastic and having a yield stress of tY = 50 MPa. Assume that the material becomes fully plastic.

Solution Fully plastic torque is applied. TP =

2p 2p t c3 = (50)(106)(0.0013) = 0.105 N # m 3 Y 3

Ans.

Ans: TP = 0.105 N # m 432

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5–128. A solid shaft is subjected to the torque T, which causes the material to yield. If the material is elastic plastic, show that the torque can be expressed in terms of the angle of twist f of the shaft as T = 43 TY11 - f3Y>4f32, where TY and fY are the torque and angle of twist when the material begins to yield.

Solution gY gL = L r rY

f = rY =

gYL f

(1)

When rY = c, f = fY From Eq. (1), c =

gYL fY

(2)

Dividing Eq. (1) by Eq. (2) yields: rY fY = c f

(3)

Use Eq. 5-26 from the text. T =

r3Y p tY 2p tYc3 (4 c3 - r3Y) = a1 )b 6 3 4 c3

Use Eq. 5-24, TY = T =

p t c3 from the text and Eq. (3) 2 Y

f3Y 4 TY a 1 b 3 4 f3

QED

433

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5–129.  The solid shaft is made of an elastic perfectly plastic material. Determine the torque T needed to form an elastic core in the shaft having a radius of rY = 20 mm. If the shaft is 3 m long, through what angle does one end of the shaft twist with respect to the other end? When the torque is removed, determine the residual stress distribution in the shaft and the permanent angle of twist.

80 mm T

T

t (MPa) 160

Solution 0.004

Elastic-Plastic Torque: T = =

p tY 1 4c 3 - r3Y 2 6

p(160)(106) 6

3 4 1 0.043 2

- 0.023 4

= 20776.40 N # m = 20.8 kN # m

g (rad)

Ans.

Angle of Twist: f =

gY 0.004 L = a b(3) = 0.600 rad = 34.4° rY 0.02

Ans.

When the reverse T = 20776.4 N # m is applied, G = f′ =

160(106) 0.004 TL = JG

= 40 GPa 20776.4(3)

p 4 9 2 (0.04 )(40)(10 )

= 0.3875 rad

The permanent angle of twist is, f r = f - f′ Ans.

= 0.600 - 0.3875 = 0.2125 rad = 12.2° Residual Shear Stress: (t′)r = c =

20776.4(0.04) Tc = = 206.67 MPa p 4 J 2 (0.04 )

(t′)r = 0.02 m =

20776.4(0.02) Tc = = 103.33 MPa p 4 J 2 (0.04 )

(tr)r = c = - 160 + 206.67 = 46.7 MPa Ans.

(tr)r = 0.02 m = - 160 + 103.33 = - 56.7 MPa

Ans: T = 20.8 kN # m, f = 34.4°, (tr)max = 56.7 MPa, f r = 12.2° 434

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5–130. The shaft is subjected to a maximum shear strain of 0.0048 rad. Determine the torque applied to the shaft if the material has strain hardening as shown by the shear stress–strain diagram. 50 mm 2 in.

T

Solution

t (MPa) (ksi)

From the shear - strain diagram,

84 12

rY 0.05 = ; 0.0006 0.0048

426

rY = 0.00625 m

From the shear stress–strain diagram, 0.0006

0.0048

g (rad)

 42(106 )  9 τ 1 = =  ρ 6.72(10 ) ρ  0.00625 

τ 2 − 42(106 ) 84(106 ) − 42(106 ) = ρ − 0.00625 0.05 − 0.00625 τ 2 0.96(109 )ρ + 36(106 ) = c

T = 2π ∫ τ ρ 2d ρ 0

= 2π ∫

0.00625 m

0

50 mm

[6.72(109 ) ρ ]ρ 2d ρ + 2π ∫

0.05 m

0.00625 m

= 2π [1.68(109 ) ρ 4 ]

0.00625 m 0

[0.96(109 ) ρ + 36(106 )]ρ 2d ρ

+ 2π [0.24(109 ) ρ 4 + 12(106 ) ρ 3 ]

0.05 m

84 MPa

42 MPa

0.00625 m

3

= 18.84(10 ) N= ⋅ m 18.8 kN ⋅ m

Ans.

0.00625 m

0.05 m

Ans: T  18.8 kN · m 435

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5–131. A solid shaft having a diameter of 50 mm is made of of elastic-plastic material a yield of elastic-plastic material having ahaving yield stress of tY =stress 112 MPa 3 tand 16 ksimodulus 2 ksi. and shear of G = 12110 of G =modulus 84 GPa. Determine the torque Y =shear Determine torquean required develop elastic corea required to the developo elastic to core in the an shaft having in the shaft having a diameter in. Also, what is the diameter of 25 mm. Also, what is of the1plasstic torque? plastic torque?

Solution Use Eq. 5-26 from the text: T =

p tY π [112(106 )] [4(0.0253 ) − 0.01253 ] (4 c3 - rY 3) = 6 6

= 3.551(10 3 ) N= ⋅ m 3.55 kN ⋅ m

Ans.

Use Eq. 5-27 from the text: TP =

2p 2π [112(106 )](0.0253 ) t c3 = 3 Y 3

= 3.665(10 3 ) N= ⋅ m 3.67 kN ⋅ m

Ans.

Ans. T = 3.55 kN ⋅ m, TP = 3.67 kN ⋅ m 336

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*5–132. The hollow shaft has the cross section shown and is made of an elastic perfectly plastic material having a yield shear stress of tY . Determine the ratio of the plastic torque Tp to the maximum elastic torque TY .

c c 2

Solution Maximum Elastic Torque: In this case, the torsion formula is still applicable. tY =

TY c J

TY =

J t c Y

= =

c 4 p 4 Jc - a b R tY 2 2 c

15 3 pc tY 32

Plastic Torque: Using the general equation, with t = tY , c

TP = 2ptY

r2dr Lc>2 c

= 2ptY ¢ =

r3 ≤2 3 c>2

7 pc 3tY 12

The ratio is 7 pc 3tY TP 12 = = 1.24 TY 15 3 pc tY 32

Ans.

Ans: TP = 1.24 TY 437

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150 mm

5–133. The hollow shaft has inner and outer diameters of 60 mm and 80 mm, respectively. If it is made of an elasticperfectly plastic material, which has the t-g diagram shown, determine the reactions at the fixed supports A and C.

450 mm B

C 15 kNm

A

t (MPa)

Solution

120

Equation of Equilibrium. Refering to the free - body diagram of the shaft shown in Fig. a, ©Mx = 0; TA + TC - 15 A 103 B = 0

(1)

Elastic Analysis. It is required that fB>A = fB>C. Thus, the compatibility equation is fB>A = fB>C TCLBC TALAB = JG JG TA (0.45) = TC(0.15) TC = 3TA

(2)

Solving Eqs. (1) and (2), TA = 3750 N # m

TC = 11 250N # m

The maximum elastic torque and plastic torque in the shaft can be determined from p A 0.044 - 0.034 B J 2 T (120) A 106 B = 8246.68 N # m TY = tY = D c 0.04

co

TP = 2ptY

Lci

r2dr

= 2p(120) A 106 B ¢

0.04 m

r3 = 9299.11 N # m ≤` 3 0.03 m

Since TC 7 TY, the results obtained using the elastic analysis are not valid. Plastic Analysis. Assuming that segment BC is fully plastic, TC = TP = 9299.11 N # m = 9.3 kN # m

Ans.

Substituting this result into Eq. (1), TA = 5700 N # m = 5.70 kN # m

Ans.

438

0.0016

g (rad)

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5–133. Continued Since TA 6 TY, segment AB of the shaft is still linearly elastic. Here, 120 A 106 B = 75GPa. G = 0.0016 fB>C = fB>A =

fB>C =

gi L ; ci BC

5700.89(0.45) TALAB = 0.01244 rad = p JG A 0.044 - 0.034 B (75) A 109 B 2 0.01244 =

gi (0.15) 0.03

gi = 0.002489 rad Since gi 7 gY, segment BC of the shaft is indeed fully plastic.

Ans. TA = 5.70 kN # m, TC = 9.3 kN # m 439

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5–134. The 2-m-long tube is made of an elastic perfectly plastic material as shown. Determine the applied torque T, which subjects the material at the tube’s outer edge to a shear strain of gmax = 0.006 rad. What would be the permanent angle of twist of the tube when this torque is removed? Sketch the residual stress distribution in the tube.

T

60 mm

50 mm

Solution

t (MPa)

Plastic Torque: The tube will be fully plastic if gi Ú gY = 0.003 rad. From Fig. a, gi 0.006 = ; 0.05 0.06

gi = 0.005 rad 7 gY 0.003

The tube indeed is fully plastic. Then Tp = 2p = =

Lci

c0

tYr2 dp

2 pt ( C 30 - C 3i ) 3 Y 2 p 3 180 ( 106 ) 4 ( 0.063 - 0.053 ) 3

= 34.306 ( 103 ) N # m = 34.3 kN # m

Ans.

Angle of Twist: fP = Here, G =

gmax C0

L = a

0.006 b(2) = 0.2 rad 0.06

180 ( 106 ) tY = = 60 ( 109 ) Pa. When a reverse torque of gY 0.003

T ′P = 34.306 ( 103 ) N # m is applied f′P =

180

34.306 ( 103 ) (2) TP ′L = = 0.1085 rad p JG ( 0.064 - 0.054 ) 3 60 ( 109 ) 4 2

Thus, the permanent angle of twist is

         f r = f p - f′p = 0.2 - 0.1085 = (0.09151 rad) a

180° b = 5.24° Ans. p rad

440

g (rad)

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5–134. Continued

Residual Shear Stress:

( t′P ) o =

( t′P ) o =

34.306 ( 103 ) (0.06) T′PC = 195.29 ( 106 ) Pa = 195.29 MPa = p J ( 0.064 - 0.054 ) 2 34.306 ( 103 ) (0.05) T′Pr = = 162.74 ( 106 ) Pa = 162.74 MPa p J 4 4 ( 0.06 - 0.05 ) 2

Thus (tr)o = - tY + ( t′p ) o = - 180 + 195.29 = 15.3 MPa

Ans.

(tr)i = - tY + ( t′P ) i = - 180 + 162.74 = -17.3 MPa

Ans.

The sketch of the residual shear stress distribution is shown in Fig. b.

Ans: TP = 34.3 kN # m, f r = 5.24°, (tr)o = 15.3 MPa, (tr)i = - 17.3 MPa 441

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5–135. The shaft is made of an elastic perfectly plastic material as shown. Plot the shear-stress distribution acting along a radial line if it is subjected to a torque of T = 20 kN # m. What is the residual stress distribution in the shaft when the torque is removed?

T

40 mm

t (MPa) 170

Solution 0.00227

Elastic - Plastic Torque: The maximum elastic torque is

g (rad)

p p t C3 = 3 170 ( 106 ) 4 ( 0.043 ) = 17.09 ( 103 ) N # m 2 Y 2

TY =

And the plastic torque is     Tp =

2 2 p t C 3 = p 3 170 ( 106 ) 4 ( 0.043 ) = 22.79 ( 103 ) N # m 3 Y 3

Since TY 6 T 6 TP, the shaft exhibits elastic-plastic behavior. T =

         20 ( 103 ) =

ptY ( 4C 3 - r3Y ) 6 p 3 170 ( 106 ) 4 6

3 4 ( 0.043 ) - r3Y 4

          rY = 0.03152 m = 31.5 mm Residual shear stress: When the reverse torque T′ = 20 kN # m is applied,      t′C =

         t ′rY =

20 ( 103 ) (0.04) T′C = = 198.94 ( 106 ) Pa = 198.94 MPa p J 4 ( 0.04 ) 2 20 ( 103 ) (0.03152) T′rY = = 156.76 ( 106 ) Pa = 156.76 MPa p J 4 ( 0.04 ) 2

Thus,    (tr)c = -tY + t′C = - 170 + 198.94 = 28.9 MPa

Ans.

   (tr)rY = - tY + t′rY = - 170 + 156.76 = - 13.2 MPa

Ans.

The sketch of the residual shear stress distribution is shown in Fig. b.

Ans: (tr)c = 28.9 MPa, (tr)rY = - 13.2 MPa 442

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*5–136. The tube has a length of 2 m and is made of an elastic perfectly plastic material as shown. Determine the torque needed to just cause the material to become fully plastic. What is the permanent angle of twist of the tube when this torque is removed?

80 mm

60 mm t (MPa) 130

Solution Plastic Torque:

0.005

Tp = 2p

= =

Lci

g (rad)

co

tY r2 dp

2 pt ( C 30 - C 3i ) 3 Y 2 p 3 130 ( 106 ) 4 ( 0.083 - 0.063 ) 3

= 80.592 ( 103 ) N # m = 80.6 KN # m

Ans.

Angle of Twist: Here, rY = Ci = 0.06 m. fP =

gY 0.005 L = a b(2) = 0.16667 rad rY 0.06

130 ( 106 ) tY = = 26 ( 109 ) Pa. When a reverse torque of T P′ = 80.592 gY 0.005 N # m is applied,

Here, G =

( 103 )

f′P =

80.592 ( 103 ) (2) T′P L = = 0.14095 rad p JG ( 0.084 - 0.064 ) 3 26 ( 109 ) 4 2

Thus, the permanent angle of twist is

         f r = f p - f′p = 0.16667 - 0.14095          = (0.02571 rad) a

180° b = 1.47° p rad

Ans.

Ans: TP = 80.6 kN # m, f r = 1.47° 443

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5–137. The tube has a length of 2 m and is made of an elastic perfectly plastic material as shown. Determine the torque needed to just cause the material to become fully plastic. What is the permanent angle of twist of the tube when this torque is removed?

100 mm

60 mm t (MPa) 350

Solution Plastic Torque:

0.007

TP = 2p = =

Lci

g (rad)

co

tY r2 dr

2p tY 3 1 c o - c 3i 2 3 2p(350)(106) 3

( 0.053 - 0.033 )

= 71837.75 N # m = 71.8 kN # m

Ans.

Angle of twist: fr =

gY L rY

= a

Where rY = ci = 0.03 m

0.007 b(2) = 0.4667 rad 0.03

When a reverse TP = 71837.75 N # m is applied. G = f P′ =

350(106) 0.007

= 50 GPa

71837.75(2) TPL = = 0.3363 rad p JG ( 0.054 - 0.034 ) 50 ( 109 ) 2

Permanent angle of twist: f r = f P - f P′ = 0.4667- 0.3363 Ans.

= 0.1304 rad = 7.47°

Ans: TP = 71.8 kN # m, f r = 7.47° 444

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5–138. A torque is applied to the shaft having a radius of 80 mm. If the material obeys a shear stress–strain relation of t = 500 g ¼ MPa, determine the torque that must be applied to the shaft so that the maximum shear strain becomes 0.008 rad.

80 mm T t (MPa)

Solution

0.008

t-r Function: First, relate g to r. g = Then

g (rad)

r r gmax = a b(0.008) = 0.1r c 0.08

t = 500 ( 106 ) (0.1r)1>4 = 281.17 ( 106 ) r1>4 Torque: T = 2p

= 2p

Lo

Lo

c

2

tr dp

0.08m

6

9

281.17 ( 10 ) r 4 dp

= 1.7666 ( 109 ) a

4 13>4 0.08 m r b` 13 0

= 148.02 ( 103 ) N # m = 148 kN # m

Ans.

Ans: T = - 148 kN # m 445

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5–139. A tubular shaft has an inner diameter of 60 mm, an outer diameter of 80 mm, and a length of 1 m. It is made of an elastic perfectly plastic material having a yield stress of tY = 150 MPa. Determine the maximum torque it can transmit. What is the angle of twist of one end with respect to the other end if the inner surface of the tube is about to yield? G = 75 GPa.

1m

30 mm 40 mm

Solution Plastic Torque: The plastic torque is the maximum torque a circular shaft can transmit. Co

Tp = 2p

LCi

= 2p tY a = =

2

tY p dp p3 3

b`

Co Ci

2 p tY ( C 30 - C 3i ) 3 2 p 3 150 ( 106 ) 4 ( 0.043 - 0.033 ) 3

= 11.624 ( 103 ) N # m = 11.6 kN # m

Ans.

Angle of Twist: The yield strain at the inner surface of the tube can be determined from tY = G gY ;

150 ( 106 ) = 75 ( 109 ) gY gY = 0.002 rad

Then, the angle of twist is f =

gY rY

L = a

0.002 b(1) 0.03

= (0.06667 rad) a

180° b = 3.82° p rad

Ans.

Ans: TP = 11.6 kN # m, f = 3.82° 446

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*5–140. The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diameter segment. Determine the radius of the elastic core produced in the smaller diameter segment. Neglect the stress concentration at the fillet.

80 mm

T

75 mm T

Solution Maximum Elastic Torque: For the larger diameter segment, the torsion formula can be applied. TYc tY = J tY c

tYJ         T = = C

p ( 0.044 ) d 2 = 32.0 ( 10-6 ) p tY 0.04

Elastic - Plastic Torque: For the smaller diameter segment, the plastic torque can be determined from Tp = 2p

L0

c

tY r2 dp

=

2p 3 tc 3 Y

=

2p t ( 0.03753 ) 3 Y

= 35.16 ( 10-6 ) p tY Since T 6 TP, the smaller segment is not fully plastic T =

p tY ( 4c 3 - rY3 ) 6

32.0 ( 10-6 ) p tY =

p tY 3 4 ( 0.03753 ) - r3Y 4 6

rY = 0.02665 m

Ans.

= 26.7 mm

Ans: rY = 26.7 mm 447

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5–141. The shear stress–strain diagram for a solid 50-mm-­diameter shaft can be approximated as shown in the figure. Determine the torque required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 3 m long, what is the corresponding angle of twist?

 (MPa) 125

50

0.0025

0.010

 (rad)

Solution g =

r g c max

gmax = 0.01 When g = 0.0025 r = =

cg gmax 0.025(0.0025) 0.010

= 0.00625

50(106) t - 0 = r - 0 0.00625 t = 8000(106)(r) t - 50(106) r - 0.00625

=

125(106) - 50(106) 0.025 - 0.00625

t = 4000(106)(r) + 25(106) T = 2p = 2p

L0

c

L0

0.00625

tr2 dr 8000(106)r3 dr 0.025

+ 2p

[4000(106)r + 25(106)]r2 dr L0.00625

T = 3269 N # m = 3.27 kN # m f =

Ans.

gmax 0.01 L = (3) c 0.025 Ans.

= 1.20 rad = 68.8°

Ans: T = 3.27 kN # m, f = 68.8° 448

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5–142. The shear stress–strain diagram for a solid 50-mm-diameter shaft can be approximated as shown in the figure. Determine the torque T required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 1.5 m long, what is the corresponding angle of twist?

T

1.5 m T t (MPa) 125

Solution

50

Strain Diagram: rg 0.0025

=

0.025 ; 0.01

0.0025

rg = 0.00625 m

0.010

g (rad)

Stress Diagram: t1 =

50(106) r = 8(109) r 0.00625

t2 - 50(106) 125(106) - 50(106) = r - 0.00625 0.025 - 0.00625 t2 = 4 A 109 B r + 25 A 106 B The Ultimate Torque: c

T = 2p

L0

t r2dr 0.00625 m

= 2p

L0

8 A 109 B r3 dr 0.025 m

+ 2p

L0.00625 m

9 6 C 4 A 10 B r + 25 A 10 B D r2dr

m = 2p C 2 A 109 B r4 D |0.00625 0

+ 2p B 1 A 109 B r4 +

25(106)r3 0.025 m R 2 3 0.00625 m

= 3269.30 N # m = 3.27 kN # m

Ans.

Angle of Twist: f =

gmax 0.01 L = a b (1.5) = 0.60 rad = 34.4° c 0.025

Ans.

Ans. T = 3.27 kN # m, f = 34.4° 449

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5–143. The shaft consists of two sections that are rigidly connected. If the material is elastic plastic as shown, determine the largest torque T that can be applied to the shaft. Also, draw the shear-stress distribution over a radial line for each section. Neglect the effect of stress concentration.

T 1 in. 25 mm

0.75 in. 20 mm T

Solution t (ksi) (MPa)

0.75 in. diameter segment will be fully plastic. From Eq. 5-27 of the text: 20-mm

12 84

2π τY c3 T = Tp = 3 2π = [84(106 )](0.013 ) 3

0.005

= 175.93 N ⋅ = m 176 N ⋅ m

Ans.

84 MPa

g (rad)

57.3 MPa

For 125-mm – in. diameter diametersegment: segment: tmax =

175.93(0.0125) Tc = 57.34(106 ) N−m 2 = π 4 J (0.0125 ) 2

= 57.3 6.75 MPa ksi 6< ttYY

Ans. T = Tp = 176 N ⋅ m 450

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*5–144. A steel alloy core is bonded firmly to the copper alloy tube to form the shaft shown. If the materials have the t -g diagrams shown, determine the torque resisted by the core and the tube.

450 mm A

100 mm 60 mm B

15 kNm

t (MPa) 180

Solution Equation of Equilibrium. Referring to the free-body diagram of the cut part of the assembly shown in Fig. a, ΣMx = 0; Tc + Tt - 15 1 103 2 = 0

(1) 180 1 10

6

Elastic Analysis. The shear modulus of steel and copper are Gst = 36 1 106 2 and G ∞ = = 18 GPa. Compatibility requires that 0.002 fC = ft

0.0024

2

= 75 GPa

Tc

1 0.034 2 (75) 1 109 2

=

Tc = 0.6204Tt

1 0.054

- 0.034 2 (18) 1 109 2

(2)

Tc = 5743.05 N # m

The maximum elastic torque and plastic torque of the core and the tube are 1 3 1 pc (tY)st = p 1 0.033 2 (180) 1 106 2 = 7634.07 N # m 2 2 2 3 2 (TP)c = pc (tY)st = p 1 0.033 2 (180) 1 106 2 = 10 178.76 N # m 3 3 (TY)c =

and

p 1 0.054 - 0.034 2 J 2 (TY)t = tY = D T c (36) 1 106 2 d = 6152.49 N # m c 0.05 (TP)t = 2p(tY) ∞

Lci

co

r2dr = 2p(36) 1 106 2 ¢

t (MPa)

Copper Alloy

Solving Eqs. (1) and (2), Tt = 9256.95 N # m

Steel Alloy

0.002

Tt p 2

r3 0.05 m ≤` = 7389.03 N # m 3 0.03 m

Since Tt 7 (TY)t, the results obtained using the elastic analysis are not valid.

451

g (rad)

36

TcL TtL = JcGst JtG ∞ p 2

0.0024

g (rad)

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5–144. Continued

Plastic Analysis. Assuming that the tube is fully plastic, Tt = (TP)t = 7389.03 N # m = 7.39 kN # m

Ans.

Substituting this result into Eq. (1), Tc = 7610.97 N # m = 7.61 kN # m

Ans.

Since Tc 6 (TY)c, the core is still linearly elastic. Thus, f t = f tc = ft =

gi L; ci

TcL = JcGst

7610.97(0.45) p 4 9 2 (0.03 )(75)(10 )

0.3589 =

= 0.03589 rad

gi (0.45) 0.03

gi = 0.002393 rad Since gi 7 (gY) ∞ = 0.002 rad, the tube is indeed fully plastic.

Ans: Tt = 7.39 kN # m, Tc = 7.61 kN # m 452

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R5–1. The shaft is made of A992 steel and has an allowable shear stress of tallow = 75 MPa. When the shaft is rotating at 300 rpm, the motor supplies 8 kW of power, while gears A and B withdraw 5 kW and 3 kW, respectively. Determine the required minimum diameter of the shaft to the nearest millimeter. Also, find the rotation of gear A relative to C.

300 mm

300 mm

C

B

Solution A

Applied Torque: The angular velocity of the shaft is v = a300

rev 1 min 2prad ba ba b = 10p rad>s min 60 s 1 rev

Thus, the torque at C and gear A are TC =

8(103) PC = = 254.65 N # m v 10p

TA =

5(103) PA = = 159.15 N # m v 10p

Internal Loading: The internal torque developed in segment BC and AB of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: By inspection, segment BC is critical. tallow =

254.65(d2 ) TBC c ;  75(106) = p d 4 J 2 (2 ) d = 0.02586 m Ans.

Use  d = 26 mm Angle of Twist: Using d = 26 mm, f A>C = Σ =

TiLi TBCLBC TABLAB = + JiGi JG JG 0.3

(159.15 p 4 9 2 (0.013 )(75)(10 )

+ 254.65) Ans.

= 0.03689 rad = 2.11°

Ans: Use d = 26 mm, f A>C = 2.11° 453

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R5–2. The shaft is made of A992 steel and has an allowable shear stress of tallow = 75 MPa. When the shaft is rotating at 300 rpm, the motor supplies 8 kW of power, while gears A and B withdraw 5 kW and 3 kW, respectively. If the angle of twist of gear A relative to C is not allowed to exceed 0.03 rad, determine the required minimum diameter of the shaft to the nearest millimeter.

300 mm

300 mm

C

B

Solution A

Applied Torque: The angular velocity of the shaft is v = a300

rev 1 min 2prad ba ba b = 10p rad>s min 60 s 1 rev

Thus, the torque at C and gear A are TC =

8(103) PC = = 254.65 N # m v 10p

TA =

5(103) PA = = 159.15 N # m v 10p

Internal Loading: The internal torque developed in segment BC and AB of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: By inspection, segment BC is critical. tallow =

254.65(d2 ) TBC c ;  75(103) = p d 4 J 2 (2 ) d = 0.02586

Angle of Twist: f A>C = Σ 0.03 =

TiLi TBCLBC TABLAB = + JiGi JG JG 0.3

p d 4 9 2 ( 2 ) (75)(10 )

(159.15 + 254.65)

d = 0.02738 m(controls) Ans.

Use d = 28 mm

Ans: Use d = 28 mm 454

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R5–3. The A-36 steel circular tube is subjected to a torque of 10 kN # m. Determine the shear stress at the mean radius r = 60 mm and calculate the angle of twist of the tube if it is 4 m long and fixed at its far end. Solve the problem using Eqs. 5–7 and 5–15 and by using Eqs. 5–18 and 5–20.

r  60 mm 4m

t  5 mm 10 kNm

Solution We show that two different methods give similar results: Shear Stress: Applying Eq. 5-7, ro = 0.06 + tr = 0.06 m =

0.005 = 0.0625 m 2

Tr = J

ri = 0.06 -

10(103)(0.06) p 4 2 (0.0625

- 0.05754)

0.005 = 0.0575 m 2

= 88.27 MPa

Applying Eq. 5-18, tavg =

10(103) T = = 88.42 MPa 2 t Am 2(0.005)(p)(0.062)

Angle of Twist: Applying Eq. 5-15, f = =

TL JG 10(103)(4)

p 4 2 (0.0625

- 0.05754)(75.0)(109)

= 0.0785 rad = 4.495° Applying Eq. 5-20, f = = =

=

TL ds 2 4AmG L t TL ds 4A2mG t L

Where

L

ds = 2pr

2pTLr 4A2mG t 2p(10)(103)(4)(0.06) 4[(p)(0.062)]2 (75.0)(109)(0.005)

= 0.0786 rad = 4.503° Rounding to three significant figures, we find t = 88.3 MPa

Ans.

f = 4.50°

Ans. Ans: t = 88.3 MPa, f = 4.50° 455

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*R5–4. A portion of an airplane fuselage can be approximated by the cross section shown. If the thickness of its 2014-T6-aluminum skin is 10 mm, determine the maximum wing torque T that can be applied if tallow = 4 MPa. Also, in a 4-m-long section, determine the angle of twist.

0.75 m

T 2m

Solution tavg

0.75 m

T = 2tAm

4(106) =

T 2(0.01)[(p)(0.75)2 + 2(1.5)]

T = 381.37(103) = 381 kN # m f =

f =

Ans.

ds TL 4A2mG L t 381.37(103)(4) 2

2

9

4[(p(0.75) + 2(1.5)) 27(10 )]

f = 0.542(10-3) rad = 0.0310°

c

4 + 2p(0.75) 0.010

d

Ans.

Ans: T = 381 kN # m, f = 0.0310° 456

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R5–5. The material of which each of three shafts is made has a yield stress of tY and a shear modulus of G. Determine which shaft geometry will resist the largest torque without yielding. What percentage of this torque can be carried by the other two shafts? Assume that each shaft is made from the same amount of material and that it has the same cross-sectional area A.

A

60 A

A 60

60

Solution For circular shaft: 1

A = p c 2; tmax =

T =

c = a

Tc , J

A 2 b p

         tY = =

p c3 t = 2 Y

Tc a 4 c 2

3 A 2 p( p )

tY

2 1

Tcir = 0.2821 A2 tY For the square shaft: 1

A = a2;   a = A2 tmax =

4.81 T 4.81 T ;  tY = 3 a3 A2 3

T = 0.2079 A2tY For the triangular shaft: A =

1 (a)(a sin 60°); 2

tmax =

20 T ; a3

tY =

1

a = 1.5197A2 20 T 3

(1.5197)3A2

3

T = 0.1755A2 tY ­­ Ans.

The circular shaft will resist the largest torque. For the square shaft: , =

0.2079 (100,) = 73.7, 0.2821

Ans.

For the triangular shaft: , =

0.1755 (100,) = 62.2 , 0.2821

Ans.

Ans: The circular shaft will resist the largest torque. For the square shaft: 73.7%, For the triangular shaft: 62.2% 457

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R5–6. Segments AB and BC of the assembly are made from 6061-T6 aluminum and A992 steel, respectively. If couple forces P = 15 kN are applied to the lever arm, determine the maximum shear stress developed in each segment. The assembly is fixed at A and C.

P

1.2 m A

0.8 m

0.8 m E

al

1.2 m

100 mm

Solution

B

Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, ©Mx = 0;

(1)

TA + TC - 15 (1.6) = 0

D

st 100 mm

C

P

Compatibility Equation: It is required that fB>A = fB>C TA LAB TC LBC = JGal JGst

TAL

J[26(109 )]

=

TC L

J[75(109 )]

TA = 0.3467 TC

(2)

Solving Eqs. (1) and (2),

TC = 17.82 kN ⋅ m

TA = 6.178 kN ⋅ m

Maximum Shear Stress: (tmax)AB =

TA c [6.178(10 3 )](0.05) 2 31.5 MPa = 31.47(106 ) N−m = = π (0.054 ) J 2

Ans.

(tmax)BC =

TC c [17.82(10 3 )](0.05) 2 90.8 MPa = 90.77(106 ) N−m = = π (0.054 ) J 2

Ans.

Ans. (τ max ) AB = 31.5 MPa

(τ max )BC = 90.8 MPa 458

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R5–7. Segments AB and BC of the assembly are made from 6061-T6 aluminum and A992 steel, respectively. If the allowable shear stress for the aluminum is (tallow)al = 90 MPa and for the steel (tallow)st = 120 MPa, determine the maximum allowable couple forces P that can be applied to the lever arm. The assembly is fixed at A and C.

P

1.2 m A

0.8 m

0.8 m E 1.2 m

100 mm

Solution

B

Equilibrium: Referring to the free-body diagram of the assembly shown in Fig. a, ©Mx = 0;

TA + TC - P(1.6) = 0

(1)

D 100 mm

C

P

Compatibility Equation: It is required that fB>A = fB>C TCLBC TA LAB = JGal JGst

TAL

J[26(109 )]

=

TC L

J[75(109 )]

TA = 0.3467 TC

(2)

Solving Eqs. (1) and (2),

= TC 1.1881 = P TA 0.4119 P Allowable Shear Stress:

= (τ allow )al

TAc (0.4119 P )(0.05) = ; 90(106 ) π (0.054 ) J 2 3 = ) N 42.9 kN = P 42.90(10

(τ allow )st =

(1.1881P)(0.05) TC c ; 120(106 ) = π (0.054 ) J 2 3 = ) N 19.8 kN (contπols) P 19.83(10 =

Ans.

Ans: P = 19.8 kN 459

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*R5–8. The tapered shaft is made from 2014-T6 aluminum alloy, and has a radius which can be described by the equation > r = 0.02(1 + x 3 2) m, where x is in meters. Determine the angle of twist of its end A if it is subjected to a torque of 450 N # m.

x

r = 0.02(1 + x3/2) m 4m

450 Nm x

Solution

A

T = 450 N # m fA =

Tdx = L0 L JG

4

4

450 dx p 4 2 (0.02) (1

3 2

4

9

+ x ) (27)(10 )

= 0.066315

L0 (1 + x2)4 dx

3

Evaluating the integral numerically, we have f A = 0.066315 [0.4179] rad Ans.

= 0.0277 rad = 1.59°

Ans: f A = 1.59° 460

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R5–9. The 60-mm-diameter shaft rotates at 300 rev>min. This motion is caused by the unequal belt tensions on the pulley of 800 N and 450 N. Determine the power transmitted and the maximum shear stress developed in the shaft. 300 rev/min

100 mm

Solution

450 N

rev 2p rad 1 min v = 300 c d = 10 p rad>s min 1 rev 60 s

800 N

T + 450(0.1) - 800(0.1) = 0 T = 35.0 N # m

Ans.

P = Tv = 35.0(10p) = 1100 W = 1.10 kW tmax =

35.0(0.03) Tc = p = 825 kPa 4 J 2 (0.03 )

Ans.

Ans: P = 1.10 kW, tmax = 825 kPa 461

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6–1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft.

B

A

250 mm

800 mm

24 kN

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

Ans: x = 0.25- , V = -24, M = -6

329 329

CH 06.indd 329

1/18/11 2:21:30 PM

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6–2. The dead-weight loading along the centerline of the airplane wing is shown. If the wing is fixed to the fuselage at A, determine the reactions at A, and then draw the shear and moment diagram for the wing.

3.75 250kN/m lb/ft

15 kN 3000 lb

6 kN/m 400 lb/ft

A

Support Reactions: + c ©Fy = 0;

0.9 3 ftm

kNlb 1575000

–4.5 – 5.625 – 1.6875 – A-y =A0y = 0 1.00– 15 - 3+ 75 + 15 - 1.25 - 0.375

Ay = 48.1875 kN = 48.2 kN a + ©MA = 0;

0.6 m 2 ft

2.4 8 ftm

5.625 kN

Ans.

4.5 kN 15 kN

1.6875 kN 0.75 m

4.5(2.3) + 15(1.5) – 75(0.9) + 1.6875(0.5) ++MM 0 0 + 5.625(0.75) 1.25(2.5) ++0.375(1.667) AA= =

1.6 m

MA = 18.583 29.5875kip kN# ft · m==18.6 29.6kip kN# ·ftm

Ans.

+ ©F = 0; : x

Ans.

Ax = 0

0.5 m 0.4 m

75 kN

V (kN)

Shear and Moment Diagram:

0.6 m 0.8 m

M (kN · m)

53.0

29.6

48.2 2.4 –4.5

3

x (m) 3.9

2.4

3

x (m) 3.9

–3.6

–19.5 – 22.0 –16.0

Ans: Ay = 48.2 kN, Ax = 0 351 351

CH 06.indd 351

1/18/11 2:22:08 PM

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6–3. Draw the shear and moment diagrams for the overhang beam.

8 kN/m

C

A B 4m

2m

Ans: x = 1.5, V = 0, M = 9, x = 4- , V = -20, M = -16 331 331

CH 06.indd 331

1/18/11 2:21:33 PM

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*6–4. Express the shear and moment in terms of x for 0 6 x 6 3 m and 3 m 6 x 6 4.5 m, and then draw the shear and moment diagrams for the simply supported beam.

300 N/m

A

B 3m

1.5 m

Solution Support Reactions: Referring to the free-body diagram of the entire beam shown in Fig. a, a+ ΣMA = 0;   

By(4.5) -

1 1 (300)(3)(2) - (300)(1.5)(3.5) = 0 2 2

By = 375 N + c ΣFy = 0;    Ay + 375 -

1 1 (300)(3) - (300)(1.5) = 0 2 2

Ay = 300 N Shear and Moment Function: For 0 … x 6 3 m, we refer to the free-body diagram of the beam segment shown in Fig. b. + c ΣFy = 0;   300 -

1 (100x)x - V = 0 2

V = {300 - 50x2} N a+ ΣM = 0;

M +

Ans.

1 x (100x)xa b - 300x = 0 2 3

M = e 300x -

50 3 x fN#m 3

Ans.

When V = 0, from the shear function, 0 = 300 - 50x2

x = 26 m

Substituting this result into the moment equation, M|x = 26 m = 489.90 N # m

For 3 m 6 x … 4.5 m, we refer to the free-body diagram of the beam segment shown in Fig. c. + c ΣFy = 0;   V + 375 -

1 3200(4.5 - x) 4(4.5 - x) = 0 2

V = e 100(4.5 - x)2 - 375 f N

a+ ΣM = 0; 375(4.5 - x) -

Ans.

1 4.5 - x [200(4.5 - x)](4.5 - x)a b - M = 0 2 3

M = e 375(4.5 - x) -

100 (4.5 - x)3 f N # m  Ans. 3

Shear and Moment Diagrams: As shown in Figs. d and e.

Ans: For 0 … x 6 3 m, V = {300 - 50x2} N, M = e 300x -

50 3 x f N # m, 3

For 3 m 6 x … 4.5 m, V = e 100(4.5 - x)2 - 375 f N, M = e 375(4.5 - x) -

464

100 (4.5 - x)3 f N # m 3

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6–5. Draw the shear and moment diagrams for the simply supported beam.

4 kN M � 2 kN�m A

B 2m

2m

2m

Ans: x = 2- , V = 1, M = 2, x = 4- , V = 1, M = 6 329 329

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6–6. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Also, express the shear and moment in the shaft as a function of x within the region 125 mm 6 x 6 725 mm.

1500 N 800 N A

B

x 125 mm

600 mm 75 mm

Solution + c ΣFy = 0;

815.63 - 800 - V = 0 Ans.

V = 15.6 N a+ ΣM = 0;

M + 800(x - 0.125) - 815.63 x = 0 M = (15.6x + 100) N # m

Ans.

Ans: V = 15.6 N, M = (15.6x + 100) N # m 466

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6–7. Express the internal shear and moment in terms of x for 0 … x 6 L>2, and L>2 6 x 6 L, and then draw the shear and moment diagrams.

w0

A

B x L 2

Solution Support Reactions: Referring to the free-body diagram of the entire beam shown in Fig. a, a+ ΣMA = 0;  By(L) By = + c ΣFy = 0;   Ay + Ay =

1 L 5 w a ba L b = 0 2 0 2 6

5 wL 24 0

L 5 1 w L - w0 a b = 0 24 0 2 2 w0L 24

Shear and Moment Function: For 0 … x 6

L , we refer to the free-body diagram of 2

the beam segment shown in Fig. b. + c ΣFy = 0;   

w0L - V = 0 24 V =

M a+ ΣM = 0;    M =

w0L 24

Ans.

w0L x = 0 24 w0L x 24

Ans.

L 6 x … L, we refer to the free-body diagram of the beam segment shown in 2 Fig. c.

For

w0L 1 w0 1 + c ΣFy = 0;    - c (2x - L) d c (2x - L) d - V = 0 24 2 L 2 V = a+ ΣM = 0;    M + M =

w0 c L2 - 6(2x - L)2 d 24L

Ans.

w0 1 w0 1 1 c (2x - L) d c (2x - L) d c (2x - L) d x = 0 2 L 2 6 24L w0 c L2x - (2x - L)3 d 24L

Ans.

467

L 2

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6–7. Continued

When V = 0, the shear function gives 0 = L2 - 6(2x - L)2  x = 0.7041L Substituting this result into the moment equation, M|x = 0.7041L = 0.0265w0L2 Shear and Moment Diagrams: As shown in Figs. d and e.

Ans:

w0L w0L L :V = ,M = x, 2 24 24 w0 L For 6 x … L: V = 3 L2 - 6(2x - L)2 4 , 2 24L w0 M = 3 L2x - (2x - L)3 4 24L For 0 … x 6

V

w0L

0.704 L

24 0

0.0208 w0L2

5 wL 0 24

0.0265 w0L2

0.5 L 0.704 L

468

x

0.5 L



M

L

L

x

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*6–8. Draw the shear and moment diagrams for the beam. Hint: The 100-kN load must be replaced by equivalent loadings at point C on the axis of the beam.

15 kN kip 75 20 kip 100 kN

1 ftm 0.25

A 75 kN

ft 14 m

100 kN 0.25 m

1m

1m

C ft 14 m

B ft 14 m

1m

75 kN 25 kN · m

100 kN 100 kN

58.33 kN

16.67 kN

V (kN) 58.3 2

3

1

x (m)

–16.67 M (kN-m)

58.3

41.7 16.7

1

2

3

x (m)

333 333

CH 06.indd 333

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6–9. The load binder is used to support a load. If the force applied to the handle is 225 N, determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC.

T1

A

C B 225 N 300 mm 75 mm

a + ©MC = 0;

225(375) - T1 (75) = 0 Ans.

T1 = 1125 N + T ©Fy = 0;

T2

225 - 1125 + T2 = 0 T2 = 900 N

Ans.

-

Ans: T1 = 1125 N, T2 = 900 N V (N )

900 x ⫺225

M (N⭈m )

x

⫺67 . 5

463

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6–10. Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.

P

Support Reactions:

P

A

D

B

C

From the FBD of segment BD a + ©MC = 0; + c ©Fy = 0; + ©F = 0; : x

By (a) - P(a) = 0 Cy - P - P = 0

By = P

a

a

a

a

Cy = 2P

Bx = 0

From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;

P(2a) - P(a) - MA = 0

MA = Pa

P - P = 0 (equilibrium is statisfied!)

Ans: x = 3a- , V = -P, M = -Pa 335 335

CH 06.indd 335

1/18/11 2:21:47 PM

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6–11. The engine crane is used to support the engine, which has a weightof 6 kN. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown.

A

3 ftm 0.9

5 ftm 1.5 B

C

ft 1.24 m

a + ©MA = 0;

4 6(2.4) = = 0 0; FB(0.9) (3) -– 1200(8) 5 A

+ c ©Fy = 0;

4 -Ay + (20) (4000) – 6-= 1200 0; = 0; 5

+ ©F = 0; ; x

Ax -

3 (4000) = 0; (20) = 0; 5

F kN lb FBA==204000

0.9 m

1.5 m

V (kN) 6

Ayy==102000 A kN lb

FB

6 kN –10

0.9

x (m) 2.4

M (kN . m)

Ax = 2400 Ax =lb12 kN

0.9

2.4

x (m)

–9.00

Ans:

x = 0.9-m, V = -10 kN # x = 0.9+m, V = 6 kN x = 0.9 m, M = - 9.00 kN # m

330 330

CH 06.indd 330

1/18/11 2:21:32 PM

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*6–12. Draw the shear and moment diagrams for the cantilevered beam.

1.5 300kN lb

3 kN/m 200 lb/ft

A A 26m ft

The free-body diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is x x 3 a ab =b1.5x = 33.33x w = 200 2 6

1 (3)(2) kN 2

1.5 kN

4.5 kN

4m 3

2m 3

5 kN · m

Referring to Fig. b, + c ©Fy = 0;

11 2 -300 (33.33x)(x) V = V0 = {–1.5 V = {- 300 - 16.67x2} lb (1) –1.5 – - (1.5x)(x) – V =-0   – 0.75x } kN 22

a+ © M = 0;

1 x {−1.5 x − 0.25 x 3 } kN ⋅ m M + (1.5 x)( x)   + 1.5 x = 0 M= 2 3

1.5 kN

1 (1.5x) x 2

(2)

The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively.

V (kN)

x (m)

–1.5 –4.5

M (kN · m) x (m)

–5

337 337

CH 06.indd 337

1/18/11 2:21:52 PM

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6–13. Draw the shear and moment diagrams for the beam.

M0

M0

M0

A

B

L/2

L/2

Solution Support Reactions: Referring to the FBD of the beam, Fig. a, a+ ΣMA = 0;

NB(L) + M0 - M0 - M0 = 0 NB =

a+ ΣMB = 0;

Ay(L) + M0 - M0 - M0 = 0 Ay =

+ ΣFx = 0; d

Ax = 0.

M0 L M0 L

Shear and Moment Diagram: Using the results of the support reaction, the shear and moment diagrams shown in Fig. b and c respectively can be plotted.

Ans: V = -

M0 , L

M0 L , M = M0 - a bx, L 2 M0 L For 6 x … L, M = - a bx L 2 For 0 … x 6

474

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6–14. Draw the shear and moment diagrams for the beam.

30 kN/m 2 kip/ft 45 kip�ft kNm 30

30 kN/m

B

45 kN · m

A 1.5 m V (kN)

41.25 kN 1.5

0

1.5 m

1.5 m 3

1.5 5 ftm

3.75 kN 4.5 x (m)

1.5 5 ftm

1.5 5 ftm

–3.75 –45

M (kN · m) 0

1.5

–33.75

3

5.625 x (m)

–39.375

Ans:

x = 1.5- m, V = -45 kN # x = 1.5+m, V = - 3.75 kN x = 1.5 m, M = -33.75 kN # m # x = 3 - m M = - 39.375 kN # m x = 3 + m, M = 5.625 kN # m 338 338

CH 06.indd 338

1/18/11 2:21:54 PM

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6–15. Members ABC and BD of the counter chair are rigidly connected at B and the smooth collar at D is allowed to move freely along the vertical slot. Draw the shear and moment diagrams for member ABC. Equations of Equilibrium: Referring to the free-body diagram of the frame shown in Fig. a, + c ©Fy = 0;

 150 750 lb N P�

Ay - 750 150 = 0 150 N lb Ay = 750

0.45 m 1.5 ft

0.45 m 1.5 ft D

ND(0.45) – 750(0.9) = 0

a + ©MA = 0;

B

A

N 300 N lb NDD ==1500

C 0.45 m 1.5 ft

Shear and Moment Diagram: The couple moment acting on B due to ND is

300(1.5) == 450 MB ==1500(0.45) . The 675 lb N #· ftm. Theloading loadingacting acting on on member member ABC ABC is is shown in Fig. b and the shear and moment diagrams are shown in Figs. c and d.

750 N 0.45 m

0.45 m

0.45 m

0.45 m

0.45 m V (N)

M(N · m) 337.5

750 x (m) 0.45

0.9

0.45 –337.5

x (m)

750 N

0.9 Ay = 750 N

1500 N

MB = 675 N· m

Ans: 0 … x … 0.9 m, V = 750 N # x = 0.45-m, M = - 337.5 N # m # x = 0.45 + m, M = 337.5 N # m

333 333

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*6–16. A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and moment diagrams for the pier. Assume the columns at A and B exert only vertical reactions on the pier.

60 kN 60 kN 35 kN 35 kN 35 kN 1 m 1 m 1.5 m 1.5 m 1 m 1 m

A

B

Solution

Ans: N/A 477

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6–17. Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition.

w

A

B a

wL2 wL - wx = 0 2a

+ c ©Fy = 0;

x = L -

L

L2 2a

wL2 x Mmax (+) + wx a b - a wL bx = 0 2 2a

a + ©M = 0;

Substitute x = L -

L2 ; 2a

Mmax (+) = a wL =

L2 w L2 2 wL2 b aL b aL b 2a 2a 2 2a

w L2 2 aL b 2 2a

Mmax (-) - w(L - a)

©M = 0;

Mmax (-) =

(L - a) = 0 2

w(L - a)2 2

To get absolute minimum moment, Mmax (+) = Mmax (-) w L2 2 w (L ) = (L - a)2 2 2a 2 L a =

L2 = L - a 2a L 22

Ans.

‚

Ans: L 12 x = 0, V = 0.243 wL # x = 0.243 L, V = 0 # x = 0.707 L- , V = - 0.414 wL x = 0.707 L+ , V = 0.293 wL x = 0.243 L, M = 0.0429 wL2 # x = 0.707 L, M = -0.0429 wL2

a =

342 342

CH 06.indd 342

1/18/11 2:21:57 PM

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6–18. The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam.

2 kN/m

Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, a + ©MA = 0;

3 FBC a b (2) - 2(3)(1.5) = 0 5

B

A 1.5 m

FBC = 7.5 kN + c ©Fy = 0;

C

3 Ay + 7.5 a b - 2(3) = 0 5

2m

1m

Ay = 1.5 kN

3 Shear and Moment Diagram: The vertical component of FBC is A FBC B y = 7.5a b 5 = 4.5 kN. The shear and moment diagrams are shown in Figs. c and d.

Ans: x = 0.75, V = 0, M = 0.5625, FBC = 7.5 kN, Ay = 1.5 kN 340 340

CH 06.indd 340

1/18/11 2:21:55 PM

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6–19. w0

Draw the shear and moment diagrams for the beam.

B

A L 3

L 3

L 3

Solution Support Reactions: As shown on FBD. Shear and Moment Diagram: Shear and moment at x = L>3 can be determined using the method of sections. w0 L w0 L w0 L + c ΣFy = 0; - V = 0 V = 3 6 6 w0 L L w0 L L a + ΣMNA = 0; M + a b a b = 0 6 9 3 3 M =

5w0 L2 54

Ans: VA = Mmax 489

w0 L , 3 23 w0 L2 = 216

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*6–20. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kN>m caused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and draw the shear and moment diagram for the pin.

0.4 kN/m

C

A

B

w0

w0

20 mm 60 mm 20 mm

Solution + c ΣFy = 0;

1 2(w0)(20)a b - 60(0.4) = 0 2

Ans.

w0 = 1.2 kN>m

Ans: w0 = 1.2 kN>m 493

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6–21. Draw the shear and moment diagrams for the compound beam.

5 kN 3 kN/m A

B 3m

3m

D

C 1.5 m

1.5 m

Ans: x = 3- , V = -11.5, M = -21 349 349

CH 06.indd 349

1/18/11 2:22:05 PM

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6–22. Draw the shear and moment diagrams for the simply supported beam.

10 kN

10 kN

15 kN�m A

B 2m

2m

2m

Ans:

x = 0, V = 7.5 kN # x = 2-m, V = 7.5 kN x = 2 + m, V = -2.5 kN x = 4 + m V = - 12.5 kN x = 2 m, M = 15 kN # m # x = 4 m, M = 10 kN # m # x = 6 m, M = - 15 kN # m 352 352

CH 06.indd 352

1/18/11 2:22:09 PM

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6–23. Draw the shear and moment diagrams for the beam.

18 kN/ m 12 kN/ m

A

B 3m

Solution

490

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*6–24. The footing supports the load transmitted by the two columns. Draw the shear and moment diagrams for the footing if the reaction of soil pressure on the footing is assumed to be uniform.

60 kN

60 kN

2m

0

2

4

30 kN.m 0

2

6

8

4m

2m

x(m)

30 kN.m 4

6

8

x(m)

Ans: V (kN) 30

30 2

6

8

6

8

4 ⫺30

x (m)

⫺30

M (kzN⭈m) 30 0

484

30 2

4

x (m)

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6–25. Draw the shear and moment diagrams for the overhanging beam.

45 kN/m

A

Support Reactions: Referring to the free-body diagram of the beam shown in Fig. a, a + ©MA = 0;

+ c ©Fy = 0;

1 By(6) (45)(4)(2.67) = 0 2 By = 40.0 kN

1 0 Ay + 40.0 − (45)(4) = 2 Ay = 50.0 kN

Shear and Moment Functions: For 0 … x 6 4 m, we refer to the free-body diagram of the beam segment shown in Fig. b. + c ©Fy = 0;

1 50.0 − (11.25 x)( x) − V = 0 2 V {50.0 − 5.625 x 2 } kN =

a + ©M = 0;

Ans.

1 x M + (11.25 x)( x)   − 50.0 x = 0 2 3 M= {50.0x − 1.875 x 3 } kN ⋅ m

Ans.

When V = 0, from the shear function,

= 0 50.0 − 5.625 x 2

x = 2.981 m

Substituting this result into the moment function,

M x= 99.38 kN ⋅ m =2.981 m For 4 m 6 x … 6 m, we refer to the free-body diagram of the beam segment shown in Fig. c. + c ©Fy = 0;

V + 40.0 kN = 0 V =

Ans.

− 40.0 kN

a + ©M = 0; 40.05(6 - x) - M = 0

M = {40.0(6 − x)} kN ⋅ m

Ans.

Shear and Moment Diagrams: As shown in Figs. d and e.

481

B 4m

2m

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6–26. The support at A allows the beam to slide freely along the vertical guide so that it cannot support a vertical force. Draw the shear and moment diagrams for the beam.

w B

A

L

Solution Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, L a+ ΣMB = 0;   wL a b - MA = 0 2 MA =

wL2 2

+ c ΣFy = 0;    By - wL = 0 By = wL Shear and Moment Diagram: As shown in Figs. b and c.

Ans: V L

x

wL M 2

wL 2

L

492

x

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6–27. Draw the shear and moment diagrams for the beam.

M0

M0

M0

A

B

a

Solution

481

a

a

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*6–28. Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition.

P

P

L – 2

L – 2

A

B a

Solution Support Reactions: As shown on FBD. Absolute Minimum Moment: In order to get the absolute minimum moment, the maximum positive and maximum negative moment must be equal that is Mmax( + ) = Mmax( - ). For the positive moment: a+ ΣMNA = 0;   Mmax( + ) - a2P -

3PL L ba b = 0 2a 2

Mmax( + ) = PL -

3PL2 4a

For the negative moment: a+ ΣMNA = 0;   Mmax( - ) - P(L - a) = 0 Mmax( - ) = P(L - a) Mmax( + ) = Mmax( - ) PL -

3PL2 = P(L - a) 4a

4a L - 3L2 = 4a L - 4a2 a =

23 L = 0.866L 2

Ans.

Shear and Moment Diagram:

Ans: a = 0.866L 480

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6–29. Draw the shear and moment diagrams for the beam.

w0

B L 3

A

2L 3

Solution + c ΣFy = 0;

w0L 1 w0x - a b(x) = 0 4 2 L x = 0.7071 L

a+ ΣMNA = 0;

M +

Substitute x = 0.7071L,

w0L 1 w0x x L a ax - b = 0 b(x)a b 2 L 3 4 3

M = 0.0345 w0L2

Ans: V 7w0L 36 x w0L 18

w0L 4

0.707 L M

0.0345w0L2

x 2

0.00617w0L

488

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6–30. The 700-N man sits in the center of the boat, which has a uniform width and a weight per linear foot of 450 N/m. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a uniform distributed load upward on the bottom of the boat. 2.25 m

2.25 m

450 N/m 605.55 N/m

155.55 N/m 394 2.25

0

4.50 x(m)

394 0

x(m) 2.25

4.50

Ans.

Mmax = 39 4 N # m

Ans: Mmax = 39 4 N # m V (N ) 3 50 x ⫺3 50 M ( N m)

#

3394 9 3 . 0

482

x

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6–31. Draw the shear and moment diagrams for the overhang beam.

18 kN 6 kN

A B 2m

2m

M � 10 kN�m

2m

Ans:

x = 0, V = 3.5 kN # x = 2+ m, V = -14.5 kN # x = 4 + m, V = 6 kN x = 2 m, M = 7 kN # m x = 4 m, M = -22 kN # m # x = 6 m M = 10 kN # m 350 350

CH 06.indd 350

1/18/11 2:22:06 PM

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*6–32. Draw the shear and moment diagrams for the beam.

5 kN/m

5 kN/m

B

A 4.5 m

4.5 m

From FBD(a) + c ©Fy = 0; a + ©MNA = 0;

9.375 - 0.5556x2 = 0

x = 4.108 m

M + (0.5556) A 4.1082 B a M = 25.67 kN # m

4.108 b - 9.375(4.108) = 0 3

From FBD(b) a + ©MNA = 0;

M + 11.25(1.5) - 9.375(4.5) = 0 M = 25.31 kN # m

345 345

CH 06.indd 345

1/18/11 2:21:59 PM

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6–33. The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed loading on the beam over its 0.6-m length. Draw the shear and moment diagrams for the beam if it supports a uniform loading of 30 kN/m.

30 kN/m

B A

2.4 m

0.6 m

0.3 m

0 0.3

2.7

3.3

2.7

3.3

1.5

x(m)

32.4

0 0.3

1.5

x(m)

Ans: V (kN) 36 3 6 0

0.3

M (kN m) M

1.5

490

3.3

x (m)

36 6 32.4 5

10 . 10.8 0.3

2.7

10.8 1 0 . 1.5

2.7

3.3

x (m)

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6–34. Draw the shear and moment diagrams for the simply supported beam.

10 kN

10 kN

15 kNm A

B 2m

Solution

501

2m

2m

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6–35. A short link at B is used to connect beams AB and BC to form the compound beam. Draw the shear and moment diagrams for the beam if the supports at A and C are considered fixed and pinned, respectively.

15 kN 3 kN/m

B

A

4.5 m

C

1.5 m

1.5 m

Solution Support Reactions: Referring to the free-body diagram of segment BC shown in Fig. a, a+ΣMC = 0;   15(1.5) - FB(3) = 0 FB = 7.5 kN + c ΣFy = 0;    Cy + 7.5 - 15 = 0 Cy = 7.5 kN Using the result of FB and referring to the free-body diagram of segment AB, Fig. b, 1 + c ΣFy = 0;    Ay - (3)(4.5) - 7.5 = 0 2 Ay = 14.25 kN 1 (3)(4.5)(3) - 7.5(4.5) = 0 2 MA = 54 kN # m

a+ΣMA = 0;    ­­ MA -

Shear and Moment Diagrams: As shown in Figs. c and d.

Ans: V (kN) 14.25 0

7.5 6

7.5

4.5

x (m)

7.5

M (kNm) 11.25 0

54

506

4.5

6

7.5

x (m)

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*6–36. The compound beam is fixed at A, pin connected at B, and supported by a roller at C. Draw the shear and moment diagrams for the beam.

2 kN 3 kN/m

C

B

A 3m

3m

Solution Support Reactions: Referring to the free-body diagram of segment BC shown in Fig. a, a+ΣMB = 0;   Cy(3) - 3(3)(1.5) = 0 Cy = 4.5 kN

c ΣFy = 0;   

By + 4.5 - 3(3) = 0 By = 4.5 kN

Using the result of By and referring to the free-body diagram of segment AB, Fig. b,

c ΣFy = 0;   

Ay - 2 - 4.5 = 0 Ay = 6.5 kN

a+ΣMA = 0;   MA - 2(3) - 4.5(3) = 0 MA = 19.5 kN # m

Shear and Moment Diagrams: As shown in Figs. c and d.

Ans: V (kN) 6.5 0

4.5 6 3

4.5

x (m)

4.5 M (kNm) 3.375 0

19.5

504

3

4.5

6

x (m)

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6–37. Draw the shear compound beam.

and

moment

diagrams

for

the

5 kN/m

A

B 2m

D

C 1m

1m

Solution Support Reactions: From the FBD of segment AB a+ ΣMA = 0;

By (2) - 10.0(1) = 0

By = 5.00 kN

+ c ΣFy = 0;

Ay - 10.0 + 5.00 = 0

Ay = 5.00 kN

From the FBD of segment BD a+ ΣMC = 0;

5.00(1) + 10.0(0) - Dy (1) = 0 Dy = 5.00 kN

+ c ΣFy = 0;

Cy - 5.00 - 5.00 - 10.0 = 0  Cy = 20.0 kN

+ ΣFx = 0;        Bx = 0 S From the FBD of segment AB + ΣFx = 0;     Ax = 0 S Shear and Moment Diagram:

Ans: V (kN) 5.00 0

2 1 5.00

3 10.0

503

M (kNm)

10.0 5.00 x (m) 4

2.50 0

1

3

4

2 7.50

x (m)

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6–38. The compound beam is fixed at A, pin connected at B, and supported by a roller at C. Draw the shear and moment diagrams for the beam.

600 N 400 N/m

A

C

B 2m

2m

2m

Solution Support Reactions: Referring to the free-body diagram of segment BC shown in Fig. a, a+ΣMB = 0;   Cy(2) - 400(2)(1) = 0 Cy = 400 N + c ΣFy = 0;    By + 400 - 400(2) = 0 By = 400 N Using the result of By and referring to the free-body diagram of segment AB, Fig. b, + c ΣFy = 0;    Ay - 600 - 400 = 0 Ay = 1000 N a+ΣMA = 0;   MA - 600(2) - 400(4) = 0 MA = 2800 N Shear and Moment Diagrams: As shown in Figs. c and d.

Ans: V (N) 1000 400 0

2

5

6

4

x (m)

400 M (Nm)

0

4 800

2800

502

200

2

5

6

x (m)

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6–39. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at B. Draw the shear and moment diagrams for the shaft.

400 Nm

B

A

1m

1m

1m

900 N

Solution Equations of Equilibrium: Referring to the free-body diagram of the shaft shown in Fig. a, a+ ΣMA = 0;   By(2) + 400 - 900(1) = 0 By = 250 N Ay + 250 - 900 = 0 + c ΣFy = 0;      Ay = 650 N Shear and Moment Diagram: As shown in Figs. b and c.

Ans: V (N) 650 1

0

2

3

x (m)

250

M (Nm) 650 0

494

1

400 2

3

x (m)

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*6–40. The beam is used to support a uniform load along CD due to the 6-kN weight of the crate. Also, the reaction at the bearing support B can be assumed uniformly distributed along its width. Draw the shear and moment diagrams for the beam.

0.5 m

0.75 m

2.75 m

2m C

A

D

B

Solution Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, a+ ΣMA = 0;

FB(3) - 6(5) = 0 FB = 10 kN

+ c ΣFy = 0;

10 - 6 - Ay = 0 Ay = 4 kN

Shear and Moment Diagram: The intensity of the distributed load at support B and FB 10 6 portion CD of the beam are wB = = = 20 kN>m and wCD = = 3 kN>m, 0.5 0.5 2 Fig. b. The shear and moment diagrams are shown in Figs. c and d.

Ans: V (kN)

0

2.95 2.75

6

3.25 4

6

3.25 4

6

x (m)

4

M (kNm)

0

2.95 2.75

6 11

499

10.5 11.4

x (m)

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6–41. Draw the shear and moment diagrams for the beam.

50 kN/m

50 kN/m

B A 4.5 m

4.5 m

Solution

Ans: V (kN) 112.5 x 112.5 M (kNm) 169 x

498

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6–42. Draw the shear cantilever beam.

and

moment

diagrams

for

the

2 kN

A

3 kNm 1.5 m

1.5 m

Solution Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, + c ΣFy = 0;   Ay - 2 = 0 Ay = 2 kN a+ ΣMA = 0;

MA - 3 - 2(3) = 0 MA = 9 kN # m

Shear and Moment Diagram: As shown in Figs. b and c.

Ans: V (kN) 2 0

1.5

3

1.5

3

x (m)

M (kNm)

0

3 6 9

495

x (m)

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6–43. Draw the shear and moment diagrams for the beam.

400 N/m 200 N/ m A

B x 3m

3m

Solution Support Reactions: As shown on FBD. Shear and Moment Functions: For 0 … x 6 3 m: + c ΣFy = 0;

V = 200 N

Ans.

M = {200 x} N # m

Ans.

200 - V = 0

a+ ΣMNA = 0;

M - 200 x = 0

For 3 m 6 x … 6 m:

+ c ΣFy = 0;

200 - 200(x - 3) V = e-

Set V = 0, x = 3.873 m a+ ΣMNA = 0;

M +

1 200 c (x - 3) d (x - 3) - V = 0 2 3

100 2 x + 500 f N 3

Ans.

x - 3 1 200 c (x - 3) d (x - 3)a b 2 3 3

+ 200(x - 3)a M = e-

x - 3 b - 200x = 0 2

100 3 x + 500x - 600 f N # m 9

Ans.

Substitute x = 3.87 m, M = 691 N # m

Ans: For 0 … x 6 3 m: V = 200 N, M = {200x} N # m, For 3 m 6 x … 6 m: V = e -

M = e-

100 2 x + 500 f N, 3

100 3 x + 500x - 600 f N # m 9 M (Nm)

V (N)

600 691 200 0

3.87

6

3 700

496

x (m)

0

3 3.87

6

x (m)

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*6–44. Draw the shear and moment diagrams for the beam.

w

w0

p w � w0 sin – x L

A

B L – 2

FR =

LA

dA = w0

L0

L

sin a

x

L – 2

2w0 L p x bdx = p L

355 355

CH 06.indd 355

1/18/11 2:22:13 PM

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6–45. Draw the shear and moment diagrams for the beam.

w

w�

L

A

L

w0 L dA = wdx = 2 x2 dx = FR = 3 L L0 LA L0 w0

w0

w0 2 x L2

w0

B

x

L

L

x3dx L2 L0 3L x = = = w0 L 4 dA 3 LA LA

xdA

+ c ©Fy = 0;

w0L w0x3 = 0 12 3L2

1 1>3 x = a b L = 0.630 L 4 w0L w0x3 1 a + ©M = 0; (x) a xb - M = 0 12 3L2 4 M =

w0Lx w0x4 12 12L2

Substitute x = 0.630L M = 0.0394 w0L2

Ans: x = 0.630 L, V = 0, M = 0.0394w0L2, M =

w0Lx w0x4 12 12L2

354 354

CH 06.indd 354

1/18/11 2:22:13 PM

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6–46. The truck is to be used to transport the concrete column. If the column has a uniform weight of w (force/length), determine the equal placement a of the supports from the ends so that the absolute maximum bending moment in the column is as small as possible. Also, draw the shear and moment diagrams for the column.

L a

a

Solution Support Reactions: As shown on FBD. Absolute Minimum Moment: In order to get the absolute minimum moment, the maximum positive and maximum negative moment must be equal that is Mmax(+) = Mmin(-). For the positive moment: a+ΣMNA = 0;   Mmax(+) +

wL L wL L a b a - ab = 0 2 4 2 2

Mmax(+) =

wL2 waL 8 2

For the negative moment: a a+ΣMNA = 0;   wa a b - Mmax(-) = 0 2 Mmax( - ) =

wa2 2

Mmax(+) = Mmax(-) wL2 wL wa2 a = 8 2 2 4a2 + 4La - L2 = 0 a =

- 4L { 216L2 - 4(4)( - L2) 2(4)

Ans.

a = 0.207L Shear and Moment Diagram:

Ans: a = 0.207L 507

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6–47. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution and cover the cross section.

25 mm

150 mm 20 mm 200 mm M  600 Nm 20 mm

Solution y = I =

(0.0125)(0.24)(0.025) + 2(0.1)(0.15)(0.02) 0.24(0.025) + 2(0.15)(0.02)

= 0.05625 m

1 (0.24) ( 0.0253 ) + (0.24)(0.025) ( 0.043752 ) 12 +2a

1 b(0.02) ( 0.153 ) + 2(0.15)(0.02) ( 0.043752 ) 12

= 34.53125 ( 10-6 ) m4 smax = sB = =

Mc I 600(0.175 - 0.05625) 34.53125 ( 10-6 ) Ans.

= 2.06 MPa sc =

600(0.05625) My = = 0.977 MPa I 34.53125 ( 10-6 )

Ans: smax = 2.06 MPa 513

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*6–48. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the resultant force the bending stress produces on the top board.

25 mm

150 mm 20 mm 200 mm M  600 Nm 20 mm

Solution y = I =

(0.0125)(0.24)(0.025) + 2(0.15)(0.1)(0.02) 0.24(0.025) + 2(0.15)(0.02)

= 0.05625 m

1 (0.24) ( 0.0253 ) + (0.24)(0.025) ( 0.043752 ) 12 + 2a

1 b(0.02) ( 0.153 ) + 2(0.15)(0.02) ( 0.043752 ) 12

= 34.53125 ( 10-6 ) m4 st =

600(0.05625) My = = 0.9774 MPa I 34.53125 ( 10-6 )

sb =

600(0.05625 - 0.025) My = = 0.5430 MPa I 34.53125 ( 10-6 )

F =

1 (0.025)(0.9774 + 0.5430) ( 106 ) (0.240) = 4.56 kN 2

Ans.

Ans: F = 4.56 kN 514

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6–49. Determine the moment M that will produce a 70 ksi MPa the cross-section. maximum stress of 10 onon the cross section.

0.5mm in. 12 A 12 0.5mm in.

753 mm in.

12 mm 0.5 in. B

C

753 mm in. M 25010mm in.

D 12 mm 0.5 in.

Section Properties: y =

=

©yA ©A

0.006(0.099)(0.012) + 2[0.0495(0.012)(0.075)] + 0.137(0.012)(0.25) = 0.08471 m 0.099(0.012) + 2(0.012)(0.075) + 0.012(0.25)

1 I NA =(0.099)(0.012 3 ) + (0.099)(0.012)(0.08471 − 0.006)2 12 1  + 2  (0.012)(0.0753 ) + (0.012)(0.075)(0.08471 − 0.0495)2   12  +

1 (0.012)(0.253 ) + (0.012)(0.25)(0.137 − 0.08471)2 12

= 34.2773(10 −6 ) m 4 Maximum Bending Stress: Applying the flexure formula smax =

70(106 ) =

Mc I

M (0.262 − 0.08471) 34.2773(10 −6 )

⋅ m 13.5 kN ⋅ m = M 13.53(10 3 ) N=

Ans.

Ans:

M = 13.5 kN # m 356 356

CH 06.indd 356

1/18/11 2:22:14 PM

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6–50. Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of # ft. M ==64kN kip· m.

0.5mm in. 12 A 12 0.5mm in.

753mm in.

12 0.5mm in. B

C

753 mm in. M 25010 mm in.

D 12 0.5mm in.

Section Properties: y =

=

©yA ©A

0.006(0.099)(0.012) + 2[0.0495(0.012)(0.075)] + 0.137(0.012)(0.25) = 0.08471 m 0.099(0.012) + 2(0.012)(0.075) + 0.012(0.25)

1 I NA =(0.099)(0.012 3 ) + (0.099)(0.012)(0.08471 − 0.006)2 12 1  + 2  (0.012)(0.0753 ) + (0.012)(0.075)(0.08471 − 0.0495)2    12 +

1 (0.012)(0.253 ) + (0.012)(0.25)(0.137 − 0.08471)2 12

= 34.2773(10 −6 ) m 4

Maximum Bending Stress: Applying the flexure formula smax =

Mc I

= (σ t )max

[6(10 3 )](0.262 − 0.08471) 6 = 31.03(10 = ) N−m 2 31.0 MPa 34.2773(10 −6 )

Ans.

= (σ c )max

[6(10 3 )](0.08471) 6 = 14.83(10 = ) N−m 2 14.8 MPa 34.2773(10 −6 )

Ans.

Ans: y = 0.08471 m, INA = 34.2773 (10 - 6) m4, (st) max = 31.0 MPa, (sc) max = 14.8 MPa 357 357

CH 06.indd 357

1/18/11 2:22:15 PM

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6–51. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards, A and B, of the beam.

A 25 mm

I =

1 1 (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4 12 12

150 mm 25 mm 25 mm

Bending Stress: Applying the flexure formula s =

M

D

Section Property:

B 150 mm

25 mm

My I

sE =

sD =

M(0.1) 91.14583(10 - 6) M(0.075) 91.14583(10 - 6)

= 1097.143 M

= 822.857 M

Resultant Force and Moment: For board A or B F = 822.857M(0.025)(0.2) +

1 (1097.143M - 822.857M)(0.025)(0.2) 2

= 4.800 M M¿ = F(0.17619) = 4.80M(0.17619) = 0.8457 M sc a

M¿ b = 0.8457(100%) = 84.6 % M

Ans.

Ans: 84.6, 359 359

CH 06.indd 359

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*6–52. Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of sD = 30 MPa . Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam.

A 25 mm M

D 150 mm 25 mm 25 mm

B 150 mm

25 mm

Section Property: I =

1 1 (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10 - 6 B m4 12 12

Bending Stress: Applying the flexure formula s = 30 A 106 B =

My I M(0.075) 91.14583(10 - 6)

M = 36458 N # m = 36.5 kN # m

smax =

Ans.

36458(0.1) Mc = 40.0 MPa = I 91.14583(10 - 6)

Ans.

Ans: M = 36.5 kN # m, smax = 40.0 MPa 359 359

CH 06.indd 359

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6–53. An A-36 steel strip has an allowable bending stress of 165 MPa. If it is rolled up, determine the smallest radius r of the spool if the strip has a width of 10 mm and a thickness of 1.5 mm. Also, find the corresponding maximum internal moment developed in the strip.

r

Solution Bending Stress-Curvature Relation: sallow =

200(109) 30.75(10 - 3) 4 Ec ;   165(106) = r r Ans.

r = 0.9091 m = 909 mm Moment Curvature Relation: 1 M = ; r EI

1 = 0.9091

M 200(109) c

1 (1)(0.00153) d 12

M = 61.875 N # m = 61.9 N # m

Ans.

Ans: r = 909 mm, M = 61.9 N # m 508

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6–54. If the beam is subjected to an internal moment of M = 30 kN # m, determine the maximum bending stress in the beam. The beam is made from A992 steel. Sketch the bending stress distribution on the cross section.

50 mm 50 mm 15 mm A

10 mm

M

150 mm

15 mm

Section Properties: The moment of inertia of the cross section about the neutral axis is I =

1 1 (0.1)(0.153) (0.09)(0.123) = 15.165(10 - 6) m4 12 12

Maximum Bending Stress: The maximum bending stress occurs at the top and bottom surfaces of the beam since they are located at the furthest distance from the neutral axis. Thus, c = 75 mm = 0.075 m. smax =

30(103)(0.075) Mc = 148 MPa = I 15.165(10 - 6)

Ans.

At y = 60 mm = 0.06 m, s|y = 0.06 m =

My 30(103)(0.06) = 119 MPa = I 15.165(10 - 6)

The bending stress distribution across the cross section is shown in Fig. a.

Ans: s max = 148 MPa 513

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6–55. If the beam is subjected to an internal moment of M = 30 kN . m, determine the resultant force caused by the bending stress distribution acting on the top flange A.

50 mm 50 mm 15 mm A

10 mm

M

150 mm

15 mm

Section Properties: The moment of inertia of the cross section about the neutral axis is 1 1 (0.1)(0.153) (0.09)(0.123) = 15.165(10 - 6) m4 12 12

I =

Bending Stress: The distance from the neutral axis to the top and bottom surfaces of flange A is yt = 75 mm = 0.075 m and yb = 60 mm = 0.06 m. st =

Myt 30(103)(0.075) = 148.37 = 148 MPa = I 15.165(10 - 6)

sb =

Myb 30(103)(0.06) = 118.69 = 119 MPa = I 15.165(10 - 6)

Resultant Force: The resultant force acting on flange A is equal to the volume of the trapezoidal stress block shown in Fig. a. Thus, FR =

1 (148.37 + 118.69)(106)(0.1)(0.015) 2 Ans.

= 200 296.74 N = 200 kN

Ans: FR = 200 kN 514

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*6–56. If the built-up beam is subjected to an internal moment of M = 75 kN # m, determine the maximum tensile and compressive stress acting in the beam.

150 mm 20 mm 150 mm 10 mm

150 mm

M

10 mm

Solution Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is y =

300 mm A

0.15(0.3)(0.02) + 23 0.225(0.15)(0.01)] + 230.295(0.01)(0.14) 4 Σy~A = = 0.2035 m ΣA 0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)

Thus, the moment of inertia of the cross section about the neutral axis is I = ΣI + Ad 2 =

1 (0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2 12 + 2c + 2c

1 (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d 12 1 (0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d 12

= 92.6509(10 - 6) m4

Maximum Bending Stress: The maximum compressive and tensile stress occurs at the top and bottom-most fiber of the cross section.   (smax)c =

75(103)(0.3 - 0.2035) My = = 78.1 MPa I 92.6509(10 - 6)

Ans.

  (smax)t =

75(103)(0.2035) Mc = = 165 MPa I 92.6509(10 - 6)

Ans.

Ans: (smax)c = 78.1 MPa, (smax)t = 165 MPa 515

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6–57. If the built-up beam is subjected to an internal moment of M = 75 kN # m, determine the amount of this internal moment resisted by plate A.

150 mm 20 mm 150 mm 10 mm

150 mm

M

10 mm

Solution

300 mm

Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is

y =

A

~A 0.15(0.3)(0.02) + 23 0.225(0.15)(0.01)] + 230.295(0.01)(0.14) 4 Σy = = 0.2035 m ΣA 0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)

Thus, the moment of inertia of the cross section about the neutral axis is I = I + Ad 2 =

1 (0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2 12

+ 2c + 2c

1 (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d 12 1 (0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d 12

= 92.6509(10 - 6) m4

Bending Stress: The distance from the neutral axis to the top and bottom of plate A is yt = 0.3 - 0.2035 = 0.0965 m and yb = 0.2035 m. st =

75(103)(0.0965) Myt = = 78.14 MPa (C) I 92.6509(10 - 6)

sb =

75(103)(0.2035) Myb = = 164.71 MPa (T) I 92.6509(10 - 6)

The bending stress distribution across the cross section of plate A is shown in Fig. b. The resultant forces of the tensile and compressive triangular stress blocks are­­

(FR)t =

1 (164.71)(106)(0.2035)(0.02) = 335 144.46 N 2

(FR)c =

1 (78.14)(106)(0.0965)(0.02) = 75 421.50 N 2

Thus, the amount of internal moment resisted by plate A is 2 2 M = 335144.46c (0.2035) d + 75421.50c (0.0965) d 3 3 = 50315.65 N # m = 50.3 kN # m

Ans.

Ans: M = 50.3 kN # m 516

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6–58. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards of the beam.

90 mm 20 mm 20 mm 20 mm 20 mm

D

90 mm 20 mm 20 mm

M

100 mm

Solution

100 mm

Section Properties: The moment of inertia of the beam’s cross-section about the neutral axis is 1 1 I = (0.24) ( 0.243 ) (0.18) ( 0.23 ) = 0.15648 ( 10 - 3 ) m4 12 12

20 mm

My Bending Stress: Applying the flexure formula, s = , the bending stress on points I D and E, Fig. a, is sD = sE =

M(0.1) 0.15648 ( 10 - 3 ) M(0.12) 0.15648 ( 10 - 3 )

= 639.06M = 766.87M

Resultant Force and Moment: The resultant of the stress block acting on boards A and B, Fig. a, is F =

1 (639.06M + 766.87M)(0.02)(0.24) = 3.3742M 2

The location of line of action of F is a =

1 2(766.87M) + 639.06M c d (0.02) = 0.010303 m 3 639.06M + 766.87M

Thus, the moment arm of F is

a′ = 2(0.1 + 0.010303) = 0.22061 m Then M′ = Fd = 3.3742M(0.22061) = 0.74438M %a

M′ 0.74438M b = a b(100) = 74.4, M M

Ans.

Ans: M′ = 74.4, M 517

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6–59. Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of sD = 10 MPa. Also sketch the stress distribution acting over the cross section and calculate the maximum stress developed in the beam.

90 mm 20 mm 20 mm 20 mm 20 mm

D

90 mm 20 mm 20 mm

M

100 mm

Solution

100 mm

Section Properties: The moment of inertia of the beam’s cross-section about the neutral axis is I =

20 mm

1 1 (0.24) ( 0.243 ) (0.18) ( 0.23 ) = 0.15648 ( 10 - 3 ) M4 12 12

Bending Stress: Applying the flexure formula, sD = 10 ( 106 ) =

MyD I M(0.1) 0.15648 ( 10 - 3 )

M = 15.648 ( 10 - 3 ) N # m = 15.6 kN # m

Ans.

The maximum bending stress is smax =

[15.648 ( 103 ) ](0.12) MC = = 12.0 ( 106 ) Pa = 12.0 MPa I 0.15648 ( 10 - 3 )

Ans.

The sketch of Bending stress distribution on the beam’s cross-section is shown in Fig. a.

Ans: M = 15.6 kN # m, smax = 12.0 MPa 518

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*6–60. If the beam is subjected to an internal moment of of M = 150 kN · m, determine the maximum tensile and compressive bending stress in the beam.

75 mm 3 in. 75 mm 3 in. in. 1506mm MM 2 in. 50 mm in. 37.1.5 5 mm

Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is

y =

©yA 0.1(0.15)(0.2) − 0.05[π (0.03752 )] 0.10863 m = = ©A 0.15(0.2) − π (0.03752 )

Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2

1 π = (0.15)(0.2 3 ) + 0.15(0.2)(0.10863 − 0.1)2 −  (0.03754 ) + π (0.03752 )(0.10863 − 0.05)2  12 4  = 85.4948(10 −6 ) m 4 Maximum Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section.

(σ max= )T

MC [150(10 3 )](0.10863) 6 190.60(10 = ) N−m 2 191 MPa = = I 85.4948(10 −6 )

Ans.

(σ max= )C

MC [150(10 3 )](0.2 − 0.10863) 2 = = 160.30(106 ) N−m = 160 MPa I 85.4948(10 −6 )

Ans.

Ans:

(σ max )T = 191 MPa, (σ max )C = 160 MPa 362 362

CH 06.indd 362

1/18/11 2:22:25 PM

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6–61. If the beam is made of material having an allowable tensile tensileand and compressive of )(s = allowable compressive stressstress of (sallow 24)tksi allow t = 168 MPa = 154 MPa,determine respectively, and (sallowand the determine maximum )c = (s 22allow ksi,)crespectively, the maximum allowable moment M to that be allowable internal moment internal M that can be applied thecan beam. applied to the beam.

75 mm 3 in. 75 mm 3 in. 1506mm in. MM 50 mm 2 in. 37.1.5 5 mm in.

Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is

y =

©yA 0.1(0.15)(0.2) − 0.05[π (0.03752 )] 0.10863 m = = ©A 0.15(0.2) − π (0.03752 )

Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2

1 π = (0.15)(0.2 3 ) + 0.15(0.2)(0.10863 − 0.1)2 −  (0.03754 ) + π (0.03752 )(0.10863 − 0.05)2  12 4  = 85.4948(10 −6 ) m 4 Allowable Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section. For the top edge, (sallow)c =

My ; I

154(106 ) =

M (0.2 − 0.10863) 85.4948(10 −6 )

M 144.11(10 3 ) N= = ⋅ m 144 kN ⋅ m For the bottom edge,

A smax B t =

Mc ; I

154(106 ) =

M (0.10863) 85.4948(10 −6 )

M 132.22(10 3 ) N= = ⋅ m 132 kN ⋅ m (controls!)

Ans.

Ans:

M = 132 kN # m 363 363

CH 06.indd 363

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6–62. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. If d = 90 mm, determine the absolute maximum bending stress in the beam, and sketch the stress distribution acting over the cross section.

12 kN/m d A

B 3m

1.5 m

Solution

Absolute Maximum Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax = =

Mmax c I 11.34(103)(0.045) p 4

(0.0454) Ans.

= 158 MPa

Ans: smax = 158 MPa 528

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6–63. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Determine its smallest diameter d if the allowable bending stress is sallow = 180 MPa.

12 kN/m d A

B 3m

1.5 m

Solution

Allowable Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 180 ( 106 ) =

Mmax c I 11.34(103) 1 d2 2 p 4

1 d2 24

Ans.

d = 0.08626 m = 86.3 mm

Ans: d = 86.3 mm 529

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*6–64. The pin is used to connect the three links together. Due to wear, the load is distributed over the top and bottom of the pin as shown on the free-body diagram. If the diameter of the pin is 10 mm, determine the maximum bending stress on the cross-sectional area at the center section a–a. For the solution it is first necessary to determine the load intensities w1 and w2.

3600 N

25 mm

w2

a

w2

w1

25 mm a 10 mm 40 mm

1800 N

1800 N

1 = = w2 (0.25) 1800; w2 144(10 3 ) N−m 12 = = w1(0.04) 3600; w1 90(10 3 ) N−m

0.01833 m

= = 33.0 N ⋅ m M 1800(0.01833) = I

1 π (0.0054 ) 0.49087(10 −9 ) m 4 = 4

σ max =

Mc 33.0(0.005) 2 = = 336.14(106 ) N−m = 336 MPa I 0.49087(10 −9 )

Ans.

Ans: smax = 324.45 MPa 535

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6–65. The shaft is supported by a thrust bearing at A and journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress in the shaft.

40 mm A

B 0.75 m

1.5 m 3 kN

D

C

25 mm

0.75 m 3 kN

Solution Shear and Moment Diagrams: As shown in Fig. a. Maximum Moment: Due to symmetry, the maximum moment occurs in region BC of the shaft. Referring to the free-body diagram of the segment shown in Fig. b. Section Properties: The moment of inertia of the cross section about the neutral axis is p I = (0.044 - 0.0254) = 1.7038(10 - 6) m4 4 Absolute Maximum Bending Stress: smax =

2.25(103)(0.04) Mmaxc = = 52.8 MPa I 1.7038(10 - 6)

Ans.

Ans: smax = 52.8 MPa 536

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6–66. Determine the absolute maximum bending stress in the 40-mm-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.

1800 N A

B

300 mm

1350 N

450 mm 375 mm

Mmax 506.25 N ⋅ m = = σ

Mc 506.25(0.02) 2 = = 80.57(106 ) N−m = 80.6 MPa π (0.02 4 ) I 4

V (N)

Ans.

1350

x

M (N.m)

x

Ans: s max = 52.8 MPa 533

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6–63. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is sallow = 154 MPa.

1800 N A 1350 N

B

300 mm 450 mm

375 mm

Mmax 506.25 N ⋅ m = 10.5(106 ) =

(506.25)c

V (N)

π c4 4

c = 0.01612 m

1350

x

= d 2= c 2(0.01612) = 0.03223 m = 32.2 mm

Ans. M (N.m)

x

Ans: d = 32.2 mm 534

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*6–68. A shaft is made of a polymer having an elliptical cross section. If it resists an internal moment of M = 50 N # m, determine the maximum bending stress in the material (a) using the flexure formula, where Iz = 14 p(0.08 m)(0.04 m)3, (b) using integration. Sketch a three-dimensional view of the stress distribution acting over the cross-sectional area. Here Ix = 14 p(0.08 m)(0.04 m)3.

y

y2 z2 ———  ———2  1 2 (40) (80) 80 mm M  50 Nm

z

160 mm

x

Solution (a) I =

1 1 p ab3 = p(0.08)(0.04)3 = 4.021238(10 - 6) m4 4 4 50(0.04) Mc = = 497 kPa I 4.021238(10 - 6)

smax =

Ans.

(b) M = =

smax y2dA c LA smax y22zdy c L

z = 20.0064 - 4y2 = 22(0.04)2 - y2 0.04

2

L- 0.04

0.04

2

y zdy = 4

L- 0.04

= 4c

=

2

y 2(0.04)2 - y2 dy

(0.04)4 8

(0.04)4 2

sin - 1a

sin - 1a

0.04 y 1 b - y2(0.04)2 - y2(0.042 - 2y2) d ` 0.04 8 -0.04

0.04 y b2 0.04 - 0.04

= 4.021238(10 - 6) m4 smax =

50(0.04) 4.021238(10 - 6)

Ans.

= 497 kPa

Ans: (a) smax = 497 kPa, (b) smax = 497 kPa 526

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6–69. Solve Prob. 6–65 if the moment M = 50 N # m is applied about the y axis instead of the x axis. Here Iy = 14 p (0.04 m)(0.08 m)3.

y

y2 z2 ———  ———2  1 2 (40) (80) 80 mm M  50 Nm

z

160 mm

x

Solution (a) I =

1 1 p ab3 = p(0.04)(0.08)3 = 16.085(10 - 6) m4 4 4

smax =

50(0.08) Mc = 249 kPa = I 16.085(10 - 6)

Ans.

(b) M =

LA

50 = 2 a

z(s dA) =

LA

za

smax b(z)(2y)dz 0.08

0.08 1>2 smax z2 b z2 a1 b (0.04)dz 0.04 L0 (0.08)2

50 = 201.06(10 - 6)smax

Ans.

smax = 249 kPa

Ans: (a) smax = 249 kPa, (b) smax = 249 kPa 527

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6–70. The beam is subjected to a moment of M = 40 kN # m. Determine the bending stress at points A and B. Sketch the results on a volume element acting at each of these points.

A

B

50 mm

M = 40 kNm

50 mm 50 mm

Solution I =

50 mm 50 mm 50 mm

1 1 (0.150)(0.05)3 + 2c (0.05)(0.05)3 + (0.05)(0.05)(0.05)2 d = 15.1042 ( 10 - 6 ) m4 12 12

sA =

40 ( 103 ) (0.075) Mc = 199 MPa = I 15.1042 ( 10 - 6 )

Ans.

sB =

40 ( 103 ) (0.025) My = = 66.2 MPa I 15.1042 ( 10 - 6 )

Ans.

Ans: sA = 199 MPa, sB = 66.2 MPa 523

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6–71. Determine the dimension a of a beam having a square cross section in terms of the radius r of a beam with a circular cross section if both beams are subjected to the same internal moment which results in the same maximum bending stress.

a

r

a

Section Properties: The moments of inertia of the square and circular cross sections about the neutral axis are IS =

1 a4 a A a3 B = 12 12

IC =

1 4 pr 4

Maximum Bending Stress: For the square cross section, c = a>2.

A smax B S =

M(a>2) 6M Mc = 3 = 4 IS a >12 a

For the circular cross section, c = r.

A smax B c =

Mc Mr 4M = Ic 1 4 pr3 pr 4

It is required that

A smax B S = A smax B C

4M 6M = 3 a pr3

Ans.

a = 1.677r

Ans: a = 1.677r 366 366

CH 06.indd 366

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*6–72.

A portion of the femur can be modeled as a tube diameter of of 0.375 9.5 mm having an inner diameter in. and an outer diameter 32 mm. of 1.25 in. Determine the maximum elastic static force P that can be applied to its center. Assume the bone to be roller supported at its ends. The s– P diagram for the bone mass is shown and is the same in tension as in compression.

P

ss(MPa) (ksi) 16.10 2.30 8.75 1.25 100 mm 4 in.

0.02

I=

π 4

(0.016 4 − 0.004754 ) = 51.0720(10 −9 ) m 4

= Mmax

0.05

100 mm 4 in.

P (mm/mm) (in./ in.)

0.1 m

P = (0.1) 0.05P 2

8.75 ksi MPa Require smax = 1.25 smax =

Mc I

8.75(106 ) =

(0.05P )(0.016) 51.0720(10 −9 )

= = N 559 N P 558.6

Ans.

Ans:

= P 559 N 389 389

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xle ofofthethe freight car is to wheel freight carsubjected is subjected to 6–73. The axle loadings of 20 kip. it iskN. supported by two journal bearings at wheel loadings of If100 If it is supported by two journal C and D,at determine maximum stress developed bearings C and D, the determine the bending maximum bending stress at the center thecenter axle, where diameter 5.5diameter in. developed atofthe of thethe axle, where is the is 137.5 mm.

A

C

10mm in. 250 20kN kip 100

σ= max

MC [25(10 3 )](0.06875) 2 = = 97.96(106 ) N−m = 98 MPa π (0.068754 ) I 4

B

60 in. 1500 m

D

10 250in. mm 20kN kip 100

Ans.

100 kN Mmax = 25 kN · m

0.25 m

100 kN

Ans: s max = 98.0 MPa 370 370

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6–74. The chair is supported by an arm that is hinged so it rotates about the vertical axis at A. If the load on the chair is 900 hollow tube section having the 180 N lb and and the the arm arm isis aa hollow dimensions shown, determine the maximum bending stress at section a–a.

180 900 lb N

1 in. 25 mm a A 8 in. 200 mm

c + ©M = 0;

M − 900(0.9) = 0

in. 75 3mm

a

2.5mm in. 63 0.5mm in. 13

900 N

= M 810 N ⋅ m 0.9 m

1 1 I x = (0.025)(0.0753 ) − (0.013)(0.0633 ) = 0.608022(10 −6 ) m 4 12 12

σ= max

MC 810(0.0375) 2 = 50.0 MPa = = 49.96(10 −6 ) N−m I 0.608022(10 −6 )

Ans.

Ans: s max = 11.1 MPa 389 389

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6–75. The boat has a weight of 11.5 kN and a center of gravity at G. If it rests on the trailer at the smooth contact A and can be considered pinned at B, determine the absolute maximum bending stress developed in the main strut of the trailer. Consider the strut to be a box-beam having the dimensions shown and pinned at C.

B 10.3 ft m

G

C

A D 3 ftm 0.9

5 ftm 1.5

4 ftm 1.2

1.75 in. 45 mm

1 ft 0.3 m

1.75 in. 45 mm

in. 75 3mm 1.5 mm in. 37.5

Boat:

11.5 kN

+ ©F = 0; : x

Bx = 0 –N 11.5(1.5)== 00 NAA(2.7) (9) ++ 2300(5)

a + ©MB = 0;

1.2 m

1277.78 NAA ==6.3889 kNlb + c ©Fy = 0;

1.5 m

6.3889 – 11.5 + B+y =B0y = 0 1277.78 2300 By == 5.1111 1022.22 kNlb

Assembly: –N NDD(3.0) (10) ++ 11.5(2.7) 2300(9) = = 00

a + ©MC = 0;

N 2070kN lb NDD ==10.35 + c ©Fy = 0;

2070–-11.5 2300 Cyy ++10.35 = 0= 0 Cy == 1.15 230 kN lb

1 1 I = (0.045)(0.0753 ) − (0.0375)(0.0453 ) =1.2973(10 −6 ) m 4 12 12

σ= max

MC [5.75(10 3 )](0.0375) = = 166.21(106 ) N− = m 2 166 MPa I 1.2973(10 −6 )

Ans.

11.5 kN 2.7 m 3.0 m

5.1111 kN

6.3889 kN

1.8 m 0.9 m 10.35 kN V (kN)

1.2 m

1.15 kN

3.9611 –1.15

–6.3889 M (kN · m)

1.38

Ans: s max = 166 MPa

–5.75

372 372

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*6–76. The steel beam has the cross-sectional area shown. Determine the largest intensity of the distributed load w0 that it can support so that the maximum bending MPa. stress in the beam does not exceed sallow = 160 22 ksi.

w0

39 m ft

39 m ft

200 mm 9 in. 6 mmin. 0.25 300in. mm 12 6 mmin. 0.25

6 mm 0.25 in.

Support Reactions. The FBD of the beam is shown in Fig. a. The shear and moment diagrams are shown in Fig. a and b, respectively. As 3wo.o. indicated on the moment diagram, Mmax = 27w The moment of inertia of the cross-section about the neutral axis is

I=

1 1 (0.2)(0.312 3 ) − (0.194)(0.33 ) = 69.6888(10 −6 ) m 4 12 12

Hence, c/ = 0.156 m. Thus, sallow =

Mmax c ; I

160(106 ) =

(3wo− )(0.156)

69.6888(10 −6 )

3 = ) N−m 23.8 kN−m = wo 23.83(10

Ans.

1 w0(6) 2

V

1.5w0 6 x (m) 3 3m Ay = 1.5w0

–1.5w0

3m By = 1.5w0 M 3w0

x (m) 3

6

Ans:

= wo 23.8 kN−m 391 391

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6–77. The steel beam has the cross-sectional area kN/m, determine the shown. If w w00 ==30 the maximum maximum bending 2 kip>ft, stress in the beam.

w0

ft 39 m

ft 39 m

200 mm 9 in. 60.25 mmin. 12 in. 300 mm

0.25 in. 6 mm

60.25 mmin.

The FBD of the beam is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c, respectively. As 54 kN kip ·# m. ft. indicated on the moment diagram, Mmax = 90 The moment of inertia of the I cross-section about the bending axis is

= I

1 1 (0.2)(0.312 3 ) − (0.194)(0.33 ) 12 12

= 69.6888(10 −6 ) m 4 Hence, c/ = 0.156 m. Thus, smax =

=

Mmax c I

[90(0.33 )](0.156) 69.6888(10 −6 )

6 = 201.47(10 = ) N−m 2 201 MPa

Ans.

1 (30)(6) kN  2

V (kN)

45 6 x (m) 3 3m Ay = 45 kN

–45

3m By = 45 kN M (kN · m) 90

x (m) 3

6

Ans: I = 69.6888 (10- 6) m4, s max = 201 MPa 392 392

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6–78. If the beam is subjected to a moment of M = 100 kN # m, determine the bending stress at points A, B, and C. Sketch the bending stress distribution on the cross section.

A M

300 mm

30 mm 30 mm B 150 mm

C

Solution Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is ~ 0.015(0.03)(0.3) + 0.18(0.3)(0.03) Σ yA y = = = 0.0975 m ΣA 0.03(0.3) + 0.3(0.03)

150 mm

Thus, the moment of inertia of the cross section about the neutral axis is I =

1 1 (0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 + (0.03)(0.33) 12 12 + 0.03(0.3)(0.18 - 0.0975)2

= 0.1907(10 - 3) m4 Bending Stress: The distance from the neutral axis to points A, B, and C is yA = 0.33 - 0.0975 = 0.2325 m, yB = 0.0975 m, and yC = 0.0975 - 0.03 = 0.0675 m. sA =

100(103)(0.2325) MyA = = 122 MPa (C) I 0.1907(10 - 3)

Ans.

sB =

100(103)(0.0975) MyB = = 51.1 MPa (T) I 0.1907(10 - 3)

Ans.

sC =

100(103)(0.0675) MyC = = 35.4 MPa (T) I 0.1907(10 - 3)

Ans.

Using these results, the bending stress distribution across the cross section is shown in Fig. b.

Ans: sA = 122 MPa (C), sB = 51.1 MPa (T), sC = 35.4 MPa (T) 542

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6–79. If the beam is made of material having an allowable tensile and compressive stress of (sallow)t = 125 MPa and (sallow)c = 150 MPa, respectively, determine the maximum moment M that can be applied to the beam.

A M

300 mm

30 mm 30 mm B 150 mm

C

Solution Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is ~ 0.015(0.03)(0.3) + 0.18(0.3)(0.03) Σ yA y = = = 0.0975 m ΣA 0.03(0.3) + 0.3(0.03)

150 mm

Thus, the moment of inertia of the cross section about the neutral axis is I =

1 1 (0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 + (0.03)(0.33) 12 12 + 0.03(0.3)(0.18 - 0.0975)2

= 0.1907(10 - 3) m4 Allowable Bending Stress: The maximum compressive and tensile stress occurs at the top and bottom-most fibers of the cross section. For the top-most fiber, (sallow)c =

Mc I

150(106) =

M(0.33 - 0.0975) 0.1907(10 - 3)

M = 123024.19 N # m = 123 kN # m (controls)

Ans.

For the bottom-most fiber, (sallow)t =

My I

125(106) =

M(0.0975) 0.1907(10 - 3)

M = 244 471.15 N # m = 244 kN # m

Ans: M = 123 kN # m 543

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*6–80. The two solid steel rods are bolted together along their length and support the loading shown. Assume the support at A is a pin and B is a roller. Determine the required diameter d of each of the rods if the allowable bending stress is sallow = 130 MPa.

20 kN/m

80 kN

A B

2m 2m

Section Property: I = 2B

p d 4 p d 2 5p 4 a b + d2 a b R = d 4 2 4 2 32

Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 130 A 106 B =

Mmax c I 100(103)(d) 5p 32

d4 Ans.

d = 0.1162 m = 116 mm

Ans: d = 116 mm 388 388

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6–81. Solve Prob. 6–94 if the rods are rotated 90° so that both rods rest on the supports at A (pin) and B (roller).

20 kN/m

80 kN

A B

2m 2m

Section Property: I = 2B

p d 4 p 4 a b R = d 4 2 32

Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 130 A 106 B =

Mmax c I

100(103)(d) p 32

d4 Ans.

d = 0.1986 m = 199 mm

Ans: d = 199 mm 388 388

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6–82. If the compound beam in Prob. 6–42 has a square cross section of side length a, determine the minimum value of a if the allowable bending stress is sallow = 150 MPa.

Solution

Allowable Bending Stress: The maximum moment is Mmax = 7.50 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 150 1 106 2 =

Mmax c I 7.50(103) 1 a2 2 1 12

a4

Ans.

a = 0.06694 m = 66.9 mm

Ans: a = 66.9 mm 551

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6–83. If the beam in Prob. 6–28 has a rectangular cross section with a width b and a height h, determine the absolute maximum bending stress in the beam.

Solution Absolute Maximum Bending Stress: The maximum moment is Mmax = indicated on the moment diagram. Applying the flexure formula smax

Mmax c = = I

23w0 L2 h 2 216 1 3 bh 12

1 2

=

23w0 L2

23w0 L2 as 216

Ans.

36bh2

Ans: smax = 552

23w0 L2 36 bh2

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*6–84. Determine the absolute maximum bending stress in the 80-mm-diameter shaft which is subjected to the concentrated forces. There is a journal bearing at A and a thrust bearing at B.

A

0.5 m 12 kN

B

0.4 m

0.6 m 20 kN

Solution The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram,  Mmax  = 6 kN # m .

The moment of inertia of the cross section about the neutral axis is I =

p (0.044) = 0.64(10 - 6)p m4 4

Here, c = 0.04 m. Thus smax =

6(103)(0.04) Mmax c = I 0.64(10 - 6)p = 119.37(106) Pa Ans.

= 119 MPa

Ans: smax = 119 MPa 553

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6–85. Determine, to the nearest millimeter, the smallest allowable diameter of the shaft which is subjected to the concentrated forces. There is a journal bearing at A and a thrust bearing at B. The allowable bending stress is sallow = 150 MPa.

A

0.5 m 12 kN

B

0.4 m

0.6 m 20 kN

Solution The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram,  Mmax  = 6 kN # m .

The moment of inertia of the cross section about the neutral axis is I =

p d 4 pd 4 a b = 4 2 64

Here, c = d>2. Thus sallow =

Mmax c ; I

150(106) =

6(103)(d > 2) pd 4 >64

d = 0.07413 m = 74.13 mm = 75 mm

Ans.

Ans: d = 75 mm 554

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6–86. If the beam is subjected to an internal moment of M = 3 kN # m, determine the maximum tensile and compressive stress in the beam. Also, sketch the bending stress distribution on the cross section.

A

100 mm 25 mm 25 mm

Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is

M

©y~A 2[0.0375(0.025)(0.075)] + 0.0875(0.2)(0.025) + 0.15(0.025)(0.1) = y = ©A 2(0.025)(0.075) + 0.2(0.025) + 0.025(0.1) = 0.08472 m

75 mm

25 mm 75 mm 75 mm 25 mm

Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2

1 1  (0.2)(0.0253 ) + 0.2(0.025)(0.0875 − 0.08472)2 = 2  (0.025)(0.0753 ) + 0.025(0.075)(0.08472 − 0.0375)2  + 12  12  +

1 (0.025)(0.13 ) + 0.025(0.1)(0.15 − 0.08472)2 12

= 23.1554(10 −6 ) m 4 Maximum Bending Stress: The maximum compressive and tensile stress occurs at the top and bottom-most fibers of the cross section.

(σ max= )c

Mc [3(10 3 )](0.2 − 0.08472) 2 = = 14.94(106 ) N−m = 14.9 MPa I 23.1554(10 −6 )

Ans.

(σ max= )t

My [3(10 3 )](0.08472) = = 10.98(106 ) N−m 2 = 11.0 MPa I 23.1554(10 −6 )

Ans.

The bending stresses at y = 0.01528 m and y = −0.009722 m are

s|y = 0.1528 m =

My =

I

s|y = - 0.009722 m =

[3(10 3 )](0.01528) 6 = 1.979(10 = ) N−m 2 1.98 MPa (C) 23.1554(10 −6 )

My I

=

[3(10 3 )](0.009722) 6 = 1.260(10 = ) N−m 2 1.26 MPa (T) 23.1554(10 −6 )

The bending stress distribution across the cross section is shown in Fig. b.

14.9 MPa 1.98 MPa

1.26 MPa

11.0 MPa

Ans: (smax)c = 14.9 MPa, (smax)t = 11.0 MPa 538

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6–87. If the allowable tensile and compressive stress for the beam are (sallow)t = 14 MPa and (sallow)c = 21 MPa, respectively, determine the maximum allowable internal moment M that can be applied on the cross section.

A

100 mm 25 mm 25 mm M

Section Properties: The neutral axis passes through the centroid C of the cross section as shown in Fig. a. The location of C is given by y=

©y~A = ©A

2[0.0375(0.025)(0.075)] + 0.0875(0.2)(0.025) + 0.15(0.025)(0.1) 2(0.025)(0.075) + 0.2(0.025) + 0.025(0.1)

= 0.08472 m Thus, the moment of inertia of the cross section about the neutral axis is

75 mm

25 mm 75 mm 75 mm 25 mm

I = ©I + Ad2

1  1 (0.2)(0.0253 ) + 0.2(0.025)(0.0875 − 0.08472)2 = 2  (0.025)(0.0753 ) + 0.025(0.075)(0.08472 − 0.0375)2  + 12  12  +

1 (0.025)(0.13 ) + 0.025(0.1)(0.15 − 0.08472)2 12

= 23.1554(10 −6 ) m 4 Allowable Bending Stress: The maximum compressive and tensile stress occurs at the top and bottom-most fibers of the cross section. (sallow)c =

Mc ; I

21(106 ) =

M (0.2 − 0.08472) 23.1554(10 −6 )

M 4.218(10 3 ) N= = ⋅ m 4.22 kN ⋅ m For the bottom-most fiber, (sallow)t =

My ; I

21(106 ) =

M (0.08472) 23.1554(10 −6 )

M 3.826(10 3 ) N= ⋅ m 3.83 kN ⋅ m (Controls) =

Ans.

Ans:

M = 3.83 kN # m

539

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*6–88. If the beam is subjected to an internal moment of M = 3 kN . m, determine the resultant force of the bending stress distribution acting on the top vertical board A.

A

100 mm 25 mm 25 mm M

Section Properties: The neutral axis passes through the centroid C of the cross section as shown in Fig. a. The location of C is given by

75 mm

25 mm 75 mm 75 mm 25 mm

~ © yA 2[0.0375(0.025)(0.075)] + 0.0875(0.2)(0.025) + 0.15(0.025)(0.1) y = = 2(0.025)(0.075) + 0.2(0.025) + 0.025(0.1) ©A = 0.08472 m Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2

1 1  (0.2)(0.0253 ) + 0.2(0.025)(0.0875 − 0.08472)2 = 2  (0.025)(0.0753 ) + 0.025(0.075)(0.08472 − 0.0375)2  + 12  12  +

1 (0.025)(0.13 ) + 0.025(0.1)(0.15 − 0.08472)2 12

= 23.1554(10 −6 ) m 4 Bending Stress: The distance from the neutral axis to the top and bottom of board A is yt = 0.2 − 0.08472 = 0.11528 m and yb = 0.1 − 0.08472 = 0.01528 m. We have

σt =

Myt [3(10 3 )](0.11528) 2 = = 14.94(106 ) N−m = 14.94 MPa I 23.1554(10 −6 )

σb =

Myb [3(10 3 )](0.01528) 2 = = 1.979(106 ) N−m = 1.979 MPa I 23.1554(10 −6 )

Resultant Force: The resultant force acting on board A is equal to the volume of the trapezoidal stress block shown in Fig. b. Thus,

1 21.14(10 3 ) N = 21.1 kN FR = [(14.94 + 1.979)(106 )](0.025)(0.1) = 2

Ans. 0.025 m

14.94 MPa

0.1 m

1.979 MPa

Ans: FR = 21.1 kN 540

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6–89. A timber beam has a cross section which is originally square. If it is oriented as shown, determine the dimension h′ so that it can resist the maximum moment possible. By what factor is this moment greater than that of the beam without its top or bottom flattened?

h

Solution x 2h = ; h - h′ h y = h′ + I = 2e = = = =

x = 2(h - h′)

h - h′ 2h′ + h = 3 3

1 1 1 2h′ + h 2 (2h) ( h3 ) - c (2)(h - h′)(h - h′)3 + (2)(h - h′)(h - h′)a b df 12 36 2 3

1 4 1 2 h - (h - h′)4 - (h - h′)2(2h′ + h)2 3 9 9 1 4 1 h - (h - h′)2 3 3h2 + 9h′2 + 6hh′ 4 3 9 1 4 1 h - ( 3h4 + 9h′4 - 12hh′3 ) 3 9 4 hh′3 - h′4 3

smax = M = =

Mc I

I s c max 4 3 3 hh′

(1) - h′4

h′

4 smax = a hh′2 - h′3 bsmax 3

dM 8 = a hh′ - 3h′2 bsmax dh′ 3

In-order to have maximum moment, dM 8 = 0 = hh′ - 3h′2 dh′ 3 h′ =

8 h 9

Ans.

For the square beam, I = I + Ad 2 I = 2c

1 h4 1 h 2 (2h)(h)3 + (2h)(h)a b d = 36 12 3 3

537

h¿

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6–89. Continued

From Eq. (1) M =

h4 3

h

smax =

h3 s = 0.3333h3 smax 3 max

For the flattened beam: I =

4 8 3 8 4 ha hb - a hb = 0.312147 h4 3 9 9

From Eq. (1) M′ =

0.312147 h4

Factor =

8 9h

smax = 0.35117 h3 smax

0.35117 h3 smax M′ = = 1.05 M′ 0.3333 h3 smax

Ans.

Ans:

8 h, 9 Factor = 1.05 h′ =

538

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6–90. The wood beam has a rectangular cross section in the proportion shown. Determine its required dimension b if the allowable bending stress is sallow = 10 MPa.

500 N/m

A

B 2m

1.5b b

2m

Allowable Bending Stress: The maximum moment is Mmax = 562.5 N # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 10 A 106 B =

Mmax c I 562.5(0.75b) 1 12

(b)(1.5b)3 Ans.

b = 0.05313 m = 53.1 mm

Ans: b = 53.1 mm 380 380

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6–91. Determine the absolute maximum bending stress in the tubular shaft if di = 160 mm and do = 200 mm.

15 kN/m 60 kN � m d i do A

B 3m

1m

Section Property: I =

p A 0.14 - 0.084 B = 46.370 A 10 - 6 B m4 4

Absolute Maximum Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax = =

Mmaxc I

60.0(103)(0.1) 46.370(10 - 6) Ans.

= 129 MPa

Ans: s max = 129 MPa 378 378

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*6–92. The tubular shaft is to have a cross section such that its inner diameter and outer diameter are related by di = 0.8do. Determine these required dimensions if the allowable bending stress is sallow = 155 MPa.

15 kN/m 60 kN � m d i do A

B 3m

1m

Section Property: I =

0.8do 4 do 4 dl 4 p do 4 p - a b R = 0.009225pd4o Ba b - a b R = B 4 2 2 4 16 2

Allowable Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow =

Mmax c I d

155 A 10

6

Thus,

B =

60.0(103) A 2o B

0.009225pd4o

do = 0.1883 m = 188 mm

Ans.

dl = 0.8do = 151 mm

Ans.

Ans: do = 0188 mm, dl = 151 mm 379 379

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6–93. If the intensity of the load w = 15 kN>m, determine the absolute maximum tensile and compressive stress in the beam.

w A

B 6m 300 mm

150 mm

Solution Support Reactions: Shown on the free-body diagram of the beam, Fig. a. Maximum Moment: The maximum moment occurs when V = 0. Referring to the free-body diagram of the beam segment shown in Fig. b, + c ΣFy = 0;

45 - 15x = 0

x = 3m

3 Mmax + 15(3)a b - 45(3) = 0 Mmax = 67.5 kN # m 2 Section Properties: The moment of inertia of the cross section about the neutral axis is a+ ΣM = 0;

I =

1 (0.15)(0.33) = 0.1125(10 - 3) m4 36

Absolute Maximum Bending Stress: The maximum compressive and tensile stresses occur at the top and bottom-most fibers of the cross section. (smax)c =

67.5(103)(0.2) Mc = = 120 MPa (C) I 0.1125(10 - 3)

Ans.

(smax)t =

67.5(103)(0.1) My = = 60 MPa (T) I 0.1125(10 - 3)

Ans.

Ans: (smax)c = 120 MPa (C), (smax)t = 60 MPa (T) 546

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6–94. If the allowable bending stress is sallow = 150 MPa, determine the maximum intensity w of the uniform distributed load.

w A

B 6m 300 mm

150 mm

Solution Support Reactions: As shown on the free-body diagram of the beam, Fig. a. Maximum Moment: The maximum moment occurs when V = 0. Referring to the free-body diagram of the beam segment shown in Fig. b, + c ΣFy = 0;

3w - wx = 0

x = 3m

3 9 Mmax + w(3)a b - 3w(3) = 0 Mmax = w 2 2 Section Properties: The moment of inertia of the cross section about the neutral axis is a+ ΣM = 0;

I =

1 (0.15)(0.33) = 0.1125(10 - 3) m4 36

Absolute Maximum Bending Stress: Here, c =

sallow

Mc = ; I

150(106) =

2 (0.3) = 0.2 m. 3

9 w(0.2) 2 0.1125 ( 10 - 3 ) Ans.

w = 18750 N>m = 18.75 kN>m

Ans: w = 18.75 kN>m 547

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6–95. The beam has a rectangular cross section as shown. Determine the largest intensity w of the uniform distributed load so that the bending stress in the beam does not exceed smax = 10 MPa.

w

50 mm 150 mm

2m

2m

2m

Solution Absolute Maximum Bending Stress: The support reactions, shear diagram, and moment diagrams are shown in Figs. a, b, and c, respectively. From the moment diagram, the maximum moment is Mmax = 2.00 w, which occurs at the pin support and roller support. Applying the flexure formula, sabs = max

(2.00 w)(0.075) MmaxC ;  10 ( 106 ) = I 1 (0.05) ( 0.153 ) 12 w = 937.5 N>m

Ans.

Ans: w = 937.5 N>m 564

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*6–96. The beam has the rectangular cross section shown. If w = 1 kN>m, determine the maximum bending stress in the beam. Sketch the stress distribution acting over the cross section.

w

50 mm 150 mm

2m

2m

2m

Solution Absolute Maximum Bending Stress: The support reactions, shear diagram, and moment diagram are shown in Figs. a, b, and c, respectively. From the moment diagram, the maximum moment is Mmax = 2000 N # m, which occurs at the supports. Applying the flexure formula, sabs = max

MmaxC = I

2000(0.075) 1 12

(0.05) ( 0.153 )

= 10.67 ( 106 ) MPa = 10.7 MPa

Ans.

Using this result, the bending stress distribution on the beam’s cross-section shown in Fig. d can be sketched.

Ans: sabs = 10.7 MPa max 565

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6–97. The simply supported truss is subjected to the central distributed load. Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the truss. top member memberisisa pipe a pipe having an outer diameter of The top having an outer diameter of 1 in. 3 25 mm and thickness 5 mm, and themember bottom is member is a and thickness of 16 the bottom a solid rod in.,ofand 1 solid rod having a of diameter of 12 mm. having a diameter 2 in.

1.8 6 ftm

∑ yA 0 + 0.16[π (0.01252 − 0.00752 )] = = 0.11765 m ∑ A π (0.006 2 ) + π (0.01252 − 0.00752 )

= y I=

1.5 100kN/m lb/ft

π 4

(0.006 4 ) + π (0.006 2 )(0.117652 ) +

π 4

1.8 6 ftm

141.5 5.75mm in.

1.8 6 ftm

0.16 m

y

(0.01254 − 0.00754 )

2.1466(10 −6 ) m 4 + π (0.01252 − 0.00752 )(0.16 − 0.11765)2 = Mmax =1.35(2.7) − 1.35(0.45) =3.0375 kN ⋅ m

σ= max

MC [3.0375(10 3 )](0.11765 + 0.006) = I 2.1466(10 −6 )

6 = 174.96(10 = ) N−m 2 175 MPa

1.35 kN

Ans.

0.45 m

2.7 m 1.35 kN

Ans: (s max )c = 120 MPa (C), (s max )t = 60 MPa (T) 370 370

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6–98. If d = 450 mm, determine the absolute maximum bending stress in the overhanging beam.

12 kN 8 kN/m

125 mm 25 mm 25 mm 75 mm d

A B 4m

75 mm 2m

Solution Support Reactions: Shown on the free-body diagram of the beam, Fig. a. Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, Mmax = 24 kN # m. Section Properties: The moment of inertia of the cross section about the neutral axis is I =

1 1 (0.175)(0.453) (0.125)(0.33) 12 12

= 1.0477(10 - 3) m4 Absolute Maximum Bending Stress: Here, c = smax =

0.45 = 0.225 m. 2

24(103)(0.225) Mmax c = = 5.15 MPa I 1.0477(10 - 3)

Ans.

Ans: smax = 5.15 MPa 562

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6–99. If the allowable bending stress is sallow = 6 MPa, determine the minimum dimension d of the beam’s cross‑sectional area to the nearest mm.

12 kN 8 kN/m

125 mm 25 mm 25 mm 75 mm d

A B 4m

75 mm 2m

Solution Support Reactions: Shown on the free-body diagram of the beam, Fig. a. Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, Mmax = 24 kN # m. Section Properties: The moment of inertia of the cross section about the neutral axis is I =

1 1 (0.175)d 3 (0.125)(d - 0.15)3 12 12

= 4.1667(10 - 3)d 3 + 4.6875(10 - 3)d 2 - 0.703125(10 - 3)d + 35.15625(10 - 6) Absolute Maximum Bending Stress: Here, c = sallow =

d . 2

Mc ; I 24(103)

6

6(10 ) =

d 2

4.1667(10 - 3)d 3 + 4.6875(10 - 3)d 2 - 0.703125(10 - 3)d + 35.15625(10 - 6)

4.1667(10 - 3)d 3 + 4.6875(10 - 3)d 2 - 2.703125(10 - 3)d + 35.15625(10 - 6) = 0

Solving, Ans.

d = 0.4094 m = 410 mm

Ans: d = 410 mm 563

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*6–100. If the reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown, determine the maximum bending stress developed in the tie. The tie has the rectangular cross section with thickness tt ==150 6 in. mm.

15 75 kip kN 0.45 1.5 ftm

15 75 kip kN 0.45 m 1.5 ft

5 ftm 1.5

300 12 mm in. t

w

Support Reactions: Referring to the free - body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;

w(2.4)-–2(15) 2(75) = = 00 w(8) w == 62.5 3.75kN/m kip>ft

Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As

# ft·. m 7.5 kipkN indicated on the moment diagram, the maximum moment is � Mmax � = 11.25 Absolute Maximum Bending Stress: = σ max

Mmax c [11.25(10 3 )](0.075) 6 = 10.0(10 ) N−m 2 10.0 MPa = 1= I (0.3)(0.153 ) 12

75 kN

75 kN 0.45 m

0.75 m

0.75 m

Ans.

V (kN)

0.45 m 46.875

28.125

1.95 0.45

1.2

x (m) 2.4

–28.125 –46.875 w(2.4)

M (kN · m)

6.328

6.328 1.2

x (m) 1.95

0.45

–11.25

Ans:

σ max = 10.0 MPa 376 376

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6–101. The reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown. If the wood has an allowable bending stress of sallow = 1.5 determine the the required minimum thickness t of tthe 10.5ksi, MPa, determine required minimum thickness of 1 rectangular crosscross sectional areaarea of the the in. the rectangular sectional of tie thetotie tonearest the nearest 8 multiples of 5 mm.

15 kN kip 75 0.45 1.5 ftm

15 kN kip 75 0.45 m 1.5 ft

5 ftm 1.5

300 12 mm in. t

w

Support Reactions: Referring to the free-body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;

75 kN 0.45 m

w(8) - 2(15) = 0= 0 w = (2.4) – 2(75)

75 kN 0.75 m

0.75 m

0.45 m

w == 62.5 3.75kN/m kip>ft Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As

# ft·. m. 7.5 kipkN indicated on the moment diagram, the maximum moment is � Mmax � = 11.25 w(2.4)

Absolute Maximum Bending Stress:

smax =

Mc ; I

10.5(106 ) =

[11.25(10 3 )]

( 2t )

1 (0.3)(t 3 ) 12

m 146.38 mm = = t 0.14638

Ans.

Use t = 150 mm t =

V (kN) M (kN · m)

0.45

6.328

6.328

46.875

28.125

1.2

1.2

1.95

x (m) 2.4

x (m) 0.45

1.95

2.4

–28.125 –46.875

–11.25

Ans: Use t = 150 mm 377 377

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6–102. A log that is 0.6 m in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is sallow = 56 MPa, determine the required width b and height h of the beam that will support the largest load possible. What is this load?

h b 0.6 m P

2 b2 + h 2 0.6=

h2 0.36 − b2 =

2.4 m

P Mmax = (2.4) 1.2 P = 2 sallow =

2.4 m

Mmax(h2 ) Mc = 1 3 I 12 (b)(h) 0.6 m

sallow =

6 Mmax

56(106 ) =

bh2

6(1.2P) b(0.36 − b2 )

P 7.7778(106 )(0.36b − b3 ) = Set

dP = 0 gives db

dP )(0.36 − b2 ) 0 = 7.7778(106= db b = 0.3464 m Thus, from the above equations,

Ans.

b = 346.41 mm = 346 mm

Ans.

h = 0.4899 m = 490 mm

Ans.

3 P 646.63(10 = ) N 647 kN =

Ans: b = 346 mm, h = 490 mm, P = 647 kN 557

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6–103. A log that is 0.6 m in diameter is to be cut into a rectangular section for use as a simply supported beam. If the wood is sallow = 56 MPa, the allowable bending stress for determine the largest load P that can be supported if the width of the beam is b = 200 mm.

h b 0.6 m P

2 0.6= h2 + 0.2 2

2.4 m

2.4 m

P = (2.4) 1.2 P 2

sallow =

Mmax c I

56(106 ) =

0. 6

Mmax =

m

h = 0.5657 m

(1.2 P )(0.5657 2)

0.2 m

1 (0.2)(0.5657 3 ) 12

3 P 497.78(10 = ) 498 kN =

Ans.

2.4 m

Ans: P = 498 kN 558

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*6–104. The member has a square cross section and is subjected to the moment M = 850 N # m. Determine the stress at each corner and sketch the stress distribution. Set u = 45°.

B 250 mm

z

125 mm

125 mm E

A

C

M  850 Nm

u D y

Solution

My = 850 cos 45° = 601.04 N # m Mz = 850 sin 45° = 601.04 N # m

1 (0.25)(0.25)3 = 0.3255208(10 - 3) m4 12

Iz = Iy = s = -

Mzy

sA = sB = sD = sE = -

Iz

+

Myz Iy

601.04( - 0.125) -3

0.3255208(10 ) 601.04( - 0.125) 0.3255208(10 - 3) 601.04(0.125) -3

0.3255208(10 ) 601.04(0.125) -3

0.3255208(10 )

+ + + +

601.04( - 0.125) 0.3255208(10 - 3) 601.04(0.125) 0.3255208(10 - 3) 601.04( - 0.125) 0.3255208(10 - 3) 601.04(0.125) 0.3255208(10 - 3)

= 0

Ans.

= 462 kPa

Ans.

= - 462 kPa

Ans.

= 0

Ans.

The negative sign indicates compressive stress.

Ans: sA = sB = sD = sE = 566

0, 462 kPa, - 462 kPa, 0

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6–105. The member has a square cross section and is subjected to the moment M = 850 N # m as shown. Determine the stress at each corner and sketch the stress distribution. Set u = 30°.

B 250 mm

z

125 mm

125 mm E

A

C

M  850 Nm

u D y

Solution

My = 850 cos 30° = 736.12 N # m Mz = 850 sin 30° = 425 N # m Iz = Iy = s = -

1 (0.25)(0.25)3 = 0.3255208(10 - 3) m4 12

Mzy

sA = sB = sD = sE = -

Iz

+

Myz Iy

425(- 0.125) -3

0.3255208(10 ) 425(- 0.125) 0.3255208(10 - 3) 425(0.125)

+ +

-3

+

-3

+

0.3255208(10 ) 425(0.125) 0.3255208(10 )

736.12(- 0.125) 0.3255208(10 - 3) 736.12(0.125) 0.3255208(10 - 3) 736.12(- 0.125) 0.3255208(10 - 3) 736.12(0.125) 0.3255208(10 - 3)

= - 119 kPa

Ans.

= 446 kPa

Ans.

= - 446 kPa

Ans.

= 119 kPa

Ans.

The negative signs indicate compressive stress.

Ans: sA = - 119 kPa, sB = 446 kPa, sD = - 446 kPa, sE = 119 kPa 567

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6–106. Consider the general case of a prismatic beam subjected to bendingmoment components My and Mz when the x, y, z axes pass through the centroid of the cross section. If the material is linear elastic, the normal stress in the beam is a linear function of position such that s = a + by + cz. Using the equilibrium conditions 0 = 1A s dA, My = 1Azs dA, Mz = 1A - ys dA, determine the constants a, b, and c, and show that the normal stress can be determined from the equation s = [ - (Mz Iy + My Iyz)y + (MyIz + MzIyz)z]>(Iy Iz - I yz2), where the moments and products of inertia are defined in Appendix A.

y z My dA sC y Mz

x

z

Solution Equilibrium Condition: sx = a + by + cz 0 =

LA

sx dA

0 =

LA

(a + by + cz) dA

0 = a

LA

dA + b

My =

LA

z sx dA

=

LA

z(a + by + cz) dA

= a

LA

LA

z dA + b

y dA + c

LA

LA

LA

-y sx dA

=

LA

- y(a + by + cz) dA LA

ydA - b

(1)

LA

(2)

yz dA + c

Mz =

= -a

z dA

LA

y2 dA - c

z2 dA

LA

(3)

yz dA

Section Properties: The integrals are defined in Appendix A. Note that LA

y dA =

LA

z dA = 0. Thus,

From Eq. (1)

Aa = 0

From Eq. (2)

My = bIyz + cIy

From Eq. (3)

Mz = - bIz - cIyz

Solving for a, b, c: a = 0 (Since A ≠ 0) b = -¢

MzIy + My Iyz Iy Iz -

Thus,  sx = - ¢

I 2yz

Ans. ≤

Mz Iy + My Iyz Iy Iz -

I 2yz

c =

My Iz + Mz Iyz

≤y + ¢

Ans.

Iy Iz - I 2yz My Iy + MzIyz Iy Iz - I 2yz

(Q.E.D.)

≤z

Ans: a = 0, b = - ¢

568

MzIy + MyIyz IyIz - I 2yz

≤, c =

MyIz + MzIyz IyIz - I 2yz

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6–107. If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown, determine the bending stress at points A and B. The location y of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.

y M � 520 N�m 20 mm z

–y

12 5

13

B

200 mm

C 20 mm 200 mm

A

20 mm

200 mm

Internal Moment Components: Mz = -

12 (520) = - 480 N # m 13

My =

5 (520) = 200 N # m 13

Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)

y =

Ans.

= 0.057368 m = 57.4 mm Iz =

1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2 12

+

1 (0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2 12

= 57.6014 A 10 - 6 B m4

Iy =

1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4 12 12

Maximum Bending Stress: Applying the flexure formula for biaxial at points A and B s = -

Mzy

sA = -

Iz

+

Myz Iy

-480(- 0.142632) -6

57.6014(10 )

+

200( - 0.2) 0.366827(10 - 3) Ans.

= -1.298 MPa = 1.30 MPa (C) sB = -

-480(0.057368)

+

-6

57.6014(10 )

200(0.2) 0.366827(10 - 3) Ans.

= 0.587 MPa (T) Orientation of Neutral Axis: tan a =

tan a =

Iz Iy

tan u

57.6014(10 - 6) 0.366827(10 - 3)

tan ( -22.62°) Ans.

a = -3.74°

Ans: y = 57.4 mm, sA = 1.30 MPa (C), sB = 0.587 MPa (T), a = -3.74°

402 402

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*6–108. The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown. Determine maximum bending stress in the strut. The location y of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.

y M � 520 N�m 20 mm z

–y

12 5

13

B

200 mm

C 20 mm 200 mm

A

20 mm

200 mm

Internal Moment Components: Mz = -

12 (520) = - 480 N # m 13

My =

5 (520) = 200 N # m 13

Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)

y =

Ans.

= 0.057368 m = 57.4 mm Iz =

1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368 - 0.01)2 12

1 (0.04) A 0.183 B + 0.04(0.18)(0.110 - 0.057368)2 12

+

= 57.6014 A 10 - 6 B m4

Iy =

1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10 - 3 B m4 12 12

Maximum Bending Stress: By inspection, the maximum bending stress can occur at either point A or B. Applying the flexure formula for biaxial bending at points A and B s = -

Mz y

sA = -

Iz

+

My z Iy

-480( - 0.142632) -6

57.6014(10 )

+

200( - 0.2) 0.366827(10 - 3) Ans.

= - 1.298 MPa = 1.30 MPa (C) (Max) sB = -

- 480(0.057368) -6

57.6014(10 )

+

200(0.2) 0.366827(10 - 3)

= 0.587 MPa (T) Orientation of Neutral Axis: tan a =

tan a =

Iz Iy

tan u

57.6014(10 - 6) 0.366827(10 - 3)

tan ( -22.62°) Ans.

a = -3.74°

403 403

CH 06.indd 403

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6–109. The steel shaft is subjected to the two loads. If the journal bearings at A and B do not exert an axial force on the shaft, determine the required diameter of the shaft if the allowable bending stress is sallow = 180 MPa.

4 kN

30 30

4 kN

B 1.25 m

1m A

1.25 m

Solution Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. The moment diagram for each component is drawn. Allowable Bending Stress: Since all the axes through the circle’s center for a circular shaft are principal axes, then the resultant moment M = 2M2y + M2z can be used for the design. The maximum resultant moment is Mmax = 24.3302 + 0.71432 = 4.389 kN # m. Applying the flexure formula, smax = sallow = 180 ( 106 ) =

Mmaxc I

4.389 ( 103 ) 1 d2 2 p 4

1 d22 4

Ans.

d = 0.06286 m = 62.9 mm

Ans: d = 62.9 mm 571

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6–110. The 65-mm-diameter steel shaft is subjected to the two loads. If the journal bearings at A and B do not exert an axial force on the shaft, determine the absolute maximum bending stress developed in the shaft.

4 kN

30 30

4 kN

B 1.25 m

1m A

1.25 m

Solution Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. The moment diagram for each component is drawn. Maximum Bending Stress: Since all the axes through the circle’s center for a circular shaft are principal axes, then the resultant moment M = 2M2y + M2z can be used to determine the maximum bending stress. The maximum resultant moment is Mmax = 24.3302 + 0.71432 = 4.389 kN # m. Applying the flexure formula, smax = =

Mmaxc I

4.389 ( 103 ) (0.0325) p 4

( 0.03254 ) Ans.

= 163 MPa

Ans: smax = 163 MPa 572

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6–111. For the section, Iz′ = 31.7(10−6) m4, Iy′ = 114(10−6) m4, Iy′z′ = −15.8(10−6) m4. Using the techniques outlined in Appendix A, the member’s cross-sectional area has principal moments of inertia of Iz = 28.8(10−6) m4 and Iy = 117(10−6) m4, calculated about the principal axes of inertia y and z, respectively. If the section is subjected to the moment M = 15 kN # m, determine the stress at point A using Eq. 6–17.

y

y¿

10.5 60 mm M  15 kNm z¿

C

z 140 mm

Solution

A

60 mm

60 mm

80 mm 60 mm

Internal Moment Components: The y and z components are My = 15 sin 10.5° = 2.7335 kN # m  Mz = 15 cos 10.5° = 14.7488 kN # m Section Properties: Given that Iy = 117 ( 10-6 ) m4 and Iz = 28.8 ( 10-6 ) m4, the coordinates of point A with respect to the y and z axes are y = 0.06 cos 10.5° - 0.14 sin 10.5° = 0.03348 m z = - (0.06 sin 10.5° + 0.14 cos 10.5°) = -0.14859 m Bending Stress: Applying the flexure formula for biaxial bending, Myz Mzy s = + Iz Iy sA = -

14.7488 ( 103 ) (0.03348) 28.8 ( 10-6 )

+

2.7335 ( 103 ) ( - 0.14869) 117 ( 10-6 )

= - 20.62 ( 106 ) Pa Ans.

= 20.6 MPa (C)

Ans: sA = 20.6 MPa (C) 573

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*6–112. Solve Prob. 6–111 using the equation developed in Prob. 6–106.

y

y¿

10.5 60 mm M  15 kNm z¿

C

z 140 mm

Solution

A

60 mm

60 mm

80 mm 60 mm

Internal Moment Components: The y′and z′ components are Mz′ = 15 kN # m  My′ = 0 Section Properties: Given that Iy′ = 114 ( 10-6 ) m4, Iz′ = 31.7 ( 10-6 ) m4 and Iy′z′ = - 15.8 ( 10-6 ) m4, the coordinates of point A with respect to the y′ and z axes are y′ = 0.06 m  z′ = - 0.14 m Bending Stress: Using the formula, s =

- (Mz′Iy′ + My′Iy′z′)y′ + (My′Iz′ + Mz′Iy′z′)z′ Iy′Iz′ - I 2y′z′

- 3 15 ( 10 ) (114) ( 10 3

sA =

) + 0 4 (0.06) + 3 0 + 15 ( 103 ) ( - 15.8) ( 10-6 ) 4 ( - 0.14) 3 114 ( 10-6 ) 4 3 31.7 ( 10-6 ) 4 - 3 - 15.8 ( 10-6 ) 4 2 -6

= -20.64 ( 106 ) Pa

Ans.

= 20.6 MPa (C)

Ans: sA = 20.6 MPa (C) 574

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6–113. If the beam is subjected to the internal moment of M = 1200 kN # m, determine the maximum bending stress acting on the beam and the orientation of the neutral axis.

y 150 mm 150 mm

Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus,

M 300 mm 30�

My = 1200 sin 30° = 600 kN # m

150 mm

Mz = - 1200 cos 30° = - 1039.23 kN # m

x 150 mm

z

Section Properties: The location of the centroid of the cross-section is given by ©yA 0.3(0.6)(0.3) - 0.375(0.15)(0.15) = = 0.2893 m ©A 0.6(0.3) - 0.15(0.15)

y =

150 mm

The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = Iz =

1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4 12 12

1 (0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2 12 - c

1 (0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d 12

= 5.2132 A 10 - 3 B m4

Bending Stress: By inspection, the maximum bending stress occurs at either corner A or B. s = -

Mzy

sA = -

+

Iz

Myz Iy

c - 1039.23 A 103 B d (0.2893) 5.2132 A 10 - 3 B

+

600 A 103 B (0.15) 1.3078 A 10 - 3 B

= 126 MPa (T)

sB = -

c - 1039.23 A 103 B d ( -0.3107) 5.2132 A 10 - 3 B

+

600 A 103 B ( -0.15)

= - 131 MPa = 131 MPa (C)(Max.)

1.3078 A 10 - 3 B

Ans.

Orientation of Neutral Axis: Here, u = - 30°. tan a =

tan a =

Iz Iy

tan u

5.2132 A 10 - 3 B

1.3078 A 10 - 3 B

tan( -30°) Ans.

a = -66.5°

The orientation of the neutral axis is shown in Fig. b.

Ans: sB = 131 MPa (C), a = -66.5° 407 407

CH 06.indd 407

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6–114. If the beam is made from a material having an allowable tensile and compressive stress of (sallow)t = 125 MPa and (sallow)c = 150 MPa, respectively, determine the maximum allowable internal moment M that can be applied to the beam.

y 150 mm 150 mm M 300 mm

Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus, My = M sin 30° = 0.5M

30� 150 mm x 150 mm

z

Mz = -M cos 30° = - 0.8660M Section Properties: The location of the centroid of the cross section is 150 mm

0.3(0.6)(0.3) - 0.375(0.15)(0.15) ©yA y = = = 0.2893 m ©A 0.6(0.3) - 0.15(0.15) The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = Iz =

1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10 - 3 B m4 12 12

1 (0.3) A 0.63 B + 0.3(0.6)(0.3 - 0.2893)2 12 - c

1 (0.15) A 0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d 12

= 5.2132 A 10 - 3 B m4

Bending Stress: By inspection, the maximum bending stress can occur at either corner A or B. For corner A which is in tension, sA = (sallow)t = 125 A 106 B = -

Mz yA Iz

+

My zA Iy

( - 0.8660M)(0.2893) 5.2132 A 10

-3

B

+

0.5M(0.15) 1.3078 A 10 - 3 B

M = 1185 906.82 N # m = 1186 kN # m (controls)

Ans.

For corner B which is in compression, sB = (sallow)c = - 150 A 106 B = -

Mz yB Iz

+

My zB Iy

(- 0.8660M)(-0.3107) 5.2132 A 10

-3

B

M = 1376 597.12 N # m = 1377 kN # m

+

0.5M( -0.15) 1.3078 A 10 - 3 B

Ans:

M = 1186 kN # m 408 408

CH 06.indd 408

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6–115. The shaft is subjected to the vertical and horizontal loadings of two pulleys D and E as shown. It is supported on two journal bearings at A and B which offer no resistance to axial loading. Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft. Determine the required diameter d of the shaft if the allowable bending stress is sallow = 180 MPa.

y

z 1m 1m 1m 1m A D x

Solution

E

B

C 400 N

100 mm 400 N 60 mm

150 N 150 N

Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. Allowable Bending Stress: Since all the axes through the circle’s center for a circular shaft are principal axes, then the resultant moment M = 2M2y + M2z can be used for the design.The maximum resultant moment is Mmax = 24002 + 1502 = 427.2 N # m.

Applying the flexure formula smax = sallow = 180 ( 106 ) =

Mmaxc I 427.2 ( d2 )

( )

p d 4 4 2

Ans.

d = 0.02891 m = 28.9 mm

Ans: d = 28.9 mm 577

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*6–116. For the section, Iy′ = 31.7(10-6) m4, Iz′ = 114(10-6) m4, -6 4 Iy′z′ = 15.8(10 ) m . Using the techniques outlined in Appendix A, the member’s cross-sectional area has principal moments of inertia of Iy = 28.8(10-6) m4 and Iz = 117(10-6) m4, calculated about the principal axes of inertia y and z, respectively. If the section is subjected to a moment of M = 2500 N # m, determine the stress produced at point A, using Eq. 6–17.

y 60 mm

y¿ 60 mm 60 mm

M  2500 Nm z¿ 10.5 z

80 mm C 140 mm

60 mm

Solution Iz = 117(10 - 6) m4

A

Iy = 28.8(10 - 6) m4

My = 2500 sin 10.5° = 455.59 N # m

Mz = 2500 cos 10.5° = 2458.14 N # m y = - 0.06 sin 10.5° - 0.14 cos 10.5° = - 0.14859 m z = 0.14 sin 10.5° - 0.06 cos 10.5° = - 0.03348 m sA = =

- Mzy Iz

+

My z Iy

2458.14( -0.14859) -6

117(10 )

+

455.59( - 0.03348) 28.8(10 - 6)

= 2.59 MPa (T)

Ans.

Ans: sA = 2.59 MPa (T) 578

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6–117. Solve Prob. 6–116 using the equation developed in Prob. 6–106.

y 60 mm

y¿ 60 mm 60 mm

M  2500 Nm z¿ 10.5 z

=

140 mm

60 mm

Solution sA =

80 mm C

A

- (Mz′Iy′ + My′Iy′z′)y′ + (My′Iz′ + Mz′Iy′z′)z′ Iy′Iz′ - Iy′z′2 - 32500(31.7)(10 - 6) + 04( -0.14) + 30 + 2500(15.8)(10 - 6) 4( - 0.06) 31.7(10 - 6)(114)(10 - 6) - 3(15.8)(10 - 6) 4 2

Ans.

= 2.59 MPa (T)

Ans: sA = 2.59 MPa (T) 579

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6–118. If the applied distributed loading of w = 4 kN>m can be assumed to pass through the centroid of the beam’s cross-sectional area, determine the absolute maximum bending stress in the joist and the orientation of the neutral axis. The beam can be considered simply supported at A and B.

w

A

15 6m 15

10 mm 15 mm

Solution Internal Moment Components: The uniform distributed load w can be resolved into its y and z components as shown in Fig. a.

w(6 m)

B

15 mm 100 mm 100 mm

15

15 100 mm

wy = 4 cos 15° = 3.864 kN>m wz = 4 sin 15° = 1.035 kN>m wy and wz produce internal moments in the beam about the z and y axes, respectively. For the simply supported beam subjected to the uniform distributed wL2 load, the maximum moment in the beam is Mmax = . Thus, 8 (Mz)max = (My)max =

wyL2

=

8 wzL2

=

8

3.864(62) 8 1.035(62) 8

= 17.387 kN # m = 4.659 kN # m

As shown in Fig. b, (Mz)max and (My)max are positive since they are directed towards the positive sense of their respective axes. Section Properties: The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = 2c

1 1 (0.015)(0.13) d + (0.17)(0.013) = 2.5142(10 - 6) m4 12 12

1 1 (0.1)(0.23) (0.09)(0.17 3) = 29.8192(10 - 6) m4 12 12 Bending Stress: By inspection, the maximum bending stress occurs at points A and B. Iz =

s = -

(Mz)max y Iz

smax = sA = -

+

(My)max z Iy

17.387(103)(- 0.1) 29.8192 (10 - 6)

+

4.659(103)(0.05) 2.5142(10 - 6) Ans.

= 150.96 MPa = 151 MPa (T) 3

smax = sB = -

17.387(10 )(0.1) -6

29.8192 (10 )

3

+

4.659(10 )(- 0.05) 2.5142(10 - 6) Ans.

= - 150.96 MPa = 151 MPa (C) Orientation of Neutral Axis: Here, u = tan - 1 c tan a = tan a =

Iz Iy

(My)max (Mz)max

tan u

29.8192(10 - 6) 2.5142(10 - 6)

d = tan - 1 a

4.659 b = 15°. 17.387

tan 15° Ans.

a = 72.5° The orientation of the neutral axis is shown in Fig. c. 580

Ans: smax = 151 MPa, a = 72.5°

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6–119. Determine the maximum allowable intensity w of the uniform distributed load that can be applied to the beam. Assume w passes through the centroid of the beam’s crosssectional area, and the beam is simply supported at A and B. The allowable bending stress is sallow = 165 MPa.

w

A

15 6m 15

10 mm 15 mm

Solution Internal Moment Components: The uniform distributed load w can be resolved into its y and z components as shown in Fig. a.

w(6 m)

B

15 mm 100 mm 100 mm

15

15 100 mm

wy = w cos 15° = 0.9659w wz = w sin 15° = 0.2588w wy and wz produce internal moments in the beam about the z and y axes, respectively. For the simply supported beam subjected to the a uniform distributed load, the wL2 maximum moment in the beam is Mmax = . Thus, 8 (Mz)max = (My)max =

wyL2 8 wzL2 8

= =

0.9659w(62) 8 0.2588w(62) 8

= 4.3476w = 1.1647w

As shown in Fig. b, (Mz)max and (My)max are positive since they are directed towards the positive sense of their respective axes. Section Properties: The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = 2c Iz =

1 1 (0.015)(0.13) d + (0.17)(0.013) = 2.5142(10 - 6) m4 12 12

1 1 (0.1)(0.23) (0.09)(0.173) = 29.8192(10 - 6) m4 12 12

Bending Stress: By inspection, the maximum bending stress occurs at points A and B. We will consider point A. sA = sallow = 165(106) = -

(Mz)maxyA Iz

+

4.3467w( - 0.1) -6

29.8192(10 )

(My)maxzA Iy +

1.1647w(0.05) 2.5142(10 - 6) Ans.

w = 4372.11 N >m = 4.37 kN>m

Ans: w = 4.37 kN>m 581

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*6–120. The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). Determine the dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two metals. What maximum moment will this beam support if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa?

h B

50 mm

A 150 mm

Section Properties: n =

68.9(109) Eal = 0.68218 = Ebr 101(109)

bbr = nbal = 0.68218(0.15) = 0.10233 m y = 0.05 =

©yA ©A 0.025(0.10233)(0.05) + (0.05 + 0.5h)(0.15)h 0.10233(0.05) + (0.15)h Ans.

h = 0.04130 m = 41.3 mm INA =

1 (0.10233) A 0.053 B + 0.10233(0.05)(0.05 - 0.025)2 12

+

1 (0.15) A 0.041303 B + 0.15(0.04130)(0.070649 - 0.05)2 12

= 7.7851 A 10 - 6 B m4

Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =

Mc INA M(0.04130) 7.7851(10 - 6)

M = 6598 N # m = 6.60 kN # m (controls!)

Ans.

Assume failure of aluminium (sallow)al = n

Mc INA

128 A 106 B = 0.68218 c

M(0.05) 7.7851(10 - 6)

d

M = 29215 N # m = 29.2 kN # m

Ans: h = 41.3 mm, M = 6.60 kN # m (controls!) 415 415

CH 06.indd 415

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6–121. The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). If the height h = 40 mm, determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa.

h B

50 mm

A

Section Properties: For transformed section.

150 mm

68.9(109) Eal = 0.68218 = n = Ebr 101.0(109) bbr = nbal = 0.68218(0.15) = 0.10233 m y = =

©yA ©A 0.025(0.10233)(0.05) + (0.07)(0.15)(0.04) 0.10233(0.05) + 0.15(0.04)

= 0.049289 m INA =

1 (0.10233) A 0.053 B + 0.10233(0.05)(0.049289 - 0.025)2 12

+

1 (0.15) A 0.043 B + 0.15(0.04)(0.07 - 0.049289)2 12

= 7.45799 A 10 - 6 B m4

Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =

Mc INA M(0.09 - 0.049289) 7.45799(10 - 6)

M = 6412 N # m = 6.41 kN # m (controls!)

Ans.

Assume failure of aluminium (sallow)al = n

Mc INA

128 A 106 B = 0.68218 c

M(0.049289) 7.45799(10 - 6)

d

M = 28391 N # m = 28.4 kN # m

Ans:

M = 6.41 kN # m 416 416

CH 06.indd 416

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6–122. The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown. If it is subjected to a moment of M = 6.5 kN # m, determine the maximum bending stress in the brass and steel. Also, what is the stress in each material at the seam where they are bonded together? Ebr = 100 GPa, Est = 200 GPa.

A

y 50 mm 200 mm M

B z

Solution n = y = I =

x 175 mm

200(109) Est = = 2 Ebr 100(109) (350)(50)(25) + (175)(200)(150) 350(50) + 175(200)

= 108.33 mm

1 1 (0.35)(0.053) + (0.35)(0.05)(0.083332) + (0.175)(0.23) + 12 12 (0.175)(0.2)(0.041672) = 0.3026042(10 - 3) m4

Maximum stress in brass: (sbr)max =

6.5(103)(0.14167) Mc1 = = 3.04 MPa I 0.3026042(10 - 3)

Ans.

Maximum stress in steel: (sst)max =

(2)(6.5)(103)(0.10833) nMc2 = = 4.65 MPa I 0.3026042(10 - 3)

Ans.

Stress at the junction: sbr =

6.5(103)(0.05833) Mr = = 1.25 MPa I 0.3026042(10 - 3)

Ans. Ans.

sst = nsbr = 2(1.25) = 2.51 MPa

Ans: (sbr)max = 3.04 MPa, (sst)max = 4.65 MPa, sbr = 1.25 MPa, sst = 2.51 MPa 582

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6–123. The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown. If the allowable bending stress for the steel is (sallow)st = 180 MPa, and for the brass (sallow)br = 60 MPa, determine the maximum moment M that can be applied to the beam. Ebr = 100 GPa, Est = 200 GPa.

A

y 50 mm 200 mm M

B z

Solution n = y = I =

x 175 mm

200(109) Est = = 2 Ebr 100(109) (350)(50)(25) + (175)(200)(150) 350(50) + 175(200)

= 108.33 mm

1 1 (0.35)(0.053) + (0.35)(0.05)(0.083332) + (0.175)(0.23) + 12 12 (0.175)(0.2)(0.041672) = 0.3026042(10 - 3) m4

(sst)allow =

nMc2 ; I

180(106) =

(2) M(0.10833) 0.3026042(10 - 3)

M = 251 kN # m (sbr)allow =

Mc1 ; I

60(106) =

M(0.14167) 0.3026042(10 - 3)

M = 128 kN # m (controls)

Ans.

Ans: M = 128 kN # m 583

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*6–124. The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stress for for the steel is (sst)allow = 280 MPa and the allowable compressive stress for the concrete is (sconc)allow = 21 MPa, determine the maximum moment M that can be applied to the section. Assume the concrete cannot support a tensile stress. Est = 200 GPa, Econc = 26.5 GPa.

8 in. 200 mm 6 150 in. mm 8 in. 200 mm

4 in. 100 mm

M M 18 mm in. 450 in. 50 2mm 1-in. diameter 25-mm diameterrods rods

π (0.01252 )] 0.3125π (10 −3 ) m 2 = Ast 2[=

550 mm

 200  −3 −3 2   [0.3125π (10 )] = 7.4094(10 ) m  26.5 

©yA = 0;

150 mm

400 mm

 h′  (h′ + 0.05)(0.55)(0.1) +   (0.15)h′ − 7.4094(10 −3 )(0.4 − h′) = 0 2 0 0.075h′2 + 0.062409 h′ − 0.21377(10 −3 ) =

Solving for the positive root: h′ = 0.0034112 m I=

1 (0.55)(0.13 ) + (0.55)(0.1)(0.0034112 + 0.05)2 12 +

1 (0.15)(0.0034112 3 ) + 0.15(0.0034112)(0.0034112 2)2 12

+ 7.4094(10 −3 )(0.4 − 0.0034112)2 = 1.36811(10 −3 ) m 4

Assume concrete fails: (scon)allow =

My ; I

 200   M (0.4 − 0.0034112)  21(106 ) =     26.5   1.36811(10 −3 )  ⋅ m 278 kN ⋅ m M = 277.83(10 3 ) N=

Assume steel fails: (sst)allow = na

My b; I

280(106 ) =

M (0.4 + 0.0034112) 1.36811(10 −3 )

= M 127.98(10 3 ) N= ⋅ m 128 kN ⋅ m (Controls)

Ans.

Ans: M = 128 kN ⋅ m (Controls) 429 429

CH 06.indd 429

1/18/11 2:24:38 PM

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6–125. The low strength concrete floor slab is integrated with a wide-flange A-36 steel beam using shear studs (not shown) to form the composite beam. If the allowable bending stress for the concrete is (sallow)con = 10 MPa, and allowable bending stress for steel is (sallow)st = 165 MPa, determine the maximum allowable internal moment M that can be applied to the beam.

1m

100 mm

15 mm 400 mm M 15 mm 15 mm

Section Properties: The beam cross section will be transformed into Econ 22.1 that of steel. Here, Thus, = = 0.1105. n = Est 200 bst = nbcon = 0.1105(1) = 0.1105 m. The location of the transformed section is y = =

200 mm

©yA ©A 0.0075(0.015)(0.2) + 0.2(0.37)(0.015) + 0.3925(0.015)(0.2) + 0.45(0.1)(0.1105) 0.015(0.2) + 0.37(0.015) + 0.015(0.2) + 0.1(0.1105)

= 0.3222 m The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =

1 (0.2) A 0.0153 B 12

+ 0.2(0.015)(0.3222 - 0.0075)2 + + +

1 (0.015) A 0.373 B + 0.015(0.37)(0.3222 - 0.2)2 12

1 (0.2) A 0.0153 B + 0.2(0.015)(0.3925 - 0.3222)2 12 1 (0.1105) A 0.13 B + 0.1105(0.1)(0.45 - 0.3222)2 12

= 647.93 A 10 - 6 B m4

Bending Stress: Assuming failure of steel, (sallow)st =

M(0.3222) Mcst ; 165 A 106 B = I 647.93 A 10 - 6 B

M = 331 770.52 N # m = 332 kN # m

Assuming failure of concrete,

(sallow)con = n

Mccon ; I

10 A 106 B = 0.1105C

M(0.5 - 0.3222) 647.93 A 10 - 6 B

S

M = 329 849.77 N # m = 330 kN # m (controls) Ans. Ans:

M = 330 kN # m 427 427

CH 06.indd 427

1/18/11 2:24:31 PM

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6–126. The wooden section of the beam is reinforced with two steel plates as shown. Determine the maximum moment M that the beam can support if the allowable stresses for the wood and steel are (sallow)w = 6 MPa, and (sallow)st = 150 MPa, respectively. Take Ew = 10 GPa and Est = 200 GPa.

15 mm

150 mm M

15 mm

Solution Section Properties: The cross section will be transformed into that of steel as shown Ew 10 in Fig. a. Here, n = = = 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m. The Est 200 moment of inertia of the transformed section about the neutral axis is I =

100 mm

1 1 (0.1)(0.183) (0.095)(0.153) = 21.88125(10 - 6) m4 12 12

Bending Stress: Assuming failure of the steel, (sallow)st =

Mcst ; I

150(106) =

M(0.09) 21.88125(10 - 6)

M = 36 468.75 N # m = 36.5 kN # m Assuming failure of wood, (sallow)w = n

Mcw ; I

6(106) = 0.05c

M(0.075) 21.88125(10 - 6)

d

M = 35 010 N # m = 35.0 kN # m (controls)

Ans.

Ans: M = 35.0 kN # m 587

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6–127. The wooden section of the beam is reinforced with two steel plates as shown. If the beam is subjected to a moment of M = 30 kN # m, determine the maximum bending stresses in the steel and wood. Sketch the stress distribution over the cross section. Take Ew = 10 GPa and Est = 200 GPa.

15 mm

150 mm M

15 mm

Solution Section Properties: The cross section will be transformed into that of steel as shown Ew 10 in Fig. a. Here, n = = = 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m. The Est 200 moment of inertia of the transformed section about the neutral axis is I =

100 mm

1 1 (0.1)(0.183) (0.095)(0.153) = 21.88125(10 - 6) m4 12 12

Maximum Bending Stress: For the steel, (smax)st =

30(103)(0.09) Mcst = = 123 MPa I 21.88125(10 - 6)

sst  y = 0.075 m =

Ans.

30(103)(0.075) My = = 103 MPa I 21.88125(10 - 6)

For the wood, (smax)w = n

30(103)(0.075) Mcw = 0.05c d = 5.14 MPa I 21.88125(10 - 6)

Ans.

The bending stress distribution across the cross section is shown in Fig. b.

Ans: (smax)st = 123 MPa, (smax)w = 5.14 MPa 588

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*6–128. y

A wood beam is reinforced with steel straps at its top and bottom as shown. Determine the maximum bending stress developed in the wood and steel if the beam is subjected to a moment of M = 150 kN # m. Sketch the stress distribution acting over the cross section. Take Ew  = 10 GPa, Est = 200 GPa.

40 mm

400 mm M  150 kNm 40 mm x

Solution Ew 10 GPa = = 0.05. Then Est 200 GPa bst = nbw = 0.05(0.2) = 0.01 m. For the transformed section shown in Fig. a,

z

Section Properties: Here, n =

INA =

200 mm

1 1 (0.2) ( 0.483 ) (0.19) ( 0.43 ) = 0.82987 ( 10-3 ) m4 12 12

Maximum Bending Stress: Applying the flexure formula, (sst)max = (sw)max = n

150 ( 103 ) (0.24) Mcst = = 43.38 ( 106 ) Pa = 43.4 MPa I 0.82987 ( 10-3 )

150 ( 103 ) (0.2) Mcw = 0.05 c d = 1.8075 ( 106 ) Pa = 1.81 MPa I 0.82987 ( 10-3 )

Ans. Ans.

The bending stress on the steel at y = 0.2 m is sst =

150 ( 103 ) (0.2) My = = 36.15 ( 106 ) Pa = 36.2 MPa I 0.82987 ( 10-3 )

Using these results, the stress distribution on the beam’s cross-section shown in Fig. b can be sketched.

Ans: (sst)max = 43.4 MPa, (sw)max = 1.81 MPa 591

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6–129. The Douglas Fir beam is reinforced with A-992 steel straps at its sides. Determine the maximum stress in the wood and steel if the beam is subjected to a moment of Mz = 80 kN # m. Sketch the stress distribution acting over the cross section.

y 20 mm

400 mm

200 mm

20 mm

z

Solution Ew 13.1 GPa = = 0.0655. Then Est 200 GPa bst = nbw = 0.0655(0.2) = 0.0131 m. For the transformed section shown in Fig. a,

Section Properties: Here, n =

INA =

1 (0.04 + 0.0131) ( 0.43 ) = 0.2832 ( 10-3 ) m4 12

Maximum Bending Stress: Applying the flexure formula, 80 ( 103 ) (0.2) Mc = = 56.497 ( 106 ) Pa = 56.5 MPa Ans. I 0.2832 ( 10-3 ) 80 ( 103 ) (0.2) Mc = n = 0.0655 c d = 3.7006 ( 106 ) Pa = 3.70 MPa Ans. I 0.2832 ( 10-3 )

(sst)max = (sw)max

Using these results, the stress distribution on the beam’s cross-section shown in Fig. b can be sketched.

Ans: (sst)max = 56.5 MPa, (sw)max = 3.70 MPa 589

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6–130. If P = 3 kN, determine the bending stress at points A, B, and C of the cross section at section a–a. Using these results, sketch the stress distribution on section a–a.

D

600 mm

E

a 300 mm A

Solution Internal Moment: The internal moment developed at section a–a can be determined by writing the moment equation of equilibrium about the neutral axis of the cross section at a–a. Using the free-body diagram shown in Fig. a, a+ ΣMNA = 0;   3(0.6) - Ma - a = 0   Ma - a = 1.8 kN # m

B C a P

20 mm 50 mm 25 mm 25 mm 25 mm Section a – a

Here, M a - a is considered negative since it tends to reduce the curvature of the curved segment of the beam. Section Properties: Referring to Fig. b, the location of the centroid of the cross section from the center of the beam’s curvature is r =

0.31(0.02)(0.075) + 0.345(0.05)(0.025) ΣrA = = 0.325909 m ΣA 0.02(0.075) + 0.05(0.025)

The location of the neutral surface from the center of the beam’s curvature can be determined from R =

A dA ΣLA r

where A = 0.02(0.075) + 0.05(0.025) = 2.75(10 - 3) m2 dA 0.32 0.37 Σ = 0.075 ln + 0.025 ln = 8.46994(10 - 3) m r 0.3 0.32 LA Thus, R =

2.75(10 - 3) 8.46994(10 - 3)

= 0.324678 m

then e = r - R = 0.325909 - 0.324678 = 1.23144(10 - 3) m Normal Stress: sA = sB = sC =

M(R - rA) AerA M(R - rB) AerB M(R - rC) AerC

= = =

1.8(103)(0.324678 - 0.3) 2.75(10 - 3)(1.23144)(10 - 3)(0.3) 1.8(103)(0.324678 - 0.32) 2.75(10 - 3)(1.23144)(10 - 3)(0.32) 1.8(103)(0.324678 - 0.37) 2.75(10 - 3)(1.23144)(10 - 3)(0.37)

= 43.7 MPa (T)

Ans.

= 7.77 MPa (T)

Ans.

= -65.1 MPa (C)

Ans.

Ans: sA = 43.7 MPa (T), sB = 7.77 MPa (T), sC = -65.1 MPa (C) 604

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6–131. If the maximum bending stress at section a–a is not allowed to exceed sallow = 150 MPa, determine the maximum allowable force P that can be applied to the end E.

D

600 mm

E

a 300 mm A

Solution Internal Moment: The internal Moment developed at section a–a can be determined by writing the moment equation of equilibrium about the neutral axis of the cross section at a–a.

B C a P

20 mm 50 mm 25 mm 25 mm 25 mm Section a – a

a+ ΣMNA = 0;   P(0.6) - Ma - a = 0   Ma - a = 0.6P Here, Ma–a is considered positive since it tends to decrease the curvature of the curved segment of the beam. Section Properties: Referring to Fig. b, the location of the centroid of the cross section from the center of the beam’s curvature is r =

0.31(0.02)(0.075) + 0.345(0.05)(0.025) Σr~A = = 0.325909 m ΣA 0.02(0.075) + 0.05(0.025)

The location of the neutral surface from the center of the beam’s curvature can be determined from R =

A dA r

A L

Σ

where A = 0.02(0.075) + 0.05(0.025) = 2.75(10 - 3) m2 Σ

dA 0.32 0.37 = 0.075 ln + 0.025 ln = 8.46994(10 - 3) m r 0.3 0.32

A L

Thus, R =

2.75(10 - 3) 8.46994(10 - 3)

= 0.324678 m

then e = r - R = 0.325909 - 0.324678 = 1.23144(10 - 3) m Allowable Normal Stress: The maximum normal stress occurs at either points A or C. For point A, which is in tension, sallow =

M(R - rA) AerA

;   150(106) =

0.6P(0.324678 - 0.3) 2.75(10 - 3)(1.23144)(10 - 3)(0.3)

P = 10 292.09 N = 10.3 kN For point C, which is in compression, sallow =

M(R - rC) AerC

;    - 150(106) =

0.6P(0.324678 - 0.37) 2.75(10 - 3)(1.23144)(10 - 3)(0.37)

P = 6911.55 N = 6.91 kN (controls)

Ans.

Ans: P = 6.91 kN 605

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*6–132. If the beam is subjected to a moment of M = 45 kN # m, determine the maximum bending stress in the A-36 steel section A and the 2014-T6 aluminum alloy section B.

A 50 mm

M 15 mm

Solution

150 mm

B

Section Properties: The cross section will be transformed into that of steel as shown 73.1 1 109 2 Eal in Fig. a. Here, n = = = 0.3655. Thus, bst = nbal = 0.3655(0.015) = Est 200 1 109 2 0.0054825 m. The location of the transformed section is ΣyA y = = ΣA

0.075(0.15)(0.0054825) + 0.2c p 1 0.052 2 d 0.15(0.0054825) + p 1 0.052 2

= 0.1882 m

The moment of inertia of the transformed section about the neutral axis is I = ΣI + Ad 2 =

1 (0.0054825) ( 0.153 ) + 0.0054825(0.15)(0.1882 - 0.075)2 12 +

1 p 10.0542 + p 10.0522 (0.2 - 0.1882)2 4

= 18.08 110 - 62 m4

Maximum Bending Stress: For the steel, (smax)st =

45 11032 (0.06185) Mcst = = 154 MPa I 18.08 110 - 62

Ans.

For the aluminum alloy, (smax)al = n

45 11032 (0.1882) Mcal = 0.3655 C S = 171 MPa I 18.08 110 - 62

Ans.

Ans: (smax)st = 154 MPa, (smax)al = 171 MPa 596

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6–133. For the curved beam in Fig. 6–40a, show that when the radius of curvature approaches infinity, the curved-beam formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13.

Solution Normal Stress: Curved-beam formula s = s =

M(R - r)

dA LA r

where A′ =

Ar(r - R)

and R =

A

dA 1A r

=

A A′

M(A - rA′)

(1)

Ar(rA′ - A)

(2)

r = r + y rA′ = r

r dA = a - 1 + 1bdA LA r LA r + y =

LA

a

= A -

r - r - y r + y y

LA r + y

+ 1b dA (3)

dA

Denominator of Eq. (1) becomes, y

LA r + y

Ar(rA′ - A) = Ar ¢A -

dA - A≤ = -Ar

y

LA r + y

dA

Using Eq. (2), Ar(rA′ - A) = - A = A =

LA

¢

ry r + y

+ y - y≤dA - Ay

y dA - A 1A y dA - Ay dA LA r + y LA r + y

y2 y Ay A ¢ ¢ A y dA y ≤dA - A 1 y ≤ dA r LA 1 + r LA 1 + r r as

y r

S 0

Ar(rA′ - A) S

Then, Eq. (1) becomes Using Eq. (2),

A I r

Mr (A - rA′) AI Mr s = (A - rA′ - yA′) AI s =

Using Eq. (3), s =

=

dA

y2

1A y dA = 0,

But,

y

LA r + y

y Mr dA C A - ¢A S dA≤ - y AI LA r + y LA r + y y dA Mr C S dA - y AI LA r + y r LA + y 598

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6–133. (Continued)

y

y dA Mr r = C ¢ ≤dA ¢ y≤S AI LA 1 + y r LA 1 + r r As

y r

S 0 y r

LA 1 + ¢

y≤

dA = 0 and

r

Therefore,   s =

y

r LA 1 + ¢

dA

y≤ r

=

yA A dA = 1 r r

y

yA My Mr ab = AI I r

(Q.E.D)

Ans: N/A 599

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6–134. The curved member is subjected to the moment of M = 50 kN # m. Determine the percentage error introduced in the calculation of maximum bending stress using the flexure formula for straight members.

M 100 mm

M 200 mm

200 mm

Solution Straight Member: The maximum bending stress developed in the straight member smax =

50(103)(0.1) Mc = 75 MPa = I 1 (0.1)(0.23) 12

Curved Member: When r = 0.2 m, r = 0.3 m, rA = 0.2 m and rB = 0.4 m, Fig. a. The location of the neutral surface from the center of curvature of the curve member is ­R =

0.1(0.2) A = = 0.288539 m 0.4 dA 0.1ln 0.2 LA r

Then e = r - R = 0.011461 m

Here, M = 50 kN # m. Since it tends to decrease the curvature of the curved member, sB =

M(R - rB) AerB

=

50(103)(0.288539 - 0.4) 0.1(0.2)(0.011461)(0.4)

= - 60.78 MPa = 60.78 MPa (C) sA =

M(R - rA) AerA

=

50(103)(0.288539 - 0.2) 0.1(0.2)(0.011461)(0.2)

= 96.57 MPa (T) (Max.) Thus, % of error = a

96.57 - 75 b100 = 22.3% 96.57

Ans.

Ans: , of error = 22.3, 600

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6–135. The curved member is made from material having an allowable bending stress of sallow = 100 MPa. Determine the maximum allowable moment M that can be applied to the member.

M 100 mm

M 200 mm

200 mm

Solution Internal Moment: M is negative since it tends to decrease the curvature of the curved member. Section Properties: Referring to Fig. a, the location of the neutral surface from the center of curvature of the curve beam is R =

0.1(0.2) A = = 0.288539 m 0.4 dA 0.1ln 0.2 LA r

Then e = r - R = 0.3 - 0.288539 = 0.011461 m Allowable Bending Stress: The maximum stress occurs at either point A or B. For point A, which is in tension, sallow =

M(R - rA) AerA

;    100(106) =

M(0.288539 - 0.2) 0.1(0.2)(0.011461)(0.2)

M = 51 778.27 N # m = 51.8 kN # m (controls)

Ans.

For point B, which is in compression, sallow =

M(R - rB) AerB

;     - 100(106) =

M(0.288539 - 0.4) 0.1(0.2)(0.011461)(0.4)

M = 82 260.10 N # m = 82.3 kN # m

Ans: M = 51.8 kN # m 601

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*6–136. The curved beam is subjected to a bending moment of M = 900 N # m as shown. Determine the stress at points A and B, and show the stress on a volume element located at each of these points.

A C B

100 mm C

A

30�

20 mm

15 mm 400 mm

150 mm B

M

Internal Moment: M = - 900 N # m is negative since it tends to decrease the beam’s radius curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m3 r = ©

2.18875 (10 - 3) ©rA = = 0.5150 m ©A 0.00425

dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10 - 3) m 0.4 0.55 LA r

R =

A

©1A

dA r

=

0.00425 = 0.509067 m 8.348614(10 - 3)

r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m Normal Stress: Applying the curved-beam formula sA =

M(R - rA) ArA (r - R)

=

-900(0.509067 - 0.57) 0.00425(0.57)(5.933479)(10 - 3) Ans.

= 3.82 MPa (T) sB =

M(R - rB) - 900(0.509067 - 0.4) = ArB (r - R) 0.00425(0.4)(5.933479)(10 - 3) Ans.

= - 9.73 MPa = 9.73 MPa (C)

Ans: sA = 
.1B
5

sB = 9.73 MPa (C) 435 435

CH 06.indd 435

1/18/11 2:24:45 PM

2018 Pearson Education, Ltd. All rights This is protected all copyright laws as they currently laws exist.asNo portion ©© 2010 Pearson Education, Inc., Upper Saddlereserved. River, NJ. Allmaterial rights reserved. Thisunder material is protected under all copyright they currently this material be reproduced, in any form or byform any or means, without in writinginfrom the publisher. exist. Noofportion of thismay material may be reproduced, in any by any means,permission without permission writing from the publisher.

6–137. The curved beam is subjected to a bending moment of M = 900 N # m. Determine the stress at point C. A C B

100 mm C

A

30�

20 mm

15 mm 400 mm

150 mm B

M

Internal Moment: M = - 900 N # m is negative since it tends to decrease the beam’s radius of curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10 - 3) m r = ©

2.18875 (10 - 3) ©rA = = 0.5150 m ©A 0.00425

dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10 - 3) m 0.4 0.55 LA r

R =

A

©1A

dA r

=

0.00425 = 0.509067 m 8.348614(10 - 3)

r - R = 0.515 - 0.509067 = 5.933479(10 - 3) m Normal Stress: Applying the curved-beam formula sC =

M(R - rC) ArC(r - R)

=

-900(0.509067 - 0.55) 0.00425(0.55)(5.933479)(10 - 3) Ans.

= 2.66 MPa (T)

Ans: ΣA = 0.00425 m2, r = 0.5150 m, dA = 8.348614(10 - 3) m, Σ LA r sC = 2.66 MPa (T) (s max )pvc = 12.3 MPa 436 436

CH 06.indd 436

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6–138. The beam is made from three types of plastic that are identified and have the moduli of elasticity shown in the figure. Determine the maximum bending stress in the PVC.

2.5 500 kN lb

2.5 lb kN 500 PVC � 450 ksiGPa PVCEE 3.15 PVC PVC � 160 ksiGPa Escon EsconEE 1.12 EE 800 Bakelite BakeliteEBE� 5.6ksi GPa B

3 ftm 0.9

 1.12  (b= = 0.015 m bk )1 n= 1bEs   (0.075)  5.6 

4 ftm 1.2

3 ft m 0.9

in. 251mm in. 502mm 2 in. 50 mm

 3.15  n= = (b = 0.0421875 m 2 bpvc  bk )2  (0.075)  5.6 

3 in. 75 mm

∑ yA 0.025(0.075)(0.05) + 0.075(0.015)(0.05) + 0.1125(0.0421875)(0.025) = y = ∑A 0.075(0.05) + 0.015(0.05) + 0.0421875(0.025)

2.5 kN

= 0.048365 m

0.9 m

1  I =2  (0.075)(0.053 ) + 0.075(0.05)(0.048365 − 0.025)2   12 

2.5 kN

2.5 kN 0.9 m

1 + (0.015)(0.053 ) + 0.015(0.05)(0.075 − 0.048365)2 12

2.5 kN

1.2 m

0.9 m

2.5 kN

Mmax = 2.25 kN · m

1 + (0.0421875)(0.0253 ) + 0.0421875(0.025)(0.1125 − 0.048365)2 12

2.5 kN

0.025 m 0.05 m

= 7.90996(10 −6 ) m 4 (σ max ) pvc

0.05 m

Mc  3.15  [2.25(10 3 )](0.1125 − 0.048365)  = = n2   I 7.90996(10 −6 )  5.6   

0.075 m

6 = 12.26(10 = ) N−m 2 12.3 MPa

Ans.

Ans: (s max )pvc = 12.3 MPa 426 426

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6–139. The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the cross section shown. If it is subjected to a moment of M = 6.5 kN # m, determine the maximum stress in the brass and steel. Also, what is the stress in each material at the seam where they are bonded together?

A

y

100 mm

M

100 mm

B z

Solution

x 125 mm

Section Properties: For the transformed section. n =

101 ( 109 ) Ebr = = 0.505 Est 200 ( 109 )

bst = nbbr = 0.505(0.125) = 0.063125 m y = =

ΣyA ΣA 0.05(0.125)(0.1) + 0.15(0.1)(0.063125) 0.125(0.1) + 0.1(0.063125)

= 0.08355 m INA =

1 (0.125) ( 0.13 ) + 0.125(0.1)(0.08355 - 0.05)2 12 +

1 (0.063125) ( 0.13 ) + 0.063125(0.1)(0.15 - 0.08355)2 12

= 57.62060 ( 10 - 6 ) m4 Maximum Bending Stress: Applying the flexure formula (smax)st =

6.5 ( 103 ) (0.08355) My = I 57.62060 ( 10-6 ) Ans.

= 9.42 MPa (smax)br = n

6.5 ( 103 ) (0.2 - 0.08355) Mc = 0.505c d I 57.62060 ( 10-6 )

Ans.

= 6.63 MPa

Bending Stress: At seam sst = =

My I 6.5 ( 103 ) (0.1 - 0.08355) 57.62060 ( 10-6 ) Ans.

= 1.855 MPa = 1.86 MPa sbr = n

My = 0.505(1.855) = 0.937 MPa I

Ans. Ans: (smax)st = 9.42 MPa, (smax)br = 6.63 MPa, sst = 1.86 MPa, sbr = 0.937 MPa 594

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*6–140. The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the cross section shown. If the allowable bending stress for the steel is (sallow)st = 180 MPa and for the brass (sallow)br = 60 MPa, determine the maximum moment M that can be applied to the beam.

A

y

100 mm

M

100 mm

B z

Solution

x 125 mm

Section Properties: For the transformed section. 101 ( 109 ) Ebr = = 0.505 Est 200 ( 109 )

n =

bst = nbbr = 0.505(0.125) = 0.063125 m y = =

ΣyA ΣA 0.05(0.125)(0.1) + 0.15(0.1)(0.063125) 0.125(0.1) + 0.1(0.063125)

= 0.08355 m INA =

1 (0.125) ( 0.13 ) + 0.125(0.1)(0.08355 - 0.05)2 12 +

1 (0.063125) ( 0.13 ) + 0.063125(0.1)(0.15 - 0.08355)2 12

= 57.62060 ( 10 - 6 ) m4 Allowable Bending Stress: Applying the flexure formula Assume failure of steel (smax)st = (sallow)st = 180 ( 106 ) =

My I M(0.08355) 57.62060 ( 10-6 )

M = 124130 N # m = 124 kN # m Assume failure of brass (smax)br = (sallow)br = n

Mc I

60 ( 106 ) = 0.505c

M(0.2 - 0.08355) 57.62060 ( 10-6 )

M = 58792 N # m

d

= 58.8 kN # m (Controls!)

Ans.

Ans: M = 58.8 kN # m 595

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6–141. The member has a brass core bonded to a steel casing. If a couple moment of 8 kN # m is applied at its end, determine the maximum bending stress in the member. Ebr = 100 GPa, Est = 200 GPa.

8 kN�m

3m 20 mm 100 mm 20 mm 20 mm

n =

Ebr 100 = = 0.5 Est 200

I =

1 1 (0.14)(0.14)3 (0.05)(0.1)3 = 27.84667(10 - 6)m4 12 12

100 mm

20 mm

Maximum stress in steel: (sst)max =

8(103)(0.07) Mc1 = 20.1 MPa = I 27.84667(10 - 6)

Ans.

(max)

Maximum stress in brass: (sbr)max =

0.5(8)(103)(0.05) nMc2 = 7.18 MPa = I 27.84667(10 - 6)

Ans: (sst) max = 20.1 MPa 422 422

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6–142. y

The Douglas Fir beam is reinforced with A-36 steel straps at its sides. Determine the maximum stress in the wood and steel if the beam is subjected to a bending moment of Mz = 4 kN # m. Sketch the stress distribution acting over the cross section.

15 mm

350 mm

200 mm

15 mm

z

Solution Section Properties: For the transformed section, n =

13.1 ( 109 ) Ew = = 0.0655 Est 200 ( 109 )

bst = nbw = 0.0655(0.2) = 0.0131 m 1 (0.03 + 0.0131) ( 0.353 ) = 153.99 ( 10-6 ) m4 12 Maximum Bending Stress: Applying the flexure formula INA =

(smax)st =

4 ( 103 ) (0.175) Mc = = 4.55 MPa I 153.99 ( 10-6 )

(smax)w = n

Ans.

4 ( 103 ) (0.175) Mc = 0.0655c d = 0.298 MPa I 153.99 ( 10-6 )

Ans.

Ans: (smax)st = 4.55 MPa, (smax)w = 0.298 MPa 597

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6–143. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stress acting at section a–a. Sketch the stress distribution on the section in three dimensions.

100 mm 50 mm

200 mm

Solution

a

Internal Moment: Referring to the FBD of the lower segment of the curve beam sectioned through a–a, Fig. a, 4 a+ΣMo = 0;  5 a b(0.2) - M = 0  M = 0.8 kN # m = 800 N # m 5 Section Properties: Referring to Fig. a, 0.2 + 0.3 r = = 0.25 m   A = (0.05)(0.1) = 0.005 m2 2

5 kN

a

200 mm

5 4 3

100 mm 5

5 kN

4 3

r2 dA 0.3 = b>n = 0.05>n a b = 0.020273255 m r1 0.2 LA r R =

A 0.005 = = 0.2466303 m 0.020273255 dA LA r

r - R = 0.25 - 0.2466303 = 0.003369654 m Normal Stress: Here M = 800 N # m is positive, since it tends to increase the beam’s radius of curvature. Applying the curved beam formula, (sT)max =

= (sC)max =

=

M(R - r1) Ar1(r - R) 800(0.2466303 - 0.2) 0.005(0.2)(0.003369654)

= 11.07 ( 106 ) Pa = 11.1 MPa (T)

Ans.

= - 8.447 ( 106 ) Pa = 8.45 MPa (C)

Ans.

M(R - r2) Ar2(r - R) 800(0.2466303 - 0.3) 0.005(0.3)(0.003369654)

Using these results, the normal stress distribution on the beam’s cross-section shown in Fig. b can be sketched.

Ans: (sT)max = 11.1 MPa (T), (sC)max = 8.45 MPa (C) 612

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*6–144. The curved member is symmetric and is subjected to a moment of M = 900 N # m. Determine the bending stress in the member at points A and B. Show the stress acting on volume elements located at these points.

0.5mm in. 12 B 2 in. 50 mm A 1.5mm in. 37 8 in.mm 200

M

M

1 A= 0.012(0.05) + (0.025)(0.05) = 0.001225 m 2 2

r =

0.012 m

0.225(0.012)(0.05) + 0.21667  21 (0.025)(0.05) ©rA = 0.220748 m = ©A 0.001225

0.05 m 0.25 m

dA 0.25   0.025(0.25)   0.25  = 0.012 ln  0.005570667 m +  − 0.025 =  ln  LA r  0.2   0.25 − 0.2   0.2  R =

A

dA 1A r

=

0.037 m 0.2 m

0.001225 = 0.2199018 0.005570667

86.4 MPa

r - R = 0.220748 − 0.2199018 = 0.846427(10 −3 ) m s =

σA =

M(R - r)

TA = 86.4 MPa

Ar(r - R)

105 MPa

900(0.2199018 − 0.2) 6 = 86.37(10 = ) N−m 2 86.4 MPa (T) 0.001225(0.2)[0.846427(10 −3 )] 900(0.2199018 − 0.25)

σB =

−3

0.001225(0.25)[0.846427(10 )]

Ans.

= −104.50(106 ) N−m 2 = 105 MPa (C) Ans.

TB = 105 MPa

Ans: sA = 86.4 MPa (T), sB = 105 MPa (C) 437 437

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6–145. The ceiling-suspended C-arm is used to support the X-ray camera used in medical diagnoses. If the camera has a mass of 150 kg, with center of mass at G, determine the maximum bending stress at section A.

G 1.2 m A

Section Properties: r =

1.22(0.1)(0.04) + 1.25(0.2)(0.02) ©rA = = 1.235 m ©A 0.1(0.04) + 0.2(0.02)

200 mm

1.24 1.26 dA © = 0.1 ln + 0.2 ln = 6.479051 A 10 - 3 B m 1.20 1.24 LA r

40 mm

A = 0.1(0.04) + 0.2(0.02) = 0.008 m2 R =

A

dA 1A r

=

100 mm

20 mm

0.008 = 1.234749 m 6.479051 (10 - 3)

r - R = 1.235 - 1.234749 = 0.251183 A 10 - 3 B m

Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. M = - 1816.93 N # m is negative since it tends to decrease the beam’s radius of curvature. Maximum Normal Stress: Applying the curved-beam formula sA = =

M(R - rA) ArA (r - R) - 1816.93(1.234749 - 1.26) 0.008(1.26)(0.251183)(10 - 3)

= 18.1 MPa (T) sB = =

M(R - rB) ArB (r - R) -1816.93(1.234749 - 1.20) 0.008(1.20)(0.251183)(10 - 3)

= - 26.2 MPa = 26.2 MPa (C)

Ans.

(Max)

Ans: LA A = 0.008 m2, sB = 26.2 MPa (C)

r = 1.235 m, Σ

dA -3 r = 6.479051(10 ) m,

439 439

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6–146 The member has a circular cross section. If it is subjected to a moment of M = 5 kN # m, determine the stress at points A and B. Is the stress at point A′, which is located on the member near the wall, the same as that at A? Explain.

200 mm

A¿ M

A 200 mm

Solution Section Properties: Referring to Fig. a B

0.2 + 0.4 r = = 0.3 m A = pC 2 = p ( 0.12 ) = 0.01p m2 2 dA 2 = 2p 1 r - 2r 2 - C 2 LA r = 2p 1 0.3 - 20.32 - 0.12 2 = 0.1078024 m

R =

A 0.012r = = 0.291421 m 0.1078024 dA LA r

r - R = 0.3 - 0.291421 = 0.00857864 m Bending Stress: Here, M = 5 kN # m is negative since it tends to decrease the beam’s radius of curvature. Applying the curved beam formula, sA =

sB =

M(R - rA) ArA(r - R)

M(R - rB) ArB(r - R)

=

- 5 ( 103 ) (0.291421 - 0.2) (0.01p)(0.2)(0.00857864)

= -8.4805 ( 106 ) Pa = 8.48 MPa (C) Ans.

=

- 5 ( 103 ) (0.291421 - 0.4) (0.01p)(0.4)(0.00857864)

= 5.0360 ( 106 ) Pa = 5.04 MPa (T) Ans.

No! Point A′ is at the fixed support where stress concentration occurs. The application of the curved beam formula is only valid for sections sufficiently removed from the support in accordance to Saint-Venant’s principle.

Ans: sA = 8.48 MPa (C), sB = 5.04 MPa (T) No, it is not the same. 610

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6–147. The member has a circular cross section. If the allowable bending stress is sallow = 100 MPa, determine the maximum moment M that can be applied to the member.

200 mm

A¿ M

A 200 mm

Solution Section Properties: Referring to Fig. a, B

0.2 + 0.4 r = = 0.3 m 2

A = pC 2 = p ( 0.12 ) = 0.01p m2

dA = 2p 1 r - 2r 2 - C 2 2 LA r

= 2p 1 0.3 - 20.32 - 0.12 2 = 0.1078024 m

R =

A 0.01p = = 0.291421 m 0.1078024 dA LA r

r - R = 0.3 - 0.291421 = 0.00857864 m Bending Stress: Here, M is negative since it tends to decrease the beam’s radius of curvature. Applying the curve beam’s formula by assuming tension failure, sB = sallow =

M(R - rB) ArB(r - R)

;  100 ( 106 ) =

- M(0.291421 - 0.4) (0.01p)(0.4)(0.00857864)

M = 99.29 ( 103 ) N # m = 99.3 kN # m Assuming compression failure, sA = sallow =

M(R - rA) ArA(r - R)

;  - 100 ( 106 ) =

- M(0.291421 - 0.2) (0.01p)(0.2)(0.00857864)

M = 58.96 ( 103 ) N # m = 59.0 kN # m (control!) Ans.

Ans: M = 59.0 kN # m 611

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*6–148. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stresses acting at section a–a. Sketch the stress distribution on the section in three dimensions.

10 kN 30 150 mm 30 10 kN

150 mm a

200 mm

Solution

a

100 mm

Internal Moment: Referring to the FBD of the upper segment of the curved beam sectioned through a–  a, Fig. a a+ ΣM0 = 0  (10 sin 30°)(0.2) + 10 cos 30°(0.15) - M = 0  M = 2.2990 kN # m

50 mm

Section Properties: Referring to Fig. a

100 mm

0.15 + 0.25 = 0.2 m   A = 0.05(0.1) = 0.005 m2 2 r2 dA 0.25 = b>n = 0.05 m a b = 0.025541281 m r1 0.15 LA r r =

R =

A

rA

dA

=

0.005 = 0.1957615 m 0.025541281

r

r - R = 0.2 - 0.1957615 = 0.00423848 m

Normal Stress: Here, M = - 2.2990 kN # m is negative since it tends to decrease the beam’s radius of curvature. Applying curved-beam formula, (sC)max =

M(R - r1) Ar1(r - R)

=

- 2.2990 ( 103 ) (0.1957615 - 0.15)

= -33.10 ( 106 ) Pa

0.005(0.15)(0.00423848)

= - 33.10 ( 106 ) Pa = 33.1 MPa (C) (sT)max =

M(R - r2) Ar2(r - R)

=

- 2.2990 ( 103 ) (0.1957615 - 0.25) 0.005(0.25)(0.00423848)

Ans.

=

= 23.54 ( 106 ) Pa = 23.5 MPa (T)

Ans.

Using these results, the normal stress distribution on the beam’s cross-section shown in Fig. b can be sketched.

Ans: (sC)max = 33.1 MPa (C), (sT)max = 23.5 MPa (T) 607

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6–149. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stresses acting at section a–a. Sketch the stress distribution on the section in three dimensions.

a 75 mm a

50 mm

100 mm 50 mm

250 N

150 mm

Solution

250 N

Internal Moment: M = 37.5 N # m is positive since it tends to increase the beam’s 75 mm radius of curvature. Section Properties: r =

0.1 + 0.175 = 0.1375 m 2

A = 0.075(0.05) = 0.00375 m2 dA 0.175 = 0.05 ln = 0.027981 m 0.1 LA r R =

A

dA 1A r

=

0.00375 = 0.134021 m 0.027981

r - R = 0.1375 - 0.134021 = 3.479478 ( 10-3 ) m Normal Stress: Applying the curved-beam formula (smax)t = =

M(R - r1) Ar1(r - R) 37.5(0.134021 - 0.1) 0.00375(0.1)(3.479478) ( 10-3 ) Ans.

= 0.978 MPa (T) (smax)c = =

M(R - r2) Ar2(r - R) 37.5(0.134021 - 0.175) 0.00375(0.175)(3.479478) ( 10-3 ) Ans.

= - 0.673 MPa = 0.673 MPa (C)

Ans: (smax)t = 0.978 MPa (T), (smax)c = 0.673 MPa (C) 608

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6–150. If the radius of each notch on the plate is rdetermine = 12.5 mm, the largest that canThe be the determine largest moment that moment can be applied. applied. allowable bending stressisfor the=material allowableThe bending stress for the material sallow 18 ksi. is stress of sallow = 125 MPa.

362.5 m 14.5 in.

M

b5 =

25 mm 1 in.

312.5 mm 12.5 in.

M

14.5 -– 312.5 12.5 362.5 25 in. mm =51.0 22

1 b 25 2.0 = =52.0 r 0.5 12.5

r 0.5 12.5 0.04 = =50.04 h 12.5 312.5

From Fig. 6-44: K = 2.60 smax = K

Mc I

  M (0.15625)  125(106 ) = 2.60  1 (0.025)(0.31253 )  12  M 19.56(10 3 ) N= ⋅ m 19.6 kN ⋅ m =

Ans.

Ans:

K = 2.60, M = 19.6 kN # m 442 442

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6–151. The symmetric notched plate is subjected to bending. If the radius of each notch is r = 12.5 mm and the applied moment is M = 15 kN · m, determine the maximum bending stress in the plate.

= b

14.5 in. 362.5 m

M

in. 251 mm

12.5 mm in. 312.5

M

362.5 − 312.5 = 25 mm 2

1 b 25 2.0 =52.0 = r 0.5 12.5

r 0.5 12.5 0.04 = =50.04 h 12.5 312.5

From Fig. 6-44: K = 2.60 smax = K

 [15(10 3 )](0.15625)  Mc = 2.60   1 (0.025)(0.31253 ) I  12 

=

6 36.864(10 = ) N−m 2 36.9 MPa

Ans.

Ans: s max = 36.9 MPa 443 443

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*6–152. The bar is subjected to a moment of M = 100 N # m. Determine the maximum bending stress in the bar and sketch, approximately, how the stress varies over the critical section.

80 mm

10 mm

20 mm

10 mm M

M 10 mm

Solution

r 10 w 80 Stress Concentration Factor: Entering = = 4.0 and = = 0.5 into h 20 h 20 Fig. 6–43, we obtain K = 1.25. Maximum Bending Stress: smax = K

Mc I

= 1.25 £

100(0.01) 1 12 (0.01)

( 0.023 )

§

= 187.5 ( 106 ) Pa = 187.5 MPa

Ans.

The bending stress distribution across the critical section is shown in Fig. a.

Ans: smax = 187.5 MPa 613

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6–153. The allowable bending stress for the bar is sallow = 200 MPa. Determine the maximum moment M that can be applied to the bar.

80 mm

10 mm

20 mm

10 mm M

M 10 mm

Solution

r 10 w 80 Stress Concentration Factor: Entering = = 4.0 and = = 0.5 into h 20 h 20 Fig. 6–43, we obtain K = 1.25. Maximum Bending Stress: smax = sallow = K

Mc I

200(106) = 1.25 £

M(0.01) 1 12

M = 106.67 N # m

(0.01)(0.023)

§

= 107 N # m

Ans.

Ans: M = 107 N # m 614

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6–15 4. The simply supported notched bar is subjected to two forces P. Determine the largest magnitude of P that can be applied without causing the material to yield.The material mm. in. is A-36 steel. Each notch has a radius of r == 30.125

P

0.5mm in. 12 1.75 in. 42 mm

1.25 in. 30 mm

20 mm in. 500

b5 =

P

20mm in. 500

20 in. 500 mm

20 in. 500 mm

1.75 - 1.25 42 – 30 = 0.25 5 6 mm 2 2

b 60.25 = 5 2; = 2; r 0.125 3

M = 0.5 P

r 0.125 3 5 0.1 = = 0.1 h 1.25 30

0.5 m 0.5 m

From Fig. 6-44. K = 1.92 sY = K

Mc ; I

1m

0.5 m

 0.5P(0.015)   250(106 ) = 1.92  1 (0.012)(0.33 )  12  Ans.

= = P 468.75 N 469 N

Ans: P  469 N 617

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6–155. The stepped bar has a thickness of 10 mm. Determine the maximum moment that can be applied to its ends if the allowable bending stress is sallow = 150 MPa.

90 mm

M

60 mm 7.5 mm

20 mm 15 mm M

Solution Stress Concentration Factor: For the smaller section, entering

w 60 = = 3.0 h 20

r 15 = = 0.75 into Fig. 6–43, we obtain K = 1.15. For the larger section, h 20 w 90 r 7.5 = = 1.5 and = = 0.125 gives K = 1.65. h 60 h 60

and

Maximum Bending Stress: For the smaller section, Mc smax = sallow = K I 150 ( 106 ) = 1.15£

M(0.01) 1 12 (0.01)

( 0.023 )

§

M = 86.96 N # m = 87.0 N # m (controls!) For the larger section, Mc smax = sallow = K I 150(106) = 1.65£

M(0.03) 1 3 12 (0.01)(0.06 )

M = 545.45 N # m

Ans.

§

Ans: M = 87.0 N # m 619

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*6–156. If the radius of each notch on the plate is r = 10 mm, determine the largest moment M that can be applied. The allowable bending stress is sallow = 180 MPa.

M 20 mm

125 mm

Solution

b 20 Stress Concentration Factor: From the graph in the text with = = 2 and r 10 r 10 = = 0.08, then K = 2.1. h 125 Allowable Bending Stress: smax = sallow = K

165 mm

M

Mc I

180 ( 106 ) = 2.1 c

M(0.0625) 1 12 (0.02)

( 0.1253 )

d

M = 4464 N # m = 4.46 kN # m

Ans.

Ans: M = 4.46 kN # m 618

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6–157. Determine the length L of the center portion of the bar so that the maximum bending stress at A, B, and C is the same. The bar has a thickness of 10 mm.

7 mm

350 N 60 mm

A 200 mm

60 w = = 1.5 h 40

40 mm 7 mm

C L 2

B L 2

200 mm

r 7 = = 0.175 h 40

From Fig. 6-43, K = 1.5 (sA)max = K

(35)(0.02) MAc d = 19.6875 MPa = 1.5c 1 3 I (0.01)(0.04 ) 12

(sB)max = (sA)max = 19.6875(106) =

MB c I

175(0.2 + L2 )(0.03) 1 3 12 (0.01)(0.06 )

Ans.

L = 0.95 m = 950 mm

Ans:

M = 87.0 kN # m 445 445

CH 06.indd 445

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6–158. Determine the shape factor for the cross section of the H-beam.

200 mm

20 mm

20 mm

200 mm

1 1 (0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10 - 6)m4 12 12

Ix =

Mp

20 mm

C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy C2 = T2 = sY(0.01)(0.24) = 0.0024sy

Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY sY = MY = k =

MYc I sY(26.8)(10 - 6) = 0.000268sY 0.1

Mp MY

=

0.00042sY = 1.57 0.000268sY

Ans.

Ans: Ix = 26.8 (10- 6) m4, Mp = 0.00042sY, MY = 0.000268sY, k = 1.57 451 451

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6–159. Determine the shape factor for the wide-flange beam. 15 mm

20 mm 200 mm Mp 15 mm

Solution

200 mm

1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6) m4 12 12

Ix =

C1 = T1 = sY(0.2)(0.015) = 0.003 sY C2 = T2 = sY(0.1)(0.02) = 0.002 sY Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY MYc I

sY = MY = k =

sY(82.78333)(10 - 6)

Mp MY

0.115 =

= 0.000719855 sY

0.000845 sY = 1.17 0.000719855 sY

Ans.

Ans: k = 1.17 621

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*6–160. Determine the plastic moment Mp that can be supported by a beam having the cross section shown. sY = 210 MPa.

50 mm 25 mm

Mp

250 mm

1 s dA = 0

25 mm

C1 + C2 - T1 = 0

[π (0.052 − 0.0252 )][210(106 )] + [(0.25 − d)(0.025)][210(106 )] − [d(0.025)][210(106 )] = 0 d = 0.24281 m < 0.25 m (o.k.!) = M p [π (0.052 − 0.0252 )][210(106 )](0.0571903) + [0.0071903(0.025)][210(106 )](0.0035951) + [0.242810(0.025)][210(106 )](0.121405) = 225.64(10 3 ) N= ⋅ m 226 kN ⋅ m

Ans.

Ans: Mp = 226 kN # m. 626

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  The beam is made of an elastic perfectly plastic material for which sY = 250 MPa. Determine the residual stress in the beam at its top and bottom after the plastic moment Mp is applied and then released.

t h t t

Solution

b

Plastic analysis: T1 = C1 = sYbt ;  T2 = C2 = sY a Mp = sYbt(h - t) + sY a = sY 3 bt(h - t) +

h - 2t bt 2

h - 2t h - 2t b(t)a b 2 2

t (h - 2t)2 4 4

Elastic analysis: 1 1 I = bh3 (b - t)(h - 2t)3 12 12 =

1 3 bh3 - (b - t)(h - 2t)3 4 12 MY =

syI c

=

=

1 ) [bh3 - (b - t)(h - 2t)3] sY ( 12 h 2

bh3 - (b - t)(h - 2t)3 6h

sY

Shape factor: k =

[bt(h - t) + 4t (h - 2t)2]sY MP = bh3 - (b - t)(h - 2t)3 MY s 6h

=

Y

2 3h 4bt(h - t) + t(h - 2t) c 3 d 3 2 bh - (b - t)(h - 2t)

Ans.

Ans: k = 622

2 3h 4bt(h - t) + t(h - 2t) c 3 d 3 2 bh - (b - t)(h - 2t)

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6–162. The rod has a circular cross section. If it is made of an elastic perfectly plastic material, determine the shape factor.

100 mm

Solution Maximum Elastic Moment: Applying the flexure formula with s = sY and p p ( 0.14 ) = 25 ( 10-6 ) p m4, I = r4 = 4 4 s =

MY(0.1) Mc ;  sY = I 25 ( 10-6 ) p MY = 0.25 ( 10 - 3 ) psY

Plastic Moment: Referring to Fig. a, p ( 0.1 ) pr 2 b = sY c d = 5 ( 10-3 ) psY 2 2 2

C = T = sY a

Thus,

d = 2a

4(0.1) 4r 0.26667 b = 2c = m d = p 3p 3p

Mp = Cd = ( 5 ( 10-3 ) p sY ) a

Shape Factor: k =

Mp MY

=

1.3333 ( 10-3 ) sY 0.25 ( 10-3 ) psY

0.26667 b = 1.3333 ( 10-3 ) sY p Ans.

= 1.6976 = 1.70

Ans: k = 1.70 623

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6–163. The rod has a circular cross section. If it is made of an elastic perfectly plastic material where sY = 345 MPa, determine the maximum elastic moment and plastic moment that can be applied to the cross section.

100 mm

Solution Maximum Elastic Moment: Applying the flexure formula with s = sy = 345 MPa p p ( 0.14 ) = 25 ( 10-6 ) p m4, and I = r 4 = 4 4 s =

MY(0.1) Mc ;  345 ( 106 ) = I 25 ( 10-6 ) p

MY = 270.96 ( 103 ) N # m = 271 kN # m

Ans.

Plastic Moment: Referring to Fig. a, p ( 0.1 ) pr 2 b = 345 ( 106 ) c d = 1.725 ( 106 ) p N 2 2 2

C = T = sy a Thus,

d = 2a

4(0.1) 4r 0.26667 b = 2c m d = p 3p 3p

MP = Cd =

3 1.725 ( 106 ) p 4 c

0.26667 d p

= 460 ( 103 ) N # m = 460 kN # m

Ans.

Ans: MY = 271 kN # m, MP = 460 kN # m 624

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*6–164. The beam is made of an elastic perfectly plastic material for which sY = 200 MPa. If the largest moment in the beam occurs within the center section a–a, determine the magnitude of each force P that causes this moment to be (a)  the largest elastic moment and (b) the largest plastic moment.

P

P a

a 2m

2m

2m

2m

200 mm

Solution (1)

M = 2P

100 mm

a) Elastic moment I =

1 (0.1)(0.23) = 66.667(10 - 6) m4 12

sY = MY =

MYc I 200(106)(66.667)(10 - 6) 0.1

= 133.33 kN # m From Eq. (1) 133.33 = 2 P Ans.

P = 66.7 kN b) Plastic moment Mp = =

b h2 s 4 Y 0.1(0.22) 4

(200)(106)

= 200 kN # m From Eq. (1) 200 = 2 P Ans.

P = 100 kN

Ans: Elastic: P = 66.7 kN, Plastic: P = 100 kN 630

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6–165. Determine the shape factor of the beam’s cross section. in. 503 mm

Referring to Fig. a, the location of centroid of the cross-section is y =

∑ yA 0.125(0.1)(0.05) + 0.05(0.05)(0.1) = = 0.0875 m ∑A 0.1(0.05) + 0.05(0.1)

in. 1006mm

The moment of inertia of the cross-section about the neutral axis is 1 I = (0.05)(0.13 ) + 0.05(0.1)(0.0875 − 0.05)2 12

25 mm 1.5 in. 503 mm in. 25 mm 1.5 in.

1 (0.1)(0.053 ) + 0.1(0.05)(0.125 − 0.0875)2 + 12 = 19.27083(10 −6 ) m 4 Here σ max= σ Y and c= y= 0.0875 m. Thus

smax =

Mc ; I

σY =

MY (0.0875)

19.27083(10 −6 )

MY = 0.22024(10 −3 )σ Y

Referring to the stress block shown in Fig. b, sdA = 0; LA

T - C1 - C2 = 0

[d(0.05)]σ Y − [(0.1 − d)(0.05)]σ Y − [0.05(0.1)]σ Y = 0 d = 0.1 m = Since d 0.1 = m, c1 0, Fig. c. Here T= C= [0.05(0.1)]σ Y= 0.005σ Y

Thus, = (0.075) 0.005σ Y= (0.075) 0.375(10 −3 )σ Y M p T=

Thus,

Ans.

Mp 0.375(10 −3 )σ Y k == = 1.7027 = 1.70 MY 0.22024(10 −3 )σ Y

0.05 m

0.05 m

0.075 m

0.1 – d 0.1 m

Ans: k = 1.70 449 449

CH 06.indd 449

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6–166. The beam is made of elastic-perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take sYY ==250 36 MPa. ksi.

503 mm in.

Referring to Fig. a, the location of centroid of the cross-section is y =

1006mm in.

∑ yA 0.125(0.1)(0.05) + 0.05(0.05)(0.1) = = 0.0875 m ∑A 0.1(0.05) + 0.05(0.1)

The moment of inertia of the cross-section about the neutral axis is

25 mm 1.5 in. 503 mm in.

1 I = (0.05)(0.13 ) + 0.05(0.1)(0.0875 − 0.05)2 12 +

25 mm 1.5 in.

1 (0.1)(0.053 ) + 0.1(0.05)(0.125 − 0.0875)2 12

= 19.27083(10 −6 ) m 4 Here σ max= σ Y =250 MPa and c/= y= 0.0875 m. Then

smax =

Mc ; I

250(106 ) =

MY (0.0875)

19.27083(10 −6 )

Ans.

MY = 55.06(10 3 ) N ⋅ m = 55.1 kN ⋅ m

Referring to the stress block shown in Fig. b, sdA = 0; LA

T - C1 - C2 = 0

[d(0.05)][250(106 )] − [(0.1 − d)(0.05)][250(106 )] − [0.05(0.1)][250(106 )] = 0 d = 0.1 m = Since d 0.1 m, c1 0, =

Here, T= C= [0.05(0.1)][250(106 )]= 1250(10 3 ) N= 1250 kN

Thus, (0.075) (1250)(0.075) = 93.75 kN = M p T=

Ans.

0.05 m

0.05 m

0.1 – d

0.075 m 0.1 m

Ans:

MY = 55.1 kN # m, MP = 93.75 kN # m 450 450

CH 06.indd 450

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6–167. Determine the shape factor for the beam. 10 mm

15 mm 200 mm Mp 10 mm

Solution I =

200 mm

1 1 (0.2)(0.22)3 (0.185)(0.2)3 = 54.133(10 - 6) m4 12 12

C1 = sY(0.01)(0.2) = (0.002) sY C2 = sY(0.1)(0.015) = (0.0015) sY Mp = 0.002 sY(0.21) + 0.0015 sY(0.1) = 0.0005 sY MYc I

sY = MY = k =

sY(54.133)(10 - 6)

Mp MY

0.11 =

= 0.000492 sY

0.00057 sY = 1.16 0.000492 sY

Ans.

Ans: k = 1.16 629

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*6–168. The beam is made of an elastic perfectly plastic material for which sY = 250 MPa. Determine the residual stress in the beam at its top and bottom after the plastic moment Mp is applied and then released.

15 mm

20 mm 200 mm Mp 15 mm

Solution Ix =

200 mm

1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10 - 6) m4 12 12

 C1 = T1 = sY(0.2)(0.015) = 0.003 sY  C2 = T2 = sY(0.1)(0.02) = 0.002 sY Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY = 0.000845(250)(106) = 211.25 kN # m

  s ′ =

Mpc I

=

y 0.115 = ; 250 293.5

211.25(103)(0.115) 82.78333(10 - 6)

= 293.5 MPa

y = 0.09796 m = 98.0 mm Ans.

stop = sbottom = 293.5 - 250 = 43.5 MPa

Ans: stop = sbottom = 43.5 MPa 640

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6–169. The box beam is made of an elastic perfectly plastic material for which sY = 250 MPa. Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released.

25 mm

Solution

150 mm 25 mm

Plastic Moment:

25 mm 150 mm 25 mm

MP = 250 11062 (0.2)(0.025)(0.175) + 250 11062 (0.075)(0.05)(0.075) = 289062.5 N # m

Modulus of Rupture: The modulus of rupture sr can be determined using the flexure formula with the application of reverse, plastic moment MP = 289062.5 N # m. I =

1 1 (0.2) 10.232 (0.15) 10.1532 12 12

= 91.14583 ( 10 - 6 ) m4 sr =

289062.5 (0.1) MP c = = 317.41 MPa I 91.14583 110 - 62

Residual Bending Stress: As shown on the diagram. = = stop = sbot = sr - sY

Ans.

= 317.14 - 250 = 67.1 MPa

Ans: stop = sbottom = 67.1 MPa 639

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6–170. Determine the shape factor of the cross section. a a a

Solution Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first. 1 1 INA = (a)(3a)3 + (2a) ( a3 ) = 2.41667 a4 12 12

a

a

a

Applying the flexure formula with s = sY , we have sY =

MY c I

MY =

sY ( 2.41667a4 ) sYI = = 1.6111a3sY c 1.5a

Plastic Moment: Mp = sY(a)(a)(2a) + sY(0.5a)(3a)(0.5a) = 2.75a3sY Shape Factor: k =

Mp MY

=

2.75a3sY 1.6111a3sY

Ans.

= 1.71

Ans: k = 1.71 638

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6–171. Determine the shape factor for the member having the tubular cross section. 2d

d

Solution Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first. INA =

15p 4 p 4 p d 4 d - a b = d 4 4 2 64

Applying the flexure formula with s = sY , we have sY = MY =

MY c I 4 sY ( 15p sY I 15p 3 64 d ) = = d sY c d 64

Plastic Moment:

y =

ΣyA = ΣA

MP = sY a

4d pd 2 2 3p

(

pd 2 2

2

MP = MY

4 ( d2 ) 3p

pd 2 p 2 4d

2

Shape Factor: k =

) -

7 3 6 d sY 15p 3 64 d sY

b

a

p 2 4d

p 2 4d

2

b

=

14 d 9p

2

28 7 d = d 3 sY 9p 6

Ans.

= 1.58

Ans: k = 1.58 637

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*6–172.

Determine the shape factor for the member. –h 2

–h 2

Plastic analysis: T = C = MP =

h 1 bh (b) a b sY = s 2 2 4 Y

b

b h2 bh h sY a b = s 4 3 12 Y

Elastic analysis: I = 2c

1 h 3 b h3 (b)a b d = 12 2 48 3

sY A bh sYI 48 B b h2 MY = = = s h c 24 Y 2

Shape factor: k =

Mp MY

=

bh2 12

sY

bh2 24

sY

Ans.

= 2

Ans: k =

Mp MY

=

bh2 12

sY

bh2 24

sY

= 2

456 456

CH 06.indd 456

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6–173. The member is made from an elastic-plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. mm, h =6 150 250 MPa. Take b == 100 sY =sY36= ksi. 4 in., h = in., mm,

–h 2

–h 2

Elastic analysis: 1 3  4 −6 I 2= =  12 (0.1)(0.075 ) 7.03125(10 ) m   MY=

σY I c

=

b

[250(106 )][7.03125(10 −6 )] = 23.4375(10 3 ) N ⋅ m= 23.4 kN ⋅ m 0.075

Ans.

Plastic analysis:

h 3

1  T= C=  (0.1)(0.075) [250(106 )]= 937.5(10 3 ) N 2  h 3  0.15  3 ⋅ m 46.9 kN ⋅ m = M p T=   [937.5(10 )]  =  46.875(10 ) N= 3  3 

Ans.

Ans:

MY = 23.4 kN # m, MP = 46.9 kN # m 456 456

CH 06.indd 456

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6–174. Determine the shape factor of the cross section. a

2a

Solution

a

Maximum Elastic Moment: The centroid and the moment of inertia about neutral axis must be determined first. y = INA =

a 2

a 2

0.5a(a) (2a) + 2a(2a) (a) ΣyA = = 1.25a ΣA a(2a) + 2a(a) 1 (2a) ( a3 ) + 2a(a) (1.25a - 0.5a)2 12 +

1 (a) (2a)3 + a(2a) (2a - 1.25a)2 12

= 3.0833a4 Applying the flexure formula with s = sY , we have sY = MY =

MY c I sY ( 3.0833a4 ) sY I = c (3a - 1.25a) = 1.7619a3 sY

Plastic Moment: LA

sdA = 0;  T - C1 - C2 = 0 sY (d) (a) - sY (2a - d) (a) - sY (a) (2a) = 0 d = 2a MP = sY (2a) (a) (1.5a) = 3.00a3 sY

Shape Factor: k =

MP 3.00a3 sY = = 1.70 MY 1.7619a3 sY

Ans.

Ans: k = 1.70 635

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6–175. The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take a = 50 mm and sY = 230 MPa.

a

2a

Solution

a

Maximum Elastic Moment: The centroid and the moment of inertia about neutral axis must be determined first. y = INA =

a 2

a 2

0.025(0.05) (0.1) + 0.1 (0.1) (0.05) ΣyA = = 0.0625 m ΣA 0.05(0.1) + 0.1(0.05) 1 (0.1) ( 0.053 ) + 0.1(0.05) (0.0625 - 0.025)2 12 +

1 (0.05) ( 0.13 ) + 0.05(0.1) (0.1 - 0.0625)2 12

= 19.2709 ( 10-6 ) m4 Applying the flexure formula with s = sY , we have sY = MY =

MY c I 230 ( 106 ) (19.2709) ( 10-6 ) sY I = c (0.15 - 0.0625)

= 50654.8 N # m = 50.7 kN # m

Ans.

Plastic Moment: LA

sdA = 0;  T - C1 - C2 = 0 sY (d) (0.05) - sY (0.1 - d)(0.05) - sY (0.05)(0.1) = 0 d = 0.100 m MP = 230 ( 10

6

) (0.100) (0.05) (0.075)

= 86250 N # m = 86.25 kN # m

Ans.

Ans: MY = 50.7 kN # m, MP = 86.25 kN # m 636

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*6–176. The beam is made of elastic perfectly plastic material for which sY = 345 MPa. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section.

100 mm 100 mm 100 mm

Solution Maximum Elastic Moment: Applying the flexure formula with s = sy = 345 MPa 1 1 and I = (0.1) ( 0.33 ) + (0.2) ( 0.13 ) = 0.24167 ( 10-3 ) m4, 12 12 s =

100 mm 100 mm 100 mm

MY(0.15) Mc ;  345 ( 106 ) = I 0.24167 ( 10-3 )

MY = 555.83 ( 103 ) N # m = 556 kN # m

Ans.

Plastic Moment: Referring to Fig. a, C1 = T1 = sy(0.1)(0.1) = 0.01sy = 0.01 3 345 ( 106 )4 = 3.45 ( 106 ) N

C2 = T2 = sy(0.3)(0.05) = 0.015sy = 0.015 3 345 ( 106 ) 4 = 5.175 ( 106 ) N d 1 = 4(0.05) = 0.2 m  d 2 = 2(0.025) = 0.05 m Thus, MP = C1d 1 + C2d 2 =

3 3.45 ( 106 ) 4 (0.2)

+

3 5.175 ( 106 ) 4 (0.05)

= 948.75 ( 103 ) N # m = 949 kN # m

Ans.

Ans: MY = 556 kN # m, MP = 949 kN # m 631

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6–177. Determine the shape factor of the cross section. 100 mm 100 mm 100 mm

Solution Maximum Elastic Moment: Applying the flexure formula with s = sy and 1 1 I = (0.1) ( 0.33 ) + (0.2) ( 0.13 ) = 0.24167 ( 10-3 ) m4, 12 12 My(0.15) Mc s = ;  sy = My = 1.6111 ( 10 - 3 ) sy I 0.24167 ( 10-3 ) Plastic Moment: Referring to Fig. a,

100 mm 100 mm 100 mm

C1 = T1 = sy(0.1)(0.1) = 0.01 sy C2 = T2 = sy(0.3)(0.05) = 0.015 sy d 1 = 4(0.05) = 0.2 m d 2 = 2(0.025) = 0.05 m Thus, Mp = C1d 1 + C2d 2 = (0.01 sy)(0.2) + (0.015 sy)(0.05) = 2.75 ( 10-3 ) sy Shape Factor:

k =

Mp My

=

2.75 ( 10-3 ) sy 1.6111 ( 10-3 ) sy

Ans.

= 1.7069 = 1.71

Ans: k = 1.71 632

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6–178.  The plexiglass bar has a stress–strain curve that can be approximated by the straight-line segments shown. Determine the largest moment M that can be applied to the bar before it fails.

s (MPa)

20 mm M 20 mm

failure

60 40

tension

0.06 0.04 0.02

Solution

P (mm/mm)

compression

Ultimate Moment: LA

0.04

80 100

s dA = 0; C - T2 - T1 = 0

1 1 d 1 d sc (0.02 - d)(0.02) d - 40 ( 106 ) c a b(0.02) d - (60 + 40) ( 106 ) c (0.02) d = 0 2 2 2 2 2 Since

0.04 P = , then 0.02 - d = 25Pd. d 0.02- d

And since

40 ( 106 ) s s = = 2 ( 109 ) , then P = . P 0.02 2 ( 109 )

So then 0.02 - d =

23sd 2 ( 109 )

= 1.25 ( 10 - 8 ) sd.

Substituting for 0.02 - d, then solving for s, yields s = 74.833 MPa. Then P = 0.037417 mm>mm and d = 0.010334 m. Therefore, 1 C = 74.833 ( 106 ) c (0.02 - 0.010334)(0.02) d = 7233.59 N 2 T1 =

1 0.010334 b d = 5166.85 N (60 + 40) ( 106 ) c (0.02)a 2 2

1 0.010334 T2 = 40 ( 106 ) c (0.02)a b d = 2066.74 N 2 2 y1 = y2 = y3 =

2 (0.02 - 0.010334) = 0.0064442 m 3 2 0.010334 a b = 0.0034445 m 3 2

0.010334 1 2(40) + 60 0.010334 + c1 - a b da b = 0.0079225 m 2 3 40 + 60 2

M = 7233.59(0.0064442) + 2066.74(0.0034445) + 5166.85(0.0079255) = 94.7 N # m

Ans.

Ans: M = 94.7 N # m 641

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6–179. The beam is made of phenolic, a structural plastic, that has the stress–strain curve shown. If a portion of the curve can be represented by the equation s = (5(106)P)1/2 MPa, determine the magnitude w of the distributed load that can be applied to the beam without causing the maximum strain in its fibers at the critical section to exceed Pmax = 0.005 mm>mm.

w

150 mm 150 mm

2m

2m

s(MPa)

Solution

s2  5(106)P

Resultant Internal Forces: The resultant internal forces T and C can be evaluated from the volume of the stress block which is a paraboloid. When P = 0.005 mm>mm, then

P(mm/mm)

s = 25 ( 106 ) (0.005) = 158.11 MPa T = C =

2 3 158.11 ( 106 ) (0.075) 4 (0.150) = 1.1859 MN 3

3 d = 2 c (0.075) d = 0.090 m 5

Maximum Internal Moment: The maximum internal moment M = 2w occurs at the overhang support as shown on FBD. Mmax = Td 2w = 1.1859 ( 106 ) (0.090) Ans.

w = 53363 N>m = 53.4 kN>m

Ans: w = 53.4 kN>m 645

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*6–180. The stress–strain diagram for a titanium alloy can be approximated by the two straight lines. If a strut made of this material is subjected to bending, determine the moment resisted by the strut if the maximum stress reaches in. 75 3mm

a value of (a) sA and (b) sB.

M

in. 502 mm �s(MPa) (ksi) s

B

180 ss �1260 BB � 980 140 sA 

A

0.01

0.04

(in./in.) P (mm/mm)

a) Maximum Elastic Moment : Since the stress is linearly related to strain up to point A, the flexure formula can be applied. sA =

Mc I

–1260 9.375 mm

=

b)

1 (0.05)(0.075 ) [980(10 )]  12   0.0375 3

1260

⋅ m 45.9 kN ⋅ m 45.9375(10 3 ) N=

Ans.

The Ultimate Moment :

28.125 mm 9.375 mm

12.5 mm

48.05 mm

28.125 mm

1 [(1260 + 980)(106 )](0.028125)(0.05) = 1575(10 3 ) N C1 = T1 = 2 C= 2 T= 2

s (MPa) 980

6

=

9.375 mm

e (mm/mm)–980

sA I M = c

1 [980(106 )](0.009375)(0.05) = 229.6875(10 3 ) N 2

= M [1575(10 3 )](0.04805) + [229.6875(10 3 )](0.0125) = 78.54(10 3 ) N= ⋅ m 78.5 kN ⋅ m

Ans.

Note: The centroid of a trapezodial area was used in calculation of moment.

461 461

CH 06.indd 461

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6–181. The bar is made of an aluminum alloy having a stress–strain diagram that can be approximated by the straight line segments shown. Assuming that this diagram is the same for both tension and compression, determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is P max = 0.03.

�s (ksi) s(MPa) 90 630 80 560 60 420

in. M 1004mm

MPa s == 574 82 ksi

0.05

P (in./ in.) (mm/mm)

y1 = 10 mm

1 C1 = T1 = [(560 + 574)(106 )](0.008333)(0.075) = 354.375(10 3 ) N 2 yz = 41.667 mm

1 [(560 + 420)(106 )](0.031667)(0.075) = 1163.75(10 3 ) N C2 = T2 = 2 C= 3 T= 3

in. 753 mm

53.17 mm 13.33 mm

80 - 80 ss –-560 63090– 560 5=  ; ; 0.03 –-0.025 0.025 0.05 0.05– 0.025 - 0.025

0.025

91.70 mm

0.006

420 MPa

560 MPa 574 MPa

1 [420(106 )](0.01)(0.075) = 157.5(10 3 ) N 2

M = [354.375(10 3 )](0.09170) + [1163.75(10 3 )](0.05317) + [157.5(10 3 )](0.01333) = 96.48(10 3 ) N= ⋅ m 96.5 kN ⋅ m

Ans.

Note: The centroid of a trapezodial area was used in calculation of moment areas.

Ans:

s = 574 MPa, M = 96.5 kN # m 463 463

CH 06.indd 463

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6–182. A beam is made from polypropylene plastic and has a stress–strain diagram that can be approximated by the curve shown. If the beam is subjected to a maximum tensile and compressive strain of P = 0.02 mm>mm, determine the moment M.

M

s (Pa)

s  10(106)P1/ 4

100 mm

M 30 mm

P (mm/ mm)

Solution Pmax = 0.02 smax = 10 1 106 2 (0.02)1>4 = 3.761 MPa 0.02 P = 0.05 y



P = 0.4 y

s = 10 1 106 2 (0.4)1>4y1>4 M =

LA

y s dA = 2

M = 0.47716 1 106 2 M =

251 N # m

L0

L0

0.05

0.05

y(7.9527) 1 106 2 y1>4(0.03)dy

4 y5>4dy = 0.47716 1 106 2 a b(0.05)9>4 9

Ans.

Ans: M = 251 N # m 643

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R6–1. Determine the shape factor for the wide-flange beam. 20 mm

30 mm 180 mm Mp 20 mm

Solution I =

180 mm

1 1 (0.18)(0.223) (0.15)(0.183) 12 12

= 86.82(10 - 6) m4 Plastic moment: Mp = sY(0.18)(0.02)(0.2) + sY(0.09)(0.03)(0.09) = 0.963(10 - 3)sY Shape Factor: MY = k =

sY(86.82)(10 - 6) sYI = = 0.789273(10 - 3)sY c 0.11

Mp MY

=

0.963(10 - 3)sY 0.789273(10 - 3)sY

= 1.22

Ans.

Ans: k = 1.22 646

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R6–2. The compound beam consists of two segments that are pinned together at B. Draw the shear and moment ­diagrams if it supports the distributed loading shown.

w

A 2/3 L

Solution + c ΣFy = 0;

2wL 1 w 2 x = 0 27 2 L x =

a+ ΣM = 0;

M +

4 L = 0.385 L A 27

1w 1 2wL (0.385L)2a b(0.385L) (0.385L) = 0 2L 3 27

M = 0.0190 wL2

647

C

B 1/3 L

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R6–3. y

The composite beam consists of a wood core and two plates of steel. If the allowable bending stress for the wood is (sallow)w = 20 MPa, and for the steel (sallow)st = 130 MPa, determine the maximum moment that can be applied to the beam. Ew = 11 GPa, Est = 200 GPa.

z 125 mm

M

Solution

x 9

200(10 ) Est = = 18.182 Ew 11(109) 1 I = (0.80227)(0.1253) = 0.130578(10 - 3)m4 12

20 mm 75 mm 20 mm

n =

Failure of wood: Mc I

(sw)max =

M(0.0625)

20(106) =

0.130578(10 - 3)

;

M = 41.8 kN # m

Failure of steel: (sst)max = 130(106) =

nMc I 18.182(M)(0.0625) 0.130578(10 - 3)

M = 14.9 kN # m (controls)

Ans.

Ans: M = 14.9 kN # m 648

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*R6–4. A shaft is made of a polymer having a parabolic upper and lower cross section. If it resists a moment of M = 125 N # m, determine the maximum bending stress in the material (a) using the flexure formula and (b) using integration. Sketch a three-dimensional view of the stress distribution acting over the cross-sectional area. Hint: The ­moment of inertia is determined using Eq. A–3 of Appendix A.

y

100 mm y  100 – z 2/ 25 M  125 N· m z

Solution Maximum Bending Stress: The moment of inertia about y axis must be determined first in order to use flexure formula I = = 2

LA

x

y2 dA 100 mm

L0

= 20

50 mm 50 mm

y2 (2z) dy

L0

100 mm

y2 2100 - y dy

100 3 5 7 3 8 16 = 20J - y2 (100 - y)2 y (100 - y)2 (100 - y)2 R ` mm 2 15 105 0

Thus,

= 30.4762 1 106 2 mm4 = 30.4762 1 10 - 6 2 m4

smax =

125(0.1) Mc = = 0.410 MPa I 30.4762(10 - 6)

Ans.

Maximum Bending Stress: Using integration dM = 2[y(s dA)] = 2b yc a M = 125 1 103 2 = 125 1 103 2 =

smax by d (2z dy) r 100

smax 100 mm 2 y 2100 - y dy 5 L0

100 smax 3 5 7 3 8 16 J - y2(100 - y)2 y(100 - y)2 (100 - y)2 R ` mm 5 2 15 105 0

smax (1.5238) 1 106 2 5

smax = 0.410 N>mm2 = 0.410 MPa

Ans.

Ans: smax = 0.410 MPa 649

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20 20�

Determine the maximum bending stress in the R6–5. N is handle of the cable cutter cutter at at section section a–a. a–a.A Aforce forceof of225 45 lb applied to the handles. The cross-sectional area is shown in the figure.

a 100 mm 4 in.

225 45 lbN 125 mm 5 in.

75 mm 3 in. A

a

18 mm 0.75 in. 12 mm 0.50 in.

225 45 lbN

a + ©M = 0;

M − 225(0.125 + 0.1cos 20°) =0 = M 49.27 N ⋅ m

smax =

Mc = I

49.27(0.009) =

1 (0.012)(0.018 3 ) 12

0.1 m

6 76.03(10 ) N−m 2 76.0 MPa =

225 N 0.125 m

Ans.

Ans: s max = 76.0 MPa 468 468

CH 06.indd 468

1/18/11 2:25:55 PM

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R6–6. The curved beam is subjected to a bending m ­ oment of M = 85 N # m as shown. Determine the stress at points A and B and show the stress on a volume element located at these points.

M  85 Nm

100 mm

A 400 mm

A

20 mm

15 mm

B

150 mm

30 B

Solution

20 mm

r2 dA 0.42 0.57 0.59 = b ln = 0.1 ln + 0.015 ln + 0.1 ln r1 0.40 0.42 0.57 LA r = 0.012908358 m A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10 - 3) m2 R =

A LA

dA r

=

6.25(10 - 3) 0.012908358

= 0.484182418 m

r - R = 0.495 - 0.484182418 = 0.010817581 m sA =

M(R - rA) ArA(r - R)

=

85(0.484182418 - 0.59) 6.25(10 - 3)(0.59)(0.010817581)

= -225.48 kPa Ans.

sA = 225 kPa (C) sB =

M(R - rB ) ArB (r - R)

=

85(0.484182418 - 0.40) 6.25(10 - 3)(0.40)(0.010817581)

= 265 kPa (T)

Ans.

Ans: sA = 225 kPa (C), sB = 265 kPa (T) 651

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R6–7. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as 8 .m. functions of x, where 0 … x 6 1. 6 ft

40 kN 8 kip

30 kN/m 2 kip/ft 75 50 kNm kip�ft

x 1.8 6 ftm

+ c ©Fy = 0;

20 –-30x 2x –-VV= = 94 0 0

243.6 kN · m

Ans.

= V {94 − 30 x} kN c + ©MNA = 0;

1.2 4 ftm 40 kN

30 kN/m

75 kN · m

1.8 m

1.2 m

94 kN

x 94 x − 243.6 − 30 x   − M = 0 2

V (kN) 94

{−15 x 2 + 94x − 243.6} kN ⋅ m M=

40

Ans. 0 M (kN · m) 0

x (m)

1.8

3

1.8

3

x (m)

–48 –123 30x

–243.6 243.6 (kN · m) 94 kN

Ans: V  94  30x, M   15x 2  94x  243.6 654

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*R6–8. A wooden beam has a square cross section as shown. Determine which orientation of the beam provides the greatest strength at resisting the moment M. What is the difference in the resulting maximum stress in both cases?

a

a

M

a

M

a (a)

(b)

Solution Case (a): smax =

M(a>2) Mc 6M = 1 4 = 3 I a 12 (a)

Case (b): I = 2c

3 1 2 1 1 1 2 1 1 2 aba bd d = 0.08333 a4 ab c a aba ab + a aba a 3 2 22 36 22 22 22 22

smax

1 ab Ma 8.4853 M Mc 22 = = = 4 I 0.08333 a a3

Case (a) provides higher strength, since the resulting maximum stress is less for a given M and a. ∆smax =

M 8.4853 M 6M - 3 = 2.49 a 3 b 3 a a a

Ans.

Ans: Case (a), ∆smax = 2.49 a 653

M b a3

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R6–9. Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings. The bearings at A and B exert only vertical reactions on the shaft.

300 N 450 N A

B

200 mm

400 mm

300 mm

200 mm 150 N

Solution

654

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R6–10. The strut has a square cross section a by a and is subjected to the bending moment M applied at an angle u as shown. Determine the maximum bending stress in terms of a,  M, and u. What angle u will give the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.

y

a

z



x

a M

Solution Internal Moment Components: Mz = -M cos u

My = - M sin u

Section Property: Iy = Iz =

1 4 a 12

Maximum Bending Stress: By inspection, maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A smax = = =

Mzy Iz

+

My z Iy

- M cos u (a2) 1 12

a4

+

- M sin u ( - a2) 1 12

a4

6M (cos u + sin u) a3

Ans.

ds 6M = 3 ( - sin u + cos u) = 0 du a cos u - sin u = 0 Ans.

u = 45° Orientation of Neutral Axis: tan a =

Iz Iy

tan u

tan a = (1) tan (45°) Ans.

a = 45°

Ans: 6M (cos u + sin u), a3 u = 45°, a = 45° smax =

655

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7–1.  If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear stress on the web at A. Indicate the shear-stress components on a volume element located at this point.

200 mm

A

20 mm

20 mm B

V 300 mm 200 mm

Solution

20 mm

The moment of inertia of the cross-section about the neutral axis is I =

1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12

From Fig. a, QA = y′A′ = 0.16 (0.02)(0.2) = 0.64(10 - 3) m3 Applying the shear formula, tA =

20(103)[0.64(10 - 3)] VQA = It 0.2501(10 - 3)(0.02) = 2.559(106) Pa = 2.56 MPa

Ans.

The shear stress component at A is represented by the volume element shown in Fig. b.

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

656

Ans: tA = 2.56 MPa

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7–2.  If the wide-flange beam is subjected to a shear of V = 20 kN, determine the maximum shear stress in the beam.

200 mm

A

20 mm

20 mm B

V 300 mm 200 mm

Solution

20 mm

The moment of inertia of the cross-section about the neutral axis is I =

1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12

From Fig. a. Qmax = Σy′A′ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10 - 3) m3 The maximum shear stress occurs at the points along neutral axis since Q is maximum and thickness t is the smallest. tmax =

20(103) [0.865(10 - 3)] VQmax = It 0.2501(10 - 3) (0.02) = 3.459(106) Pa = 3.46 MPa

Ans.

Ans: tmax = 3.46 MPa 657

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7–3.  If the wide-flange beam is subjected to a shear of V = 20 kN, determine the shear force resisted by the web of the beam.

200 mm

A

20 mm

20 mm B

V 300 mm 200 mm

Solution

20 mm

The moment of inertia of the cross-section about the neutral axis is I =

1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10 - 3) m4 12 12

For 0 … y 6 0.15 m, Fig. a, Q as a function of y is Q = Σy′A′ = 0.16 (0.02)(0.2) +

1 (y + 0.15)(0.15 - y)(0.02) 2

= 0.865(10 - 3) - 0.01y2 For 0 … y 6 0.15 m, t = 0.02 m. Thus, t =

20(103) 3 0.865(10 - 3) - 0.01y2 4 VQ = It 0.2501(10 - 3) (0.02) =

5 3.459(106)

The sheer force resisted by the web is Vw = 2

L0

0.15 m

tdA = 2

L0

0.15 m

- 39.99(106) y2 6 Pa

3 3.459(106)

6

2

- 39.99(10 ) y

= 18.95 (103) N = 19.0 kN

4 (0.02 dy)

Ans.

Ans: Vw = 19.0 kN 658

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*7–4. If the beam is subjected to a shear of V = 30 kN, determine the web’s shear stress at A and B. Indicate the shear-stress components on a volume element located at these points. Set w = 200 mm. Show that the neutral axis is located at y = 0.2433 m from the bottom and I = 0.5382(10−3) m4.

200 mm

A

20 mm

20 mm B

V 300 mm 200 mm

Solution

20 mm

Section Properties: The location of the centroid measured from the bottom is y =

(0.01)(0.2)(0.02) + 0.22 (0.02)(0.4) + 0.430 (0.3)(0.02) 0.2 (0.02) + 0.02 (0.4) + 0.3 (0.02)

= 0.2433 m The moment of inertia of the cross-section about the neutral axis is 1 I = (0.2) ( 0.023 ) + 0.2 (0.02)(0.2433 - 0.01)2 12 1 + (0.02) ( 0.43 ) + 0.02 (0.4)(0.2433 - 0.22)2 12 1 + (0.3) ( 0.023 ) + 0.3(0.02)(0.43 - 0.2433)2 12 = 0.5382 ( 10-3 ) m4 Referring to Fig. a, QA = y′A A′A = 0.1867 [0.3(0.02)] = 1.12 ( 10-3 ) m3 QB = y′B A′B = 0.2333 [0.2(0.02)] = 0.9333 ( 10-3 ) m3 Shear Stress: Applying the shear formula, 30 ( 103 ) 3 1.12 ( 10-3 ) 4 VQA tA = = = 3.122 ( 106 ) Pa = 3.12 MPa I tA 0.5382 ( 10-3 ) (0.02) tB =

30 ( 103 ) 3 0.9333 ( 10-3 ) 4 VQB = = 2.601 ( 106 ) Pa = 2.60 MPa I tB 0.5382 ( 10-3 ) (0.02)

Ans.

Ans.

These shear stresses on the volume element at points A and B are shown in Fig. b.

Ans: tA = 3.12 MPa, tB = 2.60 MPa 659

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7–5.  If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam. Set w = 300 mm.

200 mm

A

20 mm

20 mm B

V 300 mm 200 mm

Solution

20 mm

Section Properties: The moment of inertia of the cross section about the neutral axis is I =

1 1 (0.3) ( 0.443 ) (0.28) ( 0.43 ) = 0.63627 ( 10-3 ) m4 12 12

Maximum shear stress occurs at the neutral axis. Referring to Fig. a, Qmax = Σy′A′ = 0.21[0.3(0.02)] + 0.1[0.2(0.02)] = 1.66 ( 10-3 ) m3 Maximum Shear Stress: Applying the shear formula, tmax =

30 ( 103 ) 3 1.66 ( 10-3 ) 4 VQmax = It 0.63627 ( 10-3 ) (0.02)

= 3.913 ( 106 ) Pa = 3.91 MPa

Ans.

Ans: tmax = 3.91 MPa 660

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7–6.  The wood beam has an allowable shear stress of tallow = 7 MPa. Determine the maximum shear force V that can be applied to the cross section.

50 mm

100 mm

50 mm

50 mm

200 mm V 50 mm

Solution I =

1 1 (0.2)(0.2)3 (0.1)(0.1)3 = 125(10 - 6) m4 12 12

tallow =

VQmax It

7(106) =

V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)] 125(10 - 6)(0.1) Ans.

V = 100 kN

Ans: Vmax = 100 kN 661

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7–7.  The shaft is supported by a thrust bearing at A and a journal bearing at B. If P = 20 kN, determine the absolute maximum shear stress in the shaft.

A

C

1m

Solution

B

1m

D

1m P

P 30 mm

Support Reactions: As shown on the free-body diagram of the beam, Fig. a. Maximum Shear: The shear diagram is shown in Fig. b. As indicated, Vmax = 20 kN.

40 mm

Section Properties: The moment of inertia of the hollow circular shaft about the neutral axis is p ­­­­I = (0.044 - 0.034) = 0.4375(10-6)p m4­­­ 4 Qmax ca­­­­n be computed by ta­­­king the first moment of the shaded area in Fig. c about the neutral axis. Here, y′1 =

4(0.04) 3p

=

4(0.03) 4 1 m and y′2 = = m. Thus, 75p 3p 25p

Qmax = y′1A′1 - y′2A′2 ­­­­­­

=

4 p 1 p c (0.042) d c (0.032) d = 24.667(10-6) m3 75p 2 25p 2

Shear Stress: The maximum shear stress occurs at points on the neutral axis since Q is maximum and the thickness t = 2(0.04 - 0.03) = 0.02 m is the smallest. tmax =

20(103)(24.667)(10-6) Vmax Qmax = = 17.9 MPa It 0.4375(10-6)p (0.02)

Ans.

Ans: tmax = 17.9 MPa 662

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*7–8. The shaft is supported by a thrust bearing at A and a journal bearing at B. If the shaft is made from a material having an allowable shear stress of tallow = 75 MPa, determine the maximum value for P.

A

C

1m

B

1m

1m P

P

Solution

D

30 mm

Support Reactions: As shown on the free-body diagram of the shaft, Fig. a. Maximum Shear: The shear diagram is shown in Fig. b. As indicated, Vmax = P.

40 mm

Section Properties: The moment of inertia of the hollow circular shaft about the neutral axis is p I = (0.044 - 0.034) = 0.4375(10-6)p m4 4 Qmax can be computed by taking the first moment of the shaded area in Fig. c about the neutral axis. Here, y′1 =

4(0.04) 3p

=

4(0.03) 4 1 m and y′2 = = m. Thus, 75p 3p 25p

Qmax = y′1A′1 - y′2A′2 ­

=

4 p 1 p c (0.042) d c (0.032) d = 24.667(10-6) m3 75p 2 25p 2

Shear­­­­ Stress: The maximum shear stress occurs at points on the neutral axis since Q is maximum and the thickness t = 2(0.04 - 0.03) = 0.02 m. tallow =

Vmax Qmax ; It

75(106) =

P(24.667)(10-6) 0.4375(10-6)p(0.02)

­­­­­­

P = 83 581.22 N = 83.6 kN

Ans.

Ans: P = 83.6 kN 663

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7–9. Determine the largest shear force V that the member can sustain if the allowable shear stress is tallow = 856ksi. MPa.

in. 753 mm 251mm in. V in.mm 753 mm in. 1 25

Solution

125in. mm

y =

(0.0125)(0.125)(0.025)  2[0.05(0.025)(0.05)]  0.029167 m 0.125(0.025)  2(0.025)(0.05)

I =

1 (0.125)(0.0253 ) + 0.125(0.025)(0.029167 − 0.0125)2 12

1  + 2  (0.025)(0.053 ) + 0.025(0.05)(0.05 − 0.029167)2  = 2.6367(10 −6 ) m 4 12  

0.04583 m 0.022917 m

6 3 Qmax = ©y¿A¿ = 2[(0.022917)(0.025)(0.04583)]  52.5174(10 ) m

tmax = tallow =

56(106 ) 

VQmax It

V [52.5174(10 6 )] [2.6367(10 6 )][2(0.025)]

V = 140.58(10 3 ) N  141 kN

Ans.

Ans: V = 141 kN 666

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applied shear shearforce forceVV == 18 90 kip, kN, determine the 7–10. If the applied maximum shear stress in the member. 753mm in. 251mm in. V in.mm 753mm in. 125

Solution

25 mm 1 in.

y =

(0.0125)(0.125)(0.025)  2[0.05(0.025)(0.05)]  0.029167 m 0.125(0.025)  2(0.025)(0.05)

I =

1 (0.125)(0.0253 ) + 0.125(0.025)(0.029167 − 0.0125)2 12

 1 + 2  (0.025)(0.053 ) + 0.025(0.05)(0.05 − 0.029167)2  = 2.6367(10 −6 ) m 4 12   6 3 Qmax = ©y¿A¿ = 2[(0.022917)(0.025)(0.04583)]  52.5174(10 ) m

tmax =

VQmax [90(10 3 )][52.5174(10 −6 ) m 3 ] 6 = 35.85(10 = ) N/m 3 35.9 MPa Ans. = It [2.6367(10 −6 )][2(0.025)]

0.04583 m 0.022917 m

Ans: tmax = 35.9 MPa 667

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7–11.  w

The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kN>m. Determine the maximum shear stress in the beam.

A B 3m

3m 50 mm

100 mm

Solution ­tmax =

150(103) N (0.025 m)(0.05 m)(0.05 m) VQ = 1 3 It 12 (0.05 m)(0.1 m) (0.05 m) Ans.

tmax = 45.0 MPa Because the cross section is a rectangle, then also, tmax = 1.5

150(103) N V = 1.5 = 45.0 MPa A (0.05 m)(0.1 m)

Ans.­

Ans: tmax = 45.0 MPa 666

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*7–12. A member has a cross section in the form of an equilateral triangle. If it is subjected to a shear force V, determine the maximum average shear stress in the member using the shear formula. Should the shear formula actually be used to predict this value? Explain.

a V

h

Solution I =

1 (a)(h)3 36

y h ; = x a>2 Q =

LA¿

Q = a

y =

y dA = 2 c a

2h x a

1 2 2 b (x)(y) a h - y b d 2 3 3

4h2 2x b (x2)a 1 b a 3a

t = 2x t = t =

V(4h2>3a)(x2)(1 - 2x VQ a) = 3 It ((1>36)(a)(h ))(2x) 24V(x - a2 x2) a2h

24V 4 dt = 2 2 a1 - xb = 0 a dx ah At x = y =

a 4 h 2h a a b = a 4 2

tmax =

24V a 2 a a b a1 - a b b 2 a 4 ah 4

tmax =

3V ah

Ans.

No, because the shear stress is not perpendicular to the boundary. See Sec. 7-3.

Ans. tmax = 483

3V ah

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7–13. Determine the shear stress at point B on the web of the cantilevered strut at section a–a.

2 kN 250 mm

a

250 mm

4 kN 300 mm

a

20 mm 70 mm

Solution (0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)

y = I =

(0.05)(0.02) + (0.07)(0.02)

20 mm

B

50 mm

= 0.03625 m

1 (0.05) ( 0.023 ) + (0.05)(0.02)(0.03625 - 0.01)2 12 +

1 (0.02) ( 0.073 ) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625 ( 10-6 ) m4 12

y′B = 0.03625 - 0.01 = 0.02625 m QB = (0.02)(0.05)(0.02625) = 26.25 ( 10-6 ) m3 tB =

6 ( 103 ) (26.25) ( 10-6 ) VQB = It 1.78622 ( 10-6 ) (0.02) Ans.

= 4.41 MPa

Ans: tB = 4.41 MPa 679

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7–14. Determine the maximum shear stress acting at section a–a of the cantilevered strut.

2 kN 250 mm

a

250 mm

4 kN 300 mm

a

20 mm 70 mm

Solution y = I =

(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02) (0.05)(0.02) + (0.07)(0.02)

20 mm

B

50 mm

= 0.03625 m

1 (0.05) ( 0.023 ) + (0.05)(0.02)(0.03625 - 0.01)2 12 +

1 (0.02) ( 0.073 ) + (0.02)(0.07)(0.055 - 0.03625)2 = 1.78625 ( 10-6 ) m4 12

Qmax = y′A′ = (0.026875)(0.05375)(0.02) = 28.8906 ( 10-6 ) m3 tmax =

6 ( 103 ) (28.8906) ( 10-6 ) VQmax = It 1.78625 ( 10-6 ) (0.02) Ans.

= 4.85 MPa

Ans: tmax = 4.85 MPa 680

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7–15. Determine the maximum shear stress in the T-beam at the critical section where the internal shear force is maximum.

10 kN/m

A

The FBD of the beam is shown in Fig. a,

1.5 m

3m

The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN

B

C

1.5 m

150 mm

The neutral axis passes through centroid c of the cross-section, Fig. c. ' 0.075(0.15)(0.03) + 0.165(0.03)(0.15) © y A = y = ©A 0.15(0.03) + 0.03(0.15)

150 mm

30 mm 30 mm

= 0.12 m 1 I = (0.03)(0.153) + 0.03(0.15)(0.12 - 0.075)2 12 +

1 (0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2 12

= 27.0 (10 - 6) m4 From Fig. d, Qmax = y¿A¿ = 0.06(0.12)(0.03) = 0.216 (10 - 3) m3 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =

27.5(103) C 0.216(10 - 3) D Vmax Qmax = It 27.0(10 - 6)(0.03) = 7.333(106) Pa = 7.33 MPa

Ans.

Ans. t max = 7.33 MPa 490

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*7–16.

10 kN/m

Determine the maximum shear stress in the T-beam at point C. Show the result on a volume element at this point. A

B

C

1.5 m

3m

1.5 m

150 mm

Solution

150 mm

30 mm 30 mm

using the method of sections, + c ©Fy = 0;

VC + 17.5 -

1 (5)(1.5) = 0 2

VC = - 13.75 kN The neutral axis passes through centroid C of the cross-section, ©yA 0.075 (0.15)(0.03) + 0.165(0.03)(0.15) = ©A 0.15(0.03) + 0.03(0.15)

y =

= 0.12 m I =

1 (0.03)(0.15) + 0.03(0.15)(0.12 - 0.075)2 12

+

1 (0.15)(0.033) + 0.15(0.03)(0.165 - 0.12)2 12

= 27.0 (10 - 6) m4 Qmax = y¿A¿ = 0.06 (0.12)(0.03) = 0.216 (10 - 3) m3 490 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =

13.75(103) C 0.216(10 - 3) D VC Qmax = It 27.0(10 - 6) (0.03)

= 3.667(106) Pa = 3.67 MPa

Ans.

Ans. t max = 3.67 MPa 491

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7–17. The strut is subjected to a vertical shear of V = 130 kN. Plot the intensity of the shear-stress distribution acting over the cross-sectional area, and compute the resultant shear force developed in the vertical segment AB.

B

150 mm 50 mm

A

Solution I =

V ⫽ 130 kN

150 mm

1 1 (0.05)(0.353) + (0.3)(0.053) = 0.18177083(10-3) m4 12 12

QC = y ¿A ¿ = 10.1210.05210.152 = 0.75110-32 m3

150 mm 50 mm 150 mm

QD = ©y ¿A¿ = 10.1210.05210.152 + 10.0125210.35210.0252 = 0.859375110-32 m3 t =

VQ It

1tC2t = 0.05 m =

= 10.7 MPa

1301103210.752110-32

= 1.53 MPa

0.18177083110-3210.052

1tC2t = 0.35 m = tD =

1301103210.752110-32

0.18177083110-3210.352

1301103210.8593752110-32 0.18177083110-3210.352

= 1.76 MPa

A¿ = 10.05210.175 - y2 y¿ = y +

10.175 - y2 2

=

1 10.175 + y2 2

Q = y¿A¿ = 0.02510.030625 - y22 t =

=

VQ It 13010.025210.030625 - y22 0.18177083110-3210.052

= 10951.3 - 357593.1 y2 VAB =

L

t dA

dA = 0.05 dy

0.175

=

L0.025

110951.3 - 357593.1y2210.05 dy2

0.175

=

L0.025

1547.565 - 17879.66y 2 dy 2

= 50.3 kN

Ans.

Ans: VAB = 50.3 kN 672

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7–18. Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is the maximum shear stress greater than the average shear stress acting over the cross section? c y V

Solution x = 2c 2 - y2 ;

I =

p 4 c 4

t = 2x = 22c 2 - y2 dA = 2x dy = 22c 2 - y2 dy dQ = ydA = 2y2c 2 - y2 dy c

c

3 3 2 2 Q = 2y2c - y dy = - ( c 2 - y2 ) 2 ` = ( c 2 - y2 ) 2 3 3 Ly y

2

VQ t = = It

2

V 3 23 ( c 2 - y2 ) 2 4 3

(

p 4 4c

) 1 22c - y 2 2

2

=

4V 2 ( c - y2 ) 3pc 4

The maximum shear stress occur when y = 0 tmax =

4V 3pc 2

tavg =

V V = A pc 2

The factor =

tmax = tavg

4V 3pc 2 V pc 2

=

4 3

Ans.

Ans: The factor = 671

4 3

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7–19. Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.

12 mm

60 mm V 12 mm 80 mm

20 mm

20 mm

Solution Section Properties: INA =

1 1 (0.12) ( 0.0843 ) (0.04) ( 0.063 ) 12 12

= 5.20704 ( 10 - 6 ) m4 Qmax = Σy′A′ = 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 ( 10 - 6 ) m3 Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax = =

VQmax It 20(103)(87.84)(10 - 6) 5.20704(10 - 6)(0.08) Ans.

= 4. 22 MPa

Ans: tmax = 4.22 MPa 668

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  Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow = 40 MPa.

12 mm

60 mm V 12 mm 80 mm

20 mm

20 mm

Solution Section Properties: INA =

1 1 (0.12) ( 0.0843 ) (0.04) ( 0.063 ) 12 12

= 5.20704 ( 10 - 6 ) m4 Qmax = Σy′A′ = 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 ( 10 - 6 ) m3 Allowable Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax = tallow = 40 ( 106 ) =

VQmax It V(87.84)(10 - 6) 5.20704(10 - 6)(0.08) Ans.

V = 189 692 N = 190 kN

Ans: Vmax = 190 kN 669

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7–21. Determine the maximum shear stress acting in the fiberglass beam at the section where the internal shear force is maximum.

3 kN/m 200 lb/ft

2.5 150kN/m lb/ft

D D

A A 26 m ft

26 m ft

0.6 2 ftm

100 mm 4 in.

18 mm 0.75 in.

150 mm 6 in.

120.5 mm in.

18 mm 0.75 in.

4 in. 100 mm

Solution Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, Vmax = 4.783 kN. 2.5(2) = 5 kN

1 2 (3)(2) = 3 kN

Section Properties: INA =

1 1 (0.1)(0.186 3 ) − (0.088)(0.153 ) = 28.8738(10 −6 ) m 4 12 12

1m

Qmax = ©y¿A¿

1.333 m

2.267 m

3.217 kN

4.783 kN

−3

3

= 0.084(0.1)(0.018) + 0.0375(0.012)(0.075) = 0.18495(10 ) m Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section.

V (kN) 4.783 2

Applying the shear formula

4.6

2.6

–0.217

tmax

VQmax = It =

–3.217

[4.783(10 3 )][0.18495(10 −3 )] 6 = 2.553(10 = ) N/m 3 2.55 MPa [28.8738(10 −6 )](0.012)

Ans.

0.1 m 0.018 m y2 = 0.084 m

0.075 m

y1 = 0.0375 m

0.012 m

0.075 m 0.018 m

Ans. tmax = 2.55 MPa 492

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  If the beam is subjected to a shear of V = 15 kN, determine the web’s shear stress at A and B. Indicate the shear-stress components on a volume element located at these points. Set w = 125 mm. Show that the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182(10 −3) m4.

200 mm A

30 mm

25 mm V B

250 mm 30 mm

w

Solution (0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03)

y = I =

0.125(0.03) + (0.025)(0.25) + (0.2)(0.03)

= 0.1747 m

1 (0.125)(0.033) + 0.125(0.03)(0.1747 - 0.015)2 12

+

1 (0.025)(0.253) + 0.25(0.025)(0.1747 - 0.155)2 12

+

1 (0.2)(0.033) + 0.2(0.03)(0.295 - 0.1747)2 = 0.218182(10-3) m4 12

QA = yA′A = (0.310 - 0.015 - 0.1747)(0.2)(0.03) = 0.7219(10-3) m3 QB = yA′B = (0.1747 - 0.015)(0.125)(0.03) = 0.59883(10-3) m3 tA =

15(103)(0.7219)(10-3) VQA = = 1.99 MPa It 0.218182(10-3)0.025

Ans.

tB =

15(103)(0.59883)(10-3) VQB = = 1.65 MPa­ It 0.218182(10-3)0.025

Ans.

Ans: tA = 1.99 MPa, tB = 1.65 MPa 672

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  If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam. Set w = 200 mm.

200 mm A

30 mm

25 mm V B

250 mm 30 mm

w

Solution Section Properties: I =

1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10)-6 m4 12 12

Qmax = ΣyA = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10)-3 m3 tmax =

30(10)3 (1.0353)(10)-3 VQ = = 4.62 MPa­­ It 268.652(10)-6(0.025)

Ans.

Ans: tmax = 4.62 MPa 673

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  If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam. Set w = 200 mm.

200 mm A

30 mm

25 mm V B

250 mm 30 mm

w

Solution I =

1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10) - 6 m4 12 12

Q = a tf = Vf =

0.155 + y b(0.155 - y)(0.2) = 0.1(0.024025 - y2) 2

30(10)3(0.1)(0.024025 - y2) 268.652(10)-6(0.2) L

0.155

tf dA = 55.8343(10)6

L0.125

(0.024025 - y2)(0.2 dy) 0.155

1 = 11.1669(10) c 0.024025y - y3 d 3 6

0.125

Vf = 1.457 kN

Ans.

Vw = 30 - 2(1.457) = 27.1 kN­

Ans: Vw = 27.1 kN­ 674

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7–25. Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear stress.

P

h L

b

Solution Vmax = P Mmax = PL smax =

PL ( h>2 ) Mc PLh = = I I 2I

tmax =

P ( h>2 ) (b) ( h>4 ) VQ Ph2 = = It Ib 8I

Require, smax = tmax PLh Ph2 = 2I 8I L =

h 4

Ans.

Ans: L = 675

h 4

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7–26. The steel rod is subjected to a shear of 150 kN. Determine the shear stress at point A. Show the result on a volume element at this point. 25 mm

The moment of inertia of the circular cross-section about the neutral axis (x axis) is I =

A

50 mm

p 4 p 44 r = (2 in4 p(106) mm4 (50) )== 4p 1.5625 4 4

150 k N

Q for the differential area shown in Fig. a is dQ = ydA = y (2xdy) = 2xy dy 1

1

2 , Then (4 - y–2)y22, )Then However, from the equation of the circle, x = (2500 1 2

1

dQ = 2y (2500 (4 - y–2y)22)dydy Thus, Q for the area above y is 2 in. 50 mm

Q =

Ly

= -

1

1

2 2 2y2y(2500 (4 - y2–)2ydy ) dy

3 23 in.50 mm 3 3 2 2 2 (2500 (4 - y–2)y22)` 2 0 = =(4 (2500 - y2)2– y2) 2 y 3 3 3 y

1

1

2 (4 - y–2)y22. )Thus, Here t = 2x = 2 (2500 . Thus

30 150(10 )2 D C 23 (4 -3)[y232(2500 VQ – y2) 2 ] = t = 1 1 2 2 It 4p )D C 2(4 - y6)[2(2500 1.5625p(10 – y2) 2 ] 54 2 t = (4 (2500 - y2) – yksi ) Ksi 2p 125p 3

3

Thus For point A, y = 125inmm. . Thus tA =

54 2 (4 (2500 - 12)–=252.39 ksi MPa )= 19.1 2p 125p

Ans.

The state of shear stress at point A can be represented by the volume element shown in Fig. b.

19.1 MPa

50 mm

x = (2500 – y2)1/2

Ans: P = 5.9988 kN 678

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7–27.  The beam is slit longitudinally along both sides. If it is subjected to a shear of V = 250 kN, compare the maximum shear stress in the beam before and after the cuts were made.

25 mm 200 mm V 100 mm

25 mm 25 mm

Solution Section Properties: The moment of inertia of the cross section about the neutral axis is

25 mm

25 mm 200 mm

1 1 ­ I = (0.2)(0.23) (0.125)(0.153) = 98.1771(10-6) m4 12 12 Qmax is the first moment of the shaded area shown in Fig. a about the neutral axis. Thus, Qmax = 3y′1A′1 + y′2 A′2 = 3 3 0.0375(0.075)(0.025) 4 + 0.0875(0.025)(0.2) = 0.6484375(10-3) m3

Maxim­um Shear Stress: The maximum shear stress occurs at the points on the neutral axis since Q is maximum and t is minimum. Before the cross section is slit, t = 3(0.025) = 0.075 m. tmax =

250(103)(0.6484375)(10-3) VQmax = = 22.0 MPa It 98.1771(10-6)(0.075)

Ans­.

After the cross section is slit, t = 0.025 m. (tmax)s =

250(103)(0.6484375)(10-3) VQmax = = 66.0 MPa It 98.1771(10-6)(0.025)

Ans.

Ans: tmax = 22.0 MPa, (tmax)s = 66.0 MPa 682

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*7–28. The beam is to be cut longitudinally along both sides as shown. If it is made from a material having an allowable shear stress of tallow = 75 MPa, determine the maximum allowable shear force V that can be applied before and after the cut is made.

25 mm 200 mm V 100 mm

25 mm 25 mm

Solution Section Properties: The moment of inertia of the cross section about the neutral axis is ­­­I =

25 mm

25 mm 200 mm

1 1 (0.2)(0.23) (0.125)(0.153) = 98.1771(10-6) m4 12 12

Qmax is the first moment of the shaded area shown in Fig. a about the neutral axis. Thus, Qmax = 3y′1A′1 + y′2 A′2 = 3(0.0375)(0.075)(0.025) + 0.0875(0.025)(0.2) = 0.6484375(10-3) m3 Shear Stress: The maximum shear stress occurs at the points on the neutral axis since Q is maximum and  thickness t is minimum. Before the cross section is slit, t = 3(0.025) = 0.075 m. tallow =

VQmax ; It

75(106) =

V(0.6484375)(10-3) 98.1771(10-6)(0.075) Ans.

V = 851 656.63 N = 852 kN After the cross section is slit, t = 0.025 m. tallow =

VQmax ; It

75(106) =

Vs (0.6484375)(10-3) 98.1771(10-6)(0.025) Ans.

Vs = 283 885.54 N = 284 kN­­

Ans: V = 852 kN, Vs = 284 kN 683

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7–29.  The composite beam is constructed from wood and reinforced with a steel strap. Use the method of Sec. 6.6 and calculate the maximum shear stress in the beam when it is subjected to a shear of V = 50 kN. Take Est = 200 GPa, Ew = 15 GPa.

10 mm V= 50 kN 300 mm

10 mm 175 mm

Solution bst = nbw = I =

15 (0.175) = 0.013125 m 200

1 1 (0.175) ( 0.323 ) (0.175 - 0.013125) ( 0.33 ) = 0.113648 ( 10-3 ) m4 12 12

Qmax = Σy′A′ = 0.075(0.013125)(0.15) + 0.155(0.175)(0.01) = 0.4189 ( 10-3 ) m3 tmax = n

50 ( 103 ) (0.4189) ( 10-3 ) VQmax 15 = a b It 200 0.113648 ( 10-3 ) (0.013125)

Ans.

= 1.05 MPa

Ans: tmax = 1.05 MPa 684

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7–30.  The beam has a rectangular cross section and is subjected to a load P that is just large enough to develop a fully plastic moment Mp = PL at the fixed support. If the material is elastic perfectly plastic, then at a distance x 6 L the moment M = Px creates a region of plastic yielding with an associated elastic core having a height 2y′. This situation has been described by Eq. 6–30 and the moment M is distributed over the cross section as shown in Fig. 6–48e. Prove that the maximum shear stress in the beam is given by tmax = 32(P>A′), where A′ = 2y′b, the cross-sectional area of the elastic core.

P x Plastic region 2y¿

h

b Elastic region

L

Solution Force Equilibrium: The shaded area indicates the plastic zone. Isolate an element in the plastic zone and write the equation of equilibrium. { ΣFx = 0;

tlong A2 + sg A1 - sg A1 = 0 tlong = 0

This proves that the longitudinal shear stress. tlong, is equal to zero. Hence the corresponding transverse stress, tmax, is also equal to zero in the plastic zone. Therefore, the shear force V = P is carried by the m­aterial only in the elastic zone. Section Properties: INA =

1 2 (b)(2y′)3 = b y′3 12 3 y′ y′2b (y′)(b) = 2 2

Qmax = y′ A′ =

Maximum Shear Stress: Applying the shear formula tmax However,

VQmax = = It

1

V1

2 3

y′2b 2

A′ = 2by′    hence tmax =

2

by′ 2 (b) 3

=

3P 4by′

3P ‚ (Q.E.D.) 2A′

Ans: N/A 685

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7–31.  The beam in Fig. 6–48f is subjected to a fully plastic moment Mp . Prove that the longitudinal and transverse shear stresses in the beam are zero. Hint: Consider an element of the beam shown in Fig. 7–4d.

Solution Force Equilibrium: If a fully plastic moment acts on the cross section, then an element of the material taken from the top or bottom of the cross section is subjected to the loading shown. For equilibrium { ΣFx = 0;

sg A1 + tlong A2 - sg A1 = 0 tlong = 0

Thus no shear stress is developed on the longitudinal or transverse plane of the element. (Q. E. D.)

Ans: N/A 686

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*7–32. 20 mm

The double T-beam is fabricated by welding the three plates together as shown. Determine the shear stress in the weld necessary to support a shear force of V = 80 kN.

150 mm V 50 mm

Solution

20 mm

y =

0.01(0.215)(0.02) + 2[0.095(0.15)(0.02)] ΣyA = = 0.059515 m ΣA 0.215(0.02) + 2(0.15)(0.02)

I =

1 (0.215) ( 0.023 ) + 0.215(0.02)(0.059515 - 0.01)2 12 + 2c

75 mm

50 mm 20 mm

1 (0.02) ( 0.153 ) + 0.02(0.15)(0.095 - 0.059515)2 d = 29.4909 ( 10-6 ) m4 12

y′ = 0.059515 - 0.01 = 0.049515 m

Q = y′A′ = 0.049515(0.215)(0.02) = 0.2129 ( 10-3 ) m3 Shear stress: t =

80 ( 103 ) (0.2129) ( 10-3 ) VQ = It 29.4909 ( 10-6 ) (2)(0.02) Ans.

= 14.4 MPa

Ans: t = 14.4 MPa 691

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  The double T-beam is fabricated by welding the three plates together as shown. If the weld can resist a shear stress tallow = 90 MPa, determine the maximum shear V that can be applied to the beam.

20 mm

150 mm V 50 mm

Solution

20 mm

y =

0.01(0.215)(0.02) + 2[0.095(0.15)(0.02)] ΣyA = = 0.059515 m ΣA 0.215(0.02) + 2(0.15)(0.02)

I =

1 (0.215) ( 0.023 ) + 0.215(0.02)(0.059515 - 0.01)2 12 + 2c

75 mm

50 mm 20 mm

1 (0.02) ( 0.153 ) + 0.02(0.15)(0.095 - 0.059515)2 d = 29.4909 ( 10-6 ) m4 12

y′ = 0.059515 - 0.01 = 0.049515 m

Q = y′A′ = 0.049515(0.215)(0.02) = 0.2129 ( 10-3 ) m3 t = 90 ( 106 ) =

VQ It V(0.2129) ( 10-3 ) 29.491 ( 10-6 ) (2)(0.02) Ans.

V = 499 kN

Ans: V = 499 kN 692

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7–34. The beam is constructed from two boards fastened in. apart. three rows rows of ofnails nailsspaced spaceds s= =502mm together with three apart. If nail can can support support aa2.25-kN 450-lb shear each nail shear force, force, determine the maximum shear force V that can be applied to the beam. The allowable shear stress for the wood is ttallow == 2.1 300MPa. psi.

s s 1.5 in. 40 mm 1.5 in. 40 mm

V

Solution

6150 in.mm

The moment of inertia of the cross-section about the neutral axis is I =

1 (0.15)(0.08 3 ) = 6.40(10 −6 ) m 4 12

Refering to Fig. a,

0.15 m −3

QA = Qmax = y¿A¿ = 0.02(0.15)(0.04) = 0.12(10 ) m

3

The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 0.15 m. tallow =

VQmax ; It

2.1(106 ) =

0.04 m

V [0.12(10 −3 )] [6.40(10 −6 )](0.15)

y = 0.02 m

3 = V 16.8(10 = ) N 16.8 kN

Shear Flow: Since there are three rows of nails,  2.25(10 3 )  F 3 qallow = 3 a b = 3   = 135(10 ) N/m s 0.05   qallow =

VQA ; I

135(10 3 ) =

V [0.12(10 −3 )] 6.40(10 −6 )

V = 7.20(10 3 ) N = 7.20 kN

Ans.

Ans. V = 7.20 kN 497

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7–35. The beam is constructed from two boards fastened together with three rows of nails. If the allowable shear stress for the the wood wood isis ttallow = 1150MPa. psi, determine the allow = maximum shear force V that can be applied to the beam. Also, find the maximum spacing s of the nails if each nail can resist 3.25 kNininshear. shear. 650 lb

s s 1.5 in. 40 mm V

1.5 in. 40 mm

6150 in.mm

Solution The moment of inertia of the cross-section about the neutral axis is I =

1 (0.15)(0.08 3 ) = 6.40(10 −6 ) m 4 12

Refering to Fig. a, QA = Qmax = y¿A¿ = 0.02(0.15)(0.04) = 0.12(10 −3 ) m 3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 0.15 m. tallow =

VQmax ; It

1(106 ) =

V [0.12(10 −3 )] [6.40(10 −6 )](0.15)

3 = V 8.00(10 = ) N 8.00 kN

Ans.

 2.25(10 3 )   9750  Since there are three rows of nails, qallow = 3  =  N/m s   s   qallow =

VQA ; I

9750 [8(10 3 )][0.12(10 −3 )] = s 6.40(10 −6 )

s = 0.0650 m = 65.0 mm

Ans.

0.15 m

0.04 m

y = 0.02 m

Ans. V = 8.00 kN, s = 65.0 mm 498

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*7–36. The beam is constructed from four boards which are nailed together. If the nails are on both sides of the beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to the end of the beam.

3 kN

A

P

B

C

2m

2m

100 mm

30 mm 150 mm

30 mm

250 mm 30 mm

30 mm

Solution Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, VAB = (P + 3) kN. Section Properties: INA =

1 1 (0.31) A 0.153 B (0.25) A 0.093 B 12 12

= 72.0 A 10 - 6 B m4

Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450 A 10 - 3 B m3 Shear Flow: There are two rows of nails. Hence the allowable shear flow is 3(2) = 60.0 kN>m. q = 0.1 VQ q = I (P + 3)(103)0.450(10 - 3) 60.0 A 103 B = 72.0(10 - 6) P = 6.60 kN

Ans.

Ans. P = 6.60 kN 507

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7–37. The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 150 a thickness 12in. mm. shearofofV V= kN is 6 in.mm andand a thickness ofof0.5 If If a ashear = 250 50 kip applied to the cross section, determine the maximum spacing 75 kip. kN. of the bolts. Each bolt can resist a shear force of 15

0.5 in. 12 mm ss

75 mm 3 in. 125in.mm A A V V

1506mm in.

Solution

0.5 in. 12 mm

N N

Section Properties: INA

375in.mm

1 1 = (0.075)(0.224 3 ) − (0.063)(0.2 3 ) 12 12 −

75 mm

1 1 (0.012)(0.053 ) + (0.024)(0.153 ) 12 12

12 mm 12 mm

= 34.8714(10 −6 ) m 4

106 mm

−3

75 mm 75 mm

62.5 mm

25 mm 25 mm 75 mm 75 mm

3

0.15165(10 ) m Q = ©y¿A¿ = 0.0625(0.012)(0.075) + 0.106(0.075)(0.012) =

Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(75)(10 3 ) 150(10 3 ) = q = . s s

12 mm 12 mm

q =

VQ I

150(10 3 ) [250(10 3 )][0.15165(10 −3 )] = s 34.8714(10 −6 ) = s 0.1380 = m 138 mm

Ans.

Ans. s = 138 mm 499

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7–38. The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 6 in.mm andand a thickness ofof0.5 If Ifthe 150 a thickness 12 in. mm. thebolts boltsare are spaced spaced at 8 in., determine the s ==200 mm, determine themaximum maximumshear shearforce force V V that that can appliedtotothe the cross section. Each can resist be applied cross section. Each bolt bolt can resist a sheara shear of force of 15 kip. force 75 kN.

12 mm 0.5 in. s

753 mm in. mm 125in. A

Solution

1506mm in.

V

N

12 mm 0.5 in. mm 375in.

Section Properties: INA =

1 1 (0.075)(0.224 3 ) − (0.063)(0.2 3 ) 12 12

75 mm

1 1 − (0.012)(0.053 ) + (0.024)(0.153 ) 12 12

12 mm 12 mm

106 mm

= 34.8714(10 −6 ) m 4 −3

75 mm 75 mm

62.5 mm

25 mm 25 mm 75 mm 75 mm

3

0.15165(10 ) m Q = ©y¿A¿ = 0.0625(0.012)(0.075) + 0.106(0.075)(0.012) =

Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is = q

3

12 mm 12 mm

2(75)(10 ) = 750(10 3 ) N/m 0.2

q = 750(10 3 ) =

VQ I V [0.15165(10 −3 )] 34.8714(10 −6 )

3 = V 172.46(10 ) N 172 kN =

Ans.

Ans. INA = 34.8714(10- 6) m4, Q = 0.15165 (10- 3) m3 V =172 kN 499

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7–39. The double-web girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. If each lb in single each fastener fastener can can support support 600 3 kN shear, determine the required spacing s of the fasteners needed to support support the the loading loading PP == 3000 15 kN. lb. Assume Assume A is pinned and B is a roller.

2 in. 50 mm 2 in. 50 mm

P s

10 in. 250 mm

A

4 ftm 1.2

B

4 ftm 1.2

2 in. 50 mm 2 in. 50 mm 6 in. 150 mm 0.5mm in. 0.5mm in. 12 12

Solution Support Reactions: As shown on FBD.

15 kN

7.5 kN. lb. Internal Shear Force: As shown on shear diagram, Vmax = 1500 Section Properties: INA =

7.5 kN

1 1 (0.174)(0.453 ) − (0.15)(0.253 ) = 1.126(10 −3 ) m 4 12 12

7.5 kN

V (kN) 7.5

−3 3 Q = y¿A¿ = 0.175(0.15)(0.1) = 2.625(10 ) m

x (m) –7.5

Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(600) 3) 1200 2(3)(10 6000 = . . q = ss s s VQ q = I 3

12 mm

150 mm

12 mm

100 mm 175 mm

125 mm

−3

6000 7.5(10 )[2.625(10 )] = s 1.126(10 −3 )

125 mm

= s 0.3432 = m 343 mm

Ans.

100 mm

Ans. s = 343 mm 501

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*7–40. The double-web girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. The allowable bending stress for the wood is MPa the allowable = andand the allowable shearshear stress stress is tallowis =tallow sallow = 56 8 ksi 3 ksi. 21the MPa. If the fasteners are s = 150 and each If fasteners are spaced s spaced = 6 in. and eachmm fastener can fastener600 canlbsupport kN indetermine single shear, determineload the support in single3 shear, the maximum maximum load P thattocan applied to the beam. P that can be applied thebebeam.

mm 250in. mm 250in.

P s

250 mm 10 in.

A

1.2 4 ftm

B

1.2 4 ftm

mm 250in. mm 250in. 6 in. 150 mm 12 12 0.5mm in. 0.5mm in.

Solution Support Reactions: As shown on FBD.

Internal Shear Force and Moment: As shown on shear and moment diagram, Vmax = 0.500P and Mmax = 2.00P. Section Properties: 1.2 m

INA

1.2 m

1 1 (0.174)(0.453 ) − (0.15)(0.253 ) = 1.126(10 −3 ) m 4 = 12 12

Q = yœ2 A¿ = 0.175(0.15)(0.1) = 2.625(10 −3 ) m 3 2.4

3.2325(10 −3 ) m 3 Qmax = ©y¿A¿ = 0.175(0.15)(0.1) + 0.1125(0.024)(0.225) =

Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the 2(3)(10 3 ) allowable shear = flow is q = 40(10 3 ) N/m 0.15 VQ q = I 40(10 3 ) =

x (m) 1.2

0.012 m

Ans.

(Controls!)

2.4

tmax = tallow =

VQmax It

0.15 m

0.012 m

0.1 m 0.175 m

Shear Stress: Assume failure due to shear stress.

21(10 ) =

0.6P

0.5P[2.625(10 −3 )]

1.126(10 −3 ) P = 34.32(10 3 ) N = 34.3 kN

6

x (m)

1.2

0.125 m 0.125 m 0.1 m

0.5P[3.2325(10 −3 )] [1.126(10 −3 )](0.024)

3 P 351.12(10 ) N 351 kN = =

Bending Stress: Assume failure due to bending stress. smax = sallow = 56(106 ) =

Mc I 0.6P(0.225) 1.126(10 −3 )

3 P 467.08(10 = = ) N 467 kN

Ans. P = 34.3 kN 502

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  A beam is constructed from three boards bolted together as shown. Determine the shear force in each bolt if the bolts are spaced s = 250 mm apart and the shear is V = 35 kN.

25 mm 25 mm 100 mm 250 mm

V

Solution y =

2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025) 2 (0.25)(0.025) + 0.35 (0.025)

I = (2) a +

350 mm

s  250 mm

= 0.18676 m

25 mm

1 b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676 - 0.125)2 12

1 (0.025)(0.35)3 + (0.025)(0.35)(0.275 - 0.18676)2 12

= 0.270236 (10 - 3) m4 Q = y′A′ = 0.06176(0.025)(0.25) = 0.386(10 - 3) m3 q =

35(0.386)(10-3) VQ = = 49.997 kN>m I 0.270236(10-3) Ans.

F = q(s) = 49.997 (0.25) = 12.5 kN

Ans: F = 12.5 kN 694

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7–42. The simply supported beam is built up from three boards by nailing them together as shown. The wood has an allowable shear stress of tallow = 1.5 MPa, and an allowable bending stress of sallow = 9 MPa. The nails are spaced at s = 75 mm, and each has a shear strength of 1.5 kN. Determine the maximum allowable force P that can be applied to the beam.

P

s

A

B 1m

1m 100 mm

Solution

25 mm

Support Reactions: As shown on the free-body diagram of the beam shown in Fig. a. Maximum Shear and Moment: The shear diagram is shown in Fig. b. As indicated, P Vmax = .­­­­­­­ 2 Section Properties: The moment of inertia of the cross section about the neutral axis is 1 1 I = (0.1)(0.253) (0.075)(0.23) 12 12 = 80.2083(10 - 6) m4 Referring to Fig. d, QB = y ′2 A′2 = 0.1125(0.025)(0.1) = 0.28125(10 - 3) m3 Shear Flow: Since there is only one row of nails, qallow =

qallow =

VmaxQB ; I

20 ( 103 ) =

1.5(103) F = = 20 (103) N>m. s 0.075

p c 0.28125 ( 10-3 ) d 2

80.2083 ( 10-6 ) P = 11417.41 N = 11.4 kN (controls)

Bending, smax = s(106) N>m2 =

Mc I

1 P2 2(0.125 m)

80.2083(10-6) m4

P = 11.550 N = 11.55 kN Shear, tmax =

VQ It

Q = (0.1125)(0.025)(0.1) + (0.05)(0.1)(0.025) = 0.40625(10-3) m3 1.5 1 106 2 =

1 P2 2 (0.40625)(10-3) m3

80.2083(10-6) m4(0.025 m)

P = 14.808 N = 14.8 kN

695

Ans.

25 mm

200 mm

25 mm

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7–42.

Continued

Ans: P = 11.4 kN 696

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7–43. The simply supported beam is built up from three boards by nailing them together as shown. If P = 12 kN, determine the maximum allowable spacing s of the nails to support that load, if each nail can resist a shear force of 1.5 kN.

P

s

A

B 1m

1m 100 mm

Solution

25 mm

Support Reactions: As shown on the free-body diagram of the beam shown in Fig. a.

25 mm

200 mm

Maximum Shear and Moment: The shear diagram is shown in Fig. b. As indicated, Vmax =

P 12 = = 6 kN. 2 2

25 mm

Section Properties: The moment of inertia of the cross section about the neutral axis is 1 1 I = (0.1)(0.253) (0.075)(0.23) 12 12 = 80.2083(10-6) m4 Referring to Fig. d, QB = y′2 A′2 = 0.1125(0.025)(0.1) = 0.28125(10-3) m3 Shear Flow: Since there is only one row of nails, qallow =

qallow =

VmaxQB ; I

1.5(1031) s

=

1.5(1031) F = . s s

6000 3 0.28125(10-3) 4 80.2083(10-61)

Ans.

s = 0.07130 m = 71.3 mm

Ans: s = 71.3 mm 697

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*7–44. The T-beam is nailed together as shown. If the nails can each each support determine the the can support aa shear shear force force of of 4.5 950 kN, lb, determine maximum shear shear force forceVVthat thatthethe beam support maximum beam cancan support and and the 1 the corresponding maximum spacing s tonearest the nearest corresponding maximum nail nail spacing s to the 8 in. multiples of 5 mm. The allowable shear stress for the wood The allowable shear stress for the wood is tallow = 450 psi. is tallow = 3 MPa.

502mm in.

ss

300 mm 12 in. ss

30012mm in.

Solution

V V

The neutral axis passes through the centroid c of the cross-section as shown in Fig. a. ' © y A 0.325(0.3)(0.05) + 0.15(0.05)(0.3) = 0.2375 m = y = ©A 0.3(0.05) + 0.05(0.3) I =

50 mm 2 in.

1 (0.05)(0.33 ) + 0.05(0.3)(0.2375 − 0.15)2 12 +

1 (0.3)(0.053 ) + 0.3(0.05)(0.325 − 0.2375)2 12

= 0.3453125(10 −3 ) m 4

Refering to Fig. a, Qmax and QA are Qmax = y1œ A1œ = 0.11875(0.05)(0.2375) = 1.41016(10 −3 ) m 3 QA = y2œ A2œ = 0.0875(0.3)(0.05) = 1.3125(10 −3 ) m 3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 0.05 m. tallow =

VQmax ; It

3(106 ) =

V [1.41016(10 −3 )] [0.3453125(10 −3 )](0.05)

3 V 36.73(10 = = ) N 36.7 kN

Here, qallow =

Ans.

F  4.5(10 3 )  =  N/m . Then s s   −3 3  3 VQA  ;  4.5(10 )  = [36.73(10 )][1.3125(10 )] qallow = I s 0.3453125(10 −3 )   = s 0.03223 = m 32.23 mm

Ans.

Use s  35 mm 0.3 m

0.3 m 0.05 m

0.05 m

0.875 m

= 0.11875 m 0.3 m

0.2375 m

0.05 m

0.05 m

Ans. V = 36.7 kN, s = 32.23 mm 503

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7–45. The nails are on both sides of the beam and each can resist a shear of 2 kN. In addition to the distributed loading, determine the maximum load P that can be applied to the end of the beam. The nails are spaced 100 mm apart and the allowable shear stress for the wood is tallow = 3 MPa.

P

2 kN/m

A

B

C

1.5 m

Solution

1.5 m

100 mm

The FBD is shown in Fig. a. 40 mm

As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of Constant value, Vmax = (P + 3) kN. The neutral axis passes through Centroid C of the cross-section as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04) = 0.14 m I =

1 (0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18 - 0.142) 12

Refering to Fig. d, Qmax = y1œ A1œ = 0.07(0.14)(0.04) = 0.392(10 - 3) m3 QA = y2œ A2œ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3 The maximum shear stress occurs at the points on Neutral axis where Q is maximum and t = 0.04 m. Vmax Qmax ; It

3(106) =

(P + 3)(103) C 0.392(10 - 3) D 53.333(10 - 6)(0.04)

P = 13.33 kN Since there are two rows of nails qallow = 2 a qallow =

Vmax QA ; I

40 000 =

200 mm 20 mm

20 mm

= 53.333(10 - 6) m4

tallow =

200 mm

2(103) F b = 2c d = 40 000 N>m. s 0.1

(P + 3)(103) C 0.32(10 - 3) D 53.333(10 - 6)

P = 3.67 kN (Controls!)

Ans.

505

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7–45. Continued

Ans. P = 3.67 kN 506

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7–46. Determine the average shear stress developed in the nails within region AB of the beam. The nails are located on each side of the beam and are spaced 100 mm apart. Each nail has a diameter of 4 mm. Take P = 2 kN.

P

2 kN/m

A

B

C

1.5 m

The FBD is shown in Fig. a. As indicated in Fig. b, the internal shear force on the cross-section within region AB is constant that is VAB = 5 kN.

1.5 m

100 mm

The neutral axis passes through centroid C of the cross section as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04)

40 mm

= 0.14 m

200 mm

1 (0.04)(0.23) + 0.04(0.2)(0.14 - 0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18 - 0.14)2 12

I =

200 mm 20 mm

20 mm

= 53.333(10 - 6) m4 Q for the shaded area shown in Fig. d is Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10 - 3) m3 Since there are two rows of nail, q = 2 a q =

VAB Q ; I

20F =

F F b = 2a b = 20F N>m. s 0.1

5(103) C 0.32(10 - 3) D 53.333(10 - 6)

F = 1500 N Thus, the average shear stress developed in each nail is

A tnail B avg =

F 1500 = = 119.37(106)Pa = 119 MPa p Anail 2 (0.004 ) 4

504

Ans.

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7–47. The beam is made from four boards nailed together as shown. If the nails can each support a shear force of 500 N, determine determinetheir theirrequired requiredspacing spacing ssand andss if the beam 100 lb., is subjected to a shear of V V ==3.5 700kN. lb.

D

s¿ s¿ s

A

C s

1 in. 25 mm

Solution

10 in. 250 mm V B

Section Properties: y =

10 in. 250 mm

1 in. 25 mm 1 in. 25 mm 2 in. 50 mm

1.5mm in. 40

©yA 0.0125(0.25)(0.025) + 0.0375(0.05)(0.75) + 0.15(0.04)(0.25) = ©A 0.25(0.025) + 0.05(0.075) + 0.04(0.25) = 0.0859375 m

INA =

1 (0.25)(0.0253 ) + 0.25(0.025)(0.0859375 − 0.0125)2 12 +

1 (0.05)(0.0753 ) + 0.05(0.075)(0.0859375 − 0.0375)2 12

+

1 (0.04)(0.253 ) + 0.04(0.25)(0.15 − 0.0859375)2 12

= 0.137712(10 −3 ) m 4 QC = y1¿A¿ = (0.0859375 − 0.0375)(0.25)(0.075) = 90.8203(10 −6 ) m 3 0.640625(10 −3 ) m 3 QD = y2¿A¿ = (0.0859375 − 0.0125)(0.25)(0.025) + [(0.0859375 − 0.0375)](0.25)(0.075) =

Shear Flow: The allowable shear flow at points C and D is qC = 500 100 , respectively. qB = s¿ VQC qC = I 500 [3.5(10 3 )][90.8203(10 −6 )] = s 0.137712(10 −3 )

500 100 and s

Ans.

= m 216 mm = s 0.2166 VQD qD = I 500 [3.5(10 3 )][0.640625(10 −3 )] = s′ 0.137712(10 −3 ) = s′ 0.03071 = m 30 mm

Ans.

Ans. s = 216 mm, s′ = 30 mm 508

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*7–48. The beam is made from three polystyrene strips that are glued together as shown. If the glue has a shear strength of 80 kPa, determine the maximum load P that can be applied without causing the glue to lose its bond.

30 mm

P 1 —P 4

40 mm 60 mm 40 mm

1 —P 4

20 mm A

B 0.8 m

1m

1m

0.8 m

Solution Maximum shear is at supports. Vmax = I = t =

3P 4 1 1 (0.02)(0.06)3 + 2c (0.03)(0.04)3 + (0.03)(0.04)(0.05)2 d = 6.68(10-6) m4 12 12

VQ ; It

80(103) =

(3P>4)(0.05)(0.04)(0.03) 6.68(10-6)(0.02)

Ans.

P = 238 N

Ans: P = 238 N 704

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7–50. The beam is subjected to a shear force of V ow at V ==25 5 kN. Determinethe theshear shearflflow atpoints pointsAAand andB. B. kip.Determine

0.5mm in. 12 C C

5 in. 125 mm 1255 mm in. 0.5 in. 12 mm

Solution y =

0.5mm in. 12 in. 502mm

©yA 0.006(0.274)(0.012) + 2[0.112(0.012)(0.2)] + 0.156(0.25)(0.012) = ©A 0.274(0.012) + 2(0.012)(0.2) + 0.25(0.012)

8 in. 200 mm V V

1 (0.274)(0.012 3 ) + 0.274(0.012)(0.092472 − 0.006)2 12 1  + 2  (0.012)(0.2 3 ) + (0.012)(0.2)(0.112 − 0.092472)2  12   +

0.5 in. 12 mm

D D

= 0.092472 m I =

A A

1 (0.25)(0.012 3 ) + 0.25(0.012)(0.156 − 0.092472)2 12

12 mm

= 54.5990(10 −6 ) m 4

B B

250 mm

12 mm 12 mm 138 mm 12 mm 50 mm

œ 0.086472 m yA = 0.092472 − 0.006 =

0.063528 m yBœ = 0.156 − 0.092472 = −3 œ QA = yA A¿ = 0.086472(0.274)(0.012) = 0.28432(10 ) −3 QB = yBœ A¿ = 0.063528(0.25)(0.012) = 0.19058(10 )

qA =

1 VQA 1 [25(10 3 )][0.28432(10 −3 )]  3 a b =   = 65.09(10 ) N/m 2 I 2  54.5990(10 −6 )  Ans.

= 65.1 kN/m

qB =

1 VQB 1 [25(10 3 )][0.19058(10 −3 )]  3 a b =   = 43.63(10 ) N/m 2 I 2  54.5990(10 −6 )  = 43.6 kN/m

Ans.

Ans. qA = 65.1 kN>m, qB = 43.6 kN>m 515

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7–51. The beam is constructed from four plates and is subjected to to aa shear shear force force of of VV == 25 5 kip kN.. Determine the maximum shear flow in the cross section.

0.5mm in. 12 C

5 in. 125 mm 1255 mm in. 0.5 in. 12 mm

0.5mm in. 12 in. 50 2mm

Solution y =

A

0.5 in. 12 mm

D 8 in. 200 mm

©yA 0.006(0.274)(0.012) + 2[0.112(0.012)(0.2)] + 0.156(0.25)(0.012) = ©A 0.274(0.012) + 2(0.012)(0.2) + 0.25(0.012)

V

B

= 0.092472 m I =

1 (0.274)(0.012 3 ) + 0.274(0.012)(0.092472 − 0.006)2 12 1  + 2  (0.012)(0.2 3 ) + (0.012)(0.2)(0.112 − 0.092472)2   12  +

1 (0.25)(0.012 3 ) + 0.25(0.012)(0.156 − 0.092472)2 12 −6

= 54.5990(10 ) m

86.47 mm

40.235 mm

4

0.36203(10 −3 ) m 3 Qmax = 0.086472(0.274)(0.012) + 0.040236(0.012)(0.086472) =

qmax =

1 VQmax 1 [25(10 3 )][0.36203(10 −3 )]  3 a b =   = 82.88(10 ) N/m 2 I 2  54.5990(10 −6 ) 

= 82.9 kN/m

Ans.

Ans. y = 0.092472 m, I = 54.5990(10-6) m4, qmax = 82.9 kN>m 515

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*7–52. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V = 150 N, determine the shear flow at points A and B. 10 mm 40 mm

B A

10 mm 30 mm

Solution y =

10 mm

2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)

I = 2c

30 mm 10 mm

= 0.027727 m

1 (0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d 12

+ 2c +

V 40 mm

1 (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d 12

1 (0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 = 0.98197(10 - 6) m4 12

yB ′ = 0.055 - 0.027727 = 0.027272 m yA ′ = 0.027727 - 0.005 = 0.022727 m QA = yA ′A′ = 0.022727(0.04)(0.01) = 9.0909(10 - 6) m3 QB = yB ′A′ = 0.027272(0.03)(0.01) = 8.1818(10 - 6) m3 qA =

150(9.0909)(10 - 6) VQA = = 1.39 kN>m I 0.98197(10 - 6)

Ans.

qB =

150(8.1818)(10 - 6) VQB = = 1.25 kN>m I 0.98197(10 - 6)

Ans.

Ans: qA = 1.39 kN>m, qB = 1.25 kN>m 709

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7–53. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V = 150 N, determine the maximum shear flow in the strut. 10 mm 40 mm 10 mm 30 mm

Solution y =

B A V 40 mm 10 mm

30 mm 10 mm

2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)

= 0.027727 m I = 2c

1 (0.03)(0.01)3 + 0.03(0.01)(0.027727 - 0.005)2 d 12

+ 2c +

1 (0.01)(0.06)3 + 0.01(0.06)(0.03 - 0.027727)2 d 12

1 (0.04)(0.01)3 + 0.04(0.01)(0.055 - 0.027727)2 12

= 0.98197(10 - 6) m4 Qmax = (0.055 - 0.027727)(0.04)(0.01) + 2[(0.06 - 0.027727)(0.01)]a = 21.3(10 - 6) m3 qmax =

-6 1 VQmax 1 150(21.3(10 )) a b = a b = 1.63 kN>m 2 I 2 0.98197(10 - 6)

0.06 - 0.0277 b 2

Ans.

Ans: qmax = 1.63 kN>m 710

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7–54. A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at points A and B.

10 mm 30 mm 10 mm

A

100 mm C

B

100 mm 10 mm 30 mm 10 mm

Solution

150 mm

V

Section Properties:

150 mm

INA =

10 mm 125 mm

1 1 (0.145) ( 0.33 ) (0.125) ( 0.283 ) 12 12

10 mm

1 + 2c (0.125) ( 0.013 ) + 0.125(0.01) ( 0.1052 ) d 12

= 125.17 ( 10 - 6 ) m4

QA = y=2 A′ = 0.145(0.125)(0.01) = 0.18125 ( 10 - 3 ) m3 QB = y=1 A′ = 0.105(0.125)(0.01) = 0.13125 ( 10 - 3 ) m3 Shear Flow: qA = =

1 VQA c d 2 I

3 -3 1 18(10 )(0.18125)(10 ) c d 6 2 125.17(10 )

Ans.

= 13033 N>m = 13.0 kN>m qB = =

1 VQB d c 2 I

3 -3 1 18(10 )(0.13125)(10 ) c d 2 125.17(10 - 6)

Ans.

= 9437 N>m = 9.44 kN>m

Ans: qA = 13.0 kN>m, qB = 9.44 kN>m 707

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7–55. A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at point C.

10 mm 30 mm 10 mm

A

100 mm C

B

100 mm 10 mm 30 mm 10 mm

Solution

150 mm

V

Section Properties:

150 mm

INA =

10 mm 125 mm

1 1 (0.145) ( 0.33 ) (0.125) ( 0.283 ) 12 12

10 mm

1 + 2c (0.125) ( 0.013 ) + 0.125(0.01) ( 0.1052 ) d 12

= 125.17 ( 10 - 6 ) m4 QC = Σy′A′

= 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02) = 0.5375 ( 10 - 3 ) m3 Shear Flow: qC = =

1 VQC c d 2 I

3 -3 1 18(10 )(0.5375)(10 ) d c 2 125.17(10 - 4)

Ans.

= 38648 N>m = 38.6 kN>m

Ans: qC = 38.6 kN>m 708

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*7–56. A shear force of V = 300 kN is applied to the box girder. Determine the shear flow at points A and B. 90 mm A

90 mm C D

200 mm B

V

100 mm

190 mm

Solution 200 mm

The moment of inertia of the cross section about the neutral axis is I =

10 mm 180 mm

1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4 12 12

10 mm

Referring to Fig. a, Fig. b, QA = y=1A=1 = 0.195(0.01)(0.19) = 0.3705(10 - 3) m3 QB = 2y=2A=2 + y=3A=3 = 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10 - 3) m3 Due to symmetry, the shear flow at points A and A′, Fig. a, and at points B and B′, Fig. b, are the same. Thus qA =

3 -3 1 VQA 1 300(10 ) 3 0.3705(10 ) 4 a b = c s 2 I 2 0.24359(10 - 3)

= 228.15(103) N>m = 228 kN>m

qB =

Ans.

3 -3 1 VQB 1 300(10 ) 3 0.751(10 ) 4 a b = c s 2 I 2 0.24359(10 - 3)

= 462.46(103) N>m = 462 kN>m

Ans.

Ans: qA = 228 kN>m, qB = 462 kN>m 705

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7–57. A shear force of V = 450 kN is applied to the box girder. Determine the shear flow at points C and D. 90 mm A

90 mm C D

200 mm B

V

100 mm

190 mm

Solution 200 mm

The moment of inertia of the cross section about the neutral axis is I =

10 mm 180 mm

1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10 - 3) m4 12 12

10 mm

Referring to Fig. a, due to symmetry AC= = 0. Thus QC = 0 Then referring to Fig. b, QD = y1= A1= + y2= A2= = 0.195(0.01)(0.09) + 0.15(0.1)(0.01) = 0.3255(10 - 3) m3 Thus, qC =

VQC = 0 I

qD =

450(103) 3 0.3255(10 - 3) 4 VQD = I 0.24359(10 - 3)

Ans.

= 601.33(103) N>m = 601 kN>m

Ans.

Ans: qC = 0, qD = 601 kN>m 706

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7–58. The H-beam is subjected to a shear of V = 80 kN. Determine the shear flow at point A. 250 mm

50 mm

Solution

V A 30 mm 300 mm

25 mm

25 mm

1 1 I = 2c (0.025) ( 0.253 ) + (0.3) ( 0.033 ) d = 65.7792 ( 10-6 ) m4 12 12 QA = y′A′ = 0.0875(0.075)(0.025) = 0.1641(10 - 3) m3 qA =

80 ( 103 ) (0.1641) ( 10 - 3 ) VQA = 200 kN>m = I 65.7792 ( 10 - 6 )

Ans.

Ans: qA = 200 kN>m 713

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7–59. The H-beam is subjected to a shear of V = 80 kN. Sketch the shear-stress distribution acting along one of its side segments. Indicate all peak values. 250 mm

50 mm

Solution

V A 30 mm 300 mm

25 mm

25 mm

1 1 I = 2c (0.025) ( 0.253 ) d + (0.3) ( 0.033 ) = 65.7792 ( 10-6 ) m4 12 12 QB = (0.070)(0.025)(0.110) = 0.1925(10 - 3) m3 tB =

80 ( 103 ) (0.1925) ( 10-3 ) VQ = 9.36 MPa = It 65.7792 ( 10-6 ) (0.025)

tB =

80 ( 103 ) 3 2(0.1925) ( 10-3 ) 4 VQ = = 1.3378 MPa It 65.7792 ( 10-6 ) (0.35)

Qmax = 2(0.07)(0.025)(0.110) + (0.0075)(0.35)(0.015) = 0.4244 ( 10-3 ) m3 tC =

80 ( 103 ) (0.4244) ( 10-3 ) VQ = = 1.47 MPa It 65.7792 ( 10-6 ) (0.35)

Ans: tmax = 9.36 MPa 714

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*7–60. The built-up beam is formed by welding together the thin plates of thickness 5 mm. Determine the location of the shear center O.

5 mm 200 mm O e

100 mm 100 mm

200 mm

Solution

300 mm

Shear Center: Referring to Fig. a and summing moments about point A, we have a+ Σ(MR)A = ΣMA;   - Pe = - (Fw)1(0.3) e =

0.3(Fw)1

(1)

P

Section Properties: The moment of inertia of the cross section about the axis of symmetry is I =

1 1 (0.005)(0.43) + (0.005)(0.23) = 30(10 - 6) m4 12 12

Referring to Fig. b, y′ = (0.1 - s) +

s = (0.1 - 0.5s) m. Thus, Q as a function of s is 2

Q = y′A′ = (0.1 - 0.5s)(0.005s) = [0.5(10 - 3) s - 2.5(10 - 3) s2] m3 Shear Flow: q =

P[0.5(10 - 3) s - 2.5(10 - 3) s2] VQ = = P(16.6667s - 83.3333s2) I 30(10 - 6)

Resultant Shear Force: The shear force resisted by the shorter web is (Fw)1 = 2

L0

0.1 m

qds = 2

L0

0.1 m

2

P(16.6667s - 83.3333s )ds = 0.1111P

Substituting this result into Eq. (1), Ans.

e = 0.03333 m = 33.3 m

Ans: e = 33.3 m 715

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7–61. The assembly is subjected to a vertical shear of V ow at V ==35 7 kN. Determine the the shear shear flflow at points points A and B and kip.Determine the maximum shear flow in the cross section.

A A

0.5mm in. 12

B B

V V

Solution in. 50 2mm

©yA 0.006(0.274)(0.012) + 2[0.081(0.012)(0.138)] + 0.156(0.174)(0.012) = ©A 0.274(0.012) + 2(0.012)(0.138) + 0.174(0.012) = 0.070641 m

y =

6 in.mm 150

1506 mm in. 120.5 mm in.

0.5mm in. 12

2 in. 50 mm

0.5mm in. 12

0.5mm in. 12

1 (0.274)(0.012 3 ) + 0.274(0.012)(0.070641 − 0.006)2 12 1  + 2  (0.012)(0.138 3 ) + (0.012)(0.138)(0.081 − 0.070641)2   12 

I =

+

1 (0.174)(0.012 3 ) + 0.174(0.012)(0.156 − 0.070641)2 12

= 34.6283(10 −6 ) m 4 QA = y1¿A1¿ = 0.06464(0.05)(0.012) = 0.28432(10 −3 ) m 3 QB = y2¿A2¿ = 0.08536(0.174)(0.012) = 0.17823(10 −3 ) m 3 Qmax = ©y¿A¿ = 0.08536(0.174)(0.012) + 2[0.039680(0.012)(0.079359)] = 0.25380(10 −3 ) m 3 q =

VQ I

qA =

[35(10 3 )][38.7845(10 −6 )] 3 = ) N/m 39.2 kN/m = 39.20(10 34.6283(10 −6 )

Ans.

qB =

1 [35(10 3 )][0.17823(10 −3 )]  3  = 90.07(10 ) N/m = 90.1 kN/m  2  34.6283(10 −6 ) 

Ans.

qmax =

1 [35(10 3 )][0.25380(10 −3 )]  3   = 128.26(10 ) N/m = 128 kN/m 2  34.6283(10 −6 ) 

Ans.

0.06464 m 0.039680 m

0.85359 m 0.079359 m

Ans. y = 0.070641 m, I = 34.6283(10- 6) m4, qA = 39.2 kN>m, qB = 90.1 kN>m, qmax = 128 kN>m 518

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7–62. The box girder is subjected to a shear of V = 15 kN. Determine the shear flow at point B and the maximum shear flow in the girder’s web AB.

A

15 mm

250 mm 15 mm V B 150 mm

Solution 1 1 I = (0.375) ( 0.283 ) (0.3) ( 0.253 ) = 0.295375 ( 10-3 ) m4 12 12

150 mm

QB = y′B A′ = 0.1325(0.375)(0.015) = 0.7453125 ( 10-3 ) m3

25 mm

25 mm

25 mm

Qmax = Σy′A′ = 0.1325(0.375)(0.015) + 3[(0.0625)(0.125)(0.025)] = 1.33125 ( 10-3 ) m3 qB =

3 -3 1 VQB 1 15 ( 10 ) (0.7453125) ( 10 ) c d = c d = 12.6 kN>m -3 3 I 3 0.295375 ( 10 )

qmax =

3 -3 1 VQmax 1 15 ( 10 ) (1.33125) ( 10 ) c d = c d = 22.5 kN>m 3 I 3 0.295375 ( 10-3 )

Ans.

Ans.

Ans: qB = 12.6 kN>m, qmax = 22.5 kN>m 717

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7–63. Determine the location e of the shear center, point O, for the thin-walled member having a slit along its section.

100 mm 100 mm O

e 100 mm

Solution Suming moments about A, Pe = 2V1 (100) + F(200)

(1)

1 I = 2 c t ( 0.23 ) d + 2 3 (0.1)(t) ( 0.12 ) 4 = 3.3333 ( 10-3 ) t m4 12 y Q1 = y1 ′A′ = (y)t = 0.5y2 t 2 Q2 = Σy′A = 0.05(0.1)(t) + 0.1(x)(t) = 0.005t + 0.1xt q1 =

P(0.5y2 t) VQ1 = 150P y2 = I 3.3333(10-3)t

q2 =

P(0.005t + 0.1xt) VQ2 = 300P(0.005 + 0.1x) = I 3.3333(10-3)t 0.1

0.1

0.1

y3 V1 = q1 dy = 150P y dy = 150P c d 3 L0 L0 2

= 0.05P

0

F =

L0

0.1

q2 dx = 300P

L0

0.1

(0.005 + 0.1x) dx 0.1

0.1x2 = 300P c 0.005x + d` 2

= 0.3P

0

From Eq. (1);

Pe = 2(0.05P)(100) + 0.3P(200) Ans.

e = 70 mm

Ans: e = 70 mm 718

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*7–64. Determine the location e of the shear center, point O, for the thin-walled member. The member segments have the same thickness t.

b

h1

h2

O e

Solution Summing moments about A, (1)

eP = bF1 I =

1 1 1 (t) (h1 ) 3 + (t) (h2 ) 3 = t (h31 + h32 ) 12 12 12

q1 =

P(h1 >2) (t) (h1 >4)

F1 =

Ph31t 2 q1 (h1 ) = 3 12I

I

=

Ph21t 8I

From Eq. (1), e = = =

b Ph31t a b P 12I h31b

(h31 + h32 ) b 1 + (h2 >h1 ) 3

Ans.

Ans: e =

719

b 1 + (h2 >h1 ) 3

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7–65. The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB. Indicate numerical values at all peaks.

A in. 1255 mm

125 mm 5 in. 45 45

0.25 in. 6 mm

Solution

B

V

Section Properties: = b

0.006 = 0.00848528 m sin 45°

= = cos 45° 0.08839 m h 0.125 1 3  4 −6 INA = 2  (0.00848528)(0.08839 ) = 0.97656(10 ) m  12  y   Q = y¿A¿ = [0.25(0.08839) + 0.5y]  0.625 −  (0.006) sin 45°   = 8.2864(10 −6 ) − 4.2426(10 −3 ) y2

Shear Flow: q = =

VQ I [10(10 3 )][8.2864(10 −6 ) − 4.2426(10 −3 )y2 ] 0.97656(10 −6 )

3 6 2 = [84.8528(10 ) − 43.4446(10 )y ] N/m

Ans.

= [84.9 − 43.4y2 ] kN/m At y = 0, q = qmax = 84.9 kN/m

Ans.

0.125 m

6 mm

0.0625

84.9 kN/m

Ans. At y = 0, q = qmax = 84.9 kN>m 517

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7–66. Determine the shear-stress variation over the cross section of the thin-walled tube as a function of elevation y and show that t max = 2V>A, where A = 2prt. Hint: Choose a differential area element dA = Rt du. Using dQ = y dA, formulate Q for a circular section from u to (p - u) and show that Q = 2R2t cos u, where cos u = 2R2 - y2>R.

ds du y

u t

R

Solution dA = R t du dQ = y dA = yR t du Here y = R sin u Therefore dQ = R2 t sin u du p-u

Q =

p-u

R2 t sin u du = R2 t(- cos u) |

Lu

u

= R2 t [ -cos (p - u) - ( - cos u)] = 2R2 t cos u dI = y2 dA = y2 R t du = R3 t sin2 u du 2p

I =

L0

2p

R3 t sin2 u du = R3 t 2p

=

t =

sin 2u R3 t [u ]冷 2 2 0

R3 t [2p - 0] = pR3 t 2

VQ V(2R2t cos u) V cos u = = It pR t pR3 t(2t)

Here cos u = t =

=

L0

(1 - cos 2u) du 2

2R2 - y2 R

V 2R2 - y2 pR2t

Ans.

tmax occurs at y = 0; therefore tmax =

V pR t

A = 2pRt; therefore tmax =

2V A

QED

Ans. t = 519

V 2R 2 - y2 pR 2t

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7–67. Determine the location e of the shear center, point O, for the beam having the cross section shown. The thickness is t. 1 — r 2

e r

O

1 — r 2

Solution I = (2)c

1 r 2 (t)(r>2)3 + (r>2)(t) a r + b d + Isemi-circle 12 4

= 1.583333t r3 + Isemi-circle p>2

Isemi-circle =

p>2 2

L-p>2

(r sin u) t r du = t r3

L-p>2

sin2 u du

p Isemi-circle = t r3 a b 2 Thus, p I = 1.583333t r3 + t r3 a b = 3.15413t r3 2 r r Q = a bt a + rb + 2 4 Lu

p>2

r sin u (t r du)

Q = 0.625 t r2 + t r2 cos u q =

VQ P(0.625 + cos u)t r2 = I 3.15413 t r3

Summing moments about A: p>2

Pe =

L-p>2

(q r du)r p>2

Pe = e =

Pr (0.625 + cos u)du 3.15413 L-p>2

r (1.9634 + 2) 3.15413

e = 1.26 r

Ans.

Ans. e = 1.26 r 525

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*7–68. Determine the location e of the shear center, point O, for the thin-walled member. The member segments have the same thickness t.

d d — 2 e

d — 2

O

d

Solution d

Summing moments about point A: (1)

Pe = F2d + F1(2d) I = 2[dt(d)2] + 2[dt(d>2)2] = q1 = F1 =

q2 = F2 =

P(dt)(d) 19 3 6 td

=

1 19 3 t(2d)3 = td 12 6

6P 19d

1 6P 3 a b(d) = P 2 19d 19 P(dt)(d>2) 19 3 6 td

=

3P 19d

1 3P 1.5P bd = a 2 19d 19

From Eq. (1): Pe = 2d a e =

3 1.5P Pb + da b 19 19

7.5 15 d = d 19 38

Ans.

Ans: e = 723

15 d 38

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7–69.  t

A thin plate of thickness t is bent to form the beam having the cross section shown. Determine the location of the shear center O.

r O e

Solution Section Properties: For the arc segment, Fig. a, y = r cos u and the area of the differential element shown shaded is dA = t ds = tr du. Then, the moment of inertia of the entire cross section about the axis of symmetry is I =

1 (t)(2r)3 + y2 dA 12 L p

=

2 3 r t + (r cos u)2 trdu 3 L0

=

2 3 r 3t r t + (cos 2u + 1) du 3 2 L0 p

= =

p 2 3 r 3t 1 r t + a sin 2u + u b ` 3 2 2 0

r 3t (4 + 3p) 6

r . Thus, Q as a function of s is 2 u r Q = y′A′ + ydA = (rt) + r cos u(trdu) 2 L L0

Referring to Fig. a, y′ =

u

=

1 2 r t + r 2t cos udu 2 L0

=

r2 t (1 + 2 sin u) 2

Shear Flow: P 3 r2 t(1 + 2 sin u) 4 VQ 3P = = (1 + 2 sin u) r 3t I (4 + 3p)r (4 + 3p) 2

q =

6

Resultant Shear Force: The shear force resisted by the arc segment is p p 3P F = qds = qrdu = (1 + 2 sin u)rdu (4 + 3p)r L L0 L0 = =

p 3P (u - 2 cos u) ` 4 + 3p 0

3P(p + 4) 4 + 3p

Shear Center: Referring to Fig. b and summing the moments about point A, a+ Σ(MR)A = ΣMA;

Pe = r

L

Pe = r c e = c

dF

3P(p + 4) 4 + 3p

3(p + 4) 4 + 3p

d

Ans: Ans.

d r 724

e = c

3(p + 4) 4 + 3p

dr

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7–70.  t

Determine the location e of the shear center, point O, for the tube having a slit along its length.

r

e

O

Solution Section Properties: dA = t ds = t r du  y = r sin u dI = y2 dA = r 2 sin2 u(t r du) = r 3t sin2 u du I = r 3t

L0

2p

2

sin u du 2p

1 - cos 2u a b du = pr 3t 2 L0 dQ = y′A′ = ydA = r sin u(t r du) = r 2 t sin u du = r 3t

Q = r2 t

L0

u

sin u du = r 2 t (1 - cos u)

Shear Flow Resultant: q = F =

P r 2t (1 - cos u) VQ P = = (1 - cos u) I pr pr 3t L0

2p

2p

qds =

P (1 - cos u) rdu L0 pr 2p

=

P (1 - cosu) du p L0

= 2P Shear Center: Summing moments about point A. Pe = Fr Pe = 2Pr Ans.

e = 2r

Ans: e = 2r 725

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R7–1. Sketch the intensity of the shear-stress distribution acting over the beam’s cross-sectional area, and determine the resultant shear force acting on the segment AB. The shear acting at the section is V = 175 kN. Show that INA = 0.3048(10−3) m4.

C

200 mm 8 in.

Solution

V B A

6150 in.mm

Section Properties: ©yA 0.1(0.2)(0.2) + 0.275(0.05)(0.15)] = 0.12763 m = ©A 0.2(0.2) + 0.05(0.15)

y =

INA =

mm 250in.

1 (0.2)(0.2 3 ) + 0.2(0.2)(0.12763 − 0.1)2 12 0.12763

1 + (0.05)(0.153 ) + (0.05)(0.15)(0.275 − 0.12763)2 12 −3

= 0.3408(10 ) m

4

375 in.mm mm 375in.

0.2 m

0.063816 + 0.5y1

0.12763 m 0.2 m

(Q.E.D)

Q1 = yœ1A¿ = (0.063816 + 0.5 y1 )(0.2)(0.12763 − y1 )

0.15 m

0.22237

2 −3 = 1.62898(10 ) − 0.1y1

0.05 m

0.11118 + 0.5y2

Q2 = y2œ A¿ = (0.11118 + 0.5 y2 )(0.05)(0.22237 − y2 ) = 1.23619(10 −3 ) − 0.25y22 Shear Stress: Applying the shear formula t = tCB

4.18 MPa 2.84 MPa

VQ , It

11.4 MPa

VQ1 [175(10 3 )][1.62898(10 −3 ) − 0.1y12 ] = = It [0.3408(10 −3 )](0.2)

(

)

= 4.18218 − 256.74y12 (106 ) At y1 = 0,

tCB = 4.18 523 psi MPa

At y1 = −0.07237 m

tCB = 2.84 355 psi MPa

tAB =

VQ2 [175(10 3 )][1.23619(10 −3 ) − 0.25y22 ] = It [0.3408(10 −3 )](0.05)

)

(

= 12.6950 − 256.74y22 (106 ) At y2 = 0.07237 m

tAB = 11.4 MPa

Resultant Shear Force: For segment AB. VAB =

L

tAB dA 0.22237 m

=

(

)

 12.6950 − 256.74y2 (106 ) (0.05dy) 2  L0.07237 m 

= 49.78(10 3 ) N = 49.8 kN

Ans.

Ans. VAB = 49.78 kN 528

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R7–2.  The T-beam is subjected to a shear of V = 150 kN. Determine the amount of this force that is supported by the web B.

200 mm

40 mm V = 150 kN 200 mm

40 mm

Solution y = I =

B

(0.02)(0.2)(0.04) + (0.14)(0.2)(0.04) 0.2(0.04) + 0.2(0.04)

= 0.08 m

1 (0.2)(0.043) + 0.2(0.04)(0.08 - 0.02)2 12 1 + (0.04)(0.23) + 0.2(0.04)(0.14 - 0.08)2 = 85.3333(10-6) m4 12

A′ = 0.04(0.16 - y) y-′ = y +

(0.16 - y) 2

=

(0.16 + y) 2

Q = y- ′A′ = 0.02(0.0256 - y2) t = V =

150(103)(0.02)(0.0256 - y2) VQ = = 22.5(106) - 878.9(106)y2 It 85.3333(10-6)(0.04) L

t dA,

dA = 0.04 dy

0.16

V =

(22.5(106) - 878.9(106)y2) 0.04 dy L-0.04 0.16

=

(900(103) - 35.156(106)y2)dy L-0.04 Ans.

= 131 250 N = 131 kN

Ans: N/A 727

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R7–3.  The member is subject to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C. The thickness of each thin-walled segment is 15 mm.

200 mm B 100 mm A C

300 mm

V  2 kN

Solution Section Properties: y = =

INA

ΣyA ΣA 0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015)

0.2(0.015) + 0.115(0.03) + 0.3(0.015) = 0.08798 m 1 = (0.2)(0.0153) + 0.2(0.015)(0.08798 - 0.0075)2 12 1 + (0.03)(0.1153) + 0.03(0.115)(0.08798 - 0.0575)2 12 1 + (0.015)(0.33) + 0.015(0.3)(0.165 - 0.08798)2 12 = 86.93913(10 - 6) m4 Ans.

QA = 0 QB = y1′A′ = 0.03048(0.115)(0.015) = 52.57705(10 - 6) m - 3 QC = Σy1′A′ = 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015) = 0.16424(10 - 3) m3 Shear Flow: qA =

VQA = 0 I

qB =

2(103)(52.57705)(10 - 6) VQB = = 1.21 kN>m I 86.93913(10 - 6)

Ans.

qC =

2(103)(0.16424)(10 - 3) VQC = = 3.78 kN>m I 86.93913(10 - 6)

Ans.

Ans: qA = 0, qB = 1.21 kN>m, qC = 3.78 kN>m 728

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*R7–4. The beam is constructed from four boards glued together at their seams. If the glue can withstand 15 75 kN/m, lb>in., what is the maximum vertical shear V that the beam can support?

753mm in.

753mm in.

Solution Section Properties:

= 35.4467(10 −6 ) m 4 Q = y¿A¿ = 0.0435(0.1)(0.012) = 52.2(10 ) m

3

12 0.5mm in.

0.1 m

0.012 m

0.0435 m

Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2[15(10 3 )] = 30(10 3 ) N/m .

30(10 3 ) =

4 in. 100 mm

0.012 m

−6

q =

12 0.5mm in.

12 0.5mm in.

753mm in.

V V

 1 1  INA = 2  (0.012)(0.249 3 ) + 2  (0.1)(0.012 3 ) + 0.1(0.012)(0.04352 )  12   12 

12 0.5mm in.

VQ I

0.012 m

0.012 m

0.1245 m

0.1245 m

V [52.2(10 −6 )] 35.4467(10 −6 )

3 = V 20.37(10 = ) N 20.4 kN

Ans.

Ans. V = 20.4 kN 531

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R7–5. Solve Prob. R7–4 if the beam is rotated 90° from the position shown. 753 mm in.

753 mm in.

12 0.5mm in.

12 0.5mm in.

753 mm in.

V

Solution

12 0.5mm in.

4 in. 100 mm

12 0.5mm in.

Section Properties: INA =

1 1 (0.249)(0.124 3 ) − (0.225)(0.13 ) = 20.8124(10 −6 ) m 4 12 12

Q = y¿A¿ = 0.056(0.249)(0.012) = 1.167328(10 −3 ) m 3

0.012 m

Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2[15(10 3 )] = 30(10 3 ) N/m . q = 30(10 3 ) =

0.012 m 0.05 m

0.056 m

0.05 m 0.012 m

0.249 m

VQ I V [1.167328(10 −3 )] 20.8124(10 −6 )

3 V 3.731(10 = ) N 3.73 kN =

Ans.

Ans. V = 3.73 kN 531

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8–1. A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p = 300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa.

Solution sallow =

pr ; 2t

12(106) =

300(103)(1.5) 2t Ans.

t = 0.0188 m = 18.8 mm

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

731

Ans: t = 18.8 mm

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8–2. A pressurized spherical tank is to be made of 12-mm-thick steel. If If itit is is subjected subjected to an internal pressure 0.5-in.-thick steel. of pp ==1.4 determine its outer radius if the maximum 200MPa, psi, determine normal stress is not to exceed 105 MPa. 15 ksi.

Solution sallow =

pr 2t

;

105(106 ) =

[1.4(106 )]ri 2(0.012)

ri = 1.80 m ro =1.80 + 0.012 =1.812 m

Ans.

Ans. ro = 1.812 m 532

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8–3. The thin-walled cylinder can be supported in one of two ways as shown. Determine the state of stress in the wall of the cylinder for both cases if the piston P causes the Thewall wallhas has aa thickness thickness of internal pressure to to be be0.5 65 MPa. psi. The 6 mmin.and mm. 0.25 andthe theinner innerdiameter diameterofofthe thecylinder cylinderisis200 8 in.

P

Solution

P

8 in. 200 mm

200 mm 8 in.

(a)

(b)

Case (a): s1 =

pr ; t

s1 =

65(4) 0.5(100) 8.33 = = 1.04 ksiMPa 0.25 6

Ans.

s2 = 0

Ans.

Case (b): s1 =

pr ; t

s1 =

65(4) 0.5(100) = 1.04 ksiMPa = 8.33 0.256

Ans.

s2 =

pr ; 2t

s2 =

65(4) 0.5(100) 4.17psi MPa == 520 2(0.25) 2(6)

Ans.

Ans. Case (a): s1 = 8.33 MPa; s2 = 0 Case (b): s1 = 8.33 MPa; s2 = 4.17 MPa 532

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*8–4. The tank of the air compressor is subjected to an MPa. If the internal diameter internal pressure of 0.63 90 psi. If the internal diameter of of the tank thickness 6 mm, the tank is is22550 in.,mm, and and the the wallwall thickness is is0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.

A

Solution Hoop Stress for Cylindrical Vessels: Since

11 r 275 ==45.8 thenthin thinwall wall = 44 > 7 10, 10,then t 0.25 6

28.9 MPa

analysis can be used. Applying Eq. 8–1 14.4 MPa

pr 90(11) 0.63(275) = 28.875 MPa = 28.9 MPa = s1 = t 0.25 6

Ans.

Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2 s2 =

0.63(275) pr = = 14.4375 MPa = 14.4 MPa 2t 2(6 )

Ans.

Ans. s1 = 28.9 MPa, s2 = 14.4 MPa 533

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8–5. P

Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a radius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall of the cylinder.

47 mm

P

Solution p =

2(103) P = 314 380.13 Pa = A p(0.0452)

s1 =

314 380.13(0.045) pr = = 7.07 MPa t 0.002

Ans. Ans.

s2 = 0

The pressure P is supported by the surface of the pistons in the longitudinal direction.

Ans: s1 = 7.07 MPa, s2 = 0 735

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8–6. Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress in the cylinder does not exceed 3 MPa. Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm.

P 47 mm

P

Solution s =

pr ; t

3(106) =

p(0.045) 0.002

p = 133.3 kPa P = pA = 133.3 1 103 2 (p)(0.045)2 = 848 N

Ans.

Ans: P = 848 N 736

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8–7. A boiler is constructed of 8-mm-thick steel plates that are fastened together at their ends using a butt joint consisting of two 8-mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress in the boiler’s plate away from the seam, (b) the circumferential stress in the outer cover plate along the rivet line a–a, and (c) the shear stress in the rivets.

a

8 mm

50 mm

0.75 m a

Solution a)

s1 =

1.35(106)(0.75) pr = = 126.56(106) = 127 MPa t 0.008

Ans.

126.56 (106)(0.05)(0.008) = s1 ′(2)(0.04)(0.008)

b)

s1 ′ = 79.1 MPa

Ans.

c) From FBD(a) + c ΣFy = 0;

Fb - 79.1(106)[(0.008)(0.04)] = 0 Fb = 25.3 kN

(tavg)b =

Fb 25312.5 - p = 322 MPa 2 A 4 (0.01)

Ans.

Ans: (a) s1 = 127 MPa, (b) s1 ′ = 79.1 MPa, (c) (tavg)b = 322 MPa 737

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*8–8. The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of longitudinal bolts per meter length at each side of the cylindrical shell. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m. Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =

pr ; t

150 A 106 B =

3 A 106 B (2) tc

tc = 0.04 m = 40 mm For the hemispherical cap, sallow =

pr ; t

150 A 106 B =

Ans. 3 A 106 B (2) 2ts

ts = 0.02 m = 20 mm Since

Ans.

r 6 10, thin-wall analysis is valid. t

Referring to the free-body diagram of the per meter length of the cylindrical portion, Fig. a, where P = pA = 3 A 106 B [4(1)] = 12 A 106 B N, we have + c ©Fy = 0;

12 A 106 B - nc(Pb)allow - nc(Pb)allow = 0 nc =

6 A 106 B

(1)

(Pb)allow

The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c

p A 0.0252 B d = 122.72 A 103 B N 4

Substituting this result into Eq. (1), nc = 48.89 = 49 bolts>meter

Ans.

Ans. tc = 40 mm, ts = 20 mm, nc = 49 bolts>meter 536

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8–9. The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of bolts for each hemispherical cap. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m.

Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =

pr ; t

150 A 106 B =

3 A 106 B (2) tc

tc = 0.04 m = 40 mm For the hemispherical cap, sallow =

pr ; t

150 A 106 B =

Ans. 3 A 106 B (2) 2ts

ts = 0.02 m = 20 mm Since

Ans.

r 6 10, thin-wall analysis is valid. t

The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c

p A 0.0252 B d = 122.72 A 103 B N 4

Referring to the free-body diagram of the hemispherical cap, Fig. b, where p P = pA = 3 A 106 B c A 42 B d = 12p A 106 B N, 4 + ©F = 0; : x

12p A 106 B ns =

ns ns (Pb)allow (Pb)allow = 0 2 2

12p A 106 B

(1)

(Pb)allow

Substituting this result into Eq. (1), ns = 307.2 = 308 bolts

Ans.

Ans. tc = 40 mm, ts = 20 mm, (Pb)allow = 122.72(103) N, ns = 308 bolts 537

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s

8–10. A wood wood pipe pipe having havingan aninner innerdiameter diameterofof0.93 m ft is bound together each having a cross-sectogether using usingsteel steelhoops hoops each having a crosstional area of 125 mm sectional area of 0.2 in22..If If the the allowable allowable stress stress for the hoops is sallow ==8412MPa, determine their their maximum spacing s along ksi, determine the section of pipe so that the pipe can resist an internal gauge pressure kPa.Assume Assumeeach each hoop hoop supports the pressure of of 28 4 psi. pressure loading acting along the length s of the pipe.

4 psi 28 kPa

428psi kPa

s

s

Solution Equilibrium for the steel Hoop: From the FBD + : ©Fx = 0;

3

[28(103)](0.95)

3

2 P − [28(10 )](0.93) = 0 P= 12.6(10 )s 0.9 m

Hoop Stress for the Steel Hoop: s1 = sallow = 84(106 ) =

P A 12.6(10 3 )s 125(10 −6 )

ss  = 833.33 33.3 in.mm = 0.833 m

Ans.

Ans. s = 0.833 m 538

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8–11. The staves or vertical members of the wooden tank are held together using semicircular hoops having a and a width of 2ofin.50 Determine the normal thickness of of 0.5 12 in. mm and a width mm. Determine the stress instress hoop AB ifAB theif the tank is subjected normal in hoop tank is subjectedtotoan an internal pressure of of 14 2 psi gauge pressure kPa and and this this loading loading is transmitted bolts are used directly to the the hoops. hoops.Also, Also,ifif0.25-in.-diameter 6-mm-diameter bolts to connect each hoop together, determine the tensile stress in each bolt at A and B. Assume hoop AB supports the pressure loading within withinaa300-mm 12-in. length lengthofofthe thetank tankas as shown.

450 mm 18 in.

150 mm 6 in. 150 mm 6 in.

300 mm 12 in. A

B 300 mm 12 in.

Solution 0.9 m

3

3

3

FR = [14(10 )][(0.9)(0.3 )] = 3.78(10 ) N ©F = 0;

3.78(10 3 ) − 2 F = 0;

F = 1.89(10 3 ) N

1.89(10 3 ) F 6 3.15(10 = ) N/m 3 3.15 MPa = = Ah 0.012(0.05)

Ans.

1.89(10 3 ) F 6 66.84(10 = ) N/m 3 66.8 MPa sb = = = π Ab 2 (0.006 ) 4

Ans.

sh =

0.3 m

Ans. sh = 3.15 MPa, sb = 66.8 MPa 538

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*8–12. A pressure-vessel head is fabricated by welding the circular plate to the end of the vessel as shown. If the vessel sustains an internal pressure of 450 kPa, determine the average shear stress in the weld and the state of stress in the wall of the vessel.

450 mm 10 mm 20 mm

Solution + c ΣFy = 0; p(0.225)2450(103) - tavg (2p)(0.225)(0.01) = 0; Ans.

tavg = 5.06 MPa 3

s1 =

450(10 )(0.225) pr = = 5.06 MPa t 0.02

Ans.

s2 =

450(103)(0.225) pr = = 2.53 MPa 2t 2(0.02)

Ans.

Ans: tavg = 5.06 MPa, s1 = 5.06 MPa, s2 = 2.53 MPa 742

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8–13. The 304 stainless steel band initially fits snugly around the smooth rigid cylinder. If the band is then subjected to a nonlinear temperature temperaturedrop dropofof¢T ΔT = = 12 nonlinear 20 sin22 u °C, °F, where u is in radians, determine the circumferential stress in the band.

1

250 10mm in.

in. 0.4 64 mm 25 mm 1 in.

u

Solution Compatibility: Since the band is fixed to a rigid cylinder (it does not deform under load), then dF - dT = 0 2p P(2pr) a¢Trdu = 0 AE L0

2pr P a b = 12 20ar E A L0 2p 6a s = 10a E c L0

2p

sin2 udu

however,

P = sc A

2p

(1 - cos 2u)du

sc = 10aE 6aE 6 −6 = 6[17(10 = )][193(109 )] 19.686(10 = ) N/m 2 19.7 MPa

Ans.

Ans. dF - dT = 0, sc = 19.7 MPa 539

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8–14. The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with a pressure p. Determine the change in the inner radius of the ring after this pressure is applied. The modulus of elasticity for the ring is E.

ro ri w p

Solution Equilibrium for the Ring: From the FBD + ΣFx = 0; 2P - 2pri w = 0 S

P = pri w

Hoop Stress and Strain for the Ring: s1 =

pri w pri P = = A (ro - ri)w ro - ri

Using Hooke’s Law P1 = However,  P1 =

pri s1 = E E(ro - ri)

2p(ri)1 - 2pri 2pri

=

(ri)1 - ri ri

(1) =

dri . ri

Then, from Eq. (1) pri dri = ri E(ro - ri) dri =

pr 2i E(ro - ri)

Ans.

Ans: dri = 744

pr 2i E(ro - ri)

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8–15. The inner ring A has an inner radius r1 and outer radius r2. The outer ring B has an inner radius r3 and an outer radius r4, and r2 7 r3. If the outer ring is heated and then fitted over the inner ring, determine the pressure between the two rings when ring B reaches the temperature of the inner ring. The material has a modulus of elasticity of E and a coefficient of thermal expansion of a.

r4

r2 r1 A

r3 B

Solution Equilibrium for the Ring: From the FBD + ΣFx = 0;2P - 2pri w = 0 S

P = pri w

Hoop Stress and Strain for the Ring: s1 =

pri w pri P = = A (ro - ri)w ro - ri

Using Hooke’s Law pri s1 = E E(ro - ri)

P1 = However,  P1 =

2p(ri)1 - 2pri 2pri

=

(ri)1 - ri ri

(1)

=

dri . ri

Then, from Eq. (1) pri dri = ri E(ro - ri) dri =

pr 2i E(ro - ri)

Compatibility: The pressure between the rings requires (2)

dr2 + dr3 = r2 - r3 From the result obtained above dr2 =

pr 22 E(r2 - r1)

dr3 =

pr 23 E(r4 - r3)

Substitute into Eq. (2) pr 22 pr 23 + = r2 - r3 E(r2 - r1) E(r4 - r3) p =

E(r2 - r3)

Ans.

r 23 r 22 + r2 - r1 r4 - r3

 Ans: p =

745

E(r2 - r3) r 23 r 22 + r2 - r1 r4 - r3

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*8–16. The cylindrical tank is fabricated by welding a strip of thin plate helically, making an angle u with the longitudinal axis of the tank. If the strip has a width w and thickness t, and the gas within the tank of diameter d is pressured to p, show that the normal stress developed along the strip is given by su = (pd>8t)(3 - cos 2u).

w u

Normal Stress: sh = s1 =

pr p(d>2) pd = = t t 2t

sl = s2 =

p(d>2) pd pr = = 2t 2t 4t

Equilibrium: We will consider the triangular element cut from the strip shown in Fig. a. Here, Ah = (w sin u)t and Thus, Al = (w cos u)t. pd pwd and (w sin u)t = sin u Fh = shAh = 2t 2 pd pwd (w cos u)t = cos u. 4t 4

Fl = slAl =

Writing the force equation of equilibrium along the x¿ axis, ©Fx¿ = 0;

c

pwd pwd sin u d sin u + c cos u d cos u - Nu = u 2 4 Nu =

pwd A 2 sin2 u + cos2 u B 4

However, sin2 u + cos2 u = 1. This equation becomes Nu = Also, sin2 u =

pwd A sin2 u + 1 B 4

1 (1 - cos 2u), so that 2 pwd Nu = (3 - cos 2u) 8

Since Au = wt, then pwd (3 - cos 2u) Nu 8 = su = Au wt su =

pd (3 - cos 2u) 8t

(Q.E.D.)

Ans. su = 542

pd (3 - cos 2u) 8t

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8–17. In order to increase the strength of the pressure vessel, filament winding of the same material is wrapped around the circumference of the vessel as shown. If the pretension in the filament is T and the vessel is subjected to an internal pressure p, determine the hoop stresses in the filament and in the wall of the vessel. Use the free-body diagram shown, and assume the filament winding has a thickness t′ and width w for a corresponding length L of the vessel.

L w t¿

p

s1 T

t s1 T

Solution Normal Stress in the Wall and Filament Before the Internal Pressure is Applied: The entire length L of wall is subjected to pretension filament force T. Hence, from equilibrium, the normal stress in the wall at this state is 2T - (s′)w (2Lt) = 0

(s′)w =

T Lt

and for the filament the normal stress is (s′)fil =

T wt′

Normal Stress in the Wall and Filament After the Internal Pressure is Applied: In order to use s1 = pr>t, developed for a vessel of uniform thickness, we redistribute the filament’s cross-section as if it were thinner and wider, to cover the vessel with no gaps. The modified filament has width L and thickness t>w>L, still with crosssectional area wt> subjected to tension T. Then the stress in the filament becomes sfil = s + (s′)fil =

pr T + (t + t′w>L) wt′

Ans.

pr T (t + t′w>L) Lt

Ans.

And for the wall, sw = s - (s′)w = Check: 2wt′sfil + 2Ltsw = 2rLp

OK

Ans:

pr T + , t + t′w>L wt′ pr T sw = t + t′w>L Lt sfil =

747

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8–18. Determine the shortest distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses in the plate at section a–a. The plate has a thickness of 10 mm and P acts along the centerline of this thickness.

300 mm a

a

200 mm

500 mm

Solution

d

sA = 0 = sa - sb 0 =

P Mc A I

0 =

P(0.1 - d)(0.1) P - 1 3 (0.2)(0.01) 12 (0.01)(0.2 )

P

P( - 1000 + 15000 d) = 0 Ans.

d = 0.0667 m = 66.7 mm

Ans: d = 66.7 mm 748

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8–19. Determine the maximum distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses on the plate at section a–a. The plate has a thickness of 20 mm and P acts along the centerline of this thickness.

200 mm

a

P d

a

Solution Internal Loadings: Consider the equilibrium of the left segment of the plate sectioned through section a–a, Fig. a, + ΣFx = 0;     N - P = 0 S a+ ΣMo = 0;   

       N = P

P(d - 0.1) - M = 0  M = P(d - 0.1)

Section Properties: For the rectangular cross section, A = 0.2(0.02) = 0.004 m2 I =

1 (0.02) ( 0.23 ) = 13.3333 ( 10-6 ) m4 12

Normal Stress: It is required that sA = 0. For the combined loadings, sA = 0 =

N Mc A I P(d - 0.1)(0.1) P 0.004 13.3333 ( 10-6 ) Ans.

d = 0.1333 m = 133 mm

Ans: d = 133 mm 749

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*8–20. The plate has a thickness of 20 mm and the force P = 3 kN acts along the centerline of this thickness such that d = 150 mm. Plot the distribution of normal stress acting along section a–a.

200 mm

a

P d

a

Solution Internal Loadings: Consider the equilibrium of the left segment of the plate sectioned through section a–a, Fig. a, + ΣFx = 0;  N - 3 = 0 S

N = 3.00 kN

a+ ΣMo = 0;  3(0.05) - M = 0  M = 0.150 kN # m Section Properties: For the rectangular cross section, A = 0.2(0.02) = 0.004 m2 I =

1 (0.02) ( 0.23 ) = 13.3333 ( 10-6 ) m4 12

Normal Stress: For the combined loadings, the normal stress at points A and B can be determined from s =

3.00 ( 103 ) 0.150 ( 103 ) (0.1) N Mc { = { A I 0.004 13.3333 ( 10-6 )

sA = 750 ( 103 ) - 1.125 ( 106 ) = - 0.375 ( 106 ) Pa = 0.375 MPa (C)

Ans.

sB = 750 ( 103 ) + 1.125 ( 106 ) = 1.875 ( 106 ) Pa = 1.875 MPa (T)

Ans.

Using similar triangles, the location of the neutral axis can be determined y 0.2 - y = ; 1.875 0.375

y = 0.1667 m

Ans: sA = 0.375 MPa (C), sB = 1.875 MPa (T) 750

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8–21.  If the load has a weight of 2700 N, determine the maximum normal stress developed on the cross section of the supporting member at section a–a. Also, plot the normal stress distribution over the cross-section.

0.5 m a

a

25 mm

Solution Section a – a

Internal Loadings: Consider the equilibrium of the free-body diagram of the bottom cut segment shown in Fig. a. a + c ©Fy = 0;

N - 2700 = 0

a + ©MC = 0;

2700(0.5) - M = 0

N = 2700 N M = 1350 N . m

Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the member are 2 = I = A π= (0.0252 ) 0.625(10 −3 )π m

π

= (0.0254 ) 0.30680(10 −6 ) m 4 4

Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =

Mc N ; A I

By observation, the maximum normal stress occurs at point B, Fig. b. Thus,

σ max = σ= B

2700 0.625(10 −3 )π

+

1350(0.025) 0.30680(10 −6 )

6 = 111.38(10 = ) N/m 2 111 MPa (T)

Ans.

For Point A,

σA =

2700 0.625(10 −3 )π

−

1350(0.025) 0.30680(10 −6 )

= − 108.63(106 ) N/m 2 = 109 MPa (C)

Ans.

Using these results, the normal stress distribution over the cross section is shown in Fig. b. The location of the neutral axis can be determined from

50 − x x = ; 108.63 111.38

x = 25.3 mm

Ans: smax = 111 MPa (T) 753

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8–22. The bearing pin supports the load of 3.5 kN. the stress stresscomponents componentsininthe the support member Determine the support member at at point The support is 0.5 thick. point A. A. The support is 12 mmin. thick.

0.75 in. 18 mm A in. 502mm 30 30

A B

B

0.5 in. 12 mm

in. 753mm

Solution ©Fx = 0;

N - 700 cos30° 30°==0;0; 3.5 cos

©Fy = 0;

3.5 sin V - 700 sin30° 30°==0;0;

a + ©M = 0;

350kN lb V ==1.75

M − 3.5(0.032 − 0.05sin 30°) =0 ;

sA =

1.25 in. 32 mm

N ==3.0311 606.218 kNlb

3.0311(10 3 ) N Mc − = A I 0.012(0.018)

700 lb 3.5 kN

= M 0.0245 KN ⋅= m 24.5 N ⋅ m 24.5(0.009)

1 (0.012)(0.018 3 ) 12

0.05 m

23.8 MPa (C) sA = −23.78(106 ) N/m 2 =

Ans.

tA = 0

Ans.

(since QA = 0)

0.032 m

3.5 kN

Ans. sA = 23.8 MPa(C), TA = 0 547

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8–23. The bearing pin supports the load of 3.5 kN. Determine the support member at the stress stresscomponents componentsininthe the support member point B. B. The support is 12 mmin. thick. at point The support is 0.5 thick.

0.75 in. 18 mm A A in. 502mm 30 30

A A B B

B B

0.5mm in. 12

in. 75 3mm

Solution ©Fx = 0;

N - 700 cos 30° 30° ==0;0; 3.5 cos

N ==3.0311 606.218 kNlb

©Fy = 0;

3.5 sin V - 700 sin 30° 30° ==0;0;

350 kN lb V ==1.75

a + ©M = 0;

3.0311(10 3 ) N Mc + + = 0.012(0.018) A I

700kN lb 3.5

= M 0.0245 KN ⋅= m 24.5 N ⋅ m

M − 3.5(0.032 − 0.05sin 30°) =0 ;

sB =

1.25 in. 32 mm

24.5(0.009)

50 mm

1 (0.012)(0.018 3 ) 12

32 mm

sB = 51.84(106 ) N/m 2 = 51.8 MPa (T)

Ans.

tB = 0

Ans.

(since QB = 0)

3.5 kN

Ans. N = 3.031 kN, V = 1.75 kN, M = 24.5 N # m, sB = 51.8 MPa (T), tB = 0 548

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*8–24. The column is built up by gluing the two boards together. Determine the maximum normal stress on the cross section when the eccentric force of P = 50 kN is applied.

P 250 mm 150 mm

150 mm 75 mm 50 mm

300 mm

Solution Section Properties: The location of the centroid of the cross section, Fig. a, is y =

0.075(0.15)(0.3) + 0.3(0.3)(0.15) ΣyA = = 0.1875 m ΣA 0.15(0.3) + 0.3(0.15)

The cross-sectional area and the moment of inertia about the z axis of the cross section are A = 0.15(0.3) + 0.3(0.15) = 0.09 m2 Iz =

1 1 (0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 + (0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2 12 12

= 1.5609(10 - 3) m4 Equivalent Force System: Referring to Fig. b, + c ΣFx = (FR)x;

- 50 = - F

ΣMz = (MR)z;

- 50(0.2125) = -M

F = 50 kN

M = 10.625 kN # m

Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =

My N + A I

By inspection, the maximum normal stress occurs at points along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A. Thus, smax =

- 50(103) 0.09

-

10.625(103)(0.2625) 1.5609(10 - 3) Ans.

= -2.342 MPa = 2.34 MPa (C)

Ans: smax = 2.34 MPa (C) 754

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8–25. The column is built up by gluing the two boards together. If the wood has an allowable normal stress of sallow = 6 MPa, determine the maximum allowable eccentric force P that can be applied to the column.

P 250 mm 150 mm

150 mm 75 mm 50 mm

300 mm

Solution Section Properties: The location of the centroid c of the cross section, Fig. a, is y =

0.075(0.15)(0.3) + 0.3(0.3)(0.15) ΣyA = = 0.1875 m ΣA 0.15(0.3) + 0.3(0.15)

The cross-sectional area and the moment of inertia about the z axis of the cross section are A = 0.15(0.3) + 0.3(0.15) = 0.09 m2 Iz =

1 1 (0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 + (0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2 12 12

= 1.5609(10 - 3) m4 Equivalent Force System: Referring to Fig. b, + c ΣFx = (FR)x;

- P = - F

ΣMz = (MR)z;

- P(0.2125) = - M

F = P M = 0.2125P

Normal Stress: The normal stress is a combination of axial and bending stress. Thus, F =

My N + A I

By inspection, the maximum normal stress, which is compression, occurs at points along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A. Thus, - 6(106) =

0.2125P(0.2625) -P 0.09 1.5609(10 - 3) Ans.

P = 128 076.92 N = 128 kN

Ans: Pmax = 128 kN 755

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8–26. The offset link supports the loading of P = 30 kN. Determine its required width w if the allowable normal stress is sallow = 73 MPa. The link has a thickness of 40 mm.

P

Solution

w

50 mm

s due to axial force: sa =

30(103) 750(103) P = = A (w)(0.04) w

s due to bending: P

30(103)(0.05 + w2)(w2) Mc sb = = 1 3 I 12 (0.04)(w) =

4500 (103)(0.05 + w2) w2

smax = sallow = sa + sb 73(106) =

4500(103)(0.05 + w2) 750(103) + w w2

73 w2 = 0.75 w + 0.225 + 2.25 w 73 w2 - 3 w - 0.225 = 0 w = 0.0797 m = 79.7 mm

Ans.

Ans. w = 79.7 mm 548

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8–27. The offset link has a width of w = 200 mm and a thickness of 40 mm. If the allowable normal stress is sallow = 75 MPa, determine the maximum load P that can be applied to the cables.

P

Solution

w

50 mm

A = 0.2(0.04) = 0.008 m

2

I =

1 (0.04)(0.2)3 = 26.6667(10 - 6) m4 12

s =

P Mc + A I

P

0.150 P(0.1) P + 75(106) = 0.008 26.6667(10 - 6) P = 109 kN

Ans.

Ans. P = 109 kN 549

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*8–28. The pliers are made from two steel parts pinned together at A. If a smooth bolt is held in the jaws and a gripping force of 50 N is applied at the handles, determine the state of stress developed in the pliers at points B and C. Here the cross section is rectangular, having the dimensions shown in the figure.

4.5 mm 0.18 in. D

50 lb N 10

5 mm 0.2 in.

2.5 0.1mm in. E D E

Q ©Fx = 0;

753 mm in.

A B

5 mm 0.2 in. 5 mm 0.2 in.

B C

C

5 mm 0.2 in.

44 mm 1.75 in.

Solution +

30

62.5 m 2.5 in.

100 mm 4 in.

N - 50 10 sin 30° == 0;0; N =N25=N5.0 lb

a+ ©Fy = 0;

50 cos 30° == 0;0; V =V43.30 N lb V - 10 = 8.660

a + ©MC = 0;

M − 50(0.075) = 0; M = 3.75 N ⋅ m

50 lb N 10

= = 50(10 −6 ) m 2 A (0.005)(0.01) = I

1 = (0.005)(0.013 ) 0.41667(10 −9 ) m 4 12

50 N 0.075 m

QB = 0 QC = y¿A¿ = (0.0025)(0.005)(0.005) = 62.5(10 −9 ) m 3 Point B: sB =

My −25 3.75(0.005) N + + = A I 50(10 −6 ) 0.41667(10 −9 ) = 44.5(106 ) N/m 2 = 44.5 MPa (T)

tB =

Ans.

VQ = 0 It

0.0025 m

Ans.

Point C: sC =

My −25 N −0.500(106 ) N/m 2 = 0.500 MPa (C) Ans. +0= + = A I 50(10 −6 )

Shear Stress : tC =

VQ 43.30[62.5(10 −9 )] 6 = 1.299(10 = ) N/m 2 1.30 MPa = It [0.41667(10 −9 ](0.005)

Ans.

Ans. sB = 44.5 MPa, tB = 0, sC = 0.50 MPa, tC = 1.30 MPa 555

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Solve Prob. 8–33 for points D and E.

8–29.

4.5 0.18mm in. D

50 lb N 10

5 mm 0.2 in.

2.5 0.1mm in. E D E

Solution a + ©MA = 0;

30

753 mm in.

A B

50.2 mm in. 50.2 mm in.

B C

C

50.2 mm in.

44 mm 1.75 in. 62.5 m 2.5 in.

80 lb N − F (0.0625) + 50(0.1) = 0; FF== 16

1004 mm in. 50 lb N 10

Point D: sD = 0

Ans.

80[(0.00125)(0.0045)(0.0025)] VQ = 1  (0.0045)(0.0053 ) (0.0045) It  12  6 = 5.333(10 = ) N/m 2 5.33 MPa

tD =

Ans.

Point E: sE =

My = I

=

3.52(0.0025) 1 (0.0045)(0.0053 ) 12 6 2

= 187.73(10 ) N/m 188 MPa

Ans.

tE = 0

0.0625 m

Ans.

0.1 m 50 N

V = 80 N M = 80(0.044) = 3.52 N · m

0.044 m

F = 80 N

Ans. sD = 0, tD = 5.33 MPa, sE = 188 MPa, tE = 0 556

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8–30. The rib-joint pliers are used to grip the smooth pipe C. If the force of 100 N is applied to the handles, determine the state of stress at points A and B on the cross section of the jaw at section a–a. Indicate the results on an element at each point.

100 N 250 mm 25 mm 25 mm

45° a a C

A

10 mm

B 20 mm

7.5 mm

Section a – a

Solution

Support Reactions: Referring to the free-body diagram of the handle shown in Fig. a, a+ ΣMD = 0;

100(0.25) - FC (0.05) = 0

FC = 500 N

Internal Loadings: Consider the equilibrium of the free-body diagram of the segment shown in Fig. b. ΣFy′ = 0;

500 - V = 0

a+ ΣMC = 0;

M - 500(0.025) = 0

V = 500 N

M = 12.5 N # m

Section Properties: The moment of inertia of the cross section about the centroidal axis is I =

1 (0.0075)(0.023) = 5(10 - 9) m4 12

Referring to Fig. c, QA and QB are QA = 0 QB = y′A′ = 0.005(0.01)(0.0075) = 0.375(10 - 6) m3 Normal Stress: The normal stress is contributed by bending stress only. Thus s =

My I

For point A, y = 0.01 m. Then sA = -

12.5(0.01) 5(10 - 9)

Ans.

= - 25 MPa = 25 MPa (C)

For point B, y = 0. Then Ans.

sB = 0

760

100 N

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8–30. Continued

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus, tA =

VQA = 0 It

Ans.

tB =

500[0.375(10 - 6)] VQB = = 5 MPa It 5(10 - 9)(0.0075)

Ans.

The state of stress of points A and B are represented by the elements shown in Figs. d and e respectively.

Ans: sA = 25 MPa (C), sB = 0, tA = 0, tB = 5 MPa 761

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8–31. y

The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point A on the cross section of the drill bit at section a–a.

400 mm a 20 N ·m x a 125 mm

Solution

y

Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s right cut segment, Fig. a. 4 ΣFx = 0; N - 150 a b = 0 5 3 ΣFy = 0; 150 a b - Vy = 0 5 ΣMx = 0; 20 - T = 0

A z

N = 120 N

T = 20 N # m

3 4 ΣMz = 0; - 150 a b(0.4) + 150 a b(0.125) + Mz = 0 5 5 Mz = 21 N # m

Section Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drill’s cross section are A = p 1 0.0052 2 = 25p 1 10 - 6 2 m2

p 1 0.0054 2 = 0.15625p 1 10 - 9 2 m4 4

J =

p 1 0.0054 2 = 0.3125p 1 10 - 9 2 m4 2

Referring to Fig. b, QA is

QA = 0 Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =

Mzy N A Iz

For point A, y = 0.005 m. Then sA =

- 120

25p 1 10

-6

2

-

21(0.005) 0.15625p 1 10 - 9 2

= - 215.43 MPa = 215 MPa (C)

768

B Section a – a

Vy = 90 N

Iz =

5 mm

Ans.

3

5 4

150 N

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8–31. Continued

Shear Stress: The transverse shear stress developed at point A is c 1 txy 2 V d

A

=

VyQA Izt

= 0

The torsional shear stress developed at point A is

Thus,

3 (txz)T 4 A

=

20(0.005) Tc = = 101.86 MPa J 0.3125p 1 10 - 9 2

1 txz 2 A

1 txy 2 A

Ans.

= 0

= c 1 txz 2 T d

A

Ans.

= 102 MPa

The state of stress at point A is represented on the element shown in Fig. c.

769

Ans: sA = 215 MPa (C), 1 txy 2 A = 0, 1 txz 2 A = 102 MPa

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*8–32. y

The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point B on the cross section of the drill bit at section a–a.

400 mm a 20 N ·m x a 125 mm

Solution

y A

Internal Loadings: Consider the equilibrium of the free-body diagram of the drill’s right cut segment, Fig. a. 4 ΣFx = 0; N - 150a b = 0 5

z

N = 120 N

3 ΣFy = 0; 150a b - Vy = 0 5

Section a – a

T = 20 N # m

3 4 ΣMz = 0; - 150a b(0.4) + 150a b(0.125) + Mz = 0 5 5 Mz = 21 N # m

Section Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drill’s cross section are A = p 1 0.0052 2 = 25p 1 10 - 6 2 m2 Iz =

p 1 0.0054 2 = 0.15625p 1 10 - 9 2 m4 2

J =

p 1 0.0054 2 = 0.3125p 1 10 - 9 2 m4 2

Referring to Fig. b, QB is

QB = y′A′ =

4(0.005) p c 1 0.0052 2 d = 83.333 1 10 - 9 2 m3 3p 2

Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =

Mzy N A Iz

For point B, y = 0. Then

sB =

- 120

B

Vy = 90 N

ΣMx = 0; 20 - T = 0

25p 1 10 - 6 2

- 0 = - 1.528 MPa = 1.53 MPa (C)

770

5 mm

Ans.

3

5 4

150 N

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*8–32. Continued

Shear Stress: The transverse shear stress developed at point B is

c 1 txy 2 V d

B

=

VyQB Izt

=

90c 83.333 1 10 - 9 2 d

0.15625p 1 10 - 9 2 (0.01)

= 1.528 MPa

The torsional shear stress developed at point B is

Thus,

c 1 txy 2 T d

1 txz 2 B

1 txy 2 B

B

=

20(0.005) Tc = = 101.86 MPa J 0.3125p 1 10 - 9 2 Ans.

= 0 = c 1 txy 2 T d

B

- c 1 txy 2 V d

B

Ans.

= 101.86 - 1.528 = 100.33 MPa = 100 MPa

The state of stress at point B is represented on the element shown in Fig. d.

771

Ans: sB = 1.53 MPa (C), 1 txz 2 B = 100 MPa

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8–33. Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN. Indicate the result as a differential volume element.

4 kN

250 mm G

375 mm

D 2m

0.75 m 100 mm

Solution

B 1m

a + ©MD = 0;

A

B 150 mm

4(0.625) - Cy (3.75) = 0

C

20 mm

15 mm

Support Reactions:

A

200 mm 20 mm

Cy = 0.6667 kN + : ©Fx = 0;

Cx - 4 = 0

Cx = 4.00 kN

Internal Forces and Moment: + : ©Fx = 0;

4.00 - N = 0

+ c ©Fy = 0;

V - 0.6667 = 0

a + ©Mo = 0;

N = 4.00 kN V = 0.6667 kN

M - 0.6667(1) = 0

M = 0.6667 kN # m

Section Properties: A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2 I =

1 1 (0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m4 12 12

QA = ©y¿A¿ = 0.11(0.15)(0.02) + 0.05(0.1)(0.015) = 0.405 A 10 - 3 B m3 Normal Stress: s = sA =

My N ; A I 4.00(103) 9.00(10 - 3)

+

0.6667(103)(0) 82.8(10 - 6)

= 0.444 MPa (T)

Ans.

Shear Stress: Applying shear formula. tA = =

VQA It

0.6667(103) C 0.405(10 - 3) D 82.8(10 - 6)(0.015)

= 0.217 MPa

Ans.

Ans. sA = 0.444 MPa (T), tA = 0.217 MPa 564

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8–34. Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN. Indicate the result as a differential volume element.

4 kN

250 mm G

375 mm

D 2m

0.75 m 100 mm

Solution

B 1m

a + ©MD = 0;

A

B 150 mm

4(0.625) - Cy (3.75) = 0

C

20 mm

15 mm

Support Reactions:

A

200 mm 20 mm

Cy = 0.6667 kN + : ©Fx = 0;

Cx - 4 = 0

Cx = 4.00 kN

Internal Forces and Moment: + : ©Fx = 0;

4.00 - N = 0

+ c ©Fy = 0;

V - 0.6667 = 0

a + ©Mo = 0;

N = 4.00 kN V = 0.6667 kN M = 0.6667 kN # m

M - 0.6667(1) = 0

Section Properties: A = 0.24(0.15) - 0.2(0.135) = 9.00 A 10 - 3 B m2 I =

1 1 (0.15) A 0.243 B (0.135) A 0.23 B = 82.8 A 10 - 6 B m 12 12

QB = 0 Normal Stress: s = sB =

My N ; A I 4.00(103) -3

9.00(10 )

-

0.6667(103)(0.12) 82.8(10 - 6)

= - 0.522 MPa = 0.522 MPa (C)

Ans.

Shear Stress: Since QB = 0, then tB = 0

Ans.

Ans. A = 9.00 (10- 3) m2, I = 82.8 (10- 6) m, QB = 0, sB = 0.522 MPa (C), tB = 0 565

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8–35. The block is subjected to the eccentric load shown. Determine the normal stress developed at points A and B. Neglect the weight of the block.

150 kN

100 mm 150 mm C

a

A

a

B

Solution Internal Loadings: Consider the equilibrium of the upper segment of the block sectioned through a–a, Fig. a. ΣFx = 0;

N + 150 = 0

ΣMy = 0;

My + 150(0.05) = 0

ΣMz = 0;

Mz - 150(0.075) = 0

N = - 150 kN

My = - 7.50 kN # m Mz = 11.25 kN # m

Section Properties: For the rectangular cross section, A = 0.1(0.15) = 0.015 m2 Iy =

1 (0.15) ( 0.13 ) = 12.5 ( 10-6 ) m4 12

Iz =

1 (0.1) ( 0.153 ) = 28.125 ( 10-6 ) m4 12

Normal Stresses: For the combined loadings, the normal stress can be determined from s =

Myz Mzy N + A Iz Iy

For point A, yA = 0.075 m and zA = 0.05 m sA =

-150 ( 103 ) 0.015

-

11.25 ( 103 ) (0.075) 28.125 ( 10-6 )

+

3 - 7.50 ( 103 ) 4 (0.05) 12.5 ( 10-6 )

= - 70.0 ( 106 ) Pa = 70.0 MPa (C)

Ans.

For point B, yB = 0.075 m and zB = - 0.05 m sB =

- 150 ( 103 ) 0.015

-

11.25 ( 103 ) (0.075) 28.125 ( 10-6 )

+

3 - 7.50 ( 103 ) 4 ( - 0.05) 12.5 ( 10-6 )

= - 10.0 ( 106 ) Pa = 10.0 MPa (C)

Ans.

Ans: sA = 70.0 MPa (C), sB = 10.0 MPa (C) 764

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*8–36. The block is subjected to the eccentric load shown. Sketch the normal-stress distribution acting over the cross section at section a–a. Neglect the weight of the block.

150 kN

100 mm 150 mm C

a

Solution Internal Loadings: Consider the equilibrium of the upper segment of the block sectioned through a–a, Fig. a. ΣFx = 0;

N + 150 = 0

ΣMy = 0;

My + 150(0.05) = 0

ΣMz = 0;

Mz - 150(0.075) = 0

N = - 150 kN

My = -7.50 kN # m Mz = 11.25 kN # m

Section Properties: For the rectangular cross section, A = 0.1(0.15) = 0.015 m2 Iy =

1 (0.15) ( 0.13 ) = 12.5 ( 10-6 ) m4 12

Iz =

1 (0.1) ( 0.153 ) = 28.125 ( 10-6 ) m4 12

Normal Stress: For the combined loadings, the normal stress can be determined from s =

Myz Mzy N + A Iz Iy

For point A, yA = 0.075 m and zA = 0.05 m sA =

- 150 ( 103 ) 0.015

-

11.25 ( 103 ) (0.075) 28.125 ( 10-6 )

+

( -7.50) ( 103 ) (0.05) 12.5 ( 10-6 )

= - 70.0 ( 106 ) Pa = 70.0 MPa (C)

Ans.

For point B, yB = 0.075 m and zB = - 0.05 m sB =

- 150 ( 103 ) 0.015

-

11.25 ( 103 ) (0.075) 28.125 ( 10-6 )

+

( -7.50) ( 103 ) ( - 0.05) 12.5 ( 10-6 )

= - 10.0 ( 106 ) Pa = 10.0 MPa (C)

Ans.

For point C, yC = - 0.075 m and zC = - 0.05 m sC =

- 150 ( 103 ) 0.015

-

11.25 ( 103 ) ( -0.075) 28.125 ( 10

-6

)

+

( -7.50) ( 103 ) ( - 0.05) 12.5 ( 10-6 )

= 50.0 ( 106 ) Pa = 50.0 MPa (T)

Ans.

For point D, yD = - 0.075 m and zD = 0.05 m sD =

- 150 ( 103 ) 0.015

-

11.25 ( 103 ) ( -0.075) 28.125 ( 10-6 )

+

( -7.50) ( 103 ) (0.05) 12.5 ( 10-6 )

= - 10.0 ( 106 ) Pa = 10.0 MPa (C)

Ans.

765

A

B

a

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*8–36. Continued

The location of neutral axis can be found using similar triangles. y 0.15 - y = ; 10.0 50.0

y = 0.025 m

z 0.1 - z = ; 10.0 50.0

z = 0.01667 m

Using these result, the normal stress distribution over the cross section shown in Fig. b can be sketched.

Ans: sA = sB = sC = sD = 766

70.0 MPa (C), 10.0 MPa (C), 50.0 MPa (T), 10.0 MPa (C)

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8–37. If the 75-kg man stands in the position shown, determine the state of stress at point A on the cross section of the plank at section a–a. The center of gravity of the man is at G. Assume that the contact point at C is smooth.

C

G

A

a B

1.5 m

30 a 600 mm

300 mm

Solution Support Reactions: Referring to the free-body diagram of the entire plank, Fig. a, a + ©MB = 0;

FC sin 30°(2.4) - 75(9.81) cos 30°(0.9) = 0 FC = 477.88 N

©Fx¿ = 0; Bx¿ - 75(9.81) sin 30° - 477.88 cos 30° = 0 Bx¿ = 781.73 N ©Fy¿ = 0; By¿ + 477.88 sin 30° - 75(9.81) cos 30° = 0 By¿ = 398.24 N Internal Loadings: Consider the equilibrium of the free-body diagram of the plank’s lower segment, Fig. b, ©Fx¿ = 0; 781.73 - N = 0

N = 781.73 N

©Fy¿ = 0; 398.24 - V = 0

V = 398.24 N

a + ©MO = 0;

M = 238.94 N # m

M - 398.24(0.6) = 0

Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the plank’s cross section are A = 0.6(0.05) = 0.03 m2 I =

1 (0.6) A 0.053 B = 6.25 A 10 - 6 B m4 12

Referring to Fig. c, QA is QA = y¿A¿ = 0.01875(0.0125)(0.6) = 0.140625 A 10 - 3 B m3 Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =

My N ; A I

For point A, y = 0.0125 m. Then sA =

238.94(0.0125) -781.73 0.03 6.25 A 10 - 6 B

= - 503.94 kPa = 504 kPa (C)

Ans.

551

600 mm

12.5 mm Section a – a and b – b

50 mm

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8–37. Continued Shear Stress: The shear stress is contributed by transverse shear stress. Thus,

tA

VQA = = It

398.24 c 0.140625 A 10 - 3 B d 6.25 A 10 - 6 B (0.6)

= 14.9 kPa

Ans.

The state of stress at point A is represented on the element shown in Fig. d.

Ans. sA = 504 kPa (C), tA = 14.9 kPa 552

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6 kip 30 kN

8–38. Determine the normal stress developed at points A and B. Neglect the weight of the block.

3 in. 75 mm

Referring to Fig. a, a

©Fx = (FR)x;

6 -– 60 12 ==FF –30

= -kN 18.0 kip FF = –90

©My = (MR)y;

My = −1.125 kN ⋅ m 30(0.0375) − 60(0.0375) = My

©Mz = (MR)z;

= Mz 2.25 kN ⋅ m 60(0.075) − 30(0.075) = Mz

12 60 kip kN

1506mm in. B

A

a

The cross-sectional area and moment of inertia about the y and z axes of the crosssection are = A 0.075(0.15) = 0.01125 m 2 = Iy

1 = (0.15)(0.0753 ) 5.27344(10 −6 ) m 4 12

= Iz

1 = (0.75)(0.153 ) 21.09375(10 −6 ) m 4 12

The normal stress developed is the combination of axial and bending stress. Thus, s =

My z Mz y F + A Iz Iy

30 kN

For point A, y = 0.075 m and z = −0.0375 m

σA =

60 kN

−90(10 3 ) [2.25(10 3 )](0.075) [−1.125(10 3 )](−0.0375) − + 0.01125 21.09375(10 −6 ) 5.27344(10 −6 )

= −8.00(106 ) N/m 2 = 8.00 MPa (C)

Ans.

0.075 m 0.0

75

For point B, y 0.075 = = m and z 0.0375 m −90(10 3 ) 0.01125

σB = −

[2.25(10 3 )](0.075) −6

21.09375(10 )

+

0.0375 m m

7.5

m

3 0.0

[−1.125(10 3 )](0.0375) 5.27344(10 −6 )

24.00 MPa (C) = −24.00(106 ) N/m 2 =

Ans.

Ans. sA = 8.00 MPa (C), sB = 240 MPa (C) 570

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30 kN 6 kip

8–39. Sketch the normal stress distribution acting over the cross section at section a–a. Neglect the weight of the block.

75 mm 3 in.

aa

1506mm in. A A

Solution Referring to Fig. a, ©Fx = (FR)x;

–30 6 -– 60 12 ==FF

FF= –90 = -kN 18.0 kip

©My = (MR)y; 30(0.0375) − 60(0.0375) = My = −1.125 kN ⋅ m My ©Mz = (MR)z; 60(0.075) − 30(0.075) = Mz 2.25 kN ⋅ m Mz = The cross-sectional area and the moment of inertia about the y and z axes of the cross-section are = A 0.075(0.15) = 0.01125 m 2 = Iy

1 = (0.15)(0.0753 ) 5.27344(10 −6 ) m 4 12

= Iz

1 (0.75)(0.153 ) 21.09375(10 −6 ) m 4 = 12

The normal stress developed is the combination of axial and bending stress. Thus, s =

Myz Mzy F + A Iz Iy

For point A, y = 0.075 m and z = −0.0375 m

σA =

−90(10 3 ) [2.25(10 3 )](0.075) [−1.125(10 3 )](−0.0375) − + 0.01125 21.09375(10 −6 ) 5.27344(10 −6 )

= −8.00(106 ) N/m 2 = 8.00 MPa (C)

For point B, y 0.075 = = m and z 0.0375 m −90(10 3 ) 0.01125

σB = −

[2.25(10 3 )](0.075) −6

21.09375(10 )

+

[−1.125(10 3 )](0.0375) 5.27344(10 −6 )

24.00 MPa (C) = −24.00(106 ) N/m 2 =

571

60 kip kN 12

B B

aa

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8–39. Continued

For point C, y = −0.075 m and z = 0.0375 m −90(10 3 ) 0.01125

σ C =−

[2.25(10 3 )](−0.075) −6

21.09375(10 )

+

[−1.125(10 3 )](0.0375) 5.27344(10 −6 )

= −8.00(106 ) N/m 2 = 8.00 MPa (C) −0.075 m and z = −0.0375 m For point D, y = = σD

−90(10 3 ) [2.25(10 3 )](−0.075) [−1.125(10 3 )](−0.0375) − + 0.01125 21.09375(10 −6 ) 5.27344(10 −6 )

6 ) N/m 2 8.00 MPa (T) = 8.00(10 =

The normal stress distribution over the cross-section is shown in Fig. b

0.0375 m

8.00 MPa

30 kN 24.0 MPa

0.0375 m 60 kN

8.00 MPa

0.075 m

0.075 m 0.0375 m

0.075 m

0.075 m

8.00 MPa

0.0375 m

Ans. A = 0.01125 m, Iy = 5.27344(10- 6) m4, Iz = 21.09375 (10 - 6) m4, sC = 8.00 MPa (C) 572

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*8–40. The frame supports the distributed load shown. Determine the state of stress acting at point D. Show the results on a differential element at this point.

4 kN/m

A

B E 1.5 m

D 1.5 m

3m

20 mm 60 mm 20 mm

D

5m

50 mm

E

3m C

Solution sD = -

8(103) 12(103)(0.03) My P = - 1 ­ 3 A I (0.1)(0.05) 12 (0.05)(0.1)

sD = - 88.0 MPa

Ans.

tD = 0

Ans.

Ans: sD = -88.0 MPa, tD = 0 772

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8–41. The frame supports the distributed load shown. Determine the state of stress acting at point E. Show the results on a differential element at this point.

4 kN/m

A

B E 1.5 m

D 1.5 m

3m

20 mm 60 mm 20 mm

D

5m

50 mm

E

3m C

Solution sE = tE =

8(103) 8.25(103)(0.03) My P = + 1 = 57.8 MPa 3 A I (0.1)(0.05) 12 (0.05)(0.1)

4.5(103)(0.04)(0.02)(0.05) VQ = 864 kPa = 1 3 It 12 (0.05)(0.1) (0.05)

Ans. Ans.

Ans: sE = 57.8 MPa, tE = 864 kPa 773

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8–42. y

The rod has a diameter of 40 mm. If it is subjected to the force system shown, determine the stress components that act at point A, and show the results on a volume element located at this point.

100 mm

x

300 mm

B

Solution Internal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a. ΣFx = 0;

Nx - 1500 = 0

ΣFy = 0;

Vy - 600 = 0

Vy = 600 N

ΣFz = 0;

Vz + 800 = 0

Vz = - 800 N

A

z

100 Nm

1500 N 600 N

800 N

Nx = 1500 N

Tx = 100 N # m

ΣMx = 0;

Tx - 100 = 0

ΣMy = 0;

My + 800(0.3) = 0

ΣMz = 0;

Mz + 600(0.3) = 0

My = - 240 N # m Mz = - 180 N # m

Section Properties: For the circular cross section, Fig. b, A = pc 2 = p ( 0.022 ) = 0.400 ( 10-3 ) p m2 p p Iy = Iz = c 4 = ( 0.024 ) = 40.0 ( 10-9 ) p m4 4 4 p 4 p J = c = ( 0.024 ) = 80.0 ( 10-9 ) p m4 2 2 4(0.02) p (QA)z = 0 (QA)y = y′A′ = c ( 0.022 ) d = 5.3333 ( 10-6 ) m3 3p 2

Normal Stress: For the combined loadings, the normal stress at point A can be determined from sx = sA = =

MyzA MzyA Nx + A Iz Iy 1500 0.400 ( 10

-3

)p

-

( -180)(0) 40.0 ( 10

-9

)p

+

( - 240)(0.02) 40.0 ( 10-9 ) p

= - 37.00 ( 106 ) Pa = 37.0 MPa (C)

Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional shear VQ stress can be obtained using the shear formula tV = and the torsion formula It Tr tT = respectively. J (txy)A = (tV)y - tT =

600 3 5.3333 ( 10-6 ) 4 40.0 ( 10

-9

) p(0.04)

-

100(0.02) 80.0 ( 10-9 ) p

= - 7.321 ( 106 ) Pa = -7.32 MPa

Ans. Ans.

(txz)A = (tV)z = 0

Using these results, the state of stress at point A can be represented by the differential volume element show in Fig. c.

774

Ans: sA = 37.0 MPa (C), (txy)A = - 7.32 MPa, (txz)A = 0

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8–43. y

Solve Prob. 8–40 for point B.

100 mm

x

300 mm

B

Solution Internal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a. ΣFx = 0;

Nx - 1500 = 0

Nx = 1500 N

ΣFy = 0;

Vy - 600 = 0

Vy = 600 N

ΣFz = 0;

Vz + 800 = 0

Vz = -800 N

A

z

100 Nm

1500 N 600 N

800 N

Tx = 100 N # m

ΣMx = 0;

Tx - 100 = 0

ΣMy = 0;

My + 800(0.3) = 0

ΣMz = 0;

Mz + 600(0.3) = 0

My = -240 N # m Mz = -180 N # m

Section Properties: For the circular cross section, A = pc 2 = p ( 0.022 ) = 0.400 ( 10-3 ) p m2 p p Iy = Iz = c 4 = ( 0.024 ) = 40.0 ( 10-9 ) p m4 4 4 p 4 p J = c = ( 0.024 ) = 80.0 ( 10-9 ) p m4 2 2 4(0.02) p (QB)z = z′A′ = c ( 0.022 ) d = 5.3333 ( 10-6 ) m3 (QB)y = 0 3p 2

Normal Stress: For the combined loadings, the normal stress at point B can be determined from sx = sB = =

MyzB MzyB Nx + A Iz Iy 1500 0.400 ( 10

-3

)p

-

( -180)( - 0.02) 40.0 ( 10

-9

)p

+

( - 240)(0) 40.0 ( 10-9 ) p

= - 27.45 ( 106 ) = 27.5 MPa (C)

Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional shear VQ stress can be obtained using the shear formula tV = and the torsion formula It Tr tT = , respectively. J (txz)B = (tV)z - tT =

- 800 35.3333 ( 10-6 )4 40.0 ( 10

-9

) p(0.04)

-

100(0.02) 80.0 ( 10-9 ) p

= - 8.807 ( 106 ) Pa = - 8.81 MPa

Ans. Ans.

(txy)B = (tV)y = 0

Using these results, the state of stress at point B, can be represented by the differential volume element shown in Fig. c.

775

Ans: sB = 27.5 MPa (C), (txz)B = - 8.81 MPa, (txy)B = 0

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*8–44. Since concrete can support little or no tension, this problem can be avoided by using wires or rods to prestress the concrete once it is formed. Consider the simply supported beam shown, which has a rectangular cross section of 450 mm by 300 mm. If concrete has a specific weight 3 3 15024lb>ft , determine of kN/m , determinethe therequired requiredtension tensionininrod rodAB, AB, which runs through the beam so that no tensile stress is developed in the concrete at its center section a–a. Neglect the size of the rod and any deflection of the beam.

a 16 in. 400 mm mm 2 in. B 50

A 4 ftm 1.2

a

4 ftm 1.2

18 in.mm 450

in. 6150in.mm 1506mm

Solution Support Reactions: As shown on FBD. Internal Force and Moment: + : ©Fx = 0;

T - N = 0

a + ©Mo = 0;

M + T (0.175) − 3.888(10 3 )(0.6) = 0

N = T

M= {2.3328(10 3 ) − 0.175T } N ⋅ m

Section Properties: A 0.3(0.45) = = 0.135 m 2 I =

1 = 0.3(0.453 ) 2.278125(10 −3 ) m 4 12

Normal Stress: Requires sA = 0 sA = 0 = 0 = T =

N Mc + A I −T [2.3328(10 3 ) − 0.175T ](0.225) + 0.135 2.278125(10 −3 ) 3 9.3312(10 = ) N 9.33 kN

Ans.

24(0.3)(0.45)(2.4) = 7.776 kN

1.2 m 3.888 kN

1.2 m 3.888 kN

3.888 kN

0.6 m

0.6 m

0.175 m

3.888 kN

Ans. = T 9.33 kN 562

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8–45. Solve Prob. 8–38 if the rod has a diameter of 12 mm. Use the transformed area method discussed in Sec 6.6. Est = 200 GPa, Ec = 25 GPa.

aa 16 in. 400 mm mm B B 250 in.

A A 4 ftm 1.2

aa

18 in.mm 450

1506 mm in. 6150 in.mm

4 ftm 1.2

Solution Support Reactions: As shown on FBD. Section Properties: 24(0.3)(0.45)(2.4) = 7.776 kN

29(103) Est 200  8.00 = 8.0556 n = = 25 3) Econ 3.6(10  π   0.79168(10 −3 ) m 2 (n − a) Ast = (8.00 − 1)   (0.012 2 ) = Acon =  4  

1.2 m

1.2 m

3.888 kN

3.888 kN

0.3(0.45) + 0.79168(10 −3 ) = A= 0.135792 m 2

y =

©yA 0.225(0.3)(0.45) + 0.4[0.79168(10 −3 )] = 0.22602 m = ©A 0.135792

0.3 m

1 I= 0.3(0.453 ) + (0.3)(0.45)(0.22602 − 0.225)2 12

0.4 m 0.05 m

+ 0.79168(10 −3 )(0.4 − 0.22602)2 = 2.30223(10 −3 ) m 4

Internal Force and Moment: 3.888 kN

+ : ©Fx = 0; a + ©Mo = 0;

T - N = 0

N = T

M + T (0.1740) − [3.888(106 )](0.6) = 0 0.6 m

3

M = {2.3328(10 ) − 0.1740T } N ⋅ m

0.6 m

0.1740 m

3.888 kN

Normal Stress: Requires sA = 0 sA = 0 =

N Mc + A I

0 =

−T [2.3328(10 3 ) − 0.1740T ](0.45 − 0.22602) + 0.135792 2.30223(10 −3 )

3 = 9.343(10 = ) N 9.34 kN T

Ans.

Ans. = 9.34 kN T 563

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8–46. The man has a mass of 100 kg and center of mass at G. If he holds himself in the position shown, determine the maximum tensile and compressive stress developed in the curved bar at section a–a. He is supported uniformly by two bars, each having a diameter of 25 mm. Assume the floor is smooth. Use the curved-beam formula to calculate the bending stress.

300 mm

G 300 mm

a

a

150 mm 0.35 m

1m

Solution Equilibrium: For the man a+ ΣMB = 0;

981(1) - 2FA(1.35) = 0

FA = 363.33 N

Section Properties: -

r = 0.15 +

0.025 = 0.1625 m 2

dA = 2p( r - 2r 2 - c 2 ) LA r = 2p(0.1625 - 20.16252 - 0.01252) = 3.02524 ( 10 - 3 ) m

A = p ( 0.01252 ) = 0.490874 ( 10-3 ) m2 R =

0.490874 ( 10-3 ) A = = 0.162259 m dA 3.02524 ( 10-3 ) A 1 r

r - R = 0.1625 - 0.162259 = 0.240741 ( 10-3 ) m

-

Internal Force and Moment: The internal moment must be computed about the neutral axis. + c ΣFy = 0;

- 363.33 - N = 0

a+ ΣMo = 0;

- M - 363.33(0.462259) = 0

N = - 363.33 N

M = - 167.95 N # m

Normal Stress: Apply the curved-beam formula. For tensile stress (st)max = =

M(R - r2) N + A Ar2( -r - R) - 363.33

0.490874 ( 10-3 )

+

- 167.95(0.162259 - 0.175) 0.490874 ( 10-3 ) (0.175)0.240741 ( 10-3 ) Ans.

= 103 MPa (T) For compressive stress, (sc)max = =

M(R - r1) N + A Ar1( -r - R) - 363.33

0.490874 ( 10-3 )

+

- 167.95(0.162259 - 0.15) 0.490874 ( 10-3 ) (0.15)0.240741 ( 10-3 ) Ans.

= -117 MPa = 117 MPa (C)

781

Ans: (st)max = 103 MPa (T), (sc)max = 117 MPa (C)

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8–47. y

The solid rod is subjected to the loading shown. Determine the state of stress at point A, and show the results on a differential volume element located at this point.

200 mm

200 mm A C 20 kN

100 kN

30 mm

B

x z

10 kN

Solution Internal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a. ΣFx = 0;

Nx - 100 = 0

ΣFy = 0;

Vy - 10 = 0

Vy = 10 kN

ΣFz = 0;

Vz + 20 = 0

Vz = - 20 kN

Nx = 100 kN

ΣMx = 0;

Tx + 20(0.03) + 10(0.03) = 0

ΣMy = 0;

My + 20(0.2) + 100(0.03) = 0

ΣMz = 0;

Mz + 10(0.4) = 0

Tx = -0.9 kN # m

My = -7.00 kN # m

Mz = - 4.00 kN # m

Section Properties: For the circular cross section, Fig. b, A = pc 2 = p ( 0.032 ) = 0.9 ( 10-3 ) p m2 p p Iy = Iz = c 4 = ( 0.034 ) = 0.2025 ( 10-6 ) p m4 4 4 p 4 p J = c = ( 0.034 ) = 0.405 ( 10-6 ) p m4 2 2 4(0.03) p (QA)z = z′A′ = c ( 0.032 ) d = 18.0 ( 10-6 ) m3 3p 2 (QA)y = 0 Normal Stress: For the combined loadings, the normal stress at point A can be determined from sx = sA = =

MyzA MzyA Nx + A Iz Iy 100 ( 103 ) 0.9 ( 10-3 ) p

-

3 - 4.00 ( 103 ) 4 (0.03) 0.2025 ( 10-6 ) p

= 224.00 ( 106 ) Pa = 224 MPa (T)

+

3 -7.00 ( 103 ) 4 (0) 0.2025 ( 10-6 ) p

Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional shear VQ stress can be obtained using the shear formula tV = and the torsion formula It Tr tT = , respectively. J (txz)A = (tV)z + tT =

- 20 ( 103 ) 3 18.0 ( 10-6 ) 4 0.2025 ( 10

-6

) p(0.06)

+

- 0.9 ( 103 ) (0.03) 0.405 ( 10-6 ) p

= - 30.65 ( 106 ) Pa = -30.7 MPa

Ans. Ans.

(txy)A = (tV)y = 0

Using these results, the state of stress at point A can be represented by the differential volume element shown in Fig. c.

782

Ans: sA = 224 MPa (T), (txz)A = -30.7 MPa, (txy)A = 0

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*8–48. y

The solid rod is subjected to the loading shown. Determine the state of stress at point B, and show the results on a differential volume element at this point.

200 mm

200 mm A C 20 kN

100 kN

30 mm

B

x z

10 kN

Solution Internal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a. ΣFx = 0;

Nx - 100 = 0

ΣFy = 0;

Vy - 10 = 0

Vy = 10 kN

ΣFz = 0;

Vz + 20 = 0

Vz = - 20 kN

Nx = 100 kN

ΣMx = 0;

Tx + 20(0.03) + 10(0.03) = 0

ΣMy = 0;

My + 20(0.2) + 100(0.03) = 0

ΣMz = 0;

Mz + 10(0.4) = 0

Tx = -0.9 kN # m

My = -7.00 kN # m

Mz = - 4.00 kN # m

Section Properties: For the circular cross section, Fig. b, A = pc 2 = p ( 0.032 ) = 0.9 ( 10-3 ) p m2 p p Iy = Iz = c 4 = ( 0.034 ) = 0.2025 ( 10-6 ) p m4 4 4 p 4 p J = c = ( 0.034 ) = 0.405 ( 10-6 ) p m4 2 2 4(0.03) p (QB)z = z′A′ = c ( 0.032 ) d = 18.0 ( 10-6 ) m3 3p 2 (QB)y = 0 Normal Stress: For the combined loadings, the normal stress at point B can be determined from sx = sB = =

MyzB MzyB Nx + A Iz Iy 100 ( 103 ) 0.9 ( 10-3 ) p

-

3 - 4.00 ( 103 ) 4 ( - 0.03) 0.2025 ( 10-6 ) p

+

= - 153.26 ( 106 ) Pa = 153 MPa (C)

3 -7.00 ( 103 ) 4 (0) 0.2025 ( 10-6 ) p

Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional shear VQ stress can be determined using the shear formula tV = and the torsion formula It Tr tT = , respectively. J (txz)B = - tT + (tV)z =

0.9 ( 103 ) (0.03) 0.405 ( 10-6 ) p

+

- 20 ( 103 ) 3 18.0 ( 10-6 ) 4 0.2025 ( 10-6 ) p(0.06)

= 11.79 ( 106 ) Pa = 11.8 MPa

Ans. Ans.

(txy)B = (tV)y = 0

Using these results, the state of stress at point B can be represented by the differential volume element shown in Fig. c. 783

Ans: sB = 153 MPa (C), (txz)B = 11.8 MPa, (txy)B = 0

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8–49. y

The solid rod is subjected to the loading shown. Determine the state of stress at point C, and show the results on a differential volume element at this point.

200 mm

200 mm A C 20 kN

100 kN

30 mm

B

x z

10 kN

Solution Internal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a. ΣFx = 0;

Nx - 100 = 0

ΣFy = 0;

Vy - 10 = 0

Vy = 10 kN

ΣFz = 0;

Vz + 20 = 0

Vz = - 20 kN

Nx = 100 kN

ΣMx = 0;

Tx + 20(0.03) + 10(0.03) = 0

ΣMy = 0;

My + 20(0.2) + 100(0.03) = 0

ΣMz = 0;

Mz + 10(0.4) = 0

Tx = -0.9 kN # m

My = -7.00 kN # m

Mz = - 4.00 kN # m

Section Properties: For the circular cross section, Fig. b, A = pc 2 = p ( 0.032 ) = 0.9 ( 10-3 ) p m2 p p Iy = Iz = c 4 = ( 0.034 ) = 0.2025 ( 10-6 ) p m4 4 4 p p J = c 4 = ( 0.034 ) = 0.405 ( 10-6 ) p m4 2 2 4(0.03) p (QC)y = y′A′ = c d c ( 0.032 ) d = 18.0 ( 10-6 ) m3 3p 2 (QC)z = 0

Normal Stress: For the combine loadings, the normal stress at point C can be determined from sx = sC = =

MyzC MzyC Nx + A Iz Iy 100 ( 103 ) 0.9 ( 10-3 ) p

-

3 - 4.00 ( 103 ) 4 (0) 0.2025 ( 10-6 ) p

+

= - 294.73 ( 106 ) Pa = 295 MPa (C)

3 -7.00 ( 103 ) 4 (0.03) 0.2025 ( 10-6 ) p

Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional shear VQ stress can be determined using the shear formula tV = and the torsion formula It Tr tT = , respectively. J (txy)C = (tV)y - tT =

10 ( 103 ) 3 18.0 ( 10-6 ) 4 0.2025 ( 10-6 ) p(0.06)

-

-0.9 ( 103 ) (0.03) 0.405 ( 10-6 ) p

= 25.94 ( 106 ) Pa = 25.9 MPa

Ans. Ans.

(txz)C = (tV)z = 0

Using these results, the state of stress at point C can be represented by the differential volume element shown in Fig. c. 784

Ans: sC = 295 MPa (C), (txy)C = 25.9 MPa, (txz)C = 0

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8–50. The post has a circular cross section of radius c. Determine the maximum radius e at which the load P can be applied so that no part of the post experiences a tensile stress. Neglect the weight of the post.

P c e

Solution Require sA = 0 sA = 0 =

  e =

P Mc + ; A I

0 =

(Pe)c -P + 2 p 4 pc c 4

c 4

Ans.

Ans: e = 785

c 4

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8–51. x

The post having the dimensions shown is subjected to the load P. Specify the region to which this load can be applied without causing tensile stress at points A, B, C, and D.

z a

a P

A B

ey

a

a

D ez C

a a

y

Solution Equivalent Force System: As shown on FBD. Section Properties: 1 A = 2a(2a) + 2J (2a)a R = 6a2 2 Iz =

1 1 1 a 2 (2a)(2a)3 + 2J (2a) a3 + (2a) aaa + b R 12 36 2 3

= 5a4 Iy = =

1 1 1 a 2 (2a)(2a)3 + 2J (2a) a3 + (2a) aa b R 12 36 2 3

5 4 a 3

Normal Stress: s =

My z Mzy N + A Iz Iy

=

Peyy Pez z -P - 5 4 2 6a 5a4 3a

=

P 30a4

1 - 5a2

- 6eyy - 18ez z 2

At point B where y = - a and z = - a, we require sB 6 0. 0 7

P 3 - 5a2 - 6( - a) ey - 18( - a) ez 4 30a4

0 7 - 5a + 6ey + 18ez

Ans.

6ey + 18ez 6 5a When

ez = 0,

ey 6

5 a 6

When

ey = 0,

ez 6

5 a 18

Repeat the same procedures for point A, C and D. The region where P can be applied without creating tensile stress at points A, B, C, and D is shown shaded in the diagram.

786

Ans: 6ey + 18ez 6 5a

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*8–52.

800 kN

The masonry pier is subjected to the 800-kN load. Determine the equation of the line y = f1x2 along which the load can be placed without causing a tensile stress in the pier. Neglect the weight of the pier.

2.25 m 2.25 m

1.5 m y 1.5 m y x

x

C A B

Solution A = 3(4.5) = 13.5 m2 Ix =

1 (3)(4.53) = 22.78125 m4 12

Iy =

1 (4.5)(33) = 10.125 m4 12

Normal Stress: Require sA = 0 sA =

Myx Mxy P + + A Ix Iy

−800(10 3 ) [800(10 3 ) y](2.25) [800(10 3 ) x](1.5) 0= + + 13.5 22.78125 10.125 0 =79.01y + 118.52 x − 59.26

y = 0.75 - 1.5 x

Ans.

y x 800(103) N y

x

y My = 800(103)x Mx = 800(103)y x

A

Ans. y = 0.75 - 1.5 x 581

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800 kN

8–53. The masonry pier is subjected to the 800-kN load. If x = 0.25 m and y = 0.5 m, determine the normal stress at each corner A, B, C, D (not shown) and plot the stress distribution over the cross section. Neglect the weight of the pier.

2.25 m 2.25 m

1.5 m y 1.5 m y x

x

A = 3(4.5) = 13.5 m2 Ix =

1 (3)(4.53) = 22.78125 m4 12

C

1 Iy = (4.5)(33) = 10.125 m4 12 s = sA =

Mxy P + + A Ix

A B

Myx Iy

-800(103) 400(103)(2.25) 200(103)(1.5) + + 13.5 22.78125 10.125

= 9.88 kPa (T) sB =

Ans.

-800(103) 400(103)(2.25) 200(103)(1.5) + 13.5 22.78125 10.125

= - 49.4 kPa = 49.4 kPa (C) sC =

Ans.

400(103)(2.25) 200(103)(1.5) -800(103) + 13.5 22.78125 10.125

= - 128 kPa = 128 kPa (C) sD =

Ans.

- 800(103) 400(103)(2.25) 200(103)(1.5) + 13.5 22.78125 10.125

= - 69.1 kPa = 69.1 kPa (C)

Ans.

Ans. sA = 9.88 kPa (T), sB = 49.4 kPa (C), sC = 128 kPa (C), sD = 69.1 kPa (C) 582

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8–54. The vertebra of the spinal column can support a maximum compressive stress of smax, before undergoing a compression fracture. Determine the smallest force P that can be applied to a vertebra, if we assume this load is applied at an eccentric distance e from the centerline of the bone, and the bone remains elastic. Model the vertebra as a hollow cylinder with an inner radius ri and outer radius ro.

P

ro



e

ri

Solution

Pero P + p 4 A (r - r 4i ) 4 0 4er0 1 d = Pc + 2 2 p(r 0 - r i ) p(r 40 - r 4i )

smax =

smax

smax =

smax =

smax =

P =

p(r 20

4er0 P d c1 + 2 2 - ri ) (r 0 + r 2i )

P(r 20 + r 2i + 4er0)

p(r 20 - r 2i )(r 20 + r 2i ) P(r 20 + r 2i + 4er0) p(r 40 - r 4i )

dmaxp(r 40 - r 4i )

Ans.

r 20 + r 2i + 4er0

Ans: P = 787

dmaxp(r 40 - r 4i ) r 20 + r 2i + 4er0

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8–55. The coiled spring is subjected to a force P. If we assume the shear stress caused by the shear force at any vertical section of the coil wire to be uniform, show that the maximum shear stress in the coil is tmax = P>A + PRr>J, where J is the polar moment of inertia of the coil wire and A is its cross-sectional area.

P

2r R

Solution Tc PRr = max on perimeter = J J tmax =

V Tc P PRr + = + A J A J

QED

P

Ans: N/A 788

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*8–56. The support is subjected to the compressive load P. Determine the maximum and minimum normal stress acting in the material. All horizontal cross sections are circular.

P r

Section Properties: d¿ = 2r + x A = p(r + 0.5x)2 p (r + 0.5x)4 4

I =

Internal Force and Moment: As shown on FBD. Normal Stress: s =

Mc N ; A I

=

0.5Px(r + 0.5x) –P ; p 4 p(r + 0.5x)2 4 (r + 0.5)

=

P –1 2x ; B R p (r + 0.5x)2 (r + 0.5x)3

sA = = sB = =

P 1 2x + B R 2 p (r + 0.5x) (r + 0.5x)3 P r + 2.5x B R p (r + 0.5x)3

[1]

P –1 2x + B R p (r + 0.5x)2 (r + 0.5x)3 P 1.5x - r B R p (r + 0.5x)3

In order to have maximum normal stress,

[2] dsA = 0. dx

dsA P (r + 0.5x)3 (2.5) - (r + 2.5x)(3)(r + 0.5x)2 (0.5) = B R = 0 p dx (r + 0.5x)6 -

Since

P (r - 2.5x) = 0 p(r + 0.5x)4

P Z 0, then p(r + 0.5x)4 r - 2.5x = 0

x = 0.400r

Substituting the result into Eq. [1] yields smax = = -

P r + 2.5(0.400r) B R p [r + 0.5(0.400r)]3 0.368P 0.368P = (C) r2 r2

Ans.

575

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*8–56. Continued

In order to have minimum normal stress,

dsB = 0. dx

dsB P (r + 0.5x)3 (1.5) - (1.5x - r)(3)(r + 0.5x)2 (0.5) = B R = 0 p dx (r + 0.5x)6 P (3r - 1.5x) = 0 p(r + 0.5x)4 Since

P Z 0, then p(r + 0.5x)4 3r - 1.5x = 0

x = 2.00r

Substituting the result into Eq. [2] yields smin =

P 1.5(2.00r) - r 0.0796P (T) B R = p [r + 0.5(2.00r)]3 r2

Ans.

Ans. smax = 576

0.0796P 0.368P (T) (C), smin = 2 r2 r

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8–57. If P = 60 kN, determine the maximum normal stress developed on the cross section of the column.

2P 150 mm 150 mm 15 mm

15 mm 15 mm P

75 mm

100 mm

100 mm 100 mm

Solution Equivalent Force System: Referring to Fig. a, + c ©Fx = A FR B x;

- 60 - 120 = - F

F = 180 kN

©My = (MR)y;

- 60(0.075) = - My

My = 4.5kN # m

©Mz = (MR)z;

-120(0.25) = - Mz

Mz = 30kN # m

Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2 Iz =

1 1 (0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4 12 12

Iy = 2 c

1 1 (0.015) A 0.23 B d + (0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4 12 12

Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, My and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress. Thus, s =

Myz Mzy N + A Iz Iy

smax = sA =

- 180 A 103 B 0.01005

-

C - 30 A 103 B D ( -0.15) 0.14655 A 10 - 3 B

= - 71.0 MPa = 71.0 MPa(C)

+

C -4.5 A 103 B D (0.1) 20.0759 A 10 - 6 B

Ans.

Ans. s max = 71.0 MPa (C) 589

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2P

8–58. Determine the maximum allowable force P, if the column is made from material having an allowable normal stress of sallow = 100 MPa .

150 mm 150 mm 15 mm

Solution

75 mm

100 mm

100 mm 100 mm

Equivalent Force System: Referring to Fig. a, + c ©Fx = (FR)x;

15 mm 15 mm P

- P - 2P = - F F = 3P

©My = (MR)y;

- P(0.075) = - My My = 0.075 P

©Mz = (MR)z;

- 2P(0.25) = - Mz Mz = 0.5P

Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2 Iz =

1 1 (0.2) A 0.33 B (0.185) A 0.273 B = 0.14655 A 10 - 3 B m4 12 12

Iy = 2 c

1 1 (0.15) A 0.23 B d + (0.27) A 0.0153 B = 20.0759 A 10 - 6 B m4 12 12

Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, My and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress, which is in compression. Thus, s =

Myz Mzy N + A Iz Iy

- 100 A 106 B = -

(- 0.5P)( - 0.15) - 0.075P(0.1) 3P + 0.01005 0.14655 A 10 - 3 B 20.0759 A 10 - 6 B

P = 84470.40 N = 84.5 kN

Ans.

Ans. P = 84.5 kN 590

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8–59. The C-frame is used in a riveting machine. If the force at the ram on the clamp at D is P = 8 kN, sketch the stress distribution acting over the section a–a.

a

a

P D

Solution x =

(0.005)(0.04)(0.01) + 0.04(0.06)(0.01) Σx-A = = 0.026 m ΣA 0.04(0.01) + 0.06(0.01)

200 mm 10 mm

A = 0.04(0.01) + 0.06(0.01) = 0.001 m2 I =

40 mm

1 (0.04)(0.013) + (0.04)(0.01)(0.026 - 0.005)2 12 1 + (0.01)(0.063) + 0.01(0.06)(0.040 - 0.026)2 = 0.4773(10 - 6) m4 12

(smax)t =

10 mm

8(103) 1.808(103)0.026 P Mx + = + A I 0.001 0.4773(10 - 6) Ans.

= 106.48 MPa = 106 MPa 3

(smax)c =

60 mm

3

8(10 ) 1.808(10 )(0.070 - 0.026) P Mc = A I 0.001 0.4773(10 - 6) Ans.

= - 158.66 MPa = -159 MPa x 70 - x = ; 158.66 106.48

x = 41.9 mm

Ans: (smax)t = 106 MPa, (smax)c = - 159 MPa 796

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*8–60. Determine the maximum ram force P that can be applied to the clamp at D if the allowable normal stress for the material is sallow = 180 MPa.

a

a

P D

Solution x =

(0.005)(0.04)(0.01) + 0.04(0.06)(0.01) ΣxA = = 0.026 m ΣA 0.04(0.01) + 0.06(0.01)

200 mm 10 mm

A = 0.04(0.01) + 0.06(0.01) = 0.001 m2 I =

s =

40 mm

1 (0.04)(0.013) + (0.04)(0.01)(0.026 - 0.005)2 12 1 + (0.01)(0.063) + 0.01(0.06)(0.040 - 0.026)2 = 0.4773(10 - 6) m4 12

60 mm

10 mm

P Mx { A I

Assume tension failure, 180(106) =

0.226P(0.026) P + 0.001 0.4773(10 - 6)

P = 13524 N = 13.5 kN Assume compression failure, - 180(106) =

0.226P(0.070 - 0.026) P 0.001 0.4773(10 - 6) Ans.

P = 9076 N = 9.08 kN (controls)

Ans: Pmax = 9.08 kN 797

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8–61. Determine the state of stress at point A on the cross section of the pipe at section a–a.

A A 0.75 in. 19 mm B B yy

xx

aa

zz

50 lb 250 N

in. 251mm Section a–a a–a Section

60° 60°

aa

10mm in. 250

Solution Internal Loadings: Referring to the free - body diagram of the pipe’s right segment, Fig. a, Vyy –-250 50 sin 60° == 00 ©Fy = 0; V

Vy = 216.51 N

60° = Vzz –-250 50 cos 60° = 00 ©Fz = 0; V

Vz = 125 N

©Mx = 0; T + (250 sin 60°)(0.300) = 0

TT= =–64.95 N·m - 519.62 lb # in

0 ©My = 0; M y − (250 cos60°)(0.250) =

Myy = = 31.25 250 lbN# ·inm M

0 ©Mz = 0; Mz + (250 sin 60°)(0.250) =

Mzz = N lb · m# in M = –54.13 - 433.01

Section Properties: The moment of inertia about the y and z axes and the polar moment of inertia of the pipe are

π

Iy = Iz = (0.0254 − 0.019 4 ) = 65.076(10 −9 )π m 4 4 J=

π 2

(0.0254 − 0.019 4 )= 0.130152(10 −6 )π m 4

Referring to Fig. b,

A Qy B A = 0 AQz BA = y1œA1œ - y2œA2œ =  

 4(0.025)   π   4(0.019)   π −6 2  3 (0.0252 ) −    2 (0.019 ) = 5.844(10 ) m π 3 3π   2    

Normal Stress: The normal stress is contributed by bending stress only. Thus, s = -

Mzy Iz

+

Myz Iy

For point A, y 0.019 = = m and z 0. Then

σA =

(−54.13)(0.019) 6 = + 0 5.030(10 = ) N/m 2 5.03 MPa (T) 65.076(10 −9 )π

Ans.

Shear Stress: The torsional shear stress developed at point A is c A txz B T d

= A

TrA 64.95(0.019) 6 3.0182(10 = ) N/m 2 3.0182 MPa == J 0.130152(10 −6 )π

597

12mm in. 300

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8–61. Continued The transverse shear stress developed at point A is c A txy B V d c A txz B V d

= 0 A

= A

Vz A Qz B A Iy t

=

125[5.844(10 −6 )] [65.076(10 −9 )π ][2(0.006)]

6 = 0.2978(10 = ) N/m 2 0.2978 MPa

Combining these two shear stress components,

A txy B A = 0

Ans.

A txz B A = c A txz B T d - c A txz B V d A

A

2.72 MPa = 3.0182 − 0.2978 =

Ans.

250 N

0.3 m 0.25 m

y1œ =

0.019 m

4(0.025)

y2œ = 0.025 m

3π

m

4(0.019) 3π

m

Ans. T = -64.95 N # m, My = 31.25 N # m, Mz = -54.13 N # m, Iy = Iz = 65.076 (10 - 9) pm4 J = 0.130152 (10 - 6) pm4 (Qz)A = 5.844 (10 - 6) m3 sA = 5.03 MPa (T), tA = 2.72 MPa 598

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8–62. Determine the state of stress at point B on the cross section of the pipe at section a–a.

A 0.75 in. 19 mm B y

x

a

z

50 lb 250 N

in. 251 mm Section a–a Section

60° 60°

a

10 mm in. 250

Solution Internal Loadings: Referring to the free - body diagram of the pipe’s right segment, Fig. a, 60° = ©Fy = 0; V Vyy –-250 50 sin 60° = 00

Vy = 216.5 = 43.30 lb VykN

Vzz –-250 cos60° 60° = = 00 ©Fz = 0; V 50 cos

Vz = 125 kN Vz = 25 lb

60°(0.300) ©Mx = 0; TT++250 50 sin 60°(12) = 0= 0

T = –64.95TN=· m - 519.62 lb # in

©My = 0; M Myy– -250 50cos cos60°(0.250) 60°(10) = =0 0

250 lb # in MyN=· m My = 31.25

©Mz = 0; M Mzz++250 50 sin 60°(10) = 0= 0 60°(0.250)

Mz N = ·-m433.01 lb # in Mz = –54.13

Section Properties: The moment of inertia about the y and z axes and the polar moment of inertia of the pipe are

π

(0.0254 − 0.019 4 ) = 65.076(10 −9 )π m 4 Iy = Iz = 4 J=

π 2

(0.0254 − 0.019 4 )= 0.130152(10 −6 )π m 4

Referring to Fig. b,

A Qz B B = 0 AQy BB = y1œA1œ - y2œA2œ =  

 4(0.025)   π   4(0.019)   π 2  3 −6 (0.0252 ) −    (0.019 ) = 5.844(10 ) m 3π   2    3π   2

Normal Stress: The normal stress is contributed by bending stress only. Thus, s = -

Mzy Iz

+

Myz Iy

For point B, y = 0 and z = −0.025 m. Then

σ B =−0 +

31.25(−0.025) 65.076(10 −9 )π

=−3.8214(106 ) N/m 2 =3.82 MPa (C)

Ans.

Shear Stress: The torsional shear stress developed at point B is c A txy B T d

= B

TrC 64.95(0.025) 6 = 3.9713(10 ) N/m 2 3.97 MPa = = J 0.130152(10 −6 )π

599

12 mm in. 300

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8–62. Continued The transverse shear stress developed at point B is c A txz B V d c A txy B V d

= 0 B

= B

Vy A Qy B B Izt

=

216.51[5.844(10 −6 )] [65.076(10 −9 )π ][2(0.006)]

6 = ) N/m 2 0.5157 MPa = 0.5157(10

Combining these two shear stress components,

A txy B B = c A txy B T d - c A txy B V d B

B

= 3.4555 MPa = 3.46 MPa = 3.9713 − 0.5157

A txz B B = 0

Ans. Ans.

250 N

0.3 m

0.25 m

y1œ =

4(0.025)

0.019 m

y2œ = 0.025 m

3π

m

4(0.019) 3π

m

Ans. sB = 3.82 MPa (C), tB = 3.46 MPa 600

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8–63. The sign is subjected to the uniform wind loading. Determine the stress components at points A and B on the 100-mm-diameter supporting post. Show the results on a volume element located at each of these points.

z 2m 1m 3m

1.5 kPa C

B A

Solution

2m

Point A:

y

x

10.5(103)(0.05) Mc sA = = = 107 MPa (T) p 4 I 4 (0.05) tA =

D

Ans.

3(103)(0.05) Tc = 15.279(106) = 15.3 MPa = p 4 J 4 (0.05)

Ans.

Point B: Ans.

sB = 0 tB =

3000(4(0.05)/3p))(12)(p)(0.05)2 VQ Tc = 15.279(106) p 4 J It 4 (0.05) (0.1) Ans.

tB = 14.8 MPa

Ans: sA = 107 MPa (T), tA = 15.3 MPa, sB = 0, tB = 14.8 MPa 792

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*8–64. z

The sign is subjected to the uniform wind loading. Determine the stress components at points C and D on the 100-mm-diameter supporting post. Show the results on a volume element located at each of these points.

2m 1m 3m

1.5 kPa C

B A

Solution

2m

Point C:

y

x

10.5(103)(0.05) Mc sC = = = 107 MPa (C) p 4 I 4 (0.05) tC =

D

Ans.

3(103)(0.05) TC = 15.279(106) = 15.3 MPa = p 4 J 2 (0.05)

Ans.

Point D: Ans.

sD = 0 tD =

3(103)(4(0.05)/3p)(12)(p)(0.05)2 VQ Tc + = 15.279(106) + = 15.8 MPa p 4 J It 4 (0.05) (0.1) Ans.

Ans: sC = 107 MPa (C), tC = 15.3 MPa, sD = 0, tD = 15.8 MPa 793

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8–65. The pin support is made from a steel rod and has a diameter of 20 mm. Determine the stress components at points A and B and represent the results on a volume element located at each of these points.

150 N

A

D C B

80 mm

Solution I =

1 (p)(0.014) = 7.85398(10-9) m4 4

QB = y- A= = QA = 0 sA =

4(0.01) 1 a b(p)(0.012) = 0.66667(10-6) m3 3p 2

12(0.01) Mc = = 15.3 MPa I 7.85398(10-9)

Ans.

tA = 0

Ans.

 sB = 0

Ans.

tB =

150(0.6667)(10-6) VQB = = 0.637 MPa It 7.85398(10-9)(0.02)

Ans.

Ans: sA = 15.3 MPa, tA = 0,  sB = 0, tB = 0.637 MPa 802

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8–66. Solve Prob. 8–65 for points C and D.

150 N

A

D C B

80 mm

Solution I =

1 (p)(0.014) = 7.85398(10-9) m4 4

QD = y′A= = QC = 0 sC =

4(0.01) 1 a b(p)(0.012) = 0.66667(10-6) m3 3p 2

12(0.01) Mc = = 15.3 MPa I 7.85398(10-9)

Ans.

tC = 0

Ans.

sD = 0

Ans.

tD =

150(0.6667)(10-6) VQD = = 0.637 MPa It 7.8539(10-9)(0.02)

Ans.

Ans: sC = 15.3 MPa, tC = 0, sD = 0, tD = 0.637 MPa 803

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x

8–67. The eccentric force P is applied at a distance ey from the centroid on the concrete support shown. Determine the range along the y axis where P can be applied on the cross section so that no tensile stress is developed in the material.

z

P

b 2

ey

Solution Internal Loadings: As shown on the free - body diagram, Fig. a. Section Properties: The cross-sectional area and moment of inertia about the z axis of the triangular concrete support are A =

1 bh 2

Iz =

b 2

2h 3

y

h 3

1 bh3 36

Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =

Mzy N A Iz

A Pey B y -P 1 1 bh bh3 2 36

s =

2P 2 A h + 18eyy B bh3

s = -

Here, it is required that sA … 0 and sB … 0. For point A, y =

(1) h , Then. Eq. (1) gives 3

2P 2 h c h + 18ey a b d 3 bh3

0 Ú -

0 … h2 + 6hey ey Ú -

For Point B, y = -

h 6

2 h. Then. Eq. (1) gives 3 0 Ú -

2P 2 2 c h + 18ey a - hb d 3 bh3

0 … h2 - 12hey ey …

h 12

Thus, in order that no tensile stress be developed in the concrete support, ey must be in the range of -

h h … ey … 6 12

Ans.

Ans. s = 601

h h 2P 2 (h + 18eyy), - … ey … 6 12 bh3

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*8–68. The bar has a diameter of 40 mm. Determine the state of stress at point A and show the results on a differential volume element located at this point.

200 mm

x

y 200 mm A B 1200 N 3 5 4

Solution Internal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a. ΣFx = 0;

Nx - 1200 = 0

z

800 N

Nx = 1200 N

3 ΣFy = 0;  Vy - 800 a b = 0  Vy = 480 N 5

4 ΣFz = 0;  Vz + 800 a b = 0  Vz = - 640 N 5 ΣMx = 0;   Tx = 0

4 ΣMy = 0;   My + 800 a b(0.2) = 0  My = - 128 N # m 5 3 ΣMz = 0;   Mz + 800a b(0.2) = 0  Mz = - 96 N # m 5

Section Properties: For the circular cross section, Fig. b,

A = pc 2 = p(0.022) = 0.4 ( 10-3 ) p m2 p p Iy = Iz = c 4 = ( 0.024 ) = 40.0 ( 10-9 ) p m4 4 4 p 4 p J = c = ( 0.024 ) = 80.0 ( 10-9 ) p m4 2 2 4(0.02) p (QA)z = z- =A= = c ( 0.022 ) d = 5.3333 ( 10-6 ) m3 3p 2 (QA)y = 0

Normal Stress: For the combined loading, the normal stress at point A can be determined from sx = sA = =

MyzA Mz yA Nx + A Iz Iy 1200 0.4 ( 10

-3

)p

-

( -96)(0.02) 40.0 ( 10

-9

)p

+

( - 128)(0) 40.0 ( 10-9 ) p

= 16.23 ( 106 ) Pa = 16.2 MPa (T)

Ans.

Shear Stress: Since Tx = 0, the shear stress in the z and y directions is contributed by transverse shear stress only which can be obtained using the shear formula, VQ tV = It - 640 35.3333 ( 10-6 )4 Ans. (txz)A = (tV)z = = - 0.6791 ( 106 ) Pa = -0.679 MPa 40.0 ( 10-9 ) p(0.04) (txy)A = (tV)y = 0 Ans. Using these results, the state of stress at point A can be represented by the volume element shown in Fig. c. 805

Ans: sA = 16.2 MPa (T), (txz)A = - 0.679 MPa, (txy)A = 0

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8–69. Solve Prob. 8–68 for point B.

200 mm

x

y 200 mm A B 1200 N 3 5 4

Solution Internal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a. ΣFx = 0;

Nx - 1200 = 0

z

800 N

Nx = 1200 N

3 ΣFy = 0;  Vy - 800 a b = 0  Vy = 480 N 5

4 ΣFz = 0;  Vz + 800 a b = 0  Vz = - 640 N 5 ΣMx = 0;   Tx = 0

4 ΣMy = 0;   My + 800 a b(0.2) = 0  My = - 128 N # m 5 3 ΣMz = 0;   Mz + 800a b(0.2) = 0  Mz = - 96 N # m 5

Section Properties: For the circular cross section, Fig. b,

A = pc 2 = p(0.022) = 0.4 ( 10-3 ) p m2 p p Iy = Iz = c 4 = ( 0.024 ) = 40.0 ( 10-9 ) p m4 4 4 p 4 p J = c = ( 0.024 ) = 80.0 ( 10-9 ) p m4 2 2 4(0.02) p (QB)y = y- =A= = c ( 0.022 ) d = 5.3333 ( 10-6 ) m3 3p 2 (QB)z = 0

Normal Stress: For the combined loading, the normal stress at point B can be determined from sx = sB = =

MyzB Mz yB Nx + A Iz Iy 1200 0.4 ( 10

-3

)p

-

( -96)(0) 40.0 ( 10

-9

)p

+

( - 128)(0.02) 40.0 ( 10-9 ) p

= - 19.42 ( 106 ) Pa = 19.4 MPa (C)

Ans.

Shear Stress: Since Tx = 0, the shear stress in z and y directions is contributed by VQ transverse shear stress only, which can be obtained using the shear formula, tV = . It 480 35.3333 ( 10-6 )4 Ans. = 0.5093 ( 106 ) Pa = 0.509 MPa (txy)B = (tV)y = 40.0 ( 10-9 ) p(0.04) (txz)B = (tV)z = 0 Ans. Using these results, the state of stress at point B can be represented by the volume element shown in Fig. c. 806

Ans: sB = 19.4 MPa (C), (txy)B = 0.509 MPa, (txz)B = 0

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8–70. The1834-in.-diameter 8-70. The mm-diametershaft shaft isis subjected subjected to to the loading shown. Determine the stress components at point A. Sketch the results on a volume element located at this point. The journal bearing at C can exert only force components Cy and Cz on the shaft, and the thrust bearing at D can exert force components Dx , Dy , and Dz on the shaft.

125 600 lb N in. 502mm

D

z

8 in. 200 mm 125 600 lb N 20 mm in. in. 500 502 mm

A C

B 10 mm in. 250

x

in. 2008 mm

y

20 mm in. 500

Solution = A = I

π

= (0.018 2 ) 81.0(10 −6 )π m 2 4

600 N

200 mm 600 N 500 mm

π

= (0.009 4 ) 1.64025(10 −9 )π m 4 4

500 mm

250 mm 500 mm 600 N

QA = 0 tA = 0

Ans. 600 N

σA

M yc 150(0.009) = − = − = −261.98(106 ) N/m 2 = 262 MPa (C) I 1.64025(10 −9 )π

150 N · m

Ans. 600 N

σ A = 262 MPa

Ans. tA = 0, sA = 262.0 MPa (C) 603

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8–71. Solve Prob. 8–70 for the stress components at point B.

D

z 125 600 lb N in. 502mm

8 in. 200 mm 125 600 lb N 20 in. mm 2 in. 50 mm 500

A C

Solution = A = I

B 10 in. 250 mm

x

8 in. 200 mm

y

20 in. 500 mm

π

= (0.018 2 ) 81.0(10 −6 )π m 2 4

π

= (0.009 4 ) 1.64025(10 −9 )π m 4 4

600 N

600 N 500 mm

 4(0.009)   π  (0.009 2 ) = 0.486(10 −6 ) m 3 QB = y¿A¿ =    π 3 2   

500 mm

250 mm 500 mm 600 N

sB = 0 tB =

200 mm

Ans.

VzQB It

=

600[0.486(10 −6 )] −6 = 3.144(10 = ) N/m 2 3.14 MPa 1.64025(10 −9 )π  (0.018)  

Ans.

600 N 150 N · m

600 N

τ B = 3.14 MPa

Ans. sB = 0, tB = 3.14 MPa 603

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80 400lbN

N. *8–72. The hook is subjected subjected to to the the force force of of 400 80 lb. Determine the state of stress at point A at section a–a. The cross section is mm. is circular circularand andhas hasaadiameter diameterofof12 0.5 in. Use the curved-beam formula to compute the bending stress.

1.5mm in. 40 45 45

The location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from

a A

B

B a

A dA © LA r

R =

A

0.046 m

= (0.006 2 ) 36.0(10 −6 )π m 2 A π= where

©

(

)

dA 2.4692(10 −3 ) m = 2pAr - 2r2 - c2 B = 2π 0.046 − 0.046 2 − 0.006 2 = LA r

0.04 m

Thus, 36.0(10 −6 )π = 2.4692(10 −3 )

= R

0.0458035 m

Then 0.0001965 m e = r - R = 0.046 − 0.0458035 =

Referring to Fig. b, I and QA are

4(0.006)

π

(0.006 4 ) 0.324(10 −9 )π m 4 = 4

= I

QA = 0 Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c, + ©F = 0; ; x

N–-400 80cos cos 45° ==00 N

N = 56.57 N =lb282.84 N

+ c ©Fy = 0;

80 sin 400 sin45° 45°-– V V == 00

V = 56.57 V =lb282.84 N

a + © Mo = 0;

= M 12.9552 N ⋅ m 0 M − (400 cos 45°)(0.0458035) =

400 N

The normal stress developed is the combination of axial and bending stress. Thus, s =

M(R - r) N + A Ae r

Here, M = 12.9552 N · m since it tends to reduce the curvature of the hook. For point A, r = 0.04 m. Then

σ = =

282.84 36.0(10 −6 )π

+

R = 0.0458035 m

12.9552(0.0458035 − 0.04) [36.0(10 −6 )π ](0.0001965)(0.04)

−6 87.08(10 = ) N/m 2 87.1 MPa (T)

Ans.

The shear stress in contributed by the transverse shear stress only. Thus t =

87.1 MPa

VQA = 0 It

Ans.

The state of strees of point A can be represented by the element shown in Fig. d.

604

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8–73. The hook is subjected to the force of 400 N. Determine the state of stress at point B at section a–a. The cross section has has aadiameter diameterof of12 0.5mm. in. Use the curved-beam formula to compute the bending stress.

80 lbN 400 1.5mm in. 40 45 45

The location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from R =

a A

A

B

B a

A dA © LA r

0.046 m

= (0.006 2 ) 36.0(10 −6 )π m 2 A π= where

©

(

)

dA 2.4692(10 −3 ) m = 2pAr - 2r2 - c2 B = 2π 0.046 − 0.046 2 − 0.006 2 = LA r

0.04 m

Thus, 36.0(10 −6 )π = 2.4692(10 −3 )

= R

0.0458035 m

Then 0.0001965 m e = r - R = 0.046 − 0.0458035 =

Referring to Fig. b, I and QA are = I

4(0.006)

π

= (0.006 4 ) 0.324(10 −9 )π m 4 4

 4(0.006)   π  (0.006 2 ) = 0.144(10 −6 ) m 3 QB = y¿A¿ =   3π   2 

Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c, + ©F = 0; ; x

N –-400 80 cos 45° = 00 45° =

N = 56.57 N =lb282.84 N

+ c ©Fy = 0;

80 45° 400sin sin 45°-–VV = = 00

V = 56.57 V lb = 282.84 N

a + ©Mo = 0;

400 N

0= M − (400 cos 45°)(0.0458035) = M 12.9552 N ⋅ m

The normal stress developed is the combination of axial and bending stress. Thus, s =

M(R - r) N + A Ae r

R = 0.0458035 m

Here, M = 12.9552 N · m since it tends to reduce the curvature of the hook. For point B, r = 0.04. m. Then = σ

282.84 36.0(10 −6 )π

+

12.9552(0.0458035 − 0.046) [36.0(10 −6 )π ](0.0001965)(0.046)

6 = 0.01068(10 = ) N/m 2 0.0107 MPa

Ans.

The shear stress is contributed by the transverse shear stress only. Thus, VQB 282.84[0.144(10 −6 )] = = t = It [0.324(10 −9 )π ](0.012)

0.0107 MPa 6 3.3345(10 = ) N/m 2 3.33 MPa

Ans.

The state of stress of point B can be represented by the element shown in Fig. d.

605

3.33 MPa

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R8–1. The block is subjected to the three axial loads shown. Determine the normal stress developed at points A and B. Neglect the weight of the block.

100 500 lb N

250 lb 1250 N

50 lb 250 N

100 in. 50 2 in.4mm mm

5 in. 125 mm

250in.mm

3 in. 75 mm 5 in. 125 mm

A

B

Solution Mz = −1250(0.0375) − 500(0.0375) + 250(0.1625) = −25.0 N ⋅ m −87.5 N ⋅ m My = −1250(0.1) − 250(0.05) + 500(0.1) = Iz =

1 1 (0.1)(0.3253 ) + (0.1)(0.0753 ) = 0.28958(10 −3 ) m 4 12 12

Iy =

1 1 (0.75)(0.2 3 ) + (0.25)(0.13 ) = 70.8333(10 −6 ) m 4 12 12

0.04 m 2 A = 0.1(0.325) + 0.1(0.075) =

x

500 N

5m

P M y M yz σ=− z + A Iz Iy −2000 0.04

− σA =

0.125 m 0.075 m 0.125 m

(−25.0)(0.0375) 0.28958(10 −3 )

+

−2000 0.04

(−25.0)(0.1625) −3

0.28958(10 )

+

0.1

m 0.0

5m

2000 N

1250 N 250 N

(−87.5)(0.1)

z

70.8333(10 −6 )

y

= −170.29(10 3 ) N/m 2 = 170 kPa (C)

σB = −

0.0

Ans.

(−87.5)(0.1) 70.8333(10 −6 )

3 97.74(10 = ) N/m 2 97.7 kPa (C)

Ans.

Ans. sA = 170 kPa (C), sB = 97.7 kPa (C) 606

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R8–2. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point E on the cross section of the frame at section a–a. Indicate the results on an element.

50 mm 25 mm

E

75 mm

Section a – a

1m

Solution Support Reactions: Referring to the free-body diagram of member BC shown in Fig. a, a+ ΣMB = 0;  F sin 45°(1) - 20(9.81)(2) = 0  F = 554.94 N + ΣFx = 0;   554.94 cos 45° - Bx = 0 S

D

0.5 m 0.5 m 1m a B C a

30 1m b 1m

b A

75 mm F 75 mm 25 mm Section b – b

Bx = 392.4 N

+ c ΣFy = 0;   554.94 sin 45° - 20(9.81) - By = 0  By = 196.2 N Internal Loadings: Consider the equilibrium of the free-body diagram of the right segment shown in Fig. b. + ΣFx = 0;   N - 392.4 = 0     N = 392.4 N S V = 196.2 N + c ΣFy = 0;   V - 196.2 = 0      a+ ΣMC = 0;   196.2(0.5) - M = 0     M = 98.1 N # m Section Properties: The cross-sectional area and the moment of inertia of the cross section are A = 0.05(0.075) = 3.75 1 10 - 3 2 m2

I =

1 (0.05) 1 0.0753 2 = 1.7578 1 10 - 6 2 m4 12

Referring to Fig. c, QE is

QE = y′A′ = 0.025(0.025)(0.05) = 31.25 1 10 - 6 2 m3

Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =

My N { A I

For point E, y = 0.0375 - 0.025 = 0.0125 m. Then sE =

392.4 3.75 1 10

-3

2

+

98.1(0.0125) 1.7578 1 10 - 6 2

Ans.

= 802 kPa

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus, tE =

196.2 3 31.25 1 10 - 6 2 4 VQA = = 69.8 kPa It 1.7578 1 10 - 6 2 (0.05)

Ans.

The state of stress at point E is represented on the element shown in Fig. d.

Ans: sE = 802 kPa, tE = 69.8 kPa 812

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R8–3. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point F on the cross section of the frame at section b–b. Indicate the results on an element.

50 mm 25 mm

E

75 mm

Section a – a

1m

Solution Support Reactions: Referring to the free-body diagram of the entire frame shown in Fig. a, a+ ΣMA = 0;  FBD sin 30°(3) - 20(9.81)(2) = 0  FBD = 261.6 N

30 1m b 1m

b A

75 mm F 75 mm 25 mm Section b – b

+ c ΣFy = 0;   Ay - 261.6 cos 30° - 20(9.81) = 0  Ay = 422.75 N + ΣFx = 0;     Ax - 261.6 sin 30° = 0 S

D

0.5 m 0.5 m 1m a B C a

Ax = 130.8 N

Internal Loadings: Consider the equilibrium of the free-body diagram of the lower cut segment, Fig. b. + ΣFx = 0;  130.8 - V = 0 S

V = 130.8 N

+ c ΣFy = 0;  422.75 - N = 0

N = 422.75 N

  M = 130.8 N # m

a+ ΣMC = 0;   130.8(1) - M = 0

Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the cross section are A = 0.075(0.075) = 5.625 1 10 - 3 2 m2

I =

1 (0.075) 1 0.0753 2 = 2.6367 1 10 - 6 2 m4 12

Referring to Fig. c, QE is

QF = y′A′ = 0.025(0.025)(0.075) = 46.875 1 10 - 6 2 m3

Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s =

My N { A I

For point F, y = 0.0375 - 0.025 = 0.0125 m. Then sF =

- 422.75

5.625 1 10

-3

2

-

130.8(0.0125) 2.6367 1 10 - 6 2

Ans.

= - 695.24 kPa = 695 kPa (C) 

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus, 130.8c 46.875 1 10 - 6 2 d VQA tA = = = 31.0 kPa It 2.6367 1 10 - 6 2 (0.075)

Ans.

The state of stress at point A is represented on the element shown in Fig. d.

Ans: sF = 695 kPa (C), tA = 31.0 kPa 813

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*R8–4. The gondola and passengers have a weight of 7.5 kN and center of gravity at G. The suspender arm AE has a square cross-sectional area of 40 mm by 40 mm, and is pin connected at its ends A and E. Determine the largest tensile stress developed in regions AB and DC of the arm.

0.4 m E

D 1.2 m

40 mm B

C

40 mm

1.7 m

A

G G

Solution Segment AB: PAB 7500 6 4.6875(10 = ) N/m 2 4.69 MPa (smax)AB = = = A 0.04(0.04)

Ans.

Segment CD: 0.4 m

sa =

PCD 7500 6 = = 4.6875(10 ) N/m 2 4.6875 MPa = A 0.04(0.04)

sb =

3000(0.02) Mc 6 = 281.25(10 ) N/m 2 281.25 MPa == 1 (0.04)(0.04 3 ) I

3000 n . m

12

(σ max )CD = σ a + σ b = 4.6875 + 281.25 = 285.9375 = MPa 286 MPa

Ans.

Ans: (smax)AB = 4.69 MPa, (s max )CD = 286 MPa 826

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R8–5. If the cross section of the femur at section a–a can be approximated as a circular tube as shown, determine the maximum normal stress developed on the cross section at section a–a due to the load of 375 N.

50 mm 2 in.

375lbN 75

aa

aa

12 mm 0.5 in. 241mm in. Section aa –– aa

F

Solution Internal Loadings: Considering the equilibrium for the free-body diagram of the femur’s upper segment, Fig. a, + c ©Fy = 0;

N –-375 75 ==00

a + ©MO = 0;

= M 18.75 N ⋅ m M − 375(0.05) = 0

M

375 N 50 mm

N 75 lb N ==375 N

Section Properties: The cross-sectional area, the moment of inertia about the centroidal axis of the femur’s cross section are A = π (0.024 2 − 0.012 2 ) = 0.432(10 −3 )π m 2 I=

π 4

(0.024 2 − 0.012 2 )= 77.76(10 −9 )π m 4

Normal Stress: The normal stress is a combination of axial and bending stress. Thus, s =

My N + A I

By inspection, the maximum normal stress is in compression. = σ max

−375 −3

0.432(10 )π

−

18.75(0.024) 77.76(10 −9 )π

= −2.118(106 ) N/m 2 = 2.12 MPa (C)

Ans.

Ans. sC = 93.7 MPa (T), tC = 0, sD = 187 MPa (C), tD = 0 613

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R8–6. A bar having a square cross section of 30 mm by 30 mm is 2 m long and is held upward. If it has a mass of 5 kg/m, determine the largest angle u, measured from the vertical, at which it can be supported before it is subjected to a tensile stress along its axis near the grip. u

2m

Solution A = 0.03(0.03) = 0.9(10 - 3) m2 I =

1 (0.03)(0.033) = 67.5(10 - 9) m4 12

Require sA = 0 sA = 0 = 0 =

P Mc + A I

98.1 sin u(0.015) - 98.1 cos u + -3 67.5(10 - 9) 0.9(10 )

0 = - 1111.11 cos u + 222222.22 sin u tan u = 0.005;

Ans.

u = 0.286°

Ans: u = 0.286° 816

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R8–7. The wall hanger has a thickness of 6 mm and is used to support the vertical reactions of the beam that is loaded as shown. If the load is transferred uniformly to each strap of the hanger, determine the state of stress at points C and D on the strap at A. Assume the vertical reaction F at this end acts in the center and on the edge of the bracket as shown.

50 kN 30 kN/m A

B

1.8 m

0.6 m 0.6 m 50 mm 50 mm 50 mm 90 mm 65 mm 75 mm

Solution a + ©MB = 0 ;

F D

C

25 mm

54(0.9) + 50(2.4) − FA (3) = 0

25 mm

FA = 56.2 kN I

1 3  4 −6 2=  12 (0.006)(0.05 ) 0.125(10 ) m

= A 2(0.006)(0.05) = 0.6(10 −3 ) m 2

At point C ,

σ C=

P 2[28.1(10 3 )] 2 93.7 MPa (T) = = 93.67(106 ) N/m= A 0.6(10 −3 )

Ans.

tC = 0

Ans.

At point D, P A

σD = −

Mc 2[28.1(10 3 )] 2[0.7025(10 3 )](0.025) − = I 0.125(10 −6 ) 0.6(10 −3 ) 187 MPa (C) = −187.33(106 ) N/m 2 =

tD = 0

Ans. Ans.

Ans: sC = 93.7 MPa (T), tC = 0, sD = - 187 MPa (C), tD = 0 832

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*R8–8. The wall hanger has a thickness of 6 mm and is used to support the vertical reactions of the beam that is loaded as shown. If the load is transferred uniformly to each strap of the hanger, determine the state of stress at points C and D on the strap at B. Assume the vertical reaction F at this end acts in the center and on the edge of the bracket as shown.

50 kN 30 kN/m A

B

1.8 m

0.6 m 0.6 m 50 mm 50 mm 50 mm 90 mm 65 mm 75 mm

Solution a + © MA = 0 ;

F D

C

25 mm

0 −54(2.1) − 50(0.6) + FB (3) =

25 mm

F B = 47.8 kN A = 2(0.006)(0.05) = 0.6(10 −3 ) m 2 I

1 3  4 −6 2=  12 (0.006)(0.05 ) 0.125(10 ) m

At point C,

σ C=

P 2[23.9(10 3 )] 2 79.7 MPa (T) = = 79.67(106 ) N/m= A 0.6(10 −3 )

Ans.

tC = 0

Ans.

At point D, P A

σD = −

Mc 2[23.9(10 3 )] 2[0.5975(10 3 )](0.025) = − I 0.125(10 −6 ) 0.6(10 −3 ) 159 MPa (C) = −159.33(106 ) N/m 2 =

Ans.

tD = 0

Ans.

Ans: sC = 79.7 MPa (T), tC = 0, sD = 159 MPa (C), tD = 0 832

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9–1. Prove that the sum of the normal stresses sx + sy = sx′ + sy′ is constant. See Figs. 9–2a and 9–2b.

Solution Stress Transformation Equations: Applying Eqs. 9–1 and 9–3 of the text. sx′ + sy′ =

sx + sy 2

+ +

sx - sy 2 sx + sy 2

cos 2u + txy sin 2u -

sx - sy 2

cos 2u - txy sin 2u (Q. E. D.)

sx′ + sy′ = sx + sy

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

819

Ans: N/A

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9–2. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

A

65 MPa

30

B 20 MPa

Solution b + ΣFx′ = 0; sx′ ∆A + 20∆A sin 30° cos 30° + 20∆A cos 30° cos 60° - 65∆A cos 30° cos 30° = 0 Ans.

sx′ = 31.4 MPa R + ΣFy′ = 0;

tx′y′ ∆A + 20∆A sin 30° sin 30° - 20∆A cos 30° sin 60° - 65∆A cos 30° sin 30° = 0 Ans.

tx′y′ = 38.1 MPa

Ans: sx′ = 31.4 MPa, tx′y′ = 38.1 MPa 820

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9–3. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

A

ksi 8 MPa

ksi 2 2MPa ksi 5 MPa 60 B

Solution Referring to Fig a, if we assume that the areas of the inclined plane AB is ¢A, then the area of the horizontal and vertical of the triangular element are ¢A cos 60° and ¢A sin 60° respectively. The forces act acting on these two faces indicated on the FBD of the triangular element, Fig. b. +Q©Fx¿ = 0;

¢Fx¿ + 2¢A sin 60° cos 60° + 5¢ A sin 60° sin 60° + 2¢A cos 60° sin 60° - 8¢A cos 60° cos 60° = 0 ¢Fx¿ = - 3.482 ¢A

+a©Fy¿ = 0;

¢Fy¿ + 2¢A sin 60° sin 60° - 5¢ A sin 60° cos 60° - 8¢A cos 60° sin 60° - 2¢A cos 60° cos 60° = 0 ¢Fy¿ = 4.629 ¢A

From the definition, sx¿ = lim¢A:0

¢Fx¿ = - 3.48 MPa ksi ¢A

tx¿y¿ = lim¢A:0

¢Fy¿ ¢A

Ans.

= 4.63 MPa ksi

Ans.

The negative sign indicates that sx¿, is a compressive stress.

Ans. sx′ = -3.48 MPa, tx′y′ = 4.63 MPa 620

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*9–4. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

90 MPa

A

35 MPa 60 30

B 50 MPa

Solution R+ ©Fy¿ = 0

¢Fy¿ - 50¢A sin 30° cos 30° - 35¢A sin 30° cos 60° + 90¢A cos 30° sin 30° + 35¢A cos 30° sin 60° = 0 ¢Fy¿ = - 34.82¢A

b+ ©Fx¿ = 0

¢Fx¿ - 50¢A sin 30° sin 30° + 35¢A sin 30° sin 60° -90¢A cos 30° cos 30° + 35¢A cos 30° cos 60° = 0 ¢Fx¿ = 49.69 ¢A

sx¿ = lim¢A:0

¢Fx¿ = 49.7 MPa ¢A

tx¿y¿ = lim¢A:0

¢Fy¿ ¢A

Ans.

= - 34.8 MPa

Ans.

The negative signs indicate that the sense of sx¿, and tx¿y¿ are opposite to the shown on FBD.

Ans. sx′ = 49.7 MPa, tx′y′ = -34.8 MPa 623

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9–5. Solve Prob. 9–6 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch.

90 MPa

A

35 MPa 60 30

Solution sx = 90 MPa sx¿ = =

sx + sy 2

sy = 50 MPa +

sx - sy 2

txy = - 35 MPa

B 50 MPa

u = - 150°

cos 2u + txy sin 2u

90 - 50 90 + 50 + cos( - 300°) + ( -35) sin ( - 300°) 2 2

= 49.7 MPa tx¿y¿ = -

sx - sy

= -a

2

Ans. sin 2u + txy cos 2u

90 - 50 b sin( - 300°) + ( -35) cos ( -300°) = - 34.8 MPa 2

Ans.

The negative sign indicates tx¿y¿ acts in - y¿ direction.

Ans. sx′ = 49.7 MPa, tx′y′ = -34.8 MPa 623

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psi 500 MPa

9–6. The state of stress at a point in a member is shown on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

B 60

A

350MPa psi 350

Solution Referring to Fig. a, if we assume that the area of the inclined plane AB is ¢A, then the areas of the horizontal and vertical surfaces of the triangular element are ¢A sin 60° and ¢A cos 60° respectively. The force acting on these two faces are indicated on the FBD of the triangular element, Fig. b +R©Fx¿ = 0;

¢Fx¿ + 500 ¢A sin 60° sin 60° + 350¢A sin 60° cos 60° + 350¢A cos 60° sin 60° = 0 ¢Fx¿ = - 678.11 ¢A

+Q©Fy¿ = 0;

¢Fy¿ + 350¢A sin 60° sin 60° - 500¢A sin 60° cos 60° -350¢A cos 60° cos 60° = 0 ¢Fy¿ = 41.51 ¢A

From the definition sx¿ = lim¢A:0 tx¿y¿ = lim¢A:0

¢Fx¿ = -678 MPa ¢A ¢Fy¿ ¢A

Ans.

= 41.5 psi MPa

Ans.

The negative sign indicates that sx¿, is a compressive stress.

Ans. sx′ = -678 MPa, tx′y′ = 41.5 MPa 621

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9–7. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

60 MPa

A

80 MPa 50 40 MPa

Solution

B

Force Equilibrium: The areas, thus the forces acting on the faces of the lower segment of the sectioned element are shown in Fig. a a + ΣFy′ = 0;

∆Fy′ + (40 ∆A sin 50°) sin 50° - (80 ∆A sin 50°) cos 50° - (40 ∆A cos 50°) cos 50° - (60 ∆A cos 50°) sin 50° = 0   ∆Fy′ = 61.99 ∆A

+ QΣFx′ = 0;

∆Fx′ + (40 ∆A sin 50°) cos 50° + (80 ∆A sin 50°) sin 50° + (40 ∆A cos 50°) sin 50° - (60 ∆A cos 50°) cos 50° = 0   ∆Fx′ = - 61.54 ∆A

Normal And Shear Stresses: For the inclined plane that has an area of ∆A, sx′ = lim S ∆A

0

tx′y′ = lim S ∆A

∆Fx′ = - 61.54 MPa = -61.5 MPa ∆A ∆Fy′ 0

∆A

= 61.99 MPa = 62.0 MPa

Ans. Ans.

The negative sign indicates that sx′ is a compressive normal stress.

Ans: sx′ = - 61.5 MPa, tx′y′ = 62.0 MPa 825

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*9–8. Solve Prob. 9–7 using the stress transformation equations developed in Sec. 9.2.

60 MPa

A

80 MPa 50 40 MPa

Solution

B

Normal And Shear Stress: In accordance with the established sign conventions, u = + 40° (Fig. a)  sx = - 80 MPa  sy = 60 MPa  txy = -40 MPa Stress Transformation Equations: sx′ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

- 80 + 60 - 80 - 60 + cos 80° + ( -40) sin 80° 2 2 Ans.

= - 61.54 MPa = -61.5 MPa tx′y′ = -

sx - sy

= -a

2

sin 2u + txy cos 2u

- 80 - 60 b sin 80° + ( - 40) cos 80° 2

Ans.

= 61.99 MPa = 62.0 MPa

These results are indicated on the sectioned element shown in Fig. b

Ans: sx′ = - 61.5 MPa, tx′y′ = 62.0 MPa 826

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9–9. Determine the stress components acting on the plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

A 80 MPa

30

30 MPa

40 MPa

Solution Force Equilibrium: The areas, thus the forces acting on the faces of the upper segment of the sectioned element are shown in Fig. a + T ΣFy′ = 0;

B

∆Fy′ + (80 ∆A sin 30°) cos 30° + (30 ∆A sin 30°) sin 30° - (30 ∆A cos 30°) cos 30° + (40 ∆A cos 30°) sin 30° = 0     ∆Fy′ = - 36.96 ∆A

+ d ΣFx′ = 0;

∆Fx′ - (30 ∆A sin 30°) cos 30° + (80 ∆A sin 30°) sin 30° - (30 ∆A cos 30°) sin 30° - (40 ∆A cos 30°) cos 30° = 0     ∆Fx′ = 35.98 ∆A

Normal And Shear Stress: For the inclined plane that has an area of ∆A, sx′ = lim = S ∆A

0

tx′y′ = lim = S ∆A

0

∆Fx′ = 35.98 MPa = 36.0 MPa ∆A ∆Fy′ ∆A

= - 36.96 MPa = - 37.0 MPa

Ans. Ans.

The negative sign indicates that tx′y′ acts in the sense opposite to that shown in the FBD.

Ans: sx′ = 36.0 MPa, tx′y′ = - 37.0 MPa 827

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9–10. Solve Prob. 9–9 using the stress transformation equation developed in Sec. 9.2.

A 80 MPa

30

30 MPa

40 MPa

Solution Normal And Shear Stress: In accordance with the established sign conventions,

B

u = + 120° (Fig. a)  sx = - 80 MPa  sy = 40 MPa  txy = -30 MPa Stress Transformation Equations: sx′ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

- 80 + 40 - 80 - 40 + a b cos 240° + ( -30) sin 240° 2 2

Ans.

= 35.98 MPa = 36.0 MPa tx′y′ = -

sx - sy

= -a

2

sin 2u + txy cos 2u

- 80 - 40 b sin 240° + ( -30) cos 240° 2

Ans.

= - 36.96 MPa = -37.0 MPa

The negative sign indicates that tx′y′ acts in the negative y′ direction. These results are indicated on the sectioned element shown in Fig. b

Ans: sx′ = 36.0 MPa, tx′y′ = -37.0 MPa 828

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9–11. Determine the equivalent state of stress on an element at the same point oriented 60° clockwise with respect to the element shown. Sketch the results on the element.

100 MPa 75 MPa 150 MPa

Solution Stress Transformation Equations: u = - 60° (Fig. a)  sx = 150 MPa  sy = 100 MPa  txy = 75 MPa We obtain, sx′ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

150 + 100 150 - 100 + cos ( - 120°) + 75 sin ( -120°) 2 2 Ans.

= 47.5 MPa sy′ = =

sx + sy 2

-

sx - sy 2

cos 2u - txy sin 2u

150 + 100 150 - 100 cos ( - 120°) - 75 sin ( -120°) 2 2 Ans.

= 202 MPa tx′y′ = = -

sx - sy 2

sin 2u + txy cos 2u

150 - 100 sin ( - 120°) + 75 cos ( -120°) 2 Ans.

= - 15.8 MPa

The negative sign indicates that tx′y′ is directed towards the negative sense of the y′ axis. These results are indicated on the element shown in Fig. b.

Ans: sx′ = 47.5 MPa, sy′ = 202 MPa, tx′y′ = -15.8 MPa 829

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*9–12. Determine the equivalent state of stress on an element at the same point oriented 60° counterclockwise with respect to the element shown. Sketch the results on the element.

100 MPa 75 MPa 150 MPa

Solution Stress Transformation Equations: u = + 60° (Fig. a)  sx = 150 MPa  sy = 100 MPa  txy = 75 MPa We obtain, sx′ = =

sx + sy 2

sx - sy

+

2

cos 2u + txy sin 2u

150 + 100 150 - 100 + cos 120° + 75 sin 120° 2 2 Ans.

= 177 MPa sy′ = =

sx + sy 2

sx - sy

-

2

cos 2u - txy sin 2u

150 + 100 150 - 100 cos 120° - 75 sin 120° 2 2 Ans.

= 72.5 MPa tx′y′ = = -

sx - sy 2

sin 2u + txy cos 2u

150 - 100 sin 120° + 75 cos 120° 2

= - 59.2 MPa



Ans.

The negative sign indicates that tx′y′ is directed towards the negative sense of the y′ axis. These results are indicated on the element shown in Fig. b.

Ans: sx′ = 177 MPa, sy′ = 72.5 MPa, tx′y′ = -59.2 MPa 830

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9–13. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

50 MPa

A

100 MPa 60

Solution

B

Force Equilibrium: The areas, thus the forces acting on the faces of the lower segment of the sectioned element, are shown in Fig. a. + bΣFy′ = 0;

∆Fy′ + (100 ∆A sin 60°) cos 60° + (50 ∆A cos 60°) sin 60° = 0   ∆Fy′ = - 64.95 ∆A

a + ΣFx′ = 0;

∆Fx′ + (100 ∆A sin 60°) sin 60° - (50 ∆A cos 60°) cos 60° = 0   ∆Fx′ = - 62.5 ∆A

Normal And Shear Stress: For the incline plane that has an area of ∆A, sx′ = lim S ∆A

0

tx′y′ = lim S ∆A

∆Fx′ = - 62.5 MPa = -62.5 MPa ∆A ∆Fy′ 0

∆A

= - 64.95 MPa = -65.0 MPa

Ans. Ans.

The negative signs indicate that sx′ is a compressive normal stress and tx′y′ acts in the sense opposite to that shown in the FBD.

Ans: sx′ = - 62.5 MPa, tx′y′ = - 65.0 MPa 831

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9–14. Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case.

200 MPa

100 MPa 300 MPa

Solution Normal And Shear Stress: In accordance with the established sign conventions, sx = 300 MPa  sy = - 200 MPa  txy = 100 MPa a) In-Plane Principal Stresses: s1,2 = =

sx + sy

{

2

B

300 + ( -200)

a

sx - sy

{

2

= 50 { 269.26

2 B

c

2

b + txy2

300 - ( - 200) 2

2

d + 1002

s1 = 319.26 MPa = 319 MPa  s2 = - 219.26 MPa = - 219 MPa

Ans.

Orientation of Principal Plane: tan 2up =

txy

=

(sx - sy)>2

100 = 0.4 [300 - ( - 200)]>2

up = 10.90° and - 79.10° Substitute the result of u = 10.90° sx′ = =

sx + sy 2

+

sx - sy 2

300 + ( -200) 2

+

cos 2u + txy sin 2u

300 - ( -200) 2

cos 21.80° + 100 sin 21.80°

= 319.26 MPa = s1 Hence,

Ans.

up1 = 10.90° = 10.9°  up2 = - 79.10° = - 79.1°

Using these results, the state of in-plane principal stress can be represented by the differential element shown in Fig. a. b) Maximum In-Plane Shear Stress: tmax

in@plane

= =

B

a

B

c

sx - sy 2

2

b + txy2

300 - ( - 200) 2

2

d + 1002 = 269.26 MPa = 269 MPa

832

Ans.

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9–14. Continued

Orientation of the Plane for Maximum In-Plane Shear Stress: tan 2us =

- ( sx - sy ) >2 txy

=

- [300 - ( - 200)]>2 100

= -2.5 Ans.

us = - 34.10° = - 34.1° and 55.90° = 55.9° Average Normal Stress: savg =

sx + sy 2

=

300 + ( - 200) 2

Ans.

= 50.0 MPa

By observing the segment of the element of principal stresses sectioned through the diagonal, Fig. b, equilibrium along y′ axis requires that tmax to act in the direction in@plane

shown. Thus, the state of maximum in-plane shear stress can be represented by the differential element shown in Fig. c.

Ans: s1 = 319 MPa, s2 = -219 MPa, up1 = 10.9°, up2 = -79.1°, tmax = 269 MPa, in@plane

us = -34.1° and 55.9°, savg = 50.0 MPa 833

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9–15. The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case.

60 MPa

30 MPa 45 MPa

Solution sy = - 60 MPa

sx = 45 MPa a)

sx + sy

s1,2 =

{

2

B

a

txy = 30 MPa

sx - sy 2

2

b + txy2

45 - ( - 60) 2 45 - 60 { a b + (30)2 2 B 2

=

Ans.

s2 = - 68.0 MPa

s1 = 53.0 MPa

Orientation of principal stress: txy 30 tan 2up = = 0.5714 = ( sx - sy ) >2 (45 - ( -60))>2

- 75.13

up = 14.87,

Use Eq. 9–1 to determine the principal plane of s1 and s2: sx= =

sx + sy

=

2

+

sx - sy 2

45 + ( -60)

+

2

cos 2u + txy sin 2u,

45 - ( -60) 2

where u = 14.87°

cos 29.74° + 30 sin 29.74° = 53.0 MPa Ans.

Therefore up1 = 14.9° and up2 = - 75.1° b)  t max

in@plane

=

B

a

sx - sy 2

= 60.5 MPa savg =

sx + sy 2

=

2

b + txy2 =

45 + ( -60) 2

B

a

45 - ( - 60) 2

2

b + 302

Ans. Ans.

= - 7.50 MPa

Orientation of maximum in-plane shear stress: tan 2us =

- ( sx - sy ) >2 txy

=

- (45 - ( -60))>2 30

= - 1.75 Ans.

us = - 30.1° and us = 59.9°

By observation, in order to preserve equilibrium along AB, tmax has to act in the direction shown. Ans: s1 = 53.0 MPa, s2 = -68.0 MPa, up1 = 14.9°, up2 = -75.1°, tmax = 60.5 MPa, in@plane

savg = -7.50 MPa, us = -30.1°, us = 59.9° 834

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*9–16. Determine the equivalent state of stress on an element at the point which represents (a) the principal stresses and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown and sketch the results on the element.

50 MPa 15 MPa

Solution Normal and Shear Stress: sx = 50 MPa

sy = 0

txy = -15 MPa

In-Plane Principal Stresses: sx + sy

s1, 2 =

2

{

B

a

sx - sy 2

2

2 b + txy

50 + 0 50 - 0 2 2 { a b + ( - 15) 2 B 2

=

= 25 { 2850

Ans.

s2 = - 4.15 MPa

s1 = 54.2 MPa

Orientation of Principal Plane: tan 2up =

txy (sx - sy)>2

=

- 15 = -0.6 (50 - 0)>2

up = -15.48° and 74.52° Substitute u = - 15.48° into sx′ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

50 + 0 50 - 0 + cos ( - 30.96°) + ( -15) sin ( -30.96°) 2 2

= 54.2 MPa = s1

Thus,

Ans.

(up)1 = - 15.5° and (up)2 = 74.5° The element that represents the state of principal stress is shown in Fig. a. Maximum In-Plane Shear Stress: tmax

in@plane

=

B

a

sx - sy 2

2

2 b + txy =

50 - 0 2 2 b + ( -15) = 29.2 MPa 2 B a

835

Ans.

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*9–16.  Continued

Orientation of the Plane of Maximum In-Plane Shear Stress: Solution tan 2us =

- (sx - sy)>2 txy

=

- (50 - 0)>2 - 15

= 1.667 Ans.

us = 29.5° and 120°

has to act in the same sense shown in Fig. b to maintain By inspection, tmax in@plane equilibrium. Average Normal Stress: savg =

sx + sy 2

=

50 + 0 = 25 MPa 2

Ans.

The element that represents the state of maximum in-plane shear stress is shown in Fig. c.

Ans: s1 = s2 = (up)1 (up)2 tmax

54.2 MPa, -4.15 MPa, = -15.5°, = 74.5°, = 29.2 MPa,

in@plane

us = 29.5° and 120°, savg = 25 MPa 836

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9–17. Determine the equivalent state of stress on an element at the same point which represents (a) the principal stress, and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown and sketch the results on the element.

75 MPa

125 MPa

50 MPa

Solution Normal and Shear Stress:

sy = - 75 MPa

sx = 125 MPa

txy = -50 MPa

In - Plane Principal Stresses: s1,2 = =

sx - sy

{

2

B

125 + ( -75) 2

a

sx - sy 2

{

B

= 25 { 212500

a

2

b + txy 2

125 - ( - 75) 2

2

b + ( -50)2 Ans.

s2 = - 86.8 MPa

s1 = 137 MPa

Orientation of Principal Plane: txy - 50 tan 2uP = = -0.5 = ( sx - sy ) >2 (125-( - 75))>2 up = -13.28° and 76.72°

Substitute u = - 13.28° into sx′ = = Thus,

sx + sy 2

+

sx - sy 2

125 + ( -75) 2

+

cos 2u + txy sin 2u

125 - ( - 75) 2

cos ( - 26.57°) + ( -50) sin ( -26.57°)

= 137 MPa = s1

( up ) 1 = - 13.3° and ( up ) 2 = 76.7°

Ans.

125 - ( - 75)>( -50) The element that represents the state of principal stress is shown in Fig. a. Maximum In - Plane Shear Stress: t max

in@plane

= C

¢

sx - sy 2

2

≤ + txy 2 = Ca

125 - ( -75) 2

2

b + 502 = 112 MPa Ans.

837

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9–17.  Continued

Orientation of the Plane of Maximum In - Plane Shear Stress: Solution tan 2us = -

( sx - sy ) >2 txy

= -

(125 - ( - 75))>2 - 50

= 2 Ans.

us = 31.7° and 122°

has to act in the same sense shown in Fig. b to maintain By inspection, t max in@plane equilibrium. Average Normal Stress: savg =

sx + sy 2

=

125 + ( -75) 2

Ans.

= 25 MPa

The element that represents the state of maximum in-plane shear stress is shown in Fig. c.

Ans: s1 = 137 MPa, s2 = -86.8 MPa, up1 = -13.3°, up2 = 76.7°, tmax

in@plane

= 112 MPa,

us = 31.7° and 122°, savg = 25 MPa 838

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9–18.  A point on a thin plate is subjected to the two stress components. Determine the resultant state of stress represented on the element oriented as shown on the right.

85 MPa 45



sy

60 MPa

txy 30



sx

85 MPa

Solution For element a: sx = sy = 85 MPa  txy = 0  u = - 45° (sx′)a = = (sy′)a = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

85 + 85 85 - 85 + cos ( - 90°) + 0 = 85 MPa 2 2 sx + sy 2

+

sx - sy 2

cos 2u - txy sin 2u

85 + 85 85 - 85 cos ( - 90°) - 0 = 85 MPa 2 2

(tx′y′)a = = -

sx - sy 2

sin 2u + txy cos 2u

85 - 85 sin ( - 90°) + 0 = 0 2

For element b: sx = sy = 0  txy = 60 MPa  u = - 60° (sx′)b =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

= 0 + 0 + 60 sin ( - 120°) = - 51.96 MPa (sy′)b =

sx + sy 2

-

sx - sy 2

cos 2u - txy sin 2u

= 0 - 0 - 60 sin ( - 120°) = 51.96 MPa (tx′y′)b = = -

sx - sy 2

sin 2u - txy cos 2u

85 - 85 sin ( -120°) + 60 cos ( - 120°) = - 30 MPa 2

sx = (sx′)a + (sx′)b = 85 + ( - 51.96) = 33.0 MPa

Ans.

sy = (sy′)a + (sy′)b = 85 + 51.96 = 137 MPa

Ans.

txy = (tx′y′)a + (tx′y′)b = 0 + ( - 30) = -30 MPa

Ans.

Ans: sx = 33.0 MPa, sy = 137 MPa, txy = - 30 MPa 839

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9–19.  Determine the equivalent state of stress on an element at the same point which represents (a) the principal stress, and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown and sketch the results on the element.

25 MPa 100 MPa

Solution Normal and Shear Stress: sx = -100 MPa

sy = 0

txy = 25 MPa

In-Plane Principal Stresses: s1, 2 = =

sx + sy

{

2

a B

sx - sy 2

2

b + txy2

- 100 + 0 - 100 - 0 2 2 { b + 25 a 2 B 2

= -50 { 23125

Ans.

s1 = 5.90 MPa  s2 = - 106 MPa Orientation of Principal Plane: tan 2up =

txy (sx - sy)>2

=

25 = - 0.5 ( - 100 - 0)>2

up = -13.28° and 76.72° Substitute u = - 13.28° into sx′ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

- 100 + 0 - 100 - 0 + cos ( - 26.57°) + 25 sin ( - 26.57°) 2 2

= -106 MPa = s2 Thus, Ans.

(up)1 = 76.7° and (up)2 = - 13.3° The element that represents the state of principal stress is shown in Fig. a. Maximum In-Plane Shear Stress: tmax

in@plane

=

B

a

sx - sy 2

2

b + txy2 =

B

a

- 100 - 0 2 2 b + 25 = 55.9 MPa Ans. 2

840

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9–19.  Continued

Orientation of the Plane of Maximum In-Plane Shear Stress: Solution tan 2us = -

(sx - sy)>2 txy

= -

( -100 - 0)>2 25

= 2 Ans.

us = 31.7° and 122°

By inspection, tmax  has to act in the same sense shown in Fig. b to maintain in@plane equilibrium. Average Normal Stress: savg =

sx + sy 2

=

- 100 + 0 = - 50 MPa 2

Ans.

The element that represents the state of maximum in-plane shear stress is shown in Fig. c.

Ans: s1 = 5.90 MPa, s2 = -106 MPa, up1 = 76.7° and up2 = -13.3°, tmax = 55.9 MPa, savg = - 50 MPa, in@plane

us = 31.7° and 122° 841

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*9–20. The stress along two planes at a point is indicated. Determine the normal stresses on plane b–b and the principal stresses.

b

45 MPa a

a 60

σb

Solution tb = - 25 = -

sx - sy 2

25 MPa b

sin 2u + txy cos 2u

(sx - 0)

sin ( - 300°) + 45 cos ( -300°) 2 sx = 109.70 MPa sx + sy

sb = =

2

sx - sy

+

2

cos 2u + txy sin 2u

109.70 - 0 109.70 + 0 + cos ( - 300°) + 45 sin ( -300°) 2 2 Ans.

sb = 121 MPa

s1,2 = =

sx + sy 2

{

B

a

sx - sy 2

2

b + txy 2

109.70 + 0 109.70 - 0 2 2 { a b + (45) 2 A 2

s1 = 126 MPa

Ans.

s2 = -16.1 MPa

Ans.

Ans: sb = 121 MPa, s1 = 126 MPa, s2 = -16.1 MPa 842

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b

9–21. The stress acting on two planes at a point is indicated. Determine the shear stress on plane a–a and the principal stresses at the point.

a

tta 45

ksi 6060MPa 60

80 MPa ksi 80 90 a

b

Solution ksi sx = 60 sin 60° = 51.962 MPa MPa txy = 60 cos 60° = 30 ksi sa = 80 =

sx + sy 2

sx - sy

+

2

51.962 + sy 2

+

cos 2u + txy sin 2u

51.962 - sy 2

cos (90°) + 30 sin (90°)

MPa sy = 48.038 ksi ta = - a = -a

sx - sy 2

b sin 2u + txy cos u

51.962 - 48.038 b sin (90°) + 30 cos (90°) 2

ta = - 1.96 MPa ksi s1, 2 = =

sx + sy 2

;

C

Ans. a

sx - sy 2

2

b + t2xy

51.962 - 48.038 2 51.962 + 48.038 ; a b + (30)2 2 C 2

ksi s1 = 80.1 MPa

Ans.

MPa s2 = 19.9 ksi

Ans.

51.962 MPa

30 MPa

Ans. sx = 61.962 MPa, txy = 30 MPa s1 = 80.1 MPa, s2 = 19.9 MPa 639

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9–22.  The state of stress at a point in a member is shown on the element. Determine the stress components acting on the plane AB.

A 50 MPa

30

28 MPa

100 MPa

Solution

B

Construction of the Circle: In accordance with the sign convention, sx = -50 MPa, sy = -100 MPa, and txy = - 28 MPa. Hence, savg =

sx + sy 2

=

- 50 + ( -100) 2

= - 75.0 MPa

The coordinates for reference points A and C are A(–50, –28) and C(–75.0, 0). The radius of the circle is R = 2(75.0 - 50)2 + 282 = 37.54 MPa.

Stress on the Rotated Element: The normal and shear stress components ( sx′ and tx′y′ ) are represented by the coordinates of point P on the circle sx′ = - 75.0 + 37.54 cos 71.76° = - 63.3 MPa

Ans.

tx′y′ = 37.54 sin 71.76° = 35.7 MPa

Ans.

Ans: sx′ = -63.3 MPa, tx′y′ = 35.7 MPa 844

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9–23.  The grains of wood in the board make an angle of 20° with the horizontal as shown. Determine the normal and shear stress that act perpendicular and parallel to the grains if the board is subjected to an axial load of 250 N.

300 mm

60 mm 250 N

250 N 20

25 mm

Solution sx =

P 250 = = 166.67 kPa A (0.06)(0.025)

sy = 0

txy = 0

u = 70° sx′ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

166.67 + 0 166.67 - 0 + cos 140° + 0 = 19.5 kPa 2 2

tx′y′ = - a

sx - sy 2

Ans.

b sin 2u + txy cos 2u

166.67 - 0 = -a b sin 140° + 0 = -53.6 kPa 2

Ans.

Ans: sx′ = 19.5 kPa, tx′y′ = - 53.6 kPa 845

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*9–24. The wood beam is subjected to a load of 12 kN. If grains of wood in the beam at point A make an angle of 25° with the horizontal as shown, determine the normal and shear stress that act perpendicular to the grains due to the loading.

12 kN 1m

2m A 25

4m 300 mm

75 mm

200 mm

Solution I =

1 (0.2)(0.3)3 = 0.45 ( 10 - 3 ) m4 12

QA = yA′ = 0.1125(0.2)(0.075) = 1.6875 ( 10 - 3 ) m3 sA =

13.714 ( 103 ) (0.075) MyA = = 2.2857 MPa (T) I 0.45 ( 10 - 3 )

tA =

6.875 ( 103 ) (1.6875) ( 10 - 3 ) VQA = = 0.1286 MPa It 0.45 ( 10 - 3 ) (0.2)

sx = 2.2857 MPa s1, 2 =

=

sx + sy

{

2

txy = - 0.1286 MPa

sy = 0

C

a

sx - sy 2

2

b + txy 2

2.2857 + 0 2.2857 - 0 2 b + ( - 0.1286)2 { a 2 C 2

s1 = 2.29 MPa

Ans.

s2 = - 7.20 kPa

Ans.

tan 2up =

txy (sx - sy)>2

=

- 0.1286 (2.2857 - 0)>2 Ans.

up = - 3.21° Check direction of principal stress: sx′ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

2.2857 + 0 2.2857 - 0 + cos ( - 6.42°) - 0.1285 sin ( -6.42) 2 2

= 2.29 MPa

Ans: s1 = 2.29 MPa, s2 = -7.20 kPa, up = - 3.21° 846

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9–25.  The internal loadings at a section of the beam are shown. Determine the in-plane principal stresses at point A. Also compute the maximum in-plane shear stress at this point.

20 mm A

200 mm 20 mm 20 mm y B

C

Solution

100 mm 10 kNm

Section Properties: For the wide flange section, Fig. a,

60 kN 80 kN

z

2

A = 0.1(0.24) - 0.08(0.2) = 0.008 m 1 1 Iz = (0.1) ( 0.243 ) (0.08) ( 0.23 ) = 61.8667 ( 10-6 ) m4 12 12 1 1 Iy = 2 c (0.02) ( 0.13 ) d + (0.2) ( 0.023 ) = 3.4667 ( 10-6 ) m4 12 12

x

500 Nm

(QA)y = 0

Internal Loadings: Here, Nx = - 80 kN, Vy = 60 kN, My = -0.5 kN # m and Mz = 10 kN # m. Normal Stress: For the combine loadings, My z Mz y N s = + A Iz Iy sA =

- 80 ( 103 ) 0.008

-

10 ( 103 ) (0.12) 61.8667 ( 10-6 )

+

3 - 0.5 ( 103 ) 4 (0.05) 3.4667 ( 10-6 )

= - 36.6080 ( 106 ) Pa = 36.61 MPa (C) Shear Stress: Since (QA)y = 0, tA = 0 Thus, the state of stress at point A can be represented by the differential element shown in Fig. b In-Plane Principal Stresses: In accordance to the sign convention sx = -36.61 MPa, sy = 0 and txy = 0 for point A, Fig. a. Since no shear stress is acting on the element, s1 = sy = 0

Ans.

s2 = sx = - 36.61 MPa = -36.6 MPa

Ans.

Maximum In-Plane Shear Stress: tmax

in@plane

= =

B

a

sx - sy 2

2

b + txy2

- 36.61 - 0 2 b + 0 B 2

a

Ans.

= 18.30 MPa = 18.3 MPa

Ans: s1 = 0, s2 = - 36.6 MPa, = 18.3 MPa tmax in@plane

847

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9–26.  Solve Prob. 9–25 for point B.

20 mm A

200 mm 20 mm 20 mm y

100 mm 10 kNm

Section Properties: For the wide flange section, Fig. a,

60 kN 80 kN

z

2

A = 0.1(0.24) - 0.08(0.2) = 0.008 m Iz =

B

C

Solution

1 1 (0.1) ( 0.243 ) (0.08) ( 0.23 ) = 61.8667 ( 10-6 ) m4 12 12

Iy = 2 c

x

500 Nm

1 1 (0.02) ( 0.13 ) d + (0.2) ( 0.023 ) = 3.4667 ( 10-6 ) m4 12 12

(QB)y = 0

Internal Loadings: Here, Nx = - 80 kN, Vy = 60 kN, My = -0.5 kN # m and Mz = 10 kN # m Normal Stress: For the combine loadings, My z Mz y N s = + A Iz Iy sB =

-80 ( 103 ) 0.008

-

10 ( 103 ) ( -0.12) 61.8667 ( 10-6 )

+

3 - 0.5 ( 103 ) 4 ( - 0.05) 3.4667 ( 10-6 )

= 16.6081 ( 106 ) Pa = 16.61 MPa (T) Shear Stress: Since (QB)y = 0, tB = 0. Thus, the state of stress at point B can be represented by the differential element shown in Fig. b In-Plane Principal Stresses: In accordance to the sign convention sx = 16.61 MPa, sy = 0 and txy = 0 for point B, Fig. a. Since no shear stress acting on the element, s1 = sx = 16.61 MPa = 16.6 MPa

Ans.

s2 = sy = 0

Ans.

Maximum In-Plane Shear Stress: tmax

in@plane

= =

B

a

sx - sy 2

2

b + txy2

16.61 - 0 2 b + 0 B 2

a

Ans.

= 8.304 MPa = 8.30 MPa

Ans: s1 = 16.6 MPa, s2 = 0, tmax = 8.30 MPa in@plane

848

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9–27.  Solve Prob. 9–25 for point C.

20 mm A

200 mm 20 mm 20 mm y

100 mm 10 kNm

Section Properties: For the wide-flange section, Fig. a A = 0.1(0.24) - 0.08(0.2) = 0.008 m

1 1 (0.1) ( 0.243 ) (0.08) ( 0.23 ) = 61.8667 ( 10-6 ) m4 12 12

Iy = 2 c

1 1 (0.02) ( 0.13 ) d + (0.2) ( 0.023 ) = 3.4667 ( 10-6 ) m4 12 12

For the area shown shaded in Fig. a,

(QC)y = y′A′ = 0.11[0.1(0.02)] = 0.22 ( 10-3 ) m3 Internal Loadings: Here, Nx = - 80 kN, Vy = 60 kN, My = -0.5 kN # m and Mz = 10 kN # m. Normal Stress: For the combine loadings, Myz Mz y N s = + A Iz Iy sC =

-80 ( 103 ) 0.008

-

10 ( 103 ) ( -0.1) 61.8667 ( 10-6 )

+

3 - 0.5 ( 103 ) 4 (0) 3.4667 ( 10-6 )

= 6.164 ( 106 ) Pa = 6.164 MPa (T) Shear Stress: Applying the shear formula, tc =

Vy(QC)y Izt

=

60 ( 103 ) 3 0.22 ( 10-3 ) 4 61.8667 ( 10-6 ) (0.02)

60 kN 80 kN

z

2

Iz =

B

C

Solution

= 10.668 ( 106 ) Pa = 10.668 MPa

849

500 Nm

x

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9–27.  Continued

Thus, the state of stress at point C can be represented by the differential element shown in Fig. b. In-Plane Principal Stress: In accordance to the sign convention, sx = 6.164 MPa, sy = 0 and txy = 10.668 MPa for point C, Fig. a. s1,2 = =

sx + sy 2

{

B

a

sx - sy 2

2

b + txy2

6.164 - 0 2 6.164 + 0 { a b + 10.6682 2 B 2

= 3.082 { 11.104

s1 = 14.19 MPa = 14.2 MPa   s2 = - 8.022 MPa = -8.02 MPa

Ans.

Maximum In-Plane Shear Stress: tmax

in@plane

= =

B

a

sx - sy 2

2

b + txy2

6.164 - 0 2 b + 10.6682 B 2

a

Ans.

= 11.104 MPa = 11.1 MPa

Ans: s1 = 14.2 MPa, s2 = -8.02 MPa, tmax = 11.1 MPa in@plane

850

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*9–28. The drill pipe has an outer diameter of 75 mm, a wall thickness of 6 in., mm,and andaa weight weight of 0.8 kN/m.IfIf itit is 50 lb>ft. thickness of 0.25 subjected to a torque and axial load as shown, determine (a) the principal stress and (b) the maximum in-plane shear stress at a point on its surface at section a.

1500 7.5 kNlb

800kNm lbft 1.2 620mft

aa

Solution

620mft

Internal Forces and Torque: As shown on FBD(a). Section Properties: A= J=

π 4

π 2

(0.0752 / 0.0632 )= 0.414(10 /3 )π m 2

(0.03754 / 0.03154 )= 1.55977(10 /6 ) m 4 7.5 kN

Normal Stress:

1.2 kN · m

/12.3(10 3 )

N

0.8(6) = 4.8 kN

= / 9.457(106 ) N/m 2 = /9.457 MPa σ= A 0.414(10 /3 )π Shear Stress: Applying the torsion formula. t =

Tc [1.2(10 3 )](0.0375) 6 28.85(10 = ) N/m 2 28.85 MPa = = J 1.55977(10 /6 )

N = 12.3 kN T = 1.2 kN · m

- 1157.5MPa psi and = 28.846 3497.5MPa psi for a) In - Plane Principal Stresses: sx = 0, sy = –9.457 and ttxy for xy = any point on the shaft’s surface. Applying Eq. 9-5. s1,2 =

=

sx + sy 2

;

C

a

sx - sy 2

2

b + t2xy 28.846 MPa

2

0 + (/9.457)  0 / (/9.457)  2 ±   + (28.85 ) 2 2 

9.457 MPa

= /4.7285 ± 29.2354

σ 1 = 24.5 MPa

Ans.

σ 2 = /34.0 MPa

Ans.

b) Maximum In - Plane Shear Stress: Applying Eq. 9-7 t

=

max in-plane

C

a

sx - sy 2

2

b + t2xy 2

 0 / (/9.457)  + (28.852 )  2 

=

= 29.2

Ans.

Ans. s1 = 24.5 MPa, s2 = -34.0 MPa, max tin@plane = 29.2 MPa 667

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9–29.  The bell crank is pinned at A and supported by a short link BC. If it is subjected to the force of 80 N, determine the principal stresses at (a) point D and (b) point E. The crank is constructed from an aluminum plate having a thickness of 20 mm.

D

50 mm

A

C

Solution

E 15 mm

Point D: A = 0.04(0.02) = 0.8 ( 10-3 ) m2 1 (0.02) ( 0.043 ) = 0.1067 ( 10-6 ) m4 12

QD = y′A′ = 0.015(0.02)(0.01) = 3 ( 10-6 ) m3 Normal stress: 7.2(0.01) My P 64 = - 0.595 MPa + = -3 A I 0.8 ( 10 ) 0.1067 ( 10-6 )

sD =

Shear stress: 48(3) ( 10-6 ) VQ = = 0.0675 MPa It 0.1067 ( 10-6 ) (0.02)

tD =

Principal stress: sx = - 0.595 MPa  sy = 0  txy = 0.0675 MPa sx + sy sx - sy 2 s1,2 = { a b + txy2 2 B 2 =

-0.595 + 0 - 0.595 - 0 2 { a b + 0.06752 2 B 2

Ans. Ans.

s1 = 7.56 kPa s2 = -603 kPa Point E: I =

1 (0.02) ( 0.053 ) = 0.2083 ( 10-6 ) m4 12

QE = y′A′ = 0.02(0.01)(0.02) = 4.0 ( 10-6 ) m3 Normal stress: sE =

5.2364(0.015) My = = 377.0 kPa I 0.2083 ( 10-6 )

Shear stress: tE =

87.273(4.0) ( 10-6 ) VQ = = 83.78 kPa It 0.2083 ( 10-6 ) (0.02)

852

B

150 mm

40 mm 10 mm 50 mm

60 mm

I =

80 N

50 mm

5

3 4

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9–29.  Continued

Principal stress: sx = 0  sy = 377.0 kPa  txy = 83.78 kPa Solution s1,2 = =

sx + sy 2

{

B

a

sx - sy 2

2

b + txy2

0 + 377.0 0 - 377.0 2 { a b + 83.782 2 B 2

Ans. Ans.

s1 = 395 kPa s2 = -17.8 kPa

Ans: Point D: s1 = 7.56 kPa, s2 = - 603 kPa, Point E: s1 = 395 kPa, s2 = - 17.8 kPa 853

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9–30.  The beam has a rectangular cross section and is subjected to the loadings shown. Determine the principal stresses at point A and point B, which are located just to the left of the 20-kN load. Show the results on elements located at these points.

20 kN 100 mm

2m

Solution Internal Forces and Moment: As shown on FBD(b). Section Properties: A = 0.1 (0.2) = 0.020 m2 1 I = (0.1) ( 0.23 ) = 66.667 ( 10-6 ) m4 12 QA = 0 QB = y- =A′ = 0.05(0.1)(0.1) = 0.50 ( 10-3 ) m3 Normal Stresses: My N s = { A I - 10.0 ( 103 )

sA =

0.020 - 10.0 ( 103 )

sB =

0.020

20.0 ( 103 ) (0.1)

-

66.667 ( 10-6 ) 20.0 ( 103 ) (0) 66.667 ( 10-6 )

= - 30.5 MPa

= - 0.500 MPa

Shear Stress: Applying the shear formula t =

VQ , lt

tA = 0 10.0 ( 103 ) 30.50 ( 10-3 )4 tB = = 0.750 MPa 66.667 ( 10-6 ) (0.1) In-Plane Principal Stresses: sx = - 30.5 MPa, sy = 0, and txy = 0 for point A. Since no shear stress acts on the element. s1 = sy = 0

Ans.

s2 = sx = - 30.5 MPa

Ans.

sx = -0.500 MPa, sy = 0, and txy = - 0.750 MPa for point B. Applying Eq. 9–5, s1,2 = =

sx + sy 2

{

B

a

sx - sy 2

10 kN

B A

2

b + txy 2

-0.500 + 0 - 0.500 - 0 2 { a b + ( -0.750)2 2 B 2

= -0.250 { 0.7906

Ans.

s1 = 0.541 MPa  s2 = - 1.04 MPa

854

2m

B A

100 mm 100 mm

50 mm 50 mm

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9–30.  Continued

Orientation of Principal Plane: Applying Eq. 9–4 for point B. tan 2up =

txy (sx - sy)>2

=

- 0.750 = 3.000 ( - 0.500 - 0)>2

uP = 35.78°  and   - 54.22° Substituting the results into Eq. 9–1 with u = 35.78° yields sx′ =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

- 0.500 + 0 - 0.500 - 0 + cos 71.56° + ( -0.750 sin 71.56°) 2 2 = - 1.04 MPa = s2 =

Hence. Ans.

up1 = - 54.2°  up2 = 35.8°

Ans: Point A, s1 = sy = 0, s2 = sx = - 30.5 MPa, Point B, s1 = 0.541 MPa, s2 = -1.04 MPa, up1 = -54.2°, up2 = 35.8° 855

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9–31. The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at point A, which is located at the top of the web. Although it is not very accurate, use the shear formula to determine the shear stress. Show the result on an element located at this point.

120 kN/m

30 kN

A 0.3 m

0.9 m

Using the method of sections and consider the FBD of the left cut segment of the bean, Fig. a + c ©Fy = 0;

V 1 2

a + ©MC = 0;

1 2

(90)(0.9) - 30 = 0

(90)(0.9)(0.3) + 30(0.9) - M = 0

M = 39.15 kN # m

150 mm

1 1 (0.15)(0.193) (0.13)(0.153) = 49.175(10 - 6) m4 12 12

Referring to Fig. b, QA = y¿A¿ = 0.085 (0.02)(0.15) = 0.255 (10 - 3) m3 The normal stress developed is contributed by bending stress only. For point A, y = 0.075 m. Thus, s =

My 39.15(103)(0.075) = 59.71(106)Pa = 59.71 MPa (T) = I 49.175(10 - 6)

The shear stress is contributed by the transverse shear stress only. Thus t =

70.5(103) C 0.255(10 - 3) D VQA = 18.28(106)Pa = 18.28 MPa = It 49.175(10 - 6) (0.02)

Here, sx = 59.71 MPa, sy = 0 and txy = 18.28 MPa. s1, 2 = =

sx + sy 2

;

C

a

sx - sy 2

2

b + txy

59.71 - 0 2 59.71 + 0 ; a b + 18.282 2 C 2

= 29.86 ; 35.01 s1 = 64.9 MPa tan 2uP =

s2 = - 5.15 MPa txy

(sx - sy)>2

uP = 15.74°

=

and

Ans.

18.28 = 0.6122 (59.71 - 0)>2 - 74.26°

Substitute u = 15.74°, sx¿ = =

sx + sy 2

+

sx - sy 2

20 mm

V = 70.5 kN

The moment of inertia of the cross - section about the bending axis is I =

A

cos 2u + txy sin 2u

59.71 - 0 59.71 + 0 + cos 31.48° + 18.28 sin 31.48° 2 2

= 64.9 MPa = s1

651

20 mm 150 mm 20 mm

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9–31. Continued Thus,

A uP B 1 = 15.7°

A uP B 2 = - 74.3°

Ans.

The state of principal stress can be represented by the element shown in Fig. d

Ans.

V = 70.5 kN, M = 39.15 kN # m, I = 49.175(10 - 6) m4, QA = 0.255(10 - 3) m3, s1 = 64.9 MPa, s2 = -5.15 MPa, (up)1 = 15.7°, (up)2 = -74.3°

652

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  A paper tube is formed by rolling a cardboard strip in a spiral and then gluing the edges together as shown. Determine the shear stress acting along the seam, which is at 50° from the horizontal, when the tube is subjected to an axial compressive force of 200 N. The paper is 2 mm thick and the tube has an outer diameter of 100 mm.

40 200 N

200 N 100 mm

Solution Normal And Shear Stresses: The normal stress is caused by the axial force only. Thus s =

N = A

- 200

p 4

( 0.12 - 0.0962 )

= - 324.81 ( 103 ) Pa = 324.81 kPa (C)

Since there is no shear force on the cross-section, t = 0 The state of stress of a point on the cross-section can be represented by the element shown in Fig. a Stress Transformation Equations: With sx = - 324.81 kPa, sy = 0, txy = 0 and u = + 40° (Fig. b), sx - sy tx′y′ = sin 2u + txy cos 2u 2 = -

( - 324.81 - 0)

sin 80° + 0 2 = 159.94 kPa = 160 kPa

Ans.

Ans: tx′y′ = 160 kPa 856

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9–33. Solve Prob. 9–31 for the normal stress acting perpendicular to the seam.

40 200 N

200 N 100 mm

Solution Normal and Shear Stresses: The normal stress is caused by the axial force only. Thus s =

N = A

- 200

p 4

( 0.12 - 0.0962 )

= - 324.81 ( 103 ) Pa = 324.81 kPa (C)

Since there is no shear force on the cross-section, t = 0 The state of stress of a point on the cross-section can be represented by the element shown in Fig. a. Stress Transformation Equations: With sx = - 324.81 kPa, sy = 0, txy = 0 and u = + 40° (Fig. b), sx + sy sx - sy sx′ = + cos 2u + txy sin 2u 2 2 ( -324.81 - 0) - 324.81 + 0 = + cos 80° + 0 2 2 Ans.

= -190.60 kPa = - 191 kPa

Ans: sx′ = - 191 kPa 857

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9–34. Determine the principal stress at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.

7.5 mm A 50 mm

7.5 mm

Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a, 7.5 mm

©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0

FCD = 2166.67 N

+ ©F = 0; : x

Bx - 2166.67 cos 30° = 0

Bx = 1876.39 N

Section a – a

+ c ©Fy = 0;

2166.67 sin 30° - 500 - By = 0

By = 583.33 N

D

20 mm

Internal Loadings: Consider the equilibrium of the free - body diagram of the arm’s left segment, Fig. b. + ©F = 0; : x

1876.39 - N = 0

N = 1876.39 N

+ c ©Fy = 0;

V - 583.33 = 0

V = 583.33 N

+ ©MO = 0;

583.33(0.15) - M = 0

M = 87.5N # m

B

A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2 1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12

Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus, sA = =

-1876.39

0.5625 A 10

-3

B

+

MyA N + A I 87.5(0.0175)

0.16367 A 10 - 6 B

= 6.020 MPa

The shear stress is caused by transverse shear stress. tA =

583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)

The share of stress at point A can be represented on the element shown in Fig. d. In - Plane Principal Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. We have s1,2 = =

sx + sy 2

;

C

¢

sx - sy 2

2

≤ + txy 2

6.020 - 0 2 6.020 + 0 ; a b + 1.5152 2 C 2

s1 = 6.38 MPa

s2 = - 0.360 MPa

Ans.

654

a C

a

0.15 m

Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are

I =

60

0.15 m

0.35 m 500 N

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9–34. Continued Orientation of the Principal Plane: tan 2uP =

txy

A sx - sy B >2

=

1.515 = 0.5032 (6.020 - 0)>2

up = 13.36° and 26.71° Substituting u = 13.36° into sx¿ = =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

6.020 + 0 6.020 - 0 + cos 26.71° + 1.515 sin 26.71° 2 2

= 6.38 MPa = s1 Thus, A uP B 1 = 13.4 and A uP B 2 = 26.71°

Ans.

The state of principal stresses is represented by the element shown in Fig. e.

Ans. s1 = 6.38 MPa, s2 = -0.360 MPa, (up)1 = 13.4°, (up)2 = 103° 655

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7.5 mm

9–35. Determine the maximum in-plane shear stress developed at point A on the cross section of the arm at section a–a. Specify the orientation of this state of stress and indicate the results on an element at the point.

A 50 mm

7.5 mm

20 mm

7.5 mm

Solution

Section a – a

Support Reactions: Referring to the free - body diagram of the entire arm shown in Fig. a,

D

©MB = 0; FCD sin 30°(0.3) - 500(0.65) = 0

FCD = 2166.67 N

+ ©F = 0; : x

Bx = 1876.39 N

Bx - 2166.67 cos 30° = 0

+ c ©Fy = 0;

2166.67 sin 30° - 500 - By = 0

60 B

By = 583.33 N

1876.39 - N = 0

N = 1876.39 N

+ c ©Fy = 0;

V - 583.33 = 0

V = 583.33 N

+ ©MO = 0;

583.33(0.15) - M = 0

M = 87.5 N # m

Section Properties: The cross - sectional area and the moment of inertia about the z axis of the arm’s cross section are A = 0.02(0.05) - 0.0125(0.035) = 0.5625 A 10 - 3 B m2 1 1 (0.02) A 0.053 B (0.0125) A 0.0353 B = 0.16367 A 10 - 6 B m4 12 12

I =

Referring to Fig. b, QA = y¿A¿ = 0.02125(0.0075)(0.02) = 3.1875 A 10 - 6 B m3 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. Thus, sA = =

MyA N + A I - 1876.39

0.5625 A 10

-3

B

+

87.5(0.0175)

0.16367 A 10 - 6 B

= 6.020 MPa

The shear stress is contributed only by transverse shear stress. tA =

583.33 C 3.1875 A 10 - 6 B D VQA = = 1.515 MPa It 0.16367 A 10 - 6 B (0.0075)

Maximum In - Plane Shear Stress: sx = 6.020 MPa, sy = 0, and txy = 1.515 MPa. tmax

in-plane

=

C

¢

sx - sy 2

2

≤ + txy 2 =

6.020 - 0 2 b + 1.5152 = 3.37 MPa B 2 a

656

Ans.

C

a

0.15 m

Internal Loadings: Considering the equilibrium of the free - body diagram of the arm’s left cut segment, Fig. b, + ©F = 0; : x

a

0.15 m

0.35 m 500 N

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9–35. Continued Orientation of the Plane of Maximum In - Plane Shear Stress: tan 2us = -

A sx - sy B >2 txy

= -

(6.020 - 0)>2 = - 1.9871 1.515

us = - 31.6° and 58.4°

Ans.

Substituting u = - 31.6° into tx¿y¿ = = -

sx - sy 2

sin 2u + txy cos 2u

6.020 - 0 sin( - 63.29°) + 1.515 cos( - 63.29°) 2

= 3.37 MPa = t max

in-plane

This indicates that t max

is directed in the positive sense of the y¿ axis on the face

in-plane

of the element defined by us = - 31.6°. Average Normal Stress: savg =

sx + sy 2

=

6.020 + 0 = 3.01 MPa 2

Ans.

The state of maximum in - plane shear stress is represented on the element shown in Fig. e.

Ans.

us = -31.6° and 58.4°, savg = 3.01 MPa, max t in@plane = 3.37 MPa 657

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*9–36. Determine the principal stresses in the cantilevered beam at points A and B.

30 mm 40 mm B A 120 mm 150 mm 4 5 3

Solution Internal Forces and Moment: As shown on FBD. Section Properties: 1 (0.12) ( 0.153 ) = 33.75 ( 10-6 ) m4 12 1 Iy = (0.15) ( 0.123 ) = 21.6 ( 10-6 ) m4 12 (QA)y = y′A′ = 0.06(0.03)(0.12) = 0.216 ( 10-3 ) m3 Iz =

(QA)z = 0 (QB)z = z′A′ = 0.04(0.04)(0.15) = 0.240 ( 10-3 ) m3 (QB)y = 0 Normal Stress: s = sA = -

Mzy Iz

Myz

+

Iy

-14.4 ( 103 ) (0.045)

+

33.75 ( 10-6 )

- 10.8 ( 103 ) (0.06) 21.6 ( 10-6 )

= -10.8 MPa sB = -

- 14.4 ( 103 ) (0.075) 33.75 ( 10-6 )

+

- 10.8 ( 103 ) ( -0.02) 21.6 ( 10-6 )

= 42.0 MPa Shear Stress: Applying the shear formula Vy(QA)y

tA = tB =

Iz t Vz(QB)z Iy t

= =

12.0 ( 103 ) 3 0.216 ( 10-3 ) 4 33.75 ( 10-6 ) (0.12)

= 0.640 MPa

- 9.00 ( 103 ) 3 0.240 ( 10-3 ) 4 21.6 ( 10-6 ) (0.15)

= -0.6667 MPa

In-Plane Principal Stress: sx = - 10.8 MPa, sy = 0 and txy = 0.640 MPa for point A. s1,2 = =

sx + sy 2

{

B

a

sx - sy 2

2

b + t2xy

-10.8 + 0 - 10.8 - 0 2 { a b + 0.6402 2 B 2

= -5.40 { 5.4378

Ans.

s1 = 37.8 kPa  s2 = - 10.8 MPa

859

1200 mm

15 kN

800 mm

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*9–36. Continued

s x = 42.0 MPa, sz = 0, and txz = 0.6667 MPa for point B. Solution s1,2 = =

sx + sy 2

{

B

a

sx - sy 2

2

b + txy2

42.0 + 0 42.0 - 0 2 { a b + 0.66672 2 B 2

= 21.0 { 21.0105

Ans.

s1 = 42.0 MPa  s2 = - 10.6 kPa

Ans: Point A, s1 = 37.8 kPa, s2 = - 10.8 MPa, Point B, s1 = 42.0 MPa, s2 = - 10.6 kPa 860

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9–37.  The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses and the maximum in-plane shear stress at point A. The bearings only support vertical reactions.

P F

F A L 2

L 2

Solution Support Reactions: As shown on FBD(a). Internal Forces and Moment: As shown on FBD(b). Section Properties: A = Normal Stress:

p 2 p d 4 p 4 d I = a b = d QA = 0 4 4 2 64 s = = sA =

N Mc { A I -F p 2 { 4 d

1 2

PL d 2 4 p 4 64 d

4 2PL a - Fb d pd 2

Shear Stress: Since QA = 0, tA = 0 In-Plane Principal Stress: sx =

4 2PL a - Fb. pd 2 d

sy = 0 and txy = 0 for point A. Since no shear stress acts on the element, s1 = sx =

4 2PL a - Fb d pd 2

Ans. Ans.

s2 = sy = 0

Maximum In-Plane Shear Stress: Applying Eq. 9–7 for point A, tmax

in@plane

=

=

H

=

£

B

4 pd 2

a

sx - sy

1 2PL d

2

2

b + txy2

- F2 - 0

2

2

≥ + 0

2 2PL a - Fb d pd 2

Ans.

Ans: 4 2PL a - F b, s2 = 0, d pd 2 2 2PL tmax = a - Fb d pd 2 in@plane s1 =

863

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9–38. The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the principal stresses acting at point A. 450 mm 300 Nm

Solution Ix = Iy = J =

p (0.025)4 = 0.306796(10 - 6) m4 4

A B 25 mm

45 Nm 800 N

p (0.025)4 = 0.613592(10 - 6) m4 2

QA = 0 60(0.025) Mx c = 4.889 MPa = I 0.306796(10 - 6)

sA = tA =

Ty c J

=

45(0.025) 0.613592(10 - 6)

sx = 4.889 MPa s1, 2 = =

sx + sy 2

= 1.833 MPa

sy = 0 ;

C

a

sx - sy 2

txy = - 1.833 MPa 2

b + txy 2

4.889 - 0 2 4.889 + 0 ; b + ( -1.833)2 a 2 C 2

s1 = 5.50 MPa

Ans.

s2 = - 0.611 MPa

Ans.

Ans. s1 = 5.50 MPa, s2 = -0.611 MPa 676

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Solve Prob. 9–47 for point B.

9–39.

450 mm

Solution Ix = Iy = J =

p (0.025)4 = 0.306796(10 - 6) m4 4

300 Nm

p (0.025)4 = 0.613592(10 - 6) m4 2

QB = yA¿ =

A B 25 mm

45 Nm 800 N

4(0.025) 1 a b p (0.0252) = 10.4167(10 - 6) m3 3p 2

sB = 0 tB =

VzQB It

-

Ty c J

=

800(10.4167)(10 - 6) -6

0.306796(10 )(0.05)

sx = 0 s1, 2 =

sy = 0 sx + sy 2

;

C

a

sx - sy 2

-

45(0.025) 0.61359(10 - 6)

= - 1.290 MPa

txy = - 1.290 MPa 2

b + txy 2

= 0 ; 2(0)2 + (- 1.290)2 s1 = 1.29 MPa

Ans.

s2 = - 1.29 MPa

Ans.

Ans. s1 = 1.29 MPa, s2 = -1.29 MPa 676

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*9–40. The wide-flange beam is subjected to the 50-kN force. Determine the principal stresses in the beam at point A located on the web at the bottom of the upper flange. Although it is not very accurate, use the shear formula to calculate the shear stress.

50 kN A B 1m

3m A

10 mm B

Solution I =

12 mm 250 mm 12 mm

200 mm

1 1 (0.2)(0.274)3 (0.19)(0.25)3 = 95.451233(10 - 6) m4 12 12

QA = (0.131)(0.012)(0.2) = 0.3144(10 - 3) m3 sA =

150(103)(0.125) My = 196.43 MPa = I 95.451233(10 - 6)

tA =

50(103)(0.3144)(10 - 3) VQA = = 16.47 MPa It 95.451233(10 - 6)(0.01)

sx = 196.43 MPa s1, 2 = =

sx + sy 2

{

txy = - 16.47 MPa

sy = 0 a B

sx - sy 2

2

2 b + txy

196.43 + 0 196.43 - 0 2 2 { a b + ( - 16.47) 2 B 2

s1 = 198 MPa

Ans.

s2 = - 1.37 MPa

Ans.

Ans: s1 = 198 MPa, s2 = -1.37 MPa 868

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9–41. Solve Prob. 9–40 for point B located on the web at the top of the bottom flange.

50 kN A B 1m

3m A

10 mm B

Solution

12 mm 250 mm 12 mm

200 mm

1 1 I = (0.2)(0.247)3 (0.19)(0.25)3 = 95.451233(10 - 6) m4 12 12 QB = (0.131)(0.012)(0.2) = 0.3144(10 - 3) sB = -

tB =

150(103)(0.125) My = = - 196.43 MPa I 95.451233(10 - 6)

50(103)(0.3144)(10 - 3) VQB = = 16.47 MPa It 95.451233(10 - 6)(0.01)

sx = -196.43 MPa s1, 2 =

sx + sy

=

txy = - 16.47 MPa

sy = 0

2

{

B

a

sx - sy 2

2

2 b + txy

- 196.43 + 0 - 196.43 - 0 2 2 { b + ( -16.47) a 2 B 2

s1 = 1.37 MPa

Ans.

s2 = -198 MPa

Ans.

Ans: s1 = 1.37 MPa, s2 = - 198 MPa 869

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9–42. The box beam is subjected to the 26-kN force that is applied at the center of its width, 75 mm from each side. Determine the principal stresses at point A and show the results in an element located at this point. Use the shear formula to calculate the shear stress.

26 kN B

13

A

2m

5

12

3m 130 mm A

130 mm

Solution I =

B

75 mm 75 mm

150 mm

1 1 (0.15) ( 0.153 ) (0.13) ( 0.133 ) = 18.38667 ( 10 - 6 ) m4 12 12

A = 0.152 - 0.132 = 5.6 ( 10 - 3 ) m2 QA = 0 tA = 0 sA = -

- 24 ( 103 ) 28.2 ( 103 ) (0.075) P Mc + = + = 111 MPa A I 5.6 ( 10 - 3 ) 18.38667 ( 10 - 6 )

s1 = 111 MPa

Ans.

s2 = 0

Ans.

Ans: s1 = 111 MPa, s2 = 0 870

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9–43. Solve Prob. 9–42 for point B.

26 kN B

13

A

2m

5

12

3m 130 mm A

130 mm

Solution I =

B

75 mm 75 mm

150 mm

1 1 (0.15) ( 0.153 ) (0.13) ( 0.133 ) = 18.38667 ( 10 - 6 ) m4 12 12

A = 0.152 - 0.132 = 5.6 ( 10 - 3 ) m2 QB = (0.07)(0.15)(0.01) + 2(0.0325)(0.065)(0.01) = 0.14725 ( 10 - 3 ) m3 sB = tB =

24 ( 103 ) P = = - 4.286 MPa A 5.6 ( 10 - 3 )

10 ( 103 ) (0.14725) ( 10 - 3 ) VQB = = 4.004 MPa It 18.38667 ( 10 - 6 ) (0.02)

sx = -4.286 MPa s1.2 = =

sx + sy 2

{

txy = - 4.004 MPa

sy = 0 B

a

sx - sy 2

2

b + txy2

-4.286 + 0 - 4.286 - 0 2 a b + { 2 B 2 txy (sx - sy)>2

Ans.

s2 = - 6.68 MPa

s1 = 2.40 MPa tan 2uP =

( - 4.0042 )

=

- 4.004 ( - 4.286 - 0)>2

uP = 30.9°  or   - 59.1° Use Eq. 9–1, uP1 = - 59.1°

uP2 = 30.9°

Ans: s1 = 2.40 MPa, s2 = - 6.68 MPa 871

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*9–44. Solve Prob. 9–2 using Mohr’s circle.

Solution sx + sy 2

=

0 + 65 = 32.5 MPa 2

R = 2(32.5)2 + (20)2 = 38.1608 f = tan-1

20 = 31.6075° 32.5

a = 120° - 31.6075° = 88.392° sx′ = 32.5 - 38.1608 cos 88.392° = 31.4 MPa

Ans.

tx′y′ = 38.1608 sin 88.392° = 38.1 MPa

Ans.

Ans: sx′ = 31.4 MPa, tx′y′ = 38.1 MPa 872

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9–45.

Solve Prob. 9–4 using Mohr’s circle. 90 MPa

A

35 MPa 60 30

B 50 MPa

Solution sx = 90 MPa sx + sy 2

=

sy = 50 MPa

txy = - 35 MPa

A(90, - 35)

90 + 50 = 70 2

R = 2(90 - 70)2 + (35)2 = 40.311 Coordinates of point B: f = tan - 1 a

35 b = 60.255° 20

c = 300° - 180° - 60.255° = 59.745° sx¿ = 70 - 40.311 cos 59.745° = 49.7 MPa

Ans.

tx¿ = - 40.311 sin 59.745° = - 34.8 MPa

Ans.

Ans. sx′ = -388 psi, tx′y′ = 455 psi 679

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9–46.

Solution Construction of the Circle: In accordance to the sign convention, sx = 0, sy = −500 MPa and txy = 350 MPa. Hence savg =

sx + sy 2

=

0 + (-500) = -250 MPa 2

The coordinates for reference point A and center C of the circle are A(0, 350)

C(-250, 0)

The radius of the circle is R = 2[0 - (−250)]2 + 3502 = 430.12 MPa Using the results, the circle shown in Fig. a can be constructed. Stress on the Rotated Element: Plane AB can be represented by an element rotating clockwise through an angle θ = 60°, Fig. b. The normal and shear stress components on plane AB A sx¿ and tx¿y¿ B are represented by coordinates of point P on the circle. From the geometry shown on the circle  350  c tan /1  = =  54.46°  250  = β 180° / c / 2= θ 180° / 54.46° / 120° = 5.538° Then

σ x′ = /250 / 430.12 coθ 5.538° = /678 MPa

Ans

tx¿y¿ 430.12 = = θin 5.538° 41.5 MPa

Ans

Ans: MPa, tx¿y¿ 41.5 MPa σ x′ =/ 678 = 682

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9–47. Solve Prob. 9–11 using Mohr’s circle.

Solution Construction of the Circle: u = -60°, sx = 150 MPa, sy = 100 MPa, txy = 75 MPa. Thus sx + sy 150 + 100 savg = = = 125 MPa 2 2 The coordinates of the reference point A and center C of the circle are A(150, 75)

C(125, 0)

Thus, the radius of the circle is R = CA = 2(150 - 125)2 + (75)2 = 79.06 MPa

Normal and Shear Stress on Rotated Element: Here u = 60° clockwise. By rotating the radial line CA clockwise 2u = 120°, it coincides with the radial line OP and the coordinates of reference point P(sx′, tx′y′) represent the normal and shear stresses on the face of the element defined by u = -60°. sy ′ can be determined by calculating the coordinates of point Q. From the geometry of the circle, Fig. (a), sin a =

75 , a = 71.57°, b = 120° + 71.57° - 180° = 11.57° 79.06

sx′ = 125 - 79.06 cos 11.57° = 47.5 MPa

Ans.

tx′y′ = - 79.06 sin 11.57° = - 15.8 MPa

Ans.

sy′ = 125 + 79.06 cos 11.57° = 202 MPa

Ans.

The results are shown in Fig. (b).

Ans: sx′ = 47.5 MPa, tx′ y′ = - 15.8 MPa, sy′ = 202 MPa 875

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*9–48. Solve Prob. 9–15 using Mohr’s circle.

Solution sx + sy 2

=

45 - 60 = - 7.5 MPa 2

R = 2(45 + 7.5)2 + (30)2 = 60.467 MPa s1 = 60.467 - 7.5 = 53.0 MPa

Ans.

s2 = -60.467 - 7.5 = - 68.0 MPa

Ans.

2up1 = tan-1

30 (45 + 7.5) Ans.

up1 = 14.9°   counterclockwise t max

in@plane

Ans.

= 60.5 MPa

Ans.

savg = - 7.50 MPa 30 (45 + 7.5) = 30.1°   clockwise

2us1 = 90° - tan-1 us1

Ans.

Ans: s1 = 53.0 MPa, s2 = -68.0 MPa, up1 = 14.9° counterclockwise, t max = 60.5 MPa, in@plane

savg = -7.50 MPa, us1 = 30.1° clockwise 876

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9–49. Solve Prob. 9–16 using Mohr’s circle.

Solution Construction of Circle: sx = 50 MPa, sy = 0, txy = -15 MPa. Thus, savg =

sx + sy 2

=

50 + 0 = 25 MPa 2

Ans.

The coordinates of reference point A and center C of the circle are A(50, - 15)

C(25, 0)

Thus, the radius of the circle is R = CA = 2(50 - 25)2 + ( - 15)2 = t max

in@plane

= 29.15 MPa

See Fig. (a).

a) Principal Stress: s1 = 54.2 MPa, sin 2a =

15 , 29.15

s2 = - 4.15 MPa

Ans.

a = 15.5°

Ans.



See Fig. (b).

Ans: s1 = 54.2 MPa, s2 = -4.15 MPa, up = - 15.5°, savg = 25 MPa, t max

in@plane

a = 15.5° 877

= 29.2 MPa, us = 29.5°­,

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9–50. Mohr’s circle for the state of stress is shown in Fig. 9–17a. Show that finding the coordinates of point P (sx′, tx′y′) on the circle gives the same value as the stress transformation Eqs. 9–1 and 9–2.

Solution

R = sx= =

C

C aa

B(sy, - txy)

A(sx, txy)

c sx - a

sx + sy

sx + sy

+

2

u′ = 2uP - 2u

2

C

a

2

sx + sy

b d + t2xy =

sx - sy 2

2

2

C

a

b, 0b

sx - sy 2

b + t2xy cos u′

2

b + t2xy (1)

(2)

cos (2uP - 2u) = cos 2uP cos 2u + sin 2up sin 2u From the circle: sx -

cos 2uP = sin 2uP =

(3)

sx - sy 2 2 txy

+ t2xy

sx - sy 2 2

+ t2xy

41

41

sx + sy 2

2

2

(4)

Substitute Eq. (2), (3) and into Eq. (1) sx′ = tx′y′ =

sx + sy 2

C

a

+

sx - sy 2

sx - sy 2

QED

cos 2u + txy sin 2u

2

b + t2xy sin u′

(5)

sin u′ = sin (2uP - 2u)

(6)

= sin 2uP cos 2u - sin 2u cos 2uP Substitute Eq. (3), (4), (6) into Eq. (5), tx′y′ = -

sx - sy 2

QED

sin 2u + txy cos 2u

Ans: N/A­ 878

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9–51.  etermine (a) the principal stresses and (b) the maximum D in-plane shear stress and average normal stress. Specify the orientation of the element in each case.

80 MPa

60 MPa

Solution Construction of The Circle: In accordance sx = 0, sy = -80 MPa and txy = 60 MPa. Hence,

to

the

sign

sx + sy

0 + ( - 80) = = - 40.0 MPa 2 2 The coordinates for reference point A and center of circle C are savg =

convention,

Ans.

A(0, 60)  C( -40.0, 0) Thus, the radius of the circle is R = CA = 2[0 - ( -40.0)]2 + (60 - 0)2 = 72.11 MPa

Using these results, the circle shown in Fig. a can be constructed. a) In-plane Principal Stresses: The coordinates of points B and D on the circle represent s1 and s2, respectively. s1 = -40.0 + 72.11 = 32.11 MPa = 32.1 MPa

Ans.

s2 = -40.0 - 72.11 = -112.11 MPa = - 112 MPa

Ans.

879

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9–51.  Continued

Orientation of Principal Plane: From the shaded triangle on the circle, 60 tan 2up1 = = 1.5 40.0 2up1 = 56.31° up1 = 28.15° = 28.2° (counterclockwise)

Ans.

Using these results the state of in-plane principal stresses can be represented by the differential element shown in Fig. b. b) Maximum In-plane Shear Stress: Represented by the coordinates of point E on the circle. tmax = R = 72.11 MPa = 72.1 MPa Ans. in@plane

Orientation of The Plane For Maximum in-plane Shear Stress: From the circle, 40.0 tan 2us = = 0.6667 60 2us = 33.69° us = 16.845° = 16.8° (clockwise)

Ans.

Using these results, the state of maximum in-plane shear stress can be represented by the differential element shown in Fig. c. us = - 16.8°

Ans: savg = -40.0 MPa, s1 = 32.1 MPa, s2 = -112 MPa, up1 = 28.2°, tmax = 72.1 MPa, in@plane

us = -16.8° 880

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9 ksi MPa

*9–52. Determine the equivalent state of stress if an element is oriented 30° clockwise from the element shown. Show the result on the element.

4 ksi MPa

In accordance to the established sign convention, sx = –6 - 6MPa, ksi, ssyy = ksi and = 99 MPa, and MPa. Thus, . Thus, txy = 4 ksi savg =

sx + sy 2

=

6 ksi MPa

-6 + 9 = 1.50 MPa ksi 2

Then, the coordinates of reference point A and C are A( -6, 4)

C(1.5, 0)

The radius of the circle is MPa R = CA = 2( -6 - 1.5)2 + 42 = 8.50 ksi Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a, a = tan - 1 a

4 b = 28.07° 6 + 1.5

b = 60° - 28.07° = 31.93°

Then, ksi sx¿ = 1.5 - 8.50 cos 31.93° = - 5.71 MPa

Ans.

ksi tx¿y¿ = - 8.5 sin 31.95° = - 4.50 MPa

Ans.

ksi sy¿ = 8.71 MPa

Ans.

The results are shown in Fig. b.

8.71 MPa

(MPa)

4.50 MPa

5.71 MPa

(MPa)

Ans. sx′ = −5.71 MPa, tx′y′ = -4.50 MPa, sy′ = 8.71 MPa 685

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2 MPa ksi

9–53. Determine the equivalent state of stress if an element is oriented 20° clockwise from the element shown.

MPa 3 ksi

Solution

4 MPa ksi

MPa, Construction of the Circle: In accordance with the sign convention, sx = 3 ksi , sy = - 2 MPa, txy = −4 MPa. Hence, savg =

sx + sy 2

=

3 + ( -2) = 0.500 ksi MPa 2

The coordinates for reference points A and C are A(3, - 4)

(MPa)

C(0.500, 0)

The radius of the circle is R = 2(3 - 0.500)2 + 42 = 4.717 MPa ksi

(MPa)

Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinate of point P on the circle, sy¿, can be determined by calculating the coordinates of point Q on the circle. sx¿ = 0.500 + 4.717 cos 17.99° = 4.99 ksi MPa

3.99 MPa

Ans. 4.99 MPa

ksi tx¿y¿ = - 4.717 sin 17.99° = - 1.46 MPa

Ans.

sy¿ = 0.500 - 4.717 cos 17.99° = - 3.99 MPa ksi

1.46 MPa

Ans.

Ans. sx′ = 4.99 MPa, tx′y′ = -1.46 MPa, sy′ = -3.99 MPa 684

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700 kPa psi

9–54. Draw Mohr’s circle that describes each of the following states of stress.

44MPa ksi 40 MPa

600 kPa psi

(a)

Solution a) Here, sx = 600 KPa, sy = 700 KPa and txy = 0. Thus, savg =

sx + sy 2

=

600 + 700 = 650 KPa 2

Thus, the coordinate of reference point A and center of circle are A(600, 0)

C(650, 0)

Then the radius of the circle is R = CA = 650 - 600 = 50 KPa The Mohr’s circle represents this state of stress is shown in Fig. a. MPa and Thus, txytxy==00. b) Here, sx = 0, sy = 4 ksi and . Thus, savg =

sx + sy 2

=

0 + 4 = 2 MPa ksi 2

Thus, the coordinate of reference point A and center of circle are A(0, 0)

C(2, 0)

Then the radius of the circle is R = CA = 2 - 0 = 2 psi MPa c) Here, sx = sy = 0 and txy = - 40 MPa. Thus, savg =

sx + sy 2

= 0

Thus, the coordinate of reference point A and the center of circle are A(0, - 40)

C(0, 0)

Then, the radius of the circle is R = CA = 40 MPa The Mohr’s circle represents this state of stress shown in Fig. c

693

(b)

(c)

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9–54. Continued

(KPa)

(KPa)

(MPa)

(MPa)

694

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250 MPa

9–55. Determine the equivalent state of stress for an element oriented 60° counterclockwise from the element shown. Show the result on the element.

400 MPa

In accordance to the established sign convention, sx = - 560 MPa, sy = 250 MPa and txy = - 400 MPa. Thus, savg =

sx + sy 2

=

560 MPa

- 560 + 250 = - 155 MPa 2

Then, the coordinate of reference points A and C are A(- 560, -400)

C( - 155, 0)

The radius of the circle is R = CA = 3 C - 560 - ( -155) D 2 + ( - 400)2 = 569.23 MPa Using these results, the circle shown in Fig. a can be constructed. Referring to the geometry of the circle, Fig. a a = tan - 1 a

400 b = 44.64° 560 - 155

b = 120° - 44.64° = 75.36°

Then, sx¿ = - 155 - 569.23 cos 75.36° = - 299 MPa

Ans.

tx¿y¿ = 569.23 sin 75.36° = 551 MPa

Ans.

sy¿ = - 155 + 569.23 cos 75.36° = - 11.1 MPa

Ans.

The results are shown in Fig. b.

Ans. savg = -155 MPa, R = 569.23 MPa, sx′ = -299 MPa, tx′y′ = 551 MPa, sy′ = -11.1 MPa 686

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*9–56.  etermine (a) the principal stresses and (b) the maximum D in-plane shear stress and average normal stress. Specify the orientation of the element in each case.

20 MPa

100 MPa 40 MPa

Solution Construction of the Circle: In accordance to the sx = 100 MPa, sy = 20 MPa, and txy = - 40 MPa. Hence savg =

sx + sy 2

=

sign

convention,

100 + 20 = 60.0 MPa 2

Ans.

The coordinates for reference point A and center of circle C are A(100, - 40)

C(60.0, 0)

Thus, the radius of the circle is R = CA = 2(100 - 60.0)2 + ( - 40 - 0)2 = 4022 MPa

Using these results, the circle shown in Fig. a can be constructed.

a) In-Plane Principal Stresses: The coordinates of points B and D on the circle represent s1 and s2 respectively. s1 = 60.0 + 4022 = 116.57 MPa = 117 MPa

Ans.

s2 = 60.0 - 4022 = 3.4315 MPa = 3.43 MPa

Ans.

Orientation of Principal Plane: From the shaded triangle on the circle, tan 2up1 =

40 = 1  2up1 = 45.0° 100 - 60

up1 = 22.5° (Clockwise)

Ans.

Using these results, the state of in-plane principal stress can be represented by the differential element shown in Fig. b. b) Maximum In-Plane Shear Stress: Represented by the coordinates of point E on the circle. t max

in@plane

Ans.

= - R = - 40 22 MPa = - 56.6 MPa

Orientation of the Plane for Maximum In-Plane Shear Stress: From the circle, tan 2us =

100 - 60.0 = 1  2us = 45.0° 40

us = 22.5° (Counterclockwise)

Ans.

Using these results, the state of maximum in-plane shear stress can be represented by the differential element shown in Fig. c.

Ans: savg = 60.0 MPa, s1 = 117 MPa, s2 = 3.43 MPa, t max = -56.6 MPa, us = 22.5°, in@plane

up1 = 22.5° (Clockwise) 886

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9–57. Determine the principal stress, the maximum inplane shear stress, and average normal stress. Specify the orientation of the element in each case.

120 MPa psi

300 MPa psi

Solution B(0, - 120)

A(300, 120)

C(150, 0)

R = 2(300 - 150)2 + 1202 = 192.094

(MPa)

s1 = 150 + 192.094 = 342 psi MPa

Ans.

psi s2 = 150 - 192.094 = - 42.1 MPa

Ans. (MPa)

120 = 0.8 300 - 150

tan 2uP =

uP1 = 19.3° (Counterclockwise)

Ans.

savg = 150 psi MPa

Ans.

tmax

Ans.

in-plane

150 MPa 342 MPa

192 MPa 150 MPa

= 192 psi MPa

tan 2us =

42.1 MPa

300 - 150 = 1.25 120

us = 25.7°

(Clockwise)

Ans. R = 192.094, s1 = 342 MPa, s2 = -42.1 MPa, uP = 19.3° (Counterclockwise), max savg = 150 MPa, tin@plane = 19.2 MPa, us = 25.7° (Clockwise) 691

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9–58.  etermine (a) the principal stresses and (b) the maximum D in-plane shear stress and average normal stress. Specify the orientation of the element in each case.

50 MPa

30 MPa

Solution A(0, -30)

B(50, 30)

C(25, 0)

R = CA = 2 ( 25 - 0 ) 2 + 302 = 39.05 s1 = 25 + 39.05 = 64.1 MPa

Ans.

s2 = 25 - 39.05 = -14.1 MPa

Ans.

tan 2uP =

30 = 1.2 25 - 0

uP = 25.1°

Ans.

savg = 25.0 MPa

Ans.

t max

in@plane

Ans.

= R = 39.1 MPa

tan 2us =

25 - 0 = 0.8333 30 Ans.

us = -19.9°

Ans: s1 = 64.1 MPa, s2 = -14.1 MPa, uP = 25.1°, savg = 25.0 MPa, t max = 39.1 MPa, in@plane

us = -19.9° 888

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9–59.  etermine (a) the principal stresses and (b) the maximum D in-plane shear stress and average normal stress. Specify the orientation of the element in each case.

200 MPa 100 MPa 150 MPa

Solution A( -150, 100)

B(200, - 100)

C(25, 0)

R = CA = 2(150 + 25)2 + 1002 = 201.556 100 = 0.5714 150 + 25

tan 2uP =

uP = -14.9°

Ans.

s1 = 25 + 201.556 = 227 MPa

Ans.

s2 = 25 - 201.556 = -177 MPa

Ans.

= R = 202 MPa

Ans.

t max

in@plane

Ans.

savg = 25 MPa tan 2us =

150 + 25 = 1.75 100 Ans.

us = 30.1°

Ans: uP = - 14.9°, s1 = 227 MPa, s2 = -177 MPa, t max = 202 MPa, in@plane

savg = 25 MPa, us = 30.1° 889

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*9–60. Determine the principal stress, the maximum in-plane shear stress, and average normal stress. Specify the orientation of the element in each case.

30 MPa

45 MPa 50 MPa

Solution A(45, - 50)

B(30, 50)

C(37.5, 0)

R = CA = CB = 27.52 + 502 = 50.56 a)

tan 2uP =

50 7.5

s1 = 37.5 + 50.56 = 88.1 MPa

Ans.

s2 = 37.5 - 50.56 = -13.1 MPa

Ans.

2uP = 81.47°

uP = 40.7°

(Clockwise)

Ans.

b) t

max in-plane

= R = 50.6 MPa

Ans.

savg = 37.5 MPa

Ans.

2us = 90 - 2uP us = 4.27°

(Counterclockwise)

Ans.

Ans. s1 = 88.1 MPa, s2 = -13.1 MPa, up = 40.7° max tin@plane = 50.6 MPa, savg = 37.5 MPa, us = 4.27° 691

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9–61. Draw Mohr’s circle that describes each of the following states of stress.

5 MPa

20 MPa

20 MPa

5 MPa

(b)

(a)

18 MPa (c)

905

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9–62. The grains of wood in the board make an angle of 20° with the horizontal as shown. Determine the normal and shear stresses that act perpendicular and parallel to the grains if the board is subjected to an axial load of 250 N.

300 mm

60 mm 250 N

250 N 20

25 mm

Solution sx =

P 250 = = 166.67 kPa A (0.06)(0.025)

R = 83.33 Coordinates of point B: sx′ = 83.33 - 83.33 cos 40° sx′ = 19.5 kPa

Ans.

tx′y′ = - 83.33 sin 40° = -53.6 kPa

Ans.

Ans: sx′ = 19.5 kPa, tx′y′ = - 53.6 kPa 892

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9–63. The rotor shaft of the helicopter is subjected to the tensile force and torque shown when the rotor blades provide the lifting force to suspend the helicopter at midair. If the shaft has a diameter of 150 mm, determine the principal stress and maximum in-plane shear stress at a point located on the surface of the shaft.

225 kN 15 kN⭈m

Solution Internal Loadings: Considering the equilibrium of the free-body diagram of the rotor shaft’s upper segment, Fig. a, ©Fy = 0;

N - 225 = 0

N = 225 kN

©My = 0;

T - 15 = 0

T = 15 kN.m

Section Properties: The cross-sectional area and the polar moment of inertia of the rotor shaft’s cross section are A π= = (0.0752 ) 5.625(10 /3 )π m 2 J =

π

= (0.0754 ) 49.7010(10 /6 ) m 4 2

Normal and Shear Stress: The normal stress is contributed by axial stress only. = σ

N 225(10 3 ) 2 = = 12.73(106 ) N/m= 12.73 MPa A 5.625(10 /3 )π.45

The shear stress is contributed by the torsional shear stress only. [15(10 3 )](0.075) 6 22.64(10 = ) N/m 2 22.64 MPa t = = 49.7010(10 /6 ) The state of stress at point A is represented by the element shown in Fig. b. 6.366

22.64 MPa 22.64 12.73 MPa

915

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9–63. Continued Construction of the Circle: sx = 0, sy = 12.73 MPa, and txy = 22.64 MPa. Thus, savg =

sx + sy 2

=

0 + 12.73 = 6.366 MPa 2

The coordinates of reference point A and the center C of the circle are A(0, 22.64)

C(6.366, 0)

Thus, the radius of the circle is R = CA = 2(0 - 6.366)2 + 22.642 = 23.51 MPa Using these results, the circle is shown in Fig. c. In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = 6.366 + 23.51 = 29.9 MPa

Ans.

s2 = 6.366 - 23.51 = - 17.1 MPa

Ans.

Maximum In-Plane Shear Stress: The state of maximum shear stress is represented by the coordinates of point E, Fig. a. tmax

in-plane

= R = 23.5 MPa

Ans.

Ans. tmax = 23.5 MPa, s1 = 29.9 MPa, s2 = - 17.1 MPa in@plane 916

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*9–64.  he frame supports the triangular distributed load shown. T Determine the normal and shear stresses at point D that act perpendicular and parallel, respectively, to the grains. The grains at this point make an angle of 35° with the horizontal as shown.

900 N/ m

B

35 2.4 m

Support Reactions and Internal Loadings: Referring to the FBD of the entire beam, Fig. a, 1 a+ ΣMC = 0;   c (900)(3) d (1) - By(3) = 0  By = 450 N 2

1.5 m

100 mm A

Section Properties: For the rectangular cross section, Fig. c, 1 (0.1) ( 0.23 ) = 66.6667 ( 10-6 ) m4 12

QD = ∼ y ′A′ = 0.0625[0.1(0.075)] = 0.46875 ( 10-3 ) m3 Normal Stress: The normal stress is contributed by bending stress only. Applying the flexure formula, 388.8(0.025) MD yD = = 145.8 ( 103 ) Pa = 145.8 kPa (T) I 66.6667 ( 10-6 )

Shear Stress: The shear stress is contributed by the transverse shear stress only. Applying the shear formula, 414 3 0.46875 ( 10-3 ) 4 VQ = = 29.11 ( 103 ) Pa = 29.11 kPa It 66.6667 ( 10-6 ) (0.1)

Using these results, the state of stress at point D can be represented by the differential element shown in Fig. d. Construction of The Circle: In accordance to the sign convention, sx = 145.8 kPa, sy = 0 txy = 29.11 kPa and u = 55° (Counterclockwise, Fig. e). Hence, savg =

sx + sy 2

=

145.8 + 0 = 72.9 kPa 2

The coordinates of reference point A and center of circle C are A(145.8, 29.11)

C(72.9, 0)

The radius of the circle is R = CA = 2(145.8 - 72.9)2 + (29.11 - 0)2 = 78.497 kPa

897

50 mm

30 mm

1 a+ ΣMD = 0;  MD + c (720)(2.4) d (0.8) - 450(2.4) = 0  MD = 388.8 N # m 2

  tD =

0.6 m

45

E

Referring to the FBD of the left segment of the sectioned beam, Fig. b, 1 + c ΣFy = 0;  450 - (720)(2.4) - VD = 0  VD = -414 N 2

  sD =

C

3m

Solution

  I =

D

D 75 mm 100 mm

200 mm

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*9–64. Continued

Using these results, the circle shown in Fig. f, can be constructed. Stress on the Inclined Plane: The normal stress and shear stress on the inclined plane are represented by the coordinates of point P(sx′, tx′y′) on the circle which can be established by rotating radial line CA 2u = 110° counterclockwise to coincide with radial line CP, Fig. f. Here,



a = tan-1 a

29.11 b = 21.77° 145.8 - 72.9

f = 110° - 21.77° = 88.23°

Then sx′ = 72.9 + 78.497 cos 88.23° = 75.32 kPa = 75.3 kPa

Ans.

tx′y′ = - 78.497 sin 88.23° = -78.46 kPa = - 78.5 kPa

Ans.

Ans: sx′ = 75.3 kPa, tx′y′ = - 78.5 kPa 898

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9–65.  he frame supports the triangular distributed load shown. T Determine the normal and shear stresses at point E that act perpendicular and parallel, respectively, to the grains. The grains at this point make an angle of 45° with the horizontal as shown.

900 N/ m

B

35 2.4 m

C

0.6 m

3m

Solution Support Reactions and Internal Loadings: Referring to the FBD of the entire beam, Fig. a, 1 a+ ΣMC = 0;   c (900)(3) d (1) - By(3) = 0  By = 450 N 2

1.5 m

A = 0.05(0.1) = 5.00 ( 10-3 ) m2 Normal Stress: The normal stress is contributed by axial load only. NE - 450 = = - 90.0 kPa A 5.00 ( 10-3 )

Using this result, the state of stress at point E can be represented by the differential element shown in Fig. c. Construction of the Circle: In accordance to the sign convention, sx = 0, sy = -90.0 kPa, txy = 0 and u = 135° (Counterclockwise, Fig. d). Hence, 2

=

0 + ( - 90.0) 2

= - 45.0 kPa

The coordinates of reference point A and center of circle C are A(0, 0)

C( - 45.0, 0)

The radius of the circle is R = CA = 0 - ( - 45.0) = 45.0 kPa Using the results, the circle shown in Fig. e can be constructed. Stress on the Inclined Plane: The normal stress and shear stress on the inclined plane are represented by the coordinates of point P(sx′, tx′y′) on the circle which can be established by rotating the radial line CA 2u = 270° counterclockwise to coincide with radial line CP, Fig. e. Thus, sx′ = - 45.0 kPa

Ans.

tx′y′ = 45.0 kPa

Ans.

899

100 mm A

Section Properties: For the rectangular cross-section,

sx + sy

50 mm

30 mm

+ c ΣFy = 0;   - 450 - NE = 0  NE = - 450 N

savg =

45

E

Referring to the FBD of the upper segment of the sectioned column, Fig. b,

  sE =

D

D 75 mm 100 mm

200 mm

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9–65. Continued

Ans: sx′ = - 45.0 kPa, tx′y′ = 45.0 kPa 900

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9–66. Determine the principal stresses and the maximum in-plane shear stress that are developed at point A. Show the results on an element located at this point. The rod has a diameter of 40 mm.

450 N

Solution

150 mm

Using the method of sections and consider the FBD of the member’s upper cut segment, Fig. a, + c ©Fy = 0;

450 - N = 0

a + ©MC = 0;

100 mm A

150 mm

N = 450 N

450(0.1) - M = 0

B

M = 45 N # m

A = p(0.022) = 0.4(10 - 3)p m2 I =

450 N

p (0.024) = 40(10 - 9)p m4 4

The normal stress is the combination of axial and bending stress. Thus, s =

My N + A I

For point A, y = C = 0.02 m. s =

45 (0.02) 450 + = 7.520 MPa -3 0.4(10 )p 40(10 - 9)p

Since no transverse shear and torque is acting on the cross - section t = 0 The state of stress at point A can be represented by the element shown in Fig. b. In accordance to the established sign convention sx = 0, sy = 7.520 MPa and txy = 0. Thus savg =

sx + sy 2

=

0 + 7.520 = 3.760 MPa 2

Then, the coordinates of reference point A and the center C of the circle are A(0, 0)

C(3.760, 0)

Thus, the radius of the circle is R = CA = 3.760 MPa Using this results, the circle shown in Fig. c can be constructed. Since no shear stress acts on the element, s1 = sy = 7.52 MPa

s2 = sx = 0

Ans.

The state of principal stresses can also be represented by the element shown in Fig. b. The state of maximum in - plane shear stress is represented by point B on the circle, Fig. c. Thus. tmax

in-plane

= R = 3.76 MPa

Ans.

712

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9–66. Continued From the circle, 2us = 90° us = 45° (counter clockwise)

Ans.

The state of maximum In - Plane shear stress can be represented by the element shown in Fig. d.

Ans. s1 = 7.52 MPa, s2 = 0, max tin@plane = 3.76 MPa, us = 45° (Counterclockwise) 713

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9–67.  etermine the principal stresses, the maximum in-plane D shear stress, and average normal stress. Specify the orientation of the element in each case.

20 MPa 80 MPa

30 MPa

Solution In accordance to the established sign convention, sx = 30 MPa, sy = -20 MPa and txy = 80 MPa. Thus, savg =

sx + sy 2

=

30 + ( -20) 2

Ans.

= 5 MPa

Then, the coordinates of reference point A and the center C of the circle is A(30, 80)

C(5, 0)

Thus, the radius of circle is given by R = CA = 2(30 - 5)2 + (80 - 0)2 = 83.815 MPa

Using these results, the circle shown in Fig. a, can be constructed. The coordinates of points B and D represent s1 and s2 respectively. Thus s1 = 5 + 83.815 = 88.8 MPa

Ans.

s2 = 5 - 83.815 = - 78.8 MPa

Ans.

Referring to the geometry of the circle, Fig. a tan 2(uP)1 =

80 = 3.20 30 - 5 Ans.

uP = 36.3° (Counterclockwise)

The state of maximum in-plane shear stress is represented by the coordinate of point E. Thus t max

in@plane

Ans.

= R = 83.8 MPa

From the geometry of the circle, Fig. a, tan 2us =

30 - 5 = 0.3125 80 Ans.

us = 8.68° (Clockwise)

The state of maximum in-plane shear stress is represented by the element in Fig. c.

894

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9–67. Continued

Ans: savg = 5 MPa, s1 = 88.8 MPa, s2 = -78.8 MPa, uP = 36.3° (Counterclockwise), t max = 83.8 MPa, in@plane

us = 8.68° (Clockwise) 895

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*9–68. The thin-walled pipe has an inner diameter of 12 and aa thickness thickness of of 0.025 0.6 mm. 0.5mm in. and in. If If itit is is subjected subjected to to an an internal internal pressure pressure of of 3.5 500MPa psi and the axial tension tension and and torsional loadings shown, determine the principal stress at a point on the surface of the pipe.

1000 N 200 lb

1000 N 200 lb 30 20 Nm lbft

30 20 Nm lbft

Solution Section Properties:

= A = J

π 4

π 2

(0.0066 2 / 0.006 2= ) 7.56(10 /6 )π m 2

35.0 MPa

(0.0066 4 / 0.006 4= ) 0.94479(10 /9 ) m 4

59.60 MPa 209.6 MPa

60.25 r  10, thin wallwall analysis is valid. Normal Stress: Since = , thin analysis is valid. = 10 t 0.6 0.025 pr 1000 [3.5(106 )](0.006) N + + = / 6 A 2t 2(0.0006) 7.56(10 )π

R = 209.96

slong =

59.60

= 59.60(106 ) N/m 2 = 59.60 MPa shoop =

pr t

=

[3.5(106 )](0.006) 6 = 35.0(10 = ) N/m 2 35.0 MPa 0.0006

209.6 (MPa)

47.3 (MPa)

Shear Stress: Applying the torsion formula, t =

30(0.0066) Tc 6 = 209.57(10 = ) N/m 2 209.57 MPa = J 0.94479(10 /9 )

Construction of the Circle: In accordance with the sign convention sx = 59.60 7.350 MPa, ksi, sy = 35.0 MPa, and txy = -209.57 MPa. Hence, savg =

sx + sy 2

=

59.60 7.350++35.0 5.00 = 47.30 MPa 22

The coordinates for reference points A and C are

A(59.60 / 209.57)

C (47.30, 0)

The radius of the circle is

R=

(59.60 / 47.30)2 + 209.57 2 = 209.93 MPa

In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively.

σ 1 =47.30 + 209.93 =257 MPa

Ans.

/163 MPa σ2 = 47.30 / 209.93 =

Ans.

Ans. s1 = 257 MPa, s2 = -163 MPa 698

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9–69. Determine the principal stress at point A on the cross section of the hanger at section a–a. Specify the orientation of this state of stress and indicate the result on an element at the point.

0.75 m

0.75 m a

250 mm a

Solution

900 N

Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, + ©F = 0; : x

900 - N = 0

N = 900 N

+ c ©Fy = 0;

V - 900 = 0

V = 900 N

a + ©MO = 0;

900(1) - 900(0.25) - M = 0

M = 675 N # m

Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2 I =

1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12

Referring to Fig. b, QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3

Normal and Shear Stress: The normal stress is a combination of axial and bending stresses. Thus, sA =

MyA 675(0.025) N 900 + = + = 9.074 MPa -3 A I 1.4 A 10 B 1.7367 A 10 - 6 B

The shear stress is caused by the transverse shear stress. tA =

900 C 18.875 A 10 - 6 B D VQA = = 0.9782 MPa It 1.7367 A 10 - 6 B (0.01)

The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 9.074 MPa, sy = 0, and txy = 0.9782 MPa. Thus, savg =

sx + sy 2

=

9.074 + 0 = 4.537 MPa 2

The coordinates of reference points A and the center C of the circle are A(9.074, 0.9782)

0.5 m

C(4.537, 0)

Thus, the radius of the circle is R = CA = 2(9.074 - 4.537)2 + 0.97822 = 4.641 MPa Using these results, the circle is shown in Fig. d.

708

b

250 mm

b

900 N 5 mm

25 mm A

100 mm

5 mm 50 mm

5 mm

Sections a – a and b – b

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9–69. Continued In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 4.537 + 4.641 = 9.18 MPa

Ans.

s2 = 4.537 - 4.641 = - 0.104 MPa

Ans.

Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 =

0.9782 = 0.2156 9.074 - 4.537

A uP B 1 = 6.08° (counterclockwise)

Ans.

The state of principal stresses is represented on the element shown in Fig. e.

Ans. N = 900 N, V = 900 N, M = 675 N # m, A = 1.4(10- 3) m2, I = 1.7367(10- 6) m4, s1 = 9.18 MPa, s2 = -0.104 MPa, (up)1 = 6.08° (Counterclockwise) 709

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9–70. Determine the principal stress at point A on the cross section of the hanger at section b–b. Specify the orientation of the state of stress and indicate the results on an element at the point.

0.75 m

0.75 m a

250 mm a

Solution

900 N

Internal Loadings: Considering the equilibrium of the free - body diagram of the hanger’s left cut segment, Fig. a, + c ©Fy = 0;

V - 900 - 900 = 0

a + ©MO = 0;

900(2.25) + 900(0.25) - M = 0

M

Referring to Fig. b. QA = 2y1œ A1œ + y2œ A2œ = 2[0.0375(0.025)(0.005)] + 0.0475(0.005)(0.04) = 18.875 A 10 - 6 B m3

Normal and Shear Stress: The normal stress is contributed by the bending stress only. MyA 2250(0.025) = = 32.39 MPa I 1.7367 A 10 - 6 B

The shear stress is contributed by the transverse shear stress only. 1800 C 18.875 A 10 - 6 B D VQA = = 1.956 MPa It 1.7367 A 10 - 6 B (0.01)

The state stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 32.39 MPa, sy = 0, and txy = 1.956 MPa. Thus, savg =

sx + sy 2

=

32.39 + 0 = 16.19 MPa 2

The coordinates of reference point A and the center C of the circle are A(32.39, 1.956)

C(16.19, 0)

Thus, the radius of the circle is R = CA = 2(32.39 - 16.19)2 + 1.9562 = 16.313 MPa Using these results, the cricle is shown in Fig. d.

710

b

5 mm 25 mm A

100 mm

5 mm 50 mm

5 mm

Sections a – a and b – b

1 1 (0.05) A 0.13 B (0.04) A 0.093 B = 1.7367 A 10 - 6 B m4 12 12

tA =

250 mm

= 2250 N # m

A = 0.05(0.1) - 0.04(0.09) = 1.4 A 10 - 3 B m2

sA =

b

900 N

V = 1800 N

Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the hanger’s cross section are

I =

0.5 m

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9–70. Continued In - Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively. s1 = 16.19 + 16.313 = 32.5 MPa

Ans.

s2 = 16.19 - 16.313 = - 0.118 MPa

Ans.

Orientaion of Principal Plane: Referring to the geometry of the circle, Fig. d, tan 2 A uP B 1 =

A uP B 1 = 3.44°

1.956 = 0.1208 32.39 - 16.19 Ans.

(counterclockwise)

The state of principal stresses is represented on the element shown in Fig. e.

Ans. s1 = 32.5 MPa, s2 = -0.118 MPa, (up)1 = 3.44° (Counterclockwise) 711

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9–71. The ladder is supported on the rough surface at A and by a smooth wall at B. If a man weighing 675 N stands upright at C, determine the principal stresses in one of the legs at point D. Each leg is made from a 25 mm-thick board having a rectangular cross section. Assume that the total weight of the man is exerted vertically on the rung at C and is shared equally by each of the ladder’s two legs. Neglect the weight of the ladder and the forces developed by the man’s arms.

B

C 75 mm D 25 mm

3.6 m 75 mm 25 mm 25 mm

1.5 m D 1.2 m

Solution A

= A 0.025(0.075) = 1.875(10 /3 ) m 2 I

1.5 m

1 = 0.025(0.0753 ) 0.87891(10 /6 ) m 4 12

′A′ 0.025(0.025)(0.025) = y= Q = 15.625(10 /6 ) m 3 D

σD =

/N My /348.98 (47.93)(0.0125) / = / A I 1.875(10 /3 ) 0.87891(10 /6 )

= /867.78(10 3 ) N/m 2 = /867.78 KPa tD =

VQD 39.94[15.625(10 /6 )] 3 = 28.40(10 = ) N/m 2 28.40 KPa = It [0.87891(10 /6 )](0.025)

A(/867.78, / 28.40)

B(0, 28.40)

C (/433.89, 0)

(/867.78 / 433.89)2 + 28.40 2 = 434.82 KPa R= CA =

σ1 = /433.89 + 434.82 = 0.929 KPa

Ans.

σ2 = /433.89 / 434.82 = /869 KPa

Ans.

97..40 kPa 28.40 kPa 867.78 kPa

A(−867.48, −28.40)

C(−433.89, 0) B(0, 28.40)

Ans. s1 = 0.929 kPa, s2 = -869 kPa 928

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*9–72. A spherical pressure vessel has an inner radius of 1.5 m and a wall thickness of 12 mm. Draw Mohr’ Draw Mohr’s circle for the state of stress at a point on the vessel and explain the significance of the result. The vessel is subjected to an 0.56 MPa.

Solution Normal Stress: s1 = s2 =

pr 80(5)(12) 0.56(1.5)(1000) = = 4.80ksi 35 MPa 2t 2(0.5) 2(12)

Mohr’s circle: A(35, 0)0) A(4.80,

B(35, 0) 0) C(35, 0) B(4.80, C(4.80, 0)

Regardless of the orientation of the element, the shear stress is zero and the state of stress is represented by the same two normal stress components.

35 MPa

35 MPa

(35, 0)

703

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9–73.  he cylindrical pressure vessel has an inner radius of 1.25 m T and a wall thickness of 15 mm. It is made from steel plates that are welded along the 45° seam. Determine the normal and shear stress components along this seam if the vessel is subjected to an internal pressure of 8 MPa.

45 1.25 m

Solution sx =

8(1.25) pr = = 333.33 MPa 2t 2(0.015)

sy = 2sx = 666.67 MPa A(333.33, 0)

B(666.67, 0)

C(500, 0)

333.33 + 666.67 = 500 MPa 2

Ans.

tx′y′ = - R = 500 - 666.67 = - 167 MPa

Ans.

sx′ =

Ans: sx′ = 500 MPa, tx′y′ = -167 MPa 904

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9–74.  etermine the normal and shear stresses at point D that act D perpendicular and parallel, respectively, to the grains. The grains at this point make an angle of 30° with the horizontal as shown. Point D is located just to the left of the 10-kN force.

10 kN 100 mm D 30 A 1m 1m D 100 mm

300 mm

100 mm

Solution Using the method of section and consider the FBD of the left cut segment, Fig. a + c ΣFy = 0; a+ ΣMC = 0;

5 - V = 0

V = 5 kN

M - 5(1) = 0

M = 5 kN # m

The moment of inertia of the rectangular cross section about the neutral axis is I =

1 (0.1)(0.33) = 0.225(10 - 3) m4 12

Referring to Fig. b, QD = y′A′ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m. Then s =

5(103)(0.05) My = = 1.111 MPa (T) I 0.225(10 - 3)

The shear stress is contributed by the transverse shear stress only. Thus, t =

5(103)(0.001) VQD = = 0.2222 MPa It 0.225(10 - 3)(0.1)

The state of stress at point D can be represented by the element shown in Fig. c In accordance with the established sign convention, sx = 1.111 MPa, sy = 0 and txy = - 0.2222 MPa. Thus, savg =

sx + sy 2

=

1.111 + 0 = 0.5556 MPa 2

Then, the coordinate of reference point A and the center C of the circle are A(1.111, - 0.2222)

C(0.5556, 0)

Thus, the radius of the circle is given by R = 2(1.111 - 0.5556)2 + ( - 0.2222)2 = 0.5984 MPa

Using these results, the circle shown in Fig. d can be constructed. Referring to the geometry of the circle, Fig. d, a = tan - 1 a

0.2222 b = 21.80° 1.111 - 0.5556

b = 180° - (120° - 21.80°) = 81.80°

905

B 2m C

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9–74. Continued

Then sx′ = 0.5556 - 0.5984 cos 81.80° = 0.4702 MPa = 470 kPa

Ans.

tx′y′ = 0.5984 sin 81.80° = 0.5922 MPa = 592 kPa

Ans.

Ans: sx′ = 470 kPa, tx′y′ = 592 kPa 906

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9–75. Determine the principal stress at point D, which is located just to the left of the 10-kN force.

10 kN 100 mm D 30 A 1m 1m D 100 mm

300 mm

100 mm

Solution Using the method of section and consider the FBD of the left cut segment, Fig. a, + c ΣFy = 0; a+ ΣMC = 0;

5 - V = 0

V = 5 kN M = 5 kN # m

M - 5(1) = 0 I =

1 (0.1)(0.33) = 0.225(10 - 3) m4 12

Referring to Fig. b, QD = y′A′ = 0.1(0.1)(0.1) = 0.001 m3 The normal stress developed is contributed by bending stress only. For point D, y = 0.05 m s =

5(103)(0.05) My = = 1.111 MPa (T) I 0.225(10 - 3)

The shear stress is contributed by the transverse shear stress only. Thus, t =

5(103)(0.001) VQD = = 0.2222 MPa It 0.225(10 - 3)(0.1)

The state of stress at point D can be represented by the element shown in Fig. c. In accordance with the established sign convention, sx = 1.111 MPa, sy = 0, and txy = - 0.2222 MPa. Thus, savg =

sx + sy 2

=

1.111 + 0 = 0.5556 MPa 2

Then, the coordinate of reference point A and center C of the circle are A(1.111, - 0.2222)

C(0.5556, 0)

Thus, the radius of the circle is R = CA = 2(1.111 - 0.5556)2 + ( - 0.2222)2 = 0.5984 MPa

Using these results, the circle shown in Fig. d can be constructed.

In-Plane Principal Stresses: The coordinates of points B and D represent s1 and s2, respectively. Thus, s1 = 0.5556 + 0.5984 = 1.15 MPa

Ans.

s2 = 0.5556 - 0.5984 = - 0.0428 MPa

Ans.

907

B 2m C

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9–75. Continued

Referring to the geometry of the circle, Fig. d, tan (2uP)1 =

0.2222 = 0.4 1.111 - 0.5556 Ans.

(uP)1 = 10.9° (Clockwise) The state of principal stresses is represented by the element shown in Fig. e.

Ans: s1 = 1.15 MPa, s2 = -0.0428 MPa 908

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*9–76. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate force of of 400 75 lb, while subjected to aa force N, determine the principal stress in the material on the cross section at point C.

400 75 lbN B

A 100 mm 4 in.

753mm in.

C

Solution

0.2 in. 5 mm 0.3mm in. 7.5

Internal Forces and Moment: As shown on FBD Section Properties: = I

0.4mm in. 10 0.4mm in. 10

400 N 100 mm

1 = (0.0075)(0.02 3 ) 5(10 /9 ) m 4 12

M = 40000 N · mm

′A′ (0.0075)(0.0075)(0.005) = y= = 0.28125(10 /6 ) m 3 Q C

V = 400 N 7.5 mm

Normal Stress: Applying the flexure formula.

5 mm

7.5 mm

sC = -

5 mm

My (/40)(0.005) = 40(106 ) N/m 2 = 40 MPa = / I 5(10 /9 )

10 mm

Shear Stress: Applying the shear formula. 3 MPa

VQC 400[0.28125(10 /6 )] 6 = 3(10 = ) N/m 2 3 MPa tC = = It [5(10 /9 )](0.0075)

40 MPa

4.6875 ksi, Construction of the Circle: In accordance with the sign convention, sx = 40 MPa, . Hence, sy = 0, and txy = 0.3516 3 MPa.ksi Hence, savg =

sx + sy 2

20

40 +0 + 0 4.6875 = = 2.34375 ksi  20 MPa 2 2

3 (MPa)

The coordinates for reference points A and C are A(40, 3) 0.3516) C(20, 0) C(2.34375, 0) A(4.6875, The radius of the circle is

40

2 (40 – 20)-2 +2.34375) 32 = 20.224 MPa 2 = 2.3670 ksi R = 2(4.6875 + 0.3516

(MPa)

In - Plane Principal Stress: The coordinates of point B and D represent s1 and s2, respectively. s1 = 20 + 20.224 = 40.2 MPa

Ans.

s2 = 20 – 20.224 = –0.224 MPa

Ans.

Ans. s1 = 40.2 MPa, s2 = -0.224 MPa 702

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5 ksi MPa

9–77. Draw the three Mohr’s circles that describe each of the following states of stress.

Solution

MPa 3 ksi

(a) Here, smin = 0, sint = 3 MPa ksi and ksi. The andssmax = 55MPa. Thethree three Mohr’s Mohr’s circle max = of this state of stress are shown in Fig. a (b) Here, smin = 0, sint = 140 MPa and smax = 180 MPa. The three Mohr’s circle of this state of stress are shown in Fig. b

s(MPa)

t(MPa)

714

180 MPa

140 MPa (a)

(b)

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9–78. Draw the three Mohr’s circles that describe the following state of stress.

Solution –300 andssmax MPa. The three Mohr’s circle this Here, smin = 300 MPa, psi, ss . The three Mohr’s circle forfor this max== 400 psi intint==00and state of stress is shown in Fig. a.

400 400 psi MPa

s(MPa)

t(MPa)

Ans: s min = -300 MPa, sint = 0, s max = 400 MPa 714

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9–79. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.

z

y

x

90 MPa

80 MPa

Solution For y – z plane: A(0, - 80)

B(90, 80)

C(45, 0)

R = 2452 + 802 = 91.79 s1 = 45 + 91.79 = 136.79 MPa s2 = 45 - 91.79 = - 46.79 MPa Thus, sInt = 0

tabs

max

=

Ans.

sMax = 137 MPa

Ans.

sMin = -46.8 MPa

Ans.

136.79 - ( - 46.79) smax - smin = = 91.8 MPa 2 2

Ans.

Ans. sint = 0, s max = 137 MPa, s min = -46.8 MPa, abs = 91.8 MPa tmax 715

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*9–80. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.

z

yy

xx

Solution

120 MPa psi 70 70 MPa psi

Mohr’s circle for the element in y - z plane, Fig. a, will be drawn first. In accordance andttyz to the established sign convention, sy = 30 MPa, psi, ssz z== 120 120MPa psi and =7070MPa. psi. Thus yz = savg =

sy + sz 2

=

30 MPa psi

30 + 120 = 75 MPa psi 2

Thus the coordinates of reference point A and the center C of the circle are A(30, 70)

C(75, 0)

120 MPa

Thus, the radius of the circle is psi R = CA = 2(75 - 30)2 + 702 = 83.217 MPa 70 MPa

Using these results, the circle shown in Fig. b.

30 MPa

The coordinates of point B and D represent the principal stresses From the results, smax = 158.22 MPa = 158 MPa

smin = -8.22 MPa = −8.22 MPa

Ans.

sint = 0 MPa Using these results, the three Mohr’s circle are shown in Fig. c, From the geometry of the three circles, tabs

max

=

158.22 - (-8.217) smax - smin = 83.2 MPa = 2 2

Ans.

8.217 MPa

s(MPa)

158.22 MPa

s(MPa)

t(MPa) t(MPa)

Ans. sint = 0, s max = 158 MPa, s min = -8.22 MPa, abs = 83.2 MPa tmax 716

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9–81. Determine the principal stresses and the absolute maximum shear stress.

z

y

x

150 MPa

120 MPa

Solution For x - z plane: R = CA = 2(120 - 60)2 + 1502 = 161.55 s1 = 60 + 161.55 = 221.55 MPa

s2 = 60 - 161.55 = -101.55 MPa s1 = 222 MPa t abs = max

Ans.

s2 = - 102 MPa

221.55 - ( - 101.55) smax - smin = = 162 MPa 2 2

Ans.

Ans: s1 = 222 MPa, s2 = - 102 MPa, tabs = 162 MPa max

919

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z

9–82. The stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.

y

x

Solution Mohr’s circle for the element in x - z plane, Fig. a, will be drawn first. In accordance and ttxz Thus to the established sign convention, sx = –2 - 2MPa, ksi, ss = 88MPa. ksi. Thus xz = z z== 00 and savg =

sx + sz 2

=

ksi 22MPa ksi 88MPa

-2 + 0 = - 1 ksi MPa 2

Thus, the coordinates of reference point A and the center C of the circle are A( -2, 8)

C( -1, 0) 8 MPa

Thus, the radius of the circle is 2 MPa

R = CA = 2[-2 - ( -1)]2 + 82 = 265 MPa ksi Using these results, the circle in shown in Fig. b, The coordinates of points B and D represent s1 and s2, respectively.

σ 1 =/1 + 65 =7.062 MPa σ 2 =/1 / 65 =/9.062 MPa From the results obtained, ksi sint = 0 MPa

smax = 7.06 MPa ksi

smin = - 9.06 MPa ksi

Ans.

Using these results, the three Mohr’s circles are shown in Fig, c. From the geometry of the cricle, tabs

max

=

smax - smin 7.062 / (/9.062) = 8.06 MPa = 2 2

Ans. smin = 9.06 smax = 7.06

s(MPa)

s(MPa)

t(MPa)

t(MPa)

Ans. sint = 0 MPa, s max = 7.06 MPa, abs = 8.06 MPa s min = -9.06 MPa, tmax

717

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9–83. z

Determine the principal stresses and the absolute maximum shear stress.

y

x

2.5 ksi

4 ksi 5 ksi

Solution For y - z plane: A(5, - 4)

B( - 2.5, 4) 2

C(1.25, 0)

2

R = 23.75 + 4 = 5.483

s1 = 1.25 + 5.483 = 6.733 ksi s2 = 1.25 - 5.483 = - 4.233 ksi Thus,

savg = t abs = max

s1 = 6.73 ksi

Ans.

s2 = - 4.23 ksi

Ans.

6.73 + ( -4.23) 2

= 1.25 ksi

6.73 - ( -4.23) smax - smin = = 5.48 ksi 2 2

Ans.

Ans: s1 = 6.73 ksi, s2 = - 4.23 ksi, tabs = 5.48 ksi max

920

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*9–84. The state of stress at a point is shown on the element. Determine the principal stress and the absolute maximum shear stress.

z

xx

Solution

yy

2.5 kPa ksi

4 kPa ksi

For y - z plane:

5 kPa ksi

A(5, - 4)

B( - 2.5, 4)

C(1.25, 0)

R = 23.752 + 42 = 5.483 ksi s1 = 1.25 + 5.483 = 6.733 kPa s2 = 1.25 - 5.483 = - 4.233 kPa ksi Thus, Ans.

sMax = 6.73 kPa

Ans.

sInt = 0 sMin = -4.23 kPa savg = tabs

max

Ans.

6.733 + (-4.233) = 1.25 kPa ksi 2

25 kPa 4 kPa 5 kPa

6.733 - (-4.233) smax - smin = = = 5.48 kPa 2 2

Ans.

Ans. sint = 0, s max = 6.73 MPa, s min = -4.23 MPa, abs = 5.48 MPa tmax 718

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9–85.  The solid shaft is subjected to a torque, bending moment, and shear force. Determine the principal stresses at points A and B and the absolute maximum shear stress. 300 Nm

450 mm A B 25 mm

45 Nm 800 N

Solution Internal Forces and Moment: As shown on FBD. Section Properties: p ( 0.0254 ) = 0.306796 ( 10-6 ) m4 Iz = 4 p ( 0.0254 ) = 0.613592 ( 10-6 ) m4 J = 2

( QA ) v = 0 ( QB ) y = y′A′ =

4(0.025) 1 c (p) ( 0.0252 ) d = 10.417 ( 10-6 ) m3 3p 2

Normal Stress: Applying the flexure formula, s = sA = sB = -

Mz y Iz -60.0(0.025) 0.306796 ( 10-6 ) - 60.0(0) 0.306796 ( 10-6 )

= 4.889 MPa = 0

Shear Stress: Applying the torsion formula for point A, tA =

45.0(0.025) Tc = = 1.833 MPa J 0.613592 ( 10-6 )

The transverse shear stress in the y direction and the torsional shear stress can be obtained using shear formula and torsion formula. tV =

Tr VQ and ttwist = , respectively. It J tB = (tV)y - ttwist =

800 3 10.417 ( 10-6 ) 4

0.306796 ( 10-6 ) (0.05) = -1.290 MPa

-

45.0(0.025) 0.613592 ( 10-6 )

Construction of the Circle: sx = 4.889 MPa, sz = 0, and txz = -1.833 MPa for point A. Hence, savg =

sx + sz 2

=

4.889 + 0 = 2.445 MPa 2

The coordinates for reference points A and C are A (4.889, - 1.833) and C(2.445, 0).

921

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9–85.  Continued

The radius of the circle is Solution R = 2(4.889 - 2.445)2 + 1.8332 = 3.056 MPa

sx = sz = 0 and txy = - 1.290 MPa for point B. Hence, savg =

sx + sz 2

= 0

The coordinates for reference points A and C are A(0, - 1.290) and C(0, 0) The radius of the circle is R = 1.290 MPa In-Plane Principal Stresses: The coordinates of points B and D represent s1 and s2, respectively. For point A s1 = 2.445 + 3.056 = 5.50 MPa s2 = 2.445 - 3.056 = -0.611 MPa For point B, s1 = 0 + 1.290 = 1.29 MPa s2 = 0 - 1.290 = -1.290 MPa Three Mohr’s Circles: From the results obtained above, the principal stresses for point A are s1 = 5.50 MPa  s2 = - 0.611 MPa Ans. And for point B, s1 = 1.29 MPa  s2 = - 1.29 MPa

Ans.

Absolute Maximum Shear Stress: From point A, t abs = max

5.50 - ( - 0.611) smax - smin = = 3.06 MPa 2 2

Ans.

1.29 - ( -1.29) smax - smin = = 1.29 MPa 2 2

Ans.

For point B, t abs = max

Ans: s1 = 5.50 MPa, s2 = -0.611 MPa, s1 = 1.29 MPa, s2 = -1.29 MPa, t abs = 3.06 MPa, max

t abs = 1.29 MPa max

922

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9–86.  The frame is subjected to a horizontal force and couple moment. Determine the principal stresses and the absolute maximum shear stress at point A. The cross-sectional area at this point is shown.

3m 400 N 350 Nm 2m 150 mm A 150 mm

Solution I =

1 (0.05) ( 0.153 ) = 14.0625 ( 10-6 ) m4 12

50 mm A

1m

50 mm

QA = 0.05(0.05)(0.05) = 0.125 ( 10-3 ) m3 sA = tA =

450(0.025) Mx = - 800 kPa = I 14.0625 ( 10-6 )

400(0.125) ( 10-3 ) VQA = = 71.11 kPa It 14.0625 ( 10-6 ) (0.05)

A(0, 71.11)  B( - 800, - 71.11)  C( -400, 0) R = 34002 + 71.112 = 406.272 s1 = -400 + 406.272 = 6.72 kPa s2 = -400 - 406.272 = - 806 kPa Thus, s1 = 6.27 kPa s2 = -806 kPa 6.27 - ( -806.27) smax - smin t abs = = = 406 kPa max 2 2

Ans. Ans. Ans.

Ans: s1 = 6.27 kPa, s2 = - 806 kPa, t abs = 406 kPa max

923

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9–87. Determine the principal stress and absolute maximum shear stress developed at point B on the cross section of the bracket at section a–a.

12 300in.mm

Internal Loadings: Considering the equilibrium of the free - body diagram of the 150 6 in.mm bracket’s upper cut segment, Fig. a, aa

+ c ©Fy = 0;

3 N - 500 2.5 a b = 0 5

N = 1.5 300kN lb

+ ©F = 0; ; x

4 2.5 a b = 0 V - 500 5

V = 400 lb 2.0 kN

4 33 ©MO = 0; M - 500 (12) - –500 (6) = 0= 0 2.5aa b (0.150) 2.5 aa bb(0.300) 5 55

5 3 3 5 4 4

aa

500 2.5lb kN

0.5 in. 12in. mm 0.25 6 mm AA BB 0.25 in. 6 mm

0.25 in. 6 mm

mm 36 in. mm 1.536in.1.5 Section Sectiona a– –a a

M = 0.75 6000kN lb #·in m

0.3 m

Section Properties: The cross - sectional area and the moment of inertia about the centroidal axis of the bracket’s cross section are A = 0.012(0.072) / 0.006(0.06) = 0.504(10 /3 ) m 2 I=

1 1 (0.012)(0.072 3 ) / (0.006)(0.06 3 ) = 0.265248(10 /6 ) m 4 12 12

0.15 m 2.5 kN

Referring to Fig. b, QB = 0 Normal and Shear Stress: The normal stress is a combination of axial and bending stress. N A

σB = +

MxB /1.5(10 3 ) [0.75(10 3 )](0.036) = + I 0.504(10 /3 ) 0.265248(10 /6 )

0.033 m

6 = 98.82(10 = ) N/m 2 98.82 MPa

0.015 m

Since QB = 0, tB = 0. The state of stress at point B is represented on the element shown in Fig. c. In - Plane Principal Stresses: Since no shear stress acts on the element, s1 = 10.91 ksi 98.8 MPa

0.006 m 0.012 m 0.006 m

s2 = 0

0.03 m

0.03 m

0.006 m

Three Mohr’s Circles: Using these results, smax = 10.91 ksi 98.8 MPa

sint = smin = 0

Ans.

Absolute Maximum Shear Stress: tabs

max

=

smax - smin 98.82 / 0 = 49.4 MPa = 2 2

Ans.

98.82 MPa

Ans. smax = 98.8 MPa, sint = smin = 0 abs = 49.4 MPa tmax

724

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*9–88. Determine the principal stress and absolute maximum shear stress developed at point A on the cross section of the bracket at section a–a.

300 12 mm in.

150 mm 6 in. a

3

a 12 0.5mm in. B

5 4

6 mm 0.25 in. A

6 mm 0.25 in.

6 mm 0.25 in. 36 mm 36 mm 1.5 in.1.5

in. Section a – a

Solution Internal Loadings: Considering the equilibrium of the free - body diagram of the bracket’s upper cut segment, Fig. a, + c ©Fy = 0;

3 2.5 b = 0 N - 500a 5

N = 1.5 300kN lb

+ ©F = 0; ; x

4 V - 500 2.5 a b = 0 5

V = 400 lb 2.0 kN

44 33 ©MO = 0; M - 500 (12) - –500 = 0= 0 2.5aa b (6) (0.150) 2.5 aa bb(0.300) 55 55

M = 6000 lb # ·inm 0.75 kN

Section Properties: The cross - sectional area and the moment of inertia of the bracket’s cross section are

A = 0.012(0.072) / 0.006(0.06) = 0.504(10 /3 ) m 2 I=

1 1 (0.012)(0.072 3 ) / (0.006)(0.06 3 ) = 0.265248(10 /6 ) m 4 12 12

Referring to Fig. b. QA = x1œ A1œ + x2œ A2œ = 0.015(0.006)(0.03) + 0.033(0.012)(0.006) = 5.076(10 /6 ) m 3 Normal and Shear Stress: The normal stress is

1.5(10 3 )

N A

σA = = = /2.976(106 ) N/m 2 = /2.976 MPa /3 0.504(10 )

The shear stress is contributed by the transverse shear stress. tA =

VQA It

=

[2.0(10 3 )][5.076(10 /6 )] 6 = 6.379(10 = ) N/m 2 6.379 MPa [0.265248(10 /6 )](0.006)

The state of stress at point A is represented by the element shown in Fig. c. Construction of the Circle: sx = 0, sy = - 2.976 MPa, and txy = 6.379 MPa. Thus, savg =

sx + sy 2

=

0 + (–2.976) = -1.488 MPa 2

The coordinates of reference point A and the center C of the circle are

A(0, / 6.379)

C (/1.488, 0)

Thus, the radius of the circle is

R = CA =

[0 / (/1.488)]2 + 6.379 2= 6.5502 MPa 722

2.5 500kN lb

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*9–88. Continued Using these results, the cricle is shown in Fig. d. In - Plane Principal Stresses: The coordinates of reference point B and D represent s1 and s2, respectively.

σ1 = /1.488 + 6.5502 = 5.062 MPa σ2 = /1.488 / 6.5502 = /8.038 MPa Three Mohr’s Circles: Using these results, 5.06psi MPa sint = 0 smin = –8.04 MPa smax = 583 926 psi

Ans.

Absolute Maximum Shear Stress: tabs

max

=

smax - smin 5.062 / (/8.038) = 6.55 MPa = 2 2

Ans.

0.3 m

0.033 m 0.015 m 0.006 m

0.15 m

0.012 m 0.006 m

2.5 kN

0.03 m

0.03 m

0.006 m

1.488

6.379 MPa (MPa) 2.976 MPa

6.379

R = 6.5502 t (MPa)

Ans. s max = 98.8 MPa, sint = s min = 0, abs = 49.4 MPa tmax 723

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R9–1.

50 N

The wooden strut is subjected to the loading shown. Determine the principal stresses that act at point C and specify the orientation of the element at this point. The strut is supported by a bolt (pin) at B and smooth support at A.

50 N 60 C

100 mm

40 N

40 N B

A 50 mm

25 mm

200 mm 200 mm 200 mm 200 mm 100 mm 100 mm

Solution QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 1 (0.025)(0.13) = 2.0833(10 - 6) m4 12

I =

Normal stress: sC = 0 Shear stress: VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025)

t =

Principal stress: sx = sy = 0; s1,2 =

txy = - 26.4 kPa

sx + sy 2

;

C

a

sx - sy 2

2

b + t2 xy

= 0 ; 20 + (26.4)2 s1 = 26.4 kPa

s2 = - 26.4 kPa

;

Ans.

Orientation of principal stress: tan 2up =

txy (sx - sy)

= - q

2 up = + 45° and -45° Use sx¿

Eq. 9-1 to determine the principal sx + sy sx - sy = + cos 2u + txy sin 2u 2 2

plane

of

s1

and

s2

u = up = - 45° sx¿ = 0 + 0 + (- 26.4) sin(-90°) = 26.4 kPa Therefore, up1 = - 45°;

up2 = 45°

Ans.

Ans. QC = 31.25(10- 6) m3, I = 2.0833(10- 6) m4, t = 26.4 kPa, s1 = 26.4 kPa, s2 = - 26.4 kPa, up1 = -45°; up2 = 45° 736

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R9–2.

50 N

The wooden strut is subjected to the loading shown. If grains of wood in the strut at point C make an angle of 60° with the horizontal as shown, determine the normal and shear stresses that act perpendicular and parallel to the grains, respectively, due to the loading. The strut is supported by a bolt (pin) at B and smooth support at A.

50 N 60 C

100 mm

40 N

40 N B

A 25 mm

50 mm 200 mm 200 mm 200 mm 200 mm 100 mm 100 mm

Solution QC = y¿A¿ = 0.025(0.05)(0.025) = 31.25(10 - 6) m3 I =

1 (0.025)(0.13) = 2.0833(10 - 6) m4 12

Normal stress: sC = 0 Shear stress: t =

VQC 44(31.25)(10 - 6) = 26.4 kPa = It 2.0833(10 - 6)(0.025)

Stress transformation: sx = sy = 0; sx¿ =

sx + sy 2

+

sx - sy 2

txy = - 26.4 kPa;

u = 30°

cos 2u + txy sin 2u

= 0 + 0 + (- 26.4) sin 60° = - 22.9 kPa tx¿y¿ = -

sx - sy 2

Ans.

sin 2u + txy cos 2u

= - 0 + ( -26.4) cos 60° = - 13.2 kPa

Ans.

Ans. sx′ = -22.9 kPa, tx′y′ = -13.2 kPa 737

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R9–3.

A 50 MPa

The state of stress at a point in a member is shown on the element. Determine the stress components acting on the plane AB.

30

28 MPa

100 MPa

Solution

B

Construction of the Circle: In accordance with the sign convention, sx = - 50 MPa, sy = - 100 MPa, and txy = - 28 MPa. Hence, savg =

sx + sy 2

=

-50 + (- 100) = - 75.0 MPa 2

The coordinates for reference points A and C are A(–50, –28) and C(–75.0, 0). The radius of the circle is R = 2(75.0 - 50)2 + 282 = 37.54 MPa. Stress on the Rotated Element: The normal and shear stress components A sx¿ and tx¿y¿ B are represented by the coordinates of point P on the circle sx¿ = - 75.0 + 37.54 cos 71.76° = - 63.3 MPa

Ans.

tx¿y¿ = 37.54 sin 71.76° = 35.7 MPa

Ans.

Ans. sx′ = -63.3 MPa, tx′y′ = 35.7 MPa 732

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*R9−4. The crane is used to support the 1500 N load. Determine the principal stresses acting in the boom at points A and B. The cross section is rectangular and has a width of 150 mm and a thickness of 75 mm. Use Mohr's circle.

0.015 m

Solution 0.075 m

= A 0.075(0.15) = 0.01125 m 2 I QB

1 = (0.075)(0.153 ) 21.09375(10 /6 ) m 4 12 0.0375(0.075)(0.075) = 0.2109375(10 /3 ) m 3

QA = 0 Point A : P A

σA = / /

2250(0.075) My /2560.66 = / I 0.01125 21.09375(10 /6 )

0.2276 MPa

= /8.228(106 ) N/m 2 = /8.23 MPa

0.1414 MPa

τA = 0 σ1 = 0

σ 2 = /8.23 MPa

Ans

A(−0.2276, −0.1414)

Point B : P A

/2560.66 = /0.2276(106 ) N/m 2 = /0.2276 MPa 0.01125

σB = / = τB =

C(−0.1138, 0)

B(0, 0.1414)

/3

VQB 1060.66[0.2109375(10 )] 2 = 0.1414 MPa = = 0.1414(106 ) N/m It [21.09375(10 /6 )](0.075)

A(0.2276, / 0.1414)

B(0, 0.1414)

C (/0.1138, 0)

0.1815 R = 0.1138 2 + 0.1414 2 =

σ1 = /0.1138 + 0.1815 = 0.06772 MPa = 67.72 KPa

Ans

σ2 = /0.1138 / 0.1815 = /0.2953 MPa = /295 KPa

Ans

Ans: For point A :1
0, 2
8.23 MPa, For point B:1
67.7 kPa, 2
295 kPa 945

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30 MPa

R9–5. Determine the equivalent state of stress on an element at the same point which represents (a) the principal stresses, and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown and sketch the results on the element.

10 MPa

Solution Normal and Shear Stress: sx = 0

sy = -30 MPa

txy = -10 MPa

In-Plane Principal Stresses: s1, 2 =

sx + sy ;

2

A

a

sx - sy 2

2

2 b + txy

2

0 + (/30)  0 / (/30)  ±  + (/10)2 2 2  

=

= - 15 ; 2325 s1 = 3.03 MPa

s2 = - 33.0 MPa

Ans.

Orientation of Principal Plane: tan 2up =

txy (sx - sy)>2

=

- 10 = - 0.6667 [0 - (- 30)]>2

up = -16.845° and 73.155° Substituting u = - 16.845° into sx + sy

sx¿ =

=

sx - sy +

2

2

cos 2u + txy sin 2u

0 + ( - 30) 0 - ( -30) + cos ( - 33.69°) - 10 sin ( -33.69°) 2 2

= 3.03 MPa = s1 Thus, (up)1 = - 16.8° and (up)2 = 73.2°

Ans.

The element that represents the state of principal stress is shown in Fig. a. Maximum In-Plane Shear Stress: tmax

in-plane

=

a

B

sx - sy 2

2

2  0 / (/30)  2 = 18.0 MPa b + txy2 =   + (/10) 2 

946

Ans.

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R9–5. Continued Orientation of the Plane of Maximum In-Plane Shear Stress: tan 2us = -

(sx - sy)>2 [0 / (/30)] 2 = / = 1.5 txy /10

Ans.

us = 28.2° and 118° By inspection, tmax equilibrium.

in-plane

has to act in the same sense shown in Fig. b to maintain

Average Normal Stress: savg =

sx + sy 2

=

0 + ( - 30) = - 15 MPa 2

Ans.

The element that represents the state of maximum in-plane shear stress is shown in Fig. c.

Ans: s1 = 3.03 MPa, s2 = - 33.0 MPa, up1 = - 16.8° and up2 = 73.2°, tmax = 18.0 MPa, savg = - 15 MPa, us = 28.2° in-plane

947

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R9–6. The propeller shaft of the tugboat is subjected to the compressive force and torque shown. If the shaft has an inner diameter of 100 mm and an outer diameter of 150 mm, determine the principal stresses at a point A located on the outer surface. 10 kN

A 2 kN·m

Solution Internal Loadings: Considering the equilibrium of the free-body diagram of the propeller shaft’s right segment, Fig. a, ΣFx = 0; 10 - N = 0

N = 10 kN

T = 2 kN # m

ΣMx = 0; T - 2 = 0

Section Properties: The cross-sectional area and the polar moment of inertia of the propeller shaft’s cross section are A = p 1 0.0752 - 0.052 2 = 3.125p 1 10 - 3 2 m2

J =

p 1 0.0754 - 0.054 2 = 12.6953125p 1 10 - 6 2 m4 2

Normal and Shear Stress: The normal stress is a contributed by axial stress only. sA =

10 1 103 2 N = = - 1.019 MPa A 3.125p 1 10 - 3 2

The shear stress is contributed by the torsional shear stress only. tA =

2 1 103 2 (0.075) Tc = = 3.761 MPa J 12.6953125p 1 10 - 6 2

The state of stress at point A is represented by the element shown in Fig. b. Construction of the Circle: sx = - 1.019 MPa, sy = 0, and txy = -3.761 MPa. Thus, savg =

sx + sy 2

=

- 1.019 + 0 = - 0.5093 MPa 2

The coordinates of reference point A and the center C of the circle are

A( -1.019, -3.761)

C( -0.5093, 0)

Thus, the radius of the circle is

R = CA = 2[ - 1.019 - ( - 0.5093)]2 + ( - 3.761)2 = 3.795 MPa

Using these results, the circle is shown in Fig. c.

In-Plane Principal Stress: The coordinates of reference points B and D represent s1 and s2, respectively. s1 = -0.5093 + 3.795 = 3.29 MPa

Ans.

s2 = -0.5093 - 3.795 = - 4.30 MPa

Ans.

933

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R9–6.  Continued

Ans: s1 = 3.29 MPa, s2 = - 4.30 MPa 934

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R9–7.

1200 6 kNlb

800 4 kNlb

The box beam is subjected to the loading shown. Determine the principal stress in the beam at points A and B.

6 in. 150 mm A A B 8200 6 in. B in. 150 mm mm 8 in. 200 mm

Solution

A A B B 3 ft 0.9 m

2.5 ft 2.5 ft 0.75 m 0.75 m

5 ft 1.5 m

Support Reactions: As shown on FBD(a). 6 kN

4 kN

Internal Forces and Moment: As shown on FBD(b). Section Properties:

0.9 m

1 1 I = (0.2)(0.2 3 ) / (0.15)(0.153 ) = 91.1458(10 /6 ) m 4 12 12

1.5 m

8.2 kN

1.5 m 1.8 kN

QA = QB = 0 Normal Stress: Applying the flexure formula.

σ= /

My I [/0.45(10 3 )](0.1)

σA = / = 0.4937(106 ) N/m 2 = 0.494 MPa /6 91.1458(10 )

[/0.45(10 3 )](/0.075)

σB = / /6

91.1458(10 )

= /0.3703(106 ) N/m 2 = /0.370 MPa

Shear Stress: Since QA = QB = 0, then tA = tB = 0. 0 for point A. Since In - Plane Principal Stress: sx = 61.71 , sys= , and 0.494 psi MPa, andttxyxy == 0. y =00, no shear stress acts on the element, s1 = sx = 61.7 0.494psi MPa

Ans.

s2 = sy = 0

Ans.

sxx == –0.370 - 46.29MPa, psi, ssyy == 00,, and txy = 00 for for point point B. B. Since Since no no shear stress acts on the xy = element, s1 = sy = 0

Ans.

- 46.3 psi s2 = sx = –0.370 MPa

Ans.

6 kN 4.2 kN 0.45 kN · m 0.75 m

1.5 m 1.8 kN

0.494 MPa

0.370 MPa

Ans. s1 = sx = 0.494 MPa, s2 = sy = 0; s1 = sy = 0, s2 = sx = -0.370 MPa 735

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** R9–8. A A

The clamp exerts a force of 0.75 kN on the boards at G. Determine the axial force in each screw, AB and CD, and then compute the principal stresses at points E and F. Show the results on properly oriented elements located at these points. The section through EF is rectangular and is 25 mm wide, 40 mm deep.

FCD = 1.6875 kN

+ c ©Fy = 0;

1.6875 0.75-– F FAB 350 - –150 = 00 FABF= = 200kN lb AB = AB0.9375

Ans.

1 = (0.025)(0.04 3 ) 0.1333(10 /6 ) m 4 12

QE = 0

′A′ 0.014(0.025)(0.012) Q = y= = 4.2(10 /6 ) m 3 F My , I

(/37.5)(0.02)

σE = 5.625(106 ) N/m 2 = 5.625 MPa / = /6 0.1333(10 )

/37.5(10 3 )(8) 133333

(/37.5)(0.008)

6 σF = / /= 2.25(10 = ) N/m 2 2.25 MPa /6

0.1333(10 ) VQ , It

Shear Stress: Applying the shear formula t = tE = tF =

[0.9375(10 /3 )](0) [0.1333(10 /6 )](0.025)

=0

[0.9375(10 /3 )][4.2(10 /6 )] [0.1333(10 /6 )](0.025)

6 = 1.18125(10 = ) N/m 2 1.18125 MPa

MPa, 0, and forpoint pointE. E. Since no In - Plane Principal Stress: sx = 5.625 , sy s 800 psi =y 0= and txytxy==00.for shear stress acts upon the element. psiMPa s1 = sx = 800 5.625

Ans.

s2 = sy = 0

Ans.

sx = 2.25 MPa, sy = 0, and txy = 1.1825 MPa for point F. Applying Eq. 9-5 s1,2 =

sx + sy 2

;

C

a

sx - sy 2

2

b + t2xy

2

2.25 + 0  2.25 / 0  2 = ±   + 1.18125 2 2   = 1.125 ± 1.63125

σ 1 = 2.76 MPa

σ 2 = /0.506 MPa

Ans.

729

0.75lbkN 150

D D 40 1.5mm in. 40 1.5mm in.

Section Properties:

Normal Stress: Applying the flexure formula s = -

B B

Ans.

Internal Forces and Moment: As shown on FBD(b).

= I

FF

40 mm

FCD(0.080) – 0.75(180) = 0

G G

12 mm 0.5 in. E E

Support Reactions: FBD(a). a+ © MB = 0;

0.75 kN 150 lb

C C

100 mm 4 in.

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*R9–8. Continued

Orientation of Principal Plane: Applying Eq. 9-4 for point F, tan 2up =

txy

A sx - sy B >2

= θ p 23.20°

and

=

1.18125 = 1.05 (2.25 / 0) 2

/ 66.80°

Substituting the results into Eq. 9-1 with u = 23.20° yields sx¿ =

sx + sy 2

+

sx - sy 2

cos 2u + txy sin 2u

2.25 + 0 2.25 / 0 + coθ 46.40° + 1.18125θin 46.40° 2 2

=

= 2.76 MPa = σ 1 Hence,

θ p1 = 23.2°

θ p2 = /66.8°

Ans.

0.75 kN 0.08 m

0.1 m

0.9375 kN 0.04 m

0.9375 kN 37.5 N · m 0.025 m

0.014 m

0.012 m 0.02 m 0.02 m

1.18125 MPa 5.625 MPa 2.25 MPa

0.506 MPa 2.76 MPa 5.625 MPa

730

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Ans: 2 sx¿ = - 16.5 MPa, 952

tx¿y¿ = 2.95 MPa

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10–1.  Prove that the sum of the normal strains in perpendicular directions is constant, i.e., Px + Py = Px′ + Py′.

Solution Px′ = Py′ =

Px + Py 2 Px + Py 2

+ -

Px - Py 2 Px - Py 2

cos 2u + cos 2u -

gxy 2 gxy 2

sin 2u

(1)

sin 2u

(2)

Adding Eq. (1) and Eq. (2) yields: (Q.E.D.)

Px′ + Py′ = Px + Py = constant

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

Ans: N/A

938

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10–2.  The state of strain at the point on the arm has components of Px = 200 (10-6), Py = - 300 (10-6), and gxy = 400(10-6). Use the strain transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of 30° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.

y x

Solution In accordance with the established sign convention, Px = 200(10 - 6), Px′ =

Px + Py

+

2

= c

Py = - 300(10 - 6) Px - Py 2

200 + ( - 300) 2

+

gxy

cos 2u +

2

gxy = 400(10 - 6) sin 2u

200 - ( -300) 2

cos 60° +

= 248 (10 - 6) gx′y′ 2

= -a

Px - Py 2

b sin 2u +

gxy 2

u = 30°

400 sin 60° d (10 - 6) 2

cos 2u

gx′y′ = e - 3 200 - ( - 300) 4 sin 60° + 400 cos 60° f(10 - 6) = - 233(10 - 6)

Py′ =



Px + Py

= c

2

-

Px - Py 2

200 + ( - 300) 2

= - 348(10 - 6)

-

cos 2u -

gxy 2

200 - ( - 300) 2

Ans.

Ans.

sin 2u cos 60° -

400 sin 60° d (10 - 6) 2

Ans.

The deformed element of this equivalent state of strain is shown in Fig. a.

Ans: Px′ = 248(10 - 6), gx′y′ = - 233(10 - 6), Py′ = -348(10 - 6) 939

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10–3.  The state of strain at the point on the pin leaf has components of Px = 200(10-6), Py = 180(10-6), and gxy = - 300(10-6). Use the strain transformation equations and determine the equivalent in-plane strains on an element oriented at an angle of u = 60° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.

y

x

Solution Normal Strain and Shear Strain: In accordance with the sign convention, Px = 200 ( 10-6 )

Py = 180 ( 10-6 )

gxy = - 300 ( 10-6 )

u = + 60° Strain Transformation Equations: Applying Eqs. 10–5, 10–6, and 10–7, Px′ =

Px + Py 2

= a

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

200 + 180 200 - 180 - 300 + cos 120° + sin 120° b ( 10-6 ) 2 2 2

= 55.1 ( 10-6 ) gx′y′ 2

= -

Px - Py 2

sin 2u +

gxy 2

Ans.

cos 2u

gx′y′ = [ - (200 - 180) sin 120° + ( -300) cos 120°] ( 10-6 ) = 133 ( 10-6 ) Py′ =

Px + Py

= a

2

-

Ans.

Px - Py 2

cos 2u -

gxy 2

sin 2u

200 + 180 200 - 180 - 300 cos 120° sin 120° b ( 10-6 ) 2 2 2

= 325 ( 10-6 )

Ans.

Ans: Px′ = 55.1 ( 10-6 ) , gx′y′ = 133 ( 10-6 ) , Py′ = 325 ( 10-6 ) 940

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*10–4. Solve Prob. 10–3 for an element oriented u = 30° clockwise.

y

x

Solution Normal Strain and Shear Strain: In accordance with the sign convention, Px = 200 ( 10-6 )

Py = 180 ( 10-6 )

gxy = -300 ( 10-6 )

u = -30° Strain Transformation Equations: Applying Eqs. 10–5, 10–6, and 10–7, Px′ =

Px + Py 2

= c

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

200 + 180 200 - 180 -300 + cos ( - 60°) + sin ( -60°) d ( 10-6 ) 2 2 2

= 325 ( 10-6 )

gx′y′ 2

= -

Px - Py 2

sin 2u +

gxy 2

Ans.

cos 2u

gx′y′ = [ - (200 - 180) sin ( - 60°) + ( - 300) cos ( - 60°)] ( 10-6 ) = -133 ( 10-6 ) Py′ =

Px + Py

= a

2

-

Ans.

Px - Py 2

cos 2u -

gxy 2

sin 2u

200 + 180 200 - 180 -300 cos ( - 60°) sin ( -60°) b ( 10-6 ) 2 2 2

= 55.1 ( 10-6 )

Ans.

Ans: Px′ = 325 ( 10-6 ) , gx′y′ = - 133 ( 10-6 ) , Py′ = 55.1 ( 10-6 ) 941

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10–5.  The state of strain at the point on the leaf of the caster assembly has components of Px = - 400(10-6), Py = 860(10-6), and gxy = 375(10-6). Use the strain transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 30° counterclockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.

y x

Solution Normal Strain and Shear Strain: In accordance with the sign convention, Px = - 400 ( 10-6 )

Py = 860 ( 10-6 )

gxy = 375 ( 10-6 )

u = + 30° Strain Transformation Equations: Applying Eqs. 10–5, 10–6, and 10–7, Px′ =

Px + Py 2

= a

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

-400 + 860 - 400 - 860 375 + cos 60° + sin 60° b ( 10-6 ) 2 2 2

= 77.4 ( 10-6 ) gx′y′ 2

= -

Px - Py 2

sin 2u +

gxy 2

Ans.

cos 2u

gx′y′ = [ - ( - 400 - 860) sin 60° + 375 cos 60°] ( 10-6 ) = 1279 ( 10-6 ) Py′ =

Px + Py

= a

2

-

Ans.

Px - Py 2

cos 2u -

gxy 2

sin 2u

-400 + 860 - 400 - 860 375 cos 60° sin 60° b ( 10-6 ) 2 2 2

= 383 ( 10-6 )

Ans.

Ans: Px′ = 77.4 ( 10-6 ) , gx′y′ = 1279 ( 10-6 ) , Py′ = 383 ( 10-6 ) 942

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10–6.  y

The state of strain at a point on the bracket has components of Px = 150(10-6), Py = 200(10-6), gxy = -700(10-6). Use the strain transformation equations and determine the equivalent in-plane strains on an element oriented at an angle of u = 60° counterclockwise from the original position. Sketch the deformed element within the x–y plane due to these strains.

x

Solution Px = 150 (10 - 6)  Py = 200 (10 - 6)  gxy = - 700 (10 - 6)  u = 60° Px′ =

Px + Py 2

= c

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

150 + 200 150 - 200 - 700 b sin 120° d 10 - 6 + cos 120° + a 2 2 2

= -116 (10 - 6)

Py′ =

Px + Py 2

= c

-

Px - Py 2

cos 2u -

gxy 2

sin 2u

150 + 200 150 - 200 - 700 cos 120° - a b sin 120° d 10 - 6 2 2 2

= 466 (10 - 6)

gx′y′ 2

= -

Px - Py

gx′y′ = 2c -

2

sin 2u +

gxy 2

Ans.

Ans.

cos 2u

150 - 200 - 700 sin 120° + a b cos 120° d 10 - 6 = 393 (10 - 6) 2 2

Ans.

Ans: Px′ = -116(10 - 6), Py′ = 466(10 - 6), gx′y′ = 393(10 - 6) 943

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10–7.  Solve Prob. 10–6 for an element oriented u = 30° clockwise.

y

x

Solution Px = 150 (10 - 6)  Py = 200 (10 - 6)  gxy = - 700 (10 - 6)  u = -30° Px′ =

Px + Py 2

= c

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

150 + 200 150 - 200 - 700 b sin ( -60°) d 10 - 6 + cos ( - 60°) + a 2 2 2

= 466 (10 - 6)

Py′ =

Px + Py 2

= c

-

Px - Py 2

cos 2u -

gxy 2

sin 2u

150 + 200 150 - 200 - 700 cos ( - 60°) - a b sin ( -60°) d 10 - 6 2 2 2

= - 116 (10 - 6)



gx′y′ 2

= -

Px - Py

gx′y′ = 2c -

2

sin 2u +

gxy 2

Ans.

Ans.

cos 2u

150 - 200 - 700 sin ( - 60°) + cos ( - 60°) d 10 - 6 2 2

= -393(10 - 6)

Ans.

Ans: Px′ = 466(10 - 6), Py′ = - 116(10 - 6), gx′y′ = -393(10 - 6) 944

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*10–8. The state of strain at the point on the spanner wrench has components of Px = 260(10-6), Py = 320(10-6), and gxy = 180(10-6). Use the strain transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.

y x

Solution a) In-Plane Principal Strain: Applying Eq. 10–9, P1, 2 =

Px + Py

{

2

= c

B

a

Px - Py 2

2

b + a

gxy 2

b

2

260 + 320 260 - 320 2 180 2 a b + a b d ( 10-6 ) { 2 B 2 2

= 290 { 94.87 P1 = 385 ( 10-6 )

P2 = 195 ( 10-6 )

Ans.

Orientation of Principal Strain: Applying Eq. 10–8, tan 2up =

gxy Px - Py

=

180 ( 10-6 ) (260 - 320) ( 10-6 )

up = - 35.78°

and

= - 3.000

54.22°

Use Eq. 10–5 to determine which principal strain deforms the element in the x′ direction with u = - 35.78°. Px′ =

Px + Py

= c

2

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

260 + 320 260 - 320 180 + cos ( - 71.56°) + sin ( - 71.56°) d ( 10-6 ) 2 2 2

= 195 ( 10-6 ) = P2 Hence,

up1 = 54.2°

and

Ans.

up1 = - 35.8°

b) Maximum In-Plane Shear Strain: Applying Eq. 10–11, g max Px - Py 2 gxy 2 in@plane = a b + a b 2 B 2 2 g

max in@plane

= 2c

260 - 320 2 180 2 b + a b d ( 10-6 ) B 2 2 a

= 190 ( 10-6 )

Ans.

945

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*10–8.  Continued

Orientation of Maximum In-Plane Shear Strain: Applying Eq. 10–10, tan 2us = -

Px - Py gxy

= -

260 - 320 = 0.3333 180

us = 9.22°

and

Ans.

- 80.8°

Normal Strain and Shear Strain: In accordance with the sign convention, Px = 260 ( 10 - 6 )

Py = 320 ( 10 - 6 )

The proper sign of g gx′y′ 2

= -

max in@plane

Px - Py 2

gxy = 180 ( 10 - 6 )

can be determined by substituting u = 9.22° into Eq. 10–6. sin 2u +

gxy 2

cos 2u

gx′y′ = [ -(260 - 320) sin 18.44° + 180 cos 18.44°] ( 10-6 ) = 190 ( 10-6 ) Average Normal Strain: Applying Eq. 10–12, Pavg =

Px + Py 2

= c

260 + 320 d ( 10-6 ) = 290 ( 10-6 ) 2

Ans.

Ans: P1 = 385 ( 10-6 ) , P2 = 195 ( 10-6 ) , up1 = 54.2°, up1 = -35.8°, g max = 190 ( 10-6 ) in@plane

us = 9.22°

and

Pavg = 290 ( 10-6 ) 946

- 80.8°,

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10–9.  The state of strain at the point on the member has components of Px = 180(10-6), Py = - 120(10-6), and gxy = -100(10-6). Use the strain transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.

y

x

Solution a) In accordance with the established sign convention, Px = 180(10 - 6), Py = - 120(10 - 6) and gxy = - 100(10 - 6). Px + Py

P1, 2 =

{

2

= b

Px - Py 2

180 + ( - 120) 2

1 30

=

Aa

B

{ 158.11 2 (10 ) -6

P1 = 188(10 - 6) tan 2uP =

{

gxy Px - Py

3 180 and



= c

Px - Py

+

2

2

b

2

180 - ( - 120) 2

2

d + a

-100 2 b r (10 - 6) 2 Ans.

-6

- 100(10 )

=

Substitute u = - 9.217°, Px + Py

c

gxy

P2 = - 128(10 - 6)

uP = - 9.217°

Px′ =

2

b + a

2

180 + ( - 120) 2

+

- ( - 120) 4 (10 - 6)

= - 0.3333

80.78°

cos 2u +

gxy 2

sin 2u

180 - ( -120) 2

cos ( - 18.43°) +

-6

= 188(10 ) = P1

-100 sin ( -18.43) d (10 - 6) 2

Thus, (uP)1 = - 9.22°

Ans.

(uP)2 = 80.8°

The deformed element is shown in Fig (a). b) Pavg =

180(10 - 6) + ( - 120)(10 - 6) 2

g max

in@plane

2 g max

in@plane

=

B

a

= b2

tan 2us = - a

Px - Py

B

2 c

Px - Py gxy

2

b + a

gxy 2

180 - ( - 120) 2

b = -c

3 180

us = 35.78° = 35.8° and

b

= 30(10 - 6)

Ans.

2

2

d + a

- 100 2 -6 -6 b r(10 ) = 316 1 10 2 2

- ( - 120) 4 (10 - 6) - 100(10 - 6)

Ans.

s = 3 Ans.

- 54.22° = - 54.2°

Ans: P1 = 188(10 - 6), P2 = - 128(10 - 6), (uP)1 = -9.22°, (uP)2 = 80.8° Pavg = 30(10 - 6) g max = 316 1 10 - 6 2 , in@plane

us = 35.8° and 947

-54.2°

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10–10.  The state of strain at the point on the support has components of Px = 350(10-6), Py = 400(10-6), gxy = - 675(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.

Solution

P

a)

P1, 2 =

=

Px + Py

{

2

A

a

Px - Py 2

2

b + a

gxy 2

b

2

350 + 400 350 - 400 2 - 675 2 { b + a b a 2 A 2 2

P1 = 713 (10 - 6) tan 2up =

Ans.

gxy Px - Py

=

P2 = 36.6 (10 - 6)

Ans.

- 675 (350 - 400) Ans.

up1 = 133° b)

(gx′y′)max 2

(gx′y′)max 2

=

=

A

a

A

a

Px - Py 2

gxy 2 2 b + a 2 b

350 - 400 2 - 675 2 b + a b 2 2

(gx′y′)max = 677(10 - 6) Pavg =

Px + Py

tan 2us =

2

=

350 + 400 = 375 (10 - 6) 2

- (Px - Py) gxy

Ans.

=

Ans.

350 - 400 675 Ans.

us = -2.12°

Ans: (a) P1 = 713(10-6), P2 = 36.6(10 - 6), up1 = 133° (b) g max = 677(10 - 6), Pavg = 375(10 - 6), in@plane

us = -2.12° 948

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10–11.  Due to the load P, the state of strain at the point on  the bracket has components of Px = 500(10-6), Py = 350(10-6), and gxy = -430(10-6). Use the strain transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 30° clockwise from the original position. Sketch the deformed element due to these strains within the x–y plane.

P y

x

Solution Normal Strain and Shear Strain: In accordance with the sign convention, Px = 500 ( 10-6 ) u = -30°

Py = 350 ( 10-6 )

gxy = -430 ( 10-6 )

Strain Transformation Equations: Px′ =

Px + Py 2

= c

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

500 + 350 500 - 350 -430 + cos ( - 60°) + sin ( -60°) d ( 10-6 ) 2 2 2

= 649 ( 10-6 ) gx′y′ 2

= -

Px - Py 2

sin 2u +

gxy 2

Ans.

cos 2u

gx′y′ = [ -(500 - 350) sin ( -60°) + ( - 430) cos ( - 60°)] ( 10-6 ) = -85.1 ( 10-6 ) Py′ =

Px + Py

= a

2

-

Ans.

Px - Py 2

cos 2u -

gxy 2

sin 2u

500 + 350 500 - 350 -430 cos ( - 60°) sin ( -60°) b ( 10-6 ) 2 2 2

= 201 ( 10-6 )

Ans.

Ans: Px′ = 649 ( 10-6 ) , gx′y′ = - 85.1 ( 10-6 ) , Py′ = 201 ( 10-6 )

949

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*10–12. The state of strain on an element has components Px = -400(10-6), Py = 0, gxy = 150(10-6). Determine the equivalent state of strain on an element at the same point oriented 30° clockwise with respect to the original element. Sketch the results on this element.

y

gxy dy 2 x gxy 2

Solution

Pxdx

dx

Strain Transformation Equations: Px = - 400(10 - 6)  Py = 0  gxy = 150(10 - 6)  u = -30° We obtain Px′ =

Px + Py 2

= c

+

Px - Py 2

gxy

cos 2u +

2

sin 2u

- 400 + 0 - 400 - 0 150 + cos ( -60°) + sin ( - 60°) d (10 - 6) 2 2 2

= - 365(10 - 6) gx′y′ 2

= -a

ex - ey 2

b sin 2u +

gxy 2

Ans.

cos 2u

gx′y′ = [ - ( -400 - 0) sin ( -60°) + 150 cos ( -60°)](10 - 6) = - 271(10 - 6) Py′ =

Px + Py

= c

2

-

Ans.

Px - Py 2

cos 2u -

gxy 2

sin 2u

- 400 + 0 - 400 - 0 150 cos ( -60°) sin ( - 60°) d (10 - 6) 2 2 2

= - 35.0(10 - 6)

Ans.

The deformed element for this state of strain is shown in Fig. a.

Ans: Px′ = -365(10 - 6), gx′y′ = -271(10 - 6), Py′ = -35.0(10 - 6) 950

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10–13.  y

The state of plane strain on the element is Px = - 300(10-6), Py = 0, and gxy = 150(10-6). Determine the equivalent state of strain which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element.

gxy dy 2 x gxy 2

Solution

dx

In-Plane Principal Strains: Px = - 300 1 10 - 6 2 , Py = 0, and gxy = 150 1 10 - 6 2 . We obtain Px + Py

P1, 2 =

{

2

= C

¢

B

Px - Py 2

2

≤ + ¢

gxy 2

≤

2

- 300 + 0 - 300 - 0 2 150 2 { ¢ ≤ + ¢ ≤ S 1 10 - 6 2 2 C 2 2

= ( -150 { 167.71) 1 10 - 6 2

P1 = 17.7 1 10 - 6 2

P2 = - 318 1 10 - 6 2

Ans.

Orientation of Principal Strain: tan 2up =

gxy Px - Py

150 1 10 - 6 2

=

( - 300 - 0) 1 10 - 6 2

= - 0.5

uP = -13.28° and 76.72°

Substituting u = - 13.28° into Eq. 9–1, Px′ =

Px + Py 2

= c

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

- 300 + 0 - 300 - 0 150 + cos ( - 26.57°) + sin ( - 26.57°) d 1 10 - 6 2 2 2 2

= -318 1 10 - 6 2 = P2

Thus,

1 uP 2 1

= 76.7° and 1 uP 2 2 = - 13.3°

Ans.

The deformed element of this state of strain is shown in Fig. a. Maximum In-Plane Shear Strain: g max Px - Py 2 gxy 2 in@plane = ¢ ≤ + ¢ ≤ 2 B 2 2 g max

in@plane

= J2

- 300 - 0 2 150 2 -6 -6 b + a b R 1 10 2 = 335 1 10 2 A 2 2 a

Ans.

Orientation of the Maximum In-Plane Shear Strain: tan 2us = - ¢

Px - Py gxy

us = 31.7° and 122°

≤ = -C

( -300 - 0) 1 10 - 6 2 150 1 10 - 6 2

S = 2 Ans.

951

Pxdx

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10–13.  Continued

The algebraic sign for g max Solution

in@plane

gx′y′ 2

= -¢

Px - Py 2

≤ sin 2u +

when u = us = 31.7° can be obtained using gxy 2

cos 2u

gx′y′ = [ - ( - 300 - 0) sin 63.43° + 150 cos 63.43°] 1 10 - 6 2 = 335 1 10 - 6 2

Average Normal Strain: Pavg =

Px + Py 2

= a

- 300 + 0 b 1 10 - 6 2 = - 150 1 10 - 6 2 2

Ans.

The deformed element for this state of strain is shown in Fig. b.

Ans: P1 = 17.7(10 - 6), P2 = -318(10 - 6), up1 = 76.7° and up2 = -13.3°, g max = 335(10 - 6), us = 31.7° and 122°, in@plane

Pavg = -150(10 - 6) 952

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10–14.  The state of strain at the point on a boom of a shop crane has components of Px = 250(0-6 ), Py = 300(10-6), and gxy = - 180(10-6 ). Use the strain transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case, specify the orientation of the element and show how the strains deform the element within the x–y plane.

y

Solution a) In-Plane Principal Strain: Applying Eq. 10–9, Px + Py

P1, 2 =

2

= J

{

B

a

Px - Py 2

2

b + a

gxy 2

b

2

250 + 300 250 - 300 2 - 180 2 -6 { a b + a b R 1 10 2 2 A 2 2

= 275 { 93.41

P1 = 368 ( 10-6 )     P2 = 182 ( 10-6 )

Ans.

Orientation of Principal Strain: Applying Eq. 10–8, tan 2uP =

gxy Px - Py

=

- 180(10 - 6) (250 - 300)(10 - 6)

uP = 37.24° and

= 3.600

- 52.76°

Use Eq. 10–5 to determine which principal strain deforms the element in the x′ direction with u = 37.24°. Px′ =

Px + Py

= c

2

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

250 + 300 250 - 300 - 180 + cos 74.48° + sin 74.48° d 1 10 - 6 2 2 2 2

= 182 ( 10-6 ) = P2 Hence,

Ans.

up1 = -52.8° and up2 = 37.2° b) Maximum In-Plane Shear Strain: Applying Eq. 10–11, g max Px - Py 2 gxy 2 in@plane = a b + a b 2 B 2 2 g max

in@plane

= 2J

250 - 300 2 - 180 2 -6 b + a b R 1 10 2 2 2 A a

= 187 1 10 - 6 2

Ans.

953

x

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10–14.  Continued

Orientation of the Maximum In-Plane Shear Strain: Applying Eq. 10–10, Solution tan 2us = -

Px - Py gxy

= -

250 - 300 = - 0.2778 - 180 Ans.

us = - 7.76° and 82.2°

The proper sign of g max can be determined by substituting u = -7.76° into in@plane Eq. 10–6. gx′y′ 2

= -

Px - Py 2

sin 2u +

gxy 2

cos 2u

gx′y′ = 5- [250 - 300] sin ( -15.52°) + ( - 180) cos ( -15.52°)6 1 10 - 6 2 = - 187 1 10 - 6 2

Normal Strain and Shear Strain: In accordance with the sign convention, Px = 250 1 10 - 6 2

Py = 300 1 10 - 6 2

gxy = - 180 1 10 - 6 2

Average Normal Strain: Applying Eq. 10–12, Pavg =

Px + Py 2

= c

250 + 300 d 1 10 - 6 2 = 275 1 10 - 6 2 2

Ans.

Ans: P1 = 368(10 - 6), P2 = 182(10 - 6), up1 = -52.8° and up2 = 37.2°, g max = 187(10 - 6), us = -7.76° and 82.2°, in@plane

Pavg = 275 (10 - 6) 954

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*10–16.  y

The state of strain on the element has components Px = -300(10-6), Py = 100(10-6), gxy = 150(10-6). Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding elements for these states of strain with respect to the original element.

Pydy

dy

gxy 2 dx

Solution In-Plane Principal Strains: Px = - 300(10 - 6), Py = 100(10 - 6), and gxy = 150(10 - 6). We obtain Px + Py

P1, 2 =

{

2

= c

A

a

Px - Py 2

2

b + a

gxy 2

b

2

150 2 - 300 + 100 - 300 - 100 2 { d (10 - 6) b + a a 2 A 2 2 b

= ( -100 { 213.60)(10 - 6)

P1 = 114(10 - 6)    P2 = - 314(10 - 6)

Ans.

Orientation of Principal Strains: tan 2up =

gxy Px - Py

=

150(10 - 6) ( -300 - 100)(10 - 6)

= - 0.375

up = -10.28° and 79.72°



Substituting u = - 10.28° into Px′ =

Px + Py 2

= c

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

- 300 + 100 - 300 - 100 150 + cos ( - 20.56°) + sin ( -20.56°) d (10 - 6) 2 2 2

= - 314(10 - 6) = P2 Thus,

Ans.

(up)1 = 79.7° and (up)2 = - 10.3° The deformed element for the state of principal strain is shown in Fig. a. Maximum In-Plane Shear Strain: g max gxy 2 Px - Py 2 in@plane = Aa 2 b + a 2 b 2 g max

in@plane

= c2

A

gxy 2

a

150 2 - 300 - 100 2 d (10 - 6) = 427(10 - 6) b + a 2 2 b

955

Ans.

Pxdx

x

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*10–16.  Continued

Orientation of Maximum In-Plane Shear Strain: tan 2us = - a

Px - Py gxy

b = -c

( - 300 - 100)(10 - 6) 150(10 - 6)

us = 34.7° and 125°

The algebraic sign for g max

in@plane

gx′y′ 2

= -a

Px - Py 2

d (10 - 6) = 2.6667

Ans.

when u = us = 34.7° can be determined using

b sin 2u +

gxy 2

cos 2u

gx′y′ = [ -( -300 - 100) sin 69.44° + 150 cos 69.44°](10 - 6) = 427(10 - 6) Average Normal Strain: Pavg =

Px + Py 2

= a

- 300 + 100 b(10 - 6) = - 100(10 - 6) 2

Ans.

The deformed element for this state of maximum in-plane shear strain is shown in Fig. b.

Ans: P1 = 114(10 - 6), P2 = -314(10 - 6), (up)1 = 79.7° and (up)2 = -10.3°, g max = 427(10 - 6), us = 34.7° and 125°, in@plane Pavg = -100(10 - 6) 956

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10–17.  Solve Prob. 10–3 using Mohr’s circle.

Solution Construction

of

the

Circle:

In

accordance with the sign gxy ex = 200 ( 10 - 6 ) , ey = 180 ( 10 - 6 ) , and = - 150 ( 10 - 6 ) . Hence, 2 ex + ey 200 + 180 eavg = = a b (1 0 - 6 ) = 1 9 0 (1 0 - 6 ) 2 2

convention,

The coordinates for reference points A and C are A(200, -150) ( 10 - 6 )

C(190, 0) ( 10 - 6 )

The radius of the circle is R =

( 2(200 - 190)2 + 1502 )( 10 - 6 ) = 150.33 ( 10 - 6 )

Strain on The Inclined Element: The normal and shear strain aex′ and

represented by coordinates of point P on the circle. eg′ calculating the coordinates of point Q on the circle.

b are 2 can be determined by

ex′ = (190 - 150.33cos 26.19°) ( 10 - 6 ) = 55.1 ( 10 - 6 ) gx′y′ 2

gx′y′

Ans.

= (150.33sin 26.19°) ( 10-6 )

gx′y′ = 133 ( 10-6 )

Ans.

ey′ = (190 + 150.33cos 26.19°) ( 10 - 6 ) = 325 ( 10 - 6 )

Ans.

Ans: Px′ = 55.1 ( 10-6 ) , gx′y′ = 133 ( 10-6 ) , Py′ = 325 ( 10-6 ) 957

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10–18.  Solve Prob. 10–4 using Mohr’s circle.

Solution Construction

of

the

Circle:

accordance with the sign gxy ex = 200 ( 10 - 6 ) , ey = 180 ( 10 - 6 ) , and = - 150 ( 10 - 6 ) . Hence, 2 eavg =

ex + ey 2

= a

In

convention,

200 + 180 b ( 10 - 6 ) = 190 ( 10 - 6 ) 2

The coordinates for reference points A and C are A(200, -150) ( 10-6 )

C(190, 0) ( 10-6 )

The radius of the circle is R =

( 2(200 - 190)2 + 1502 )( 10 - 6 ) = 150.33 ( 10 - 6 )

Strain on the Inclined Element: The normal and shear strain aex′ and

represented by coordinates of point P on the circle. ey′ calculating the coordinates of point Q on the circle.

b are 2 can be determined by

ex′ = (190 + 150.33cos 26.19°) ( 10 - 6 ) = 325 ( 10 - 6 ) gx′y′ 2

gx′y′

Ans.

= - (150.33sin 26.19°) ( 10-6 )

gx′y′ = -133 ( 10-6 )

Ans.

ey′ = (190 - 150.33cos 26.19°) ( 10-6 ) = 55.1 ( 10-6 )

Ans.

Ans: Px′ = 325 ( 10-6 ) , gx′y′ = -133 ( 10-6 ) , Py′ = 55.1 ( 10-6 ) 958

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10–19.  Solve Prob. 10–5 using Mohr’s circle.

Solution Construction of the Circle: In accordance ex = -400 ( 10 - 6 ) , ey = 860 ( 10 - 6 ) and gxy

with

the

sign

convention

= 187.5 ( 10 - 6 ) . Hence,

2

eavg =

ex + ey 2

= a

- 400 + 860 b ( 10 - 6 ) = 230 ( 10 - 6 ) 2

The coordinates for reference points A and C are A( -400, 187.5) ( 10 - 6 )

C(230, 0) ( 10 - 6 )

The radius of the circle is R =

( 2(400 + 230)2 + 187.52 )( 10 - 6 ) = 657.31 ( 10 - 6 )

Strain on the Inclined Element: The normal and shear strain aex′ and

gx′y′

b are 2 represented by the coordinates of point P on the circle. ey′ can be determined by calculating the coordinates of point Q on the circle. ex′ = (230 - 657.31cos 76.57°) ( 10 - 6 ) = 77.4 ( 10 - 6 ) gx′y′ 2

Ans.

= (657.31sin 76.57°) ( 10 - 6 )

gx′y′ = 1279 ( 10 - 6 ) ey′ = (2 3 0 + 6 5 7 .3 1 cos 7 6 .5 7 °)(1 0

Ans. -6

) = 3 8 3 (1 0

-6

)

Ans.

Ans: Px′ = 77.4 ( 10-6 ) , gx′y′ = 1279 ( 10-6 ) , Py′ = 383 ( 10-6 ) 959

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*10–20.  Solve Prob. 10–8 using Mohr’s circle.

Solution Construction of the Circle: In accordance ex = 260 ( 10 - 6 ) , ey = 320 ( 10 - 6 ) , and gxy 2

with

the

sign

convention

= 90 ( 10 - 6 ) . Hence, eavg =

ex + ey 2

= a

260 + 320 b ( 10 - 6 ) = 290 ( 10 - 6 ) 2

Ans.

The coordinates for reference points A and C are

A(260, 90) ( 10 - 6 )

C(290, 0) ( 10 - 6 )

The radius of the circle is R =

( 2(290 - 260)2 + 902 )( 10 - 6 ) = 94.868 ( 10 - 6 )

In-Plane Principal Strain: The coordinates of points B and D represent e1 and e2, respectively. e1 = (290 + 94.868) ( 10 - 6 ) = 385 ( 10 - 6 ) e2 = (2 9 0 - 9 4 .8 6 8 )(1 0

-6

) = 1 9 5 (1 0

Ans. -6

)

Ans.

Orientation of Principal Strain: From the circle, tan 2uP2 =

90 = 3.000 290 - 260

2uP2 = 71.57°

2uP1 = 180° - 2uP2 uP1 =

180° - 71.57° = 54.2°  (Counterclockwise) 2

Ans.

Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in@plane

2 g

max in@plane

= R = 94.868 ( 10 - 6 ) = 190 ( 10 - 6 )

Ans.

Orientation of Maximum In-Plane Shear Strain: From the circle, tan 2us =

290 - 260 = 0.3333 90

Ans: Ans.

us = 9.22°  (Counterclockwise)

eavg = 290 ( 10 - 6 ) , e1 = 385 ( 10 - 6 ) , e2 = 195 ( 10 - 6 ) , uP1 = 54.2° (Counterclockwise), g max = 190 ( 10 - 6 ) , in@plane

us = 9.22° (Counterclockwise) 960

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10–21.  Solve Prob. 10–7 using Mohr’s circle.

Solution Px = 150(10 - 6)  Py = 200(10 - 6)  gxy = - 700(10 - 6)  

gxy 2

= -350(10 - 6)

u = - 30°  2u = - 60° A(150, -350);

C(175, 0)

R = 2(175 - 150)2 + ( - 350)2 = 350.89

Coordinates of point B:

Px′ = 350.89 cos 34.09° + 175 = 466(10 - 6) gx′y′ 2

Ans.

= -350.89 sin 34.09°

gx′y′ = - 393(10 - 6)

Ans.

Coordinates of point D: Py′ = 175 - 350.89 cos 34.09° = - 116(10 - 6)

Ans.

Ans: Px′ = 466(10 - 6), gx′y′ = - 393(10 - 6), Py′ = -116(10 - 6) 961

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10–22.  The strain at point A on the bracket has components Px = 300(10-6 ), Py = 550(10-6 ), gxy = - 650(10-6 ), Pz = 0. Determine (a) the principal strains at A in the x9y plane, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain.

y A

x

Solution Px = 300(10 - 6) Py = 550(10 - 6) gxy = - 650(10 - 6)

gxy 2

= -325(10 - 6)

A(300, -325)10 - 6 C(425, 0)10 - 6 R = a)

3 2(425

- 300)2 + ( - 325)2 4 10 - 6 = 348.2(10 - 6)

P1 = (425 + 348.2)(10 - 6) = 773(10 - 6)

Ans.

P2 = (425 - 348.2)(10 - 6) = 76.8(10 - 6)

Ans.

b) g max

in@plane

c) g max

in@plane

2

Ans.

= 2R = 2(348.2)(10 - 6) = 696(10 - 6)

=

773(10 - 6) 2

Ans.

; g abs = 773(10 - 6) max

Ans: (a) P1 = 773(10 - 6), P2 = 76.8(10 - 6), (b) gmax = 696(10 - 6), in@plane

-6 g (c) abs = 773(10 ) max

962

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10–23.  The strain at point A on a beam has components Px = 450(10-6), Py = 825(10-6), gxy = 275(10-6), Pz = 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x–y plane, and (c) the absolute maximum shear strain.

A

Solution Px = 450(10 - 6)

Py = 825(10 - 6)

A(450, 137.5)10 - 6

gxy = 275(10 - 6)

gxy 2

= 137.5(10 - 6)

C(637.5, 0)10 - 6

R = [ 2(637.5 - 450)2 + 137.52]10 - 6 = 232.51(10 - 6)

a)

P1 = (637.5 + 232.51)(10 - 6) = 870(10 - 6)

Ans.

P2 = (637.5 - 232.51)(10 - 6) = 405(10 - 6)

Ans.

g max

= 2R = 2(232.51)(10 - 6) = 465(10 - 6)

Ans.

870(10 - 6)

Ans.

b) in@plane

c) g abs

max

2

=

2

; g abs = 870(10 - 6) max

Ans: (a) P1 = 870(10 - 6), P2 = 405(10 - 6), (b) g max = 465(10 - 6), in@plane

(c) g abs = 870(10 - 6) max

963

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*10–24.  The strain at point A on the pressure-vessel wall has components Px = 480(10-6), Py = 720(10-6), -6 gxy = 650(10 ). Determine (a) the principal strains at A, in the x9y plane, (b) the maximum shear strain in the x9y plane, and (c) the absolute maximum shear strain.

y A

x

Solution Px = 480(10 - 6)

Py = 720(10 - 6)

A(480, 325)10 - 6

C(600, 0)10 - 6

gxy = 650(10 - 6)

gxy 2

= 325(10 - 6)

R = ( 2(600 - 480)2 + 3252 )10 - 6 = 346.44(10 - 6)

a)

P1 = (600 + 346.44)10 - 6 = 946(10 - 6)

Ans.

P2 = (600 - 346.44)10 - 6 = 254(10 - 6)

Ans.

b) g max

in@plane

= 2R = 2(346.44)10 - 6 = 693(10 - 6)

Ans.

c) g abs

max

2

=

946(10 - 6) 2

;

Ans.

g abs = 946(10 - 6) max

Ans: P1 = 946(10 - 6), P2 = 254(10 - 6), g max -6 , in@plane = 693(10 ) g abs = 946(10 - 6) max

964

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10–25.  The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: Pa = -200(10-6), Pb = 300(10-6), and Pc = 250(10-6). Determine the in-plane principal strains. a

b

c

45 45 45

Solution Strain Rosettes (45°): Applying the equation in the text with Pa = -200 ( 10-6 ) , Pb = 300 ( 10-6 ) , Pc = 250 ( 10-6 ) , ua = 135°, ub = 90° and uc = 45°. 300 ( 10-6 ) = Px cos2 90° + Py sin2 90° + gxy sin 90° cos 90° Py = 300 ( 10-6 ) - 200 ( 10-6 ) = Px cos2 135° + 300 ( 10-6 ) sin2 135° + gxy sin 135° cos 135° Px - gxy = - 700 ( 10-6 ) 250 ( 10

-6

(1)

) = Px cos 45° + 300 ( 10 ) sin 45° + gxy sin 45° cos 45° -6

2

2

Px + gxy = 200 ( 10-6 )

(2)

Solving Eqs. (1) and (2) Px = - 250 ( 10-6 )    gxy = 450 ( 10-6 ) Construction of The Circle: With Px = - 250 ( 10-6 ) , Py = 300 ( 10-6 ) and gxy = 225 ( 10-6 ) , 2 Pavg =

Px + Py 2

= a

- 250 + 300 b ( 10-6 ) = 25.0 ( 10-6 ) 2

The coordinates of reference point A and center of circle C are A( -250, 225) ( 10-6 )    C(25.0, 0) ( 10-6 ) The radius of the circle is R = CA = 3 2( - 250 - 25)2 + (225 - 0)2 4 ( 10-6 ) = 355.32 ( 10-6 )

Using these results, the circle shown in Fig. a can be constructed.

In-Plane Principal Strain: The coordinates of points B and D represent P1 and P2 respectively. P1 = (25.0 + 355.32) ( 10-6 ) = 380.32 ( 10-6 ) = 380 ( 10-6 ) P2 = (25.0 - 355.32) ( 10

-6

Ans.

) = - 330.32 ( 10 ) = -330 ( 10 ) -6

-6

Ans.

Ans: P1 = 380 ( 10-6 ) , P2 = - 330 ( 10-6 ) 965

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10–26.  The 45° strain rosette is mounted on the surface of  a pressure vessel. The following readings are obtained for  each gage: Pa = 475(10-6), Pb = 250(10-6), and Pc = -360(10-6). Determine the in-plane principal strains.

a 45 45 c

b

Solution Strain Rosettes (45°): Applying the equations in the text with Pa = 475 ( 10 - 6 ) , Pb = 250 ( 10 - 6 ) , Pc = - 360 ( 10 - 6 ) , ua = 0°, ub = - 45°, and uc = -90°. 475 ( 10-6 ) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0° Px = 475 ( 10-6 ) 250 ( 10-6 ) = 475 ( 10-6 ) cos2 ( - 45° ) + Py sin2 ( - 45° ) + gxy sin ( -45° ) cos ( -45° ) 250 ( 10-6 ) = 237.5 ( 10-6 ) + 0.5 Py - 0.5 gxy 0.5Py - 0.5 gxy = 12.5 ( 10 - 6 ) -360 ( 10

-6

(1)

) = 475 ( 10 ) cos ( -90°) + Py sin ( -90°) + gxy sin ( - 90°) cos ( -90°) -6

2

2

Py = -360 ( 10-6 ) gxy = - 385 ( 10-6 )

From Eq. (1),  

Therefore,      Px = 475 ( 10-6 ) Construction of the Circle: With Pavg =

Px + Py 2

= a

Py = - 360 ( 10-6 )

gxy 2

= - 192.5 ( 10-6 ) and

475 + ( - 360) 2

gxy = -385 ( 10-6 )

b ( 10-6 ) = 57.5 ( 10-6 )

The coordinates for reference points A and C are A(475, - 192.5) ( 10-6 )

C(57.5, 0) ( 10-6 )

The radius of the circle is R =

12(475

- 57.5)2 + 192.52 2 ( 10-6 ) = 459.74 ( 10-6 )

In-Plane Principal Strain: The coordinates of points B and D represent P1 and P2, respectively. P1 = (57.5 + 459.74) ( 10 - 6 ) = 517 ( 10 - 6 )

Ans.

P2 = (57.5 - 459.74) ( 10 - 6 ) = - 402 ( 10 - 6 )

Ans.

Ans: P1 = 517 ( 10 - 6 ) , P2 = -402 ( 10 - 6 ) 966

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10–27.  The 60° strain rosette is mounted on the surface of the bracket. The following readings are obtained for each gage: Pa = -780(10-6), Pb = 400(10-6), and Pc = 500(10-6). Determine (a) the principal strains and (b) the maximum in-plane shear strain and associated average normal strain. In each case show the deformed element due to these strains.

b c 60 60

Solution Strain Rosettes (60°): Applying the equations in the text with Pa = -780 ( 10-6 ) , Pb = 400 ( 10-6 ) , Pc = 500 ( 10-6 ) , ua = 0°, ub = 60°, and uc = 120°, Px = Pa = - 780 ( 10-6 ) Py = =

1 (2P + 2Pc - Pa) 3 b 1 [2(400) + 2(500) - ( - 780)] ( 10-6 ) 3

= 860 ( 10-6 ) gxy = =

2 23 2

23

(Pb - Pc) (400 - 500) ( 10-6 )

= - 115.47 ( 10-6 ) Construction of the Circle: With Px = - 780 ( 10-6 ) , Py = 860 ( 10-6 ) , and gxy = -57.735 ( 10-6 ) . 2 Px + Py - 780 + 860 Pavg = = a b ( 10-6 ) = 40.0 ( 10-6 ) Ans. 2 2 The coordinates for reference points A and C are A( -780, - 57.735) ( 10-6 )

C(40.0, 0) ( 10-6 )

The radius of the circle is R = a)

( 2(780 + 40.0)2 + 57.7352 )( 10-6 ) = 822.03 ( 10-6 )

In-Plane Principal Strain: The coordinates of point B and D represent P1 and P2, respectively. P1 = (40.0 + 822.03) ( 10-6 ) = 862 ( 10-6 )

Ans.

P2 = (40.0 - 822.03) ( 10

Ans.

-6

) = - 782 ( 10 ) -6

967

a

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10–27. Continued

Orientation of Principal Strain: From the circle, Solution tan 2up2 =

57.735 = 0.07041 2up2 = 4.03° 780 + 40

2up1 = 180° - 2up2 up1 =

180° - 4.03° = 88.0°  (Clockwise) 2

Ans.

b) Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in@plane

2 g

max in@plane

= - R = - 822.03 ( 10-6 ) = - 1644 ( 10-6 )

Ans.

Orientation of Maximum In-Plane Shear Strain: From the circle, tan 2us =

780 + 40 = 14.2028 57.735

us = 43.0°  (Clockwise)

Ans.

Ans: Pavg = 40.0 ( 10-6 ) , P1 = 862 ( 10-6 ) , P2 = -782 ( 10-6 ) , up1 = 88.0°  (Clockwise), = -1644 ( 10-6 ) , g max in@plane

us = 43.0°  (Clockwise) 968

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*10–28. The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from each gage: Pa = 800(10-6), Pb = 520(10-6), Pc = -450(10-6). Determine the in-plane principal strains.

c

b

45 45 a

Solution Pa = 800 ( 10-6 )

Pb = 520 ( 10-6 )

Pc = -450 ( 10-6 )

ua = -45°

ub = 0°

uc = 45°

Pb = Px cos2 ub + Py sin2 ub + gxy sin ub cos ub 520 ( 10-6 ) = Px cos2 0° + Py sin2 0° + gxy sin 0° cos 0° Pa = 520 ( 10-6 ) Pa = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua 800 ( 10-6 ) = Px cos2 ( -45°) + Py sin2 ( -45°) + gxy sin ( - 45°) cos ( - 45°) 800 ( 10-6 ) = 0.5Px + 0.5Py - 0.5gxy

(1)

Pc = Px cos2 uc + Py sin2 uc + gxy sin uc cos uc - 450 ( 10-6 ) = Px cos2 45° + Py sin2 45° + gxy sin 45° cos 45° - 450 ( 10-6 ) = 0.5Px + 0.5Py + 0.5gxy

(2)

Subtract Eq. (2) from Eq. (1) 1250 ( 10-6 ) = - gxy gxy = -1250 ( 10-6 ) Py = - 170 ( 10-6 ) gxy 2

= -625 ( 10-6 )

A(520, -625)10-6

C(175, 0)10-6

R = 3 2(520 - 175)2 + 6252 4 10-6 = 713.90 ( 10-6 ) P1 = (175 + 713.9)10-6 = 889 ( 10-6 ) P2 = (175 - 713.9)10

-6

= - 539 ( 10

-6

Ans.

)

Ans.

Ans: P1 = 889 ( 10-6 ) , P2 = - 539 ( 10-6 ) 969

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10–30.  For the case of plane stress, show that Hooke’s law can be written as sx =

E E (Px + nPy), sy = (Py + nPx) 2 (1 - n ) (1 - n2)

Solution Generalized Hooke’s Law: For plane stress, sz = 0. Applying Eq. 10–18, Px =

1 1 s - n sy 2 E x

nEPx =

1 sx

- n sy 2 n

nEPx = n sx - n2 sy Py =

(1)

1 (s - n sx) E y (2)

E Py = - n sx + sy Adding Eq. (1) and Eq. (2) yields. nE Px + E Py = sy - n2 sy sy = Substituting sy into Eq. (2)

E 1 nPx + Py 2 1 - n2

E Py = - nsx + sx = = =

(Q.E.D.)

E 1 n Px + Py 2 1 - n2

E 1 n Px + Py 2 2

n (1 - n )

-

EPy n

EnPx + EPy - EPy + EPy n2 n(1 - n2) E (Px + n Py) 1 - n2

(Q.E.D.)

Ans: N/A 970

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10–31.  Use Hooke’s law, Eq. 10–18, to develop the strain tranformation equations, Eqs. 10–5 and 10–6, from the stress tranformation equations, Eqs. 9–1 and 9–2.

Solution Stress Transformation Equations: sx′ =

sx + sy 2

tx′y′ = sy′ =

+

sx - sy 2

sx + sy 2

-

sx - sy

(1)

cos 2u + txy sin 2u

2

(2)

sin 2u + txy cos 2u sx - sy

(3)

cos 2u - txy sin 2u

2

Hooke’s Law: Px =

n sy sx E E

(4)

Py =

sy - n sx + E E

(5) (6)

txy = G gxy G =

E 2(1 + n)

(7)

From Eqs. (4) and (5) Px + Py = Px - Py =

(1 - n)(sx + sy)

(8)

E (1 + n)(sx - sy)

(9)

E

From Eqs. (6) and (7) txy =

E g 2(1 + n) xy

(10)

From Eq. (4) Px′ =

n sy′ sx′ E E

(11)

Substitute Eqs. (1) and (3) into Eq. (11) Px′ =

(1 - n)(sx + sy) 2E

+

(1 + n)(sx - sy) 2E

cos 2u +

(1 + n)txy sin 2u E

(12)

By using Eqs. (8), (9) and (10) and substitute into Eq. (12), Px′ =

Px + Py 2

+

Px - Py 2

cos 2u +

gxy 2

(Q.E.D.)

sin 2u

971

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10–31. Continued

From Eq. (6). Solution tx′y′ = Ggx′y′ =

E g 2(1 + n) x′y′

(13)

Substitute Eqs. (13), (6) and (9) into Eq. (2), E(Px - Py) E E gx′y′ = sin 2u + g cos 2u 2(1 + n) 2(1 + n) 2(1 + n) xy gx′y′ 2

= -

(Px - Py) 2

sin 2u +

gxy 2

(Q.E.D.)

cos 2u

Ans: N/A 972

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*10–32. A bar of copper alloy is loaded in a tension machine and it is determined that Px = 940110-62 and s 14 ksi, sy s=y = 0, 0,szsz= =0.0.Determine sxx == 100 MPa, Determine the the modulus modulus of elasticity, Ecu, and the dilatation, ecu, of the copper. ncu = 0.35.

Solution ex =

1 [s - v(sy + sz)] E x

−6 = 940(10 )

1 [100(106 ) − 0.35(0 + 0)] Ecu

9 = Ecu 106.38(10 = ) Pa 106 GPa

= ecu

Ans.

1 − 2v 1 − 2(0.35) (u x + u = [100(106 ) += 0 + 0] 0.282(10 3 ) y + u z) E 106.38(109 )

Ans.

Ans. Ecu = 106 GPa, ecu = 0.282(10)−3 772

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10–33.  A rod has a radius of 10 mm. If it is subjected to an axial load of 15 N such that the axial strain in the rod is Px = 2.75(10-6), determine the modulus of elasticity E and the change in the rod’s diameter. n = 0.23.

Solution sx =

15 = 47.746 kPa, p(0.01)2

Px =

1 [s - n(sy + sz)] E x

2.75(10 - 6) =

sy = 0,

sz = 0

1 [47.746(103) - 0.23(0 + 0)] E Ans.

E = 17.4 GPa Py = Pz = - nPx = - 0.23(2.75)(10 - 6) = - 0.632(10 - 6) ∆d = 20( - 0.632(10 - 6)) = - 12.6(10 - 6) mm

Ans.

Ans: E = 17.4 GPa, ∆d = -12.6(10 - 6) mm 974

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10–34. The principal strains at a point on the aluminum fuselage of a jet aircraft are P1 = 780110-62 and P2 = 400110-62. Determine the associated principal stresses at the point 10110 2 ksi, vnalal = = 0.33. point in in the thesame sameplane. plane.EalEal= = 70 3GPa, Hint: See Prob. 10–30.

Solution Plane stress, s3 = 0 See Prob 10-30, s1 = = s2 = =

E (e1 + ve2) 1 - v2 70(109 ) 1 − 0.332

[780(10 −6 ) + 0.33(400)(10 −6 )] = 71.64(106 ) Pa = 71.6 MPa

Ans.

E (e2 + ve1) 1 - v2 70(109 ) 1 − 0.332

[400(10 −6 ) + 0.33(780)(10 −6 )] = 51.64(106 ) Pa = 51.6 MPa

Ans.

Ans: s1 = 71.6 MPa, s2 = 51.6 MPa 772

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10–35. The cross section of the rectangular beam is subjected to the bending moment M. Determine an expression for the increase in length of lines AB and CD. The material has a modulus of elasticity E and Poisson’s ratio is n.

C B

D

h A M b

Solution For line AB, sz = ey = -

12My My My = 1 = 3 I b h3 12 b h v sz

=

E

¢LAB = =

h 2

L0

12 v My E b h3 h

ey dy =

2 12 v M y dy 3 E b h L0

3vM 2Ebh

Ans.

For line CD, sz = ex = -

M h2 Mc 6M = - 1 = - 2 3 I bh b h 12 v sz E

=

6vM E b h2

¢LCD = ex LCD = =

6vM (b) E b h2

6vM E h2

Ans.

Ans. sz = ∆LAB ∆LCD

778

12 My 3

, Py =

bh 3 nM = , 2Ebh 6 nM = Eh2

12 nMy Ebh3

,

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*10–36. The spherical pressure vessel has an inner ­ diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is ­ observed to increase in length by 0.012 mm when the ­ vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est = 200 GPa and nst = 0.3.

20 mm

Solution r 1000 = = 100 7 10, the thin wall analysis is valid to t 10 determine the normal stress in the wall of the spherical vessel. This is a plane stress problem where smin = 0 since there is no load acting on the outer surface of the wall.

Normal Stresses: Since

smax = slat =

p(1000) pr = = 50.0p 2t 2(10)

(1)

Normal Strains: Applying the generalized Hooke’s Law with Pmax = Plat =

0.012 = 0.600 1 10 - 3 2 mm>mm 20

Pmax =

1 3 s - n (slat + smin) 4 E max

0.600 1 10 - 3 2 =

1 [50.0p - 0.3 (50.0p + 0)] 200(109) Ans.

p = 3.4286 MPa = 3.43 MPa

From Eq. (1) smax = slat = 50.0(3.4286) = 171.43 MPa Maximum In-Plane Shear (Sphere’s Surface): Mohr’s circle is simply a dot. As the result, the state of stress is the same, consisting of two normal stresses with zero shear stress regardless of the orientation of the element. tmax

in@plane

Ans.

= 0

Absolute Maximum Shear Stress: t abs = max

smax - smin 171.43 - 0 = = 85.7 MPa 2 2

Ans.

Ans: p = 3.43 MPa tmax = 0 in@plane

t abs = 85.7 MPa max

977

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10–37. Determine the bulk modulus for each of the following materials: (a) rubber, E Er == 2.8 0.4MPa, ksi, nvr r == 0.48, and 3 (b) glass, E GPa, vg =ng0.24. Egg = = 56 8110 2 ksi, = 0.24.

Solution a) For rubber: Kr =

Er 0.4 2.8 = = 23.3 MPa 3 (1 - 2 vr) 3[1 - 2(0.48)]

Ans.

b) For glass: Kg =

Eg 3 (1 - 2 vg)

=

56 3) 8(10 = 35.9 5.13 GPa (103) ksi 3[1 - 2(0.24)]

Ans.

Ans. (a) Kr = 23.3 MPa, (b) Kg = 35.9 GPa 775

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10–38. The strain in the x direction at point A on the steel beam is measured and found to be Px = - 100110-62. Determine the applied load P. What is the shear strain gxy at point A? Est = 200 GPa, vst = 0.3.

yy 3 in. 75 mm xx

A A 3 ftm 0.9

4 ftm 1.2

75 mm

Solution Ix =

7 ftm 2.1

12 mm 200 mm 12 mm

A 12 mm

1 1 (0.15)(0.224 3 ) − (0.138)(0.2 3 ) = 48.4928(10 −6 ) m 4 12 12

P P

150 mm

QA = 0.106(0.15)(0.012) + 0.0685(0.012)(0.063) = 0.242586(10 −3 ) m 3 6 = u E= ε x [200(109 )][100(10 −= )] 20(106 ) N/m 2

s =

My , I

20(106 ) =

(0.45P)(0.037) 48.4928(10 −6 )

P = 58.24(10 3 ) N = 58.2 kN tA =

VQ 0.5[58.24(10 3 )][0.242586(10 −3 )] = 12.14(106 ) Pa = It [48.4928(10 −6 )](0.012)

G =

29(10 200 3) E  76.92 GPa = 2(1 ++ 0.3) 2(1 + v) 0.3)

gxy =

Ans.

txy 12.14(106 ) −3 = 0.1578(10 = ) rad 0.158(10 −3 ) rad = G 76.92(109 )

0.9 m

1.2 m

Ans.

2.1 m

M = 0.45P 0.9 m

Ans. P = 58.2 kN, gxy = 0.158(10-3) rad 777

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10–39. The principal strains in a plane, measured experimentally at a point on the aluminum fuselage of a jet aircraft, are P1 = 630(10-6) and P2 = 350(10-6). If this is a case of plane stress, determine the associated principal stresses at ksi EalE=al 10(10 at the the point pointininthe thesame sameplane. plane. = 70 3)GPa and nal = 0.33.

Solution Normal Stresses: For plane stress, s3 = 0. Normal Strains: Applying the generalized Hooke’s Law. e1 =

1 C s - v (s2 + s3) D E 1

6 630(10 −= )

1 70(109 )

[u 1 − 0.33(u 2 + 0)]

44.1(106= ) u 1 − 0.33u 2

e2 =

[1]

1 C s - v (s1 + s3) D E 2

6 350(10 −= )

1 70(109 )

[u 2 − 0.33(u 1 + 0)]

24.5(106= ) u 2 − 0.33u 1

[2]

Solving Eqs.[1] and [2] yields: 6 6 u 2 43.83(10 = = ) Pa 43.8 MPa u 1 58.56(10 = = ) Pa 58.6 MPa

Ans.

Ans: s1 = 58.6 MPa, s2 = 43.8 MPa 782

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*10–40. The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled in. from the top cooled itit isis 0.012 0.3 mm of the cavity. If the top of the cavity is covered and the temperature is increased by 110°C, 200°F, determine the stress components sx , sy , and sz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).

z

0.012 in. 0.3 mm

4 in. 100 mm 4 in. 100 mm

6 in. 150 mm

yy

Solution Normal Strains: Since the aluminum is confined at its sides by a rigid container and 0.3 0.012 = 0.002. allowed to expand in the z direction, ex = ey = 0; whereas ez = 6 150 Applying the generalized Hooke’s Law with the additional thermal strain, ex = = 0

1 C s - v(sy + sz) D + a¢T E x 1 68.9(109 )

[u x − 0.35(u y + u z )] + [24(10 −6 )](110)

u x − 0.35u y − 0.35u z + 181.896(106 ) 0= ey = = 0

[1]

1 C s - v(sx + sz) + a¢T E y 1 68.9(109 )

[u y − 0.35(u x + u z )] + [24(10 −6 )](110)

u y − 0.35u x − 0.35u z + 181.896(106 ) 0= ez = = 0.002

xx

[2]

1 C s - v A sx + sy B D + a¢T E z 1 68.9(109 )

[u z − 0.35(u x + u z )] + [24(10 −6 )](110)

0 = u z − 0.35u x − 0.35u z + 44.096(106 )

[3]

Solving Eqs.[1], [2] and [3] yields:

ux = uy = −487.23(106 ) N/m 2 = −487 MPa

Ans.

uz = −385.16(106 ) N/m 2 = −385 MPa

Ans.

Ans. σx = σy = −487 MPa, σz = −385 MPa 788

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10–41. The smooth rigid-body cavity is filled with liquid cooled itit isis 0.012 0.3 mm 6061-T6 aluminum. When cooled in. from the top of the cavity. If the top of the cavity is not covered and the determine the strain stress temperature is increased by 110°C, 200°F, determine components Px , Py , and Pz in the aluminum. Hint: Use Eqs. 10–18 with an additional strain term of a¢T (Eq. 4–4).

z

0.3 mm 0.012 in.

100 mm 4 in. 100 mm 4 in.

150 mm 6 in.

y

Solution Normal Strains: Since the aluminum is confined at its sides by a rigid container, then ex = ey = 0

x

Ans.

and since it is not restrained in z direction, sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ex =

1 C s - v A sy + sz B D + a¢T E x 1

= 0

68.9(109 )

[u x − 0.35(u y + u z )] + [24(10 −6 )](110)

u x − 0.35u y − 0.35u z + 181.896(106 ) 0= ey =

1 C s - v(sx + sz) D + a¢T E y 1

= 0

[1]

68.9(109 )

[u y − 0.35(u x + u z )] + [24(10 −6 )](110)

0= u y − 0.35u x − 0.35u z + 181.896(106 )

[2]

Solving Eqs. [1] and [2] yields:

u x = u y = −279.84(106 ) N/m 2 ez = =

1 C s - v A sx + sy B D + a¢T E z 1 68.9(109 )

{0 − 0.35[−279.84(106 ) + (−279.84)(106 )]} + [24(10−6 )](110)

−3 = ) 0.00548 = 5.483(10

Ans.

Ans. Px = Py = 0, Pz = 5.483(10- 3) 789

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10–42. The block is fitted between the fixed supports. If the jointcan canresist resista amaximum maximum shear stress tallow 14 glued joint shear stress of tof = 2=ksi, allow MPa, determine the temperature rise that willthe cause thetojoint determine the temperature rise that will cause joint fail. to fail. Take E = 70 GPa, v = 0.2, and α = 11(10−6 )/°C.

40

Hint: Use Eq. 10−18 with an additional strain term of a¢T (Eq. 4–4).

Solution Normal Strain: Since the block is confined along the y direction by the rigid frame, then ey = 0 and sx = sz = 0. Applying the generalized Hooke’s Law with the additional thermal strain, ey = = 0

1 C s - v(sx + sz) D + a¢T E y 1 70(109 )

0.77ΔT

[u y − 0.2(0 + 0)] + 11(10 −6 )(∆T )

−770(10 3 )∆T uy = Construction of the Circle: In accordance with the sign convention. sx = 0, sy = −770(10 3 )∆T and txy = 0. Hence.

u x + u y 0 + [−770(10 3 )∆T ] u avg = = = −385(10 3 )∆T 2

2

The coordinates for reference points A and C are A (0, 0) and C (−385(10 3 )∆T, 0). The radius of the circle is R =

{0 − [−385(10 3 )∆T ]}2 + = 0 385(10 3 )∆T

Stress on The inclined plane: The shear stress components tx¿y¿, are represented by the coordinates of point P on the circle. tx¿y¿ = 385(10 3 )∆ = T sin 80° 379.15(10 3 )∆T

0.385(103)ΔT

Allowable Shear Stress: R = 0.385(103)ΔT

tallow = tx¿y¿ = 14(106 ) 379.15(10 3 )∆T ∆T = 36.92 = ° 36.9°

Ans.

Ans. ∆T = 36.92°C 787

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10–43. Two strain gauges a and b are attached to a plate made from a material having a modulus of elasticity of E = 70 GPa and Poisson’s ratio n = 0.35. If the gauges give a reading of Pa = 450110-62 and Pb = 100110-62, determine the intensities of the uniform distributed load wx and wy acting on the plate. The thickness of the plate is 25 mm.

wy

b y

45

a

Solution Normal Strain: Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0 and ub = 45°, we have

z

wx

x

ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua

450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0 ex = 450 A 10 - 6 B

eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub

100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0

ey = - 250 A 10 - 6 B

Generalized Hooke’s Law: This is a case of plane stress. Thus, sz = 0. ex =

1 C s - v A sy + sz B D E x

450 A 10 - 6 B =

1

70 A 109 B

C sy - 0.35 A sy + 0 B D

sx - 0.35sy = 31.5 A 106 B ey =

(1)

1 C s - v A sx + sz B D E y

- 250 A 10-6 B =

1

70 A 109 B

C sy - 0.35 A sy + 0 B D

sy - 0.35sx = - 17.5 A 106 B

(2)

Solving Eqs. (1) and (2), sy = - 7.379 A 106 B N>m2

sx = 28.917 A 106 B N>m2

Then, wy = syt = - 7.379 A 106 B (0.025) = - 184 N>m

Ans.

wx = sxt = 28.917 A 106 B (0.025) = 723 N>m

Ans.

Ans. wy = -184 kN>m, wx = 723 kN>m 785

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*10–44. Two strain gauges a and b are attached to the surface of the plate which is subjected to the uniform distributed load wx = 700 kN/m and wy = −175 kN/m. If the gauges give a reading of ε a = 450(10−6) and ε b = 100(10−6), determine the modulus of elasticity E, shear modulus G, and Poisson’s ratio v for the material.

wy

b y

Normal Stress and Strain: The normal stresses along the x, y, and z axes are sx =

700 A 103 B 0.025

sy = -

45

a

= 28 A 106 B N>m2

175 A 103 B 0.025

= - 7 A 106 B N>m2

z

wx

x

sz = 0 (plane stress) Since no shear force acts on the plane along the x and y axes, gxy = 0. With ua = 0° and ub = 45°, we have ea = ex cos2 ua + ey sin2 ua + gxy sin ua cos ua 450 A 10 - 6 B = ex cos2 0° + ey sin2 0° + 0

ex = 450 A 10 - 6 B

eb = ex cos2 ub + ey sin2 ub + gxy sin ub cos ub

100 A 10 - 6 B = 450 A 10 - 6 B cos2 45° + ey sin2 45° + 0 ey = - 250 A 10 - 6 B

Generalized Hooke’s Law: ex =

1 C s - v A sy + sz B D E x

450 A 10 - 6 B =

1 B 28 A 106 B - v C - 7 A 106 B + 0 D R E

450 A 10 - 6 B E - 7 A 106 B v = 28 A 106 B ey =

(1)

1 [s - v(sx + sz)] E y

- 250 A 10 - 6 B =

1 b - 7 A 106 B - v C 28 A 106 B + 0 D r E

250 A 10 - 6 B E - 28 A 106 B v = 7 A 106 B

(2)

Solving Eqs. (1) and (2), E = 67.74 A 109 B N>m2 = 67.7 GPa

Ans.

v = 0.3548 = 0.355

Ans.

Using the above results, G =

67.74 A 109 B E = 2(1 + v) 2(1 + 0.3548)

= 25.0 A 109 B N>m2 = 25.0 GPa

Ans. Ans. E = 67.7 GPa, v = 0.355, G = 25.0 GPa 786

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10–45.  A material is subjected to principal stresses sx and sy. Determine the orientation u of the strain gage so that its reading of normal strain responds only to sy and not sx. The material constants are E and n.

y

u x

Solution sx =

sx + sy

+

2

sx - sy

cos 2u + txy sin 2u

2

Since txy = 0, sn =

sx + sy 2

+

sx - sy 2

cos 2u

cos 2u = 2 cos2 u - 1 sn =

sy sx sx sy + + (sx - sy ) cos2 u + 2 2 2 2

= sy (1 - cos2 u) + sx cos2 u = sx cos2 u + sy sin2 u sn

+ 90°

=

sx + sy 2

-

sx - sy 2

cos 2u

=

sx - sy sx sy b(2 cos2 u - 1) + - a 2 2 2

=

sx sx sy sy + - (sx - sy ) cos2 u + 2 2 2 2

= sx (1 - cos2 u) + sy cos2 u = sx sin2 u + sy cos2 u Pn = =

1 (s - n sn + 90°) E n 1 (s cos2 u + sy sin2 u - n sx sin2 u - n sy cos2 u) E x

If Pn is to be independent of sx, then cos2 u - n sin2 u = 0 u = tan - 1 a

1 2n

or

tan2 u = 1>n Ans.

b

Ans: u = tan - 1 a 986

1 2n

b

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10–46. The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce the bending stress that would occur if flat ends were used. The bending stresses at the seam where the caps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder, respectively. This requires the radial expansion to be the same for both the hemispheres and cylinder. Show that this ratio is tc/th = (2 − v)/(1 − v). Assume that the vessel is made of the same material and both the cylinder and hemispheres have the same inner radius. If the cylinder is to have a thickness of 12 mm, what is the required of the hemispheres? Take v = 0.3.

tc th r

Solution For cylindrical vessel: s1 =

pr ; tc

e1 =

1 [s - v (s2 + s3)] E 1

=

s2 =

pr 2 tc s3 = 0

vpr pr 1 1 pr a b = a1 - vb E tc 2 tc E tc 2

d r = e1 r =

p r2 1 a1 - vb E tc 2

(1)

For hemispherical end caps: s1 = s2 = e1 = =

pr 2 th

1 [s - v (s2 + s3)] ; E 1

s3 = 0

vpr pr 1 pr a b = (1 - v) E 2 th 2 th 2 E th

d r = e1 r =

p r2 (1 - v) 2 E th

(2)

Equate Eqs. (1) and (2): p r2 p r2 1 a 1 - vb = (1 - v) E tc 2 2 E th 2 (1 - 12 v) tc 2 - v = = th 1 - v 1 - v th =

QED

(12) (1 - 0.3) (0.5) (1 - v) tc = = 4.94 0.206mm in. 2 - v 2 - 0.3

Ans.

Ans. th = 4.94 mm 781

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10–47. A thin-walled cylindrical pressure vessel has an inner radius r, thickness t, and length L. If it is subjected to an internal pressure p, show that the increase in its inner radius is dr = rP1 = pr211 - 12 n2>Et and the increase in its length is ¢L = pLr112 - n2>Et. Using these results, show that the change in internal volume becomes dV = pr211 + P12211 + P22L - pr2L. Since P1 and P2 are small quantities, show further that the change in volume per unit volume, called volumetric strain, can be written as dV>V = pr12.5 - 2n2>Et.

Solution Normal stress: pr ; t

s1 =

s2 =

pr 2t

Normal strain: Applying Hooke’s law e1 = =

1 [s - v (s2 + s3)], E 1

vpr pr 1 1 pr a b = a1 - vb E t 2t Et 2

d r = et r = e2 = =

s3 = 0

p r2 1 a1 - vb Et 2

1 [s - v (s1 + s3)], E 2

QED s3 = 0

vpr pr 1 1 pr a b = a - vb E 2t t Et 2

¢L = e2 L =

pLr 1 a - vb Et 2

V¿ = p(r + e1 r)2 (L + e2L) ;

QED V = p r2 L

dV = V¿ - V = pr2 (1 + e1)2 (1 + e2)L - pr2 L

QED

(1 + e1)2 = 1 + 2 e1 neglect e21 term (1 + e1)2 (1 + e2) = (1 + 2 e1)(1 + e2) = 1 + e2 + 2 e1 neglect e1 e2 term dV = 1 + e2 + 2 e1 - 1 = e2 + 2 e1 V =

2pr pr 1 1 a - vb + a1 - vb Et 2 Et 2

=

pr (2.5 - 2 v) Et

QED

791

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10–48. The rubber block is confined in the U-shape smooth rigid block. If the rubber has a modulus of elasticity E and Poisson’s ratio n, determine the effective modulus of elasticity of the rubber under the confined condition.

P

Solution Generalized Hooke’s Law: Under this confined condition, ex = 0 and sy = 0. We have ex = 0 =

1 C s - v A sy + sz B D E x 1 (s - vsz) E x

sx = vsz

(1)

ez =

1 C s - v A sx + sy B D E z

ez =

1 [s - v(sx + 0)] E z

ez =

1 (s - vsx) E z

(2)

Substituting Eq. (1) into Eq. (2), ez =

sz E

A 1 - v2 B

The effective modulus of elasticity of the rubber block under the confined condition can be determined by considering the rubber block as unconfined but rather undergoing the same normal strain of ez when it is subjected to the same normal stress sz, Thus, sz = Eeff ez Eeff =

sz ez

=

sz sz E

A1 - v B 2

=

E 1 - v2

Ans.

Ans. Eeff =

792

E 1 - v2

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10–49. Initially, gaps between the A-36 steel plate and the rigid constraint are as shown. Determine the normal stresses sx and sy developed in the plate if the temperature To solve, solve, add add the thermal is increased by by ¢T ΔT == 100°F. 55°C. To strain a¢T to the equations for Hooke’s Law.

yy 0.0015 mm in. 0.0375

6 in. 150 mm

8 in. 200 mm

Solution

0.0025 0.0625 in. mm xx

Generalized Hooke’s Law: Since there are gaps between the sides of the plate and the rigid constraint, the plate is allowed to expand before it comes in contact with the constraint. dy dx 0.0625 0.0025 0.0015 0.0375 ey = = = 0.3125 A 10 - 3 B and = = 0.25 A 10 - 3 B . Thus, ex = Lx 8 Ly 6 200 150 However, the plate is allowed to have free expansion along the z direction. Thus, sz = 0. With the additional thermal strain term, we have ex =

1 c s - v A sy + sz B d + a¢T E x

−3 = 0.3125(10 )

1 200(109 )

[u x − 0.32(u y + 0)] + [12(10 −6 )](55)

u x − 0.32u y = −69.5(106 ) ey =

(1)

1 C s - v A sx + sz B D + a¢T E y

−3 = 0.25(10 )

1 200(109 )

− [u y − 0.32(u x + 0)] + [12(10 6 )](55)

u y − 0.32u x = −82(106 )

(2)

Solving Eqs. (1) and (2),

ux = −106.66(106 ) N/m 2 = 107 MPa (C)

Ans.

uy = −116.13(106 ) N/m 2 = 116 MPa (C)

Ans.

Since sx 6 sY and sy 6 sY, the above results are valid.

Ans. sx = 107 MPa (C), sy = 116.1 MPa (C) 784

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10–50.  y

The steel shaft has a radius of 15 mm. Determine the torque T in the shaft if the two strain gages, attached to the surface of the shaft, report strains of Px′ = - 80(10-6) and Py′ = 80(10-6). Also, determine the strains acting in the x and y directions. Est = 200 GPa, nst = 0.3.

T

y¿

x¿ 45

x T

Solution Px′ = -80(10 - 6)

Py′ = 80(10 - 6) Ans.

Pure shear Px = Py = 0 2

2

Px ′ = Px cos u + Py sin u + gxy sin u cos u u = 45° - 80(10 - 6) = 0 + 0 + gxy sin 45° cos 45° gxy = - 160(10 - 6)

Ans.

Also, u = 135° 80(10 - 6) = 0 + 0 + g sin 135° cos 135° gxy = - 160(10 - 6) G =

200(109) E = = 76.923(109) 2(1 + n) 2(1 + 0.3)

t = Gg = 76.923(109)(160)(10 - 6) = 12.308(106) Pa tJ T = = c

p 12.308(106) a b(0.015)4 2 = 65.2 N # m 0.015

Ans.

Ans: Px = Py = 0, gxy = -160(10 - 6), T = 65.2 N # m 991

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10–51. y

The shaft has a radius of 15 mm and is made of L2 tool steel. Determine the strains in the x′ and y′ direction if a torque T = 2 kN  #  m is applied to the shaft.

T

y¿

x¿ 45

x T

Solution t =

2(103)(0.015) Tc = p = 377.26 MPa 4 J 2 (0.015 )

Stress-Strain Relationship: gxy =

txy G

= -

377.26(106) 75.0(109)

= - 5.030(10 - 3) rad

This is a pure shear case, therefore, Px = Py = 0 Applying Eq. 10–15, Px ′ = Px cos2 ua + Py sin2 ua + gxy sin ua cos ua Here ua = 45° Px ′ = 0 + 0 - 5.030(10 - 3) sin 45° cos 45° = -2.52(10 - 3) Px ′ = - 2.52(10 - 3)

Ans.

-3

Ans.

Py ′ = 2.52(10 )

Ans: ex′ = -2.52(10 - 3), ey′ = 2.52(10 - 3) 992

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*10–52. The A-36 steel pipe is subjected to the axial loading of 60 kN. Determine the change in volume of the material after the load is applied.

30 mm 40 mm 60 kN

60 kN

0.5 m

Solution Normal Stress: The pipe is subjected to uniaxial load. Therefore, N sv = sz = 0 and sx = . A Dilatation: Applying Eq. 10–23. dV 1 - 2v = (sx + sy + sz) V E dV 1 - 2v N = a b V E A

dV =

1 - 2v N a b V  However, V = AL E A

dV = a = c

1 - 2v b NL E

1 - 2(0.32) 200 ( 109 )

d (60) ( 103 ) (0.5)

= 54.0 ( 10-9 ) m3 = 54.0 mm3

Ans.

Ans: dV = 54.0 mm3 993

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10–53. Air is pumped into the steel thin-walled pressure vessel at C. If the ends of the vessel are closed using two pistons connected by a rod AB, determine the increase in the diameter of the pressure vessel when the internal gage pressure is 5 MPa. Also, what is the tensile stress in rod AB if it has a diameter of 100 mm? The inner radius of the vessel is 400 mm, and its thickness is 10 mm. Est = 200 GPa and nst = 0.3.

C 400 mm A

B

Solution Circumferential Stress: s =

5(400) pr = = 200 MPa t 10

Note: longitudinal and radial stresses are zero. Circumferential Strain: P =

200(106) s = = 1.0(10 - 3) E 200(109)

∆d = P d = 1.0(10 - 3)(800) = 0.800 mm

Ans.

For rod AB: + ΣFx = 0;   TAB - 5(106)a p b(0.82 - 0.12) = 0 d 4 TAB = 2474 kN

sAB =

2474(103) TAB = p = 315 MPa 2 AAB 4 (0.1 )

Ans.

Ans: ∆d = 0.800 mm, sAB = 315 MPa 994

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10–54. Determine the increase in the diameter of the pressure vessel in Prob. 10–53 if the pistons are replaced by walls connected to the ends of the vessel.

C 400 mm A

B

Solution Principal Stress: 5(400) pr = = 200 MPa; t 10 1 s2 = s1 = 100 MPa 2 s1 =

s3 = 0

Circumferential Strain: 1 1 P1 = [s - n(s2 + s3)] = [200(106) - 0.3{100(106) + 0}] E 1 200(109) = 0.85(10 - 3) ∆d = P1 d = 0.85(10 - 3)(800) = 0.680 mm

Ans.

Ans: ∆d = 0.680 mm 995

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10–55. A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to an internal pressure p. Show that the increase in the volume within the vessel is ∆V = (2ppr 4 >Et)(1 - n). Use a small-strain analysis.

Solution s1 = s2 =

pr 2t

s3 = 0 1 (s - ns2) E 1 pr P1 = P2 = (1 - n) 2t E 1 P3 = ( - n(s1 + s2)) E n pr P3 = tE P1 = P2 =

4pr 3 3 4p 4pr 3 ∆r 3 V + ∆V = (r + ∆r)3 = a1 + b 3 3 r where ∆V V V, ∆r V r V =

V + ∆V = PVol =

4p r 3 ∆r a1 + 3 b 3 r

∆V ∆r = 3a b V r

Since P1 = P2 = PVol = 3P1 =

2p(r + ∆r) - 2p r 2p r

=

∆r r

3pr (1 - n) 2t E

∆V = VPVol =

2pp r 4 (1 - n) Et

(Q.E.D.)

Ans: N/A 996

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*10–56. The thin-walled cylindrical pressure vessel of inner radius r and thickness t is subjected to an internal pressure p. If the material constants are E and n, determine the strains in the circumferential and longitudinal directions. Using these results, calculate the increase in both the diameter and the length of a steel pressure vessel filled with air and having an internal gage pressure of 15 MPa. The vessel is 3 m long, and has an inner radius of 0.5 m and a thickness of 10 mm. Est = 200 GPa, nst = 0.3.

3m

0.5 m

Solution Normal Stress: pr pr s1 = s2 = t 2t

s3 = 0

Normal Strain: 1 Pcir = [s - n(s2 + s3)] E 1 npr pr 1 pr a b = (2 - n) E t 2t 2Et

=

Plong = =

Ans.

1 [s - n(s1 + s3)] E 2

npr pr 1 pr a b = (1 - 2n) E 2t t 2Et

Ans.

Numerical Substitution: Pcir =

15(106)(0.5) 2(200)(109)(0.01)

(2 - 0.3) = 3.1875 (10 - 3)

∆d = Pcir d = 3.1875 (10 - 3)(1000) = 3.19 mm Plong =

15(106)(0.5) 2(200)(109)(0.01)

Ans.

(1 - 2(0.3)) = 0.75(10 - 3)

∆L = Plong L = 0.75 (10 - 3)(3000) = 2.25 mm

Ans.

Ans:

pr (2 - n), 2E t pr Plong = (1 - 2n), 2E t ∆d = 3.19 mm, ∆L = 2.25 mm Pcir =

997

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10–57. Estimate the increase in volume of the pressure vessel in Prob. 10–56.

3m

0.5 m

Solution By basic principles, ∆V = p(r + ∆r)2(L + ∆L) - p r 2 L = p(r 2 + ∆r 2 + 2 r ∆r)(L + ∆L) - p r 2 L = p(r 2L + r 2 ∆L + ∆r 2L + ∆r 2 ∆L + 2 r ∆r L + 2 r ∆r ∆L - r 2 L) = p(r 2 ∆L + ∆r 2 L + ∆r 2 ∆L + 2 r ∆r L + 2 r ∆r ∆L) Neglecting the second order terms, ∆V = p(r 2 ∆L + 2 r ∆r L) From Prob. 10–56, ∆L = 0.00225 m

∆r =

∆d = 0.00159375 m 2

∆V = p[(0.52)(0.00225) + 2(0.5)(0.00159375)(3)] = 0.0168 m3

Ans.

Ans: ∆V = 0.0168 m3 998

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10–58. z

A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Assuming that Px = 0 and Py = 0, determine the factor by which the stiffness of the material, or the apparent modulus of elasticity, will be increased when a load is applied, if n = 0.3 for the material.

P

x

y

Solution Normal Strain: Since the material is confined in a rigid cylinder. Px = Py = 0. Applying the generalized Hooke’s Law, Px =

1 3 s - n(sy + sz) 4 E x

(1)

1 3 s - n(sx + sz) 4 E y

(2)

  0 = sx - n(sy + sz) Py =

  0 = sy - n(sx + sz) Solving Eqs. (1) and (2) yields: sx = sy =

n s 1 - n z

Thus, Pz = = = = =

1 3 s - n(sx + sy) 4 E z

1 n n cs - na s + s bd E z 1 - n z 1 - n z sz E

c1 -

2n2 d 1 - n

sz 1 - n - 2n2 c d E 1 - n

sz (1 + n)(1 - 2n) c d E 1 - n

Thus, when the material is not being confined and undergoes the same normal strain of Pz, then the required modulus of elasticity is E′ =

sz Pz

=

The increase factor is k =

1 - n E (1 - 2n)(1 + n) E′ 1 - n = E (1 - 2n)(1 + n) =

1 - 0.3 [1 - 2(0.3)](1 + 0.3) Ans.

= 1.35

Ans: k = 1.35 999

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10–59. A material is subjected to plane stress. Express the distortion energy theory of failure in terms of sx, sy, and txy.

Solution Maximum distortion energy theory: (s21 - s1 s2 + s22) = s2Y s1, 2 =

sx + sy

Let a =

2

{

sx + sy 2

s1 = a + b;

B

a

(1)

sx - sy

and b =

2 B

s2 = a - b

s21 = a2 + b2 + 2 a b;

a

2

b + t2xy

sx - sy 2

2

b + t2xy

s22 = a2 + b2 - 2 a b

s1 s2 = a2 - b2 From Eq. (1) (a2 + b2 + 2 a b - a2 + b2 + a2 + b2 - 2 a b) = sY2 (a2 + 3 b2) = s2Y (sx + sy)2 4

+ 3

(sx - sy)2 4

+ 3 t2xy = s2Y

s2x + s2y - sxsy + 3 t2xy = s2Y

Ans.

Ans: s2x + s2y - sxsy + 3t2xy = s2y 1000

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*10–60. A material is subjected to plane stress. Express the maximum shear stress theory of failure in terms of sx, sy, and txy. Assume that the principal stresses are of different algebraic signs.

Solution Maximum shear stress theory: (1)

|s1 - s2| = sY s1,2 =

sx + sy

 2 s1 - s2 2 = 2 From Eq. (1) 4 ca

{

2

sx - sy 2 2

B

a

B

a

sx - sy 2

sx - sy 2

2

2

b + t2xy

b + t2xy

2

b + t2xy d = s2Y

(sx - sy) + 4 t2xy = s2Y

Ans.

Ans: 2 (sx - sy) + 4 t2xy = s2Y 1001

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10–61. A bar with a square cross-sectional area is made of a material having a yield stress of of s 120MPa. ksi. If the bar sYY==840 is subjected to a bending moment of 75 10 kip kN #·in., m, determine the required size of the bar according to the maximumdistortion-energy theory. Use a factor of safety of 1.5 with respect to yielding.

Solution Normal and Shear Stress: Applying the flexure formula,

u =

Mc [10(10 3 )](a 2) 60(10 3 ) = = 1 a4 I a3 12 60(106) a3

In-Plane Principal Stress: Since no shear stress acts on the element 60(10 3 )

u= 1 u= x

a3

s2 = sy = 0

Maximum Distortion Energy Theory: s21 - s1 s2 + s22 = s2allow 2

 60(10 3 )   840(106 )  0 0 − + =     3  a   1.5 

2

a = 0.047495 m = 47.5 mm

Ans.

Ans. a = 47.5 m 796

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10–62. Solve Prob. 10–64 using the maximum-shearstress theory.

Solution Normal and Shear Stress: Applying the flexure formula,

u =

Mc [10(10 3 )](a 2) 60(10 3 ) = = 1 a4 I a3 12

In-Plane Principal Stress: Since no shear stress acts on the element. 60(10 3 )

u= 1 u= x

a3

s2 = sx = 0

Maximum Shear Stress Theory: |s2| = 0 6 sallow =

840 120 = 560 80.0MPa ksi  1.5

(O.K!)

|s1| = sallow 60(10 3 ) a3

=

840(106 ) 1.5

a = 0.047495 m = 47.5 mm

Ans.

Ans. s = 797

60(106) a3

, a = 47.50 mm

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10–63. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T.

Solution Principal stresses: s1 =

Me c ; I

ud =

1 + v 2 (s1 - s1 s2 + s22) 3E

s2 = 0

1 + v M2e c2 a 2 b 3E I

(ud)1 =

(1)

Principal stress: s1, 2 = s1 =

s + 0 s - 0 2 3 ; a b + t 2 A 2

s s2 + + t2; 2 A4

s2 =

s s2 + t2 2 A4

Distortion Energy: Let e a =

s s2 + t2 b = 2 A4

s12 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b s22 - s1 s2 + s22 = 3 b2 + a2 Apply s = (ud)2 = =

Mc ; I

t =

Tc J

1 + v 1 + v s2 3s2 a + 3 t2 b + (3 b2 + a2) = 3E 3E 4 4 1 + v 2 1 + v M2 c2 3 T2 c2 (s + 3 t2) = a 2 + b 3E 3E I J2

(2)

798

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10–63. Continued

Equating Eq. (1) and (2) yields: (1 + v) Me c2 1 + v M2 c2 3T2 c2 b a 2 b = a 2 + 3E 3E I I J2 M2e 2

I

=

M1 3 T2 + 2 I J2

M2e = M1 + 3 T2 a

I 2 b J

For circular shaft I = J

p 4 p 2

c4 c4

=

1 2

1 2 Hence, M2e = M2 + 3 T2 a b 2 Me =

A

M2 +

3 2 T 4

Ans.

Ans. Me = 799

A

M2 +

3 2 T 4

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*10–64. Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with a circular cross section, would cause the same maximum shear stress as the combination of an applied moment M and torque T. Assume that the principal stresses are of opposite algebraic signs.

Solution Bending and Torsion: Mc Mc 4M ; = p 4 = I c p c3 4

s =

t =

Tc Tc 2T = p 4 = J c p c3 2

The principal stresses:

s1, 2 = = tabs max

sx + sy 2

;

A

a

sx - sy 2

2 b + txy = 2

4M pc3

+ 0 2

4M

;

¢pc

3

Q

- 0 2

2

≤ + a

2T 3

pc

b

2

2 2M ; 2M2 + T2 p c3 p c3

= s1 - s2 = 2 c

2 2M2 + T2 d p c3

(1)

Pure bending: s1 = tabs max

Me c 4 Me Mc = p 4 = ; I c p c3 4

= s1 - s2 =

s2 = 0

4 Me

(2)

p c3

Equating Eq. (1) and (2) yields: 4 Me 4 2M2 + T2 = p c3 p c3 Me = 2M2 + T2

Ans.

Ans. Me = 2M2 + T2 801

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10–65. Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment M and torque T.

Solution t =

Te c J

Principal Stress: s1 = t s2 = -t ud =

1 + n 2 (s1 - s1 s2 + s22) 3E

(u d)1 =

1 + n 1 + n 3 T 2e c 2 ( 3 t2) = a b 3E 3E J2

Bending Moment and Torsion: s =

Mc ; I

t =

Tc J

Principal Stress: s1, 2 =

s - 0 2 s + 0 2 a b + t { B 2 2 2

s1 =

s s + t2 ; + B 4 2

Let a =

s 2

b =

s2 =

s2 + t2 B 4

s2 s + t2 B 4 2

s21 = a2 + b2 + 2 a b s1 s2 = a2 - b2 s22 = a2 + b2 - 2 a b

s21 - s1 s2 + s22 = 3 b2 + a2 ud =

1 + n 2 (s1 - s1 s2 + s22) 3E

(u d)2 = =

1 + n 1 + n 3 s2 s2 (3 b2 + a2) = a + 3t2 + b 3E 3E 4 4

c 2(1 + n) M2 1 + n 2 3 T2 (s + 3 t2) = a 2 + b 3E 3E I J2

(u d)1 = (u d)2

c 2(1 + n) 3 Te2 c 2(1 + n) M2 3 T2 = a + b 3E 3E J2 I2 J2

1006

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10–65.  Continued

For circular shaft J = I Te = Te =

p 4 2c p 4 4c

= 2

J 2 M2 + T2 B I2 3 4 2 M + T 2 B3

Ans.

Ans: Te = 1007

4 2 M + T2 B3

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10–66. An aluminum alloy 6061-T6 is to be used for a solid drive shaft shaft such such that that itit transmits transmits 33 40 kW hp at 2400 rev>min. Using a factor of safety of 2 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum-shear-stress theory.

Solution v = a 2400 T =

2p rad 1 min rev ba ba b = 80 p rad>s min rev 60s

3 40 (550) 33(10 ) (12) 3300 3300 P  = = N · m lb # in. 80p v 80 p (8p) p

Tc J

Applying t = t =

A 3300 8p p B c p 2

c4

=

6600 p33cc33 8p

The principal stresses: s1 = t =

6600 ; 8p p2 2cc33

s2 = - t =

6600 8p p2 2cc33

Maximum shear stress theory: Both principal stresses have opposite sign, hence,

` s1 - s2 ` =

sY ; F.S.

37 (1036)) 255(10 6600 2 a 22 33b = ` ` 2 8p p cc

c = 0.010945 0.4166 in.m = 10.945 mm d = 21.89 0.833 mm in.

Ans.

Ans. v = 80 p rad>s, T = d = 21.89 mm 795

3300 N # m, 8p

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10–67. Solve Prob. 10–61 using the maximum-distortionenergy theory.

Solution v = a 2400 T =

2p rad 1 min rev ba ba b = 80 p rad>s min rev 60s

3 (12) 40 (550) 33(10 ) 3300 3300 P = = N · mlb.in.  v 80 p (8p) p 80p

Applying t = t =

A 3300 8p p B c p 2

c4

=

Tc J 6600 8p p2 3cc33

The principal stresses: s1 = t =

6600 ; 8p p2 2cc33

s2 = - t = -

6600 6600 2 2 33 8p p cc

The maximum distortion-energy theory: s21 - s1 s2 + s22 = a

sY 2 b F.S.

255(10 37(1036)) 2 6600 2 b 3 B 22 33R = a 2 p cc 8p c = 0.0104328 0.3971 in. m = 10.4328 mm d = 20.87 0.794 mm in.

Ans.

Ans. d = 20.87 mm 795

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*10–68. If the material is machine steel having a yield stress of sY = 700 MPa, determine the factor of safety with respect to yielding if the maximum shear stress theory is considered.

50 MPa

80 MPa

Solution smax = 80 MPa

smin = - 50 MPa

tabs =

80 - ( -50) smax - smin = = 65 MPa 2 2

tmax =

sY 700 = = 350 MPa 2 2

F.S. =

tmax 350 = = 5.38 tabs 65

max

Ans.

max

Ans: F.S. = 5.38 1010

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10–69. The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N # m and an axial compressive force of 2 kN. Determine if it fails according to the maximum normal stress theory. The ultimate stress of the concrete is sult = 28 MPa.

2 kN

500 Nm

500 Nm

Solution

2 kN

A =

p (0.05)2 = 1.9635(10 - 3) m2 4

J =

p (0.025)4 = 0.61359(10 - 4) m4 2

s =

2(103) P = = 1.019 MPa A 1.9635(10 - 3)

t =

500(0.025) Tc = = 20.372 MPa J 0.61359(10 - 6)

sx = 0 s1, 2 = s1, 2 =

sy = - 1.019 MPa sx + sy 2

{

B

a

sx - sy 2

txy = 20.372 MPa 2

b + txy 2

0 - ( -1.019) 2 0 - 1.018 { a b + 20.3722 2 B 2

s1 = 19.87 MPa

s2 = - 20.89 MPa

Failure criteria: |s1| 6 salt = 28 MPa

OK

|s2| 6 salt = 28 MPa

OK Ans.

No.

Ans: No 1011

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10–70. A bar with a circular cross-sectional area is made of SAE 1045 carbon steel having a yield stress of If the barbaris issubjected s sY = 150 1000ksi. MPa. If the subjectedtoto aa torque torque of Y = 30 3.75kip kN# in. · mand anda abending bendingmoment momentofof567 kip kN #·in., m, determine the required diameter of the bar according to the maximum-distortion-energy theory. Use a factor of safety of 2 with respect to yielding.

Solution Normal and Shear Stresses: Applying the flexure and torsion formulas.

u = t =

Mc [7(10 3 )](d 2) 224(10 3 ) = = π (d 2)4 I π d3 4 [3.75(10 3 )](d 2) 60(10 3 ) Tc = = π (−d 2)4 J π d3 2

The critical state of stress is shown in Fig. (a) or (b), where

ux =

224(10 3 )

txy =

sy = 0

π d3

60(10 3 )

π d3

In - Plane Principal Stresses : Applying Eq. 9-5, s1,2 =

sx + sy 2

;

A

a

sx - sy 2

2 b + txy 2

2  224   2243 − 0  2  π d3 + 0  60   3 d π   = ± +   (10 )  2   π d 3    2      112 127.06  3 =  3±  (10 ) π d3   πd

 239.06  3  (10 )  π d3 

u1 = 

 15.06  3  (10 )  π d3 

u2 = −

Maximum Distortion Energy Theory : s21 - s1s2 + s22 = s2allow 2 2  1000(106 )   239.06   239.06  15.06   15.06  3 2 − − + − (10 ) =           2  π d 3   π d 3  π d 3   π d 3   

d = 0.05397 m = 54.0 mm

2

Ans.

Ans. d = 54.0 m 812

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10–71. The plate is made of hard copper, which yields at sY 105MPa. ksi. Using the maximum-shear-stress theory, Y ==735 determine the tensile stress sx that can be applied to the plate if a tensile stress sy = 0.5sx is also applied.

sy  0.5sx

sx

Solution s1 = sx

s2 =

1 s 2 x

|s1| = sY 735 ksi MPa sx = 105

Ans.

Ans. sx = 735 MPa 810

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*10–72. Solve Prob. 10–80 using the maximum-distortionenergy theory.

sy  0.5sx

sx

Solution s1 = sx s2 =

sx 2

s21 - s1 s2 + s22 = s2Y s2x -

s2x s2x + = 735 2 2 4 Ans.

sx = 848.7 MPa = 849 MPa

Ans. sx = 849 MPa 810

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10–73. The state of stress acting at a critical point on the seat frame of an automobile during a crash is shown in the figure. Determine the smallest yield stress for a steel that can be selected for the member, based on the maximumshear-stress theory.

25 ksi 175 MPa 80 ksi 560 MPa

Solution Normal and Shear Stress: In accordance with the sign convention. s MPa sxx ==560 80 ksi

sys=y =00

txy== 25 175ksi MPa txy

In - Plane Principal Stress: Applying Eq. 9-5. s1,2 = = =

sx + sy 2

;

A

a

sx - sy 2

2 b + txy 2

560-– 0 2 80 80 ++ 0 560 ; 2522 a b + 175 2 A 2 280 ± 330.19

u 1 = 610.19 MPa

u 2 = −50.19 MPa

Maximum Shear Stress Theory: s1 and s2 have opposite signs so |s1 - s2| = sY 610.19 − (−50.19) =u Y = u Y 660.38 = MPa 660 MPa

Ans.

Ans. sY = 660 MPa 811

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10–74. Solve Prob. 10–82 using the maximum-distortionenergy theory.

Solution

25 MPa ksi 175

Normal and Shear Stress: In accordance with the sign convention. MPa sxx ==560 80 ksi

sys= y =00

80 MPa ksi 560

175ksi MPa ttxyxy == 25

In - Plane Principal Stress: Applying Eq. 9-5. s1,2 = =

sx + sy 2

;

A

a

sx - s 2 2 b + txy 2

560 80 -– 0 2 80 ++ 0 560 2522 ; a b + 175 2 A 2

330.189 = 280 40 ;; 47.170

u 1 = 610.19 MPa

u 2 = −50.19 MPa

Maximum Distortion Energy Theory: s21 - s1s2 + s22 = s2Y

610.19 2 − (610.19)(−50.19) + (−50.19)2 =u Y2 = u Y 636.77 = MPa 637 MPa

Ans.

Ans. sY = 637 MPa 811

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10–75. The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum distortion energy theory. The yield stress for the steel is sY = 700 MPa.

200 MPa 150 MPa

500 MPa

Solution Normal And Shear Stresses: In accordance with the sign convention, sx = 500 MPa  sy = - 200 MPa  txy = 150 MPa In-Plane Principal Stresses: s1, 2 = =

sx + sy 2

{

C

500 + ( - 200)

2 = 150 { 380.79

a

sx - sy

{

2

C

c

2

b + txy2

500 - ( - 200) 2

2

d + 1502

     s1 = 530.79 MPa   s2 = - 230.79 MPa Maximum Distortion Energy Theory: This theory gives s12 - s1s2 + s22 = 530.792 - 530.79( - 230.79) + ( -230.79)2 = 457500 6 sY 2 = 490000 Based on the results obtained above, the material will not yield according to the maximum distortion energy theory. Ans.

Ans: No 1018

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*10–76. Solve Prob. 10–75 using the maximum shear stress theory.

200 MPa 150 MPa

500 MPa

Solution Normal And Shear Stresses: In accordance with the sign convention, sx = 500 MPa  sy = - 200 MPa  txy = 150 MPa In-Plane Principal Stresses: s1, 2 = =

sx + sy 2

{

C

500 + ( -200) 2

a

sx - sy

{

= 150 { 380.79

2

C

c

2

b + txy2

500 - ( -200) 2

2

d + 1502

s1 = 530.79 MPa   s2 = - 230.79 MPa Maximum Shear-Stress Theory: Here, s1 and s2 have opposite signs. So

0 s1 - s2 0

= 530.79 - ( - 230.79) = 761.58 MPa 7 sY = 700 MPa

Based on the results obtained above, the material yield according to the maximum shear-stress theory. Ans.

Ans: Yes 1019

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10–77. If the A-36 steel pipe has outer and inner diameters of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-shear-stress theory.

900 N 150 mm A

100 mm

200 mm

Solution Internal Loadings. Considering the equilibrium of the free - body diagram of the post’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0

Vy = 0 T = - 360 N # m

©Mx = 0; T + 900(0.4) = 0

©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz =

p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4

J =

p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2

Normal Stress and Shear Stress. The normal stress is contributed by bending stress. Thus, sY = -

MyA 90(0.015) = = - 42.31MPa Iz 10.15625p A 10 - 9 B

The shear stress is contributed by torsional shear stress. t =

360(0.015) Tc = = 84.62 MPa J 20.3125p A 10 - 9 B

The state of stress at point A is represented by the two - dimensional element shown in Fig. b. In - Plane Principal Stress. sx = - 42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 = =

sx + sz 2

;

A

a

sx - sz 2

2 b + txz 2

- 42.31 - 0 2 - 42.31 + 0 2 ; a b + 84.62 2 A 2

= ( -21.16 ; 87.23) MPa s1 = 66.07 MPa

s2 = - 108.38 MPa

805

200 mm

900 N

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10–77. Continued

Maximum Shear Stress Theory. s1 and s2 have opposite signs. This requires |s1 - s2| = sallow 66.07 - (- 108.38) = sallow sallow = 174.45 MPa The factor of safety is F.S. =

sY 250 = = 1.43 sallow 174.45

Ans.

Ans. F.S. = 1.43 806

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10–78. If the A-36 steel pipe has an outer and inner diameter of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximum-distortion-energy theory.

900 N 150 mm A

100 mm

200 mm

Solution Internal Loadings: Considering the equilibrium of the free - body diagram of the pipe’s right cut segment Fig. a, ©Fy = 0; Vy + 900 - 900 = 0

Vy = 0 T = - 360 N # m

©Mx = 0; T + 900(0.4) = 0

©Mz = 0; Mz + 900(0.15) - 900(0.25) = 0 Mz = 90 N # m Section Properties. The moment of inertia about the z axis and the polar moment of inertia of the pipe’s cross section are Iz =

p A 0.0154 - 0.014 B = 10.15625p A 10 - 9 B m4 4

J =

p A 0.0154 - 0.014 B = 20.3125p A 10 - 9 B m4 2

Normal Stress and Shear Stress. The normal stress is caused by bending stress. Thus, sY = -

MyA 90(0.015) = = - 42.31MPa Iz 10.15625p A 10 - 9 B

The shear stress is caused by torsional stress. t =

360(0.015) Tc = = 84.62 MPa J 20.3125p A 10 - 9 B

The state of stress at point A is represented by the two -dimensional element shown in Fig. b. In - Plane Principal Stress. sx = - 42.31 MPa, sz = 0 and txz = 84.62 MPa. We have s1, 2 = =

sx + sz 2

;

A

a

sx - sz 2

2 b + txz 2

- 42.31 - 0 2 - 42.31 + 0 2 ; a b + 84.62 2 A 2

= ( - 21.16 ; 87.23) MPa s1 = 66.07 MPa

s2 = - 108.38 MPa

807

200 mm

900 N

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10–78. Continued

Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 66.072 - 66.07( -108.38) + ( -108.38)2 = sallow 2 sallow = 152.55 MPa Thus, the factor of safety is F.S. =

sY 250 = = 1.64 sallow 152.55

Ans.

Ans. F.S. = 1.64 808

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10–79. If the 50-mm diameter shaft is made from brittle sult= 350 material having having an anultimate ultimatestrength strengthofofsult = 50MPa ksi for both tension and compression, determine if the shaft fails according to the maximum-normal-stress theory. Use a factor of safety of 1.5 against rupture.

Solution Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are

kN · ft m 46 kip

150 kN 30 kip

A π= = (0.0252 ) 0.625(10 −3 )π m 2 J =

π

(0.0254 ) 0.1953125(10 −6 )π m 4 = 2

The normal stress is caused by axial stress. 150(10 3 )

N

u= = − = −76.39(106 ) N/m 2 = −76.39 MPa A 0.625(10 −3 )π

76.39 MPa

The shear stress is contributed by torsional shear stress.

= τ

244.46 MPa

Tc [6(10 3 )](0.025) 2 = = 244.46(106 ) N/m= 244.46 MPa J 0.1953125(10 −6 )π

The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. u x = −76.39 MPa, u y = 0 and τ xy = −244.46 MPa. We have s1, 2 =

sx + sy 2

;

A

a

sx - sy 2

2 b + txy 2

2

=

−76.39 + 0  −76.39 − 0  2 ±   + (−244.46) 2 2  

= −30.20 ± 247.43

u 1 = 209.23 MPa

u 2 = −285.63 MPa

Maximum Normal-Stress Theory. sallow =

sult 50 350 233.3 ksi MPa = = 33.33 F.S. 1.5 1.5

u 1 = 209.23 < u allow = 233.33 MPa

(O.K.)

u 2 = 285.63 > u allow = 233.33 MPa

(N.G.)

Based on these results, the material fails according to the maximum normal-stress theory.

Ans: Yes 803

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*10–80. If the 50-mm diameter shaft is made from cast iron having tensile and compressive ultimate strengths of (s1s )t 2=t =350 )c2c= =525 1s 50 MPa ksi and (s 75 MPa, ksi, respectively, ultult ultult determine if the shaft fails in accordance with Mohr’s failure criterion.

Solution Normal Stress and Shear Stresses. The cross-sectional area and polar moment of inertia of the shaft’s cross-section are

46kip kN· ·ftm

= A π= (0.0252 ) 0.625(10 −3 )π m 2

30 150kip kN

π

= (0.0254 ) 0.1953125(10 −6 )π m 4 2 The normal stress is caused by axial stress. = J

150(10 3 )

N

u= = − = −76.39(106 ) N/m 2 = −76.39 MPa A 0.625(10 −3 )π The shear stress is contributed by torsional shear stress. = τ

Tc [6(10 3 )](0.025) 2 = = 244.46(106 ) N/m= 244.46 MPa J 0.1953125(10 −6 )π

The state of stress at the points on the surface of the shaft is represented on the element shown in Fig. a. In-Plane Principal Stress. u x = −76.39 MPa, u y = 0 and τ xy = −244.46 MPa. We have s1, 2 =

sx + sy 2

;

A

a

sx - sy 2

2 b + txy 2

2

=

−76.39 + 0  −76.39 − 0  2 ±   + (−244.46) 2 2  

= −30.20 ± 247.43

u 1 = 209.23 MPa

u 2 = −285.63 MPa

s2 (MPa)

Mohr’s Failure Criteria. As shown in Fig. b, the coordinates of point A, which represent the principal stresses, are located inside the shaded region. Therefore, the material does not fail according to Mohr’s failure criteria.

350

350

s1 (MPa) 76.4 MPa

285.63 525

244.5 MPa

525 209.23

804

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10–81. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximum-shearstress theory.

60 MPa

40 MPa

70 MPa

Solution In accordance to the established sign convention, sx = 70 MPa, sy = - 60 MPa and txy = 40 MPa. s1, 2 = =

sx + sy 2

;

A

a

sx - sy 2

2 b + txy 2

70 + ( - 60) 70 - ( - 60) 2 2 ; c d + 40 2 A 2

= 5 ; 25825 s1 = 81.32 MPa

s2 = - 71.32 MPa

In this case, s1 and s2 have opposite sign. Thus, |s1 - s2| = |81.32 - (-71.32)| = 152.64 MPa 6 sY = 250 MPa Based on this result, the steel shell does not yield according to the maximum shear stress theory.

Ans: No 802

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10–82. The components of plane stress at a critical point on an A-36 steel shell are shown. Determine if failure (yielding) has occurred on the basis of the maximumdistortion-energy theory.

60 MPa

40 MPa

70 MPa

Solution In accordance to the established sign convention, sx = 70 MPa, sy = - 60 MPa and txy = 40 MPa. s1, 2 = =

sx + sy 2

;

A

a

sx - sy 2

2 b + txy 2

70 + ( -60) 70 - ( - 60) 2 2 ; c d + 40 2 A 2

= 5 ; 25825 s1 = 81.32 MPa

s2 = - 71.32 MPa

s1 2 - s1 s2 + s2 2 = 81.322 - 81.32(-71.32) + (-71.32)2 = 17,500 6 sY 2 = 62500 Based on this result, the steel shell does not yield according to the maximum distortion energy theory.

Ans: No 802

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10–83. The yield stress for heat-treated beryllium copper is sY = 900 MPa. If this material is subjected to plane is sY = 130 ksi. If this material is subjected to plane stress stress and elastic failure occurs when one principal stress is and elastic failure occurs when one principal stress is 145 ksi, 1000 MPa, what is the smallest magnitude of the other what is the smallest magnitude of the other principal stress? principal stress? Use the maximum-distortion-energy Use the maximum-distortion-energy theory. theory.

Solution Maximum Distortion Energy Theory : With s1 = 1000 145 ksi , MPa, s21 - s1 s2 + s22 = s2Y 1000 s2222 = 900 145s2 2++ s 13022 1452 2-– 1000s s22 –-1000s 145s2 ++ 190000 4125 == 00

u2 =

−(−1000) ± (1000)2 − 4(1)(190000) 2(1)

= 500 ± 244.95 Ans.

Choose the smaller root, s2 = 255.05 MPa = 255 MPa

Ans: s2 = 255 MPa 810

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*10–84. The state of stress acting at a critical point on a wrench is shown. Determine the smallest yield stress for steel that might be selected for the part, based on the maximum distortion energy theory.

150 MPa 300 MPa

Solution Normal And Shear Stress: In accordance with the sign convention, sx = 300 MPa  sy = 0  txy = 150 MPa In-Plane Principal Stress: s1, 2 = =

sx + sy 2

{

C

a

sx - sy 2

2

b + txy2

300 + 0 300 - 0 2 { a b + 1502 2 C 2

= 150 { 15022

s1 = 362.13 MPa   s2 = - 62.13 MPa Maximum Distortion Energy Theory: The theory gives s12 - s1s2 + s22 = sY 2 362.132 - 362.13( - 62.13) + ( -62.13)2 - sY 2 Ans.

sY = 396.86 MPa = 397 MPa

Ans: sY = 397 MPa 1027

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10–85. The state of stress acting at a critical point on a wrench is shown in the figure. Determine the smallest yield stress for steel that might be selected for the part, based on the maximum shear stress theory. 150 MPa 300 MPa

Solution Normal And Shear Stresses: In accordance to the sign convention,

sx = 300 MPa   sy = 0   txy = 150 MPa

In-Plane Principal Stress: s1, 2 = =

sx + sy 2

{

C

a

sx - sy 2

2

b + txy2

300 + 0 300 - 0 2 { a b + 1502 2 C 2

= 150 { 15022

s1 = 362.13 MPa   s2 = - 62.13 MPa Maximum Shear Stress Theory: Here, s1 and s2 have opposite signs. So

0 s1 - s2 0 = sY

0 362.13 - ( - 62.13) 0 = sY

Ans.

sY = 424.26 MPa = 424 MPa

Ans: sY = 424 MPa 1028

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10–86. The shaft consists of a solid segment AB and a hollow segment BC, which are rigidly joined by the coupling at B. If the shaft is made from A-36 steel, determine the ­maximum torque T that can be applied according to the maximum shear stress theory. Use a factor of safety of 1.5  against yielding.

A

80 mm B

T C

80 mm 100 mm

Solution

T

Shear Stress: This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = 1 0.054 - 0.044 2 = 1.845p 1 10 - 6 2 m4. Thus, 2 T(0.05) Tch = = 8626.28T Jh 1.845p 1 10 - 6 2

(tmax)h =

For the solid segment, Js =

p 1 0.044 2 = 1.28p 1 10 - 6 2 m4. Thus, 2

(tmax)s =

T(0.04) Tcs = = 9947.18T Js 1.28p 1 10 - 6 2

By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 = =

sx + sy 2

{

C

¢

sx - sy 2

2

≤ + t2xy

0 + 0 0 - 0 2 { ¢ ≤ + (9947.18T)2 2 C 2

s1 = 9947.18T

s2 = - 9947.18T

Maximum Shear Stress Theory. sallow =

sY 250 = = 166.67 MPa F.S. 1.5

Since s1 and s2 have opposite sings, |s1 - s2| = sallow 9947.18T - ( -9947.18T) = 166.67 1 106 2 T = 8377.58 N # m = 8.38 kN # m

Ans.

Ans: T = 8.38 kN # m 1029

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10–87. The shaft consists of a solid segment AB and a ­hollow segment BC, which are rigidly joined by the ­coupling at B. If the shaft is made from A-36 steel, determine the maximum torque T that can be applied according to the maximum distortion energy theory. Use a factor of safety of 1.5 against yielding.

A

80 mm B

T C

80 mm 100 mm

Solution

T

Shear Stress. This is a case of pure shear, and the shear stress is contributed by p torsion. For the hollow segment, Jh = 1 0.054 - 0.044 2 = 1.845p 1 10 - 6 2 m4. Thus, 2 T(0.05) Tch = = 8626.28T Jh 1.845p 1 10 - 6 2

(tmax)h =

For the solid segment, Js =

p 1 0.044 2 = 1.28p 1 10 - 6 2 m4. Thus, 2

(tmax)s =

T(0.04) Tcs = = 9947.18T Js 1.28p 1 10 - 6 2

By comparision, the points on the surface of the solid segment are critical and their state of stress is represented on the element shown in Fig. a. In - Plane Principal Stress. sx = sy = 0 and txy = 9947.18T. We have s1,2 = =

sx + sy 2

{

C

¢

sx - sy 2

2

≤ + t2xy

0 + 0 0 - 0 2 { ¢ ≤ + (9947.18T)2 2 C 2

s1 = 9947.18T

s2 = - 9947.18T

Maximum Distortion Energy Theory. sallow =

sY 250 = = 166.67 MPa F.S. 1.5

Then, s1 2 - s1s2 + s2 2 = sallow 2 (9947.18T)2 - (9947.18T)( - 9947.18T) + ( - 9947.18T)2 = T = 9673.60 N # m = 9.67 kN # m

3 166.67 1 106 2 4 2

Ans.

Ans: T = 9.67 kN # m 1030

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*10–88. The principal stresses acting at a point on a thin-walled cylindrical pressure vessel are s1 = pr>t, s2 = pr>2t, and s3 = 0. If the yield stress is sY , determine the maximum value of p based on (a) the maximum shear stress theory and (b) the maximum distortion energy theory.

Solution a) Maximum Shear Stress Theory: s1 and s2 have the same signs, then |s2| = sY

`

|s1| = sY

`

pr ` = sY 2t pr ` = sY t

p =

2t s r Y

p =

t s (Controls!) r Y

Ans.

b) Maximum Distortion Energy Theory: s21 - s1 s2 + s22 = s2Y a

pr 2 pr pr pr 2 b - a ba b + a b = s2Y t t 2t 2t p =

2t

23r

Ans.

sY

Ans: (a) p = (b) p =

1031

t s, r y 2t 23r

sy

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10–89. The gas tank has an inner diameter of 1.50 m and a wall thickness of 25 mm. If it is made from A-36 steel and the tank is pressured to 5 MPa, determine the factor of safety against yielding using (a) the maximum-shear-stress theory, and (b) the maximum-distortion-energy theory.

Solution (a) Normal Stress. Since

0.75 r = = 30 7 10, thin - wall analysis can be used.We have t 0.025

s1 = sh =

5(0.75) pr = = 150 MPa t 0.025

s2 = slong =

pr 5(0.75) = = 75 MPa 2t 2(0.025)

Maximum Shear Stress Theory. s1 and s2 have the sign. Thus, |s1| = sallow sallow = 150 MPa The factor of safety is F.S. =

sY 250 = = 1.67 sallow 150

Ans.

(b) Maximum Distortion Energy Theory. s1 2 - s1s2 + s2 2 = sallow 2 1502 - 150(75) + 752 = sallow 2 sallow = 129.90 MPa The factor of safety is F.S. =

sY 250 = = 1.92 sallow 129.90

Ans.

Ans: F.S. = 1.67, F.S. = 1.92 818

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10–90. The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using (a) the maximum shear stress theory, and (b) maximum distortion energy theory. Apply a factor of safety of 1.5 against yielding.

Solution (a) Normal Stress: Assuming that thin-wall analysis is valid, we have s1 = sh =

5 1 106 2 (0.75) 3.75 1 106 2 pr = = t t t

s2 = slong =

5 1 106 2 (0.75) 1.875 1 106 2 pr = = 2t 2t t

Maximum Shear Stress Theory: sallow =

250 1 106 2 sY = = 166.67 1 106 2 Pa FS. 1.5

s1 and s2 have the same sign. Thus,  s1  = sallow 3.75 1 106 2 t

= 166.67 1 106 2

Ans.

t = 0.0225 m = 22.5 mm Since

r 0.75 = = 33.3 7 10, thin-wall analysis is valid. t 0.0225

(b) Maximum Distortion Energy Theory: sallow = Thus,

250 1 106 2 sY = = 166.67 1 106 2 Pa F.S. 1.5

s1 2 - s1s2 + s2 2 = sallow 2 C

3.75 1 106 2 t

3.2476 1 106 2 t

2

S - C

3.75 1 106 2 t

SC

1.875 1 106 2 t

S + C

1.875 1 106 2 t

= 166.67 1 106 2

2

S = c 166.67 1 106 2 d

Ans.

t = 0.01949 m = 19.5 mm Since

2

r 0.75 = = 38.5 7 10, thin-wall analysis is valid. t 0.01949

Ans: (a) t = 22.5 mm, (b) t = 19.5 mm 1033

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10–91. The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300KN lb # ft, 1500kN lb # ·ft, 3.45 · m,aabending bendingmoment moment of of 2.25 m,and and an an axial thrust of 12.5 2500kN. lb. If If the the yield yield points for tension and shear are sY 100MPa ksi and stYY ==350 50 MPa, ksi, respectively, s respectively,determine determine the Y ==700 required diameter of the shaft using the maximum-shearstress theory. 3.45lbft kNm 2300

2.25 lbft kNm 1500

12.5 kN 2500 lb

Solution

P

p 4 c 4

I =

A = p c2

Mc

J =

 12.5(10 3 )

uA = + + = − A I  π c 2 τ =

p 4 c 2

 12.5(10 3 ) 9(10 3 )  2.25(10 3 )c     + = − π c4 π c3   π c 2  4 4

Tc 3.45(10 3 )c 6.9(10 3 ) = = π c4 J π c3 2

s1,2 =

sx + sy 2

;

A

a

sx - sy 2

2 b + txy 2

2

 12.5(10 3 )c + 9(10 3 )   12.5(10 3 )c + 9(10 3 )   6.9(10 3 )  = − ± +      3 2π c 3 2π c 3       π c

2

(1)

Assume s1 and s2 have opposite signs:

u1 − u2 = uY 2

2

 12.5(10 3 )c + 9(10 3 )   6.9(10 3 )  2  + 700(106 )    = 3 2π c 3   π c   490(109 )π 2c 6 − 156.25c 2 − 225c − 271.44 = 0 By trial and error: c = 0.019621 m Substitute c into Eq. (1): 6 = u 1 155.20(10 = ) N/m 2 155.20 MPa

u2 = −544.81(106 ) N/m 2 = −544.81 MPa s1 and s2 are of opposite signs

OK

Therefore,

= d 2(0.019621) = 0.039242 = m 39.2 mm

Ans.

Ans. d = 39.2 mm 817

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*10–92. If the material is machine steel having a yield stress of sY = 750 MPa, determine the factor of safety with respect to yielding using the maximum distortion energy theory.

200 MPa 500 MPa

30

Solution Maximum Distortion Energy Theory: Here, s1 = 200 MPa and s2 = -500 MPa. The theory gives s12 - s1s2 + s22 = s2allow 2002 - (200)( - 500) + ( - 500)2 = s2allow sallow = 624.50 MPa The factor of safety is F.S. =

sY 750 = = 1.20 sallow 624.50

Ans.

Ans: F.S. = 1.20 1035

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10–93. If the material is machine steel having a yield stress of sY = 750 MPa, determine the factor of safety with respect to yielding if the maximum shear stress theory is considered.

450 MPa

150 MPa

Solution Maximum Shear Stress Theory: Here, s1 = 150 MPa and s2 = -450 MPa since s1 and s2 have opposite signs, then

0 s1 - s2 0 = sallow

0 150 - ( - 450) 0 = sallow sallow = 600 MPa

The factor of safety is F.S. =

sY 750 = = 1.25 sallow 600

Ans.

Ans: F.S. = 1.25 1036

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R10–1. In the case of plane stress, where the in-plane principal strains are given by P1 and P2, show that the third principal strain can be obtained from - n ( P1 + P2 ) P3 = (1 - n) where n is Poisson’s ratio for the material.

Solution Generalized Hooke’s Law: In the case of plane stress, s3 = 0. Thus, P1 =

1 (s - ns2) E 1

(1)

P2 =

1 (s - ns1) E 2

(2)

n (s + s2) E 1

(3)

P3 = -

Solving for s1 and s2 using Eqs. (1) and (2), we obtain s1 =

E(P1 + nP2) 1 - n2

s2 =

E(P2 + nP1) 1 - n2

Substituting these results into Eq. (3), P3 = P3 = P3 = P3 = -

E(P2 + nP1) n E(P1 + nP2) c + d 2 E 1-n 1 - n2 (P1 + P2) + n(P1 + P2) n c d 1 - n 1 + n (P1 + P2)(1 + n) n c d 1 - n 1 + n n (P + P2) 1 - n 1

(Q.E.D.)

Ans.

Ans: N/A 1037

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R10–2. The plate is made of material having a modulus of elasticity E = 200 GPa and Poisson’s ratio n = 13 . Determine the change in width a, height b, and thickness t when it is subjected to the uniform distributed loading shown.

a  400 mm

2 MN/m t  20 mm

3 MN/m

y

Solution

b  300 mm

Normal Stress: The normal stresses along the x, y, and z axes are sx =

3(106) 0.02

sy = -

= 150 MPa

2(106) 0.02

x

z

= - 100 MPa

sz = 0 Generalized Hooke’s Law: Px = =

1 c s - n(sy + sz) d E x

1 1 e 150(106) - c - 100(106) + 0 d f 3 200(109)

= 0.9167(10 - 3) Py = =

1 c s - n(sx + sz) d E y

1 1 e - 100(106) - c 150(106) + 0 d f 9 3 200(10 )

= - 0.75(10 - 3) Pz = =

1 c s - n(sx + sy) d E z

1 1 e 0 - c 150(106) + ( -100)(106) d f 9 3 200(10 )

= - 83.33(10 - 6)

Thus, the changes in dimensions of the plate are da = Pxa = 0.9167(10 - 3)(400) = 0.367 mm

Ans.

-3

db = Pyb = -0.75(10 )(300) = - 0.225 mm

Ans.

dt = Pzt = - 83.33(10 - 6)(20) = - 0.00167 mm

Ans.

The negative signs indicate that b and t contract.

Ans: da = 0.367 mm, db = -0.255 mm, dt = -0.00167 mm 1038

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R10–3. If the material is machine steel having a yield stress of sY = 500 MPa, determine the factor of safety with respect to yielding if the maximum shear stress theory is considered.

100 MPa

150 MPa

Solution Here, the in plane principal stresses are

s1 = sy = 100 MPa s2 = sx = - 150 MPa

Since s1 and s2 have the same sign, F.S. =

sy  s1 - s2 

=

500 = 2  100 - ( - 150) 

Ans.

Ans: F.S. = 2 1039

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*R10–4. The components of plane stress at a critical point on a thin steel shell are shown. Determine if yielding has occurred on the basis of the maximum distortion energy theory. The yield stress for the steel is sY = 650 MPa.

340 MPa 65 MPa

55 MPa

Solution sx = - 55 MPa sy = 340 MPa txy = 65 MPa s1, 2 = =

sx + sy 2

{

A

a

sx - sy 2

2

2 b + txy

- 55 - 340 2 - 55 + 340 2 { a b + 65 2 A 2

s1 = 350.42 MPa s2 = - 65.42 MPa

(s1 2 - s1s2 + s2 ) = [350.422 - 350.42( -65.42) + ( - 65.42)2] = 150 000 6 s2Y = 422 500

OK Ans.

No. 

Ans: No 1040

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R10–5. The 60° strain rosette is mounted on a beam. The following readings are obtained for each gage: Pa = 600(10-6), Pb = -700(10-6), and Pc = 350(10-6). Determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case show the deformed element due to these strains.

a 60 60 b

60

Solution c

Strain Rosettes (60º): Applying Eq. 10–15 with Px = 600 1 10 - 6 2 ,

Pb = -700 1 10 - 6 2 , Pc = 350 1 10 - 6 2 , ua = 150°, ub = -150° and uc = -90°, 350 1 10 - 6 2 = Px cos2 ( - 90°) + Py sin2 ( -90°) + gxy sin ( -90°) cos ( -90°) Py = 350 1 10 - 6 2

600 1 10 - 6 2 = Px cos2 150° + 350 1 10 - 6 2 sin2 150° + gxy sin 150° cos 150° 512.5 1 10 - 6 2 = 0.75 Px - 0.4330 gxy - 700 1 10

-6

2

= Px cos ( - 150°) + 350 1 10 2

- 787.5 1 10 - 6 2 = 0.75Px + 0.4330 gxy

-6

2 sin ( - 150°)

Solving Eq. (1) and (2) yields Px = - 183.33 1 10

2

-6

2

(1)

+ gxy sin ( -150°) cos ( -150°) (2)

gxy = -1501.11 1 10

-6

2

Construction of the Circle: With Px = - 183.33 1 10 - 6 2 , Py = 350 1 10 - 6 2 , and gxy = - 750.56 1 10 - 6 2 . 2 Pavg =

Px + Py 2

= a

- 183.33 + 350 b 1 10 - 6 2 = 83.3 1 10 - 6 2 2

Ans.

The coordinates for reference points A and C are A( -183.33, - 750.56) 1 10 - 6 2

The radius of the circle is

C(83.33, 0) 1 10 - 6 2

R = a 2(183.33 + 83.33)2 + 750.562 b 1 10 - 6 2 = 796.52 1 10 - 6 2

a)

In-Plane Principal Strain: The coordinates of points B and D represent P1 and P2, respectively. P1 = (83.33 + 796.52) 1 10 - 6 2 = 880 1 10 - 6 2

Ans.

P2 = (83.33 - 796.52) 1 10 - 6 2 = - 713 1 10 - 6 2

Ans.

Orientation of Principal Strain: From the circle, tan 2uP1 =

750.56 = 2.8145 2uP2 = 70.44° 183.33 + 83.33

2uP1 = 180° - 2uP2 uP =

180° - 70.44° = 54.8° (Clockwise) 2

Ans.

1041

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R10–5.  Continued

b) Maximum In-Plane Shear Strain: Represented by the coordinates of point E on the circle. g max in@plane = -R = - 796.52 1 10 - 6 2 2 g max

in@plane

= - 1593 1 10 - 6 2

Ans.

Orientation of Maximum In-Plane Shear Strain: From the circle. tan 2us =

183.33 + 83.33 = 0.3553 750.56 Ans.

us = 9.78° (Clockwise)

Ans: Pavg = 83.3(10 - 6), P1 = 880(10 - 6), P2 = -713(10 - 6), up = 54.8° (clockwise), g max = -1593(10 - 6), in@plane

us = 9.78° (clockwise) 1042

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R10–6. y

The state of strain at the point on the bracket has components Px = 350(10-6), Py = - 860(10-6), -6 gxy = 250(10 ). Use the strain transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of u = 45° clockwise from the original position. Sketch the deformed element within the x–y plane due to these strains.

Solution

x

-6

-6

Px = 350(10 ) Px ′ =

Px + Py 2

= c

+

-6

Py = - 860(10 ) Px - Py 2

cos 2u +

gxy = 250(10 )

gxy 2

u = -45°

sin 2u

350 - ( -860) 350 - 860 250 + cos ( -90°) + sin ( -90°) d (10 - 6) 2 2 2

= -380(10 - 6) Py′ =

Px + Py

= c

2

-

Px - Py 2

cos 2u -

gxy 2

sin 2u

350 - ( -860) 350 - 860 250 cos ( - 90°) sin ( - 90°) d (10 - 6) 2 2 2

= -130(10 - 6) gx′y′ 2

= -

Px - Py

gx′y′ = 2c- a

2

sin 2u +

350 - ( - 860) 2

Ans.

Ans.

g cos 2u 2 b sin (- 90°) +

250 cos (-90°)d (10 - 6) = 1.21(10 - 3) Ans. 2

Ans: Px′ = -380(10 - 6), Py′ = - 130(10 - 6), gx′y′ = 1.21(10 - 3) 1043

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R10–7. A strain gauge forms an angle of 45° with the axis of the 50mm diameter shaft. If it gives a reading of P = −200(10−6) when the torque T is applied to the shaft, determine the magnitude of T. The shaft is made from A-36 steel. T 45

Solution Shear Stress. This is a case of pure shear, and the shear stress developed is p contributed by torsional shear stress. Here, J = A 0.0254 B = 0.1953125p A 10 - 6 B m4. 2 Then t =

T

0.128 A 106 B T T(0.025) Tc = = p J 0.1953125p A 10 - 6 B

The state of stress at points on the surface of the shaft can be represented by the element shown in Fig. a. Shear Strain: For pure shear ex = ey = 0. We obtain, ea = ex cos2ua + ey sin2ua + gxysin ua cos ua - 200 A 10 - 6 B = 0 + 0 + gxy sin 45° cos 45°

gxy = - 400 A 10 - 6 B

Shear Stress and Strain Relation: Applying Hooke’s Law for shear, txy = Ggxy -

0.128 A 106 B T p

= 75 A 109 B C - 400 A 10 - 6 B D

T = 736 N # m

Ans.

Ans:

T = 736 N # m

823

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*R10–8. A differential element is subjected to plane strain that has the following components; Px = 950(10-6), Py = 420(10-6), gxy = - 325(10-6). Use the strain-transformation equations and determine (a) the principal strains and (b) the maximum in-plane shear strain and the associated average strain. In each case specify the orientation of the element and show how the strains deform the element.

Solution Px + ey

P1, 2 =

2

= c

{

Aa

Px - ey 2

2

2 b + gxy

950 - 420 2 950 + 420 - 325 2 -6 { a b + a b d (10 ) 2 A 2 2 P1 = 996(10 - 6)

Ans.

P2 = 374(10 - 6)

Ans.

Orientation of P1 and P2: gxy

tan 2uP =

Px - Py

=

- 325 950 - 420

uP = - 15.76°, 74.24° Use Eq. 10–5 to determine the direction of P1 and P2. Px′ =

Px + Py 2

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

u = uP = -15.76° Px′ = b

( -325) 950 + 420 950 - 420 + cos ( - 31.52°) + sin ( -31.52°) r(10 - 6) = 996(10 - 6) 2 2 2

uP1 = - 15.8°

Ans.

uP2 = 74.2°

Ans.

b) g max

in@plane

2 g max

in@plane

Pavg =

=

B

= 2c

a

2

2

b + a

gxy 2

b

2

950 - 420 2 - 325 2 b + a b d (10 - 6) = 622(10 - 6) 2 2 B

Px + Py 2

Px - Py

a

= a

950 + 420 b(10 - 6) = 685(10 - 6) 2

Ans. Ans.

1046

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*R10–8.  Continued

Orientation of gmax: tan 2uP =

- (Px - Py) gxy

=

- (950 - 420) - 325 Ans.

uP = 29.2° and uP = 119° Use Eq. 10–6 to determine the sign of g max

in@plane

gx′y′ 2

= -

Px - Py 2

sin 2u +

gxy 2

:

cos 2u

u = us = 29.2° gx′y′ = 2c

- (950 - 420) 2

gxy = -622(10 - 6)

sin (58.4°) +

- 325 cos (58.4°) d (10 - 6) 2

Ans: P1 = 996(10 - 6), P2 = 374(10 - 6), up1 = - 15.8, up2 = 74.2, g max

in@plane

= 622(10 - 6),

Pavg = 685(10 - 6), us = 29.2° and 119° 1047

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R10–9. The state of strain at the point on the bracket has components Px = - 130(10-6), Py = 280(10-6), -6 gxy = 75(10 ). Use the strain transformation equations to determine (a) the in-plane principal strains and (b) the maximum in-plane shear strain and average normal strain. In each case specify the orientation of the element and show how the strains deform the element within the x–y plane.

y

Solution Px = - 130(10 - 6)

Py = 280(10 - 6)

gxy = 75(10 - 6)

x

a) Px + Py

P1, 2 =

{

2

= c

B

a

Px - Py 2

2

b + a

gxy 2

b

2

- 130 + 280 - 130 - 280 2 75 2 { a b + a b d (10 - 6) 2 B 2 2

P1 = 283(10 - 6)

P2 = - 133(10 - 6)

Ans.

Ans.

Orientation of P1 and P2: tan 2up =

gxy Px - Py

up = - 5.18°

=

75 - 130 - 280

and

84.82°

Use Eq. 10–5 to determine the direction of P1 and P2: Px′ =

Px + Py 2

+

Px - Py 2

cos 2u +

gxy 2

sin 2u

u = up = - 5.18° Px′ = c

- 130 + 280 - 130 - 280 75 + cos ( - 10.37°) + sin ( -10.37°) d (10 - 6) = -133(10 - 6) 2 2 2

Therefore up1 = 84.8°

Ans.

Ans.

up2 = - 5.18°

1048

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R10–9.  Continued

b) g max

in@plane

2 g max

in@plane

Pavg =

=

B

= 2c

a

Px - Py 2

2

a

= a

Orientation of gmax:



gxy 2

b

2

- 130 - 280 2 75 2 b + a b d (10 - 6) = 417(10 - 6) B 2 2

Px + Py

tan 2us =

2

b + a

- 130 + 280 b(10 - 6) = 75.0(10 - 6) 2

- (Px - Py) gxy

=

Ans.

- ( - 130 - 280) 75 Ans.

us = 39.8° and us = 130°

Use Eq. 10–16 to determine the sign of g max

in@plane

gx′y′ 2

= -

Px - Py 2

Ans.

sin 2u +

gxy 2

cos 2u;

:

u = us = 39.8°

gx′y′ = -( -130 - 280) sin (79.6°) + (75) cos (79.6°) = 417(10 - 6)

Ans: P1 = 283(10 - 6), P2 = -133(10 - 6), up1 = 84.8°, up2 = -5.18°, g max = 417(10 - 6), in@plane

Pavg = 75.0(10 - 6), us = 39.8° and 130° 1049

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R10–10. The state of plane strain on the element is Px = 400(10-6), Py = 200(10-6), and gxy = - 300(10-6). Determine the equivalent state of strain, which represents (a) the principal strains, and (b) the maximum in-plane shear strain and the associated average normal strain. Specify the orientation of the corresponding element at the point with respect to the original element. Sketch the results on the element.

y

Pydy dy

gxy 2 dx

Solution Construction of the gxy = - 150 1 10 - 6 2 . Thus, 2 Px + Py

Pavg =

2

= a

Circle:

Px = 400 1 10 - 6 2 ,

Py = 200 1 10 - 6 2 ,

400 + 200 b 1 10 - 6 2 = 300 1 10 - 6 2 2

and

Ans.

The coordinates for reference points A and the center C of the circle are A(400, -150) 1 10 - 6 2

The radius of the circle is

C(300, 0) 1 10 - 6 2

R = CA = 2(400 - 300)2 + ( - 150)2 = 180.28 1 10 - 6 2

Using these results, the circle is shown in Fig. a.

In-Plane Principal Stresses: The coordinates of points B and D represent P1 and P2, respectively. Thus, P1 = (300 + 180.28) 1 10 - 6 2 = 480 1 10 - 6 2

Ans.

P2 = (300 - 180.28) 1 10 - 6 2 = 120 1 10 - 6 2

Ans.

Orientation of Principal Plane: Referring to the geometry of the circle, tan 2 1 up 2 1 =

1 up 2 1

150 = 1.5 400 - 300 Ans.

= 28.2° (clockwise)

The deformed element for the state of principal strains is shown in Fig. b. Maximum In-Plane Shear Stress: The coordinates of point E represent Pavg and g max . Thus in@plane g max in@plane

2 g max

in@plane

= -R = - 180.28 1 10 - 6 2

Ans.

= - 361 1 10 - 6 2 

Orientation of the Plane of Maximum In-plane Shear Strain: Referring to the geometry of the circle, tan 2us =

gxy 2

400 - 300 = 0.6667 150 Ans.

uS = 16.8° (counterclockwise)

The deformed element for the state of maximum In-plane shear strain is shown in Fig. c.

1050

x Pxdx

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R10–10.  Continued

Ans: P1 = 480(10 - 6), P2 = 120(10 - 6), up1 = 28.2° (clockwise), g max = -361(10 - 6), in@plane

us = 16.8° (counterclockwise), Pavg = 300(10 - 6) 1051

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11–1. The beam is made of timber that has an allowable bending stress of sallow = 6.5 MPa and an allowable shear stress of tallow = 500 kPa. Determine its dimensions if it is to be rectangular and have a height-to-width ratio of 1.25. Assume the beam rests on smooth supports.

8 kN/m

2m

4m

2m

Solution Ix =

1 (b)(1.25b)3 = 0.16276b4 12

Qmax = y′A′ = (0.3125b)(0.625b)(b) = 0.1953125b3 Assume Bending Moment Controls: Mmax = 16 kN # m Mmax c I 3 16(10 )(0.625b) 6.5(106) = 0.16276b4 sallow =



b = 0.21143 m = 211 mm

Ans.

h = 1.25b = 264 mm

Ans.

Check Shear: Qmax = 1.846159(10 - 3) m3 I = 0.325248(10 - 3) m4 tmax =

16(103)(1.846159)(10 - 3) VQmax = = 429 kPa 6 500 kPa‚ It 0.325248(10 - 3)(0.21143)

OK

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

1052

Ans: b = 211 mm, h = 264 mm

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11–2. Select the lightest-weight W310 steel wide-flang safely support support the loading loading beam from Appendix B that will safely where PP==306 kN. Theallowable allowablebending bendingstress stress kip. The shown, where is isallow and allowable = 22 ksiand sallow s = 150 MPa the the allowable shear shear stress isstress tallow is = tallow = 12 ksi. 84 MPa.

P P

P P

9 ftm 2.7

Solution

6 ft 1.8 m

1.86 ft m

From the Moment Diagram, Fig. a, Mmax = 81 54 kN kip ·# ft m.. Sreq¿d =

=

Mmax sallow

81(10 3 ) 150(106 )

−3 = 0.54(10 = ) m 3 540(10 3 ) mm 3

Select W310 W12 *26 39

3 3 in3, d = 12.22 andmm, tw =tw 0.230 547(10 ) mm , d =in310 = 5.84in. mm] C Sx[S=x =33.4 D

kN. Provide the the shear-stress check 7.5 kip From the shear diagram, Fig. a, Vmax = 37.5 . Provide check 12 * for W W310  26 39., tmax = =

Vmax tw d 37.5(10 3 ) = 20.71(106 ) N/m 2 0.00584(0.310)

20.7 ksi MPa6< ttallow (O.K!) 12MPa ksi (O.K!) = 2.67 allow ==84 Hence W12 *26 W310 39

Use

Ans.

30 kN

30 kN

1.8 m

2.7 m

1.8 m

67.5 kN

7.5 kN

V (kN) 37.5

7.5 2.7

x (m) 4.5

6.3

–30 M (kN · m) 2.7

4.5

6.3

x (m)

–13.5

–81

Ans: Use W310 * 39 837

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P P

11–3. Select the lightest-weight W360 steel wide-flang beam from Appendix B that will safely support the loading shown, where P = 60 kN. The allowable bending stress is sallow = 150 MPa and the allowable shear stress is τallow = 84 MPa.

P P

2.7 9 ftm

1.8 6 ftm

1.86 ft m

Solution Mmax sallow

Sreq¿d = =

162(10 3 ) 150(106 )

−3 = 1.08(10 = ) m 3 1080(10 3 ) mm 3 3 1280(10 , d in = 354 = 9.4in. mm] in3, d3)=mm 13.66 and mm, tw = tw0.305 C Sx[S=x =62.7 D

W360*43 79 Select W14

75 kip kN.. Provide the shear-stress check From the shear diagram, Fig. a, Vmax = 15 for W360 , W14 *43 79, Vmax tw d

tmax =

75(10 3 ) = 22.54(106 ) N/m 2 0.0094(0.354)

=

12MPa ksi‚ (O.K!) = 3.60 22.5 ksi MPa6< ttallow allow ==84 Hence, W14 *43 W360 79

Use

Ans.

60 kN

60 kN

1.8 m

2.7 m

1.8 m

135 kN

15 kN

V (kN) 75

15 2.7

x (m) 4.5

6.3

–60

M (kN · m) 2.7

4.5

6.3

x (m)

–27

−162

Ans: Use W360 * 79 838

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P

*11–4. Determine the minimum width of the beam to 1 nearest multiples safely the support loadingthe of the nearest 5 mm thatsupport will safely 4 in. thatofwill bending bending stress is stress P = 8 kip. ksi sallowis=s24 loading of PThe = 40allowable kN. The allowable = allow and MPa the allowable shear stress is tstress ksi. = 100 MPa. 168 and the allowable shear tallow allow =is15

1.8 6 ftm

1.8 m 6 ft

150 6mm in.

B B

A A 1.8 m

Solution

40 kN

Beam design: Assume moment controls. sallow

Mc = ; I

168(106 ) =

40 kN

40

1 b(0.153 ) 12

0 1.8 –40

3.6

x(m)

Ans. M (kN · m)

Check shear: VQ [40(10 3 )][0.0375(0.115)(0.075)] = It  1 (0.115)(0.153 ) (0.115)  12 

=

80 kN

V (kN)

[72.0(10 3 )](0.075)

b = 0.1143 m = 114.3 mm Use b = 115 mm

tmax =

1.8 m

0

72.0

1.8

3.6

x(m)

6 3.478(10= ) N/m 2 3.48 MPa < 100 MPa OK

Ans: b = 114.3 mm 840

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25 kN 5 kip

11–5. Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the machine loading = 168 shown. The The allowable allowablebending bendingstress stressisissallow = 24MPa ksi and sallow MPa. the allowable shear stress is tallow == 100 14 ksi.

25 kN 5 kip

25 kN 5 kip

25 kN 5 kip

Solution 0.6 2 ftm

# ft. kN · m. Bending Stress: From the moment diagram, Mmax = 45 30.0 kip Assume bending controls the design. Applying the flexure formula.

0.6 2 ftm

0.6 2 ftm

0.6 2 ftm

0.6 2 ftm

25 kN 25 kN 25 kN 25 kN

Sreq¿d

Mmax = sallow

45(10 3 ) −3 = 0.2679(10 = ) m 3 267.9(10 3 ) mm 3 168(106 )

= Select

W12 W310*16 24

0.6 m 0.6 m 0.6 m 0.6 m 0.6 m 50 kN

Sxx = 17.1 in33), d = 3,11.99 in.,mm, tw =tw0.220 in.mm) = 281(10 mm d = 305 = 5.59 A(S B

50 kN

V (kN)

V 24wide wide-for the W310 W12 *16 tw d kN = 50 10.0 kip

50

Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax

25 1.8 0.6

2.4

3.0

x (m)

1.2 –25

tmax

Vmax = tw d

50(10 3 ) == 0.00559(0.305)

–50

M (kN · m)

6

29.33(10 ) N/m

2 45

45

30

= 27.8 MPa 6 tallow = 100 MPa (O.K!)

30 x (m)

0.6

Hence, Use

W12 W310*16 24

1.2

1.8

2.4

3.0

Ans.

Ans: Use W310 * 24 833

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11–6. The compound beam is made from two sections, which are pinned together at B. Use Appendix B and select the lightest-weight wide-flange beam that would be safe for each section sectionif if allowable bending is = sallow = thethe allowable bending stressstress is sallow 24 ksi 168 and the allowable shear is 14 tallow 100 beam MPa. and MPa the allowable shear stress is tstress ksi.=The allow = The beam supports a pipe loading 6 kN andshown. 9 kN as supports a pipe loading of 1200 lb andof1800 lb as shown.

1800 9 kNlb

1200 6 kNlb

A

C

B 6 ftm 1.8

6 ftm 1.8

8 ft 2.4 m

10 3 mft

Solution Bending Stress: From the moment diagram, Mmax = 28.8 kN # m for member AB. Assuming bending controls the design, applying the flexure formula. Sreq¿d =

Mmax sallow

Select

4 kN

6 1.8

9

x (m)

3.6 –4

M (kN · m)

12(10 3 ) −6 = 71.43(10 = ) m 3 71.43(10 3 ) mm 3 168(106 )

12 1.8

3 91.2(10 , din., = 150 = 4.32 in3, d3)=mm 5.90 tw =mm, 0.17twin. A Sx(S=x =5.56 B mm)

W150 W6 *  9 14

3m

5

Mmax = sallow

=

2.4 m

V (kN)

kN ·m.# ft. For member BC, Mmax = 12 8.00 kip Sreq¿d

9 kN

5 kN

28.8 kN · m

Sxx =179(10 = 10.9 in3)3,mm d =3, 9.87 in., mm, tw = t0.19 in. B mm) d = 251 A(S w = 4.83

W10 *12 W250 18

5 kN

1.8 m 1.8 m

28.8(10 3 ) −3 = = 0.1714(10 = ) m 3 171.4(10 3 ) mm 3 168(106 ) Select

6 kN 11 kN

x (m) 3.6 –9

6

9

V W250*12 18widewide for the W10 tw d kN. flange section for member AB. From the shear diagram, Vmax = 11 2.20 kip. Shear Stress: Provide a shear stress check using t =

tmax =

Vmax tw d

11(10 3 ) 6 = = 9.073(10 = ) N/m 2 9.07 MPa 0.00483(0.251) (O.K!) 14 MPa ksi (O.K!) = 9.07 1.17 MPa ksi 6< ttallow allow ==100 Use

W250 * 18

Ans.

For member BC (W6 = = 1.00 kip. * 9),14), Vmax (W150 Vmax 5 kN tmax =

Vmax tw d

5(10 3 ) 6 = ) N/m 2 7.72 MPa == 7.716(10 0.00432(0.15) = 7.72 0.997MPa ksi fton = 154 MPa and the allowable shear stress stress is is ssallow and the allowable shear = 22 ksi allow selectthe thelightest lightest wide-flange wide-flange section from tallow = 84 MPa,select Appendix B that will safely support the load.

kN/m 1.2020 kip/ ft

1.2 m 4 ft

3 mft 10

1.8 6 ftm b

Solution

Bending Stress: From the moment diagram, Mmax = 66.825 kN# · m. Assuming bending controls the design and applying the flexure formula.

0.5 in.

66.825(10 3 ) = 154(106 )

=

9 in. 0.5 in.

Mmax = sallow

Sreq d

0.5 in.

−3 0.4339(10 = ) m 3 433.9(10 3 ) mm 3

3 W12 W360*22 33 = (Sx 475(10= ) mm 3 , d 349 mm, tw 5.84 mm) =

Select

V for the W12 W360*22 33wide wide-tw d 33 kN. = 6.60 kip.

Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax Vmax tw d

tmax =

33(10 3 ) == 0.00584(0.349)

6 16.19(10 = ) N/m 2 16.2 MPa

16.2 ksi MPa6< ttallow (O.K!) = 2.06 12MPa ksi (O.K!) allow ==84 W12 *22 W360 33

Use

Hence,

Ans.

20(3) = 60 kN

1.2 m 1.5 m

1.5 m

1.8 m

33 kN

27 kN

V (kN) 33 4.2 1.2

6

x (m)

2.85 –27

M (kN · m)

66.825 48.6

39.6 x (m) 1.2

2.85

4.2

6

Ans: Use W360 * 33 831

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*11–8. If the bearing pads at A and B support only vertical forces, determine the greatest magnitude of the uniform distributed loading w that can be applied to the beam. sallow = 15 MPa, tallow = 1.5 MPa.

w

A

B 1m

1m 150 mm 25 mm 150 mm 25 mm

Solution The location of c, Fig. b, is y =

0.1625(0.025)(0.15) + 0.075(0.15)(0.025) ©yA = ©A 0.025(0.15) + 0.15(0.025) = 0.11875 m

I = +

1 (0.025)(0.153) + (0.025)(0.15)(0.04375)2 12 1 (0.15)(0.0253) + 0.15(0.025)(0.04375)2 12

= 21.5820(10- 6) m4 Referring to Fig. b, Qmax = y¿A¿ = 0.059375 (0.11875)(0.025) = 0.17627(10- 3) m3 Referring to the moment diagram, Mmax = 0.28125 w. Applying the Flexure formula with C = y = 0.11875 m, sallow =

Mmax c ; I

15(106) =

0.28125w(0.11875) 21.5820(10- 6)

W = 9.693(103) N>m Referring to shear diagram, Fig. a, Vmax = 0.75 w. tallow =

Vallow Qmax ; It

1.5(106) =

0.75w C 0.17627(10 - 3) D

21.5820(10- 6)(0.025)

W = 6.122(103) N>m = 6.12 kN>m (Control!)

Ans.

Ans: W = 6.12 kN>m (Control!) 835

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11–9. Select the lightest W360 wide-flange beam from Appendix B that can safely support the loading. The beam has an allowable normal stress of sallow = 150 MPa and an allowable shear stress of tallow = 80 MPa. Assume there is a pin at A and a roller support at B.

50 kN 40 kN/m

A

4m

B

2m

Solution Shear and Moment Diagram: As shown in Fig. a. Bending Stress: Referring to the moment diagram, Fig. a, Mmax = 100 kN # m. Applying the flexure formula, Srequired =

100(103) Mmax = sallow 150(106)

= 0.6667(10 - 3) m3 = 666.67(103) mm3 Select W360 * 45(Sx = 688(103) mm3, d = 352 mm and t w = 6.86 mm) Shear Stress: Referring to the shear diagram, Fig. a, Vmax = 105 kN. We have tmax =

105(103) Vmax = tw d 6.86(10 - 3)(0.352) (OK)

= 43.48 MPa 6 tallow = 80 MPa

Ans.

Hence, use W360 * 45

Ans: Use W360 * 45 1060

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11–10. Investigate if the W250 * 58 beam can safely support the loading. The beam has an allowable normal stress  of sallow = 150 MPa and an allowable shear stress of tallow = 80 MPa. Assume there is a pin at A and a roller support at B.

50 kN 40 kN/m

A

4m

B

2m

Solution Shear and Moment Diagram: As shown in Fig. a. Bending Stress: Referring to the moment diagram, Fig. a, Mmax = 100 kN # m. For a W250 * 58 section, Sx = 693(103) mm3 = 0.693(10 - 3) m4. Applying the flexure formula, 100(103) Mmax smax = = = 144.30 MPa 6 sallow = 150 MPa (OK) Sx 0.693(10 - 3) Shear Stress: Referring to the shear diagram, Fig. a, Vmax = 105 kN. For a W250 * 58 section, d = 252 mm and t w = 8.00 mm. We have tmax =

105(103) Vmax = tw d 8.00(10 - 3)(0.252) (OK)

= 52.08 MPa 6 tallow = 80 MPa

Ans.

The W250 * 58 can safely support the loading.

Ans: Yes, it can 1061

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w

11–11. The box beam has an allowable bending stress of sallow = 10 MPa and an allowable shear stress of tallow = 775 kPa. Determine the maximum intensity w of the distributed loading that it can safely support. Also, determine the maximum safe nail spacing for each third of the length of the beam. Each nail can resist a shear force of 200 N.

30 mm 250 mm 30 mm 150 mm 30 mm

6m

Solution Section Properties: I =

1 1 (0.21) A 0.253 B (0.15) A 0.193 B = 0.1877 A 10 - 3 B m4 12 12

QA = y1 ¿A¿ = 0.11(0.03)(0.15) = 0.495 A 10 - 3 B m3 Qmax = ©y¿A¿ = 0.11(0.03)(0.15) + 0.0625(0.125)(0.06) = 0.96375 A 10 - 3 B m3 Bending Stress: From the moment diagram, Mmax = 4.50w. Assume bending controls the design. Applying the flexure formula. sallow = 10 A 106 B =

Mmax c I 4.50w (0.125) 0.1877 (10 - 3)

w = 3336.9 N>m Shear Stress: Provide a shear stress check using the shear formula. From the shear diagram, Vmax = 3.00w = 10.01 kN. tmax =

=

Vmax Qmax It 10.01(103) C 0.96375(10 - 3) D 0.1877(10 - 3)(0.06)

= 857 kPa 7 tallow = 775 kPa (No Good!) Hence, shear stress controls. tallow = 775 A 103 B =

Vmax Qmax It 3.00w C 0.96375(10 - 3) D 0.1877(10 - 3)(0.06)

w = 3018.8 N>m = 3.02 kN>m

Ans.

Shear Flow: Since there are two rows of nails, the allowable shear flow is 2(200) 400 q = . = s s Ans: w = 3.02 kN>m, s = 16.7 mm, s = 50.2 mm 849

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*11–12. Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading shown. The allowable bending stress is sallow = 150 MPa and the allowable shear stress is tallow = 84 MPa.

5 kip 25 kN 18 kN kip ft 27 m B

A 12 3.6ftm

6 ft m 1.8

Solution

# ft. 72 kip kN ·m. From the moment diagram, Fig. a, Mmax = 48 Sreq¿d =

=

Mmax sallow

25 kN

3

72(10 ) = 150(106 )

−3 0.48(10 = ) m 3 480(10 3 ) mm 3

= 480(10 26.18 in3)3 mm3

27 kN · m 3.6 m

1.8 m

The choices are

W310 × 39 W360 × 39 W410 × 39 The section that has shortest depth is preferable, so

V (kN)

Select W310 W 14  * 22 29.0 in3)3,mm d =3, 13.74 in.mm andand tw =tw0.230 39 [S d = 310 =5.84in. mm] C Sxx ==547(10 D

20

5 kN. kip. Provide the shear stress check for Vmax From the shear diagram, Fig. a, V max ==25 W 310 * 39, tmax =

20 kN

45 kN

1.8

x (m) 5.4

–25 M (kN · m)

Vmax twd

1.8

5.4 x (m)

25(10 3 ) = = 0.00584(0.3101)

6

2

–27

13.81(10 = ) N/m 13.8 MPa

12MPa ksi‚ (O.K!) = 1.58 13.8 ksi MPa6< ttallow allow ==84 Use

–72

W14 W310*22 39

Ans.

Ans: Use W310  39 852

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11–13. The simply supported beam is made of timber that has an allowable bending stress of sallow = 6.72 MPa and an allowable shear stress of tallow = 0.525 MPa. Determine the dimension b if it is to be rectangular and have a height-towidth ratio of 1.25.

5 kip/ft 75 kN/m

B

A 6 ftm 1.8

6 ft 1.8 m

1.25 b

Solution

b

1 (b)(1.25b)3 = 0.16276b4 12

I =

Sreq¿d =

I 0.16276b4 = 0.26042b3 = c 0.625b

Assume bending moment controls: Mmax = 81 60 kN kip ·# ft m Mmax Sreq¿d

sallow =

81(10 3 )

6.72(106 ) =

0.26042b3

= b 0.35904 = m 359.04 mm Check shear: tmax =

1.5[67.5(10 3 )] 1.5V = 0.6283(106 ) N/m 2 = (0.35904)[1.25(0.35904)] A

=

0.628 MPa > 0.525 MPa (NO Good!)

Shear controls: tallow =

1.5V ; A

0.525(106 ) =

1.5[67.5(10 3 )] b(1.25b)

= b 0.3928 = m 393 mm

Ans.

75 kN/m

1.8 m

67.5 kN

1.8 m

67.5 kN

67.5 kN M = 81 kN · m

1.2 m

0.6 m

67.5 kN V (kN) 67.5

1.8

3.6 x (m)

0

–67.5 M (kN · m)

0

81

x (m) 1.8

3.6

Ans: b = 393 mm 842

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11–14. The beam is used in a railroad yard for loading and unloading cars. If the maximum anticipated hoist load is 60 kip, kN, select the lightest-weight steel wide-flange wide-flange section 12 from Appendix B that will safely support the loading. The hoist travels along the bottom flange of the beam, 0.3 m,and and has has negligible negligible size. size. Assume the beam 1 ft m… ≤xx …≤ 7.5 25 ft, is pinned to the column at B and roller supported at A. stress is sallow =allow 168 =MPa and and the The allowable allowablebending bending stress is s 24 ksi allowable shear stress is tallow = 84=MPa. the allowable shear stress is tallow 12 ksi.

x

8 mft 27

A

B

60kip kN 12

4.5 15 m ft

C

Solution Maximum moment occurs when load is in the center of beam. lb ·# ft Mmax = (6 (30kip)(13.5 kN)(4 m)ft)= = 12081kN m Sreq'd = =

Mmax

σ allow 120(10 3 ) = 168(106 )

−3 = 0.7143(10 ) m 3 714.3(10 3 ) mm 3

3 3 in3, d 13.84 tw mm, = 0.270 Select a W410 W 14 *  30, 46 Sx S=x 42.0 = 774(10 )= mm , d =in,403 tw = in. 6.99 mm

57.75kip kN At x = 0.3 1 ft,m, V V==11.56 t =

V 57.75(10 3 ) = = Aweb 0.00699(0.403)

6 = 20.50(10 ) N/m 2 20.50 MPa < 84 MPa

46 Use W410 W14 *30

Ans.

60 kN 4m

30 kN

4m

30 kN

Ans: Use W410 * 46 841

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1500 lb 7.5 kN

11–15. The beam is constructed from three boards as shown. If each nail can support a shear shear force force of of 1.5 300 kN, lb, determine the maximum allowable spacing of the nails, s, s¿, s– , for regions AB, BC, and CD to the nearest 5 mm respectively. Also, if the allowable bending stress is sallow = 10 MPa and the allowable shear stress is tallow = 1 MPa, determine if it can safely support the load.

500kN lb 2.5 s¿

s

A

C

B 6 ftm 1.8

ft 1.86 m 10 in. 250 mm

4 in. 100 mm

Solution

∑ yA 0.3(0.25)(0.1) + 2[0.125(0.05)(0.25)] = ∑A 0.25(0.1) + 2(0.05)(0.25)

D

6 ftm 1.8

10 in. 250 mm

The neutral axis passes through centroid c of the beam’s cross-section. The location of c, Fig. b, is

= y

s¿¿

2 in. 50 mm

2 in. 50 mm

7.5 kN

2.5 kN

= 0.2125 m The moment of inertia of the beam’s cross-section about the neutral axis is 1.8 m

1  I =  (0.05)(0.253 ) + 0.05(0.25)(0.2125 − 0.125)2   12  +

1.8 m 12.5 kN

1 (0.25)(0.13 ) + 0.25(0.1)(0.3 − 0.2125)2 12

1.8 m 2.5 kN

V (kN)

= 0.53385(10 −3 ) m 4

5 2.5 1.8

x (m)

Referring to Fig. b,

QA =

−3

= 0.0875(0.25)(0.1) = 2.1875(10 ) m

=

3.6

5.4

M (kN · m)

3

Referring to the moment diagram, Fig. a, Mmax = 9000 13.5 kN Applyingflexure flexure . Applying lb # ·ftm. formula with c = y = 0.2125 mm, smax

5.4

–7.5

3 −3 Qmax = 2y2œ A2œ = 2[(0.10625)(0.05)(0.2125)] = 2.2578(10 ) m

y1œ A1œ

3.6

Mmax c = I

1.8

x (m) –4.5 –13.5

[13.5(10 3 )](0.2125) 0.53385(10 −3 )

= 5.374(106 ) N/m 2

(O.K!) = MPa 1.50 ksi = 671.70 psi m 843

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w

11–19. The simply supported beam is composed of two W310 33 sections sections built built up up as shown. Determine Determine if the beam W12 *22 kN/m.. The allowable will safely support a loading of w == 30 2 kip>ft and the the allowable allowable shear bending stress stress isis ssallow 22 MPa ksi and allow==150 MPa. stress is ttallow . = 100 14 ksi allow =

24 7.2ftm

Solution Section properties:

7.2(30) = 216 KN

4 4 6 For W310 33 (d(d== 12.31 313 mm ) in mm = 156 W 12  * 22 in. Ix = Ix65(10

ttww ==6.60 mm mm2in ) 2) 0.260 in. A =A4180 = 6.48 3.6 m 108 kN

2   313   I= 334.76(106 ) mm 4 2 65(106 ) + 4180   =  2    334.76(106) 802.98 I 6 = ) mm 3 1.0695(10 −3 ) m 3 = = 65 1.0695(10 S = = c 12.31 313

3.6 m 108 kN

V (kN) 108

Bending stress: –108

Mmax 194.4(10 3 ) = = 181.77(106 ) N/m 2 σ max = S 1.0695(10 −3 ) = 150 = 181.8 sallow 26.5 ksi MPa 7 s>allow = 22 ksi MPa

M (kN · m)

Ans.

No, the beam fails due to bending stress criteria. Check shear: tmax =

194.4

(Neglect area of flanges.)

Vmax 108(10 3 ) = = 26.14(106 ) N/m 2 Aw 2(0.0066)(0.313) = 26.1 MPa < τallow = 100 MPa OK

Ans: The beam fails. 843

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*11–20. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. If P = 5 kN and the shaft is made from steel having an allowable normal stress of sallow = 150 MPa and an allowable shear stress of tallow = 85 MPa, determine the required minimum wall thickness t of the shaft to the nearest millimeter to safely support the load.

B

A

1m

0.5 m t

40 mm

P

Solution Shear and Moment Diagram: As shown in Fig. a, Bending Stress: From the moment diagram, Fig. a, Mmax = 2.5 kN # m. The moment of inertia of the shaft about the bending axis is I =

p (0.044 - ri 4). Applying the 4

flexure formula, sallow =

Mmax c l

150(106) =

2.5(103)(0.04) p (0.044 - ri 4) 4

ri = 0.03617 m = 36.17 mm Thus, t = r0 - ri = 40 - 36.17 = 3.83 mm Use t = 4 mm

Ans.

Shear Stress: Using this result, ri = 0.04 - 0.004 = 0.036 m. Then Qmax = 4(0.04) p 4(0.036) p c (0.042) d c (0.0362) d = 11.5627(10 - 6) m3, Fig. b, and I = 3p 2 3p 2 p 4 4 (0.04 - 0.036 ) = 0.69145(10 - 6) m4. Referring to the shear diagram, Fig. a, 4 Vmax = 5 kN. tmax =

VmaxQmax 5(103)[11.5627(10 - 6)] = It 0.69145(10 - 6)(0.008)

= 10.45 MPa 6 tallow = 85 MPa (OK)

Ans: Use t = 4 mm 1094

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11–21. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. If the shaft is made from steel having an allowable normal stress of sallow = 150 MPa and allowable shear stress of tallow = 85 MPa , determine the maximum allowable force P that can be applied to the shaft. The thickness of the shaft’s wall is t = 5 mm.

B

A

1m

0.5 m t

40 mm

P

Solution Shear and Moment Diagram: As shown in Fig. a, Bending Stress: The moment of inertia of the shaft about the bending axis is p I = (0.044 - 0.0354) = 0.8320(10 - 6) m4. Referring to the moment diagram, 4 Fig. a, Mmax = 0.5P. Applying the flexure formula, sallow =

Mmax c l

150(106) =

0.5P(0.04) 0.8320(10 - 6)

P = 6240.23N = 6.24 kN

Ans.

Shear Stress: Qmax =

4(0.04) p 4(0.035) p c (0.042) d c (0.0352) d 3p 2 3p 2

= 14.0833(10 - 6) m3 Referring to the shear diagram, Fig. a, Vmax = P = 6240.23 N. Applying the shear formula, tmax =

Vmax Qmax 6240.23[14.0833(10 - 6)] = It 0.8320(10 - 6)(0.01)

= 10.56 MPa 6 tallow = 85 MPa (OK!)

Ans: P = 6.24 kN 1095

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60 kN/m 4 kip/ft

11–22. Determine the minimum depth h of the beam to of 5safely mm that will the safely support the the nearest 18multiples support loading shown. in. that will loading shown. The allowable sallow The allowable bending stressbending is sallowstress and= 147 the = 21isksi MPa and the allowable tallow = 70 MPa. The allowable shear stress isshear tallowstress The beam has a = 10isksi. beam hasthickness a uniform of 75 mm. uniform ofthickness 3 in.

h

A

B 3.6 m 12 ft

1.8 6 ftm

Solution The section modulus of the rectangular cross-section is = S

108 kN

1 (0.075)(h 3 ) 12

I = C

h /2

= 0.0125h2

From the moment diagram, Mmax = 72 kip 97.2 kN# ft· .m. Sreq¿d = 0.0125h2 =

3.6 m 27 kN

Mmax sallow

0.9 m 0.9 m 135 kN

V (kN)

3

97.2(10 )

108

6

147(10 )

h = 0.229995 m = 230 mm

3.6

–27

Use

Use h  230 mm h = 9 18 in

Ans. M (kN · m)

kN. Referring to Fig. b, 24 kip From the shear diagram, Fig. a, Vmax = 108 . Referring b,  0.23   0.23  −3 3 Qmax = y¿A¿ =   (0.075)   = 0.4959375(10 ) m and  4   2  1 = (0.075)(0.233 ) 76.04375(10 −6 ) m 4 . Provide the shear stress check by providing 12 shear formula, Vmax Qmax tmax = It I

=

x (m) 5.4

5.4

3.6

x (m)

–97.2

0.075 m

[108(10 3 )][0.4959375(10 −3 )] [76.04375(10 −6 )](0.075)

6 = ) N/m 2 9.39 MPa < τallow = 70 MPa (O.K!) = 9.391(10

Ans: Use h = 230 mm 848

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11–23. The spreader beam AB is used to slowly lift the 13.5 kN pipe that is centrally located on the straps at C and D. If the beam is a W310 * 67, determine if it can safely support the load. The allowable bending stress is s  allow = 154 MPa and the allowable shear stress is tallow = 84 MPa.

13.5 kN

1.5 m A 0.3 m

B 1.8 m

0.9 m C

Solution = Fh

1.8 m

0.9 m D

6750 = 12150 N tan 29.055°

[S 948(10 3 ) mm 3= tw For W310 × 67 = , d 306 mm, = = σ max tmax =

8.51 mm]

Mmax 7897.5 6 = ; σ max = 8.331(10 = ) N/m 2 8.33 MPa < 154 MPa OK S 0.948(10 −3 ) Vmax ; Aweb

t =

6750 = 2.592(106 ) N/m 2 = 2.59 MPa < 84 MPa OK (0.00851)(0.306)

Yes.

Ans.

Ans: Yes. 1080

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*11–24. Determine the maximum uniform loading w the W310 W12 *14 21beam beam will will support support ifif the the allowable bending stress is is ssallow 22 MPa ksi and andthe theallowable allowable shear shear stress stress is allow ==150 ttallow == 84 12MPa. ksi.

w w 3 mft 10 310mft

Solution From the moment diagram, Fig. a, Mmax = 2.53125 28.125 ww. . For , SS in33,) mm3, ForW12 W310*1421, 244(10 x x== 14.9 d 303 = mm and tw 5.08 mm. sallow = 150(106 ) =

Mmax S 2.53125w 0.244(10 −3 )

3 = w 14.46(10 = ) N/m 14.5 kN/m

Ans.

From the shear diagram, Fig. a, Vmax = 2.25(14.46) = 32.53 kN. Provide a shear stress check on W12 , W310*14 21, tmax =

Vmax tw d

32.53(10 3 ) 6 2 = = 21.14(10 ) N/m 0.00508(0.303) = 3.06 12MPa ksi (O.K) 21.1 ksi MPa6< ttallow ==84 (O.K) W(3)

4.5 m

1.5 m

0.75W

2.25W V (kN)

2.25W 6

3 2.25

x (m) –0.75W

M (kN · m) 2.53125W 2.25W

x (m) 2.25 3

6

Ans: = w 14.5 kN/m 846

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11–25. Determine if the W360  33 beam will safely The allowable allowable bending kip>ft.The support a loading of of ww == 1.5 25 kN/m. and the allowable shearstress stress = 22 ksi stress isisss = 150 MPa the allowable shear is allow allow tallow = MPa. 12 ksi. tisallow = 84

w 3m 10 ft 3 mft 10

Solution 3 W14 *22 in3,3)d mm = 13.74 andmm 0.23 Frommm. theFrom moment tw =and For W360 33, SxSx==29.0 475(10 d in = 349 twin=. 5.84 the diagram, diagram, Fig. a, Mmax kip # ft. kN · m. moment Fig. =a, 42.1875 Mmax = 63.28125

smax = =

Mmax S 63.28125(10 3 ) = 133.22(106 ) N/m 2 0.475(10 −3 )

= 133 MPa < σallow = 150 MPa (O.K!) From the shear diagram, Fig. a, Vmax = 11.25 56.25 kip kN.. tmax = =

Vmax tw d

56.25(10 3 ) = 0.00584(0.349)

27.60(106 ) N/m 2

27.6 ksi MPa6< tallow (O.K!) = 3.56 12MPa ksi (O.K!) allow ==84 Based on the investigated results, we conclude that W360 W14 *22 33can cansafely safelysupport support the loading. 25(3) = 75 kN

4.5 m

1.5 m

18.75 kN

56.25 kN V (kN)

56.25 6

3 2.25

x (m) –18.75

M (kN · m) 63.28125

56.25

x (m) 2.25 3

6

Ans: Yes 847

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11–26. The simply supported beam supports a load of P = 16 kN. Determine the smallest dimension a of each timber if the allowable bending stress for the wood is sallow = 30 MPa and the allowable shear stress is tallow = 800 kPa . Also, if each bolt can sustain a shear of 2.5 kN, determine the spacing s of the bolts at the calculated dimension a.

P

3m

3m s

a a a

Solution Section Properties: I =

1 (a)(2 a)3 = 0.66667 a4 12

Qmax = y¿A¿ =

a (a)(a) = 0.5 a3 2

Assume bending controls. sallow =

Mmaxc ; I

30(106) =

24(103)a 0.66667 a4

a = 0.106266 m = 106 mm

Ans.

Check shear: tmax =

VQ 8(103)(0.106266>2)(0.106266)2 = 531 kPa 6 tallow = 800 kPa = It 0.66667(0.1062664)(0.106266)

OK

Bolt spacing: q =

8(103)(0.106266>2)(0.1062662) VQ = 56462.16 N>m = I 0.66667(0.1062664)

s =

2.5(103) = 0.04427 m = 44.3 mm 56462.16

Ans.

Ans: a = 106 mm , s = 44.3 mm 1104

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11–27. Select the lightest W360 shape section from Appendix B that can safely support the loading acting on the overhanging beam. The beam is made from steel having an allowable normal stress of sallow = 150 MPa and an allowable shear stress of tallow = 80 MPa.

50 kN 40 kN/m

4m

2m

Solution Shear and Moment Diagram: As shown in Fig. a. Bending Stress: Referring to the moment diagram, Fig. a, Mmax = 100 kN # m. Applying the flexure formula, Srequired =

100(103) Mmax = sallow 150(106)

= 0.6667(10 - 3) m3 = 666.67(103) mm3 Select W360 * 45(Sx = 688(103) mm3, d = 352 mm and t w = 6.86 mm) Shear Stress: Referring to the shear diagram, Fig. a, Vmax = 105 kN. We have tmax =

105(103) Vmax = tw d 6.86(10 - 3)(0.352)

= 43.48 MPa 6 tallow = 80 MPa (OK) Hence, use W360 * 45

Ans.

Ans: Use W360 * 45 1083

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*11–28. Investigate if a W250 * 58 shape section can safely support the loading acting on the overhanging beam. The beam is made from steel having an allowable normal stress of sallow = 150 MPa and an allowable shear stress of tallow = 80 MPa.

50 kN 40 kN/m

4m

2m

Solution Shear and Moment Diagram: As shown in Fig. a, Bending Stress: Referring to the moment diagram, Fig. a, Mmax = 100 kN # m. For a W250 * 58 section, Sx = 693(103) mm3 = 0.693(10 - 3) m4. Applying the flexure formula, smax =

100(103) Mmax = 144.30 MPa 6 sallow = 150 MPa (OK) = Sx 0.693(10 - 3)

Shear Stress: Referring to the shear diagram, Fig. a, Vmax = 105 kN. For a W250 * 58 section, d = 252 mm and tw = 8.00 mm. We have tmax =

105(103) Vmax = tw d 8.00(10 - 3)(0.252)

= 52.08 MPa 6 tallow = 80 MPa (OK) The W250 * 58 can safely support the loading.

Ans.

Ans: Yes. 1084

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11–29.  The beam is to be used to support the machine, which exerts the forces of 27 kN and 36 kN as shown. If the maximum bending stress is not to exceed sallow = 154 MPa, determine the required width b of the flanges.

b 12.5 mm

12.5 mm 27 kN 175 mm

36 kN

12.5 mm

1.8 m

2.4 m

1.8 m

Solution Section Properties:

I =

1 1 (b)(0.2 3 ) − (b − 0.0125)(0.1753 ) 12 12

= 0.220052(10 −3 )b + 5.58268(10 −6 ) S=

I 0.220052(10 −3 )b + 5.58268(10 −6 ) = c 0.1

6

4.2

= 2.20052(10 −3 )b + 55.8268(10 −6 ) Sreq’d =

Mmax sallow −3

53.46

−6

2.20052(10 )b + 55.8268(10 ) =

59.94(10 3 )

1.8

x (m)

59.94

4.2

6

x (m)

154(106 )

b 0.15151 = = m 152 mm

Ans.

Ans: b = 151.5 mm. 1103

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11–30. The steel beam has an allowable bending stress sallow = 140 MPa and an allowable shear stress of tallow = 90 MPa. Determine the maximum load that can safely be supported. 2m

20 mm 150 mm

P

Solution

50 mm

Section Properties: y =

2m

(10)(120)(20) + (95)(150)(20) 120(20) + 150(20)

50 mm 20 mm

= 57.22 mm

Qmax = y′A′ = (0.05638)(0.02)(0.170 - 0.05722) = 0.127168 ( 10-3 ) m3 I =

1 (0.12) ( 0.023 ) + 0.12(0.02)(0.05722 - 0.01)2 + 12 1 (0.02) ( 0.153 ) + 0.15(0.02)(0.095 - 0.05722)2 = 15.3383 ( 10-6 ) m4 12

S =

15.3383 ( 10-6 ) I = = 0.136005 ( 10-3 ) m3 c (0.170 - 0.05722)

For Moment: M = sallow S 2P = 140 ( 106 ) (0.136005) ( 10-3 ) Ans.

P = 9520 N = 9.52 kN  (Controls) For Shear: It V ≟ tallow a b Qmax P = 90 ( 106 ) a

15.3383 ( 10-6 ) (0.02) 0.127168 ( 10-3 )

b = 217106 = 217 kN

Ans: P = 9.52 kN 1081

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11–31. Determine the variation in the width w as a function of x for the cantilevered beam that supports a concentrated force P at its end so that it has a maximum bending stress sallow throughout its length. The beam has a constant thickness t.

w ––0 2

x w ––0 2

w t L

Solution

P

Section Properties: I =

1 (w)(t 3) 12

sallow = At x = L, sallow =

S =

I = c

1 3 12 (w)(t )

t>2

=

wt 2 6

M Px =  S w t 2 >6

(1)

PL  w0 t 2 >6

(2)

Equate Eqs. (1) and (2), Px PL = w t 2 >6 w0 t 2 >6

w =

w0 x L

Ans.

Ans: w = 1082

w0 x L

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*11–32. w

The tapered beam supports a uniform distributed load w. If it is made from a plate and has a constant width b, determine the absolute maximum bending stress in the beam.

h0

h0

2 h0 L 2

L 2

Solution Section Properties: h - h0 h0 = L; x 2 I = S =

2 h = h0 a x + 1b L

3 1 3 2 bh0 a x + 1b 12 L

I = c

(

)

1 3 2 3 12 bh0 L x + 1 h0 2 2 Lx + 1

(

)

=

2 1 2 2 bh0 a x + 1b 6 L

Bending Stress: s =

M = S

w 2 2 (Lx - x ) 1 2 2 2 6 bh0 L x + 1

(

)

=

3w Lx - x2 £ § bh20 ( L2 x + 1 ) 2

(1)

2 2 2 2 2 ds 3w ( Lx + 1 ) (L - 2x) - (Lx - x )(2) ( Lx + 1 )( L ) = £ § = 0 dx bh20 ( L2 x + 1 ) 4

2 4 a x + 1b(L - 2x) - ( Lx - x2 ) = 0 ; L L

x =

L 4

Hence, from Eq. (1),

3w L ( 4 ) - ( 4 ) wL2 £ 2 L § = 2 2 bh0 ( L ( 4 ) + 1 ) 4bh20 L

smax =

L 2

Ans.

Ans: smax =

1083

WL2 4bh20

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11–33. The tapered beam supports the concentrated force P at its center. Determine the absolute maximum bending stress in the beam. The reactions at the supports are vertical.

P

L

2r0

r0

L

Solution Moment Function: As shown on FBD(b). Section Properties: r - r0 r0 r0 =    r = (x + L) x L L I =

4 pr 40 p r0 c (x + L) d = (x + L)4 4 L 4L4

I S = = c

pr 40 (x 4L4 r0 L (x

+ L)4

+ L)

=

pr 30 4L3

(x + L)3

Bending Stress: Applying the flexure formula, s =

M = S

p 2x pr 30 (x 4L3

+ L)3

=

2PL3x + L)3

(1)

pr 30(x

In order to have the absolute maximum bending stress,

ds = 0. dx

3 2 2PL3 (x + L) (1) - x(3)(x + L) (1) ds = c d = 0 3 6 dx pr 0 (x + L)

x = Substituting x =

L 2

L into Eq. (1) yields 2 smax =

8PL 27pr 30

Ans.

Ans: smax = 1084

8PL 27pr 30

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11–34. w0

The beam is made from a plate that has a constant thickness b. If it is simply supported and carries the distributed loading shown, determine the variation of its depth h as a function of x so that it maintains a constant maximum bending stress sallow throughout its length.

A

h

h0

C

B

x L –– 2

L –– 2

Solution Support Reactions: As shown on the free-body diagram of the entire beam, Fig. a. Moment Function: The distributed load as a function of x is w0 w = x L>2

w =

2w0 x L

The free-body diagram of the beam’s left cut segment is shown in Fig. b. Considering the moment equilibrium of this free-body diagram, a+ ΣMO = 0;

M + M =

x 1 2w0 1 J x R x¢ ≤ - w0Lx = 0 2 L 3 4 w0 1 3L2x - 4x3 2 12L

Section Properties: At position x, the height of the beam’s cross section is h. Thus 1 bh3 12

I = Then

1 bh3 I 12 1 S = = = bh2 c h>2 6 Bending Stress: The maximum bending stress smax as a function of x can be obtained by applying the flexure formula.

smax

At x =

w0 1 3L2x - 4x3 2 w0 M 12L = = = 1 3L2x - 4x3 2 ‚ S 1 2 2bh2L bh 6

(1)

L , h = h0. From Eq. (1), 2 w0L2

smax =

(2)

2bh0 2

Equating Eqs. (1) and (2), w0 2

2bh L h =

1 3L2x

h0

L3>2

- 4x3 2 =

1 3L2x

w0L2 2bh0 2

- 4x3 2 1>2

Ans. Ans: h = 1085

h0 L3>2

(3L2x - 4x3)1>2

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11–35. Determine the variation in the depth d of a cantilevered beam that supports a concentrated force P at its end so that it has a constant maximum bending stress sallow throughout its length. The beam has a constant width b0.

P x h

d L

b0

Solution Section Properties: I =

1 b d 3; 12 0

S =

I = c

1 3 12 b 0 d d 2

=

b0 d 2 6

Maximum Bending Stress: sallow =

M Px 6Px = = 2 d S b0 d 2 b0 6

At x = L, sallow =

(1)

d = h

6PL b0 h2

(2)

Equating Eqs. (1) and (2), 6Px 6PL = 2 b0 d b0 h2 d = h

x AL

Ans.

Ans: d = h 1086

x AL

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*11–36. Determine the variation of the radius r of the cantilevered beam that supports the uniform distributed load so that it has a constant maximum bending stress smax throughout its length.

w r0 r

L

x

Solution Moment Function: As shown on FBD. Section Properties: I =

p 4 r 4

I = c

S =

p 4

r4 r

=

p 3 r 4

Bending Stress: Applying the flexure formula, smax

M = = S smax =

wx2 2 p 3 4r

2wx2 pr 3

(1)

At x = L, r = r0. From Eq. (1), smax =

2wL2 pr 30

(2)

Equating Eq. (1) and (2) yields r3 =

r 30 L2

x 2

Ans.

Ans: r3 = 1087

r 30 L2

x2

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11–37. The tapered beam supports a uniform distributed load w. If it is made from a plate that has a constant width b0, determine the absolute maximum bending stress in the beam.

w

y h x L

Solution Section Properties: y x = ; h L

h

h x L

y =

b0

b0 h3 3 1 h 3 I = (b0)a xb = x 12 L 12L3 I S = = c

b 0 h3 3 x 12L3 = h 2L x

b0 h2 6L2

x2

Bending Stress: smax =

M = S

w 2

x2 2

b0 h 6L2

x2

=

3wL2 b0 h2

Ans.

The bending stress is independent of x. Therefore, the stress is constant throughout the span.

Ans: smax = 1088

3wL2 b0h2

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11–38. Determine the variation in the width b as a function of x for the cantilevered beam that supports a uniform distributed load along its centerline so that it has the same maximum bending stress sallow throughout its length. The beam has a constant depth t.

b —0 2 b —0 2 b — 2 w L

Solution

x t

Section Properties: I =

1 b t3 12

S =

I = c

1 12

b t3 t 2

=

t2 b 6

Bending Stress: sallow

M = = S

wx2 2 2

t 6b

=

3wx2 t 2b

(1)

At x = L, b = b0 sallow =

3wL2 t 2b0

(2)

Equating Eqs. (1) and (2) yields: 3wx2 3wL2 = 2 2 t b t b0 b =

b0 L2

x 2

Ans.

Ans: b = 1089

b0 L2

x2

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11–39. z

The tubular shaft has an inner diameter of 15 mm. Determine to the nearest millimeter its minimum outer diameter if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft. Use an allowable shear stress of tallow = 70 MPa, and base the design on the maximum shear stress theory of failure.

100 mm B 500 N

150 mm

A 200 mm

I =

500 N

150 mm

x

Solution

y

100 mm

p 4 p (c - 0.00754) and J = (c 4 - 0.00754) 4 2

tallow =

Aa

tallow =

Aa

t2allow = t2allowa

sx - sy 2

2

2 b + txy

Mc 2 + Tc 2 a b b J 2I

M2c 2 T 2c 2 + 4I 2 J2

4T 2 c 4 - 0.00754 2 4M2 b = + 2 c p p2

c 4 - 0.00754 2 = 2M2 + T 2 p tallow c

c 4 - 0.00754 2 = 2752 + 502 c p(70)(106) c 4 - 0.00754 = 0.8198(10 - 6)c Solving, c = 0.0103976 m d = 2c = 0.0207952 m = 20.8 mm Ans.

Use d = 21 mm­­­

Ans: Use d = 21 mm 1090

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*11–40. z

Determine to the nearest millimeter the minimum diameter of the solid shaft if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft. Base the design on the maximum distortion energy theory of failure with sallow = 150 MPa.

100 mm B 500 N

150 mm

A 200 mm

s1, 2 =

500 N

150 mm

x

Solution

y

100 mm

sx s2x { + t2xy 2 A4

Let a =

sx s2x ,b = + t2xy 2 A4

s1 = a + b, s2 = a - b Require s21 - s1s2 + s22 = s2allow

a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = sallow a2 + 3b2 = s2allow sx2 sx2 + 3a + t2xy b = s2allow 4 4

sx2 + 3t2xy = s2allow

Mc 2 Tc 2 a p 4 b + 3 a p 4 b = s2allow 2c 4c 1

c

6

ca

c6 =

4M 2 2T 2 b + 3 a b d = s2allow p p 16

s2allowp2

c = a = c

12T 2 s2allowp2

M2 +

4 s2allowp2 4

( 4M + 3T ) b 2

(150(106))2(p)

d = 2c = 0.0181 m

2

1 6

1 6

(4(75)2 + 3(50)2) d ∞ 0.009025 m 2 Ans.

Use d = 19 mm.

Ans: Use d = 19 mm 1091

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11–41. The 50-mm-diameter shaft is supported by journal bearings at A and B. If the pulleys C and D are subjected to the loadings shown, determine the absolute maximum bending stress in the shaft.

A

C D

400 mm 400 mm

300 N 300 N

150 N 150 N B

400 mm

Solution Internal Moment Components: The shaft is subjected to two bending moment components Mz and My. Bending Stress: Since all the axes through the centroid of the circular cross section of the shaft are principal axes, then the resultant moment M = 2My 2 + Mz 2 can be used to determine the maximum bending stress. The maximum bending moment occurs at C (x = 0.4 m). Then, Mmax = 2402 + 1602 = 164.92 N # m. smax = =

Mmax c I

164.92(0.025) p (0.0254) 4 Ans.

= 13.439 MPa = 13.4 MPa

Ans: smax = 13.4 MPa 1092

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11–42. The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use the maximum-shear-stress theory of failure with tallow = 60 MPa.

z

100 mm T 250 mm C 50 mm 75 mm

150 mm

B y

A

x

Solution

100 mm Fz  1.5 kN

From the free - body diagrams: T = 100 N # m

Ans.

Critical section is at support A. c = c

1

1

3 3 2 2 22252 + 1502 d 2M2 + T2 d = c 6 p tallow p(60)(10 )

= 0.01421 m d = 2c = 0.0284 m = 28.4 mm Use d = 29 mm

Ans.

Ans:

T = 100 N # m, Use d = 29 mm 868

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11–43. The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exert only y and z components of force on the shaft, determine the equilibrium torque T at gear C and then determine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Use the maximum-distortionenergy theory of failure with sallow = 80 MPa.

z

100 mm T 250 mm C 50 mm 75 mm

150 mm

B y

A

Solution

x

100 mm Fz  1.5 kN

From the free-body diagrams: T = 100 N # m

Ans.

Critical section is at support A. s1, 2 =

sx s2x 2 ; 2 A 4 + txy sx s2x 2 ,b = A 4 + txy 2

Let a =

s1 = a + b, s2 = a - b Require, s21 - s1 s2 + s22 = s2allowa2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4 s2x + 3t2xy = s2allow Mt 2 Tc 2 a p 4 b + 3 a p 4 b = s2allow 4 c 2 c 1 4M 2 2T 2 ca b + 3 a b d = s2allow 4 p p c c4 =

12T2 2 M + s2allow p2 s2allow p2 16

c = a = c

4 s2allow p2

(4M2 + 3T2) b

4 (80(106))2(p)2

1 2

(4(225)2 + 3(150)2) d

1 2

= 0.01605 m d = 2c = 0.0321 m = 32.1 mm

Ans:

Use d = 33 mm

Ans.

869

T = 100 N # m, Use d = 33 mm

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*11–44.  The two pulleys attached to the shaft are loaded as shown. If the bearings at A and B exert only vertical forces on the shaft, determine the required diameter of the shaft to the nearest mm using the maximum-distortion energy theory. sallow = 469 MPa.

A

D

C 150 mm 600 mm

150 mm 900 mm

300 mm

540 N 1350 N

B

540 N

1350 N

Solution Section just to the left of point C is the most critical. Both states of stress will yield the same result. sa, b =

Let

s s 2 2 ; a b + t 2 A 2

s s 2 = A and a b + t2 = B 2 A 2

sa2 = (A + B)2

sb2 = (A - B)2

sa sb = (A + B)(A - B) sa2 - sa sb + sb2 = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 s2 s2 + 3a + t2 b 4 4

=

= s2 + 3t2 sa2 - sa sb + sb2 = s2allow s2 + 3t2 = s2allow

(1)

s =

Mc Mc 4M = p 4 = I p c3 4c

t =

Tc Tc 2T = p 4 = J c p c3 2

From Eq (1) 12T2

16M2 p2c6 c = a

+

p2c6

= s2allow 1/6

 16(9452 ) + 12(121.52 )  16M2 + 12T2 1>6 b =   2 2 p sallow π 2 [469(106 )]2  

c = 0.013718 m = = 0.027435 m = 27.44 mm d 2= c 2(0.013718) Use d = 28 mm

Ans. Ans:

Use d = 28 mm 1119

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11–45. The bearings at A and D exert only y and z components of force on the shaft. If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the maximum-shearstress theory of failure.

z

350 mm D 400 mm 200 mm

Solution

B A

Critical moment is at point B: M = 2(473.7)2 + (147.4)2 = 496.1 N # m

y

C 75 mm

Fz  2 kN

Fy  3 kN 50 mm

x

T = 150 N # m c = a

1>3 1>3 2 2 2496.12 + 1502 b 2M2 + T2 b = a = 0.0176 m 6 p tallow p(60)(10 )

c = 0.0176 m = 17.6 mm d = 2c = 35.3 mm Use d = 36 mm

Ans.

Ans: Use d = 36 mm 872

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11–46. The bearings at A and D exert only y and z components of force on the shaft. If tallow = 60 MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use the maximumdistortion-energy theory of failure. sallow = 130 MPa.

z

350 mm D 400 mm 200 mm

Solution

B A

The critical moment is at B. M = 2(473.7)2 + (147.4)2 = 496.1 N # m

y

C 75 mm

Fz  2 kN

Fy  3 kN 50 mm

x

T = 150 N # m Since, sa, b =

Let

s s 2 2 ; 2 Aa 2 b + t

s = A 2

and

s 2 2 = B a A 2b + t

s2a = (A + B)2

s2b = (A - B)2

sa sb = (A + B)(A - B) s2a - sa sb + s2b = A2 + B2 + 2AB - A2 + B2 + A2 + B2 - 2AB = A2 + 3B2 =

s2 s2 + 3a + t2 b 4 4

= s2 + 3t2 s2a - sasb + s2b = s2allow s2 + 3t2 = s2allow s =

Mc Mc 4M = p 4 = I pc3 4 c

t =

Tc Tc 2T = p 4 = 3 J c pc 2

(1)

From Eq (1) 12T2 16M2 + = s2allow p2c4 p2c4 c = a = a

16M2 + 12T2 1>6 b p2s2allow 16(496.1)2 + 12(150)2 p2((130)(104))2

b

1>4

= 0.01712 m

d = 2c = 34.3 mm

Ans. Ans: Use d = 34.3 mm 873

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R11–1. The cantilevered beam has a circular cross section. If it supports a force P at its end, determine its radius y as a function of x so that it is subjected to a constant maximum bending stress sallow throughout its length.

y y

x

Solution P

Section Properties: I =

p 4 y 4

S =

p 4 I p 4y = = y3 c y 4

sallow =

M Px = p 3 S 4y 1

y = c

3 4P xd p sallow

Ans.

Ans: y = c 1098

1>3 4P xd p sallow

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R11–2. Draw the shear and moment diagrams for the shaft, and then determine its required diameter to the nearest millimeter if sallow = 140 MPa and tallow = 80 MPa. The bearings at A and B exert only vertical reactions on the shaft.

1500 N 800 N A

B

125 mm

600 mm 75 mm

Solution Bending Stress: From the moment diagram, Mmax = 111 N # m. Assume bending controls the design. Applying the flexure formula. sallow = 140 A 106 B =

Mmax c I 111 A d2 B p 4

A d2 B 4

d = 0.02008 m = 20.1 mm d = 21 mm

Use

Ans.

Shear Stress: Provide a shear stress check using the shear formula with I =

p A 0.01054 B = 9.5466 A 10 - 9 B m4 4

Qmax =

4(0.0105) 1 c (p)(0.0105)2 d = 0.77175 A 10 - 6 B m3 3p 2

From the shear diagram, Vmax = 1484 N. tmax =

=

Vmax Qmax It

1484 C 0.77175(10 - 6) D 9.5466(10 - 9)(0.021)

= 5.71 MPa 6 tallow = 80 MPa (O.K!)

Ans: Use d = 21 mm 874

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R11–3. z

The journal bearings at A and B exert only x and z components of force on the shaft. Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings without exceeding an allowable shear stress of tallow = 80 MPa. Use the maximum shear stress theory of failure.

Fx  5 kN A 75 mm x

50 mm

150 mm 350 mm

Fz  7.5 kN 250 mm

Solution

Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m 1

B y

1

3 3 2 2 c = c 2M2 + T 2 d = c 21274.752 + 3752 d = 0.0219 m p tallow p(80)(106)

d = 2c = 0.0439 m = 43.9 mm

Ans.

Use d = 44 mm.

Ans: Use d = 44 mm 1100

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*R11–4. z

The journal bearings at A and B exert only x and z components of force on the shaft. Determine the shaft’s diameter to the nearest millimeter so that it can resist the loadings. Use the maximum distortion energy theory of failure with sallow = 200 MPa.

Fx  5 kN A 75 mm x

50 mm

150 mm 350 mm

Fz  7.5 kN 250 mm

Solution

Maximum resultant moment M = 212502 + 2502 = 1274.75 N # m s1, 2 =

B y

sx s2x 2 { 2 A 4 + txy

sx s2x 2 ,b = 2 A 4 + txy

Let a =

s1 = a + b,

s2 = a - b

Require, s21 - s1 s2 + s22 = s2allow a2 + 2ab + b2 - [a2 - b2] + a2 - 2ab + b2 = s2allow a2 + 3b2 = s2allow s2x s2x + 3a + t2xy b = s2allow 4 4

s2x + 3t2xy = s2allow

Mc 2 Tc 2 a p 4 b + 3 a p 4 b = s2allow 2 c 4 c 1

c6

Ja

c6 =

4M 2 2T 2 b + 3 a b R = s2allow p p 16

s2allow p

c = c = c

M2 + 2

4 s2allow p

12T 2 s2allow p2

(4M2 + 3 T 2) d 2

4

(200(106))2(p)

1 6

(4(1274.75)2 + 3(375)2) d 2

1 6

= 0.0203 m = 20.3 mm

d = 40.6 mm Use Ans.

d = 41 mm

Ans: Use d = 41 mm 1101

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R11–5. Draw the shear and moment diagrams for the beam. Then select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loading. Take sallow = 150 MPa and tallow = 84 MPa.

3 kip/ft 50 kN/m 1.5 kip 2.25 kN  ftm A B 12 ft 3.6 m

6 ftm 1.8

Solution

# ftAssume Bending Stress: From the moment diagram, Mmax = 18.0 kip . Assumebending bending 27 kN · m. controls the design. Applying the flexure formula. Sreq¿d =

27(10 3 ) −3 = 0.18(10 = ) m 3 180(10 3 ) mm 3 150(106 )

= Select

Mmax sallow

3 3 3 W310 21 (SxA= ) in mm 303in., mmtwand tw =5.08 W10 *12 Sx244(10 = 10.9 , d, d= =9.87 = 0.19 in. B mm)

V 21wide wide-for the W310 W10 *12 twd kN = 45 9.00 kip

Shear Stress: Provide a shear stress check using t = flange section. From the shear diagram, Vmax tmax = =

Vmax tw d 45(10 3 ) = 29.24(106 ) N/m 2 0.00508(0.303)

29.2 ksi MPa62) 1 L2 2 TL PL2 = = GJ GJ 4GJ

L PL3 (vA)2 = u a b = 2 8GJ

vA = vD + (vA)1 + (vA)2 =

PL3 PL3 PL3 + + 24EI 24EI 8GJ

= PL3 a

1 1 + b T 12EI 8GJ

Ans.

Ans: vA = PL3 a 1218

1 1 + bT 12EI 8GJ

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*12–96. The framework consists of two A-36 steel cantilevered beams CD and BA and a simply supported beam CB. If each beam is made of steel and has a moment 4 6 principal axisaxis of Ixof= Ix118 , determine of inertia inertiaabout aboutitsits principal = in 46(10 ) mm4, the deflection the center beamG CB. determine the at defl ection at G theofcenter of beam CB.

A A

75 kip kN 15 B B

D D C C 4.8 m 16 ft

G G 82.4 ft m

82.4 ft m

Solution 75 kN

PL3 37.5(4.8 3 ) 1382.4 kN ⋅ m 3 T ∆C = = = 3EI EI 3EI ′G ∆=

PL3 75(4.8 3 ) 172.8 kN ⋅ m 3 = = 48 EI 48 EI EI

T

37.5 kN

∆G = ∆C + ∆′G = =

1382.4 172.8 1555.2 kN ⋅ m 3 + = EI EI EI 1555.2(10 3 ) N ⋅ m 3 = 0.1690 = m 169 mm T [200(109 ) N/m 2 ][46(10 −6 ) m 4 ]

Ans.

Ans: ∆G = 169 mm T 988

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12–97. The relay switch consists of a thin metal strip or armature AB that is made of red brass C83400 and is attracted to the solenoid S by a magnetic field. Determine the smallest force F required to attract the armature at C in order that contact is made at the free end B. Also, what should the distance a be for this to occur? The armature is fixed at A and has a moment of inertia of I = 0.18(10-12) m4.

B 2 mm 50 mm a

C S

Solution Elastic Curve: As shown.

50 mm A

Method of Superposition: Using the table in Appendix C, the required slopes and displacements are uC = ∆C =

2 F(0.052) PLAC 0.00125F m2 = = 2EI 2EI EI 3 41.667 ( 10-6 ) F m3 F(0.053) PLAC = = 3EI 3EI EI

(1)

∆ B = ∆ C + uCLCB = =

41.667 ( 10-6 ) F EI

+

0.00125 ( 10-6 ) F EI

(0.05)

104.167 ( 10-6 ) F m3

(2)

EI

Required the displacement ∆ B = 0.002 m. From Eq. (2), 0.002 =

104.167 ( 10-6 ) F 101 ( 109 ) (0.18) ( 10-12 ) Ans.

F = 0.349056 N = 0.349 N From Eq. (1), a = ∆C =

41.667 ( 10-6 ) (0.349056) 101 ( 109 ) (0.18) ( 10-12 )

= 0.800 ( 10-3 ) m = 0.800 mm

Ans.

Ans: F = 0.349 N, a = 0.800 mm 1220

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12–98. P

Determine the moment M0 in terms of the load P and dimension a so that the deflection at the center of the shaft is zero. EI is constant.

M0

M0 a

a 2

a 2

a

Solution Elastic Curve: The elastic curves for the concentrated load and couple moment are drawn separately as shown. Method of Superposition: Using the table in Appendix C, the required slope and displacement are (∆ C)1 = (∆ C)2 = (∆ C)3 =

=

=

Pa3 T 48EI M0 x ( x2 - 3Lx + 2L2 ) 6EIL M0 (a2)

a 2 a c a b - 3(a)a b + 2a2 d 6EIa 2 2

M0 a2 c 16EI

Require the displacement at C to equal zero. ( + c )  ∆ C = 0 = (∆ C)1 + (∆ C)2 + (∆ C)3 0 = M0 =

M0 a2 M0 a2 Pa3 + + 48EI 16EI 16EI

Pa 6

Ans.

Ans: M0 = 1221

Pa 6

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12–99. Determine the reactions at the supports A and B, then draw the shear and moment diagram. EI is constant. Neglect the effect of axial load.

P B

A

1– 3L

Solution L MA + By L - P a b - MB = 0 3

a+ ΣMA = 0;

(1) (2)

Ay + By - P = 0

+ c ΣFy = 0;

Moment Functions: M1(x1) = Byx1 - MB M2(x2) = Ay x2 - MA Slope and Elastic Curve: EI

d 2v = M(x) dx2

For M1(x) = Byx1 - MB; EI

EI

d 2v1 dx 12

= By x1 - MB

By x12 dv2 = - MBx1 + C1 dx2 2

EIv1 =

By x 13 6

-

(3)

MBx 12 + C1x + C2 2

(4)

For M2(x) = Ay x2 - MA EI EI

d 2v2 dx22 dv2 dx22

EIv2 =

= Ayx2 - MA Ayx 22

=

- MA x2 + C3

(5)

MAx 23 + C3x2 + C4 2

(6)

2

Ay x 23 6

-

Boundary Conditions: At x1 = 0,

dv1 = 0 dx1

From Eq. (3), 0 = 0 - 0 + C1;

C1 = 0

At x1 = 0, v1 = 0 From Eq. (4), 0 = 0 - 0 + 0 + C2; C2 = 0 Similarly, from Eqs. (5) and (6), C3 = C4 = 0.

1222

2– 3L

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12–99.  Continued

At x1 =

dv1 dv2 2 1 = . L, x 2 = L, v1 = v2 and dx1 dx2 3 3

From Eqs. (4) and (6), 3 2 3 2 By 2 Ay 1 MB 2 MA 1 a Lb a Lb = a Lb a Lb 6 3 2 3 6 3 2 3

(7)

8ByL - 36MB = AyL - 9MA From Eqs. (3) and (5),

2 2 By 2 Ay 1 2 1 a L b - MB a L b = a L b + MAa L b 2 3 3 2 3 3

(8)

4ByL - 12MB = - AyL + 6MA

Solving Eqs. (1), (2), (7) and (8) simultaneously, Ay = MA = By = MB =

20 P 27

Ans.

4 PL 27

Ans.

7 P 27

Ans.

2 PL 27

Ans.

Ans: Ax = Bx = 0, 20 Ay = P, 27 4 MA = PL, 27 7 By = P, 27 2 MB = PL 27 1223

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*12–100. P

Determine the reactions at the supports, then draw the shear and moment diagram. EI is constant.

A

B L

Solution Support Reactions: FBD(a). + ΣFx = 0 ; S

Ans.

Ax = 0

+ c ΣFy = 0;   By - Ay - P = 0

(1)

a+ ΣMB = 0;    Ay L - MA - PL = 0

(2)

Moment Functions: FBD(b) and (c). M(x1) = -Px1 M(x2) = MA - Ay x2 Slope and Elastic Curve: EI

d 2v = M(x) dx2

For M(x1) = -Px1 , EI EI

d 2v1 dx21

= - Px1

dv1 P = - x21 + C1 dx1 2

EI v1 = -

(3)

P 3 x1 + C1x1 + C2 6

(4)

For M(x2) = MA - Ay x2 , EI EI

d 2v2 dx22

= MA - Ay x2

Ay dv2 = MA x2 x22 + C3 dx2 2

EI v2 =

(5)

Ay MA 2 x2 x32 + C3x2 + C4 2 6

(6)

1224

L

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*12–100.  Continued

Boundary Conditions: v2 = 0 at x2 = 0. From Eq. (6), C4 = 0 dv2 = 0 at x2 = 0. From Eq. (5), C3 = 0 dx2 v2 = 0 at x2 = L. From Eq. (6), 0 =

3 MAL2 AyL 2 6

(7)

Solving Eqs. (2) and (7) yields, MA =

PL 2

Ay =

3P 2

Ans.

Substituting the value of Ay into Eqs. (1), By =

5P  2

Ans.

Note: The other boundary and continuity condition can be used to determine the constants C1 and C2 which are not needed here.

Ans: PL , 2 3P Ay = , 2 5P By = 2 MA =

1225

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12–101. Determine the reactions at the supports A, B, and C, then draw the shear and moment diagrams. EI is constant.

P

A L 2

Solution Support Reactions: FBD(a). + ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay + By + Cy - 2P = 0

Ans. (1)

L 3L By L + Cy (2L) - P a b - P a b = 0 2 2

a+ ΣMA = 0;

(2)

Moment Function: FBD(b) and (c). M(x1) = Cy x1 M(x2) = Cy x2 - Px2 +

PL 2

Slope and Elastic Curve: EI

d 2v = M(x) dx2

For M(x1) = Cy x1, EI EI

d 2v1

= Cyx1

dx21

Cy dv1 = x21 + C1 dx1 2 Cy

EI v1 =

6

x31 + C1x1 + C2

For M(x2) = Cyx2 - Px2 + EI EI

(3)

d 2v2 dx22

(4)

PL , 2

= Cyx2 - Px2 +

PL 2

Cy dv2 P PL = x22 - x22 + x + C3 dx2 2 2 2 2

EI v2 =

Cy 6

x32 -

(5)

P 3 PL 2 x2 + x2 + C3x2 + C4 6 4

(6)

Boundary Conditions: v1 = 0 at x1 = 0. From Eq. (4), C2 = 0 dv2 Due to symmetry, = 0 at x2 = L. From Eq. (5), dx2 0 =

Cy L2 2

-

PL2 PL2 + + C3 2 2

C3 = -

Cy L2 2 1226

P

B L 2

C L 2

L 2

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12–101.  Continued

v2 = 0 at x2 = L. From Eq. (6), 0 =

Cy L3 6

-

Cy L2 PL3 PL3 + + abL + C4 6 4 2 C4 =

Cy L3 3

-

PL3 12

Continuity Conditions: At x1 = x2 =

dv2 L dv1 , = . From Eqs. (3) and (5), 2 dx1 dx2 Cy L 2 Cy L 2 Cy L2 P L 2 PL L a b + C1 = a b - a b + a b 2 2 2 2 2 2 2 2 2 C1 =

At x1 = x2 =

Cy L2 PL2 8 2

L , v = v2. From Eqs. (4) and (6), 2 1

Cy L 3 Cy L2 L PL2 a b + a ba b 6 2 8 2 2 =

Cy L 3 Cy L2 L Cy L3 P L 3 PL L 2 PL3 a b - a b + a b + aba b + 6 2 6 2 4 2 2 2 3 12 Cy =

5 P 16

Ans.

Substituting Cy into Eqs. (1) and (2), By =

11 P 8

Ay =

5 P 16

Ans.

Ans: Ax = 0, Cy = 1227

5 11 5 P, By = P, Ay = P 16 8 16

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12–102. P

Determine the reactions at the supports A and B, then draw the shear and moment diagrams. EI is constant. A

B L – 2

L – 2

Solution + ΣFx = 0;  Ax = 0  S

Ans.

+ c ΣFy = 0  Ay + By - P = 0 (1)

Ay = P - By a+ ΣMA = 0  MA + By (L) - P(L>2) = 0 MA =

PL - ByL 2

(2)

Bending Moment M(x): M(x) = - ( - By) 8x - 09 - P h x -

Elastic Curve and Slope:

L L i = By x - P h x - i 2 2

d 2v = M(x) dx2 d 2v L EI 2 = By x - P h x - i 2 dx 2 B x y dv P L 2 EI = - h x - i + C 1 dx 2 2 2 By x3 P L 3 EIv = - h x - i + C1x + C3 6 6 2 EI

(3) (4)

Boundary Conditions:

v = 0  at  x = 0 From Eq. (4) C1 = 0 v = 0  at  x = L From Eq. (4) ByL3 PL3 0 = + C1L 6 48

(5)

dv = 0  at  x = L dx From Eq. (3), 0 =

ByL2 2

-

PL2 + C 1 8

(6)

Solving Eqs. (5) and (6) yields; By =

5 P 16

Substitute: By = Ay =

11 P 16

Ans.

C1 =

- PL3 32

5 P into Eqs. (1) and (2), 16 Ans.

MA =

Ans:

3PL 16

Ans.

1228

By =

5 11 3PL P, Ay = P, MA = 16 16 16

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12–103. Determine the reactions at the supports A and B, then draw the shear and moment diagrams. EI is constant.

w

A

B L

Solution d ΣFa = 0;  Ax = 0

Ans.

+ c ΣFy = 0;  Ay + By - wL = 0

(1)

L a+ ΣMA = 0;  MA + ByL - wL a b = 0 2 x a+ ΣMNA = 0;  By(x) - wx a b - M(x) = 0 2 wx2 M(x) + Byx 2 2 d v EI 2 = M(x) dx d 2v wx2 EI 2 = By x 2 dx 2 B x y dv wx3 EI = + C 1 dx 2 6 By x3 wx4 EIv = + C1x + C2 6 24

(2)

(3) (4)

Boundary Conditions: At x = 0, v = 0 From Eq. (4), 0 = 0 - 0 + 0 + C2;    C2 = 0 dv At x = L, = 0 dx From Eq. (3), ByL2

wL3 + C 1 2 6 At x = L, v = 0 0 =

(5)

-

From Eq. (4), 0 =

ByL3 6

-

wL4 + C1L 24

(6)

Solving Eqs. (5) and (6) yields: 3wL 8 wL3 C1 = 48

Ans.

By =

Substituting By into Eqs. (1) and (2) yields: Ay =

5wL 8

wL2 MA = 8

1229

Ans.

Ans:

Ans.

3wL , 8 5wL wL2 Ay = , MA = 8 8 Ax = 0, By =

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*12–104. P

Determine the moment reactions at the supports A and B. EI is constant.

P

A

B a

a L

Solution Support Reactions: FBD(a). a+ ΣMB = 0; Pa + P(L - a) + MA - Ay L - MB = 0 (1)

PL + MA - Ay L - MB = 0 Moment Functions: FBD(b) and (c). M(x1) = Ay x1 - MA

M(x2) = Ay x2 - Px2 + Pa - MA

Slope and Elastic Curve: EI

d 2v = M(x) dx2

For M(x1) = Ay x1 - MA, EI EI

d 2v1 dx21

= Ay x1 - MA

Ay dv1 = x21 - MA x1 + C1 dx1 2

EI v1 =

Ay 6

(2)

MA 2 x1 + C1x1 + C2 2

x31 -

(3)

For M(x2) = Ay x2 - Px2 + Pa - Ma, EI EI

d 2v = Ay x2 - Px2 + Pa - MA dx22

Ay dv2 P = x22 - x22 + Pax2 - MA x2 + C3 dx2 2 2

EI v2 =

Ay 6

(4)

MA 2 P 3 Pa 2 x2 + x2 x2 + C3 x2 + C4 6 2 2

x32 -

(5)

Boundary Conditions: dv1 = 0 at x1 = 0. From Eq. (2), C1 = 0 dx1 v1 = 0 at x1 = 0. From Eq. (3), C2 = 0 Due to symmetry, 0 =

dv2 L = 0 at x2 = . From Eq. (4), dx2 2

Ay L 2 P L 2 L L a b - a b + Paa b - MAa b + C3 2 2 2 2 2 2

C3 = -

Ay L2 8

+

MA L PL2 PaL + 8 2 2

1230

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*12–104.  Continued

Due to symmetry, Ay a2

dv1 dv2 = at x1 = a and x2 = L - a. From Eqs. (2) and (4), dx1 dx2

- MA a = -

2

Ay 2

(L - a)2 +

+ MA (L - a) +

Ay L2

2

- Ay a2 -

3Ay L 8

+ Ay aL +

8

P (L - a)2 - Pa(L - a) 2 -

MA L PL2 PaL + 8 2 2

MAL 3PL2 3PaL 3Pa2 + + = 0 8 2 2 2

(6)

Continuity Conditions: At x1 = x2 = a, Ay a2

- MA a

2 =

dv1 dv2 . From Eqs. (2) and (4), = dx1 dx2

Ay a2 2

-

Ay L2 MA L Pa2 PL2 PaL + Pa2 - MA a + + 2 8 8 2 2

Ay L2 MA L Pa2 PL2 PaL + + = 0 2 8 8 2 2

(7)

Solving Eqs. (6) and (7) yields, MA =

Pa (L - a) L

Ans.

Ay = P Substitute the value of MA and Ay obtained into Eqs. (1), MB =

Pa (L - a) L

Ans.

Ans: Pa (L - a), L Pa MB = (L - a) L MA =

1231

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12–105. M0

Determine the reactions at the supports A and B, then draw the moment diagram. EI is constant.

A

B L

Solution Support Reactions: FBD(a). + ΣFx = 0;  Ax = 0 S

Ans.

+ c ΣFy = 0; Ay - By = 0

(1)

a+ ΣMB = 0; M0 - Ay L + MB = 0

(2)

Moment Function: FBD(b) a+ ΣMNA = 0; M(x) + M0 - Ay x = 0 M(x) = Ay x - M0 Slope and Elastic Curve: EI EI

d 2v = M(x) dx2

d 2v = Ay x - M0 dx2

Ay dv = x2 - M0x + C1 dx 2

EI

Ay

EIv =

6

x3 -

(3)

M0 2 x + C1x + C2 2

(4)

Boundary Conditions: At x = 0, v = 0. From Eq. (4), C2 = 0 At x = L, 0 =

dv = 0. From Eq. (3), dx

Ay L2 2

(5)

- M0 L + C1

At x = L, v = 0. From Eq. (4), 0 =

Ay L3 6

-

M0 L2 + C1 L 2

(6)

Solving Eqs. (5) and (6) yields, Ay =

3M0 2L

C1 =

M0 L 4

Ans.

Substituting Ay, into Eqs. (1) and (2) yields: By =

3M0 2L

MB =

M0  2

Ans.

1232

Ans: Ax = 0, 3M0 Ay = , 2L 3M0 By = , 2L M0 MB = 2

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12–106. w0

Determine the reactions at the support A and B. EI is constant. A

B L

Solution Support Reactions: FBD(a). + ΣFx = 0;        Ax = 0 S

Ans.

w0L = 0 2 w0L L a b = 0 a+ΣMA = 0;    By L + MA 2 3 + c ΣFy = 0;    Ay + By -

(1) (2)

Moment Function: FBD(b).

1 w0 x a xb x a b + By x = 0 2 L 3 w0 3 M(x) = By x x 6L Slope and Elastic Curve:

a+ ΣMNA = 0;   - M(x) -

d 2v = M(x) dx2 2 w0 3 d v EI 2 = By x x 6L dx By w0 4 dv EI = x2 x + C 1 dx 2 24L By w0 5 EIv = x3 x + C1x + C2 6 120L EI

(3) (4)

Boundary Conditions: At x = 0,  v = 0.  From Eq. (4),  C2 = 0 At x = L, 0 =

dv = 0.    From Eq. (3), dx ByL2 2

C1 =

w0L3 + C1 24

-

ByL2 2

+

w0L3 24

At x = L,  v = 0.   3

0 =

ByL 6

From Eq. (4),

ByL2 w0L w0L3 + a+ bL 120 2 24 4

-

By =

w0L 10

Ans.

Substituting By into Eqs. (1) and (2) yields, Ay =

2w0L 5

MA =

w0L2 15

Ans. Ans: Ax = 0, By = 1233

w0L 2w0L w0L2 , Ay = , MA = 10 5 15

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12–107. w

Determine the reactions at roller support A and fixed support B. A L 3

B 2L 3

Solution Equations of Equilibrium: Referring to the free-body diagram of the entire beam, Fig. a, + ΣFx = 0;    Bx = 0 Ans. S + c ΣFy = 0;

(1)

Ay + By - wL = 0

L 2 wL a b - Ay a L b - MB = 0 a+ ΣMB = 0; 2 3 wL2 2 MB = - AyL (2) 2 3 Moment Functions: Referring to the free-body diagram of the beam’s segment, Fig. b, a+ ΣMO = 0;

L x L M(x) + wx a b + w a b ¢x + ≤ - Ay x = 0 2 3 6

w 2 wL wL2 x x 2 3 18 Equations of Slope and Elastic Curves: M(x) = Ay x -

d 2v = M(x) dx2 2 d v w 2 wL wL2 EI 2 = Ayx x x 2 3 18 dx Ay dv w wL wL2 EI = x2 x3 x2 x + C 1 (3) dx 2 6 6 18 Ay w 4 wL 3 wL2 2 EIv = x3 x x x + C1x + C2 (4) 6 24 18 36 Boundary Conditions: At x = 0, v = 0. Then Eq. (4) gives 2 dv 0 = 0 - 0 - 0 - 0 + 0 + C2 C2 = 0 At x = L, = 0. Then Eq. (3) gives 3 dx Ay 2 2 w 2 3 wL 2 2 wL2 2 0 = a Lb a Lb a Lb a L b + C1 2 3 6 3 6 3 18 3 EI

C1 =

2AyL2 13wL3 81 9

(5)

2 L, v = 0. Then Eq. (4) gives 3 Ay 2 3 w 2 4 wL 2 3 wL2 2 2 2 0 = a Lb a Lb a Lb a L b + C1a L b 6 3 24 3 18 3 36 3 3 2AyL2 wL3 C1 = 18 27

At x =

(6)

Solving Eqs. (5) and (6), Ay =

17wL wL Ans. C1 = Substituting the result of Ay into Eqs. (1) and (2), 24 324

By =

7wL 24

MB =

wL2 36

Ans.

The shear and moment diagrams are shown in Figs. c and d, respectively. 1234

Ans: Bx = 0, 17wL Ay = , 24 7wL By = , 24 wL2 MB = 36

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*12–108. Determine the moment reactions at the supports A and B, then draw the shear and moment diagrams. Solve by expressing the internal moment in the beam in terms of Ay and MA. EI is constant.

w

B

A L

Solution M(x) = Ayx - MA -

wx2 2

Elastic Curve and Slope: d 2v wx2 = M(x) = A x M y A 2 dx2 2 3 Ayx dv wx EI = - MAx + C1 dx 2 6 EI

Ayx3

EIv =

6

-

(1)

MAx2 wx4 + C1x + C2 2 24

(2)

Boundary Conditions: dv = 0 at x = 0 dx From Eq. (1) C1 = 0 v = 0 at x = 0 From Eq. (2) C2 = 0 dv = 0 at x = L dx From Eq. (1) 0 =

AyL2 2

- MAL -

wL3 6

(3)

v = 0 at x = L

From Eq. (2) 0 =

AyL3 6

-

MAL2 wL4 2 24

(4)

Solving Eqs. (3) and (4) yields: Ay = MA =

wL 2 wL2 12

Ans.

Due to symmetry: MB =

wL2 12

Ans.

Ans:

wL2 , 12 2 wL MB = 12 MA =

1235

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12–109.  The beam has a constant E1I1 and is supported by the fixed wall at B and the rod AC. If the rod has a cross-sectional area A2 and the material has a modulus of elasticity E2, determine the force in the rod.

C w

L2

B

A L1

Solution + c ΣFy = 0

TAC + By - wL1 = 0

c + ΣMB = 0

TAC(L1) + MB MB =

wL1 2 = 0 2

(1)

wL1 2 - TACL1 2

(2)

Bending Moment M(x): M(x) = TACx -

wx2 2

Elastic Curve and Slope: EI

d 2v wx2 = M(x) = TACx 2 2 dx

EI

TACx2 dv wx3 = + C1 dx 2 6

EIv =

(3)

TACx3 wx4 + C1x + C2 6 24

(4)

Boundary Conditions: v =

TACL2 A2E2

x = 0

From Eq. (4) -E1I1a

TACL2 b = 0 - 0 + 0 + C2 A2E2

C2 = a

v = 0

-E1I1L2 bTAC A2E2 at

x = L1

From Eq. (4) 0 =

TACL1 3 wL1 4 E1I1L2 + C1L1 T 6 24 A2E2 AC dv = 0 at x = L1 dx

(5)

From Eq. (3) 0 =

TACL1 2 wL1 3 + C1 2 6

(6)

Solving Eqs. (5) and (6) yields: TAC =

3A2E2wL1

Ans:

4

8 1 A2E2L1 3 + 3E1I1L2 2

Ans. 1236

TAC =

3A2E2wL41 8(A2E2L31 + 3E1I1L2)

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12–110.  The beam is supported by a pin at A, a roller at B, and a post having a diameter of 50 mm at C. Determine the support reactions at A, B, and C. The post and the beam are made of the same material having a modulus of elasticity E = 200 GPa, and the beam has a constant moment of inertia I = 255(106) mm4.

15 kN/m

A 1m 6m

B

C 6m

Solution Equations of Equilibrium: Referring to the free-body diagram of the entire beam, Fig. a, + ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay + By + FC - 15(12) = 0

a+ ΣMB = 0;

15(12)(6) - FC(6) - Ay(12) = 0

Ans. (1)

2Ay + FC = 180 (2) Moment Functions: Referring to the free-body diagram of the beam’s segment, Fig. b, x M(x) + 15xa b - Ayx = 0 2

a+ ΣMO = 0;

M(x) = Ayx - 7.5x2 Equations of Slope and Elastic Curves: EI

d 2v = M(x) dx2

d 2v = Ayx - 7.5x2 dx2 Ay dv EI = x2 - 2.5x3 + C1 dx 2 Ay EIv = x3 - 0.625x4 + C1x + C2 6 EI

(3) (4)

Boundary Conditions: At x = 0, v = 0. Then Eq. (4) gives 0 = 0 - 0 + 0 + C2 At x = 6 m, v = - ∆ C = -

E 3 255 1 10 - 6 2 4 a -

C2 = 0 FC(1) FCLC 1600FC = . Then Eq. (4) gives = p ACE pE 2 1 0.05 2 E 4

Ay 1600FC b = 1 63 2 - 0.625 1 64 2 + C1(6) pE 6

C1 = 135 - 6Ay - 0.02165FC dv Due to symmetry, = 0 at x = 6 m. Then Eq. (3) gives dx 0 =

Ay 2

1 62 2

- 2.5 1 63 2 + 135 - 6Ay - 0.02165FC

(5)

12Ay - 0.02165FC = 405

Solving Eqs. (2) and (5),

FC = 112.096 kN = 112 kN

Ay = 33.95 kN = 34.0 kN

Ans. Ans:

Substituting these results into Eq. (1), Ans.

By = 33.95 kN = 34.0 kN 1237

Ax = 0, FC = 112 kN, Ay = 34.0 kN, By = 34.0 kN

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12–111.  Determine the moment reactions at the supports A and B. EI is constant.

w

B

A L – 2

L – 2

Solution uB>A = 0 =

- MA 1 AyL 1 -wL2 L a b(L) + a b(L) + a ba b 2 EI EI 3 8EI 2 0 =

t B>A = 0 = 0 =

2

- MA -

wL2 48

(2)

- MA 1 AyL 1 -wL2 L L L L a b(L)a b + a b(L)a b + a ba ba b 2 EI 3 EI 2 3 8EI 2 8

AyL 6

AyL

-

MA wL2 2 384

(3)

Solving Eqs. (2) and (3) yields: Ay = MA =

3wL 32 5wL2 192

c+ Σ MB = 0;  MB +

Ans. 3wL 5wL2 wL L (L) a b = 0 32 192 2 4 MB =

11wL2 192

Ans.

Ans: MA = 1238

5wL2 11wL2 , MB = 192 192

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*12–112. The rod is fixed at A, and the connection at B consists of a roller constraint which allows vertical displacement but resists axial load and moment. Determine the moment reactions at these supports. EI is constant.

w

B

A L

Solution Support Reaction: FBD(a). a + ©MA = 0;

MB + MA - wL a

L b = 0 2

[1]

Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for MB and the uniform distributed load acting on a cantilever beam are shown. Moment-Area Theorems: Since both tangents at A and B are horizontal (parallel), uB>A = 0. uB>A = 0 = a

MB 1 wL2 b (L) + a b (L) EI 3 2EI

MB =

wL2 6

Ans.

Substituting MB into Eq.[1], MA =

wL2 3

Ans.

Ans: MA =

1009

2 wL2 M = wL ; B 6 3

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12–113. Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment. EI is constant.

P a

L

Solution (tA>B)1 =

2(L - a) -P(L - a)2(2L + a) 1 -P(L - a) a b (L - a) aa + b = 2 EI 3 6EI

(tA>B)2 =

A yL3 2L 1 A yL a b (L) a b = 2 EI 3 3EI

tA>B = 0 = (tA>B)1 + (tA>B)2 0 =

A yL3 -P(L - a)2(2L + a) + 6EI 3EI Ay =

P(L - a)2(2L + a) 2L3

Require: |M1| = |M2| Pa(L - a)2(2L + a) 2L3

=

Pa(L - a)(L + a) 2L2

a2 + 2La - L2 = 0 a = 0.414L

Ans.

Ans: a = 0.414L 1010

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12–114.  Determine the reactions at the supports A and B, then draw the shear and moment diagrams. EI is constant.

P

P

A

B L – 3

L – 3

L – 3

Solution (t B>A)1 =

(t B>A)2 =

1 - PL L 2L 2L 1 - 2PL 2L L 4L 2PL3 a ba ba + b + a ba ba + b = 2 3EI 3 3 9 2 3EI 3 3 9 9EI ByL3 1 ByL 2L b(L)a b = a 2 EI 3 3EI

t B>A = 0 = (t B>A)1 + (t B>A)2 0 =-

ByL3 2PL3 + 9EI 3EI By =

2 P 3

Ans.

From the free-body diagram, PL 3 4 Ay = P 3

Ans.

MA =

Ans. Ans.

Ax = 0

Ans: By = 1241

2 PL 4 P, MA = , Ay = P, Ax = 0 3 3 3

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12–115.  Determine the reactions at the supports. EI is constant.

M0

M0 B

A a

a

a

Solution Support Reaction: FBD(a). + S ΣFx = 0;

Ax = 0

+ c ΣFy = 0;

- By + Ay = 0

(1)

a+ ΣMA = 0;

- By (3a) + MA = 0

(2)

Ans.

Elastic Curve: As shown. M>EI Diagrams: M>EI diagrams for By and M0 acting on a cantilever beam are drawn. Moment - Area Theorems: From the elastic curve, t B>A = 0. t B>A = 0 =

M0 3By a 2 1 a ab (3a) a b (3a) + a b (a) aa + b 2 EI 3 EI 2 By =

M0 6a

Ans.

Substituting By into Eqs. (1) and (2) yields, Ay =

M0 M0    MA = 6a 2

Ans.

Ans: Ax = 0, M0 By = , 6a M0 Ay = , 6a M0 MA = 2 1242

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*12–116. Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant. Support B is a thrust bearing.

P A

B

Solution + ©F = 0; : x + c ©Fy = 0; a+ ©MA = 0;

L 2

L

Support Reactions: FBD(a). Bx = 0

C

L 2

Ans.

-A y + By + Cy - P = 0 By (L) + Cy (2L) - Pa

[1]

3L b = 0 2

[2]

Elastic Curve: As shown. M/EI Diagrams: M/EI diagrams for P and By acting on a simply supported beam are drawn separately. Moment-Area Theorems: (tA>C)1 = =

1 3PL 3L 2 3L 1 3PL L 3L L a ba ba ba b + a ba ba + b 2 8EI 2 3 2 2 8EI 2 2 6 7PL3 16EI

(tA>C)2 =

By L By L3 1 ab (2L)(L) = 2 2EI 2EI

(tB>C)1 =

1 PL L 2 L PL L L a ba ba ba b + a ba ba b 2 8EI 2 3 2 4EI 2 4 +

=

1 3PL L L L a ba ba + b 2 8EI 2 2 6

5PL3 48EI

By L By L3 1 L b (L)a b = (tB>C)2 = a 2 2EI 3 12EI tA>C

By L3 7PL3 = (tA>C)1 + (tA>C)2 = 16EI 2EI

tB>C = (tB>C)1 + (tB>C)2 =

By L3 5PL3 48EI 12EI

From the elastic curve, tA>C = 2tB>C By L3 By L3 5PL3 7PL3 = 2a b 16EI 2EI 48EI 12EI By =

11P 16

Ans.

Substituting By into Eqs. [1] and [2] yields, Cy =

13P 32

Ay =

Ans:

3P 32

Ans.

1012

Ay =

3P 13P ; By = 11P ; Cy = 32 32 16

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12–117.  Determine the reactions at the journal bearing supports A, B, and C of the shaft, then draw the shear and moment diagrams. EI is constant.

A

C

B

1m

1m 400 N

1m

1m 400 N

Solution Support Reactions: FBD(a). + c ΣFy = 0; a+ ΣMA = 0;

(1)

Ay + By + Cy - 800 = 0

(2)

By (2) + Cy (4) - 400(1) - 400(3) = 0

Method of Superposition: Using the table in Appendix C, the required displacements are vB = = = = vB > =

Pbx 1 L2 - b2 - x2 2 6EIL

400(1)(2) 6EI(4)

1 42

366.67 N # m3 EI

- 12 - 22 2 T

By 1 43 2 1.3333By m3 PL3 = = 48EI 48EI EI

c

The compatibility condition requires ( + T)

0 = 2vB ′ + vB ″ 0 = 2a

1.3333By 366.67 b + ab EI EI

Ans.

By = 550 N

Substituting By into Eqs. (1) and (2) yields, Ay = 125 N

Ans.

Cy = 125 N

Ans: By = 550 N, Ay = 125 N, Cy = 125 N 1244

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12–118.  Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.

w

A

C

B L

L

Solution Support Reactions: FBD(a). + ΣFx = 0; S

Ans.

Ax = 0

 + c ΣFy = 0;

Ay + By + Cy - 2wL = 0

(1)

a+ ΣMA = 0;

By(L) + Cy(2L) - (2wL)(L) = 0

(2)

Method of Superposition: Using the table in Appendix C, the required displacements are vB ′ =

5w(2L)4 5wL4AC 5wL4 = = T 384EI 384EI 24EI

vB″ =

By(2 L)3 By L3 PL3AC = = c 4 8 EI 4 8 EI 6 EI

The compatibility condition requires ( + T)

0 = vB ′ + v B ″ 0 =

ByL3 5wL4 + ab 24EI 6EI

By =

5wL 4

Ans.

Substituting the value of By into Eqs. (1) and (2) yields, Cy = Ay =

3wL 8

Ans.

Ans: Ax = 0, 5wL By = , 4 3wL Cy = 8 1245

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12–119.  Determine the reactions at the supports, then draw the shear and moment diagrams. EI is constant.

w

A

C

B L

L

Solution 5w(2L)4

∆ =

768EI

∆′ =

By(2L)4 48EI

=

5wL4 T 48EI ByL3

=

6EI

c

Require:

(+T) 0 =

0 = ∆ - ∆′

ByL3 5wL 48EI 6EI

By =

4

5 wL c  8

c+ ΣMA = 0;

+ c ΣFy = 0;

Ans. L 5 wLa b - wL(L) - Cy(2L) = 0 2 8 Cy = -

wL wL = T 16 16

- wL -

wL 5 + wL + Ay = 0 16 8

Ay =

Ans.

7 wL c  16

Ans.

Ans: 5 wL c , 8 wL Cy = T, 16 7 Ay = wL c 16 By =

1246

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*12–120. Determine the reactions at the supports A and  B. EI is constant.

w0

A

B L

Solution Support Reactions: FBD(a). + ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay + By -

a+ ΣMA = 0;

ByL + MA -

Ans. w0L = 0 2

(1)

w0L L a b = 0 2 3

(2)

Method of Superposition: Using the table in Appendix C, the required displacements are vB ′ =

ByL3 w0 L4 T      vB ″ = c 30EI 3EI

The compatibility condition requires ( + T)

0 = vB ′ + v B ″ 3

0 =

ByL w0 L4 + ab 30EI 3EI

By =

w0L 10

Ans.

Substituting By into Eqs. (1) and (2) yields, Ay =

2w0L 5

MA =

w0L2 15

Ans.

Ans: Ax = 0, w0L By = , 10 2w0L Ay = , 5 2 w0L MA = 15 1247

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12–121. Determine the reactions at the supports A and B. EI is constant.

P

A

Solution

B L

Referring to the FBD of the beam, Fig. a + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

L 2

Ans.

By - P - A y = 0 A y = By - P

(1)

3 a+ ©MA = 0; -MA + By L - P a L b = 0 2 MA = By L -

3 PL 2

(2)

Referring to Fig. b and the table in appendix, the necessary deflections are computed as follow: yP =

Px 2 (3LAC - x) 6EI

=

P(L2) 3 c3a Lb - L d 6EI 2

=

7PL3 12EI

yBy =

T

By L3 PL3AB c = 3EI 3EI

The compatibility condition at support B requires that (+ T )

0 = vP + vBy 0 =

-By L3 7PL3 + a b 12EI 3EI

By =

7P 4

Ans.

Substitute this result into Eq (1) and (2) Ay =

3P 4

MA =

PL 4

Ans.

Ans: By = 1015

7P

, Ay =

3P PL , MA = 4 4

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12–122.  Determine the reactions at the supports A and B. EI is constant.

M0 A

B L

Solution Referring to the FBD of the beam, Fig. a + ΣFx = 0;   Ax = 0 S + c ΣFy = 0;

Ans.

By - P - Ay = 0 (1)

Ay = By - P 3 a+ ΣMA = 0; - MA + By L - P a L b = 0 2 MA = By L -

3 PL 2

(2)

Referring to Fig. b and the table in appendix, the necessary deflections are computed as follow: vP = = = vBy =

Px2 (3LAC - x) 6EI P(L2) 6EI

3 c 3a L b - L d 2

7PL3 T 12EI

By L3 PL3AB = c 3EI 3EI

The compatibility condition at support B requires that ( + T)

0 = vP + vBy 0 =

- By L3 7PL3 + a b 12EI 3EI

By =

7P 4

Ans.

Substitute this result into Eqs. (1) and (2) Ay =

3P 4

MA =

PL 4

Ans.

Ans: Ax = 0, 7P By = , 4 3P Ay = , 4 PL MA = 4 1249

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12–123.  Determine the reactions at the supports A and B. EI is constant.

w

B

A L 2

L 2

Solution Support Reactions: FBD(a). + ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay + By -

a+ ΣMA = 0;

By(L) + MA - a

Ans. wL = 0 2

(1)

wL L b a b = 0 2 4

(2)

Method of Superposition: Using the table in Appendix C, the required displacements are vB ′ =

ByL3 7wL4 PL3 T      vB ″ = = c 384EI 3EI 3EI

The compatibility condition requires ( + T)

0 = vB ′ + v B ″ 3

0 =

ByL 7wL4 + ab 384EI 3EI

By =

7wL 128

Ans.

Substituting By into Eqs. (1) and (2) yields Ay =

57wL 128

MA =

9wL2 128

Ans.

Ans: Ax = 0, 7wL By = , 128 57wL Ay = , 128 2 9wL MA = 128 1250

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*12–124. Before the uniform distributed load is applied to the beam, there is a small gap of 0.2 mm between the beam and the post at B. Determine the support reactions at A, B, and C. The post at B has a diameter of 40 mm, and the moment of inertia of the beam is I = 875(106) mm4. The post and the beam are made of material having a modulus of elasticity of E = 200 GPa.

30 kN/m

A 1m 6m

B

0.2 mm

C

6m

Solution Equations of Equilibrium: Referring to the free-body diagram of the beam, Fig. a, + ΣFx = 0; S

Ax = 0

+ c ΣFy = 0;

Ay + FB + Cy - 30(12) = 0

(1)

a+ ΣMA = 0;

FB(6) + Cy(12) - 30(12)(6) = 0

(2)

Ans.

Method of Superposition: Referring to Fig. b and the table in the appendix, the necessary deflections are (vB)1 = (vB)2 =

5(30) 1 124 2 5wL4 8100 kN # m3 = = T 384EI 384EI EI FB 1 123 2 36FB PL3 = = c 48EI 48EI EI

The deflection of point B is vB = 0.2 1 10 - 3 2 +

FB(1) FBLB = 0.2 1 10 - 3 2 + T AE AE

The compatibility condition at support B requires

1+T2

vB = (vB)1 + (vB)2 0.2 1 10 - 3 2 +

FB (1) AE

0.2 1 10 - 3 2 E + FB

p 1 0.042 2 4

+

=

36FB 8100 + ab EI EI

FB 36FB 8100 = A I I 36FB

875 1 10 - 6 2

=

FB = 219.78 kN = 220 kN

8100 875 1 10 - 6 2

-

0.2 1 10 - 3 2 3 200 1 109 2 4 1000

Ans.

Substituting the result of FB into Eqs. (1) and (2), Ans.

Ay = Cy = 70.11 kN = 70.1 kN

Ans: Ax = 0, FB = 220 kN, Ay = Cy = 70.1 kN 1251

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12–125.  The fixed supported beam AB is strengthened using the simply supported beam CD and the roller at F which is set in place just before application of the load P. Determine the reactions at the supports if EI is constant.

P A

B

C L — 4

L — 4

D

F

L — 4

L — 4

Solution dF = Deflection of top beam at F d′F = Deflection of bottom beam at F dF = d′F (+ T)

(P - Q)(L3) 48EI

(P - Q)L 48

-

-

2M 1 L2 2

Q 1 L2 2 L 2 c L2 - a b d = 6 EIL 2 48EI

3

QL 1 3 M = 6 4 48(8)

(1)

8PL - 48M = 9QL uA = u′A + u″A = 0 -

c+

(P - Q)L2 ML ML + = 0 6EI 3EI 16EI (2)

8M = (P - Q)L

Solving Eqs. (1) and (2): M = QL>16 Q = 2P>3 S = P>3 R = P>6 M = PL>24 Thus, MA = MB =

1 PL 24

Ans.

Ay = By =

1 P 6

Ans.

Cy = Dy =

1 P 3

Ans. Ans.

Dx = 0

Ans: MA = MB = Cy = Dy = 1252

1 1 PL, Ay = By = P, 24 6

1 P, Dx = 0 3

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12–126.  The beam has a constant E1I1 and is supported by the fixed wall at B and the rod AC. If the rod has a cross-sectional area A2 and the material has a modulus of elasticity E2, determine the force in the rod.

C L2

w

B

A L1

Solution ( ∆A ) ′ = dA =

wL41 ; 8E1I1

∆A =

TACL2 A2E2

TACL31 3E1I1

By Superposition: ∆ A = (∆ A)′ - dA

( + T)

TACL2 TACL31 wL41 = A2E2 8E1I1 3E1I1 TAC a

L31 wL41 L2 + b = A2E2 3E1I1 8E1I1

TAC =

3wA2E2L41

Ans.

8(3E1I1L2 + A2E2L31)

Ans: TAC =

1253

3wA2E2L41 8(3E1I1L2 + A2E2L31)

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12–127.  The beam is supported by the bolted supports at its ends. When loaded these supports initially do not provide an actual fixed connection, but instead allow a slight rotation a before becoming fixed after the load is fully applied. Determine the moment at the supports and the maximum deflection of the beam.

P

L — 2

L — 2

Solution u - u′ = a PL2 ML ML = a 16EI 3EI 6EI ML = a M =

PL2 - a b(2EI) 16EI

PL 2EI a 8 L

∆ max = ∆ - ∆ ′ = ∆ max = ∆ max =

Ans. M(L2 ) PL3 - 2c 3 L2 - (L>2)2 4 d 48EI 6EIL

PL3 L2 PL 2EIa b a 48EI 8EI 8 L PL3 aL + 192EI 4

Ans.

Ans: M = 1254

PL 2EI PL3 aL a, ∆ max = + 8 L 192EI 4

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*12–128. The 25-mm-diameter A-36 steel shaft is supported by unyielding bearings A and C. The bearing B rests unyielding bearings at Aatand C. The bearing at B at rests on a on a simply supported steel wide-fl angehaving beamahaving a simply supported steel wide-flange beam moment 6 4 4 195(10 ) mm . If the belt loads on moment of inertia of I = of inertia of I = 500 in . If the belt loads on the pulley are the pulley are 2 kN each, the vertical reactions at 400 lb each, determine thedetermine vertical reactions at A, B, and C. A, B, and C.

3 ftm 0.9 5 ftm 1.5

A

2 ftm 0.6 5 ftm 1.5 B

For the shaft:

( ∆ b )= 1 ( ∆ b= )2

2400 kN

 4(10 )(0.9)(1.5)  2 1782 2 2 ↓ )   (3 − 0.9 − 1.5 = EI s 6EI s (3)   3

lb

400 2 kN lb C

5 ftm 1.5

3

By (3 ) 0.5625By = ↑ EI s 48 EI s

For the beam: 3 3B B 20.833B A 10 Byy(3 ) 0.5625B y y ¢b = = 48EI EI 48EI EI bb b b

Compatibility condition: + T ¢ b = (¢ b)1 - (¢ b)2 20.833B 20.833B 1782 0.5625Byy 13200 0.5625B yy =  EIbb EIss EI EI EI s s

π

= (0.01254 ) 19.175(10 −9 ) m 4 4

Is =

0.5625By 0.5625By 1782 = − E[195(10 −6 )] E[19.175(10 −9 )] E[19.175(10 −9 )] = = N 3.17 kN By 3167.69

Ans.

Form the free-body digram, kN 243 lb A y = 1.22

Ans.

Cy = 0.384 kN

Ans.

4 kN

4 kN

0.9 m 1.22 kN

0.6 m

1.5 m

4 kN 3.17 kN

0.3838 kN

Ans:

kN; Ay 1.22 kN; Cy = 0.384 kN = By 3.17= 1025

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12–129.  The beam is made from a soft linear elastic material having a constant EI. If it is originally a distance ∆ from the surface of its end support, determine the length a that rests on this support when it is subjected to the uniform load w0, which is great enough to cause this to happen.

w0 

a L

Solution The curvature of the beam in region BC is zero, therefore there is no bending moment in the region BC. The reaction R is at B where it touches the support. The slope is zero at this point and the deflection is ∆ where ∆ =

w0(L - a)4 8EI

ux = 0 =

-

R(L - a)3 3EI

w0(L - a)3 6EI

-

R(L - a)2 2EI

Thus, R = ∆ =

w0(L - a) 3 w0(L - a)4 (72EI) 1

L - a = a a = L - a

72∆EI b4 w0 1

72∆EI b4 w0

Ans.

Ans: a = L - a 1256

72∆EI 1/4 b w0

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12–130. If the temperature of the 75-mm-diameter post CD is increased by 60°C, determine the force developed in the post. The post and the beam are made of A-36 steel, and the moment of inertia of the beam is I = 255(106) mm4.

3m

3m

A

C

B

3m D

Solution Method of Superposition. Referring to Fig. a and the table in the Appendix, the necessary deflections are (vC)1 =

FCD A 33 B PLBC 3 9FCD c = = 3EI 3EI EI

(vC)2 = (uB)2LBC =

3FCD (3) 9FCD MOLAB c (LBC) = (3) = 3EI 3EI EI

The compatibility condition at end C requires

A+cB

vC = (vC)1 + (vC)2 =

9FCD 9FCD 18FCD c + = EI EI EI

Referring to Fig. b, the compatibility condition of post CD requires that dFCD + vC = dT dFCD =

(1)

FCD (3) FCD LCD = AE AE

dT = a¢TL = 12 A 10 - 6 B (60)(3) = 2.16 A 10 - 3 B m Thus, Eq. (1) becomes 3FCD 18FCD + = 2.16 A 10 - 3 B AE EI 18FCD 3FCD + = 2.16 A 10 - 3 B C 200 A 109 B D -6 p 255 10 2 A B A 0.075 B 4 Ans.

FCD = 6061.69N = 6.06 kN

Ans: FCD = 6.06 kN 1026

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12–131.  The rim on the flywheel has a thickness t, width b, and specific weight g. If the flywheel is rotating at a constant rate of v, determine the maximum moment developed in the rim. Assume that the spokes do not deform. Hint: Due to symmetry of the loading, the slope of the rim at each spoke is zero. Consider the radius to be sufficiently large so that the segment AB can be considered as a straight beam fixed at both ends and loaded with a uniform centrifugal force per unit length. Show that this force is w = btgv2r>g.

A

t

B

v r

Solution Centrifugal Force: The centrifugal force acting on a unit length of the rim rotating at a constant rate of v is g btgv2r w = mv2r = bt a bv2r = g g

(Q.E.D)

Elastic Curve: Member AB of the rim is modeled as a straight beam with both of its ends fixed and subjected to a uniform centrifugal force w. Method of Superposition: Using the table in Appendix C, the required displacements are uB ′ = vB ′ =

wL3 6EI

uB ″ =

MBL EI

uB ‴ =

2

4

wL c 8EI

vB ″ =

MBL c 2EI

vB ‴ =

ByL2 2EI ByL3 3EI

T

Compatibility requires 0 = uB ′ + uB ″ + uB ‴ 0 =

ByL2 MBL wL3 + + ab 6EI EI 2EI

0 = wL2 + 6MB - 3ByL

(+c)

(1)

0 = vB ′ + v B ″ + v B ‴ 3

0 =

ByL MBL2 wL4 b + + a8EI 2EI 3EI

0 = 3wL2 + 12MB - 8ByL

(2)

Solving Eqs. (1) and (2) yields

Due to symmetry,  

By =

wL 2

MB =

wL2 12

Ay =

wL 2

MA =

wL2 12

Maximum Moment: From the moment diagram, the maximum moment occurs at btgv2r pr the two fixed end supports. With w = and L = ru = , g 3 Mmax

wL2 = = 12

btgv2r pr 2 g 3

( )

12

=

p2btgv2r 3 108g

Ans.

Ans: Mmax = 1258

p2btgv2r 3 108g

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*12–132. The box frame is subjected to a uniform distributed loading w along each of its sides. Determine the moment developed in each corner. Neglect the deflection due to axial load. EI is constant.

E

A

B

L

w

Solution Elastic Curve: In order to maintain the right angle and zero slope (due to symmetrical loading) at the four corner joints, the box frame deforms into the shape shown when it is subjected to the internal uniform distributed load. Therefore, member AB of the frame can be modeled as a beam with both ends fixed.

C

D L

Method of Superposition: Using the table in Appendix C, the required displacements are uB ′ =

wL3 6EI

uB ″ =

MBL EI

uB ‴ =

vB ′ =

wL4 c 8EI

vB ″ =

MBL2 c 2EI

vB ‴ =

ByL2 2EI ByL3 3EI

T

Compatibility conditions require 0 = uB ′ + uB ″ + uB ‴ 2

0 =

ByL MBL wL3 + + ab 6EI EI 2EI

0 = wL2 + 6MB - 3ByL

(+c)

(1)

0 = vB ′ + v B ″ + v B ‴ 3

0 =

ByL MBL2 wL4 + + ab 8EI 2EI 3EI

0 = 3wL2 + 12MB - 8ByL

(2)

Solving Eqs. (1) and (2) yields By = MB =

wL 2 wL2 12

Ans.

Ans: MB = 1259

wL2 12

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R12–1. The shaft supports the two pulley loads shown. Using discontinuity functions, determine the equation of the elastic curve. The bearings at A and B exert only vertical reactions on the shaft. EI is constant.

x A

12 in. 0.3 m

B

12 in. 0.3 m

36 in. 0.9 m

70 lb 350 N 180 N lb 900

Solution M x   x6x0.6 M = 900 - 180 6xx-0 07 (1387.5) - (- 277.5) 6 x 0.3 - 12 7 350 - 70 -  24 7 M M = 900x - 180x+ 1387.5 277.5  6 xx - 0.3 12  7 -350 706xx-0.6 247

0.3 m 0.3 m 900 N 1387.5 N

Elastic curve and slope: EI

d2v M = M = 900x - 180x+ 1387.5 277.5  6 xx - 0.3 12  7 -350 706xx-0.6 247 dx2

EI

dv 693.75 6  xx   x- 240.67 2 2++ C C11 = 450x - 90x2 2+138.75 - 0.3 12  7 2 2-175 35(x dx

350 N

0.9 m 137.5 N

EIv = 150x - 30x3 3+46.25 - 11.67 6 - 0.6 24  73 3+ +C1C x C+2 C2 (1) 231.256xx-12 0.3733 58.333 x x1+ Boundary conditions: v = 0

at

xx == 0.3 12 m in,

From Eq. (1) 0 = 4.05 - 51,840 + 12C  0.3C C2C2 1 1 + 0.3C11 + C C22 =  51 4.05 12C 840 v = 0

at

(2)

x == 1.5 60 m in.

From Eq.(1) 0 = 506.25 - 6 480 000 + 5 114 880 -  544 3201  + C60C  399.6  42.525 1.5C 2 1 + C2 1.5C11 + C C22 =  1909440 149.175 60C

(3)

Solving Eqs. (2) and (3) yields: C1  = 120.94 38 700 v =

C22  = 32.23 - 412 560 C

1 [ -150x3 + 2318x - 0.39 3 EI

- 58.38x - 0.69 3 + 121x - 32.2] N # m3

Ans.

Ans: 1 [ -150x3 + 2318x - 0.39 3 EI - 58.38x - 0.69 3 + 121x - 32.2] N # m3

v =

1027

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R12–2. The shaft is supported by a journal bearing at A, which exerts only vertical reactions on the shaft, and by a thrust bearing at B, which exerts both horizontal and vertical reactions on the shaft. Draw the bending-moment diagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft’s centerline. Determine the equations of the elastic curve using the coordinates x1 and x2. EI is constant.

400 N 80 lb A

x1

100 mm 4 in.

400 N 80 lb 300 12mm in.

EI

dx21

80 N · m

= 133.33x 26.67x1 1 133.33 N

133.33 N M (N · m)

dv1 66.67 21 + C1 = 13.33x dx1

(1)

EIv11=22.22 4.44x31 + C1x1 + C2 EIv

(2)

EI

40

–40

133.33x - 26.67x For M2 (x) =  2 2 EI

d2v2 dx22

x2

300 12mm in.

For M1 (x) = 133.33x 26.67 x11 d2v1

B

100 mm 4 in.

=  - 26.67x 133.33x 2 2 M(x) = 133.33x2

dv2 66.67 22 + C3 EI = - 13.33x dx2

(3)

EIv2 = −22.22 x23

(4)

133.33 N

+ C 3 x2 + C 4

M(x) = 133.33x1

133.33 N

Boundary conditions: v1 = 0

at

x1 = 0

at

x2 = 0

From Eq.(2) C2 = 0 v2 = 0 C4 = 0 Continuity conditions: dv1 dv2 = dx1 dx2

at

0.300 m x1 = x2 = 12

From Eqs. (1) and (3) 19206  + C11  = (6 - ( - 1920  C+3) C3) C1 = - C3 v1 = v2

at

(5)

0.300 m x1 = x2 = 12

= 0.6 - 7680 12C 7680 + 0.3C 12C11  0.6   +0.3C 3 3 4 C3 - C1 = 1280

(6)

Solving Eqs. (5) and (6) yields: C3 = 2 N ⋅ m2

−2 N ⋅ m 2 C1 =

Ans:

v1 =

1 (22.2x13 - 2x1)N # m3 EI

Ans.

v1 =

1 (22.2x31 - 2x1)N # m3, EI

v2 =

1 ( - 22.2x32 + 2x2) N # m3 EI

Ans.

v2 =

1 ( - 22.2x32 + 2x2) N # m3 EI

1028

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R12–3.  Determine the moment reactions at the supports A and B. Use the method of integration. EI is constant.

w0

A

B L

Solution Support Reactions: FBD(a). + c ΣFy = 0;

Ay + By -

w0L = 0 2

(1)

w0L L ByL + MA - MB a b = 0 2 3

a+ ΣMA = 0;

(2)

Moment Function: FBD(b).

M(x) = Byx -

(b)

x 1 w0 - M(x) - a xbx a b - MB + Byx = 0 2 L 3

a+ ΣMNA = 0;

w0 3 x - MB 6L

Slope and Elastic Curve: EI

d 2v = M(x) dx2

EI

w0 3 d 2v = Byx x - MB 6L dx2

EI

By w0 4 dv = x2 x - MBx + C1 dx 2 24L

EI v =

By 6

x3 -

At x = 0, v = 0. At x = L, 0 =

From Eq. (4),

dv = 0. dx

By L2 2

-

(3)

w0 5 MB 2 x x + C1x + C2 120L 2

Boundary Conditions: dv At x = 0, = 0 From Eq. (3), dx

(4)

C1 = 0 C2 = 0

From Eq. (3),

w0L3 - MBL 24

0 = 12By L - w0 L2 - 24MB At x = L, v = 0. 3

0 =

By L 6

-

(a)

(5)

From Eq. (4), w0 L4 MB L2 120 2

0 = 20By L - w0 L2 - 60MB

(6)

1262

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R12–3.  Continued

Solving Eqs. (5) and (6) yields, MB = By =

w0 L2 30

Ans.

3w0L 20

Substituting By and MB into Eqs. (1) and (2) yields, MA =

w0L2 20

Ay =

7w0 L 20

Ans.

Ans: MB = 1263

w0L2 w0L2 , MA = 30 20

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*R12–4. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the slope at A and the maximum deflection. Use the method of integration. EI is constant.

w

C

B

A

x1

x2 L

Solution Elastic Curve and Slope: EI

d 2v = M(x) dx2 - wx11 2

For M1(x) = EI EI

d 2v1 dx21

dv1 - wx31 = + C 1 dx1 6

EI

dx22

(2)

- wLx2 2

For M2(x) = d 2v2

(1)

- wx41 + C1x1 + C2 24

EIv1 =

EI

- wx21 2

=

=

- wLx2 2

- wLx23 dv2 = + C 3 dx2 4

EIv2 =

(3)

- wLx32 + C3x2 + C4 12

(4)

Boundary Conditions: v2 = 0  at  x2 = 0 From Eq. (4): C4 = 0 v2 = 0  at  x2 = L From Eq. (4): 0 = C3 =

- wL4 + C3L 12 wL3 12

v1 = 0  at  x1 = L From Eq. (2): 0 = -

wL4 + C1L + C2 24

(5)

1264

L

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*R12–4.  Continued

Continuity Conditions: dv1 dv2 =   at  x1 = x2 = L dx1 - dx2 From Eqs. (1) and (3) -

wL3 wL3 wL3 + C1 = - a+ b 6 4 12

C1 =

wL3 3

Substitute C1 into Eq. (5) C2 = -

7wL4 24

dv1 w = (2L3 - x31) dx1 6EI dv2 w = (L3 - 3Lx22) dx2 12EI uA = v1 =

dv1 dv2 ` = ` dx1 x1 = L dx2 x2

(6)

=L

=

wL3 6EI

Ans.

w ( - x41 + BL3x1 - 7L4) 24EI

(v1)max =

- 7wL4 24EI

Ans.

(x1 = 0)

The negative sign indicates downward displacement. v2 =

wL 2 (L x2 - x32)         (7) 12EI

(v2)max occurs when

Ans.

dv2 = 0 dx2

From Eq. (6) L3 - 3Lx22 = 0 x2 =

L

Ans:

23

Substitute x2 into Eq. (7), (v2)max =

wL4

(8)

1823EI

vmax = (v1)max = -

7wL4 24EI

wL3 , 6EI w v1 = ( - x41 + BL3x1 - 7L4), 24EI wL 2 v2 = (L x2 - x32),   12EI 7wL4 vmax = 24EI uA =

(9)

Ans.

1265

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R12–5.  Determine the maximum deflection between the supports A and B. Use the method of integration. EI is constant.

w

C

B

A

x1

x2 L

Solution Elastic Curve and Slope: EI

d 2v = M(x) dx2 - wx21 2

For M1(x) = EI EI

d 2v1

dv1 - wx31 = + C 1 dx1 6

EI

=

dx22

(2)

- wLx2 2

For M2(x) = d 2v2

(1)

- wx41 + C1x1 + C2 24

EIv1 =

EI

- wx21 2

=

dx21

- wLx2 2

dv2 - wLx22 = + C 3 dx2 4

EIv2 =

(3)

- wLx32 + C3x2 + C4 12

(4)

Boundary Conditions: v2 = 0  at  x2 = 0 From Eq. (4): C4 = 0 v2 = 0  at  x2 = L From Eq. (4): 0 = C3 =

- wL4 + C3L 12 wL3 12

v1 = 0  at  x1 = L From Eq. (2): 0 = -

wL4 + C1L + C2 24

(5)

1266

L

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R12–5.  Continued

Continuity Conditions: dv1 dv2 =   at  x1 = x2 = L dx1 - dx2 From Eqs. (1) and (3) -

wL3 wL3 wL3 + C1 = - a+ b 6 4 12

C1 =

wL3 3

Substitute C1 into Eq. (5) C2 = -

7wL4 24

dv1 w = (2L3 - x31) dx1 6EI dv2 w = (L3 - 3Lx22) dx2 12EI uA = v1 =

(6)

dv1 dv2 wL3 ` = ` = dx1 x1 = L dx2 x2 = L 6EI

w ( - x41 + 8L3x1 - 7L4) 24EI

(v1)max =

- 7wL4 24EI

(x1 = 0)

The negative sign indicates downward displacement. v2 =

wL 2 (L x2 - x32) 12EI

(v2)max occurs when

(7)

dv2 = 0 dx2

From Eq. (6) L3 - 3Lx22 = 0 x2 =

L 23

Substitute x2 into Eq. (7), (v2)max =

wL4

Ans.

18 23EI

Ans: (v2)max =

1267

wL4 1823EI

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R12–6.  Determine the slope at B and the deflection at C. Use the moment-area theorems. EI is constant.

P A

B

a

a

C

a

Solution Support Reaction and Elastic Curve: As shown. M , EI Diagram: As shown. Moment-Area Theorems: uB>D =

1 Pa Pa2 a b(a) = 2 2EI 4EI

Due to symmetry, the slope at point D is zero. Hence, the slope at B is uB =  uB>D  =

Pa2 4EI

Ans.

Pa2 Pa3 (a) = c 4EI 4EI

Ans.

The displacement at C is ∆ C = uB LBC =

Ans: uB = 1268

Pa2 Pa3 , ∆C = c 4EI 4EI

R12–7.  Determine the reactions, then draw the shear and moment diagrams. Use the moment-area theorems. EI is constant.

B

A

1m

1m

C

2m

200 N

Solution (t B>A)1 = (t C>A)1 = (t B>A)2 = (t C>A)2 =

1 3PL L L L 1 PL L L PL L L 5PL3 a ba ba + b + a ba ba b + a ba b = 2 8EI 2 2 6 2 8EI 2 3 4EI 2 4 48EI 1 3PL L 3L L 1 3PL 3L 7PL3 a ba ba + b + a ba b(L) = 2 8EI 2 2 6 2 8EI 2 16EI

- ByL3 1 -ByL L a b(L) a b = 2 2EI 3 12EI - ByL3 1 -ByL a b(2L)(L) = 2 2EI 2EI

2t B>A = t C>A

2[(t B>A)1 + (t B>A)2] = (t C>A)1 + (t C>A)2 2c

-ByL3 - ByL3 7PL3 5PL3 + a bd = + a b 48EI 12EI 16EI 2EI

By =

11 P 16

Thus, By =

11 (200) = 138 N c  16

Ans.

As shown on the free-body diagram Ay = 81.3 N c 

Ans.

Cy = 18.8 N T 

Ans.

Ans: By = 138 N c , Ay = 81.3 N c , Cy = 18.8 NT 1269

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*R12–8. w

Using the method of superposition, determine the magnitude of M0 in terms of the distributed load w and dimension a so that the deflection at the center of the beam is zero. EI is constant.

M0

M0 a

a

a

Solution (∆ C)1 =

5wa4 T 384EI

(∆ C)2 = (∆ C)3 =

M0a2 c 16EI

∆ C = 0 = (∆ C)1 + (∆ C)2 + (∆ C)3 + c     0 = M0 =

5wa2 48

M0a2 - 5wa4 + 384EI 8EI Ans.

Ans: M0 = 1270

5wa2 48

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*12–9. Beam ABC is supported by beam DBE and fixed at C. Determine the reactions at B and C. The beams are made of the same material having a modulus of elasticity E = 200 GPa, and the moment of inertia of both beams is I = 25.0(106) mm4.

a

a

A D

B

a 4 ft 2 m

A B a

DE

Equation of Equilibrium. Referring to the free-body diagram of the beam, Fig. a, + ©F = 0; : x

Cx = 0

+ c ©Fy = 0;

By + Cy - 9(6) = 0

a+ ©MC = 0;

9(6)(3) - By(4) - MC = 0

Ans. (1)

MC = 162 - 4By

(2)

Method of superposition: Referring to Fig. b and the table in the appendix, the deflections are By A 63 B 4.5By PLDE 3 = = T 48EI 48EI EI

(vB)1 = =

(vB)2 =

9 A 42 B wx2 A x2 - 4Lx + 6L2 B = C 4 2 - 4(6)(4) + 6 A 62 B D 24EI 24EI 816 kN # m3 T EI By A 4 3 B 21.3333By PLBC 3 c = = 3EI 3EI EI

The compatibility condition at support B requires that

A+TB

vB = (vB)1 + (vB)2 4.5By EI

C

a

=

21.3333By 816 + ab EI EI

By = 31.59 kN = 31.6 kN

Ans.

Substituting the result of By into Eqs. (1) and (2), MC = 35.65 kN # m = 35.7 kN # m

Ans.

Cy = 22.41 kN = 22.4 kN

Ans.

Ans:

Cx = 0, By = 31.6 kN, MC = 35.7 kN # m, Cy = 22.4 kN 1035

C

4m 6 ft AC B 6 in.

Section a – a 3m

Solution

vB =

E

4 ft 6 ft 3 in.

D

a

9100 kN/m lb/ft

Section a – a

E 3m

13–1.  Determine the critical buckling load for the column. The material can be assumed rigid.

P k L 2

L 2

Solution F1 = k(L u);

k

A

L F2 = k a u b 2

L a+ ΣMA = 0;  P(u)(L) - (F1L) - F2 a b = 0 2 L 2 P(u)(L) - kL2 u - k a b u = 0 2

Require:

Pcr = kL +

kL 5kL = 4 4

Ans.

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

1272

Ans: Pcr =

5KL 4

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13–2.  The column consists of a rigid member that is pinned at its bottom and attached to a spring at its top. If the spring is unstretched when the column is in the vertical position, determine the critical load that can be placed on the column.

P k

B

L

Solution

A

a+ ΣMA = 0;  PL sin u - (kL sin u)(L cos u) = 0 P = kL cos u Since u is small  

cos u = 1 Ans.

Pcr = kL

Ans: Pcr = kL 1273

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13–3. The leg in (a) acts as a column and can be modeled (b) by the two pin-connected members that are attached to a torsional spring having a stiffness k (torque兾rad). Determine the critical buckling load. Assume the bone material is rigid.

P

L — 2

Solution a + ©MA = 0;

k

-P(u) a

L b + 2ku = 0 2

L — 2

Require: Pcr =

4k L

Ans. (a)

(b)

Ans: Pcr =

1040

4k L

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*13–4. Rigid bars AB and BC are pin connected at B. If the spring at D has a stiffness k, determine the critical load Pcr that can be applied to the bars.

P A a B a k

Solution Equilibrium: The disturbing force F can be related to P by considering the equilibrium of joint A and then the equilibrium of member BC. Joint A (Fig. b) + c ΣFy = 0;

D a C

FAB cos f - P = 0

FAB =

P cos f

Member BC (Fig. c) ΣMC = 0; F(a cos u) -

P P cos f (2a sin u) sin f(2a cos u) = 0 cos f cos f

F = 2P(tan u + tan f) Since u and f are small, tan u ≅ u and tan f ≅ f. Thus, (1)

F = 2P(u + f) Also, from the geometry shown in Fig. a, 2au = af

f = 2u

Thus Eq. (1) becomes F = 2P(u + 2u) = 6Pu Spring Force: The restoring spring force Fsp can be determined using the spring formula, Fsp = kx, where x = au, Fig. a. Thus, Fsp = kx = kau Critical Buckling Load: When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring spring force Fsp. F = Fsp 6Pcru = kau Pcr =

ka 6

Ans.

Ans: Pcr = 1275

ka 6

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13–5.  A 2014-T6 aluminum alloy column has a length of 6 m and is fixed at one end and pinned at the other. If the cross-sectional area has the dimensions shown, determine the critical load. sY = 250 MPa.

300 mm

10 mm

10 mm 200 mm 10 mm

Solution Section Properties: For the cross-section shown A = 0.3(0.22) - 0.29(0.2) = 8.00 ( 10-3 ) m2 Ix =

1 1 (0.3) ( 0.223 ) (0.29) ( 0.23 ) = 72.8667 ( 10-6 ) m4 12 12

Iy = 2c

1 1 (0.01) ( 0.33 ) d + (0.2) ( 0.013 ) = 45.0167 ( 10-6 ) m4 (controls) 12 12

Critical Buckling Load: K = 0.7 for the column with one end fixed and the other end pinned. For 2014-T6 aluminium alloy, E = 73.1 GPa. Applying Euler’s formula, Pcr =

=

p2EI (KL)2 p2 3 73.1 ( 109 ) 4 3 45.0167 ( 10-6 ) 4 [0.7(6)]2

= 1.8412 ( 106 ) N = 1.84 MN

Ans.

Critical Stress: Euler’s formula valid only if scr 6 sY . For 2014-T6 aluminium alloy, sY = 414 MPa. scr =

1.8412 ( 106 ) Pcr = 230.15 MPa 6 sY . = A 8.00 ( 10-3 )

(O.K!)

Ans: Pcr = 1.84 MN 1276

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13–6.  Solve Prob. 13–5 if the column is pinned at its top and bottom.

300 mm

10 mm

10 mm 200 mm 10 mm

Solution Section Properties: For the cross-section shown A = 0.3(0.22) - 0.29(0.2) = 8.00 ( 10-3 ) m2 Ix =

1 1 (0.3) ( 0.323 ) (0.29) ( 0.23 ) = 72.8667 ( 10-6 ) m4 12 12

Iy = 2c

1 1 (0.01) ( 0.33 ) d + (0.2) ( 0.013 ) = 45.0167 ( 10-6 ) m4 12 12

(controls!)

Critical Buckling Load: K = 1.0 for the column pinned at both ends. For 2014-T6 aluminium alloy, E = 73.1 GPa. Applying Euler’s formula, Pcr =

=

p2EI (KL)2 p2 3 73.1 ( 109 ) 4 3 45.0167 ( 10-6 ) 4 [1.0(6)]2

= 902.17 ( 103 ) N = 902 kN

Ans.

Critical Stress: Euler’s formula valid only if scr 6 sY . For 2014-T6 aluminium alloy, sY = 414 MPa. scr =

902.17 ( 103 ) Pcr = 112.77 MPa 6 sY  = A 8.00 ( 10-3 )

(O.K!)

Ans: Pcr = 902 kN 1277

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13–7. The W360  57 column is made of A-36 steel and is fixed fi xed supported at its base. If it is subjected to an axial load 15kN, kip,determine determinethe thefactor factorof ofsafety safety with with respect to of P == 75 buckling.

P

6 mft 20

Solution From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W360 W14 *38 57are are 6 I y 11.1(10 ) mm 4 11.1(10 −6 ) m 4 = A 7200 mm 2 7.2(10 −3= ) m2 = =

The column is fixed at its base and free at top, k = 2. Here, the column will buckle about the weak axis (y axis). For A36 steel, E = 200 GPa and αY = 250 MPa. Applying Euler’s formula, Pcr =

p2EIy 2

(KL)

=

π 2 [200(109 )][11.1(10 −6 )]

3 = = 152.16(10 ) N 152.16 kN [2(6)]2

Thus, the factor of safety with respect to buckling is F.S =

Pcr 152.16 = = 2.03 75 P

Ans.

The Euler’s formula is valid only if scr 6 sY . scr =

Pcr 152.16(10 3 ) 6 = 21.13(10= ) N−m 2 21.1 MPa < σ Y = A 7.2(10 −3 ) = 250 MPa O.K.

Ans: F.S. = 2.03 1044

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*13–8. The W360  57 column is made of A-36 steel. Determine the critical load if its bottom end is fixed supported and its top is free to move about the strong axis and is pinned about the weak axis.

6m 20 ft

Solution From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W360 W14 *38 57are are = A 7200 = mm 2 7.2(10 −3 ) m 2 6 6 −6 = ) mm 4 11.1(10 −6 ) m 4 I y 11.1(10 = = ) mm 4 160(10= ) m4 I x 160(10

The column is fixed at its base and free at top about strong axis. Thus, kx = 2. For E ==200 GPa3)and 250 A36 steel, E 29.0(10 ksi sand 36 ksi. g s g =MPa. Pcr =

p2EIx (KxLx)

2

=

π 2 [200(109 )][160(10 −6 )]

6 = 2.193(10 = ) N 2.193 MN [2(6)]2

The column is fixed at its base and pinned at top about weak axis. Thus, ky = 0.7. Pcr =

p2EIy 2

(KyLy)

=

π 2 [200(109 )][11.1(10 −6 )] [0.7(6)]2

6 = 1.242(10 = ) N 1.24 MN

(Control)

Ans.

The Euler’s formula is valid only if scr 6 sY . scr =

Pcr 1.242(106 ) 6 == 172.51(10 = ) N−m 2 172.51 MPa < σ Y A 7.2(10 −3 ) = 250 MPa O.K.

Ans: Pcr = 1.24 MN (Control) 1044

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13–9.  A steel column has a length of 9 m and is fixed at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. Est = 200 GPa, sY = 250 MPa.

200 mm

10 mm

10 mm 150 mm 10 mm

Solution Section Properties: A = 0.2(0.17) - 0.19(0.15) = 5.50 ( 10-3 ) m2 Ix =

1 1 (0.2) ( 0.173 ) (0.19) ( 0.153 ) = 28.44583 ( 10-6 ) m4 12 12

Iy = 2c

1 1 (0.01) ( 0.23 ) d + (0.15) ( 0.013 ) = 13.34583 ( 10-6 ) m4 (Controls!) 12 12

Critical Buckling Load: K = 0.5 for fixed support ends column. Applying Euler’s formula, Pcr =

=

p2EI (KL)2 p2 (200) ( 109 ) (13.34583) ( 10-6 ) [0.5(9)]2 Ans.

= 1300919 N = 1.30 MN Critical Stress: Euler’s formula is only valid if scr 6 sY . scr =

Pcr 1300919 = = 236.53 MPa 6 sY = 250 MPa A 5.50 ( 10-3 )

(O.K!)

Ans: Pcr = 1.30 MN 1280

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13–10.  A steel column has a length of 9 m and is pinned at its top and bottom. If the cross-sectional area has the dimensions shown, determine the critical load. Est = 200 GPa, sY = 250 MPa.

200 mm

10 mm

10 mm 150 mm 10 mm

Solution Section Properties: A = 0.2(0.17) - 0.19(0.15) = 5.50 ( 10-3 ) m2 Ix =

1 1 (0.2) ( 0.173 ) (0.19) ( 0.153 ) = 28.44583 ( 10-6 ) m4 12 12

Iy = 2c

1 1 (0.01) ( 0.23 ) d + (0.15) ( 0.013 ) = 13.34583 ( 10-6 ) m4 (Controls!) 12 12

Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula, Pcr =

=

p2EI (KL)2 p2 (200) ( 109 ) (13.34583) ( 10-6 ) [1(9)]2 Ans.

= 325229.87 N = 325 kN Critical Stress: Euler’s formula is only valid if scr 6 sY . scr =

Pcr 325229.87 = = 59.13 MPa 6 sY = 250 MPa A 5.50 ( 10-3 )

(O.K!)

Ans: Pcr = 325 kN 1281

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13–11. The A-36 steel angle has a cross-sectional area of 2 2 A == 1550 mm anda aradius radiusofofgyration gyrationabout aboutthe the xx axis axis of and A 2.48 in and about the the yy axis axisofofry r=y 21.975 rrxx == 31.5 1.26 mm in. and = 0.879mm. in. The smallest radius radius of gyration gyration occurs occurs about about the the zz axis and is smallest rrzz == 16.1 0.644mm. in. If the angle is to be used as a pin-connected 3-m long column, 10-ft-long column, determine determine the the largest largest axial load that can be applied through its centroid C without causing it to buckle.

y z C

x

y

x z

Solution The least radius of gyration: = = mm 0.0161 m πz 16.1

scr =

=

p2E

2 A KL r B

;

controls.

K = 1.0

π 2 [200(109 )]

6 = 56.85(10 = ) N−m 2 56.85 MPa < σ Y = 250 MPa 2  1.0(3)   0.0161 

−3 6 = )] 88.12(10 3= ) N 88.1 kN Pcπ σ= cπA [56.85(10 )][1.55(10=

O.K.

Ans.

Ans: Pcr = 88.1 kN 1045

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  The deck is supported by the two 40-mm-square columns. Column AB is pinned at A and fixed at B, whereas CD is pinned at C and D. If the deck is prevented from sidesway, determine the greatest weight of the load that can be applied without causing the deck to collapse. The center of gravity of the load is located at d = 2 m. Both columns are made from Douglas Fir.

d G A

C

4m

4m 5m

Solution D

a+ ΣMC = 0;    FAB(5) - W(3) = 0

B

FAB = 0.6 W + c ΣFy = 0;  FCD + 0.6 W - W = 0 FCD = 0.4 W Column CD: Pcr =

1 ) (0.04)4 p2(13.1) ( 109 )( 12 p2EI = = 0.4 W (KL)2 (1(4))2

W = 4.31 kN Column AB: Pcr =

1 ) (0.04)4 p2(13.1) ( 109 )( 12 p2EI = = 0.6 W (KL)2 (0.7(4))2

W = 5.86 kN Thus, Ans.

W = 4.31 kN

Ans: W = 4.31 kN 1293

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  The deck is supported by the two 40-mm-square columns. Column AB is pinned at A and fixed at B, whereas CD is pinned at C and D. If the deck is prevented from sidesway, determine the position d of the center of gravity of the load and the load’s greatest magnitude without causing the deck to collapse. Both columns are made from Douglas Fir.

d G A

C

4m

4m 5m

Solution D

Column CD: Pcr =

FCD =

B

p2EI (KL)2 1 ) (0.04)4 p2(13.1) ( 109 )( 12

(1.0(4))2

= 1.7239 ( 103 ) N

Column AB: Pcr =

FAB =

p2EI (KL)2 1 ) (0.04)4 p2(13.1) ( 109 )( 12

(0.7(4))2

= 3.5181 ( 103 ) N

Thus, + c ΣFy = 0;  1.7239 ( 103 ) + 3.5181 ( 103 ) - W = 0 W = 5.2420 ( 103 ) N = 5.24 kN

Ans.

a+ ΣMA = 0;  5.2420 ( 103 ) (d) - 1.7239 ( 103 ) (5) = 0 Ans.

d = 1.64 m

Ans: W = 5.24 kN, d = 1.64 m 1294

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13–14. Determine the maximum force P that can be applied to the handle so that the A-36 steel control rod BC does not buckle. The rod has a diameter of 25 mm.

P 350 mm A 250 mm

45

Solution

C

B

800 mm

Support Reactions: a + ©MA = 0;

P(0.35) - FBC sin 45°(0.25) = 0 FBC = 1.9799P

Section Properties: A =

p A 0.0252 B = 0.15625 A 10 - 3 B p m2 4

I =

p A 0.01254 B = 19.17476 A 10 - 9 B m4 4

Critical Buckling Load: K = 1 for a column with both ends pinned. Appyling Euler’s formula, Pcr = FBC = 1.9799P =

p2EI (KLBC)2

p2(200)(109) C 19.17476(10 - 9) D [1(0.8)]2

Ans.

P = 29 870 N = 29.9 kN Critical Stress: Euler’s formula is only valid if scr 6 sg. scr =

1.9799(29 870) Pcr = 120.5 MPa 6 sg = 250 MPa = A 0.15625(10 - 3)p

O.K.

Ans: P = 29.9 kN 1049

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  Determine the maximum load P the frame can support without buckling member AB. Assume that AB is made of steel and is pinned at its ends for y–y axis buckling and fixed at its ends for x–x axis buckling. Est = 200 GPa, sY = 360 MPa.

3m

A

P

y 50 mm x 50 mm

x

4m

50 mm 6m

y

C

Solution

B

3 + ΣFx = 0;   - FAC a b + P = 0 S 5 FAC = + c ΣFy = 0;  FAB -

5 P 3

5 4 Pa b = 0 3 5

FAB =

4 P 3

Iy =

1 (0.10)(0.05)3 = 1.04167 ( 10-6 ) m4 12

Ix =

1 (0.05)(0.10)3 = 4.16667 ( 10-6 ) m4 12

Pcr =

p2 EI (KL)2

x–x axis buckling: Pcr =

p2(200) ( 109 ) (4.16667) ( 10-6 ) (0.5(6))2

= 914 kN

y–y axis buckling: Pcr =

p2(200) ( 109 ) (1.04167) ( 10-6 ) (1(6))2

= 57.12 kN

y–y axis buckling controls 4 P = 57.12 3 Ans.

P = 42.8 kN Check: scr =

57.12 ( 103 ) P = = 11.4 MPa 6 sY  A (0.1)(0.05)

OK

Ans: P = 42.8 kN 1284

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laced together to *13–16. The two two steel steelchannels channelsare aretotobebe laced together form a 9-m bridge bridge columncolumn assumedassumed to be pin connected to form a long 30-ft-long to be pin at its ends.atEach channel a cross-sectional area of connected its ends. Each has channel has a cross-sectional 2 6 2 A = 1950 andinmoments of inertia Ix = 21.60(10 ) mm area of Amm and moments of inertia = 3.10 in44, Ix = 55.4 64 4 centroid C of areaisis located located in The. The centroid C of itsitsarea 0.382 in) .mm IIyy == 0.15(10 the figure. Determine the proper distance d between the centroids of the channels so that buckling occurs about the x–x and y¿ –y¿ axes due to the same load. What is the value of this critical load? Neglect the effect of the lacing. 3 Est == 200 GPa, sY s=Y350 MPa. E 29110 2 ksi, = 50 ksi.

yy

y¿ y¿

0.269 in. 6.5 mm

1.231 30 mmin.

xx

C C

C C

xx

d d yy

y¿ y¿

Solution Ix

−6 2[21.60(10 = ) m 4 ] 43.2(10 −6 ) m 4

2[0.15(10 −6 ) + 1.95(10 −3 )(d 2)2 ] = 0.3(10 −6 ) + 0.975(10 −3 )d 2 Iy = In order for the column to buckle about x - x and y - y at the same time, Iy must be equal to Ix Iy = Ix 0.3(10 −6 ) + 0.975(10 −3 )d 2 = 43.2(10 −6 ) = d 0.20976 = m 210 mm

Ans.

Check: d 7 2(1.231) 2.462 in. 2(30) = 60= mm Pcr =

O.K.

p2 EI π 2 [200(109 )][43.2(10 −6 )] = 2 (KL) [1(9)]2

6 = 1.053(10 = ) N 1.05 MN

Ans.

Check stress: P A

cπ σ= cπ =

1.053(106 ) 6 = 269.94(10= ) N−m 2 270 MPa < σ Y 2[1.95(10 −3 )]

Therefore, Euler’s formula is valid.

Ans: Pcr = 1.05 MN 1046

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13–17. The W250 W10 * 45 67 is made of A992 steel and is used a length 15 m. ft. If its ends are assumed as a column that has length ofof 4.55 pin supported, and it is subjected to an axial load of 100 500 kip, kN, determine the factor of safety with respect to buckling.

P

4.55 15 ftm

Solution Critical Buckling Load: 6 = = ) mm 4 22.2(10 −6 ) m 4 foπ a W250 × 67 wide flange section and I y 22.2(10 K = 1 for pin supported ends column. Applying Euler’s formula,

Pcr

P

p2EI = (KL)2

=

π 2 [200(109 )][22.2(10 −6 )] [1(4.55)]2

3 = 2116.70(10 = ) N 2116.70 kN

Critical Stress: Euler’s formula is only valid if scr 6 sY. A = 8560 mm2 = 8.56(10−3) m2 for the W250 W10 *45 67wide-flange wide-flangesection. section.

σ= cπ

Pcπ 2116.70(10 3 ) = = 247.28(106 ) N−m 2 A 8.56(10 −3 )

= 247.28 MPa < σ Y = 345 MPa

O.K.

Factor of Safety: = F.S.

Pcπ 2116.70 = = 4.23 A 500

Ans.

Ans: F.S. = 4.23 1050

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13–18. The W250 * 67 is made of A-36 steel and is used as a column that has length a length 15 m. ft. If the ends of the ofof 4.55 column are fixed fixed supported, can the column support the critical load without yielding?

P P

15 ftm 4.55

Solution Critical Buckling Load: 6 = I y 22.2(10 = ) mm 4 22.2(10 −6 ) m 4 foπ a W250 × 67 wide flange section and K = 0.5 for fixed ends support column. Applying Euler’s formula,

Pcr = =

P P

p2EI (KL)2

π 2 [200(109 )][22.2(10 −6 )] [0.5(4.55)]2

6 = ) N 8.47 MN = 8.4668(10

Critical Stress: Euler’s formula is only valid if scr 6 sY. A = 8560 mm2 = 8.56(10−3) m2 for the W250  67 wide-flange section. P 8.4668(106 ) = 989.11(106 ) N−m 2 σ cπ = cπ = A 8.56(10 −3 ) = 989 MPa < σ Y = 345 MPa (No!)

Ans.

The column will yield before the axial force achieves the critical load Pcr and so Euler’s formula is not valid.

1050

Ans: scr = 345 MPa 1No!2

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13–19. The 50-mm-diameter C86100 bronze rod is fixed supported at A and has a gap of 2 mm from the wall at B. Determine the increase in temperature ΔT that will cause the rod to buckle. Assume that the contact at B acts as a pin.

A

B 1m 2 mm

Solution Section Properties: p ( 0.052 ) = 0.625p ( 10-3 ) m2 4

A = I =

p ( 0.0254 ) = 97.65625p ( 10-9 ) m4 4

Compatibility Condition: This requires

+) (S

0.002 = dT + dF 0.002 = 17 ( 10-6 ) (∆T)(1) -

F(1) 0.625p ( 10-3 ) (103) ( 109 )

F = 3438.08∆T - 404480.05 Critical Buckling Load: K = 0.7 for a column with one end fixed and the other end pinned. Applying Euler’s formula, Pcr = F =

3438.08∆T - 404 480.05 =

p2EI (KL)2 p2 (103) ( 109 ) 3 97.65625p ( 10-9 ) 4 [0.7(1)]2

∆T = 302.78°C = 303°C

Ans.

Critical Stress: Euler’s formula is only valid if scr 6 sY . Pcr = 3438.08(302.78) - 404 480.05 = 636 488.86 N scr =

Pcr 636 488.86 = 324.2 MPa 6 sY = 345 MPa = A 0.625p ( 10-3 )

(O.K!)

Ans: ∆T = 303°C 1283

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*13–20. An A992 steel W200 * 46 column of length 9 m is fixed at one end and free at its other end. Determine the allowable axial load the column can support if F.S. = 2 against buckling.

Solution Section Properties: From the table listed in the appendix, the cross-sectional area and moment of inertia about the y axis for a W200 * 46 are A = 5890 mm2 = 5.89(10 - 3) m2 Iy = 15.3(106) mm4 = 15.3(10 - 6) m4 Critical Buckling Load: The column will buckle about the weak (y) axis. Applying Euler’s formula, Pcr =

p2 E Iy (KL)2

=

p2[200(109)][15.3(10 - 6)] [2(9)]2

= 93.21 kN

Thus, the allowable centric load is Pallow =

Pcr 93.21 = = 46.61 kN = 46.6 kN F.S. 2

Ans.

Critical Stress: Euler’s formula is valid only if scr 6 sY . scr =

93.21(103) Pcr = = 15.83 MPa 6 sY = 345 MPa A 5.89(10 - 3)

(O.K.)

Ans: Pallow = 46.6 kN 1287

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13–21. The 3-m wooden rectangular column has the dimensions shown. Determine loadload if theif ends dimension Determinethe thecritical critical the are assumed to be connected. ksi, Ew E=w 1.6110 ends are assumed to pin be pin connected. = 12 32GPa, sYY ==35 5 ksi. s MPa.

10 ft 3m in. 1004mm

Solution

2 in. 50 mm

Section Properties:

A 0.1(0.05) = 0.005 m 2 = Ix =

1 = (0.05)(0.13 ) 4.1667(10 −6 ) m 4 12

Iy =

1 = (0.1)(0.053 ) 1.04167(10 −6 ) m 4 (Controls!) 12

Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s formula,. Pcr =

=

p2EI (KL)2

π 2 [12(109 )][1.04167(10 −6 )] [1(3)]2

3 = ) N 13.7 kN = 13.71(10

Ans.

Critical Stress: Euler’s formula is only valid if scr 6 sY .

σ= cπ

Pcπ 13.71(10 3 ) 2 = 2.74 MPa < σ Y = 35 MPa = = 2.742(106 ) N−m A 0.005

O.K.

Ans: A = 5000 mm2, Ix = 4.1667 (106) mm4, Iy = 1.041667 (106) mm4, Pcr = 13.7 kN 1048

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13–22. The 3-m column has the dimensions shown. Determine the critical load if the bottom is fixed and the top is pinned. Ew == 12 1.6110 = 5 ksi. GPa,32sksi, 35Y MPa. Y =s

10 ft 3m

Solution

in. 1004mm

Section Properties:

2 in. 50 mm

A 0.1(0.05) = 0.005 m 2 = Ix =

1 = (0.05)(0.13 ) 4.1667(10 −6 ) m 4 12

Iy =

1 = (0.1)(0.053 ) 1.04167(10 −6 ) m 4 (Controls!) 12

Critical Buckling Load: K = 0.7 for column with one end fixed and the other end pinned. Applying Euler’s formula. Pcr =

=

p2EI (KL)2

π 2 [12(109 )][1.04167(10 −6 )] [0.7(3)]2

3 = ) N 28.0 kN = 27.98(10

Ans.

Critical Stress: Euler’s formula is only valid if scr 6 sY .

σ= cπ

Pcπ 27.98(10 3 ) 2 = 5.60 MPa < σ Y = 35 MPa = = 5.595(106 ) N−m A 0.005

O.K.

Ans: Pcr = 28.0 kN 1048

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A

13–23. If load C has a mass of 500 kg, determine the required minimum diameter of the solid L2-steel rod AB to the nearest mm so that it will not buckle. Use F.S. = 2 against buckling. 45° 4m

D

Solution Equilibriun. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - 500(9.81) cos 45° = 0

FAB = 13 400.71 N

60°

C

B

Section Properties. The cross-sectional area and moment of inertia of the solid rod are A =

p 2 d 4

I =

p d 4 p 4 a b = d 4 2 64

Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 13400.71(2) = 26801.42 N. Applying Euler’s formula,

Pcr =

p2EIy (KL)2

26801.42 =

p2 C 200 A 109 B D c

p 4 d d 64

[1(4)]2

d = 0.04587 m = 45.87 mm Use d = 46 mm

Ans.

Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =

Pcr 26801.42 = = 16.13 MPa 6 sY = 703 MPa p A A 0.0462 B 4

O.K.

Ans: Use d = 46 mm 1061

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*13–24. If the diameter of the solid L2-steel rod AB is 50 mm, determine the maximum mass C that the rod can support without buckling. Use F.S. = 2 against buckling.

A

45° 4m

D

Solution Equilibrium. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. ©Fy¿ = 0; FAB sin 15° - m(9.81) cos 45° = 0

FAB = 26.8014m

60°

C

B

Section Properties. The cross-sectional area and moment of inertia of the rod are A = I =

p A 0.052 B = 0.625 A 10 - 3 B pm2 4 p A 0.0254 B = 97.65625 A 10 - 9 B pm4 4

Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 26.8014m(2) = 53.6028m. Applying Euler’s formula, Pcr =

p2EIy (KL)2

53.6028m =

p2 c 200 A 109 B d c 97.65625 A 10 - 9 B p d [1(4)]2

m = 706.11 kg = 7.06 kg

Ans.

Critical Stress. Euler’s formula is valid only if scr 6 sY. scr =

53.6028(706.11) Pcr = = 19.28 MPa 6 sY = 703 MPa A p 0.625 A 10 - 3 B

O.K.

Ans: m = 7.06 kg 1062

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H

13–25. The members of the truss are assumed to be pin connected. If member GF is an A-36 steel rod having a diameter of 50 mm, determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle.

G

F

E

12 ft 3m A 16 ft 4m

Solution

16 ft 4m P P

D

C

B

16 ft 4m P P

Support Reactions: As shown on FBD(a). Member Forces: Use the method of sections [FBD(b)]. + ©MB = 0;

FGF (12) P(16) (3) – P(4) =0 = 0

FGF = 1.3333P (C)

Section Properties:

π

= (0.052 ) 0.625(10 −3 )π m 2 4

= A = I

π

= (0.0254 ) 97.65625(10 −9 )π m 4 4

Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Euler’s formula, Pcr = FGF = 1.3333P =

p2EI (KLGF)2

π 2 [200(109 )][97.65625(10 −9 )π ] [1(4)]2

3 P 28.39(10 ) N 28.4 kN = =

Ans.

Critical Stress: Euler’s formula is only valid if scr 6 sY .

σ= cπ

Pcπ 1.3333[28.39(10 3 )] = = 19.28(106 ) N−m 2 A (0.625)(10 −3 )π

= 19.3 MPa m

Ans.

(controls)

y -y axis K = 0.5, 4.0w =

L = 2m

p2 (200)(109)(20)(10 - 9) [(0.5)(2)]2

w = 9870 N>m = 9.87 kN>m Check: scr =

4(5552) Pcr = = 37.0 MPa 6 sg A (0.02)(0.03)

O.K.

Ans: w = 5.55 kN>m 1054

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*13–28. Determine if the frame can support a load of w = 6 kN>m if the factor of safety with respect to buckling of member AB is 3. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Est = 200 GPa, sY = 360 MPa.

w

C

B

1.5 m

0.5 m

2m 30 mm x

Solution

20 mm y

y

Check x- x axis buckling: Ix =

1 (0.02)(0.03)3 = 45.0(10 - 9) m4 12

K = 1.0 Pcr =

x

30 mm

A

L = 2m

p2(200)(109)(45.0)(10 - 9) p2EI = (KL)2 ((1.0)(2))2

Pcr = 22.2 kN a + ©MC = 0;

FAB(1.5) - 6(2)(1) = 0 FAB = 8 kN

Preq’d = 8(3) = 24 kN 7 22.2 kN Ans.

No, AB will fail.

Ans: No, AB will fail. 1055

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  A 6061-T6 aluminum alloy solid circular rod of length 4 m is pinned at both of its ends. If it is subjected to an axial load of 15 kN and F.S. = 2 against buckling, determine the minimum required diameter of the rod to the nearest mm.

Solution Section Properties: The cross-sectional area and moment of inertia of the solid rod are p 2 p d 4 p 4 d      I = a b = d 4 4 2 64

A =

Critical Buckling Load: The critical buckling load is Pcr = Pallow (F.S.) = 15(2) = 30 kN Applying Euler’s formula, Pcr =

p2EIy (KL)2

30(103) =

p2 c 68.9(109) d c [1(4)]2

p 4 d d 64

d = 0.06158 m = 61.58 mm Ans.

Use d = 62 mm Critical Stress: Euler’s formula is valid only if scr 6 sY . scr =

30(103) Pcr = = 9.94 MPa 6 sY = 255 MPa p A (0.0622) 4

(O.K.)

Ans: Use d = 62 mm 1305

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  A 6061-T6 aluminum alloy solid circular rod of length 4 m is pinned at one end while fixed at the other end. If it is subjected to an axial load of 15 kN and F.S. = 2 against buckling, determine the minimum required diameter of the rod to the nearest mm.

Solution Section Properties: The cross-sectional area and moment of inertia of the solid rod are p 2 p d 4 p 4 d      I = a b = d 4 4 2 64

A =

Critical Buckling Load: The critical buckling load is Pcr = Pallow (F.S.) = 15(2) = 30 kN Applying Euler’s formula, Pcr =

p2EIy (KL)2

30(103) =

p2 c 68.9(109) d c

[0.7(4)]2

p 4 d d 64

d = 0.05152 m = 51.52 mm Ans.

Use d = 52 mm Critical Stress: Euler’s formula is valid only if scr 6 sY . scr =

30(103) Pcr = = 14.13 MPa 6 sY = 255 MPa p A (0.0522) 4

(O.K.)

Ans: Use d = 52 mm 1306

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13–31. The A-36 steel bar AB has a square cross section. If it is pin connected at its ends, determine the maximum allowable load P that can be applied to the frame. Use a factor of safety with respect to buckling of 2.

C C

A A

Solution a + ©MA = 0;

FBC sin 30°(10) P(10) (3) – P(3) =0 = 0

30 B B

40 1.5mm in. 401.5 mm in.

3m 10 ft P P

FBC = 2 P + : ©Fx = 0;

40 1.5mm in.

FA - 2P cos 30° = 0 FA = 1.732 P

Buckling load: Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P

L=3m Iy =

3m

1 = (0.04)(0.04 3 ) 0.21333(10 −6 ) m 4 12

Pcr =

p2 EI (KL)2

0.464 P =

π 2 [200(109 )][0.21333(10 −6 )] [1(3)]2 Ans.

3 = P 13.51(10 = ) N 13.5 kN

N 46.79 kN = Pcr = FA(F.S.) = 1.732(13.51)(2) Check:

σ= cπ

Pcπ 46.79(10 3 ) 2 = 29.24 MPa < σ Y = = 29.24(106 ) N−m A (0.04)(0.04)

O.K.

Ans: P = 13.5 kN 1053

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  Determine the maximum allowable load P that can be applied to member BC without causing member AB to buckle. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Use a factor of safety with respect to buckling of F.S. = 3. Est = 200 GPa, sY = 360 MPa.

P 1m

1m B

C

2m

30 mm A

Solution

y

x 20 mm y x 30 mm

Support Reactions: a+ ΣMC = 0;   FAB (2) - P(1) = 0   FAB = 0.500P Section Properties: A = 0.02(0.03) = 0.600 ( 10-3 ) m2 1 (0.02) ( 0.033 ) = 45.0 ( 10-9 ) m4 12 1 Iy = (0.03) ( 0.022 ) = 20.0 ( 10-9 ) m4 12 Ix =

Critical Buckling Load: With respect to x–x axis, K = 1 (column with both ends pinned). Applying Euler’s formula, Pcr = 3FAB = 3(0.500P) =

p2EI (KL)2 p2(200) ( 109 ) 3 45.0 ( 10-9 ) 4 [1(2)]2

Ans.

P = 14804.4 N = 14.8 kN (Controls!) With respect to y–y axis, K = 0.5 (column with both ends fixed). Pcr = 3FAB = 3(0.500P) =

p2EI (KL)2 p2(200) ( 109 ) 3 20.0 ( 10-9 ) 4 [0.5(2)]2

P = 26 318.9 N

Critical Stress: Euler’s formula is only valid if scr 6 sY. scr =

3(0.5)(14 804.4) Pcr = = 37.01 MPa 6 sY = 360 MPa (O.K!) A 0.600 ( 10-3 )

Ans: P = 14.8 kN 1304

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13–33. Determine if the frame can support a load of P = 20 kN if the factor of safety with respect to buckling of member AB is F.S. = 3. Assume that AB is made of steel and is pinned at its ends for x–x axis buckling and fixed at its ends for y–y axis buckling. Est = 200 GPa, sY = 360 MPa.

P 1m

1m B

C

2m

30 mm A

Solution

y

x 20 mm y x 30 mm

Support Reactions: a+ ΣMC = 0;   FAB(2) - 20(1) = 0   FAB = 10.0 kN Section Properties: A = 0.02(0.03) = 0.600 ( 10-3 ) m2 1 (0.02) ( 0.033 ) = 45.0 ( 10-9 ) m4 12 1 Iy = (0.03) ( 0.022 ) = 20.0 ( 10-9 ) m4 12 Ix =

Critical Buckling Load: With respect to x–x axis, K = 1 (column with both ends pinned). Applying Euler’s formula, Pcr = =

p2EI (KL)2 p2(200) ( 109 ) 3 45.0 ( 10-9 ) 4 [1(2)]2

= 22 206.61 = 22.207 kN (Controls!) With respect to y–y axis, K = 0.5 (column with both ends fixed). Pcr = =

p2EI (KL)2 p2(200) ( 109 ) 3 20.0 ( 10-9 ) 4 [0.5(2)]2

= 39 478.42 N

Critical Stress: Euler’s formula is only valid if scr 6 sY. scr =

Pcr 32 127.6 = = 6.426 MPa 6 sY = 360 MPa A 5.00 ( 10-3 )

(O.K!)

Factor of Safety: The required factor of safety is 3. F.S =

Pcr 22.207 = = 2.22 6 3 (No Good!) FAB 10.0

Hence, the frame cannot support the load with the required F.S.

Ans.

Ans: No 1303

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  w

The steel bar AB has a rectangular cross section. If it is pin connected at its ends, determine the maximum allowable intensity w of the distributed load that can be applied to BC without causing AB to buckle. Use a factor of safety with respect to buckling of 1.5. Est = 200 GPa, sY = 360 MPa.

B

C 5m y

3m 20 mm A

x

30 mm x

y 20 mm

Solution Buckling Load: Pcr = FAB (F.S.) = 2.5 w(1.5) = 3.75 w 1 (0.03)(0.02)3 = 20 (10 - 9) m4 12 K = 1.0 I =

Pcr =

p2 E I (K L)2

3.75 w =

p2 (200)(109)(20)(10 - 9) [(1.0)(3)]2 Ans.

w = 1170 N>m = 1.17 kN>m Pcr = 4.39 kN Check: scr =

4.39 (103) Pcr = = 7.31 MPa 6 sY A 0.02 (0.03)

OK

Ans: w = 1.17 kN>m 1302

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45 is used as a structural A-36 steel W14  * 30 The W360 13–35. The column that can be assumed pinned at both of its ends. Determine the largest axial force P that can be applied without causing it to buckle.

P

Solution

7.5 m 25 ft

From the table in appendix, the cross-sectional area and the moment of inertia about weak axis (y-axis) for W360 W14 * 30 45 are are 6 = A 5710 = mm 2 5.71(10 −3= ) m2 I y 8.16(10 = ) mm 4 8.16(10 −6 ) m 4

Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For A36 steel, E = 200 GPa and sY = 250 MPa. Here, the buckling occurs about the weak axis (y-axis). p2EIy π 2 [200(109 )][8.16(10 −6 )] = P = P= cr [1(7.5)]2 (KL)2 3 = ) N 286 kN = 286.35(10

Ans.

Euler’s formula is valid only if scr 6 sY.

σ= cπ

Pcπ 286.35(10 3 ) 2 = = 50.14(106 ) N−m = 50.1 MPa < σ Y = 250 MPa A 5.71(10 −3 )

O.K.

Ans: A = 5710 mm2, Iy = 8.16 (106) mm4, P = 286 kN 1053

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P == 30 kN. As a 6 kip. The beam supports the load of P result, the A-36 steel member BC is subjected to a compressive load. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling. Determine the factor of safety with respect to buckling about each of these axes. *13–36.

P P 4 ftm 1.2

4 ftm 1.2

A 3 ftm 0.9

B

C x

y

Solution a + ©MA = 0;

y

375in. mm

125in. mm

x

3 FBC a b (1.2) (4) -–6000(8) 30(2.4) == 00 5 kN FBC = 100 20 kip

x- x axis buckling: Pcr = = F.S.

1 (0.025)(0.0753 ) π 2 [200(109 )]  12 p2EI  771.06(10 3 = = ) N 771.06 kN = (KL)2 [1.0(1.5)]2

771.06 = 7.71 100

1.2 m

1.2 m

Ans.

y - y axis buckling: Pcr

1 (0.075)(0.0253 ) π 2 [200(109 )]  12 p2EI  342.69(10 3 = = ) N 342.69 kN = = (KL)2 [0.5(1.5)]2

= F.S.

342.69 = 3.43 100

Ans.

Ans: x-x axis buckling: F.S. = 7.71, y-y axis buckling: F.S. = 3.43 1055

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13–37. Determine the greatest load P the frame will support without causing the A-36 steel member BC to buckle. Due to the forked ends on the member, consider the supports at B and C to act as pins for x–x axis buckling and as fixed supports for y–y axis buckling.

P P 4 ftm 1.2

A 3 ftm 0.9

B

a + © MA = 0;

y

3 in. 75 mm C x

Solution

4 ftm 1.2

y

1 in. 25 mm

x

 3 FBC   (1.2) − P(2.4) = 0  5 10 a bPP FBC = 3.33 3

x -x axis buckling: Pcr

1 (0.025)(0.0753 ) π 2 [200(109 )]  12 p2EI  771.06(10 3 = ) N 771.06 kN = = = (KL)2 [1.0(1.5)]2

1.2 m

1.2 m

y -y axis buckling: Pcr

1 (0.075)(0.0253 ) π 2 [200(109 )]  12 p2EI  342.69(10 3 = ) N 342.69 kN = = = (KL)2 [0.5(1.5)]2

Thus,

 10    P = 342.69  3  Ans.

= = kN 103 kN P 102.81

Ans: P = 103 kN 1056

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13–38. The members of the truss are assumed to be pin connected. If member AB is an A-36 steel rod of 40 mm diameter, determine the maximum force P that can be supported by the truss without causing the member to buckle.

2m

C

E

D 1.5 m B

A 2m

Solution

P

By inspecting the equilibrium of joint E, FAB = 0. Then, the compressive force developed in member AB can be determined by analysing the equilibrium of joint A, Fig. a. + c ©Fy = 0;

3 FAC a b - P = 0 5

+ : ©Fx = 0;

5 4 P a b - FAB = 0 3 5

A = p(0.022) = 0.4(10 - 3)p m2

I =

FAC =

5 P (T) 3

FAB =

4 P(c) 3

p (0.024) = 40(10 - 9) p m4 4

Since both ends of member AB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr =

p2EI ; (KL)2

p2 C 200(109) D C 40(10 - 9)p D 4 P = 3 C 1(2) D 2

Ans.

P = 46.51(103) N = 46.5 kN

The Euler’s formula is valid only if scr 6 sg.

scr

4 (46.51)(103) Pcr 3 = 49.35(106) Pa = 49.35 MPa 6 sg = 250 MPa O.K. = = A 0.4(10 - 3)p

Ans: P = 46.5 kN 1059

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13–39. The members of the truss are assumed to be pin connected. If member CB is an A-36 steel rod of 40 mm diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle.

2m

C

E

D 1.5 m B

A 2m

Solution

P

Section the truss through a–a, the FBD of the left cut segment is shown in Fig. a. The compressive force developed in member CB can be obtained directly by writing the force equation of equilibrium along y axis. + c ©Fy = 0;

FCB - P = 0

A = p(0.022) = 0.4(10 - 3)p m2

FCB = P (C) I =

p (0.024) = 40(10 - 9)p m4 4

Since both ends of member CB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr =

p2EI ; (KL)2

P =

p2 C 200(109) D C 40(10 - 9)p D

C 1(1.5) D 2

Ans.

= 110.24(103) N = 110 kN The Euler’s formula is valid only if scr 6 sg. scr =

110.24(103) Pcr = 87.73(106) Pa = 87.73 MPa 6 sg = 250 MPa = A 0.4(10 - 3)p

O.K.

Ans: P = 110 kN 1060

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*13–40. 3m

The steel bar AB of the frame is assumed to be pin connected at its ends for y–y axis buckling. If P = 18 kN, determine the factor of safety with respect to buckling about the y–y axis. Est = 200 GPa, sY = 360 MPa.

A

P

y

x 50 mm

4m 6m

y

50 mm x 50 mm

C B

Solution Iy =

1 (0.10)(0.053) = 1.04167 (10 - 6) m4 12

Joint A: + ΣFx = 0; d

3 F - 18 = 0 5 AC FAC = 30 kN

+ c ΣFy = 0;

FAB -

4 (30) = 0 5

FAB = 24 kN Pcr =

p2(200)(109)(1.04167)(10 - 6) p2E I = = 57116 N = 57.12 kN (K L)2 [(1.0)(6)]2

F.S. =

Pcr 57.12 = = 2.38 FAB 24

Ans.

Check: scr =

57.12 (103) Pcr = = 11.4 MPa 6 sY A 0.1 (0.05)

OK

Ans: F.S. = 2.38 1311

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13–41.  w

The ideal column has a weight w (force>length) and is subjected to the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1, with the origin at the midspan. The general solution is v = C1 sin kx + C2 cos kx +  (w>(2P))x2 - (wL>(2P)) x - (wEI>P 2 ) where k 2 = P>EI.

P L

Solution Moment Functions: FBD(b). x wL wx a b - M(x) - a bx - Pv = 0 2 2

a+ ΣMo = 0; M(x) =

w 2 1 x - Lx 2 - Pv 2

(1)

Differential Equation of The Elastic Curve: EI EI

d 2v = M(x) dx2 d 2v w 2 = 1 x - Lx 2 - Pv 2 2 dx

d 2v P w + v = 1 x2 - Lx 2 2 EI 2EI dx

The solution of the above differential equation is of the form v = C1 sin a

P w 2 wL wEI P xb + x x xb + C2 cos a A EI 2P 2P A EI P2

(2)

and

w wL dv P P P P x≤ - C2 sin ¢ x≤ + x= C1 cos ¢ A EI A EI A EI P 2P dx A EI

The integration constants can be determined from the boundary conditions. Boundary Condition: At x = 0, v = 0. From Eq. (2), 0 = C2 -

At x = 0 = C1

wEI P2

C2 =

wEI P2

L dv , = 0. From Eq. (3), 2 dx P P L wEI P P L w L wL cos ¢ ≤ sin ¢ ≤ + a b A EI A EI 2 A EI 2 P 2 2P P2 A EI C1 =

wEI P L tan ¢ ≤ A EI 2 P2

1312

(3)

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13–41.  Continued

Solution Elastic Curve: v =

w EI EI x2 L EI P L P P J tan ¢ ≤ sin ¢ x≤ + cos ¢ x≤ + - x R P P A EI P A EI 2 2 P A EI 2

However, v = vmax at x = vmax = =

L . Then, 2

w EI EI L2 EI P L P L P L ≤ sin ¢ ≤ + ≤ R J tan¢ cos ¢ P P P 8 P A EI 2 A EI 2 A EI 2

wEI PL2 P L Jsec ¢ ≤ - 1R 2 A EI 2 8EI P

Maximum Moment: The maximum moment occurs at x = Mmax =

L . From Eq. (1), 2

w L2 L J - La b R - Pvmax 2 4 2

= = -

wL2 wEI PL2 P L - P b 2 Jsec¢ ≤ - 1R r 8 A EI 2 8EI P

wEI P L Jsec ¢ ≤ - 1R P A EI 2

Ans.

Ans: Mmax = 1313

wEI L P c sec a b - 1d P 2 B EI

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13–42.  The ideal column is subjected to the force F at its midpoint and the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 13–1. The  general solution is v = C1 sin kx + C2 cos kx - c 2x>k 2, where c 2 = F>2EI, k 2 = P>EI.

F P L 2

Solution Moment Functions: FBD(b). c + ΣMo = 0;

M(x) +

F x + P(v) = 0 2 M(x) = -

F x - Pv 2

(1)

Differential Equation of The Elastic Curve: EI

d 2v = M(x) dx2

EI

d 2v F = - x - Pv 2 dx2

d 2v P F + v = x EI 2EI dx2 The solution of the above differential equation is of the form v = C1 sin a

and

P P F xb + C2 cos ¢ xb x A EI A EI 2P

dv P P P P F = C1 cos ¢ x≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI 2P

The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, v = 0. From Eq. (2), C2 = 0 At x =

L dv , = 0. From Eq. (3), 2 dx 0 = C1 C1 =

Elastic Curve: v = =

F P P L cos ¢ ≤ A EI A EI 2 2P

F EI P L sec¢ ≤ 2P A P A EI 2

F F EI P L P sec¢ ≤ sin¢ x≤ x 2P A P A EI 2 A EI 2P F EI P L P J sec¢ ≤ sin¢ x≤ - x R 2P A P A EI 2 A EI

1314

(2)

(3)

L 2

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13–42.  Continued

Solution However, v = vmax at x = vmax = =

L . Then, 2

F L EI P L P L J sec¢ ≤ sin¢ ≤ - R 2P A P A EI 2 A EI 2 2 F L EI P L J tan¢ ≤ - R 2P A P A EI 2 2

Maximum Moment: The maximum moment occurs at x = Mmax = = -

L . From Eq. (1), 2

F L a b - Pvmax 2 2

FL F L EI P L - Pb J tan¢ ≤ - Rr 4 2P A P A EI 2 2

= -

F EI P L tan¢ ≤ 2 AP A EI 2

Ans.

Ans: Mmax = 1315

F EI L P tan a b 2 BP 2 B EI

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13–43.  P

The column with constant EI has the end constraints shown. Determine the critical load for the column.

L

Solution Moment Function. Referring to the free-body diagram of the upper part of the deflected column, Fig. a, a + ΣMO = 0;

M + Pv = 0

M = - Pv

Differential Equation of the Elastic Curve. EI

d 2v = M dx2

EI

d 2v = - Pv dx2

d 2v P + v = 0 EI dx2 The solution is in the form of v = C1 sin a

P P xb + C2 cos ¢ xb A EI A EI

(1)

dv P P P P = C1 cos ¢ x≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI

(2)

Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives 0 = 0 + C2 At x = L,

C2 = 0

dv = 0. Then Eq. (2) gives dx 0 = C1

P P cos ¢ L≤ A EI A EI

C1 = 0 is the trivial solution, where v = 0. This means that the column will remain straight and buckling will not occur regardless of the load P. Another possible solution is cos ¢

P L≤ = 0 A EI

np P L = A EI 2

n = 1, 3, 5

The smallest critical load occurs when n = 1, then p Pcr L = A EI 2 Pcr =

p2EI 4L2

Ans. Ans: Pcr = 1316

p2EI 4L2

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*13–44. Consider an ideal column as in Fig. 13–10c, having both ends fixed. Show that the critical load on the column is Pcr = 4p2EI>L2. Hint: Due to the vertical deflection of the top of the column, a constant moment M′ will be developed at the supports. Show that d 2v>dx2 + (P>EI)v = M′>EI. The solution is of the form v = C1 sin( 1P>EIx) + C2 cos( 1P>EIx) + M′>P.

Solution Moment Functions: M(x) = M′ - Pv Differential Equation of The Elastic Curve: EI EI

d 2v = M(x) dx2

d 2v = M′ - Pv dx2

d 2v P M′ + v = EI EI dx2

(Q.E.D.)

The solution of the above differential equation is of the form v = C1 sin a

and

P P M′ xb + C2 cos ¢ xb + A EI A EI P

(1)

P P P P dv x≤ cos ¢ x≤ - C2 sin ¢ = C1 A EI A EI A EI A EI dx

(2)

The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, v = 0. From Eq. (1), C2 = At x = 0,

M′ P

dv = 0. From Eq. (2), C1 = 0 dx

Elastic Curve: v = and

M′ P J1 - cos ¢ x≤ R P A EI

dv M′ P P = sin¢ x≤ dx P A EI A EI However, due to symmetry sinJ

dv L = 0 at x = . Then, dx 2

P L a bR = 0 A EI 2

or

P L a b = np A EI 2

where n = 1, 2, 3,...

The smallest critical load occurs when n = 1. Pcr =

4p2EI L2

(Q.E.D.)

1317

Ans: N/A

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13–45.  Consider an ideal column as in Fig. 13–10d, having one end fixed and the other pinned. Show that the critical load on the column is Pcr = 20.19EI>L2. Hint: Due to the vertical deflection at the top of the column, a constant moment M′ will be developed at the fixed support and horizontal reactive forces R′ will be developed at both supports. Show that d 2v>dx2 + (P>EI)v = (R′>EI)(L - x). The solution is of the form v = C1sin (1P>EIx) + C2 cos (1P>EIx) + (R′>P)(L - x). After application of the boundary conditions show that tan (1P>EIL) = 1P>EI L. Solve numerically for the smallest nonzero root.

Solution Equilibrium. FBD(a). Moment Functions: FBD(b). M(x) = R′(L - x) - Pv Differential Equation of The Elastic Curve: EI

d 2v = M(x) dx2

EI

d 2v = R′(L - x) - Pv dx2

d 2v P R′ + v = (L - x) 2 EI EI dx

(Q.E.D.)

The solution of the above differential equation is of the form

and

v = C1 sin a

R′ P P xb + C2 cos ¢ xb + (L - x) A EI A EI P

dv R′ P P P P = C1 cos ¢ x≤ - C2 sin ¢ x≤ dx A EI A EI A EI A EI P

The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, v = 0. From Eq. (1), C2 = At x = 0,

R′L P

dv R′ EI = 0. From Eq. (2), C1 = dx P BP

Elastic Curve: v = =

R′ EI R′L R′ P P sin¢ x≤ cos ¢ x≤ + (L - x) P AP A EI P A EI P R′ EI P P J sin¢ x≤ - L cos ¢ x≤ + (L - x) R P AP A EI A EI

1318

(1)

(2)

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13–45.  Continued

Solution However, v = 0 at x = L. Then, 0 =

EI P P sin¢ L ≤ - L cos ¢ L≤ AP A EI A EI

tan¢

P P L≤ = L A EI A EI

(Q.E.D.)

By trial and error and choosing the smallest root, we have

Then,

P L = 4.49341 A EI

Pcr =

20.19EI L2

(Q.E.D.)

Ans: N/A 1319

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13–46. T he W360  39 structural A-36 steel member is column that is assumed to be fixed at its used as 6-m-long a 20-ft-long column that is assumed to be fixed at toptop andand fixed at its bottom. If the 75-kN load is applied at If the 15-kip load is applied its fixed at its bottom. an an eccentric distance of of 25010mm, at eccentric distance in., determine the maximum column stress in the column.

75 kip kN 15

250in.mm 10

6m 20 ft

Solution W360  26 39 Section Properties for W 14 * = A 4960 = mm 2 4.96(10 −3 ) m 2 = d 353 = mm 0.353 m = πx 143 = mm 0.143 m Yielding about x- x axis: smax =

P ec KL P c 1 + 2 sec a b d; A 2 r AE A r

K = 0.5

P 75(10 3 ) 6 15.12(10 = ) N−m 2 15.12 MPa; = = A 4.96(10 −3 )

(

)

0.353 ec 0.25 2 2.157807 = = 2 0.1432 π

0.5(6) 75(10 3 ) KL P = 2π EA 2(0.143) [200(109 )][4.96(10 −3 )] secsec (0.091207)] smax = 15.12[1 1.9506[1+ +2.157807 2.178714 (0.087094)] 250 MPa = 47.9 MPa < σ Y =

O.K.

Ans.

Ans: s max = 47.9 MPa 1089

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13-47. The W360  39 structural A-36 steel member is used a column is assumed to fixed be fixed its and top used as as a column thatthat is assumed to be at itsattop and pinned its bottom. If the 75-kN is applied at pinned at itsatbottom. If the 15-kip loadload is applied at an an eccentric distance of in., 250determine mm, determine the maximum eccentric distance of 10 the maximum stress stress the column in the in column.

75 kip kN 15

250in.mm 10

6m 20 ft

Solution Section Properties for W360 W 14  * 26 39 = A 4960 = mm 2 4.96(10 −3 ) m 2 = d 353 = mm 0.353 m = πx 143 = mm 0.143 m

Yielding about x- x axis: smax =

P ec KL P b d; c 1 + 2 sec a A 2 r AE A r

K = 0.7

P 75(10 3 ) 6 15.12(10 = ) N−m 2 15.12 MPa; = = A 4.96(10 −3 )

(

)

0.353 ec 0.25 2 2.157807 = = 0.1432 π2

0.7(6) 75(10 3 ) KL P = = 0.127690 2π EA 2(0.143) [200(109 )][4.96(10 −3 )]

smax = 15.12[1 1.9506[1+ +2.157807 2.178714 (0.121931)] secsec (0.127690)] = 48.0 MPa A vs. KL>r for the column.

s (MPa) 350 200

Solution

0

Tangent Moduli: From the stress-strain diagram, (Et)1 = (Et)2 =

200 1 106 2

(350 - 200) 1 106 2 0.004 - 0.001

0.004

P (in./in.)

0 … s 6 200 MPa

= 200 GPa

0.001

0.001

= 50 GPa

200 MPa 6 s … 350 MPa

Critical Stress: Applying Engesser’s equation, scr =

p2Et

P = A

(1)

KL 2 a b r

If Et = (Et)1 = 200 GPa, Eq. (1) becomes 1.974 1 106 2 p2 3 200 1 109 2 4 P = = MPa 2 A KL KL 2 b b a a r r When scr =

P = sY = 200 MPa, this equation becomes A

200 1 106 2 =

p2 3 200 1 109 2 4 a

KL 2 b r

KL = 99.346 = 99.3 r If Et = (Et)2 = 50 GPa, Eq. (1) becomes

P = A

p2 c 50 1 109 2 d ¢

when scr =

KL 2 ≤ r

=

0.4935 1 106 2 ¢

KL 2 ≤ r

MPa

P = sY = 200 MPa, this equation gives A

200 1 106 2 =

p2 3 50 1 109 2 4 a

KL 2 b r

KL = 49.67 = 49.7 r

Using these results, the graphs of

P KL vs.   as shown in Fig. a can be plotted. A r 1347

Ans: N/A

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  The stress–strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are pinned. Assume that the load acts through the axis of the bar. Use Engesser’s equation.

s (MPa) 1100

200 0.001

Solution E1 = E2 =

200 (106) 0.001

0.007

P (mm/mm)

= 200 GPa

1100 (106) - 200 (106) 0.007 - 0.001

= 150 GPa

Section properties: I = r =

p 4 c ; A = pc 2 4 p 4 I c 0.04 4 c = = = = 0.02 m AA C p c2 2 2

Engesser’s equation:

1.0(1.5) KL = = 75 r 0.02 scr =

p2 Et

1 KrL 2 2

=

p2 Et (75)2

= 1.7546(10 - 3) Et

Assume Et = E1 = 200 GPa scr = 1.7546 (10 - 3)(200)(109) = 351 MPa 7 200 MPa Therefore, inelastic buckling occurs: Assume Et = E2 = 150 GPa scr = 1.7546 (10 - 3)(150)(109) = 263.2 MPa O.K.

200 MPa 6 scr 6 1100 MPa Critical load: Pcr = scr A = 263.2 (106)(p)(0.042) = 1323 kN

Ans.

Ans: Pcr = 1323 kN 1344

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  The stress–strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.

s (MPa) 1100

200 0.001

Solution E1 = E2 =

200 (106) 0.001

0.007

P (mm/mm)

= 200 GPa

1100 (106) - 200 (106) 0.007 - 0.001

= 150 GPa

Section properties: I = r =

p 4 c ; A = pc 2 4 p 4 I c 0.04 4 c = = = = 0.02 m 2 AA Cp c 2 2

Engesser’s equation:

0.5 (1.5) KL = = 37.5 r 0.02 scr =

p2 Et

1

2

KL 2 r

=

p2 Et (37.5)2

= 7.018385(10 - 3) Et

Assume Et = E1 = 200 GPa scr = 7.018385 (10 - 3)(200)(109) = 1403.7 MPa 7 200 MPa

NG

Assume Et = E2 = 150 GPa scr = 7.018385 (10 - 3)(150)(109) = 1052.8 MPa O.K.

200 MPa 6 scr 6 1100 MPa Critical load: Pcr = scr A = 1052.8 (106)(p)(0.042) = 5292 kN

Ans.

Ans: Pcr = 5292 kN 1345

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13–73. The stress–strain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and length of 1.5 m is made from this material, determine the critical load provided one end is pinned and the other is fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.

s (MPa) 1100

200 0.001

Solution E1 = E2 =

200 (106) 0.001

0.007

P (mm/mm)

= 200 GPa

1100 (106) - 200 (106) 0.007 - 0.001

= 150 GPa

Section properties: I = r =

p 4 c ; A = pc 2 4 p 4 I c 0.04 4 c = = = = 0.02 m AA C pc 2 2 2

Engesser’s equation:

0.7 (1.5) KL = = 52.5 r 0.02 scr =

p2 Et

1

2

KL 2 r

=

p2 Et (5 2 .5 )2

= 3 .5 8 0 8 1 (1 0 - 3 ) Et

Assume Et = E1 = 200 GPa scr = 3.58081 (10 - 3)(200)(109) = 716.2 MPa 7 200 MPa

NG

Assume Et = E2 = 150 GPa scr = 3.58081 (10 - 3)(150)(109) = 537.1 MPa O.K.

200 MPa 6 scr 6 1100 MPa Critical load: Pcr = scr A = 537.1 (106)(p)(0.042) = 2700 kN

Ans.

Ans: Pcr = 2700 kN 1346

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s (MPa)

13–74. Construct the buckling curve, P兾A versus L兾r, for a column that has a bilinear stress–strain curve in compression as shown. The column is pinned at its ends.

260

140

Solution

0.001

Tangent modulus: From the stress–strain diagram, (Et)1 =

140(106) = 140 GPa 0.001

(Et)2 =

(260 - 140)(106) = 40 GPa 0.004 - 0.001

Critical Stress: Applying Engesser’s equation, scr

p2Et P = A L 2 a b r

[1]

Substituting (Et)1 = 140 GPa into Eq. [1], we have p2 C 140(109) D P = A ALB2 r

1.38(106) P = MPa A ALB2 r

When

L P = 140 MPa, = 99.3 r A

Substitute (Et)2 = 40 GPa into Eq. [1], we have p2 C 40(109) D P = A ALB2 r

0.395(106) P = MPa A ALB2 r

When

P L = 140 MPa, = 53.1 r A

1098

0.004

P (mm/mm)

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13–75.  s (MPa)

The stress–strain diagram of the material can be approximated by the two line segments. If a bar ­having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are pinned. Assume that the load acts through the axis of the bar. Use Engesser’s equation.

550

100 0.001

Solution E1 =

E2 =

100 (106) 0.001

0.007

P (mm/mm)

= 100 GPa

550 (106) - 100 (106) 0.007 - 0.001

= 75 GPa

Section Properties: I = r =

p 4 c; 4

A = p c2

p 4 c 0.04 I 4 c = = = 0.02 m = C p c2 2 2 BA

Engesser’s Equation:

1.0 (1.5) KL = = 75 r 0.02 ­scr =

p2 Er 2 (KL r )

=

p2 Er (75)2

= 1.7546 (10 - 3) Et

Assume Et = E1 = 100 GPa scr = 1.7546 (10 - 3)(100)(109) = 175 MPa 7 100 MPa Therefore, inelastic buckling occurs: Assume Et = E2 = 75 GPa scr = 1.7546 (10 - 3) (75)(109) = 131.6 MPa 100 MPa 6 scr 6 550 MPa

OK

Critical Load: Pcr = scr A = 131.6(106)(p)(0.042) = 661 kN

Ans.

Ans: Pcr = 661 kN 1349

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*13–76. s (MPa)

The stress–strain diagram of the material can be approximated by the two line segments. If a bar ­having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.

550

100 0.001

Solution E1 = E2 =

100 (106) 0.001

0.007

P (mm/mm)

= 100 GPa

550 (106) - 100 (106) 0.007 - 0.001

= 75 GPa

Section Properties: I = r =

p 4 c; 4

A = p c2

p 4 c 0.04 I 4c = = = 0.02 m = B 2 2 2 BA pc

Engesser’s Equation:

0.5 (1.5) KL = = 37.5 r 0.02 scr =

p2 Er 2 (KL r )

=

p2 Er (37.5)2

= 7.018385 (10 - 3) Et

Assume Et = E1 = 100 GPa scr = 7.018385 (10 - 3)(100)(109) = 701.8 MPa 7 100 MPa

NG

Assume Et = E2 = 75 GPa scr = 7.018385 (10 - 3)(75)(109) = 526.4 MPa OK

100 MPa 6 scr 6 550 MPa Critical Load: Pcr = scr A = 526.4 (106)(p)(0.042) = 2.65 ( 103 ) kN

Ans.

Ans: Pcr = 2.65 ( 103 ) kN 1350

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13–77.  s (MPa)

The stress–strain diagram of the material can be approximated by the two line segments. If a bar ­having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided one end is pinned and the other is fixed. Assume that the load acts through the axis of the bar. Use Engesser’s equation.

550

100 0.001

Solution E1 = E2 =

100 (106) 0.001

0.007

P (mm/mm)

= 100 GPa

550 (106) - 100 (106) 0.007 - 0.001

= 75 GPa

Section Properties: I =

r =

p 4 c; 4

A = p c2

p 4 I c 0.04 4 c 0.02 m = A p c2 = 2 = 2 = AA

Engesser’s Equation:

0.7 (1.5) KL = = 52.5 r 0.02 scr =

p2 Er 2 (KL r )

=

p2 Er (52.5)2

= 3.58081 (10 - 3) Et

Assume Et = E1 = 100 GPa scr = 3.58081 (10 - 3)(100)(109) = 358.1 MPa 7 100 MPa

NG

Assume Et = E2 = 75 GPa scr = 3.58081 (10 - 3)(75)(109) = 268.6 MPa 100 MPa 6 scr 6 550 MPa

OK

Critical Load: Pcr = scr A = 268.6 (106)(p)(0.042) = 1.35 ( 103 ) kN

Ans.

Ans: Pcr = 1.35 ( 103 ) kN 1351

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13–78. Determine the largest length of a structural A-36 steel rod if it is fixed supported and subjected to an axial load of 100 kN. The rod has a diameter of 50 mm. Use the AISC equations.

Solution Section Properties: A = p A 0.0252 B = 0.625 A 10 - 3 B p m2 I =

p A 0.0254 B = 97.65625 A 10 - 9 B p m4 4

r =

97.65625(10 - 9)p I = = 0.0125 m AA A 0.625(10 - 3)p

Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, KL 0.5L = = 40.0L r 0.0125 AISC Column Formula: Assume a long column. sallow = 100(103) 0.625(10 - 3)p

=

12p2E

2 23 A KL r B

12p2 C 200(109) D 23(40.0L)3

L = 3.555 m KL KL 2p2E = 40.0(3.555) = 142.2 and for A–36 steel, a b = r r e A sg KL KL 2p2[200(109)] = = 125.7. Since a b … … 200, the assumption is correct. r e r A 250(106)

Here,

Thus, L = 3.56 m

Ans.

Ans: L = 3.56 m 1102

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13–79. Check if a W250  58 column can safely support an axial force of =P 250 = 1150 column is ft 6m long and axial force of P TheThe column is 20 long and is kip.kN. is pinned bothends endsand andbraced bracedagainst against its its weak weak axis axis at pinned at atboth 200 3GPa It isis made made ofofsteel steelhaving havingEE= =29(10 mid-height. It ) ksi and sY MPa. usethe theAISC AISCcolumn columndesign designformulas. formulas. s Use = 350 50 ksi. Y=

Solution Section Properties. From the table listed in the appendix, the necessary section 58are are properties for a W250 W10 *39 = A 7400 = mm 2 7.4(10 −3 ) m 2 = πx 109 = mm 0.109 m = πy 50.4 = mm 0.0504 m Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, = Lx 6= m and Ly 3 m. Thσs, 1(6)  KL  = 55.05  =   π  x 0.109 1(3)  KL  = 59.52 (contπols) =   π  y 0.0504 AISC

Column

Formulas.

2π 2 [200(109 )] = 106.21. 350(106 )

For

A-36

Since ¢

steel

¢

KL 2p2E ≤ = r c A sY

KL KL ≤ 6 ¢ ≤ , the column is an r y r c

intermediate column.

B1 sallow =

(KL>r)2 2(KL>r)c 2

R sY

3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3

59.6222 60.606 d (350) C 1c 1-  2 S(50) 2 A2(106.21 107.002 B )  5 3(59.52) (59.523) = 3(60.606)  60.6063 3 5 3 + 8(106.21)- 8(106.21 ) 3 8(107.00) 8 A 107.003 B = 159.06 MPa

Thus, the allowable force is 6 = Pallow = σ allow A [159.06(10 = )][7.4(10 −3 )] 1177.04(10 3 ) N

= 1177 = kN > P

1150 kN

O.K.

Thus, a W10 W250*39 58column columnisisadequate. adequate.

Ans: Yes 1109

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*13–80. A W200  36 A-36-steel column of 9-m length is pinned at both ends and braced against its weak axis at mid-height. Determine the allowable axial force P that can be safely supported by the column. Use the AISC column design formulas.

Solution Section Properties. From the table listed in the appendix, the necessary section are properties for a W200 are W8 *  2436

= A 4570 = mm 2 4.57(10 −3 ) m 2 πy 40.9 = = mm 0.0409 m

πx 86.8 = = mm 0.0868 m

Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, = m and Ly 4.5 m. Thσs, Lx 9=

1(9)  KL  = 103.69 =   π  x 0.0868 1(4.5)  KL  = 110.02 (contπols) =   π  y 0.0409 AISC Column Formulas. For A-36 steel ¢

KL 2p2E ≤ = r c A sY

2π 2 [200(109 )] KL KL = 125.66. Since ¢ ≤ 6 ¢ ≤ , the 6 r y r c 250(10 ) intermediate column.

B1 sallow =

(KL>r)2 2(KL>r)c 2

column

is

an

R sY

3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3

111.802 2 Cc 11- 110.02 d (250) S (36) 2 A 126.102)2 B 2(125.66  3 = 5 3(110.02) (110.02 5 111.803)  3 + ) 33 8(125.66) 8(126.10) 8(125.66 8 A 126.103 B

= 80. 67 MPa Thus, the allowable force is

6 −3 3 = )] 368.68(10 = ) N 369 kN Pallow = sallowA = [80.67(10 )][4.57(10

Ans.

Ans: Pallow = 369 kN 1108

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13–81. Using the AISC equations, select from Appendix B the lightest-weight structural A-36 steel the lightest-weight structural A-36 steel column thatcolumn is 30 ft that is 9m long and load ofends 1000are kN. The long and supports an supports axial loadan of axial 200 kip.The fixed. ends are fixed.

Solution Tπy W250× 80 = mm 0.065 m πy 65.0 = mm 2 0.0102 m 2 = A 10200 =  KL  =   π c

2π 2 E =

2π 2 [200(109 )] = 125.66 250(106 )

σY

KL 0.5(9) = = 69.23 πy 0.065 a

KL KL b 6 a b intermediate column. ry r c

b 1 - 12 B sallow =

b 53 + 38 B

A

KL r

B

KL r c

A

KL r

B

KL r c

2

R r sY

R - 18 B

A

KL r

B

KL r c

3

R r

1 86.54 2 e 1 -69.23 D f36 2 C 126.1 {1 – 12[125.66 ]2}250  114.48 MPa = 5 3 69.23 69.23 3 {3 5+ 8[125.66 ] – 18[125.66 } 1 ] 86.54 e 3 + 38 C 86.54 126.1 D - 8 C 126.1

6 )](0.0102) 1167.71(10 3 ) N = Pallow = sallow A = [114.48(10

kN > P 1000 kN = 1168 = Use

O.K.

W 8 *4880 W250

Ans.

Ans: Use W250 * 80 1107

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AISCAISC equations, select fromselect Appendix B 13–82. Using ethe equations, from the lightest-weight structural A-36 steel column thatcolumn is 24 ft Appendix B the lightest-weight structural A-36 steel long is and an axial load of kip.The ends fixed. that 7.2supports m long and supports an100 axial load of 450are kN. The ends are fixed.

Solution Section Properties: Try a W200 W8 *  2436 wide flange wide flangesection, section, 2 = mm 0.0409 m πy 40.9 A 4570 = mm 2 4.57(10 −3 ) m= =

Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, 0.5(7.2)  KL  = 88.02 =  π   y 0.0409 AISC

Column

Formula:

For

A–36

steel,

a

KL 2p2E b = r c A sg

KL KL 2π 2 [200(109 )] = 125.66. Since 6 a b , the column is an intermediate r r c 250(106 ) column. Applying Eq. 13–23,

B1 sallow =

(KL>r)2 2(KL>r)2c

R sg

3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c

2 (89.44 (88.022)) Rd (36) (250) 2(125.66 2(126.12)2) =  3 3) 3(88.02) (88.02 ) 3(89.44) (89.44 55   + 3 3) 8(125.66 33 8(125.66) 8(126.1) 8(126.1 )

Bc 1 

 100.02 MPa The allowable load is Pallow = sallowA = [100.02(106 )][4.57(10 −3 )] = 457.09(10 3 ) N = 457 101 kN kip >7PP= 450 = 100 kNkip Thus, Use

O.K.

W200 W8 * 2436

Ans.

Ans: Use W200 * 36 1106

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13–83. Determine the largest length of a W250 * 67 A992 structural steel column if it is pin supported and subjected to an axial load of 1450 kN. Est = 200 GPa, sY = 350 MPa. Use the AISC equations.

Solution Section Properties: For W200  46 wide flange section,

= mm 0.0509 m πy 50.9 = A 8560 = mm 2 8.56(10 −3 ) m 2= Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,

1(L)  KL  = 19.6464 L =  π 0.0509  y AISC Column Formula: Assume a long column, sallow =

1450(10 3 ) −3

8.56(10 )

12p2E

2 23 A KL r B

=

12π 2 [200(109 )] 23(19.6464L)2

L = 3.9688 m Heπe,

KL π

2π 2 E  KL  19.6464(3.9688) = 77.97 and foπ A992 steel,  =  σY  π c

2π 2 [200(109 )] KL  KL  106.97. Since r)2 (2(KL>r) 2c

R sY

3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c

 (19.6464L)2  6 1 −  [345(10 )] 2(106.97 2 )  1450(10 )  = 8.56(10 −3 ) 5 3(19.6464 L) (19.6464 L)3 + − 3 8(106.97) 8(106.97 3 ) 3

0 −3.80208(10 −4 )L3 + 0.016865L2 + 0.0338157L − 0.181679 = Solving by trial and error,

= L 2.4790 = m 2.48 m

Ans.

Ans: L = 2.48 m 1103

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* 13–84. Determine the largest length of a W250 * 18 structural A-36 steel section if it is pin supported and is subjected to an axial kN.Use Use the AISC equations. axial load load of of140 28 kip.

Solution For a W250  18, A 2280 = mm 2 2.28(10 −3 ) m 2 =

πy 20.1 mm 0.0201 m = =

σ=

P 140(10 3 ) 2 61.40 MPa = = 61.40(106 ) N−m= A 2.28(10 −3 )

Assume a long column: sallow =

 KL  =    π   KL  =   π c

12p2E 23(KL>r)2

12π 2 E = 23σ allow 2π 2 E =

σY

12π 2 [200(109 )] = 129.51 23[61.40(106 )] 2π 2 [200(109 )] KL  KL  > 125.66, =  6 π 250(10 )  π c

Long column. KL 129.51 = 137.4 r

 π   0.0201  L 129.51 = m 2.60 m = =   129.51 =  2.603 K  1 

Ans.

Ans: L = 2.60 m 1104

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13–85. Using the AISC equations, select from the lightest-weight structural A-36 steel column thatcolumn is 14 ft Appendix B the lightest-weight structural A-36 steel long is and load ofan40axial kip.The arekN. pinned. that 4.2supports m long an andaxial supports loadends of 200 The ends are pinned.

Solution Try, W150  22

πy 36.8 = mm 0.0368 m = A 2860 = mm 2 2.86(10 −3 ) m 2=  KL  =   π c

2π 2 E =

σY

2π 2 [200(109 )] = 106.97 345(106 )

 KL  1.0(4.2)  KL   KL  =   = 114.13,  >        πy  0.0368  πy   π c Long column sallow =

12π 2 [200(109 )] 12 p2E 6 79.06(10 = ) N−m 2 79.06 MPa = = 2 23(114.132 ) 23(KL>r)

Pallow = sallowA 6 [79.06(10 = )][2.86(10 −3 )] 226 kN > 200 kN

O.K.

Use W150 W6 *  1522

Ans.

Ans: Use W150 * 22 1104

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AISCAISC equations, select fromselect Appendix B equations, from 13–86. Using ethe the lightest-weight structural A-36 steel column thatcolumn is 12 ft Appendix B the lightest-weight structural A-36 steel long is and an axial load of kip.load The of ends fixed. that 3.6supports m long and supports an40 axial 200are kN. The ends are fixed.

Solution Tπy W150× 14 2 = mm 0.023 m πy 23.0 = mm 2 1.73(10 −3 ) m= = A 1730

 KL  =   π c

2π 2 E =

2π 2 [200(109 )] = 106.97 345(106 )

σY

KL 0.5(3.6) = = 78.26 πy 0.023 KL KL 6 a b ry r c Intermediate column

σ allow

(

) ) (

 1 KL π 2  1 − 2 (KL π )c  σ Y   = =  5 3 KL π KL π 3  1  3 + 8 (KL π )c − 8 ( KL π)c   

(

)

( (

) ) (

1 − 1 78.26 2  (345 MPa)  2 106.97  = 133.54 MPa  5 + 3 78.26 − 1 78.26 3   3 8 106.97 8 106.97 

)

Pallow = sallowA

= [133.54(106 )][1.73(10 −3 )] = 231.03(10 3 ) N

O.K.

= 231 kN > 200 kN W6 *  9 14 Use W150

Ans.

Ans: Use W150 * 14 1105

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13–87. Check if a W250  67 column can safely support axial force force of ofPP == 200 1000kip. kN.The Thecolumn column is ft 4.5long m long an axial is 15 and andpinned is pinned at both ends. madeof of steel steel having is at both of of its its ends. It Itisismade 3 E = GPa sYs= Use the the AISC column and E = 200 29(10 ) ksiand = 50MPa. ksi. Use Y 350 design formulas.

Solution Section Properties. Try W250 . From W10 *4567. From the the table table listed listed in in the the appendix, the necessary section properties are

πy 50.9 mm 0.0509 m = A 8560 = mm 2 8.56(10 −3 ) m 2 = = Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus,

1(4.5)  KL  = 88.41 =   π  y 0.0509

AISC Column Formulas. Here, a Since

a

KL KL b 6 a b , the r y r c

KL 2p2E = b = r c A sY

2π 2 [200(109 )] = 106.21. 350(106 )

column is an intermediate column.

B1 sallow =

(KL>r)2 2(KL>r)c 2

R sY

3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)c 3

2 2 89.552 88.41 S(50) 2 d (350) 2(106.21 2 A 107.00)2 B  = 5 3(88.41) (88.413) 5  3(89.552)  89.55233 3 + 8(106.21) - 8(106.21 ) 3 8(107.00) 8 A 107.003 B

cC11-

= 199.96 MPa

Thus, the allowable force is 6 −3 = Pallow σ= = )] 1026.88(10 3 ) N allow A [119.96(10 )][8.56(10

= 1027 kN > 1000 kN

O.K.

Thus, use W250 * 67

Ans.

Ans: Use W250 * 67 1111

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*13–88. A 1.5-m-long rod is used in a machine to transmit an axial compressive load of 15 kN. Determine its smallest diameter if it is pin connected at its ends and is made of a 2014-T6 aluminum alloy.

Solution Section properties: A =

p 2 d ; 4

I =

p d 4 pd4 a b = 4 2 64

pd4

I d 64 r = = = AA C p4 d2 4

σ allow =

P 15(10 3 ) 60(10 3 ) = = π d2 A π d2 4

Assume long column:

KL 1.0(1.5) 6 = = π d 4 d

σ allow

372550 MPa 60(10 3 ) 372550(106 ) = ; 2 π d2 (KL π ) (6 d)2

= = m 36.9 mm d 0.03686

Ans.

6 KL 1.0(1.5) = = 162.79 > 55 (Assσmption is coππect!) = 0.03686 π d 4

O.K.

Ans:

d = 36.9 mm 1110

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13–89.  ­ aving the Using the AISC equations, check if a column h cross section shown can support an axial force of 1500 kN. The column has a length of 4 m, is made from A992 steel, and its ends are pinned.

20 mm

350 mm

20 mm 300 mm

10 mm

Solution Section Properties: A = 0.3(0.35) - 0.29(0.31) = 0.0151 m2 Iy = ry =

1 1 (0.04) 1 0.33 2 + (0.31) 1 0.013 2 = 90.025833 1 10 - 6 2 m4 12 12 Iy

BA

=

90.02583(10 - 6)

B

0.0151

= 0.077214 m

Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a AISC

Column

1(4) KL b = = 51.80 r y 0.077214

Formula:

2p2[200(109)]

For

= 107­. Since A 345(106) column. Applying Eq. 13–23, =

a

KL 2p2E b = r c A sg

(KL>r)2 2(KL>r)2c

R sY

3(KL>r) (KL>r)3 5 + 3 8(KL>r)c 8(KL>r)3c J1 -

=

steel,

KL KL b , the column is an intermediate 6 a r r c J1 -

sallow =

A992

(51.802) 2(1072)

R (345)(106)

3(51.80) (51.803) 5 + 3 8(107) 8(1073)

= 166.1 MPa The allowable load is Pallow = sallowA = 166.1 1 106 2 (0.0151)

O.K.

= 2508 kN 7 P = 1500 kN

Ans.

Thus, the column is adequate.

Ans: Yes 1363

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13–90.  The beam and column arrangement is used in a ­railroad yard for loading and unloading cars. If the ­ maximum anticipated hoist load is 12 kip, determine if the W8 * 31 wide-flange A-36 steel column is adequate for supporting the load. The hoist travels along the bottom flange of the beam, 1 ft ≤ x ≤ 25 ft, and has negligible size. Assume the beam is pinned to the column at B and pin supported at A. The column is also pinned at C and it is braced so it will not buckle out of the plane of the loading.

x

27 ft

A

B 12 kip 15 ft

Solution For W8 * 31,

A = 9.13 in2

rx = 3.47 in.,

C

Maximum axial load occurs when x = 25 ft. (1.0)(15)(12) KL = = 51.87 r 3.47 a

2p2(29) ( 103 ) KL 2p2E bc = = = 126.1 r C sY C 36

Here 0 6 51.87 6 126.1 Intermediate Column: sallow =

=

s =

3 53

+

31

3 8

-

1 2

KL > r 2 1 (KL > r)c 2 4 sY

KL > r 1 (KL > r)c 2

-

1 8

KL > r 3 1 (KL > r)c 2 4

( 1 - 12 ( 51.87>126.1 ) 2 ) 36 = 18.2 ksi 5 53 + 3 ( 38 ) (51.87>126.1) 4 - 3 18 ( 51.87>126.1 ) 3 4 6

P 11.11 = = 1.22 ksi 6 18.2 ksi A 9.13

OK

Ans.

Yes, the column is adequate.

Ans: Yes 1364

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13–91. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is pin connected at its ends.

3 kNlb 600 bb

5b 5b

2.48mft

Solution Section Properties: A = b(5b) = 5b2 Iy =

1 5 4 (5b) A b3 B = b 12 12

ry =

5 4 Iy 23 12 b = = b AA C 5b2 6

3 kNlb 600

Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,

1(2.4) 8.3138  KL  =  =  3b b  π y 6 Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26. 372550 MPa σ allow = (KL π)2

3(10 3 ) 5b2

=

372550(106 ) (8.3138 b)2

= b 0.01827 = m 18.3 mm KL KL 8.3138 Heπe, = = 455.15. Since 7 55, the assumption is correct. Thus, r π 0.01827

b = 18.3 mm

Ans.

Ans: b = 18.3 mm 1114

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*13–92. The bar is made of a 2014-T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is fixed connected at its ends.

600 lb 3 kN b

5b

2.48mft

Solution Section Properties: A = b(5b) = 5b2 Iy = ry =

1 5 4 (5b) A b3 B = b 12 12 5 12

600 lb 3 kN

4

b Iy 23 = = b 2 AA C 5b 6

Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,

0.5(2.4) 4.1569  KL  = =  3b π b  y 6 Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply Eq. 13–26.

σ allow = 3(10 3 ) 5b

2

=

372550 MPa (KL π)2 372550(106 ) (4.1569 b)2

= b 0.01292 = m 12.9 mm KL KL 4.1569 Heπe, = = 321.84. Since 7 55, the assumption is correct. r π 0.01292 Thus,

b = 12.9 mm

Ans.

Ans:

b = 12.9 mm 1115

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13–93. The 2014-T6 aluminum hollow section has the cross section shown. If the column is 3 m long and is fixed at both ends, determine the allowable axial force P that can be safely supported by the column.

4 in. 100 mm

Solution Section Properties. in. 753 mm

A =π (0.052 − 0.03752 ) =3.4361(10 −3 ) m 2 I= = π

π 4

(0.054 − 0.03754 ) = 3.3556(10 −6 ) m 4

I = A

3.3556(10 −6 ) = 0.03125 m 3.4361(10 −3 )

Slenderness Ratio. For a column fixed at both of its ends, K = 0.5. Thus,

0.5(3) KL = = 48 0.03125 π 2014-T6 Aluminum Alloy Column Formulas. Since 12 6

KL 6 55, the column can r

be classified as an intermediate column.



 KL  

σ allow =  212 − 1.59    MPa  π   = [212 − 1.59(48)] MPa = 135.68 MPa Thus, the allowable load is

Pallow = σ allow A = [135.68(106 )][3.4361(10 −3 )] 3 = 466.21(10 = ) N 466 kN

Ans.

Ans: Pallow = 466 kN 1118

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*13–94. The 2014-T6 aluminum hollow section has the cross section shown. If the column is fixed at its base and pinned at its top, and is subjected to the axial force P == 500 P 100kN, kip,determine determinethe themaximum maximum length length of the column for it to safely support the load.

4 in. 100 mm

Solution Section Properties.

in. 753 mm

A =π (0.052 − 0.03752 ) =3.4361(10 −3 ) m 2 I= = π

π 4

(0.054 − 0.03754 ) = 3.3556(10 −6 ) m 4

I = A

3.3556(10 −6 ) = 0.03125 m 3.4361(10 −3 )

Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Thus,

KL 0.7 L = = 22.4 L 0.03125 π 2014-T6 Aluminum Alloy Column Formulas. Assuming an intermediate column,



 KL  

=  212 − 1.59  σ allow   MPa  π   500(10 3 ) = [212 − 1.59(22.4L)](106 ) 3.4361(10 −3 ) = L 1.8668 = m 1.87 m

Ans.

KL KL = Heπe, 22.4(1.8668) = 41.8. Since 12 < < 55, the assumptions of an π π intermediate column is correct.

Ans: L = 1.87 m 1119

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13–95. The 2014-T6 aluminum column of 3-m length has the cross section shown. If the column is pinned at both ends and braced against the weak axis at its mid-height, determine the allowable axial force P that can be safely supported by the column.

15 mm

170 mm

15 mm

15 mm 100 mm

Solution Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Ix =

1 1 (0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10 - 6 B m4 12 12

Iy = 2 c

1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12

31.86625 A 10 Ix = AA C 5.55 A 10 - 3 B

-6

rx = ry =

B

= 0.07577

2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B

Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m and Ly = 1.5 m. Thus, a

(1)(3) KL b = = 39.592 r x 0.07577

a

(1)(1.5) KL b = = 70.009 (controls) r y 0.02143

2014-T6 Alumimum Alloy Column Formulas. Since a

KL b 7 55, the column can r y

be classified a long column, sallow = D

= C

372550 a

KL 2 b r

372550 70.0092

T MPa

S MPa

= 76.011 MPa Thus, the allowed force is Pallow = sallowA = [76.011(106 )] [5.55(10- 3)] = 421.86(103) N = 422 kN

Ans.

Ans: A = 5.55 (10- 3) m2, Ix = 31.86625 (10- 6) m4, Iy = 2.5478 (10 - 6) m4, Pallow = 422 kN 1116

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*13–96. The 2014-T6 aluminum column has the cross section shown. If the column is pinned at both ends and subjected to an axial force P = 100 kN, determine the maximum length the column can have to safely support the loading.

15 mm

170 mm

15 mm

15 mm 100 mm

Solution Section Properties. A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2 Iy = 2 c ry =

1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4 12 12

2.5478 A 10 - 6 B Iy = = 0.02143 m AA C 5.55 A 10 - 3 B

Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, a

1(L) KL = 46.6727L b = r y 0.02143

2014-T6 Alumimum Alloy Column Formulas. Assuming a long column, sallow = D 100 A 103 B

5.55 A 10

-3

B

372550 KL 2 b a r

T MPa

372550 = C S A 106 B Pa (46.6727L)2 Ans.

L = 3.0809 m = 3.08 m Since a

KL b = 46.6727(3.0809) = 143.79 7 55 , the assumption is correct. r y

Ans: L = 3.08 m 1117

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13–97. The tube is 0.25 6 mm in.thick, thick, isis made made of of aa 2014-T6 aluminum alloy, and is fixed at its bottom and pinned at its top. Determine the largest axial load that it can support.

P P y y 6 in. 150 mm 150 mm6 in. y y x x x

x

3 m10 ft

Solution Section Properties: P P

A= 0.15(0.15) − 0.138(0.138) = 3.456(10 −3 ) m 2 1 1 11.9647(10 −6 ) m 4 I = (0.15)(0.153 ) − (0.138)(0.138 3 ) = 12 12 = π

I = A

11.9647(10 −6 ) = 0.058839 m 3.456(10 −3 )

Slenderness Ratio: For a column fixed at one end and pinned at the other end, K = 0.7. Thus,

KL 0.7(3) = = 35.69 π 0.058839 Aluminium (2014 –- T6 alloy) Column Formulas: Since 12 6

KL 6 55, the r

column is classified as an intermediate column. Applying Eq. 13–25,



 KL  

=  212 − 1.59  σ allow   MPa  π   = [212 − 1.59(35.69)] = 155.25 MPa The allowable load is 6 −3 3 = Pallow σ= = )] 536.55(10 = ) N 537 kN allow A [155.25(10 )][3.456(10

Ans.

Ans: Pallow = 537 kN 1120

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*13–98. The tube is 6 mm thick, is made of a 2014-T6 aluminum alloy, and is fixed connected at its ends. Determine the largest axial load that it can support.

x

P x

150 mm 6 in. y y

P y y 150 mm 6 in. x x

3 m 10 ft

Solution Section Properties: P

A= 0.15(0.15) − 0.138(0.138) = 3.456(10 −3 ) m 2

P

1 1 11.9647(10 −6 ) m 4 I = (0.15)(0.153 ) − (0.138)(0.138 3 ) = 12 12 = π

I = A

11.9647(10 −6 ) = 0.058839 m 3.456(10 −3 )

Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus,

KL 0.5(3) = = 25.49 π 0.058839 Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6

KL 6 55, the r

column is classified as an intermediate column. Applying Eq. 13–25,



 KL  

=  212 − 1.59  σ allow   MPa  π   = [212 − 1.59(25.49)] = 171.46 MPa The allowable load is 6 −3 3 = Pallow σ= = )] 592.58(10 = ) N 593 kN allow A [171.46(10 )][3.456(10

Ans.

Ans: Pallow = 593 kN 1121

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13–99. The tube mmin.thick, tube is is 60.25 thick,is ismade madeof ofa 2014-T6 aluminum alloy and is pin connected at its ends. Determine the largest axial load it can support.

P P y y 6 in. 150 mm 150 mm6 in. y y x x x

x

3 m10 ft

Solution Section Properties:

A= 0.15(0.15) − 0.138(0.138) = 3.456(10 −3 ) m 2

P P

1 1 11.9647(10 −6 ) m 4 I = (0.15)(0.153 ) − (0.138)(0.138 3 ) = 12 12 = π

11.9647(10 −6 ) = 0.058839 m 3.456(10 −3 )

I = A

Slenderness Ratio: For a column pinned as both ends, K = 1. Thus,

KL 1(3) = = 50.99 π 0.058839 Aluminum (2014 – T6 alloy) Column Formulas: Since 12 6

KL 6 55, the r

column is classified as an intermediate column. Applying Eq. 13–25,



 KL  

=  212 − 1.59  σ allow   MPa  π   = [212 − 1.59(50.99)] = 130.93 MPa

The allowable load is 6 −3 3 = Pallow σ= = )] 452.49(10 = ) N 452 kN allow A [130.93(10 )][3.456(10

Ans.

Ans: Pallow = 452 kN 1122

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*13–100.

PP

The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine its greatest allowable length if it supports an KN. axial load of P = 10 2 kip.

xx 2 in. 50 mm yy

yy xx in. 1004 mm

LL

Solution Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus,

KL 2(L) = = 40 L 0.05 d NFPA Timber Column Formulas: Assume a long column. Apply Eq. 13–29,

σ allow =

3725 (KL d)2

MPa

10(10 3 ) 3725(106 ) = 0.05(0.1) (40L)2 L = 1.079 m KL KL = 40(1.079) = 43.16. Since 26 6 Heπe, 6 50, the assumption is correct. d d Thus,

L = 1.08 m

Ans.

Ans:

L = 1.08 m 1126

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13–101. The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine the largest allowable axial load P that it can support if it has a length L = 1.2 m.

PP xx 2 in. 50 mm yy

yy xx in. 1004 mm

LL

Solution Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus,

KL 2(1.2) = = 48.0 d 0.05 NFPA Timber Column Formulas: Since 26 6

KL 6 50, it is a long column. Apply d

Eq. 13–29,

σ allow = =

3725 (KL d)2 3725 48.0 2

MPa

MPa

= 1.6168 MPa

The allowable axial force is 6 3 Pallow σ= = 8.084(10 = ) N 8.08 kN = allow A [1.6168(10 )][0.05(0.1)]

Ans.

Ans: Pallow = 8.08 kN 1126

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13–102.  A rectangular wooden column has the cross s­ ection shown. If a = 3 in. and the column is subjected to an axial force of P = 15 kip, determine the maximum length the column can have to safely support the load. The column is pinned at its top and fixed at its base.

a

2a

Solution Slenderness Ratio: For a column fixed at its base and pinned at its top, K = 0.7. Then, KL 0.7L = = 0.2333L d 3 NFPA Timber Column Formula: Assuming an intermediate column, sallow = 1.20c 1 -

1 KL>d 2 a b d ksi 3 26.0

15 1 0.2333L 2 = 1.20c 1 - a b d 3(6) 3 26.0

Ans.

L = 106.68 in. = 8.89 ft

KL KL = 0.2333(106.68) = 24.89. Since 11 6 6 26, the assumption is d d correct.

Here,

Ans: L = 8.89 ft 1376

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13–103. The wooden column shown is formed by gluing the150 6 in. * 0.512in. boards. IfIfthe together the mm mm boards. thecolumn columnis is pinned P == 100 20 kip, at both ends and is subjected to an axial load P kN, determine the required number of boards needed to form the column in order to safely support the loading.

P 0.5mm in. 12

6 in. 150 mm

9 ftm 2.7

Solution Slenderness Ratio. For a column pinned at both of its ends, K = 1. If the number of the boards required is n and assuming that n(0.012) < 0.15 m. Then, d = n(0.012). Thus,

P

KL 1(2.7) 225 = = d n(0.012) n NFPA Timber Column Formula. Assuming an intermediate column,



2 1  KL d     MPa 3  26.0  

σ= 8.28 1 −  allow 

 1  1.8 d  2  6 100(10 3 ) = 8.28 1 −    (10 ) [n(0.012)](0.15)  3  26.0   n2 − 6.70961n − 24.963 = 0 Solving for the positive root,

n = 9.373 n Use n = 10 9

Ans.

KL KL 216 KL 225 , 26, = = 22.5. Since n(0.012) = 10(0.012) = 0.12 m < 0.15 m and 11  6 26  d 9 10 d the assumptions made are correct. Here,

Ans: Use n = 10 1125

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*13–104. The timber column has a square cross section and is assumed to be pin connected at its top and bottom. If it supports an axial load of 250 kN, determine its smallest side dimension a to the nearest multiples of 15 mm. Use the NFPA formulas. 14 ft 4.2 m aa

Solution Section properties: sallow = s =

A = a2

3 50 P 250(10 ) = 2 2 A a a

Assume long column:

σ allow =

3725 MPa

250(10 3 ) a

2

(KL d)2 =

3725(106 ) [1.0(4.2) a]2

a = 0.1855 m KL 1.0(4.2) KL = = 22.6, < 26 d d 0.1855

Assumption NG

Assume intermediate column:



1  KL d   3  26.0 

σ= 8.28 1 −  allow 

2

 MPa 

 1  1.0(4.2) a  2  6 250(10 3 ) = 8.28 1 −    (10 ) 2 a  3  26.0   = = a 0.19721 m 197.21 mm KL KL 1(4.2) 21.3, 11 < = = = < 26 d d 0.19721

Assumption O.K.

1 mm Use aa  = 200 7 in. 2

Ans.

Ans: Use a  200 mm 1124

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a

13–105. A rectangular wooden column has the cross section shown. shown. IfIfthe thecolumn columnis is6 ft 1.8long m long and subjected and subjected to an to anforce axialofforce of kip, P =determine 75 kN, determine the minimum required axial the required P = 15 1 minimum a of its cross-sectional area to16 the dimension dimension a of its cross-sectional area to the nearest in. nearest of 5can mm so that the column can safely so that multiples the column safely support the loading. The support the loading. The ends. column is pinned at both ends. column is pinned at both

2a

Solution Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then,

KL 1(1.8) 1.8 = = d a a NFPA Timber Column Formula. Assuming an intermediate column,



2 1  KL d     MPa 3  26.0  

σ= 8.28 1 −  allow 

 1  1.8 d  2  6 75(10 3 ) = 8.28 1 −    (10 ) 2a(a)  3  26.0   a = 0.07827 m = 78.27 mm Use aa = 80 3 in. Use mm

Ans.

KLKL KL 1.8 . Since 11 6 , the theassumption assumptionisiscorrect. correct. Since 11 < 6 r)2 23(152.67 2 )

smax = (sa)allow =

44.18(106 ) =

KL 2π 2 [200(109 )]  KL  = 125.66,  < 200  < πy 250(106 )  π c

My c P + A Iy

P + 200(10 3 ) 7.2(10 −3 )

+

[ P(0.4)](0.172 2) 11.1(10 −6 )

3 P 5.067(10 = ) N 5.07 kN =

Ans.

Ans: P = 5.07 kN 1131

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13–113. The W250  67 structural A-36 steel column is fixed at its bottom and free at its top. If it is subjected to a load of P == 10 2 kip, kN, determine if it is safe based on the AISC equations of Sec. 13.6 and Eq. 13–30.

200 40 kN kip

P 400 mm 16 in.

3 mft 10

Solution Section Properties for W250 : W10 *45 67: = A 8560 = mm 2 8.56(10 −3 ) m 2

= d 257 = mm 0.257 m

6 4 πy 50.9 = mm 0.0509 m = I y 22.2(10 = ) mm 4 22.2(10 −6 ) m=

= = bf 204 mm 0.204 m

Allowable stress method: KL 2(3) = = 117.88 πy 0.0509  KL  =   π c

2π 2 E =

2π 2 [200(109 )] = 125.66 250(106 )

σY

KL KL 6 a b r r c

(sa)allow Ú (sa)allow =

= σ max

c1 5 3

+

A

2

1 (KL>r) 2 (KL>r)2

3 KL>r 8 KL>rc

My c P + A Iy

210(10 3 ) 8.56(10 −3 )

+

B -

c

d sY

1 (KL>r)3 8 (KL>rc)3

=

1 119.4 2 B D 36 [1 C– 12 (117.88 ) 2](250) 2 A 126.1 125.66  73.10 MPaksi = 10.37 55 3 3117.88 1 117.88 ] – [ 18 A 119.4 ]3 3 A 119.4 33 ++8 [8125.66 136.1 B 8-125.66 126.1 B

[10(10 3 )](0.4)(0.204 2) 22.2(10 −3 )

6 = 42.91(10 = ) N−m 2 42.91 MPa

O.K.

Since σ allow > σ max the colσmn is safe.

Ans.

Yes.

Ans: Yes 1132

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W10 * 45 A-36-steel W250 13–114. The A-36-steel 67 column column isis fixed fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using an interaction formula. The allowable bending stress is (sb)allow = 100 15 ksi. MPa.

30012mm in. xx yy

P yy

xx

7.2ftm 24

Solution Section Properties. From the table listed in the appendix, the section properties for a W250 67are are W10 *45

= A 8560 = mm 2 8.56(10 −3 ) m 2

πx 110 = = mm 0.11 m

6 = I y 22.2(10 = ) mm 4 22.2(10 −6 ) m 4

πy 50.9 = = mm 0.0509 m

bf 204 = = mm 0.204 m Slenderness Ratio. Here, Lx = 7.2 m and for a column fixed at its base and free at its top, Kx = 2. Thus,

2(7.2)  KL  = 130.91 (contπols)  =  0.11  π x Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 7.2 m. Then,

0.7(7.2)  KL  = 99.02 =   π  y 0.0509 Allowable Stress. The allowable stress will be determined using the AISC column 2π 2 E 2π 2 [200(109 )]  KL  = = 125.66. Since formulas. For A-36 steel, =  σY 250(106 )  π c KL KL a b 6 a b 6 200, the column is classified as a long column. r c r x 12p2E 23(KL>r)2

sallow =

=

12π 2 [200(109 )] = 60.10(106 ) N−m 2 23(130.912 )

P(0.3 m) and Interaction Formula. Bending is about the weak axis. Here, M = P(12) and bf 0.204 = = 0.102 m c = 2 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow

P 8.56(10 −3 ) 6

60.10(10 )

+

[ P(0.3)](0.102) [8.56(10 −3 )](0.0509 2 ) 100(106 )

3 = ) N 63.5 kN = P 63.53(10

σa

=

(σ a )allow

Ans.

63.53(10 3 ) 8.56(10 −3 ) = 0.123 < 0.15 60.10(106 )

O.K.

Ans: P = 63.5 kN 1134

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W12 * 50 A-36-steel W310 13–115. The A-36-steel 74 column column isis fixed fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the P kN is applied to the column, eccentric force P = 75 15 kip investigate if the column is adequate to support the loading. Use the allowable stress method.

in. 30012mm x y

P y

x

24 ftm 7.2

Solution Section Properties. From the table listed in the appendix, the section properties for W12 *50 a W310 74are are

= A 9480 = mm 2 9.48(10 −3 ) m 2

= πx 132 = mm 0.132 m

6 = I y 23.4(10 = ) mm 4 22.3(10 −6 ) m 4

= πy 49.7 = mm 0.0497 m

= bf 205 = mm 0.205 m Slenderness Ratio. Here, Lx = 7.2 m and foπ a colσmn fixed at its base and pinned at its top, Kx = 2. Thus,

2(7.2)  KL  = 109.09 (contπols)  =   π  x 0.132 Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 7.2 m. Then,

0.7(7.2)  KL  = 101.41 =  π   y 0.0497 Allowable Stress. The allowable stress will be determined using the AISC column

2π 2 E 2π 2 [200(109 )]  KL  = = 125.66. Since formulas. For A-36 steel, =  σY 250(106 )  π c KL KL a b 6 a b , the column can be classified as an intermediate column. r x r c

B1 sallow =

(KL>r)2 2(KL>r)C 2

R sY

3(KL>r) (KL>r)3 5 + 3 8(KL>r)C 8(KL>r)C 3

1 109.09 2 C 1c 1-a b a b d6)(250) 125.66 2 2 .10  3 109.09 1 109.09 3 = 5 ba b 111.20 a b a3 b 5  a3(111.20) 3 + 8 125.668 125.66 3 3 8(126.10) 8 A 126.10 B

 81.55 MPa

= 75(0.3) 15(12)  = 22.5 180 kip Maximum Stress. Bending is about the weak axis. Since, M  kN #· in m. bf 0.205 and= c = = 0.1025 m 2 2 smax =

75(10 3 ) [22.5(10 3 )](0.1025) P Mc + + = A I 9.48(10 −3 ) 23.4(10 −6 ) = 106.47(106 ) N−m 2 = 106.47 MPa

Since smax 7 sallow, the W310 * 74 column is inadequate according to the allowable stress method.

1135

"OT The column is not adequate.


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*13–116. The A-36-steel W250  67 column is fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method.

30012mm in. xx yy

P yy

xx

7.2ftm 24

Solution Section Properties. From the table listed in the appendix, the section properties for a W250 W10 *45 67are are

A 8560 = mm 2 8.56(10 −3 ) m 2 =

πx 110 mm 0.11 m = =

6 I y 22.2(10 ) mm 4 22.2(10 −6 ) m 4 = =

πy 50.9 mm 0.0509 m = =

bf 204 = mm 0.204 m = Slenderness Ratio. Here, Lx = 7.2 m and for a column fixed at its base and free at its top, Kx = 2. Thus,

2(7.2)  KL  = 130.91 (contπols)  =  0.11  π x Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 7.2 m. Then,

0.7(7.2)  KL  = 99.02 =   π  y 0.0509 Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A-36 steel,

 KL  =   π c

2π 2 E =

σY

KL KL 2π 2 [200(109 )] b 6 a b 6 200, = 125.66. Since a r c r x 250(106 )

the column is classified as a long column. sallow =

=

12p2E 23(KL>r)2

12π 2 [200(109 )] 6 = 60.10(10 = ) N−m 2 60.10 MPa 23(130.912 )

Maximum Stress. Bending is about the weak axis. Since M = P(12) and P(0.3 m) and bf 0.204 c = = = 0.102 m 2 2 sallow =

Mc P + A I

6 = 60.10(10 )

P −3

8.56(10 )

+

[P(0.3)](0.102) 22.2(10 −6 )

3 = P 40.19(10 = ) N 40.2 kN

Ans.

Ans: P = 40.2 kN 1133

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13–117. The A-36-steel W310 W12 * 50 74 column column is is fixed fixed at its base. Its top is constrained to move along the x–x axis but free to rotate about and move along the y–y axis. If the eccentric force P  = 65 15 kip kN is applied to the column, investigate if the column is adequate to support the loading. Use the interaction formula. The allowable bending stress is (sb)allow = 15 100ksi. MPa

in. 30012mm x y

P y

x

24 ftm 7.2

Solution Section Properties. From the table listed in the appendix, the section properties for W12 *50 a W310 74are are

A 9480 = mm 2 9.48(10 −3 ) m 2 = 6

−6

4

πx 132 = mm 0.132 m = 4

πy 49.7 mm 0.0497 m = I y 23.4(10 ) mm 22.3(10 ) m = = = bf 205 mm 0.205 m = = Slenderness Ratio. Here, Lx = 7.2 m and foπ a colσmn fixed at its base KL and pinned at its top, Thus, Kx = 2.2(288) a b

2(7.2)  KL  = 109.09 (contπols)  =  π   x 0.132 Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 7.2 m. Thσs,

0.7(7.2)  KL  = 101.41 =  π   y 0.0497

2p2 C 29 A 103 B The D allowable stress will KLbe determined KL Allowable Stress. using the AISC column = 126.10. Since , the column can be 2π 2 E 2π 2 [200(109 )]  KL  = = 125.66. Since formulas. For steel, =  classified as an A-36 intermediat σY 250(106 )  π c KL KL (KL>r)2 b 6 a b , the columnC can a 1 - be classifiedSas s an intermediate column. r x r c

bf 8.08 = 180 kip # Formula. in. and c = Bending = = 4.04 in. Interaction 2 2is about the weak axis. Here M  65(0.3)  19.5 kN · m bf 0.205 c = = 0.1025 m and = 2 2 (sa)allow (sb)allow P>A Mc>Ar2 65(10 3 ) 9.48(10 −3 ) [19.5(10 3 )](0.1025) [9.48(10 −3 )](0.0497 2 ) + + = P 14.57 kip 14.6 ki (sa=)allow (sb)=allow 81.55(106 ) 100(106 )

= 0.9376 < 1

σa

= (σ a )allow

65(10 3 ) 9.48(10 −3 ) = 0.0841 < 0.15 81.55(106 )

O.K. O.K.

Thus, a W12 * 50 column is adequate according to the interaction formula. Thus, a W310  74 column is adequate according to the interaction formula.

1136

Ans: The column is adequate.

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P P

13–118. A 4.8-m-long column is made of aluminum alloy 2014-T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and Eq. 13–30.

A 4.25mm in. 106 x in. 120.5 mm y 8 in. 200 mm

y x 200 8 in.mm

12 mm 0.5 in.

12 mm 0.5 in.

Solution Section properties:

A= 0.012(0.2) + 2[0.2(0.012)] = 7.2(10 −3 ) m 2 Ix =

1 1 (0.2)(0.224 3 ) − (0.188)(0.2 3 ) = 61.9904(10 −6 ) m 4 12 12

1 1  I y =(0.2)(0.012 3 ) + 2  (0.012)(0.2 3 ) = 16.0288(10 −6 ) m 4 12 12   πy =

Iy = A

16.0288(10 −6 ) = 0.04718 m 7.2(10 −3 )

Allowable stress method:

KL 0.5(4.8) KL = = 50.87, 12 < < 55 πy πy 0.04718 

 KL  

=  212 − 1.59  σ allow   MPa  π   = [212 − 1.59(50.87)] = 131.12 MPa smax = sallow = 6 131.12(10 = )

Mx c P + A Ix

P −3

7.2(10 )

+

[ P(0.106)](0.112) 61.9904(10 −6 )

3 = = ) N 397 kN P 396.86(10

Ans.

Ans: P = 397 kN 1137

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P P

13–119. A 4.8-m-long column is made of aluminum alloy 2014T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and the interaction formula with (σb)allow = 140 MPa.

A 4.25mm in. 106 x in. 120.5 mm y 8 in. 200 mm

y x 200 8 in.mm

12 mm 0.5 in.

12 mm 0.5 in.

Solution Section properties:

A= 0.012(0.2) + 2[0.2(0.012)] = 7.2(10 −3 ) m 2 Ix =

1 1 (0.2)(0.224 3 ) − (0.188)(0.2 3 ) = 61.9904(10 −6 ) m 4 12 12

1 1  I y =(0.2)(0.012 3 ) + 2  (0.012)(0.2 3 ) = 16.0288(10 −6 ) m 4 12  12  πy =

16.0288(10 −6 ) = 0.04718 m 7.2(10 −3 )

Iy = A

Allowable stress method:

KL KL 0.5(4.8) = = 50.87, 12 < < 55 πy πy 0.04718 

 KL  

=  212 − 1.59  σ allow   MPa  π   = [212 − 1.59(50.87)] = 131.12 MPa

σ= a

P P = = 138.89 P A 7.2(10 −3 )

σb =

Mc [P(0.106)](0.112) = = 191.51P Ix 61.9904(10 −6 ) sa sb + = 1.0 (sa)allow (sb)allow

138.89 P 6

131.12(10 )

+

191.51(106 ) 140(106 )

= 1.0

3 = P 412.00(10 = ) N 412 kN

Ans.

Ans: P = 412 kN 1138

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*13–120. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the allowable stress method. The thickness of the wall for the section is t = 12 mm.

in. 150 6mm

P

753 mm in. in. 150 6mm

8 ft m 2.4

Solution Section Properties.

A =0.15(0.075) − 0.126(0.051) =4.824(10 −3 ) m 2 1 1 I x = (0.075)(0.153 ) − (0.051)(0.126 3 ) =12.592152(10 −6 ) m 4 12 12 1 1 I y = (0.15)(0.0753 ) − (0.126)(0.0513 ) =3.880602(10 −6 ) m 4 12 12 = πx

Ix = A

12.592152(10 −6 ) = 0.05109 m 4.824(10 −3 )

= πy

Iy = A

3.880602(10 −6 ) = 0.02836 m 4.824(10 −3 )

Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus,

2(2.4)  KL  = 169.24 =  π 0.02836  y Allowable Stress. Since a

KL b 7 55, the column can be classified as a long r y

column.

= σ allow

372550 MPa 372550 MPa 13.01 MPa = = (KL π )2 169.24 2

Maximum Stress. Bending occurs about the strong axis so that M ==P(0.150 P(6) and m) and 0.15 c = 0.075 m = 2 sallow =

13.01(106 ) =

P Mc + A I

P 4.824(10 −3 )

+

[ P(0.15)](0.075) 12.592152(10 −6 )

P = 11.82(10 3 ) N = 11.8 kN

Ans.

Ans: P = 11.8 kN 1139

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13–121. The 2014-T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the interaction formula. The allowable bending stress is (sbb)allow == 200 The thickness ofof the wall forfor thethe section is 30 ksi. MPa. The thickness the wall section tis =t =0.5 12in. mm.

1506mm in. 753 mm in.

P

150 6mm in.

Solution

82.4 ft m

Section Properties.

A =0.15(0.075) − 0.126(0.051) =4.824(10 −3 ) m 2 1 1 I x = (0.075)(0.153 ) − (0.051)(0.126 3 ) =12.592152(10 −6 ) m 4 12 12 1 1 I y = (0.15)(0.0753 ) − (0.126)(0.0513 ) =3.880602(10 −6 ) m 4 12 12 = πx

Ix = A

12.592152(10 −6 ) = 0.05109 m 4.824(10 −3 )

= πy

Iy = A

3.880602(10 −6 ) = 0.02836 m 4.824(10 −3 )

Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus,

2(2.4)  KL  = 169.24 =   π  y 0.02836 Allowable Stress. Since a

KL b 7 55, the column can be classified as a long r y

column.

= σ allow

372550 MPa 372550 MPa 13.01 MPa = = (KL π)2 169.24 2

Maximum Stress. Bending occurs about the strong axis so that M ==P(0.150 P(6) and m) and M = P(6) and 0.15 = c = 0.075 m 2 P>A Mc>Ar2 + = 1 (sa)allow (sb)allow P 4.824(10 −3 ) 13.01(106 )

+

P(0.15)(0.075) [4.824(10 −3 )](0.05109 2 ) 200(106 )

3 = P 49.01(10 = ) N 49.0 kN

= 1 Ans.

Ans: P = 49.0 kN 1140

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13–122. Determine if the column can support the eccentric compressive load of 7.5 kN. Assume that the ends are pin connected. Use the NFPA equations in Sec. 13.6 and Eq. 13–30.

7.5 kN 300 mm 75 mm

40 mm

1.8 m

Solution = A 0.3(0.04) = 0.012= m2 Ix

1 = (0.04)(0.33 ) 90.0(10 −6 ) m 4 12 7.5 kN

d = 0.04 m 1.0(1.8) KL = 45 = 0.04 d 26 6

KL 6 50 d

= (σ a )allow smax =

3725 MPa 3725 MPa = = 1.8395 MPa (KL d)2 452

Mx c P + A Ix

7.5(10 3 ) [7.5(10 3 )](0.075)(0.15) = + 0.012 90.0(10 −6 ) = 1.5625(106 ) N−m 2 = 1.5625 MPa (sa)allow 7 smax The column is adequate. Yes.

Ans.

Ans: Yes. 1424

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13–123. Determine if the column can support the eccentric compressive load of 7.5 kN. Assume that the bottom is fixed and the top is pinned. Use the NFPA equations in Sec. 13.6 and Eq. 13–30.

7.5 kN 300 mm 75 mm

40 mm

1.8 m

Solution 2 = = 0.012 m A 0.3(0.04) = Ix

1 = (0.04)(0.33 ) 90.0(10 −6 ) m 4 12

7.5 kN

d = 0.04 m 0.7(1.8) KL = 31.5 = 0.04 d 26 6

KL 6 50 d

= (σ a )allow

smax =

3725 MPa 3725 MPa = = 3.754 MPa (KL d)2 31.52

Mxc 7.5(10 3 ) [7.5(10 3 )](0.075)(0.15) P + = + A Ix 0.012 90.0(10 −6 )

1.5625(106 ) N−m 2 = 1.5625 MPa = (sa)allow 7 smax The column is adequate. Yes.

Ans.

Ans: Yes 1425

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*13–124. The 250-mm-diameter utility pole supports the transformer that has a weight of 3 kN and center of gravity at G. If the pole is fixed to the ground and free at its top, determine if it is adequate according to the NFPA equations of Sec. 13.6 and Eq. 13–30.

G

15375 in.mm

18 ft 5.4 m

Solution KL 2(5.4) = = 43.2 0.25 d 26
Ar2 P>A + = 1 (sa)allow (sb)allow

P 0.0152 70.40(106 )

+

[ P(0.038)](0.076) 0.0152(0.04388 2 ) 126(106 )

= 1

3 P 582.16(10 = = ) N 582 kN

Ans.

Ans: P = 582 kN 1142

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R13–1. If the torsional springs attached to ends A and C of the rigid members AB and BC have a stiffness k, determine the critical load Pcr.

P

L 2

Solution

Member AB + c ©Fy = 0; a + ©MA = 0;

By - P = 0

(1)

By a

(2)

L L sin u b + Bx a cos u b - M = 0 2 2

Member BC a + ©MC = 0; -By a

L L sin u b + Bx a cos u b + M = 0 2 2

(3)

Solving Eqs. (1), (2), and (3), we obtain Bx = 0

By =

2M L sin u

M =

PL sin u 2

Since u is very small, the small angle analysis gives sin u ⬵ u. Thus, PL u 2

(4)

Torslonal Spring Moment. The restoring couple moment Msp can be determined using the torsional spring formula, M = ku. Thus, Msp = ku Critical Buckling Load. When the mechanism is on the verge of bucklling M must equal Msp. M = Msp Pcr L u = ku 2 Pcr =

A

B

Equilibrium. When the system is given a slight lateral disturbance, the configuration shown in Fig. a is formed. The couple moment M can be related to P by considering the equilibrium of members AB and BC.

M =

k

2k L

Ans.

1149

L 2 k

C

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R13–1. Continued

Ans: 2M PL ,M = sin u, L sin u 2 2k PL u, Pcr = M = 2 L

Bx = 0, By =

1150

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R13–2. Determine the maximum intensity w of the uniform distributed load that can be applied on the beam without causing the compressive members of the supporting truss to buckle. The members of the truss are made from A-36-steel rods having a 60-mm diameter. Use F.S. = 2 against buckling.

w

B A

2m

C

3.6 m

1.5 m

D

Solution Equilibrium. The force developed in member BC can be determined by considering the equilibrium of the free-body diagram of the beam AB, Fig. a. 3 w(5.6)(2.8) - FBC a b (5.6) = 0 FBC = 4.6667w 5

a + ©MA = 0;

The Force developed in member CD can be obtained by analyzing the equilibrium of joint C, Fig. b, + c ©Fy = 0;

FAC a

5 3 b - 4.6667w a b = 0 13 5

+ : ©Fx = 0;

4 12 4.6667wa b + 7.28a b w - FCD = 0 5 13

FAC = 7.28w (T) FCD = 10.4533w (C)

Section Properties. The cross-sectional area and moment of inertia of the solid circular rod CD are A = p A 0.032 B = 0.9 A 10 - 3 B p m2

I =

p A 0.034 B = 0.2025 A 10 - 6 B p m4 4

Critical Buckling Load. Since both ends of member CD are pinned, K = 1. The critical buckling load is Pcr = FCD (F.S.) = 10.4533w(2) = 20.9067w Applying Euler’s formula, Pcr =

p2EI (KL)2

20.9067w =

p2 C 200 A 109 B D C 0.2025 A 10 - 6 B p D [1(3.6)]2

w = 4634.63 N>m = 4.63 kN>m

Ans.

Critical Stress: Euler’s formula is valid only if scr 6 sY. scr =

20.907(4634.63) Pcr = = 34.27 MPa 6 sY = 250 MPa A 0.9 A 10 - 3 B p

O.K.

Ans: w = 4.63 kN>m 1151

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R13–3.  A steel column has a length of 5 m and is free at one end and fixed at the other end. If the cross-sectional area has the dimensions shown, determine the critical load. Est = 200 GPa, sY = 360 MPa.

10 mm 10 mm

60 mm 80 mm

Solution Section Properties: A = 0.06 (0.01) + 2 (0.06)(0.01) = 1.80(10 - 3) m2 y =

0.005 (0.06)(0.01) + 2[0.03 (0.06)(0.01)] ΣyA = = 0.02167 m ΣA 0.06 (0.01) + 2 (0.06)(0.01)

Ix =

1 (0.06)(0.01)3 + 0.06 (0.01)(0.02167 - 0.005)2 12 + [

Iy =

1 (0.01)(0.06)3 + 0.01 (0.06)(0.03 - 0.02167)2] = 0.615 (10 - 6) m4  (controls) 12

1 1 (0.06)(0.08)3 (0.05)(0.06)3 = 1.66 (10 - 6) m4 12 12

Critical Load: Pcr = =

p2EI ; (KL)2

K = 2.0

p2 (200)(109)(0.615)(10 - 6) [2.0 (5)]2 Ans.

= 12140 N = 12.1 kN

Check Stress: scr =

Pcr 12140 = = 6.74 MPa 6 sY = 360 MPa A 1.80 (10 - 3)

Hence, Euler’s equation is still valid.

Ans: Pcr = 12.1 kN 1404

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*R13–4. The A-36-steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its mid-height along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the interaction formula. The allowable bending stress is (sb)allow = 100 MPa.

600 mm 10 kN

4.5 m

Solution Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10 - 3 B m2 Ix = 71.1 A 10 B mm = 71.1 A 10 6

4

-6

d = 266 mm = 0.266 m

Bm

rx = 112 mm = 0.112 m

4

4.5 m

ry = 35.1 mm = 0.0351 mm

Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus, a

0.7(9) KL b = = 56.25 r x 0.112

Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4..5 m. Then, a

Allowable =

C

1(4.5) KL b = = 128.21 (controls) r y 0.0351

Axial

2p2 C 200 A 109 B D 250 A 106 B

Stress.

For

= 125.66. Since a

A–36

steel,

a

KL 2p2E b = r c B sY

KL KL b 6 a b 6 200, the column can be r c r y

classified as a long column. sallow =

12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 23(KL>r)2 23(128.21)2

Interaction Formula. Bending is about the strong axis. Here, P = 10 + 40 = 50 kN, 0.266 d M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 P>A Mc>Ar2 + = (sa)allow (sb)allow

50 A 103 B n 5.70 A 10 - 3 B 62.657 A 106 B

+

= 0.5864 6 1 sa = (sa)allow

50 A 103 B n 5.7 A 10 - 3 B 62.657 A 106 B

24 A 103 B (0.133) n C 5.70 A 10 - 3 B A 0.1122 B D 100 A 106 B

O.K.

= 0.140 6 0.15

O.K.

Thus, a W250 * 45 column is adequate according to the interaction formula.

1154

40 kN

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R13–5. If the A-36 steel solid circular rod BD has a diameter of 50 mm, determine the allowable maximum force P that can be supported by the frame without causing the rod to buckle. Use F.S. = 2 against buckling.

100 mm A

B

1.2 m 0.9 m

C 0.9 m

D

Solution

P

Equilibrium. The compressive force developed in BD can be determined by considering the equilibrium of the free-body diagram of member ABC, Fig. a. a + © MA = 0;

4 FBD   (0.9) − P(1.9) + P(0.1) = 0 5

FBD= 2.5P

Section Properties. The cross-sectional area and moment of inertia of BD are 2 = A π= (0.0252 ) 0.625(10 −3 )π m = I

π

= (0.0254 ) 97.65625(10 −9 )π m 4 4

Critical Buckling Load. Since BD is pinned at both of its ends, K = 1. The critical buckling load is Pcr = FBD(F.S.) = 2.5P(2) = 5P The length of BD is L = Pcr =

5P =

0.9 2 + 1.2 2 = 1.5 m . Applying Euler’s for,mula

p2EI (KL)2

π 2 [200(109 )][97.65625(10 −9 )π ] [1.0(1.5)]2

3 = = P 53.83(10 ) N 53.8 kN

Ans.

Critical Stress. Euler’s formula is valid only if scr 6 sY.

σ= cπ

Pcπ 5[53.83(10 3 )] = = 137.08(106 ) N−m 2 A 0.625(10 −3 )π

= 137.08 = MPa < σ Y 250 MPa

(O.K.)

4 3

Ans: P = 53.8 kN 1433

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112 mm

R13–6. If P = 75 kN, determine the required minimum diameter of the A992 steel solid circular rod BD to the nearest mm. Use F.S. = 2 against buckling.

A

B

1.2 m 0.9 m

C 0.9 m

D

Solution Equilibrium. The compressive force developed in BD can be determined by considering the equilibrium of the free-body diagram of member ABC, Fig. a,

4 FBD   (0.9) − 75(1.9) + 75(0.1)= 0  5

a+ © MA = 0;

P

FBC= 187.5 kN

Section Properties. The cross-sectional area and moment of inertia of BD are A =

p 2 d 4

I =

p 4 p d 4 a b = d 4 2 64

Critical Buckling Load. Since BD is pinned at both of its ends, K = 1. The critical buckling load is Pcr = FBD(F.S.) = 187.5(2) = 375 kN The length of BD is L = Pcr =

375(10 3 ) =

0.9 2 + 1.2 2 = 2.5 m. Applying Euler’s formula,

p2EI (KL)2

π 2 [200(109 )]

( 64π d 4 )

[1.0(1.5)]2

= d 0.05432 m 54.32 mm = Use d = 55 mm

Ans.

Critical Stress. Euler’s formula is valid only if scr 6 sY.

σ= cπ

Pcπ = A

375(10 3 ) = 157.84(106 ) N−m 2 π (0.0552 ) 4

= 157.85 = MPa < σ Y 345 MPa

(O.K.)

4 3 Ans: Use d = 55 mm

1434

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R13–7.  The steel pipe is fixed supported at its ends. If it is 4 m long and has an outer diameter of 50 mm, determine its required thickness so that it can support an axial load of P = 100 kN without buckling. Est = 200 GPa, sY = 250 MPa.

P

4m

Solution I =

p (0.0254 - r i 4 ) 4

Critical Load: Pcr =

p2EI ; (K L)2

100(103) =

P

K = 0.5

p2(200)(109)[p4 (0.0254 - r i4)] [0.5(4)]2

ri = 0.01908 m = 19.1 mm t = 25 mm - 19.1 mm = 5.92 mm

Ans.

Check Stress: s =

100(103) Pcr = = 122 MPa 6 sY = 345 MPa A p(0.0252 - 0.01912)

(OK)

Ans: t = 5.92 mm 1408

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*R13–8. The W200 * 46 wide-flange A992-steel column can be considered pinned at its top and fixed at its base. Also, the column is braced at its mid-height against weak axis buckling. Determine the maximum axial load the column can support without causing it to buckle.

6m

Solution Section Properties: From the table listed in the appendix, the section properties for a W200 * 46 are A = 5890 mm2 = 5.89 1 10 - 3 2 m2

Iy = 15.3 1 106 2 mm4 = 15.3 1 10 - 6 2 m4

6m

Ix = 45.5 1 106 2 mm4 = 45.5 1 10 - 6 2 m4

Critical Buckling Load: For buckling about the strong axis, Kx = 0.7 and Lx = 12 m. Since the column is fixed at its base and pinned at its top,

Pcr =

p2EIx (KL)x 2

=

p2 c 200 1 109 2 d c 45.5 1 10 - 6 2 d [0.7(12)]2

= 1.273 1 106 2 N = 1.27 MN

For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides a support equivalent to a pin. Applying Euler’s formula,

Pcr =

p2EIy (KL)y 2

=

p2 c 200 1 109 2 d c 15.3 1 10 - 6 2 d [1(6)]2

= 838.92 kN = 839 kN (controls) Ans.

Critical Stress: Euler’s formula is valid only if scr 6 sY . scr =

838.92 1 103 2 Pcr = = 142.43 MPa 6 sY = 345 MPa A 5.89 1 10 - 3 2

(O.K.)

Ans: Pcr = 839 kN 1409

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R13–9.  P

The wide-flange A992 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine the maximum force P that can be applied at A without causing it to buckle or yield. Use a factor of safety of 3 with respect to buckling and yielding.

20 mm A 10 mm 100 mm 4m

100 mm

150 mm A

10 mm 100 mm

10 mm

Solution Section Properties: ΣA = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 - 3) m2 ~A 0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) Σx = ΣA 4.5(10 - 3)

x =

= 0.06722 m 1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12

Iy =

+

1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12

+

1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 12

= 20.615278 (10 - 6) m4 1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) 12 12 12

Ix =

= 7.5125 (10 - 6) m4 Iy

ry =

BA

=

20.615278(10 - 6) B

4.5 (10 - 3)

= 0.0676844

Buckling about x–x axis: p2(200)(109)(7.5125)(10 - 6) p2 EI = 2 (KL) [2.0(4)]2

Pcr =

= 231.70 kN scr =

(controls)

231.7 (103) Pcr = = 51.5 MPa 6 sg = 345 MPa A 4.5 (10 - 3)

Yielding about y–y axis: smax =

P P ec KL b d; c 1 + 2 sec a A 2r B EA r

e = 0.06722 - 0.02 = 0.04722 m

0.04722 (0.06722) ec = = 0.692919 r2 0.06768442

2.0 (4) P P KL = 1.96992 (10 - 3) 2P = 9 -3 B EA 2(0.0676844) B 200 (10 )(4.5)(10 ) 2r

345(106)(4.5)(10 - 3) = P[1 + 0.692919 sec (1.96992 (10 - 3) 2P)]

By trial and error: Hence,

P = 434.342 kN Pallow =

231.70 = 77.2 kN 3

Ans. 1410

Ans: Pallow = 77.2 kN

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R13–10.  P

The wide-flange A992 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine if the column will buckle or yield when the load P = 10 kN is applied at A. Use a factor of safety of 3 with respect to buckling and yielding.

20 mm A 10 mm 100 mm 4m

100 mm

150 mm A

10 mm 100 mm

10 mm

Solution Section Properties: ΣA = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10 - 3) m2 x =

~A 0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) Σx = = 0.06722 m ΣA 4.5 (10 - 3)

Iy =

Ix = ry =

1 (0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2 12 +

1 (0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2 12

+

1 (0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 = 20.615278 (10 - 6) m4 12

1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) = 7.5125 (10 - 6) m4 12 12 12 Iy BA

=

20.615278 (10 - 6) B

4.5 (10 - 3)

= 0.0676844 m

Buckling about x–x axis: Pcr =

p2(200)(109)(7.5125)(10 - 6) p2 EI = = 231.70 kN 2 (KL) [2.0(4)]2

scr =

231.7 (103) Pcr = = 51.5 MPa 6 sg = 345 MPa A 4.5 (10 - 3)

Pallow =

(O.K.)

Pcr 231.7 = = 77.2 kN 7 P = 10 kN FS 3

Hence the column does not buckle. Yielding about y–y axis: smax = P =

P ec KL P c 1 + 2 sec a bd A 2r B EA r

e = 0.06722 - 0.02 = 0.04722 m

10 = 3.333 kN 3

3.333 (103) P = = 0.7407 MPa A 4.5 (10 - 3) 0.04722 (0.06722) ec = = 0.692919 2 r (0.067844)2 2.0 (4) 3.333 (103) KL P = = 0.113734 2 r BE A 2(0.0676844) B 200 (109)(4.5)(10 - 3)

smax = 0.7407 [1 + 0.692919 sec (0.113734)] = 1.26 MPa 6 sg = 345 MPa Hence the column does not yield! Ans.

No. 1411

Ans: It does not buckle or yield

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14–1.  A material is subjected to a general state of plane stress. Express the strain energy density in terms of the elastic constants E, G, and n and the stress components sx, sy, and txy.

sy txy

sx

Solution Strain Energy Due to Normal Stresses: We will consider the application of normal stresses on the element in two successive stages. For the first stage, we apply only sx on the element. Since sx is a constant, (Ui)1 =

s2x s2x V dV = 2E Ln 2E

When sy is applied in the second stage, the normal strain Px will be strained by vsy Px ′ = -nPy = . Therefore, the strain energy for the second stage is E s2y

(Ui)2 =

Ln 2E

=

Ln 2E

¢

J

s2y

Since sx and sy are constants, (Ui)2 =

+ sx Px ′ ≤dV + sxa -

nsy E

b R dV

V (s2y - 2nsx sy) 2E

Strain Energy Due to Shear Stresses: The application of txy does not strain the element in a normal direction. Thus, from Eq. 14–11, we have (Ui)3 =

t2xy

Ln 2G

dV =

t2xy V 2G

The total strain energy is Ui = (Ui)1 + (Ui)2 + (Ui)3 =

t2xy V s2x V V + (s2y - 2nsx sy) + 2E 2E 2G

=

t2xy V V (s2x + s2y - 2nsx sy) + 2E 2G

and the strain energy density is t2xy Ui 1 = (s2x + s2y - 2nsx sy) + V 2E 2G

Ans.

These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.

1412

Ans: t2xy Ui 1 2 = (sx + s2y - 2nsxsy) + V 2E 2G

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14–2.  The strain-energy density for plane stress must be the same whether the state of stress is represented by sx, sy, and txy, or by the principal stresses s1 and s2. This being the case, equate the strain–energy expressions for each of these two cases and show that G = E>[2(1 + n)].

Solution U =

1 2 1 n (s2x + s2y) - sxsy + t R dV 2 E E 2 G xy Ln

U =

1 n J (s21 + s22) s s R dV E 1 2 Ln 2 E

J

Equating the above two equations yields. 1 n 1 2 1 n (s2 + s2y) ss + t = (s2 + s22) s s 2E x E x y 2 G xy 2E 1 E 1 2 However, s1, 2 = Thus,

1

s21

+

s22

sx + sy

2

=

2

{

s2x

s2y

and also

+

A

a

sx - sy

+ 2

2 t2xy

(1)

2

2 b + txy

s1 s2 = sxsy - t2xy

Substitute into Eq. (1) 1 n 1 2 1 n n 2 1 s2x + s2y 2 - sxsy + t = (s2 + s2y + 2t2xy) ss + t 2E E 2 G xy 2E x E x y E xy t2xy 1 2 n 2 txy = + txy 2G E E 1 1 n = + 2G E E 1 1 = (1 + n) 2G E G =

E 2(1 + n)

QED

Ans: N/A 1413

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14–3.  The A-36 steel bar consists of two segments, one of circular cross section of radius r, and one of square cross section. If the bar is subjected to the axial loading of P, determine the dimensions a of the square segment so that the strain energy within the square segment is the same as in the circular segment.

P a a

2L L

P

Solution Axial Strain Energy: Applying Eq. 14–16 to the circular segment gives (Ui)c =

P2(2L) N2Lc P2L = = 2 2AE pr 2E 2 ( pr ) E

Applying Eq. 14–16 to the square segment gives (Ui)s = Require

N2Ls P2L P2L = = 2 2AE 2a2E 2( a )E

       (Ui)c = (Ui)s P2L P2L = 2 pr E 2a2E a =

p r A2

Ans.

Ans: a = 1414

p r A2

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*14–4. Determine the torsional strain energy in the A992 steel shaft. The shaft has a radius of 50 mm.

8 kNm 0.6 m 0.6 m 12 kNm

Solution Internal Torque: As shown on the FBD in Fig. a and b. p ( 0.054 ) = Torsional Strain Energy: With polar moment of inertia J = 2 -6 4 3.125 ( 10 ) p m and G = 75 GPa for A992 steel, T 2L 2GJ 1 = 5 3 12.0 ( 103 ) 4 2 (0.6) + 2GJ

Ui = Σ

= =

0.1632 ( 109 ) N2 # m3

3 20.0 ( 103 ) 4 2 (0.6) 6

GJ 0.1632 ( 109 ) 75 ( 109 ) 3 3.125 ( 10-6 ) p 4

= 221.65 J

Ans.

= 222 J

Ans: Ui = 222 J 1415

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14–5. If P = 50 kN, determine the total strain energy stored in the truss. Each member has a diameter of 50 mm and is made of A992 steel.

D

1.2 m

A

C B 0.9 m

Solution

0.9 m P

Normal Forces. The normal forces developed in each member of the truss can be determined using the method of joints. Joint B (Fig. a) + ©F = 0; : x

FBC - FAB = 0

+ c ©Fy = 0;

FBD - 50 = 0

(1) FBD = 50 kN (T) (max)

Joint D (Fig. b) + : ©Fx = 0;

 3  3 FAD = 0   − FCD   =  5  5

+ c ©Fy = 0;

  4  2  F    − 50 = 0   5 

FAD = FCD = F = 31.25 kN (C)

 3 31.25   − FBC = 0  5

FBC = 18.75 kN (T)

FAD = FCD = F

Joint C (Fig. c) + : ©Fx = 0;

Using the result of FBC, Eq. (1) gives FAB = 18.75 kN (T)

π

= (0.052 ) 0.625(10 −3 )π m 2 and L CD = 2 0.92 + 1.2 2 = 1.5 m 4

Axial Strain Energy. A =

(Ui)a = ©

4

4

3

N2L 2 AE 1

=

5

2[0.625(10 −3 )π ][200(109 )]

3

{[50(103 )]2 (1.2) + 2[31.25(103 )]2 (1.5) }

+ 2[18.75(10 3 )]2 (0.9)

= 8.356 N= ⋅ m 8.36 J

Ans.

This result is only valid if s 6 sY. We only need to check member BD since it is subjected to the greatest normal force sBD =

5

4

3

FBD 50(10 ) 6 25.46(10 = ) N−m 2 25.46 MPa < σ Y = = A 0.625(10 −3 )π = 345 MPa (O.K.)

1450

5

3

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D

14–6. Determine the maximum force P and the corresponding maximum total strain energy that can be stored in the truss without causing any of the members to have permanent deformation. Each member of the truss has a diameter of 50 mm and is made of A-36 steel.

1.2 m

A

C B 0.9 m

0.9 m P

Solution Normal Forces. The normal force developed in each member of the truss can be determined using the method of joints. Joint B (Fig. a) + ©F = 0; : x

FBC - FAB = 0

+ c ©Fy = 0;

FBD - P = 0

(1) FBD = P(T)

Joint D (Fig. b) + ©F = 0; : x

 3  3 FAD = 0   − FCD   =  5  5

+ c ©Fy = 0;

  4  2 F   − P = 0   5 

FAD = FCD = F = 0.625P (C)

 3 0.625   − FBC = 0  5

FBC = 0.375P (T)

FAD = FCD = F

Joint C (Fig. c) + ©F = 0; : x

Using the result of FBC, Eq. (1) gives FAB = 0.375P (T)

π

= (0.052 ) 0.625(10 −3 )π m 2 and LCD = 2 0.92 + 1.2 2 = 1.5 m 4 Member BD is critical since it is subjected to greatest normal force. Thus, Axial

Strain

A = Energy.

sY =

FBD A

250(106 ) =

P 0.625(10 −3 )π

3 ) N 491 kN = P = 490.87(10

Ans.

Using the result of P, 3 F= BD 490.87(10 ) N

(Ui)a = ©

3 F= AD F= CD 306.80(10 ) N

3 F= F= BC AB 184.08(10 ) N

}

{

1 N2L [490.87(10 3 )]2 (1.2) + 2[306.80(10 3 )]2 (1.5) + 2[184.08(10 3 )]2 (0.9) = 2AE 2[0.625(10 −3 )π ][200(109 )] = 805.34 = N ⋅ m 805 J

Ans.

Ans: P = 491 kN, Ui = 0.805 J 1451

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  Using bolts of the same material and cross-sectional area, two possible attachments for a cylinder head are shown. Compare the strain energy developed in each case, and then explain which design is better for resisting an axial shock or impact load.

L1

(a) L2

Solution Case (a) Ua =

N2L1 2AE

Ans. (b)

Case (b) Ub =

N2L2 2AE

Ans.

Since Ub 7 Ua, i.e., L2 7 L1, the design for case (b) is better able to absorb energy. Ans.

Case (b)

Ans: N2L1 N2L2 , Ub = 2AE 2AE Since Ub 7 Ua, i.e., L2 7 L1, the design for case (b) is better able to absorb energy. Ua =

1416

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  The shaft assembly is fixed at C. The hollow segment BC has an inner radius of 20 mm and outer radius of 40 mm, while the solid segment AB has a radius of 20 mm. Determine the torsional strain energy stored in the shaft. The shaft is made of 2014-T6 aluminum alloy. The coupling at B is rigid.

600 mm 20 mm

600 mm C 40 mm B 60 Nm

Solution

A 20 mm 30 Nm

Internal Torque: Referring to the free-body diagram of segment AB, Fig. a, ΣMx = 0;

TAB = - 30 N # m

TAB + 30 = 0

Referring to the free-body diagram of segment BC, Fig. b, ΣMx = 0;

TAB = -90 N # m

TBC + 30 + 60 = 0

p Torsional Strain Energy: Here, JAB = 1 0.024 2 = 80 1 10 - 9 2 p m4 and JBC = 2 p 1 0.044 - 0.024 2 = 1200 1 10 - 9 2 p m4, 2 (Ui)t = Σ =

TBC 2LBC TAB 2LAB T 2L = + 2GJ 2GJAB 2GJBC ( -30)2(0.6)

2 3 27 1 10

9

2 4 3 80 1 10 2 p 4 -9

= 0.06379 J = 0.0638 J

+

( - 90)2(0.6) 2 3 27 1 10

9

2 4 3 1200 1 10 - 9 2 p 4

Ans.

Ans: Ui = 0.0638 J 1421

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14–9. Determine the total axial and bending strain energy in the A992 steel beam. A = 2850 mm2, I = 28.9(106) mm4.

6 kN/m

50 kN

6m

Solution Support Reactions and Internal Loadings: The support reactions and the necessary moment function are shown on the FBD in Figs. a and b, respectively. Strain Energy: For the axial load and bending, Ui = (Ua)i + (Ub)i L

2

=

N2L M dx + 2EA L0 2EI

=

3 - 50 ( 103 ) 4 2(6)

= =

2EA

7.50 ( 109 ) N2 # m EA

+ +

L0

200 ( 10

) 3 2.85 ( 10 ) 4

= 64.41 J = 64.4 J

-3

3 ( 6.00x

-

1 6

x3 )( 103 ) 4 2dx

2EI

0.29623 ( 109 ) N2 # m3

7.50 ( 109 ) 9

6m

EI +

0.29623 ( 109 ) 200 ( 109 ) 3 28.9 ( 10-6 ) 4

Ans.

Ans: Ui = 64.4 J 1422

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14–10. Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 40 mm.

12 kNm 6 kNm

0.5 m

8 kNm

0.4 m 0.6 m

Solution Internal Torsional Moment: As shown on FBD. Torsional Strain Energy: With polar moment p 4 -6 4 J = 1 0.04 2 = 1.28 1 10 2 p m . Applying Eq. 14–22 gives 2

of

inertia

T 2L Ui = a 2GJ = = =

1 3 80002 (0.6) + 20002 (0.4) + 2GJ 45.0(106) N2 # m3

1 - 100002 2 (0.5) 4

GJ 45.0(106) 75(10 )[1.28(10 - 6) p] 9

Ans.

= 149 J

Ans: Ui = 149 J 1420

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14–11. Determine the torsional strain energy in the A992 steel shaft. The shaft has a radius of 40 mm.

8 kNm 0.6 m

30 kNm/m 0.6 m 0.8 m

Solution Internal Torque: As shown on the FBD in Fig. a, b and c. p ( 0.044 ) = Torsional Strain Energy: With polar moment of inertia J = 2 -6 4 1.28 ( 10 ) p m and G = 75 GPa for A992 steel, L

Ui = = = =

T 2dx L0 2GJ 1 e 2GJ L0

0.8 m

3 30 ( 103 ) x 4 2 dx

0.3264 ( 109 ) N2 # m3

+

3 24.0 ( 103 ) 4 2 (0.6)

+

3 16.0 ( 103 ) 4 2 (0.6) f

GJ 0.3264 ( 109 ) 75 ( 109 ) 3 1.28 ( 10-6 ) p 4

Ans.

= 1082.25 J = 1.08 kJ

Ans: Ui = 1.08 kJ 1419

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  If P = 60 kN, determine the total strain energy stored in the truss. Each member has a cross-sectional area of 2.5(103) mm2 and is made of A-36 steel.

2m B C

1.5 m

D A

Solution

P

Normal Forces: The normal force developed in each member of the truss can be determined using the method of joints. Joint A (Fig. a) + ΣFx = 0; S

FAD = 0

+ c ΣFy = 0;

FAB - 60 = 0

FAB = 60 kN (T)

Joint B (Fig. b) 3 FBDa b - 60 = 0    FBD = 100 kN (C) 5

+ c ΣFy = 0;

4 100 a b - FBC = 0    FBC = 80 kN (T) 5

+ ΣFx = 0; S Axial

Strain

2

Energy:

2

22 + 1.5 = 2.5 m (Ui)a = Σ =

N2L 2AE

A = 2.5 1 103 2 mm2 = 2.5 1 10 - 3 2 m2

1 2 3 2.5 1 10 - 3 2 4 3 200 1 109 2 4

c 3 60 1 103 2 4 2 (1.5) +

= 43.2 J

and

LBD =

3 100 1 103 2 4 2 (2.5) +

3 80 1 103 2 4 2 (2) d

Ans.

This result is only valid if s 6 sY . We only need to check member BD since it is subjected to the greatest normal force sBD =

100 1 103 2 FBD = = 40 MPa 6 sY = 250 MPa A 2.5 1 10 - 3 2

O.K.

Ans: Ui = 43.2 J 1417

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  Determine the maximum force P and the corresponding maximum total strain energy stored in the truss without causing any of the members to have permanent deformation. Each member has the cross-sectional area of 2.5(103) mm2 and is made of A-36 steel.

2m B C

1.5 m

D A

Solution

P

Normal Forces: The normal force developed in each member of the truss can be determined using the method of joints. Joint A (Fig. a) + ΣFx = 0; S

FAD = 0

+ c ΣFy = 0;

FAB - P = 0

FAB = P (T)

Joint B (Fig. b) 3 FBD a b - P = 0 5

+ c ΣFy = 0;

4 1.6667P a b - FBC = 0 5

+ ΣFx = 0; S

FBD = 1.6667P (C) FBC = 1.3333P(T)

Axial Strain Energy: A = 2.5 1 103 2 mm2 = 2.5 1 10 - 3 2 m2. Member BD is critical since it is subjected to the greatest force. Thus, sY =

FBD A

250 1 106 2 =

P = 375 kN Using the result of P FAB = 375 kN

1.6667P 2.5 1 10 - 3 2

FBD = 625 kN

Ans.

FBC = 500 kN

Here, LBD = 21.52 + 22 = 2.5 m. (Ui)a = Σ =

N2L 2AE

1 2 3 2.5 1 10 - 3 2 4 3 200 1 109 2 4

= 1687.5 J = 1.69 kJ

c 3 375 1 103 2 4 2 (1.5) +

3 625 1 103 2 4 2 (2.5)

+

3 500 1 103 2 4 2 (2) d

Ans.

Ans: P = 375 kN, Ui = 1.69 kJ 1418

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14–14.  Consider the thin-walled tube of Fig. 5–26. Use the formula for shear stress, tavg = T>2t A m , Eq. 5–18, and the general equation of shear strain energy, Eq. 14–11, to show that the twist of the tube is given by Eq. 5–20. Hint: Equate the work done by the torque T to the strain energy in the tube, determined from integrating the strain energy for a differential element, Fig. 14–4, over the volume of material.

Solution Ui =

t2 dV Lv 2 G

but t =

T 2 t Am

Thus, Ui =

T2 dV 2 2 Lv 8 t AmG L

=

2

T2 T2 T L dV dV dA = dx = 2 2 2 2 8 Am G Lv t 8 A2m G LA t 2 L0 L 8 AmG A t

However, dA = t ds. Thus, Ui =

ds T 2L 8 A2mG L t

Ue =

1 Tf 2

Ue = Ui ds 1 T 2L Tf = 2 8 A2mG L t f =

ds TL 2 4 AmG L t

QED

Ans: N/A 1425

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14–15. Determine the bending strain-energy in the beam due to the loading shown. EI is constant.

M0 A

C B L — 2

L — 2

Solution L

Ui =

M2dx L0 2EI

=

1 c 2EI L0

=

M20L 24EI

L>2

a

L>2 2 2 - M0 M0 x1 b dx1 + a x2 b dx2] L L L0

Ans.

Note: Strain energy is always positive regardless of the sign of the moment function.

Ans: Ui = 1170

M20L 24EI

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*14–16. The beam shown is tapered along its width. If a force P is applied to its end, determine the strain energy in the beam and compare this result with that of a beam that has a constant rectangular cross section of width b and height h.

b

h

Solution Moment of Inertia: For the beam with the uniform section, I =

bh3 = I0 12

For the beam with the tapered section, I =

I0 bh3 1 b a x b A h3 B = x = x 12 L 12L L

Internal Moment Function: As shown on FBD. Bending Strain Energy: For the beam with the tapered section, applying Eq. 14–17 gives L

UI =

M2 dx L0 2EI

=

L ( -Px)2 1 dx I0 2E L0 L x

=

P2L xdx 2EI0 L0

=

P2L3 3P2 L3 = 4EI0 bh3 E

L

Ans.

For the beam with the uniform section, L

Ui =

M2dx L0 2EI L

=

1 ( -Px)2 dx 2EI0 L0

=

P3 L3 6EI0

The strain energy in the capered beam is 1.5 times as great as that in the beam having a uniform cross section. Ans.

1177

L

P

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  The steel beam is supported on two springs, each having a stiffness of k = 8 MN>m. Determine the strain energy in each of the springs and the bending strain energy in the beam. Est = 200 GPa, I = 5(106) mm4.

2 kN/m

k 1m

k 2m

1m

Solution Fsp 4.00 ( 103 ) Spring Strain Energy: The spring deforms dsp = = = 0.500 ( 10-3 ) m k 8 ( 106 ) under the applied load. 1 2 kd 2 sp 1 = 3 8 ( 106 ) 4 3 0.500 ( 10- 2 4 2 2 = 1.00 J

(Ui)sp =

Ans.

Bending Strain Energy: Applying Eq. 14–17 gives L

(Ui)b = = = =

2

M dx L0 2EI 1 c2 2EI L0

0.400 kN EI

1m

2

( - 1.00x21 ) 2 dx1 +

# m3

L0

2m

( 2x2 - x22 - 1 ) 2 dx2 d

0.400 ( 106 ) 200 ( 109 ) 3 5 ( 10-6 ) 4

Ans.

= 0.400 J

Ans: (Ui)s p = 1.00 J (Ui)b = 0.400 J 1429

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14–18. w0

Determine the bending strain energy in the simply supported beam. EI is constant.

L 2

L 2

Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, + ΣF = 0; S x

Ax = 0 1 L w La b - Ay(L) = 0 2 0 2

a+ ΣMB = 0;

Ay =

w0 L 4

Internal Moment: Referring to the FBD of the beam’s left cut segment, Fig. b, w0L w0 1 2w0 x M + c a xbx d a b x = 0 M = (3L2x - 4x3) 2 L 3 4 12L

a+ ΣM0 = 0;

Bending Strain Energy: L

(Ui)b = Σ

2

M dx = 2 L0 2EI L0 =

=

=

L>2 c

2 w0 (3L2x - 4x3) d dx 12L 2EI

w 02

144 EIL2 L0 w 02 2

144 EIL

L>2

(9L4x2 + 16x6 - 24L2x4) dx.

(3L4x3 +

16 7 24 2 5 2 L>2 x Lx ) 7 5 0

17w 02L5 10 080 EI

Ans.

Ans: Ui = 1428

17w 02L5 10 080 EI

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14–19. M0

M0

Determine the bending strain energy in the beam. EI is constant.

L

Solution Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b). Bending Strain Energy: L

Ui =

M2dx L0 2EI

=

L M0 dx 2EI L0

=

M0L 2EI

Ans.

Ans: Ui = 1427

M0L 2EI

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*14–20. Determine the strain energy in the horizontal curved bar due to torsion. There is a vertical force P acting at its end. JG is constant.

r 90

Solution

P

T = Pr(1 - cos u) Strain energy: L

Ui =

T2 ds L0 2JG

However, s = ru;

ds = rdu u

Ui =

T2rdu r = 2JG 2JG L0 L0 P2r3 2JG L0

p>2

=

P2r3 2JG L0

p>2

=

P2r3 2JG L0

p>2

= =

P2r3 3p a - 1b JG 8

p>2

[Pr(1 - cos u)]2du

(1 - cos u)2 du (1 + cos2 u - 2 cos u)du (1 +

cos 2u + 1 - 2 cos u) du 2 Ans.

Ans: Ui = 1174

P2r3 3p a - 1b JG 8

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14–21.  w0

Determine the bending strain energy in the beam. EI is constant. A

B L

Solution 2 5

3 2 L w 0L w0 x 2 M dx 1 = a b dx = 2EI 2EI 6L 504 EI L0 L0 L

Ui =

Ans.

Ans: Ui = 1432

2 5

w 0L

504 EI

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14–22.  The bolt has a diameter of 10 mm, and the arm AB has a rectangular cross section that is 12 mm wide by 7 mm thick. Determine the strain energy in the arm due to bending and in the bolt due to axial force. The bolt is tightened so that it has a tension of 500 N. Both members are made of A-36 steel. Neglect the hole in the arm.

60 mm

B 30 mm

7 mm

Solution

50 mm

Axial Strain Energy: Applying Eq. 14–16 gives (Ui)b = = = =

N2L 2AE

A

5002(0.06) 2AE

7500 N2 # m AE 7500 p 4

( 0.01 ) 3 200 ( 109 ) 4 2

= 0.477 ( 10-3 ) J

Ans.

Bending Strain Energy: Applying Eq. 14–17 gives L

(Ui)l = =

M2dx L0 2EI 1 c 2EI L0

0.05 m

2

(187.5x1) dx1 +

1.171875 N2 # m3 = EI 1.171875 = 1 200 ( 109 ) 3 12 (0.012) ( 0.0073 ) 4

L0

0.03 m

2

(312.5x2) dx2 d

Ans.

= 0.0171 J

Ans: (Ui)b = 0.477 ( 10-3 ) J (Ui)l = 0.0171 J 1433

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14–23.  Determine the bending strain energy in the cantilevered beam. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on a segment dx of the beam is displaced a distance y, where y = w( - x4 + 4L3 x - 3L4)>(24EI), the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 12 (w dx)( - y). Integrate this equation to obtain the total strain energy in the beam. EI is constant.

w dx w

dx

x L

Solution Internal Moment Function: As shown on FBD. Bending Strain Energy: a) Applying Eq. 14–17 gives L

Ui =

M2dx L0 2EI L

=

1 w 2 J c - x2 d dx R 2EI L0 2 L

b) Integrating dUi =

dUi =

dUi =

Ui =

=

=

w2 J x4 dx R 8EI L0

=

w 2 L5 40EI

Ans.

1 (wdx)( - y) 2 1 w (wdx) J 2 24EI

1 - x4

+ 4L3x - 3L4 2 R

w2 1 x4 - 4L3x + 3L4 2 dx 48EI w2 48EI L0 w 2L5 40EI

L

1 x4

- 4L3x + 3L4 2 dx

Ans.

Ans: Ui = 1434

w 2L5 40 EI

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*14–24. Determine the bending strain energy in the simply supported beam. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on the segment dx of the beam is displaced a distance y, where y = w( - x4 + 2Lx3 - L3x)>(24EI), the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 12 (w dx)( - y). Integrate this equation to obtain the total strain energy in the beam. EI is constant.

w dx w

x

dx L

Solution Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b). Bending Strain Energy: a) Applying Eq. 14–17 gives L

Ui =

M2dx L0 2EI L

=

2 1 w 2 J c (Lx - x ) d dx R 2EI L0 2 L

=

w2 J (L2x2 + x4 - 2Lx3) dx R 8EI L0

=

w 2L5 240EI

b) Integrating dUi =

Ans. 1 (wdx) ( -y) 2

dUi =

1 w (wdx) J ( -x4 + 2Lx3 - L3x) R 2 24EI

dUi =

w2 (x4 - 2Lx3 + L3x) dx 48EI L

Ui =

=

w2 (x4 - 2Lx3 + L3x) dx 48EI L0 w 2L5 240EI

Ans.

Ans: Ui = 1435

w 2L5 240EI

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14–25.  Determine the vertical displacement of joint D. AE is constant.

C

0.8L

L

D

Solution

B

0.6L

A

P

Member Forces: By inspection of joint D, member AD is a zero-force member and FCD = P(T). Applying the method of joints at C, we have 4 + c ΣFy = 0 ;     FCA - P = 0    FCA = 1.25P (C) 5 + ΣFx = 0;    FCB - 3 (1.25P) = 0 FCB = 0.750P (T) S 5 At joint A 4 (1.25P) = 0 5 Axial Strain Energy: Applying Eq. 14–16, we have + c ΣFy = 0 ;    FAB -

FBA = 1.00P (T)

N2L Ui = a 2AE 1 = 3 P2(0.8L) + ( - 1.25P)2 (L) 2AE =

+ (0.750P)2 (0.6L) + (1.00P)2(0.8L) 4

1.750P2L AE

External Work: The external work done by force P is Ue =

1 (P)(∆ D)v 2

Conservation of Energy: Ue = Ui 1 1.750P2L (P)(∆ D)v = 2 AE (∆ D)v =

3.50PL AE

Ans.

Ans: (∆ D)v = 1436

3.50PL AE

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14–26. Determine the horizontal displacement of joint A. Each bar is made of A-36 steel and has a cross-sectional 2 2 950 in mm area of 1.5 . .

2 kip 10 kN

A A

0.93mft D D 0.93mft C C

B B

Solution

4 ft 1.2 m

Member Forces: Applying the method of joints to joint at A, we have + ©F = 0; : x

4 - 10 2 == 0 F 5 AD

+ c ©Fy = 0;

FAB -

3 (12.5) == 00 (2.50) 5

FAD = 12.5 2.50 kN kip (T) 10 kN

1.50kN kip(C) (C) FAB = 7.5

At joint D 12.5 kN

+ ©F = 0; : x

4 4 F - (2.50) (12.5) == 00 5 DB 5

+ c ©Fy = 0;

3 3 ((12.5) 2.50) + (12.5) (2.50) - FDC = 0 5 5

FDB = 12.5 2.50 kN kip (C)

kN (T) FDC = 15 3.00 kip (T) Axial Strain Energy: Applying Eq. 14–16, we have N2L Ui = a 2AE =

1 [(12.5)2 (1.5) + (−7.5)2 (1.8) + (−12.5)2 (1.5) + (15)2 (0.9)] 2 AE

=

386.25 kN 2 ⋅ m AE

=

386.25(10 3 )2 N 2 ⋅ m = 2.0329 N ⋅ m [0.95(10 −3 ) m 2 ][200(109 ) N−m 2 ]

forceisis External Work: The external work done by 10-kN 2 kip force Ue =

1 (10)(10 3 )(∆ A )= 5(10 3 )(∆ A )h h 2

Conservation of Energy: Ue = Ui (¢A)h =

2.0329 5(10 3 )

= 0.4066(10 −3 ) m = 0.407 mm

Ans.

Ans: (∆D)v =

1180

3.50PL AE

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14–27. Determine the vertical displacement of joint C. AE is constant.

P B

C

0.8L

D

Solution

L

0.6L

A

Member Forces: By inspecting joint C, we notice that member BC is a zero-force member and FCD = P (C). Subsequently, consider the equilibrium of joint D, Fig. a. 4 5 + c ΣFy = 0 ;    FBD a b - P = 0    FBD = P (T) 5 4 + ΣFx = 0;     5 P a 3 b - FAD = 0   FAD = 3 P (C) S 4 5 4 Axial Strain Energy: N2L (Ui)a = a 2AE = =

2 2 1 4 5 3 3 c P2 a L b + a P b (L) + a P b a L b d 2AE 5 4 4 5

27P2L 20AE

External Work: The external work done by force P is Ue =

1 P (∆ C)v 2

Using the concept of conservation of energy, Ue = (Ui)a 1 27P2L P (∆ C)v = 2 20AE (∆ C)v =

27PL 10AE

Ans.

Ans: (∆ C)v = 1439

27PL 10AE

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*14–28. Determine the horizontal displacement of joint C. AE is constant.

P

C L

A

L/2 B

Solution Member Forces: Consider the equilibrium of joint C, Fig. a. + ΣFx = 0;    P - FAC a 23 b = 0    FAC = 2 P (T) S 2 23 2 1 1 P (C) + c ΣFy = 0 ;     FBC - a P b a b = 0  FBC = 2 23 23

Axial Strain Energy:

N2L (Ui)a = a 2AE 2 2 1 2 1 L P b (L) + a Pb a b d ca 2AE 2 23 23 2 3P L = 4AE

=

External Work: The external work done by force P is Ue =

1 P (∆ C)h 2

Using the concept of conservation of energy, Ue = (Ui)a 1 3P2L P (∆ C)h = 2 4AE (∆ C)h =

3PL 2AE

Ans.

Ans: (∆ C)h = 1437

3PL 2AE

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14–29. Determine the slope at point A of the beam. EI is constant.

A

B

C

M0 a

a

a

Solution L

Ui = = Ue =

a

a

b

2 M0 M2dx 1 = c ( -M0)2dx1 + (0)dx2 + ax3 b dx3 d 2EI L0 a L0 2EI L0 L0

2M20 a 3EI

1 1 Mu = M0 uA 2 2

Conservation of Energy: Ue = Ui 2M02a 1 M0 uA = 2 3EI uA =

4M0 a 3EI

Ans.

Ans: uA = 1443

4M0 a 3EI

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14–30. Determine the slope of the beam at the pin support A. Consider only bending strain energy. EI is constant.

M0 A

B L

Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0;    By L - M0 = 0    By =

M0 L

Internal Moment: Referring to the FBD of the beam’s right cut segment, Fig. b, a+ ΣMB = 0;    

M0 M0 x - M = 0    M = x L L

Bending Strain Energy: L

(Ui)b = =

2

M dx = L0 2EI L0 M02L 6EI

L

M0

1 L x 2 2dx 2EI

=

M02

2EIL2 L0

L

2

2

x dx =

M0

3

L x a b` 2 3 0 2EIL

External Work: The external work done by M0 is Ue =

1 Mu 2 0 A

Conservation of Energy: Ue = (Ui)b M02 L 1 M0 uA = 2 6EI uA =

M0 L b 3EI

Ans.

Ans: uA = 1446

M0L 3EI

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14–31. Determine the vertical displacement of point C of the A992 steel beam. I = 80(106) mm4.

60 kN

A

B

C 4m

2m

Solution Support Reactions and Internal Loadings: The support reactions and the necessary moment functions are shown on the FBD in Figs. a and b, respectively, Bending Strain Energy: L

(Ui)b = = = =

2

M dx L0 2EI 1 c 2EI L0

4m

2

(20.0x1) dx1 +

L0

2m

( 40.0 x22 ) dx2

1 400 3 4 m 1600 3 2 m x1 b ` + a x2 b ` d ca 2EI 3 3 0 0 6400 kN2 # m3 EI

For A992 steel, E = 200 GPa. Then (Ui)b =

6400 ( 10002 ) 200 ( 109 ) 3 80 ( 10-6 ) 4

= 400 J

External Work: The external work done by 60 kN force is Ue =

1 1 P ( ∆ C)v = 3 60 ( 103 ) ( ∆ C ) v = 30 ( 103 )( ∆ C ) v 2 2

Using the concept of conservation of energy, Ue = (Ui)b 30 ( 103 ) (∆ C)v = 400

Ans.

(∆ C)v = 0.013333 m = 13.3 mm

Ans: (∆ C)v = 13.3 mm 1440

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  The A992 steel bars are pin connected at C and D. If they each have the same rectangular cross section, with a height of 200 mm and a width of 100 mm, determine the vertical displacement at B. Neglect the axial load in the bars.

500 N B

C

D

A

E 2m

2m

3m

3m

Solution Internal Strain Energy: L

Ui =

M2 dx 1 = 2 EI L0 L0 2 EI

2m

[500 x]2 dx =

0.3333 (106) EI

External Work: Ue =

1 1 P ∆ B = (500) ∆ B = 250 ∆ B 2 2

Conservation of Energy: Ue = Ui 250 ∆ B = ∆B =

0.3333 (106) EI 1333.33 1333.33 = 1 9 EI 200 (10 ) 1 12 2 (0.1)(0.23)

= 0.1 (10 - 3) m = 0.100 mm

Ans.

Ans: ∆ B = 0.100 mm 1444

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14–33. Determine the vertical displacement of point B on the A992 steel beam. I = 80(106) mm4.

20 kN

A

C B 3m

5m

Solution 2 3 3 1.875(109) M dx 1 = c [(12.5)(103)(x1)]2dx1 + [(7.5)(103)(x2)]2dx2 d = 2EI L0 EI L0 2EI L0 L

Ui = Ue =

1 1 P∆ = (20)(103)∆ B = 10(103)∆ B 2 2

Conservation of Energy: Ue = Ui 10(103)∆ B = ∆B =

1.875(109) EI

187500 187500 = = 0.0117 m = 11.7 mm EI 200(109)(80)(10 - 6)

Ans.

Ans: ∆ B = 11.7 mm 1442

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  Determine the vertical displacement of end B of the cantilevered 6061-T6 aluminum alloy rectangular beam. Consider both shearing and bending strain energy.

150 kN 1m

a B

A

a 100 mm

Solution

300 mm

Internal Loadings: Referring to the FBD of beam’s right cut segment, Fig. a,

Section a – a

+ c ΣFy = 0;    V - 150(103) = 0    V = 150(103) N a+ ΣM0 = 0;     - M - 150(103) x = 0   M = -150(103)x Shearing Strain Energy: For the rectangular beam, the form factor is fs =

6 . 5

6 3 2 1m fsV 2dx 5 [150(10 )] dx = = 17.31 J 9 L0 2GA L0 2[26(10 )][0.1(0.3)] L

(Ui)v =

Bending Strain Energy: I = L

(Ui)b =

M2dx = L0 2EI L0

1 (0.1)(0.33) = 0.225(10 - 3) m4. We obtain 12

1m

[ - 150(103)x]2dx 2[68.9(109)][0.225(10 - 3)] = 725.689

L0

= 725.689 a = 241.90 J

1m

x2 dx

x3 1 m b` 3 0

Thus, the strain energy stored in the beam is Ui = (Ui)v + (Ui)b = 17.31 + 241.90 = 259.20 J External Work: The work done by the external force P = 150 kN is Ue =

1 1 P∆ = [150 (103)] ∆ B = 75(103) ∆ B 2 2

Conservation of Energy: Ue = Ui 75(103) ∆ B = 259.20 ∆ B = 3.456 (10 - 3) m = 3.46 mm

Ans.

Ans: ∆ B = 3.46 mm 1441

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4 kN 800 lb

14–35. The A-36 steel bars are pin connected at B. If each has a square cross section, determine the vertical displacement at B.

502 mm in. A

B

2.4 8 ftm

Solution

C

D

50 mm 2 in.

3 mft 10

4 ftm 1.2

4 kN

Support Reactions: As shown on FBD(a). Moment Functions: As shown on FBD(b) and (c). 2.4 m

1.2 m

Bending Strain Energy: Applying 14–17, we have L

Ui =

M2dx L0 2EI

=

1.6 kN

4 kN

1.2 m 4ft

1 cB = 2EI L 00

3m 5.6 kN

−4x1

3 m 10ft

2 2 dxdx ((–4x -800x 1) 1) 1 +1

+ 0

(–1.6x ( -320x 2) dx22)d dx2 R L0 2

2

−1.6x2

16.128 kN 2 ⋅ m 3 EI

1.6 kN

16.128(10 3 N)2 ⋅ m 3 = 154.829 J 9 1 (0.05 m)(0.05 m)3  [200(10 ) N−m 2 ]  12   kNlbforce External Work: The external work done by 4800 forceisis Ue =

1 3 (800)(¢ [4(10 )](Δ 2000Δ B)B)==400¢ BB 2

Conservation of Energy:

Ue = Ui 2000 ∆ B = 154.829 = ∆ B 0.07741 = m 77.41 mm

Ans.

Ans: ∆B = 77.4 mm 1186

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*14–36. The cantilevered beam has a rectangular cross-sectional area A, a moment of inertia I, and a modulus of elasticity E. If a load P acts at point B as shown, determine the displacement at B in the direction of P, accounting for bending, axial force, and shear.

A

B u P

L

Solution Strain Energy: Applying Eq. 14–15, 14–17 and 14–19, we have 2

2 L 2 L fsV dx N dx M dx + + L0 2AE L0 2EI L0 2GA L

Ui =

6 for a rectangular section. 5 L L L (P cos u ) 2 dx [(P sin u) x] 2 dx (P sin u)2 dx 6 Ui = + + 2AE 2EI 5 L0 2GA L0 L0

However, fs =

=

P2L 15 cos2 u 5L2 sin2 u 18 sin2 u a + + b 30 AE EI GA

External Work: The external work done by force P is Ue =

1 (P)(∆ B) 2

Conservation of Energy: Ue = Ui 1 P2L 15 cos2 u 5L2 sin2 u 18 sin2 u (P)(∆ B) = a + + b 2 30 AE EI GA ∆B =

PL 15 cos2 u 5L2 sin2 u 18 sin2 u a + + b 15 AE EI GA

Ans.

Ans: ∆B = 1447

PL 15 cos2 u 5L2 sin2 u 18 sin2 u a + + b 15 AE EI GA

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14–37. The pipe assembly is fixed at A. Determine the vertical displacement of end C of the assembly. The pipe has an inner diameter of 40 mm and outer diameter of 60 mm and is made of A-36 steel. Neglect the shearing strain energy.

A

800 mm

600 N

Solution Internal Loading: Referring to the free-body diagram of the cut segment BC, Fig. a, ©My = 0; My + 600x = 0

B

400 mm C

My = - 600x

Referring to the free-body diagram of the cut segment AB, Fig. b, ©Mx = 0; Mx - 600y = 0

Mx = 600y

©My = 0; 600(0.4) - Ty = 0

Ty = 240 N # m

Torsional Strain Energy. J = L

(Ui)t =

T2dx = L0 2GJ L0

Bending Strain Energy. I =

p A 0.034 - 0.024 B = 0.325 A 10 - 6 B pm4. We obtain 2

0.8 m

2 C 75 A 109 B D C 0.325 A 10 - 6 B p D

= = =

= 0.3009 J

p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B pm4. We obtain 4

L

(Ui)b =

2402 dx

1 M2dx = B 2EI L0 L0 2EI

0.4 m

0.8 m

( -600x)2 dx +

L0

A 600y)2 dy R

0.4 m 0.8 m 1 + 120 A 103 B y3 2 B 120 A 103 B x3 2 R 2EI 0 0

34 560 N2 # m3 EI 34 560 200 A 109 B c 0.1625 A 10 - 6 B p d

= 0.3385 J

Thus, the strain energy stored in the pipe is Ui = (Ui)t + (Ui)b = 0.3009 + 0.3385 = 0.6394 J External Work. The work done by the external force P = 600 N is Ue =

1 1 P¢ = (600)¢ C = 300¢ C 2 2

Conservation of Energy. Ue = Ut 300¢ C = 0.6394

¢ C = 2.1312 A 10 - 3 B = 2.13 mm

Ans. 1189

Ans: ∆C = 2.13 mm

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  Determine the vertical displacement of end B of the frame. Consider only bending strain energy. The frame is made using two A-36 steel W460 * 68 wide-flange sections.

3m B

4m

Solution Internal Loading: Using the coordinates x1 and x2, the free-body diagrams of the frame’s segments in Figs. a and b are drawn. For coordinate x1, a+ ΣMO = 0; For coordinate x2,

- M1 - 20 1 103 2 x1 = 0

M2 - 20 1 103 2 (3) = 0

a+ ΣMO = 0;

M1 = -20 1 103 2 x1

M2 = 60 1 103 2 N # m

Bending Strain Energy: L

(Ub)i =

M2dx 1 = J 2EI 2EI L0 L0

400 1 10 1 = D£ 2EI 3

6

=

2

3m

2

c - 20 1 103 2 x1 d dx1 +

3m

x1 ≥ 3 3

0

9 1 109 2 N2 # m3

+ 3.6 1 109 2 x 2

4m 0

L0

4m

20 kN A

2

c 60 1 103 2 d dx2 R

T

EI

For a W460 * 68, I = 297 1 106 2 mm4 = 297 1 10 - 6 2 m4. Then (Ub)i =

9 1 109 2

200 1 109 2 (297) 1 10 - 6 2

= 151.52 J

External Work: The work done by the external force P = 20 kN is Ue =

1 1 P∆ = c 20 1 103 2 d ∆ B = 10 1 103 2 ∆ B 2 2

Conservation of Energy:

Ue = Ui 10 1 103 2 ∆ B = 151.52

∆ B = 0.01515 m = 15.2 mm

Ans.

Ans: ∆ B = 15.2 mm 1452

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  The rod has a circular cross section with a moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Only consider the strain energy due to bending. The modulus of elasticity is E.

r A P

Solution Moment Function: a+ ΣMB = 0;   P [r (1 - cos u)] - M = 0;

M = Pr (1 - cos u)

Bending Strain Energy: S

Ui =

M2 ds L0 2EI

ds = r du p

=

M2r du r = [P r (1 - cos u)]2 du 2EI 2EI L0 L0

=

P2r 3 (1 + cos2 u - 2 cos u) du 2EI L0

u

p

=

P2r 3 p 1 cos 2u a1 + + - 2 cos u b du 2EI L0 2 2

P2r 3 3 cos 2u P2r 3 3 3pP2r 3 a + - 2 cos u b du = a pb = 2EI L0 2 2 2EI 2 4EI p

=

Conservation of Energy: Ue = Ui ; ∆A =

1 3pP2r 3 P ∆A = 2 4EI

3pPr 3 2EI

Ans.

Ans: ∆A = 1448

3pPr 3 2EI

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  The rod has a circular cross section with a moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Only consider the strain energy due to bending. The modulus of elastcity is E. r

A P

Solution Bending Strain Energy: Applying 14–17 with ds = rdu , we have s

Ui =

M2ds L0 2EI p

=

1 (Pr sin u)2 rdu 2EI L0

=

P2r 3 2 sin udu 2EI L0

=

P2r 3 (1 - cos 2u) du 4EI L0

=

pP2r 3 4EI

p

p

External Work: The external work done by force P is Ue =

1 (P)(∆ A) 2

Conservation of Energy: Ue = Ui 1 pP2r 3 (P)(∆ A) = 2 4EI ∆A =

pPr 3 2EI

Ans.

Ans: ∆A = 1449

pPr 3 2EI

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14–41. z

The rod has a circular cross section with a polar moment of inertia J and moment of inertia I. If a vertical force P is applied at A, determine the vertical displcement at this point. Consider the strain energy due to bending and torsion. The material constants are E and G.

y r P

Solution T = Pr(1 - cos u); M = Pr sin u Torsion Strain Energy:

x

s

Ui =

A

T 2 ds T 2 rdu = L0 2GJ L0 2GJ u

p

=

r [Pr(1 - cos u)]2 du 2GJ L0

=

P2 r 3 (1 + cos2 u - 2 cos u)du 2GJ L0 p

P2 r 3 cos 2u + 1 a1 + - 2 cos u bdu 2GJ L0 2 p

= =

3P2r 3 p 4GJ

Bending Strain Energy: s

Ui =

M2ds L0 2EI M2r du r = [Pr sin u]2 du 2EI 2EI L0 L0 u

=

p

1 - cos 2u P2 r 3 P2 r 3 p a bdu = 2EI L0 2 4EI p

=

Conservation of Energy:

Ue = Ui 1 3P2 r 3 p P2 r 3 p P∆ A = + 2 4GJ 4EI ∆A =

Pr 3 p 3 1 a + b 2 GJ EI

Ans.

Ans: ∆A = 1451

Pr 3 p 3 1 a + b 2 GJ EI

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14–42.  A bar is 4 m long and has a diameter of 30 mm. Determine the total amount of elastic energy that it can absorb from an impact loading if (a) it is made of steel for which Est = 200 GPa, sY = 800 MPa, and (b) it is made from an aluminum alloy for which Eal = 70 GPa, sY = 405 MPa.

Solution a) PY =

800(106) sY = = 4(10 - 3) m>m E 200(109)

ur =

1 1 (s )(P ) = (800)(106)(N>m2)(4)(10 - 3) m>m = 1.6 MJ>m3 2 Y Y 2

V =

p (0.03)2(4) = 0.9(10 - 3)p m2 4

Ui = 1.6(106)(0.9)(10 - 3)p = 4.52 kJ

Ans.

b) PY =

405(106) sY = = 5.786(10 - 3) m>m E 70(109)

Ur =

1 1 (s )(P ) = (405)(106)(N>m2)(5.786)(10 - 3) m>m = 1.172 MJ>m3 2 Y Y 2

V =

p (0.03)2 (4) = 0.9(10 - 3)p m3 4

Ui = 1.172(106)(0.9)(10 - 3)p = 3.31 kJ

Ans.

Ans: (a) Ui = 4.52 kJ (b) Ui = 3.31 kJ 1453

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14–43.  Determine the diameter of a red brass C83400 bar that is 8 ft long if it is to be used to absorb 800 ft # lb of energy in tension from an impact loading. No yielding occurs.

Solution Elastic Strain Energy: The yielding axial force is PY = sY A. Ui =

(sYA)2L s2YAL N2L = = 2AE 2AE 2E

Substituting, we have Ui = 0.8(12) =

s2YAL 2E 11.42 3 p4 (d 2) 4 (8)(12) 2[14.6(103)]

Ans.

d = 5.35 in.

Ans: d = 5.35 in. 1454

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*14–44. Determine the speed v of the 50-Mg mass when it is just over the top of the steel post, if after impact, the maximun stress developed in the post is 550 MPa. The post has a length of L = 1 m and a cross-sectional area of 0.01 m2. Est = 200 GPa, sY = 600 MPa. L

Solution The Maximum Stress: smax =

Pmax A

550 (106) = ∆ max = =

Pmax ; 0.01

Pmax k

Pmax = 5500 kN Here k =

5500 (103) 2 (109)

0.01(200) (109) AE = = 2 (109) N>m L 1

= 2.75 (10 - 3) m

Conservation of Energy: Ue = Ui 1 1 mv2 + W ∆ max = k ∆ 2max 2 2 1 1 (50) (103) (v2) + 50 (103) (9.81) [2.75 (10 - 3)] = (2) (109) [2.75 (10 - 3) ]2 2 2 Ans.

v = 0.499 m>s

Ans: v = 0.499 m>s 1455

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14–45. Rods AB and AC have a diameter of 20 mm and are made of 6061-T6 aluminum alloy. They are connected to the rigid collar which slides freely along the vertical guide rod. Determine the maximum height h from which the 50-kg block D can be dropped without causing yielding in the rods when the block strikes the collar.

D h

A 400 mm 30⬚

Solution

30⬚

400 mm

C

B

Equilibrium. Referring to the free-body diagram of joint A, Fig. a + ©Fx = 0; :

FAB sin 30° - FAC sin 30° = 0

FAB = FAC = F

+ c ©Fy = 0;

2F cos 30° - PA = 0

PA = 1.732F

(1)

Compatibility Equation. From the geometry shown in Fig. b

dF = ¢ A cos 30° F(0.4) = ¢ A cos 30° p (0.022)[68.9(109)] 4 ¢ A = 21.3383(10 - 9)F Thus, the equivalent spring constant for the system can be determined from PA = k¢ A 1.732F = k[21.3383(10 - 9)F] k = 81.171(106) N>m Maximum Stress. The maximum force that can be developed in members AB and AC is Fmax = sYA = 255(106 ) c

p (0.022) d = 80.11(103) N 4

From Eq. (1), (PA)max = 1.732Fmax = 1.732[80.11(103)] = 138.76(103) N Then, (¢ A)max =

138.76(103) (PA)max = 1.7094(10 - 3) m = k 81.171(106)

Conservation of Energy. mg[h + (¢ A)max] =

1 k(¢ A)max2 2

50(9.81)[h + 1.7094(10 - 3)] =

1 [81.171(106)][1.7094(10 - 3)]2 2

h = 0.240 m

Ans.

1490

Ans: h = 0.240 m

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14–46. Rods AB and AC have a diameter of 20 mm and are made of 6061-T6 aluminum alloy. They are connected to the rigid collar A which slides freely along the vertical guide rod. If the 50-kg block D is dropped from height h = 200 mm above the collar, determine the maximum normal stress developed in the rods.

D h

A

Equilibrium. Referring to the free-body diagram of joint A, Fig. a + : ©Fx = 0;

FAB sin 30° - FAC sin 30° = 0

+ c ©Fy = 0;

2F cos 30° - PA = 0

FAB = FAC = F

PA = 1.732F

400 mm 30⬚

30⬚

400 mm

(1)

Compatibility Equation. From the geometry shown in Fig. b

B

C

dF = ¢ A cos 30° F(0.4) p (0.022)[68.9(109)] 4

= ¢ A cos 30°

¢ A = 21.3383(10 - 9)F Thus, the equivalent spring constant for the system can be determined from PA = k¢ A 1.732F = k[21.3383(10 - 9)F] k = 81.171(106) N>m Conservation of Energy. mg[h + (¢ A)max] =

1 k(¢ A)max2 2

50(9.81)[0.2 + (¢ A )max ] =

1 [81.171(106)](¢ A)max2 2

40.5855(106)(¢ A)max2 - 490.5(¢ A)max - 98.1 = 0 Solving for the positive root, (¢ A)max = 1.5608(10 - 3) m Then, (PA)max = k(¢ A)max = 81.171(106)[1.5608(10 - 3)] = 126.69(103) N Maximum Stress. From Eq. (1), (PA)max = 1.732Fmax Fmax = 73.143(103) N Thus, the maximum normal stress developed in members AB and AC is (smax)AB = (smax)AC =

Fmax 73.143(103) = = 232.82 MPa = 233 MPa p A (0.022) 4

Since (smax)AB = (smax)AC < sY = 255 MPa, this result is valid.

1489

Ans.

Ans: (smax)AB = (smax)AC = 233 MPa

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14–47. A steel cable having a diameter of 16 mm wraps over a drum and is used to lower an elevator having a over a drum and is used to lower an elevator having a weight mass of 400 kg. The elevator is 45 m below the drum and is descending at the constant rate of 0.6 m/s when the drum suddenly stops. Determine the maximum stress developed in the cable when this occurs. Est = 200 GPa, when σY = 350 MPa. 45 150mft

Solution = k

2 9 π AE  4 (0.016 ) [200(10 )] = = 893.61(10 3 ) N−m 45 L

Ue = Ui 1 1 mv2 + mg ∆ max = k∆ 2max 2 2 1 1 = ∆ max (400)(0.6 2 ) + 400(9.81) [893.61(10 3 )]∆ 2max 2 2 446.80(10 3 )∆ 2max − 3924 ∆ max − 72 = 0 ¢max  0.017823 m k ∆ max = Pmax = [893.61(10 3 )](0.017823) = 15.927(10 3 ) N

σ= max

Pmax 15.927(10 3 ) = = 79.22(106 ) N−m 2 π (0.016 2 ) A 4 = 79.22 MPa < σ Y O.K.

Ans.

Ans: s max = 79.2 MPa 1193

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  The A-36 steel bolt is required to absorb the energy of a 2-kg mass that falls h = 30 mm. If the bolt has a diameter of 4 mm, determine its required length L so the stress in the bolt does not exceed 150 MPa. L h

Solution Maximum Stress: With ∆ st =

WL = AE

= 7.80655 ( 10- 6 ) L and sst =

W = A

2(9.81)(L) p 4

( 0.0042 ) 3 200 ( 109 ) 4

2(9.81) p 4

( 0.0042 )

= 1.56131 MPa, we have

smax = nsst where n = 1 +

150 ( 106 ) = C 1 +

B

1 + 2a

B

1 + 2a

0.03 7.80655 ( 10-6 ) L

L = 0.8504 m = 850 mm

h b ∆ st

b S 3 1.56131 ( 106 ) 4

Ans.

Ans: L = 850 mm 1460

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  The A-36 steel bolt is required to absorb the energy of a 2-kg mass that falls h = 30 mm. If the bolt has a diameter of 4 mm and a length of L = 200 mm, determine if the stress in the bolt will exceed 175 MPa. L h

Solution Maximum Stress: With ∆ st =

WL = AE

sst =

W = A

2(9.81)(0.2) p 4

( 0.0042 ) 3 200 ( 109 ) 4

2(9.81) p 4

( 0.0042 )

= 1.56131 ( 10-6 ) m

= 1.56131 MPa

Applying Eq. 14–34, we have n = 1 +

B

1 + 2a

h 0.03 b = 1 + 1 + 2a b = 197.04 ∆ st B 1.56131 ( 10-6 )

Thus, smax = nsst = 197.04(1.56131) = 307.6 MPa Ans.

Yes, smax exceeded 175 MPa.

Ans: Yes 1461

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  The A-36 steel bolt is required to absorb the energy of a 2-kg mass that falls along the 4-mm-diameter bolt shank that is 150 mm long. Determine the maximum height h of release so the stress in the bolt does not exceed 150 MPa. L h

Solution WL = AE

Maximum Stress: With ∆ st = = 1.17098 ( 10-6 ) m and sst =

W = A

2(9.81)(0.15) p 4

( 0.0042 ) 3 200 ( 109 ) 4

2(9.81) p 4

( 0.0042 )

= 1.56131 MPa,

we have smax = nsst where n = 1 +

150 ( 106 ) = C 1 +

B

1 + 2a

B

1 + 2a

h 1.17098 ( 10-6 )

h b ∆ st

b S 3 1.56131 ( 106 ) 4

h = 5.292 ( 10-3 ) m = 5.29 mm

Ans.

Ans: h = 5.29 mm 1462

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14–51. The 5-kg block is traveling with the speed of v = 4 m/s just before it strikes the 6061-T6 aluminum stepped cylinder. Determine the maximum normal stress developed in the cylinder.

300 mm

300 mm C

B 40 mm

v A

20 mm

Solution Equilibrium. The equivalent spring constant for segments AB and BC are

kAB

kBC

p A 0.022 B c 68.9 A 109 B d AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B c 68.9 A 109 B d ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3

Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC

72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC =

1 ¢ 4 AB

(1)

Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2

(2)

Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4

1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5) A 42 B = c 72.152 A 106 B d ¢ AB 2 + c 288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.9418 A 10 - 3 B m Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B c 0.9418 A 10 - 3 B d = 67.954 A 103 B N.

Thus, smax = sAB =

67.954 A 103 B FAB = = 216.30 MPa = 216 MPa p AAB A 0.022 B 4

Ans.

Since smax 6 sY = 255 MPa, this result is valid. Ans: s max = 216 MPa 1195

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300 mm

300 mm

*14–52. Determine the maximum speed v of the 5-kg block without causing the 6061-T6 aluminum stepped cylinder to yield after it is struck by the block.

C

B 40 mm

Solution

v A

20 mm

Equilibrium. The equivalent spring constant for segments AB and BC are

kAB

kBC

p A 0.022 B C 68.9 A 109 B D AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B C 68.9 A 109 B D ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3

Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC

72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC =

1 ¢ 4 AB

(1)

Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2

(2)

Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4

1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5)v2 = c 72.152 A 106 B d ¢ AB 2 + c 288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.23545 A 10 - 3 B v Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B C 0.23545 A 10 - 3 B v D = 16988.46v. Thus, smax = sAB = 255 A 106 B =

FAB AAB

16988.46v p A 0.022 B 4

v = 4.716 m>s = 4.72 m>s

Ans. Ans. v = 4.72 m/s 1196

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14–53.  The composite aluminum 2014-T6 bar is made from two segments having diameters of 7.5 mm and 15 mm. Determine the maximum axial stress developed in the bar if the 10-kg collar is dropped from a height of h = 100 mm. 7.5 mm

Solution ∆ st = Σ

WL = AE

300 mm

200 mm

10(9.81)(0.3) p 2 9 4 (0.0075) (73.1)(10 )

+

10(9.81)(0.2)

15 mm

h

p 2 9 4 (0.015) (73.1)(10 )

= 10.63181147 (10 - 6) m n = c1 +

B

smax = n sst

1 + 2a

h 0.1 b d = c1 + 1 + 2a b d = 138.16 ∆ st B 10.63181147(10 - 6)

Here sst =

10 (9.81) W = p = 2.22053 MPa 2 A 4 (0.0075 )

smax = 138.16 (2.22053) Ans.

= 307 MPa < sY = 414 MPa  OK

Ans: smax = 307 MPa 1464

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14–54.  The composite aluminum 2014-T6 bar is made from two segments having diameters of 7.5 mm and 15 mm. Determine the maximum height h from which the 10-kg collar should be dropped so that it produces a maximum axial stress in the bar of smax = 300 MPa.

7.5 mm

Solution ∆ st = Σ

WL = AE

300 mm

200 mm

10(9.81)(0.3) p 2 9 4 (0.0075) (73.1)(10 )

+

10(9.81)(0.2)

15 mm

h

p 2 9 4 (0.015) (73.1)(10 )

= 10.63181147 (10 - 6) m n = c1 + =

31

B

1 + 2a

h h b d = c1 + 1 + 2a bd ∆ st B 10.63181147(10 - 6)

+ 21 + 188114.7 h 4 Here sst =

smax = n sst 300 (106) =

31

10 (9.81) W = p = 2.22053 MPa 2 A 4 (0.0075 )

+ 21 + 188114.7 h 4 (2220530)

Ans.

h = 0.09559 m = 95.6 mm

Ans: h = 95.6 mm 1465

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14–55. The 25-kg block is failing at 0.9 m/s at the instant it is 0.6 m above the spring and post assembly. Determine the maximum stress in the post if the spring has a stiffness of k = 40 MN/m. The post has a diameter of 75 mm and a modulus of elasticity of E = 48 GPa. Assume the material will not yield.

0.93m/s ft/s

0.6 2 ftm

k k

Solution 0.6 2 ftm

Equilibrium: This requires Fsp = FP. Hence ksp ¢ sp = kP ¢ P

and

kp

∆ sp =

ksp

[1]

∆P

Conservation of Energy: The equivalent spring constant for the post is 2 9 π AE  4 (0.075 ) [48(10 )] = = 353.43(106 ) N−m kp =  0.6 L Ue = Ui 1 1 1 my2 + W(h + ¢ max) = kP ¢ 2P + ksp ¢ 2sp 2 2 2

[2]

However, ¢ max = ¢ P + ¢ sp. Then, Eq. [2] becomes 1 1 1 my2 + W A h + ¢ P + ¢ sp B = kP ¢ 2P + ksp ¢ 2sp 2 2 2

[3]

Substituting Eq. [1] into [3] yields kp 1 1 1 1 kP2k2P my2 + W ¢ h + ¢ P + ¢ P ≤ = kP¢ kpΔP2 2P+ a ¢ ΔP2 b ¢ 2P ≤ 2 ksp 2 2 2kspksp

 1 353.43(106 )  (25)(0.9 2 ) + [25(9.81)] 0.6 + ∆ p + ∆p 2 40(106 )   =

1 1 [353.43(106 )]2  2 [353.43(106 )]∆ 2p +  ∆p 2 2  40(106 ) 

1.73812(109 )∆ 2p − 2412.21∆ p − 157.275 = 0 Solving for positive root, we have

0.301503(10 −3 ) m ∆P = Maximum Stress: The maximum axial force for the post is Pmax = kp ¢ p

= [353.43(106 )][0.301503(10 −3 )] = 106.56(10 3 )

σ= max

Pmax 106.56(10 3 ) 2 = 24.1 MPa = = 24.12(106 ) N−m π (0.0752 ) A 4

Ans.

Ans: s max = 24.1 MPa 1200

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*14–56. The collar has a mass of 5 kg and falls down the titanium Ti-6A1-4V bar. If the bar has a diameter of 20 mm, determine the maximum stress developed in the bar if the weight is (a) dropped from a height of h = 1 m, (b) released from a height h ≈ 0, and (c) placed slowly on the flange at A. 20 mm 1.5 m h

Solution Maximum Stress: With ∆ st = and sst =

WL = AE

5(9.81)(1.5) p 4

( 0.022 ) 3 120 ( 109 ) 4

= 1.9516 ( 10-6 ) m

A

5(9.81) W = 0.156131 MPa and Applying Eq. 14–34, we have = p 2 A 4 ( 0.02 )

a) n = 1 +

B

1 + 2a

h 1 b = 1 + 1 + 2a b = 1013.31 ∆ st B 1.9516 ( 10-6 )

Thus, smax = nsst = 1013.31(0.156131) = 158 MPa

Ans.

b) n = 1 + Thus,

B

1 + 2a

h 0 b = 1 + 1 + 2a b = 2 ∆ st B 1.9516 ( 10-6 )

smax = nsst = 2(0.156131) = 0.312 MPa

Ans.

smax = sst = 0.156 MPa

Ans.

c)

Since all of the smax 6 sY = 924 MPa, the above analysis is valid.

Ans: a) smax = 158 MPa, b) smax = 0.312 MPa, c) smax = 0.156 MPa 1467

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14–57.  The collar has a mass of 5 kg and falls down the titanium Ti-6A1-4V bar. If the bar has a diameter of 20 mm, determine if the weight can be released from rest at any point along the bar and not permanently damage the bar after striking the flange at A. 20 mm 1.5 m h

Solution Maximum Stress: With ∆ st =

WL = AE

5(9.81)(1.5) p 4

( 0.022 ) 3 120 ( 109 ) 4

= 1.9516 ( 10-6 ) m,

A

5(9.81) W = p = 0.156131 MPa and h = hmax = 1.5 m. Applying Eq. 14–34, 2 A 4 ( 0.02 ) we have sst =

n = 1 + Thus,

B

1 + 2a

h 1.5 b = 1 + 1 + 2a b = 1240.83 ∆ st B 1.9516 ( 10-6 )

smax = nsst = 1240.83(0.156131) = 193.7 MPa Since smax 6 sY = 924 MPa, the weight can be released from rest at any position along the bar without causing permanent damage to the bar. Ans.

Ans: Yes, from any position 1468

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14–58. of 60 tonnes traveling 14–58. The tugboat tugboathas hasa mass a weight of 120 and 000islb and is forward 0.6 m/s it strikes the the 300-mm-diameter traveling at forward at 2when ft兾s when it strikes 12-in.-diameter fender post AB used to protect a bridge pier. If the post is made from treated white spruce and is assumed fixed at the river bed, determine the maximum horizontal distance the top of the post will move due to the impact. Assume the tugboat is rigid and neglect the effect of the water.

A A

3 ft 0.9 m

C C 12 m ft 3.6

B B

Solution From Appendix C: Pmax =

3EI(¢ C)max (LBC)3

Conservation of energy: 1 1 mv2 = Pmax (¢ C)max 2 2 2 1 1  3EI (∆C ) max mv2 =  2 2  (LBC )3 

  

(¢ C)max =

mv2L3BC C 3EI

(¢ C)max =

2 3 (120 3000>32.2)(2) 60(10 )(0.6)2(3.6)3 (12) m = 295.89 mm. 0.29589 = p 9 4 p 4 C 3(9.65)(10 (3)(1.40)(10)(6)(144)( 64 )(0.3)4 )(0.5)

Pmax

uC =

( )

 3(9.65)(109 ) π (0.3)4  64 =  (0.29589) 73.00(10 3 ) N   (3.6)3  

PmaxL2BC 73.00(10 3 )(3.6)2 = 0.12329 πad = π (0.3)4 2EI 2(9.65)(109 ) 64

( )

(¢ A)max = (¢ C)max + uC(LCA) (¢A)max = 295.89 + 0.12329(0.9)(10 3 ) = 406.74 mm = 407 mm

Ans.

Ans: (∆A) max = 407 mm 1204

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  The overhang beam is made of 2014-T6 aluminum. If the 75-kg block has a speed of v = 3 m>s at h = 0.75 m, determine the maximum bending stress in the beam.

v

D

h 150 mm

A B 4m

Solution

2m

C 75 mm

Equilibrium: The support reactions and the moment function for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a. Ue = Ui L

1 M2dx P∆ st = Σ 2 L0 2EI 1 1 P∆ st = c 2 2EI L0

4m

(0.5Px1)2dx1 +

L0

2m

(Px2)2dx2 d

1 P2 3 2 m 1 0.25 2 3 4 m P∆ st = £a P x1 b ` + x ` § 2 2EI 3 3 2 0 0 ∆ st =

8P EI

1 (0.075)(0.153) = 21.09375(10 - 6) m4 and E = Eal = 73.1 GPa. Then, the 12 equivalent spring constant can be determined from I =

P = k∆ st P = ka k =

8P b EI

73.1(109)[21.09375(10 - 6)] EI = 8 8

= 192.74(103) N>m Conservation of Energy: Ue = Ui 1 1 mv2 + mg(h + ∆ max) = k∆ max2 2 2 1 1 (75)(32) + 75(9.81)(0.75 + ∆ max) = c 192.74(103)d ∆ max2 2 2

96372.07∆ max2 - 735.75∆ max - 889.3125 = 0 ∆ max = 0.09996 m

Maximum Bending Stress: The maximum force on the beam is Pmax = k∆ max = 192.74(103)[0.09996] = 19.266(103) N. The maximum moment occurs at support B. Thus, Mmax = Pmax(2) = 19.266(103)(2) = 38.531(103) N # m. Applying the flexure formula, smax =

38.531(103)(0.15 >2) Mmaxc = = 137 MPa I 21.09375(10 - 6) 2

Ans. Ans: smax = 137 MPa

Since smax 6 sY = 414 MPa, this result is valid.

1477

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  The overhang beam is made of 2014-T6 aluminum. Determine the maximum height h from which the 100-kg block can be dropped from rest (v = 0), without causing the beam to yield.

v

D

h 150 mm

A B

Solution

4m

2m

C 75 mm

Equilibrium: The support reactions and the moment function for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a. Ue = Ui 2

L

1 M dx P∆ st = Σ 2 L0 2EI 1 1 P∆ st = c 2 2EI L0

4m

(0.5Px1)2 dx1 +

L0

2m

(Px2)2dx2 d

0.25 2 3 4 m 1 1 P2 3 2 m P∆ st = ca P x1 b ` + x ` d 2 2EI 3 3 2 0 0 ∆ st = I = E al

8P EI

1 (0.075)(0.153) = 21.09375(10 - 6) m4, 12 = 73.1 GPa. Then, ∆ st =

8(981) 73.1(10 )[21.09375(10 - 6) 4 9

P = 100(9.81) = 981N,

and

E =

= 5.0896(10 - 3) m

Maximum Bending Stress: The maximum force on the beam is Pmax = nP = 981n. The maximum moment occurs at support B. Thus, Mmax = Pmax(2) = (981n)(2) = 1962n. Applying the flexure formula, smax =

Mmaxc I

414(106) =

1962n(0.15>2) 21.09375(10 - 6)

n = 59.35 Impact Factor: n = 1 +

B

1 + 2a

59.35 = 1 + h = 8.66 m

B

h b ∆ st

1 + 2c

h d 5.0896(10 - 3)

Ans.

Ans: h = 8.66 m 1478

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14–61. Block C of mass 50 kg is dropped from height h = 0.9 m onto the spring of stiffness k = 150 kN>m mounted on the end B of the 6061-T6 aluminum cantilever beam. Determine the maximum bending stress developed in the beam.

C h a

Solution

100 mm B

Conservation of Energy. From the table listed in the appendix, the

a

PL . Thus, the = 3EI

equivalent spring constant for the beam is kb = 1 (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, 12

A

200 mm

3

displacement of end B under static conditions is ¢ st

I =

k

L = 3 m,

3m

Section a – a

3EI , where L3

and

E = Eal

= 68.9 GPa. Thus,

3EI kb = = L3

3 c 68.9 A 109 B d c 66.6667 A 10 - 6 B d 33

= 510.37 A 103 B N>m

Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b

150 A 103 B ¢ sp = 510.37 A 103 B ¢ b ¢ sp = 3.4025¢ b

(1)

We have, Ue = Ui mg A h + ¢ sp + ¢ b B =

1 1 kb ¢ b 2 + ksp ¢ sp 2 2 2

Substituting Eq. (1) into this equation, 50(9.81)(0.9 + 3.4025¢ b + ¢ b) =

1 1 c 510.37 A 103 B d ¢ b 2 + c150 A 103 B d(3.4025¢ b)2 2 2

1123444.90¢ b 2 - 2159.41¢ b - 441.45 = 0 Solving for the positive root, ¢ b = 0.020807 m Maximum

Stress.

The

maximum

force

Pmax = kb ¢ b = 510.37 A 10 B (0.020807) = 10.619 A 10 3

occurs

at

= 31.858 A 103 B smax =

3

on

the

beam

is

B N. The maximum moment

Mmax = Pmax L = 10.619 A 103 B (3) 0.2 N # m. Applying the flexure formula with c = = 0.1 m, 2

fixed

support

A,

where

31.858 A 103 B (0.1) Mmax c = = 47.79 MPa = 47.8 MPa I 66.6667 A 10 - 6 B

Since smax 6 sY = 255 MPa, this result is valid.

Ans. Ans: s max = 47.8 MPa

1212

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14–62. Determine the maximum height h from which 200-kg block C can be dropped without causing the 6061-T6 aluminum cantilever beam to yield. The spring mounted on the end B of the beam has a stiffness of k = 150 kN>m.

C h a

k

100 mm B

A

a 3m

Solution

200 mm Section a – a

Maximum Stress. From the table listed in the appendix, the displacement of end B PL3 under static conditions is ¢ st = . Thus, the equivalent spring constant for the 3EI 3EI 1 beam is kb = , where I = (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, L = 3 m, and 12 L3 E = Eal = 68.9 GPa. Thus, 3EI kb = = L3

3 c 68.9 A 109 B d c 66.6667 A 10 - 6 B d 33

= 510.37 A 103 B N>m

The maximum force on the beam is Pmax = kb ¢ b = 510.37 A 103 B ¢ b. The maximum moment occurs at the fixed support A, where Mmax = Pmax L = 510.37 A 103 B ¢ b(3)

= 1.5311 A 106 B ¢ b. Applying the flexure formula with smax = sY = 255 MPa and 0.2 = 0.1 m, c = 2 Mmax c I

smax = sY = 255 A 106 B =

1.5311 A 106 B ¢ b(0.1) 66.6667 A 10 - 6 B

¢ b = 0.11103 m Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b

150 A 103 B ¢ sp = 510.37 A 103 B (0.11103) ¢ sp = 0.37778 m

Conservation of Energy. Ue = Ui mg A h + ¢ sp + ¢ b B =

1 1 kb ¢ b 2 + ksp ¢ sp 2 2 2

200(9.81)(h + 0.37778 + 0.11103) =

1 1 c 510.37 A 103 B d(0.11103)2 + c150 A 103 B d(0.37778)2 2 2

h = 6.57 m

Ans. Ans: h = 6.57 m 1213

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14–63. The weight of 90 kg is dropped from a height of fromthe thetop topofofthe theA-36 A-36 steel steel beam. beam. Determine Determine the 41.2ftmfrom deflection and andmaximum maximumstress stress beam if maximum deflection in in thethe beam if the the supporting springs at A and eachhave haveaa stiffness stiffness of supporting springs at A and B Beach mm thick and 100wide. mm wide. k == 100 k 500kN/m. Thebeam beamisis75 3 in. thick and 4 in. lb>in.The

1.2 4 ftm A

B k

k

Solution

2.4 8 ftm

From Appendix C: ¢ beam =

PL3 48EI

k= beam

9 3 1 48 EI 48[200(10 )]  12 (0.1)(0.075 ) = = 305.18(10 3 ) N−m L3 4.8 3

75 mm 3 in. 100 mm 4 in.

2.4 8 ftm

From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam ∆sp =

305.18(10 3 ) 2[100(10 3 )]

∆ beam

¢ sp = 1.7700¢ 1.5259 Δbeam beam

(1)

Conservation of energy: Ue = Ui W(h + ¢ sp + ¢ beam) =

1 1 k ¢2 + 2 a bksp ¢ 2sp 2 beam beam 2

From Eq. (1): + ∆ beam ] 90(9.81)[1.2 + 1.5259 ∆ beam =

1 [305.18(10 3 )]∆ 2beam + [100(10 3 )](1.5259 ∆ beam )2 2

0 385.4185(10 3 )∆ 2beam − 2230.0985∆ beam − 1059.48 = ∆ beam = 0.05540 m = 55.40 mm From Eq. (1): ¢sp = 84.54 mm ∆ max = ∆ sp + ∆ beam Ans.

= 84.54 + 55.40 = 139.94 mm = 140 mm Fbeam = kbeam ∆ beam 3 [305.18(10 = )](0.05540) 16.91(10 3 )

Mmax =

σ max =

Fbeam L [16.91(10 3 )](4.8) = = 20.29(10 3 ) N ⋅ m 4 4 Mmax c [20.29(10 3 )](0.0375) = 1 (0.1)(0.0753 ) I 12

6

= 216.42(10 = ) N−m 2 216 MPa < σ Y

O.K.

Ans.

Ans: s max = 216 MPa

1210

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*14–64. The weight of 90 kg is dropped from a height of 1.2 m 4 ft from the of the A-36 steel beam. Determine the from the top of top the A-36 steel beam. Determine the load factor load factor n if the supporting A aand B each n if the supporting springs at A andsprings B eachat have stiffness of have stiffness Theand beam 3 in. thick k =is300 lb/in.thick k = 60akN/m. Theof beam 75 mm 100ismm wide. and 4 in. wide.

1.2 4 ftm A

Solution From Appendix C: ¢ beam =

k= beam

B k

k 2.4 8 ftm

75 mm 3 in. 100 mm 4 in.

2.4 8 ftm

PL3 48EI 9 3 1 48 EI 48[200(10 )]  12 (0.1)(0.075 ) = = 305.18(10 3 ) N−m L3 4.8 3

From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam

∆sp =

305.18(10 3 ) 2[60(10 3 )]

∆ beam

¢ sp = 2.5431 2.95¢ beam Δbeam

(1)

Conservation of energy: Ue = Ui W(h + ¢ beam + ¢ sp) =

1 1 k ¢2 + 2a bksp ¢ 2sp 2 beam beam 2

From Eq. (1):

90(9.81)[1.2 + ∆ beam + 2.5431 = ∆ beam ]

1 2 [305.18(10 3 )]∆ beam + [60(10 3 )](2.5431∆ beam )2 2

540.6359(10 3 )∆ 2beam − 3128.23∆ beam − 1059.48 = 0 0.047256 m ∆ beam = = kbeam ∆ beam Fbeam 3 [305.18(10 = )][0.047256] 14.42(10 3 ) N

= n

14.42(10 3 ) = 16.33 = 16.3 90(9.81)

smax = n(sst)max = n a M =

Ans.

Mc b I

175(16)(12) 90(9.81)(4.8) 1059.48 =8.40 kip #N in.· m 44

 1059.48(0.0375)   1 (0.1)(0.0753 )  12 

σ max = 16.33 

6 = 184.59(10 = ) N−m 2 184.59 MPa < σ Y

O.K.

Ans: s max = 184.59 MPa 1211

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simply supported supported W250 W10 * 14–65. The simply  15 22 structural A-36 steel beam lies in the horizontal plane and acts as a shock traveling toward it at absorber for the the 500-lb 250-kgblock blockwhich whichis is traveling toward it 5 ft兾s. Determine the maximum deflection of the and at 1.5 m/s. Determine the maximum deflection of beam the beam the maximum stress in in thethe beam and the maximum stress beamduring duringthe the impact. impact. The 1000 lb>in. spring has a stiffness of k ==200 kN/m.

12 m ft 3.6  1.5 5 ft/s vv  m/s

6 4 4 For W250 22:: I I=68.9 28.8(10 ) mm d in. 254−6mm in4 = W 10  * 15 d = 9.99 28.8(10 )m d 254 = mm 0.254 m =

kk

From Appendix C: PL3 48EI

¢ beam = k= beam

12 m ft 3.6

48 EI 48[200(109 )][28.8(10 −6 )] = = 740.74(10 3 ) N−m L3 7.2 3

Equilibrium (equivalent system): Fsp = Fbeam ksp ¢ sp = kbeam ¢ beam ∆sp =

740.74(10 3 ) 200(10 3 )

∆ beam

¢ sp = 3.7037 4.015¢Δ beam beam

(1)

Conservation of energy: Ue = Ui 1 1 1 mv2 = kbeam ¢ 2beam + ksp ¢ 2sp 2 2 2 From Eq. (1): 1 1 1 2 2 (250)(1.5 ) [740.74(10 3 )]∆ beam + [200(10 3 )](3.7037 ∆ beam )2 = 2 2 2 1.74211(106 )∆ 2beam = 281.25 ∆ beam = 0.01271 m = 12.7 mm

Ans.

Fbeam = kbeam ¢ beam 3 3 )][0.01271] 9.4118(10 ) N 9.41 kN = [740.74(10 = =

= Mmax = σ max

[9.4118(10 3 )](7.2) = 16.941(10 3 ) N ⋅ m 4 Mmax c [16.941(10 3 )](0.25412) = I 28.8(10 −6 )

6 ) N−m 2 74.7 MPa < σ Y = 74.71(10 =

O.K.

Ans.

Ans: ∆beam = 12.7 mm, s max = 74.7 MPa 1218

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14–66. The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. If the railcar of mass 10 Mg is traveling with a speed of v = 1.5 m>s, determine the maximum bending stress developed in the bar. The springs at A and B have a stiffness of k = 15 MN>m.

300 mm k

A

v

2m

2m

Solution Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x

2Fsp - P = 0

Fsp

P = 2

(1)

a

a

k

B

300 mm 400 mm

Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 48EI 1 equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 3 12 L and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa. kb =

48 c 73.1 A 109 B d c 0.9 A 10 - 3 B d 43

= 49.3425 A 106 B N>m

Using Eq. (1) Fsp =

P 2

ksp ¢ sp =

1 k ¢ 2 b b

1 kb 1 49.3425 A 10 ¢ ≤ ¢b = C 2 ksp 2 15 A 106 B

6

¢ sp =

B

S ¢ b = 1.64475¢ b

Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2 c ksp ¢ sp 2 d 2 2 2 Substituting Eq. (2) into this equation, 1 1 mv2 = kb ¢ b 2 + ksp (1.64475¢ b)2 2 2 1 1 mv2 = kb ¢ b 2 + 2.7052ksp ¢ b 2 2 2 1 1 c 10 A 103 B d A 1.52 B = c 49.3425 A 106 B d ¢ b 2 + 2.7052c15 A 106 B d ¢ b 2 2 2 ¢ b = 0.01313 m

1214

(2)

Section a – a

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14–66. Continued

Solution Maximum Stress. The maximum force on the bar is (Pb)max = kb ¢ b = 49.3425 A 106 B (0.01313) = 647.90 A 103 B N. The maximum moment occurs at the midspan of the bar, where Mmax = = 647.90 A 103 B N # m. Applying the flexure formula, smax =

647.90 A 103 B (4) (Pb)max L = 4 4

647.90 A 103 B (0.15) Mmax c = = 107.98 MPa = 108 MPa I 0.9 A 10 - 3 B

Ans.

Since smax 6 sY = 414 MPa, this result is valid.

Ans: s max = 108 MPa 1215

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14–67. The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. Determine the maximum speed v the 10-Mg railcar without causing the bar to yield when it is struck by the railcar. The springs at A and B have a stiffness of k = 15 MN>m.

300 mm k

A

v

2m

2m

Solution Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x

2Fsp - P = 0

Fsp

P = 2

(1)

a

a

k

B

300 mm 400 mm

Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 48EI 1 equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 3 12 L and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa.

kb =

48 c 73.1 A 109 B d c 0.9 A 10 - 3 B d 43

= 49.3425 A 106 B N>m

Using Eq. (1) Fsp =

P 2

ksp ¢ sp =

1 k ¢ 2 b b

6 1 kb 1 49.3425 A 10 B = ¢ S ¢ b = 1.64475¢ b ≤ ¢b = C 2 ksp 2 15 A 106 B

(2)

Maximum Stress. The maximum force on the bar is

(Pb)max = kb ¢ b

¢ sp

= 49.3425 A 106 B ¢ b. The maximum moment occurs at the midspan of the bar, where 49.3425 A 106 B ¢ b(4) (Pb)max L = = 49.3425 A 106 B ¢ b. Applying the flexure Mmax = 4 4 formula with smax = sY = 414 MPa, smax =

Mmax c I

414 A 106 B =

49.3425 A 106 B ¢ b (0.15) 0.9 A 10 - 3 B

¢ b = 0.050342 m

1216

Section a – a

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14–67. Continued

Solution Substituting this result into Eq. (2), ¢ sp = 0.0828 m Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2 B ksp ¢ sp 2 R 2 2 2 1 1 1 c 10 A 103 B d v2 = c 49.3425 A 106 B d A 0.0503422 B + 2 B c15 A 106 B d A 0.08282 B R 2 2 2 v = 5.75 m>s

Ans.

Ans: kb = 49.3425 (106) N>m, ∆b = 0.050342 m, v = 5.75 m>s 1217

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*14–68.

The steel beam AB acts to stop the oncoming railroad car, which has a mass of 10 Mg and is coasting towards it at v = 0.5 m>s . Determine the maximum stress developed in the beam if it is struck at its center by the car. The beam is simply supported and only horizontal forces occur at A and B. Assume that the railroad car and the supporting framework for the beam remains rigid. Also, compute the maximum deflection of the beam. Est = 200 GPa,sY = 250 MPa.

v  0.5 m/s

200 mm 200 mm

A 1m 1m B

Solution From Appendix C: ¢ st = k =

10(103)(9.81)(23) PL3 = 0.613125(10 - 3) m = 1 48EI 48(200)(104)(12 )(0.2)(0.23)

10(103)(9.81) W = 160(106) N>m = ¢ st 0.613125(10 - 3)

¢ max =

0.613125(10 - 3)(0.52) ¢ st v2 = = 3.953(10 - 3) m = 3.95 mm C g C 9.81

Ans.

W¿ = k¢ max = 160(106)(3.953)(10 - 3) = 632455.53 N M¿ =

632455.53(2) w¿L = = 316228 N # m 4 4

smax =

316228(0.1) M¿c = 237 MPa 6 sg = 1 3 I 12 (0.2)(0.2 )

Ans.

O.K.

Ans: ¢max = 3.95 mm, smax = 237 MPa 1197

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14–69. The diver weighs 750 N and, while holding 1h = 02board rigid, strikes end ofthe a wooden board diving with himself rigid,the strikes end ofdiving a wooden ft>s. Determine a downward of 4 velocity the Determine maximum (h = 0) with velocity a downward of 1.2 m/s. bending stress bending developed in developed the board. in The the maximum stress theboard board.has Thea thickness in. and of width of 1.5 ft. width ksi, Ew = 1.8110 board hasofa 1.5 thickness 40 mm and of 45032mm. sYw == 12.6 8 ksi. E GPa, sY = 56 MPa.

vv

h

Solution

1.2 4 ftm

3m 10 ft

Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy. L

M2 dx 1 P¢ = 2 L0 2EI 4 ft

10 ft

1.2 m 3m 1 11 2 (150)¢ ((1875x -375x1)12)dx ( -150x22))2 dx2dd (750)Δstst = cc dx (750x 1 1+ 2 2EI 0L0 2EI L 0

= ¢ Δst  =

70.0(10 9450 N 3· )mlb3 # ft3 EIEI

9450 1 (0.45)(0.04 3 ) [12.6(10 )]  12   9

= 0.3125 m

Conservation of Energy: The equivalent spring constant for the board is 750 W = = 2.40(10 3 ) N−m , k = ∆ st 0.3125 Ue = Ui 1 1 my2 + W¢ max = k¢ 2max 2 2 1  750  1 2 2 [2.40(10 3 )]∆ max = ∆ max   (1.2 ) + 750 2  9.81  2 2 − 750 ∆ max − 55.05 = 1.20(10 3 )∆ max 0

Solving for the positive root, we have ∆ max = 0.69135 m

Maximum Stress: The maximum force on to the beam is Pmax = k¢ max [2.40(10 3 )](0.69135) = 1.6592(10 3 ) N. The maximum moment occurs at the middle support 3 support Mmax [1.6592(10 = )](3) 4.9777(10 3 ) N ⋅ m =

σ max =

Mmax c 4.9777(10 3 )(0.02) = 1 (0.45)(0.04 3 ) I 12

6 ) N−m 2 41.5 MPa < σ Y = = 41.48(10

O.K.

Ans.

Note: The result will be somewhat inaccurate since the static displacement is so large. 750 N M(x1) = –1875x1 M(x2) = –750x2

1.2 m 1875 N 2625 N

Ans:

3m

s max = 41.5 MPa 1208

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14–70. The diver weighs 750 N and, while holding himself rigid, strikes the end of the wooden diving board. Determine the maximum height h from which he can jump onto the board so that the maximum bending stress in the wood does Theboard boardhas hasaa thickness thickness of does not notexceed exceed426 MPa. ksi. The 40 andwidth widthof of1.5 450ft.mm. = 12.632GPa. 1.5mm in. and Ew E =w1.8110 ksi.

vv

Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy.

hh

L

M2 dx 1 P¢ = 2 L0 2EI

4 ftm 1.2

10 ft 3m

1.24mft 310mft 1 11 2 (750)Δstst = cc dx  (750x dx2dd ((1875x -375x1)12)dx + ( -150x22))2 dx (150)¢ 1 1 2EI 2 2EI 0L0 0 L

Δst  ¢ = =

9450 N 3· )mlb3 # ft3 70.0(10 EIEI

9450 1 (0.45)(0.04 3 ) [12.6(109 )]  12  

= 0.3125 m

Maximum Stress: The maximum force on the beam is Pmax. The maximum moment occurs at the middle support Mmax = Pmax (3) (10)(12) = 120Pmax. = 3Pmax •

smax = 12.6(109 ) =

Mmax c I (3Pmax )(0.02) 1 (0.45)(0.04 3 ) 12

1680 N Pmax = 337.5 lb Conservation of Energy: The equivalent spring constant for the board is 750 W = = 2.40(10 3 ) N−m. The maximum displacement at the end of k = ∆ st 0.3125 1680 Pmax = = = 0.700 m. the board is ∆ max k 2.40(10 3 ) Ue = Ui W(h + ¢ max) =

1 k¢ 2max 2

1 750(h + 0.700) = [2.40(10 3 )](0.700 2 ) 2 = m 84.0 mm. = h 0.0840

Ans.

Note: The result will be somewhat at inaccurate since the static displacement is so large.

750 N M(x1) = –1875x1 M(x2) = –750x2

1.2 m 1875 N 2625 N

3m

Ans: h = 84.0 mm 1209

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14–71.  The car bumper is made of polycarbonate-polybutylene terephthalate. If E = 2.0 GPa, determine the maximum deflection and maximum stress in the bumper if it strikes the rigid post when the car is coasting at v = 0.75 m>s. The car has a mass of 1.80 Mg, and the bumper can be considered simply supported on two spring supports connected to the rigid frame of the car. For the bumper take I = 300(106) mm4, c = 75 mm, sY = 30 MPa and k = 1.5 MN>m.

Solution Equilibrium: This requires Fsp = ksp ∆ sp =

k∆ beam 2

0.9 m k

0.9 m k

v  0.75 m/s

Pbeam . Then 2 or

∆ sp =

k ∆ 2ksp beam

(1)

Conservation of Energy: The equivalent spring constant for the beam can be determined using the deflection table listed in the Appendix C. k =

48 3 2(109) 4 3 300(10 - 6) 4 48EI = = 4 938 271.6 N>m L3 1.83

Thus, Ue = Ui 1 1 1 2 mv2 = k∆ beam + 2 a ksp ∆ 2sp b 2 2 2

(2)

Substitute Eq. (1) into (2) yields

1 1 k2 2 2 mv2 = k∆ beam + ∆ 2 2 4ksp beam (4 93 8271.6)2 2 1 1 2 (1800) 1 0.752 2 = (493 8271.6) ∆ beam + ∆ beam 2 2 4[1.5(106)] ∆ beam = 8.8025 1 10 - 3 2 m

4 938 271.6 Maximum Displacement: From Eq. (1) ∆ sp = 3 8.8025 1 10 - 3 2 4 = 2[1.5(106)] 0.014490 m. ∆ max = ∆ sp + ∆ beam

= 0.014490 + 8.8025 1 10 - 3 2

Ans.

= 0.02329 m = 23.3 mm

Maximum Stress: The maximum force on the beam is Pbeam = k∆ beam = 4 938 271.6 3 8.8025 1 10 - 3 2 4 = 43 469.3 N. The maximum moment occurs at 43 469.3(1.8) Pbeam L mid-span. Mmax = = = 19 561.2 N # m. 4 4 smax =

19 561.2(0.075) Mmax c = = 4.89 MPa I 300(10 - 6)

Ans.

Since smax 6 sY = 30 MPa, the above analysis is valid.

Ans: ∆ max = 23.3 mm, smax = 4.89 MPa 1482

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*14-72. Determine thevertical displacement of point E. Each A-36 steel member has a cross-sectional area of Each A-36 steel member has a cross-sectional area of 4.5 in2. 2800 mm2.

F

E

D

6 ftm 1.8

Solution

A

n n (kN)

NN(kN)

L (m)

AB

0.6667

16.6667 3.333

2.4

26.667 213.33

BC

0.6667

16.6667 3.333

2.4

26.667 213.33

CD

0

0

1.8

00

DE

0

0

2.4

00

EF

0

0

2.4

00

AF

0

0

1.8

00

AE

–0.8333

–20.8333 –4.167

3.0

52.083 416.67

CE

–0.8333

–20.8333 –4.167

3.0

52.083 416.67

BE

0

25.00 5.00

1.8

00

8 ftm 2.4

8 ftm 2.4

nNL

5 kip 25 kN

20

)

3 33

kN

(C

.83

25 kN (T)

Member

C

B

Virtual-Work Equation: Applying Eq. 14–39, we have

0.8

2

16.6667 kN (T)

33

kN

1.8 m

(C

)

16.6667 kN (T)

2.4 m

2.4 m 25 kN

1 kN

©157.5 kN2 # m

0.8

)

N 3k

33

(C

(C

)

0

nNL 1#¢ = a AE

3k

N

3 .83

0.6667 kN (T)

0.6667 kN (T)

157.5 kN ⋅ m 2 1 kN ⋅ (∆ E )v = AE = (∆ E )v

157.5(10 3 ) = 0.28125(10 −3 ) m = 0.281 mm T [2.80(10 −3 )][200(109 )]

Ans.

Ans: = (∆ E )v

1223

0.281 mm T

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14–73. Determine the vertical displacement of point B. •14–77. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 2 Each mm A-36 2 steel member has a cross-sectional area of 4.5 in . 2800 .

F

E

D

6 ftm 1.8

Solution

A

C

B 8 ftm 2.4

8 ftm 2.4

Virtual-Work Equation: Applying Eq. 14–39, we have

5 kip 25 kN

n n (kN)

N N (kN)

LL(m)

nNL nNL

AB

0.6667

3.333 16.6667

96 2.4

213.33 26.667

BC

0.6667

3.333 16.6667

96 2.4

213.33 26.667

CD

0

0 0

72 1.8

00

DE

0

0 0

96 2.4

00

20

)

3 33

(C

kN

.83

25 kN (T)

Member

.8 20

16.6667 kN (T)

33

16.6667 kN (T)

0 0

96 2.4

00

AF

0

0 0

72 1.8

00

AE

–0.8333

–20.8333 –4.167

120 3.0

52.083 416.67

CE

–0.8333

–20.8333 –4.167

120 3.0

52.083 416.67

BE

1.00

25.00 5.00

72 1.8

2.4 m 25 kN

)

3 33

45.00 360.00

kN

(C

0.8

©202.5 kN2 # m

1.00 kN (T)

0

1.8 m

(C

)

2.4 m

EF

kN

0.6667 kN (T)

nNL 1#¢ = a AE

0.8

33

3k

N

(C

)

0.6667 kN (T)

1 kN

202.5 kN ⋅ m 2 1 kN ⋅ (∆ B )v = AE = (∆ B )v

202.5(10 3 ) = 0.3616(10 −3 ) m = 0.362 mm T [2.80(10 −3 )][200(109 )]

Ans.

Ans: (∆B)n = 0.362 mm 1222

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14–74. Determine the vertical displacement of joint A. Each A992 steel member has a cross-sectional area of 400 mm2.

B

C

2m D

A E 1.5 m

Solution Member

n

AB

1.25 - 0.75

AE

2.25

BC BD

N

L

60 kN

- 30 1.5 33.75 180 3.0 1215.00 213 1171.80

0 60 2.0 - 0.75

DE

40 kN

50 2.5 156.25

- 0.5213 ­ - 50213

BE

nNL

3m

0

- 30 3.0 67.5 Σ = 2644.30

1 # ∆ Av = Σ (∆ A)v =

nNL AE 2644.30(103)

400(10-6)(200)(109)

= 0.0331 m = 33.1 mmT

Ans.

Ans: (∆ A)v = 33.1 mmT 1483

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*14–75. Determine the vertical displacement of joint H. Each A-36 steel member has a cross-sectional area of 2 4.5 . 2. 2800inmm

J

I

12 ft

Solution

= ∆ Hv

G

F 9 ft

A

1 # ¢ Nv = ©

H

Ans.

B 6 kip

12 ft

C 8 kip

12 ft

D

E 12 ft

6 kip

nNL AE

2828.89(10 3 ) = 5.052(10 −3 ) m = 5.05 mm T [2.80(10 −3 )][200(109 )]

J

I

H

Ans.

F

G

3m A

B 4m

C 4m

30 kN

4m 40 kN

E

D 4m 30 kN

Ans: ¢ Hv = 5.05 mm 1226

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*14–76. Determine the vertical displacement of joint C. Each A-36 steel member has a cross-sectional area of 2800 mm2.

II

JJ

12 ft 4m

Solution

= ∆Cv

G G

FF 93 ftm

AA

1 # ¢ Cv = ©

H H

Ans.

BB 630kip kN

12 ft 4m

CC

12 ft 4m

840kip kN

EE

D D

12 ft 4m

630kip kN

nNL AE

2948.89(10 3 ) = 5.266(10 −3 ) m = 5.27 mm T [2.80(10 −3 )][200(109 )]

Ans.

Ans: = ∆Cv

1226

5.27 mm

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14–77. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 400 mm2.

E

D

2m

Solution

B 1.5 m

Virtual-Work Equation: Member

C

A

n

N

AB

0

BC

0

AE

0

CE

–1.25

BE

1.00

DE

0.750

–22.5 A 10

3

37.5 A 10

3

B

–22.5 A 103 B

B

–62.5 A 103 B 22.0 A 103 B 60.0 A 103 B

L

30 kN

nNL

1.5

0

1.5

0

2.5

0

2.5 2 1.5

1.5 m 20 kN

195.3125 A 103 B 40.0 A 103 B 67.5 A 103 B

© 302.8125 A 103 B N2 # m

nNL 1#¢ = a AE 1 N # (¢ B)v = (¢ B)v =

302.8125(103) N2 # m AE 302.8125(103) 0.400(10 - 3)[200(109)]

= 3.785 A 10 - 3 B m = 3.79 mm T

Ans.

Ans: (¢ B)v = 3.79 mmT 1225

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14–78. Determine the vertical displacement of point A. Each A-36 steel member has a cross-sectional area of 400 mm2.

E

D

2m

Solution

C

A B 1.5 m

Virtual-Work Equation: Member

n

AB

–0.750

BC

–0.750

AE

1.25

CE

–1.25

BE

0

DE

N

1.50

–22.5 A 103 B –22.5 A 103 B 37.5 A 103 B

–62.5 A 103 B 22.0 A 103 B 60.0 A 10

3

B

L 1.5 1.5 2.5 2.5 2 1.5

nNL

25.3125 A 103 B

1.5 m 20 kN

30 kN

25.3125 A 103 B

117.1875 A 103 B 195.3125 A 103 B 135.00 A 10

0 3

B

© 498.125 A 103 B N2 # m

nNL 1#¢ = a AE 1 N # (¢ A)v = (¢ A)v =

498.125(103) N2 # m AE 498.125(103) 0.400(10 - 3)[200(109)]

= 6.227 A 10 - 3 B m = 6.23 mm T

Ans.

Ans: 1 N # (¢ A)v =

498.125(103) N2 # m , AE

(¢ A)v = 6.23 mm T 1224

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14–79.  Determine the horizontal displacement of joint B of the truss. Each A992 steel member has a cross-sectional area of 400 mm2.

4 kN

5 kN

2m

C

B

1.5 m

A

D

Solution Member n AB

0 –1.25

AC

N L nNL 0 1.5

0

–5.00 2.5 15.625

AD

1.00 4.00 2.0 8.000

BC

1.00 4.00 2.0 8.000

CD

0.75 –2.00 1.5 –2.25 Σ = 29.375

1 # (∆ B)h = Σ (∆ B)h =

nNL AE 29.375(103)

400(10 - 6)(200)(109)

= 0.3672(10 - 3) m = 0.367 mm d

Ans.

Ans: (∆ B)h = 0.367 mm d 1490

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*14–80.  Determine the vertical displacement of joint C of the truss. Each A992 steel member has a cross-sectional area of 400 mm2.

4 kN

5 kN

2m

C

B

1.5 m

A

D

Solution Member n

N L nNL

AB

0

0 1.5

0

AC

0 –5.00 2.5

0

AD

0 4.00 2.0

0

BC

0 4.00 2.0

0

–1.00

CD

–2.00 1.5 3.00 Σ = 3.00

1 # (∆ C)v = Σ (∆ C)v =

nNL AE 3.00 (103)

400(10 - 6)(200)(109)

= 37.5(10 - 6) m = 0.0375 mmT

Ans.

Ans: (∆ C)v = 0.0375 mmT 1491

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14–81.  Determine the horizontal displacement of joint  C. Each A-36 steel member has a cross-sectional area of 400 mm2.

B

5 kN C

2m

D 10 kN

Solution

1.5 m

A

Member Real Forces N: As shown on Figure (a). Member Virtual Forces n: As shown on Figure (b). Virtual-Work Equation: Applying Eq. 14–39, we have Member AB BC

n

N

0 10.0(

L

nNL

) 2 0

103

1.00 12.5(103)

1.5          18.75(103)

CD

0 10.0(103) 2 0

AD

0       0

AC

–12.5(

0

1.5

0

) 2.5 0

103

Σ18.7 ( 103 ) N2 # m 1#∆ = Σ 1 N # (∆ C)h = (∆ C)h =

nNL AE

18.75 ( 103 ) N2 # m AE 18.75 ( 103 ) 0.400 ( 10-3 ) 3 200 ( 109 ) 4

= 0.2344 ( 10-3 ) m = 0.234 mm   d

Ans.

Ans: (∆ C)h = 0.234 mm   d 1492

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14–82.  Determine the vertical displacement of joint D. Each A-36 steel member has a cross-sectional area of 400 mm2.

B

5 kN C

2m

D 10 kN

Solution

1.5 m

A

Member Real Forces N: As shown on Figure (a). Member Virtual Forces n: As shown on Figure (b). Virtual-Work Equation: Applying Eq. 14–39, we have Member n

N

AB

1.00 10.0(

BC   

0.750 12.5(103)

CD

1.00 10.0(103)

AD AC

)

103

0       0 –1.25 –12.5(

103

)

L nNL 2

20.0(103)

1.5         14.0625(103) 2

20.0(103)

1.5

0

2.5    39.0625(103) Σ93.125 ( 103 ) N2 # m

1#∆ = Σ 1 N # (∆ D)v = (∆ D)v =

nNL AE

93.125 ( 103 ) N2 # m AE 93.125 ( 103 ) 0.400 ( 10-3 ) 3 200 ( 109 ) 4

= 1.164 ( 10-3 ) m = 1.16 mm

Ans.

T

Ans: (∆ D)v = 1.16 mm 1493

T

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14–83.  Determine the vertical displacement of joint A. The truss is made from A992 steel rods having a diameter of 30 mm.

20 kN C

D

2m

A

Solution

1.5 m

­­Members Real Force N: As indicated in Fig. a.

50(103) FBD = = 70.74 MPa 6 sY = 345 MPa, p A (0.032) 4

Member

n (N)

N (N)

L(m)

AB

- 0.75

- 15(103)

3

­AD

1.25

nNL(N2 # m) 33.75(103)

25(103) 2.5

78.125(103)

1 40(103) 2

BC

CD

80(103)

- 50(103) 2.5

156.25(103)

1.5 45(103) 1.5

101.25(103)­­

- 1.25

BD

B

20 kN

Members Virtual Force n: As indicated in Fig. b. Virtual Work Equation: Since smax =

1.5 m

Σ 449.375(103)

Then 1#∆ = Σ

nNL AE

1 N # (∆ A)v =

449.375(103) p (0.032)[200(109)] 4

(∆ A)v = 3.179(10 - 3) m = 3.18 mm T

Ans.

Ans: (∆ A)v = 3.18 mm T 1494

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*14–84. Determine the vertical displacement of joint D. The truss is made from A992 steel rods having a diameter of 30 mm.

20 kN C

D

2m

A

Solution

1.5 m

Members Real Force N: As indicated in Fig. a.

50(103) FBD = p = 70.74 MPa 6 sY = 345 MPa, A (0.032) 4

Virtual Work Equation: Since smax =

n (N)

N (N)

- 15(103) 0

AB

nNL(N2 # m)

3

0

25(10 ) 2.5

0

0

BC

40(103) 1

2

80(103)

- 50(103) 2.5

156.25(103)

0.75 45(103) 1.5

50.625(103)­­

- 1.25

CD

L(m)

3

­AD

BD

B

20 kN

Members Virtual Force n: As indicated in Fig. b.

Member

1.5 m

Σ 286.875(103)

Then 1#∆ = Σ

nNL AE

1 N # (∆ D)v =

286.875(103) p (0.032)[200(109)] 4

(∆ D)v = 2.029(10 - 3) m = 2.03 mm T

Ans.

Ans: (∆ D)v = 2.03 mm T 1495

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14–85.  Determine the horizontal displacement of joint D. Each A-36 steel member has a cross-sectional area of 300 mm2.

D

4 kN

3m

C

2 kN E

Solution

3m

Virtual-Work Equation: Applying Eq. 14–39, we have A

Member

n

N

L

nNL

AE

2.00

800(103)

3

48.0(103)

ED

2.00

800(

3

48.0(

CD

- 2.236

- 8.944(103)

3.354

67.082(103)

BC

- 2.236

-11.180(103)

3.354

83.853(103)

CE

0

- 2.00(103)

1.5

0

0

2.236(

3.354

0

AC

)

103

)

103

B 3m

)

103

Σ246.935 ( 103 ) N2 # m 1#∆ = Σ 1 N # (∆ D)h = (∆ D)h =

nNL AE

246.935 ( 103 ) N2 # m AE 246.935 ( 103 ) 0.300 ( 10-3 ) 3 200 ( 109 ) 4

= 4.116 ( 10-3 ) m = 4.12 mm  

S

Ans.

Ans: (∆ D)h = 4.12 mm   1496

S

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14–86.  Determine the horizontal displacement of joint E. Each A-36 steel member has a cross-sectional area of 300 mm2.

D

4 kN

3m

C

2 kN E

Solution

3m

Virtual-Work Equation: Applying Eq. 14–39, we have Member

n

N

L

nNL

AE

0

800 ( 103 )

3

0

ED

0

800 ( 103 )

3

0

CD

0

–8.944 ( 103 )

3.354

0

BC

- 1.118

–11.180 ( 103 )

3.354

41.926 ( 103 )

CE

- 1.00

–2.00 ( 103 )

1.5

300 ( 103 )

AC

1.118

2.236 ( 103 )

3.354

8.385( 103 )

A

B 3m

Σ53.312 ( 103 ) N2 # m 1#∆ = Σ 1 N # (∆ E)h = (∆ E)h =

nNL AE

53.312 ( 103 ) N2 # m AE 53.312 ( 103 ) 0.300 ( 10-3 ) 3 200 ( 109 ) 4

= 0.8885 ( 10-3 ) m = 0.889 mm

S

Ans.

Ans: (∆ E)h = 0.889 mm 1497

S

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14–87.  Determine the displacement at point C. EI is constant.

P

P

A

B a

a– 2

C a– 2

a

Solution 1 # ∆C =

L0

L

mM dx EI a

∆C = 2 a =

1 1 bc a x b(Px1)dx1 + EI L0 2 1 L0

23Pa3 24EI

a>2

1 (a + x2)(Pa)dx2 d 2 Ans.

Ans: ∆C = 1498

23Pa 3 24EI

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*14–88. The beam is made of southern pine for which Ep = 13 GPa. Determine the displacement at A.

15 kN 4 kN/m

A

B 1.5 m

120 mm

L

mM L0 EI 1.5

∆A = =

3m 180 mm

Solution 1 # ∆A =

C

3

1 c (x1)(15x1)dx1 + (0.5x2)(2x22 + 1.5x2)dx2 d EI L0 L0

43.875(103) 43.875 kN # m3 = = 0.0579 m = 57.9 mm Ans. 1 9 EI 13(10 )(12 )(0.12)(0.18)3

Ans: ∆ A = 57.9 mm 1499

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14–89.  Determine the displacement at point C. EI is constant.

P

A

C B a

a

Solution 1 # ∆C =

L

mM dx L0 EI a

∆C = =

a

1 c (x )(Px1)dx1 + (x2)(Px2)dx2 d EI L0 1 L0 2Pa3 3EI

Ans.

Ans: ∆C = 1500

2Pa 3 3EI

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14–90.  Determine the slope at point C. EI is constant.

P

A

C B a

a

Solution 1 # uC =

uC = =

L

mu Mdx EI L0 L0

a x1 a

1 2 Px1 dx1 EI

+

L0

a

(1)Px2 dx2 EI

Pa2 Pa2 5Pa2 + = b 3EI 2EI 6EI

Ans.

Ans: uC = 1501

5Pa 2 6EI

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14–91.  Determine the slope at point A. EI is constant.

P

A

C B a

a

Solution 1 # uA =

uA =

L

mu M dx L0 EI a a x1 1 Pa2 c a1 - b(Px1) dx1 + (0) (Px2) dx2 d = EI L0 a 6EI L0

Ans.

Ans: uA = 1502

Pa 2 6EI

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*14–92. Determine the displacement at B of the 1.5-in30-mmdiameter A-36 steel shaft.

2 ftm 0.6 2 ftm 0.6

A

ft 13 m

D

Solution 1 ⋅ ∆B = ∫

L mM

EI

0

B 140 700 lb N

dx

1.5 ftm 0.45

140 700 lb N C

1 0.6m ∆B = ∫ (0.5472x1 )(1.6264x1 )dx1 EI  0

=

0.6m

+

∫0

+

∫0

+

∫0

0.45m

1m

320 lb N 320 lb 1600 N 1600

(0.5472)(0.6 + x2 )(0.9758 + 0.2264 x2 )dx2 (0.4528x3 )(2.9731x3 )dx3

0.4528( x4 + 0.45)(1.3381 − 0.2264 x4 )dx4  

0.93401 kN ⋅ m 3 0.93401(10 3 ) = EI [200(109 )]  π4 (0.0154 )   0.11745 m = 117 mm T

Ans.

1 kN

1.6264 kN

1.4 kN

3.2 kN

2.9735 kN

0.5472 kN

0.4528 kN M= 0.5472x1

1.6264x1 0.5472 kN

1.6264 kN 0.9758 + 0.2264x2 0.6 m 1.4 kN 1.6264 kN

0.6 m M= 2.9735x

M= 0.5472(0.6 + x2)

0.5472 kN 2.9735 kN

M= 0.4528x3 0.4528 kN

1.3381 – 0.2264x4 0.45 m 2.9735 kN

M= 0.4528(x4 + 0.45) 0.45

0.4528 kN

Ans: ∆B = 117 mm T 1231

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30-mm-diameterA-36 A-36 14–93. Determine the slope of the 1.5-in-diameter steel shaft at the bearing support A.

2 ftm 0.6 2 ftm 0.6

A

Solution

ft 13 m

D L mθ M 1 ⋅θA = dx ∫0 EI

= θA

=

1  0.6m (1 − 0.3774x1 )(1.6264x1 )dx1 EI  ∫0 0.6m

+

∫0

+

∫0

+

∫0

140 700 lb N

140 700 lb N C

(0.7736 − 0.3774 x2 )(0.9758 + 0.2264 x2 )dx2

0.45m

1m

1.5 ftm 0.45

B

320 lbN 320 lbN 1600 1600

0.3774x3 (2.9735x3 )dx3

0.3774( x4 + 0.45)(1.3381 − 0.2264 x4 )dx4  

1.12668 kN ⋅ m 2 1.12668(10 3 ) = EI [200(109 )]  π4 (0.0154 )  

Ans.

0.141682 πad = 8.12° (clockwise)

1 kN · m

1.6264 kN

1.4 kN

3.2 kN

2.9735 kN

0.3774 kN 1 kN · m

0.3774 kN 1 – 0.3774x1

1.6264x1 0.3774 kN

1.6264 kN

1 kN · m

0.7736 – 0.3774x2 0.3774x3

0.9758 + 0.2264x2 0.6 m 1.4 kN 1.6264 kN

0.6 m 0.3774 kN

M= 2.9735x 2.9735 kN

0.3774 kN

x2

0.3774(x4 + 0.45) 0.45 m 0.3774 kN

1.3381 – 0.2264x4 0.45 m 2.9735 kN

Ans: uA = 8.12° (clockwise) 1232

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14–94.  The beam is made of Douglas fir. Determine the slope at C.

8 kN

A

B 1.5 m

C 1.5 m

1.5 m 180 mm

Solution

120 mm

Virtual Work Equation: For the slope at point C, 1#u =

L

muM dx L0 EI

1 kN # m # uC = 0 +

1 EI L0 +

uC = =

1.5 m

1 EI L0

(0.3333x2)(4.00x2) dx2 1.5 m

(1 - 0.3333x3)(4.00x3)dx3

4.50 kN # m3 EI 4.50(1000) 1 13.1(10 ) 3 12 (0.12)(0.183) 4 9

= 5.89 1 10 - 3 2 rad = 0.337°

Ans.

Ans: uC = 0.337° 1505

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14–95.  Determine the displacement at pulley B. The A992 steel shaft has a diameter of 30 mm.

A

4 kN

B

3 kN

0.4 m 0.4 m

Solution 1 # ∆B =

C

L

mM dx L0 EI 0.4

1 c (0.4286x1) (4.357x1) dx1 + 0.4286 (x2 + 0.4) (0.357x2 + 1.7428) dx2 EI L0 L0 L0

=

1 kN

0.3 m

0.4

∆B =

1 kN 0.3 m

0.3

(0.5714x3) (4.643x3) dx3 +

L0

0.3

0.5714 (x4 + 0.3)(1.643x4 + 1.3929) dx4 d

0.37972(103) 0.37972 kN # m3 = = 0.0478 m = 47.8 mm T EI 200(109) 1 p4 2 (0.0154)

Ans.

Ans: ∆ B = 47.8 mm T 1506

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*14–96. The A992 steel beam has a moment of inertia of I = 125(106) mm4. Determine the displacement at point D.

18 kNm

18 kNm

A

D 4m

B

3m

3m

C

4m

Solution 1 # ∆D =

L

mM dx L0 EI

∆ D = (2)

3 81(103) 1 81 kN # m3 c (0.5x2) (18) dx2 d = = EI L0 EI 200 (109)(125)(10 - 6)

= 3.24 (10 - 3) = 3.24 mm T

Ans.

Ans: ∆ D = 3.24 mm T 1507

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14–97.  The A992 steel beam has a moment of inertia of I = 125(106) mm4. Determine the slope at A.

18 kNm

18 kNm

A

D 4m

B

3m

3m

C

4m

Solution 1 # uA =

L

mu M dx L0 EI 1 126 kN # m2 c (1) (18) (dx1) + (0.1667x3) (18) dx3 d = EI L0 EI L0 4

uA = =

126 (103)

200 (109) (125) (10 - 6)

6

= 5.04 (10 - 3) rad = 0.289°

Ans.

Ans: uA = 0.289° 1508

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14–98.  The A992 structural steel beam has a moment of inertia of I = 125(106) mm4. Determine the slope at B.

18 kNm

18 kNm

A

D 4m

B

3m

3m

C

4m

Solution 1 # uB =

L0

L

uB = 0 + =

mu M dx EI 1 EI L0

6

1 - 16 x2

+ 1 2 (18)dx EI

54 (103) 54 = = 0.00216 rad = 0.124° EI 200 (109) (125 (10 - 6))

Ans.

Ans: uB = 0.124° 1509

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14–99.  Determine the displacement at point C of the shaft. EI is constant.

A

B C

L

L 2 P

Solution Real Moment Function M: As indicated in Fig. a. Virtual Moment Function m: As indicated in Fig. b. Virtual Work Equation: 1#∆ = 1 # ∆C = ∆C =

L

mM dx L0 EI L x1 P 1 c a b a x1 b dx1 + EI L0 2 2 L0

PL3 T 8EI

L>2

x2 (Px2) dx2 d Ans.

Ans: ∆C = 1510

PL3 T 8EI

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*14–100. Determine the slope at A of the shaft. EI is constant.

A

B C

L

L 2 P

Solution Real Moment Function M: As indicated in Fig. a. Virtual Moment Function M: As indicated in Fig. b. Virtual Work Equation: 1#u = 1 # uA = uA =

L

mu M dx L0 EI L x1 P 1 c a1 - b a x1 b dx1 + EI L0 L 2 L0

PL2 12EI

L>2

(0) (Px2) dx2 d Ans.

Ans: uA = 1511

PL2 12EI

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14–101.  Determine the slope of end C of the overhang beam. EI is constant.

w

C

A B

D L 2

L 2

L 2

Solution Real Moment Function M: As indicated in Fig. a. Virtual Moment Function mU: As indicated in Fig. b. Virtual Work Equation: 1#u = 1 # uC = uC =

L

mu M dx L0 EI L L>2 x1 w 1 w 3 J ¢ - ≤c 1 11Lx1 - 12x1 2 2 d dx1 + (1) ¢ x2 ≤dx2 R EI L0 L 24 3L L0

1 w J EI 24L L0

uC = -

L

1 12x1 3

13wL3 13wL3 = b 576EI 576EI

- 11Lx1 2 2 dx1 +

w 3L L0

L>2

x2 3 dx2 R

Ans.

Ans: uC = 1512

13wL3 576 EI

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14–102.  w

Determine the displacement of point D of the overhang beam. EI is constant.

C

A B

D L 2

L 2

L 2

Solution Real Moment Function M: As indicated in Fig. a. Virtual Moment Function m: As indicated in Fig. b. Virtual Work Equation: 1#∆ =

L

mM dx L0 EI

1 # ∆D =

1 J EI L0 +

∆D =

w J 48EI L0

∆D =

wL4 T 96EI

L>2

L0 L>2

¢ L>2

x1 w ≤c 1 11Lx1 - 12x1 2 2 d dx1 2 24

¢

x2 w ≤c 1 13Lx2 - 12x2 2 - L2 2 d dx2 R 2 24

1 11Lx1 2

- 12x1 3 2 dx1 +

L0

L>2

1 13Lx2 2

- 12x2 3 - L2x2 2 dx2 R

Ans.

Ans: ∆D = 1513

wL4 T 96 EI

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14–103.  Determine the slope at A of the 2014-T6 aluminum shaft having a diameter of 100 mm.

A

C

1m

B

0.5 m 0.5 m 8 kN

1m

8 kN

Solution Real Moment Function M: As indicated in Fig. a. Virtual Moment Function m: As indicated in Fig. b. Virtual Work Equation: 1#u =

L

mu M dx L0 EI

1 kN # m # uA =

1 c EI L0 +

1m

L0

(1 - 0.3333x1) (8x1) dx1 +

= =

[0.3333 (x2 +1)] 8dx2

1m

1m

uA =

L0

1m

(0.3333x3) (8x3) dx3 d

1m

1m

1 c8 (1 - 0.3333x1) dx1 + 2.6667 (x2 + 1) dx2 + 2.6667 x32 dx3 d EI L0 L0 L0

8 kN # m2 EI

8(103) 73.1(109) 3 p4 (0.054) 4

Ans.

= 0.02229 rad = 1.28°b

Ans: uA = - 1.28° 1514

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*14–104. Determine the displacement at point C of the 2014-T6 aluminum shaft having a diameter of 100 mm.

A

C

1m

B

1m

0.5 m 0.5 m 8 kN

8 kN

Solution Real Moment Function M: As indicated in Fig. a. Virtual Moment Function m: As indicated in Fig. b. Virtual Work Equation: 1#∆ =

L

mM dx L0 EI

1 # ∆C = 2 c ∆C = = =

1 c EI L0

2 c EI L0

1m

(0.5x1)(8x1)dx1 +

1m

4x12dx1 +

L0

0.5 m

7.6667 kN # m3 EI

L0

0.5 m

[0.5(x2 + 1)](8)dx2 d d

4(x2 + 1)dx2 d

7.6667(103) p 73.1(109) 3 (0.054) 4 4

Ans.

= 0.02137 m = 21.4 mm T

Ans: ∆ C = 21.4 mm T 1515

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14–105.  Determine the displacement at point C and the slope at B. EI is constant.

P

B

C L 2

L 2

Solution Real Moment Function M(x): As shown on Figure (a). Virtual Moment Functions m(x) and mU(x): As shown on Figure (b) and (c). Virtual Work Equation: For the displacement at point C, apply Eq. 14–42. 1#∆ =

L

mM dx L0 EI

1 # ∆C = 2 c ∆C =

L

2 x1 P 1 a b a x1 b dx1 d EI L0 2 2

PL3 T 48EI

Ans.

For the slope at B, apply Eq. 14–43. 1#u = 1 # uB =

uB =

L0

L

muM EI L

dx L

2 2 x1 P x2 P 1 a b a x1 b dx1 + a 1 - b a x2 b dx2 d c EI L0 L 2 L 2 L0

PL2 16EI

Ans.

Ans: PL3 T 48EI PL2 uB = 16EI ∆C =

1516

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14–106. Determine the displacement of point C of the beam made from A992 steel and having a moment of inertia of I = 22.3(106) mm4.

40 kN A

B C

1.5 m

3m

1.5 m

Solution 1 ⋅ ∆C = ∫

L mM

0

∆= C =

EI

dx

3m 1  0 + ∫ (1.5 − 0.5x2 )(20x2 )dx2 + 0   0 EI 

45 kN ⋅ m 2 45(10 3 ) = EI [200(109 )][22.3(10 −6 )] 0.01009 m = 10.1 mm ↓

Ans.

40 kN

M=0

M = 40x1

M = 20x2 M=0

20 kN 1

M = 1.5 − 0.5x2

M = −x3

1

1.5 m 1.5

Ans: ∆C = 10.1 mm 1530

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14–107. Determine the slope at B of the beam made from A992 steel and having a moment of inertia of I = 22.3(106) mm4.

40 kN A

B C

1.5 m

3m

1.5 m

Solution L mθ M

1 ⋅θB = ∫ 0

θB = =

EI

dx

1  1.5m (0)(40x1 )dx1 + EI  ∫0

3m

∫0

(1 − 0.3333x2 )(20x2 )dx2 + 0  

30 kN ⋅ m 2 30(10 3 ) = EI [200(109 )][22.3(10 −6 )] 6.7265(10−3 ) πad = 0.385°

0.3333 kN

M = 1 − 0.3333x2

Ans.

uB

0.3333 kN

0.3333 kN

Ans: uB = 0.385° 1531

uB

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*14–108. w

Determine the slope at A. EI is constant. C B L

A L

Solution L

uA =

mu M dx L0 EI L0

= 0 + 4

=

- w8 LL + EI

L

1 Lx

w L3 6

- 1 2 1 - w2 x EI

= -

w L3 24 EI

2

2

dx Ans.

Ans: uA = 1519

wL3 24 EI

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P

14–109. Determine the slope and displacement of end C of the cantilevered beam. The beam is made of a material having a modulus of elasticity of E. The moments of inertia for segments AB and BC of the beam are 2I and I, respectively.

L 2

Real Moment Function M. As indicated in Fig. a.

C

B

A

L 2

Virtual Moment Functions mu and M. As indicated in Figs. b and c. Virtual Work Equation. For the slope at C, L

1#u =

mu M dx EI L0

1 # uC = uC =

1 EI L0

L>2

1(Px1)dx1 +

1 2 EI L0

L>2

1 B P ax2 +

L b R dx2 2

5PL2 16 EI

Ans.

For the displacement at C, 1#¢ = 1 # ¢C = ¢C =

L

mM dx L0 EI 1 EI L0

L>2

x1(Px1)dx1 +

1 2EI L0

L>2

¢ x2 +

L L ≤ B P ¢ x2 + ≤ R dx2 2 2

3PL T 16EI

Ans.

Ans: uC = 1244

5PL2 3PL ,∆ = T 16EI C 16EI

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14–110. Determine the displacement at point B. The moment of inertia of the center portion DG of the shaft is 2I, whereas the end segments AD and GC have a moment of inertia I. The modulus of elasticity for the material is E.

w A

C D a

B a

G a

a

Solution Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions m(x): As shown on Fig. b. Virtual Work Equation: For the slope at point B, apply Eq. 14–42. 1#¢ =

L

L0

1 # ¢B = 2 B

mM dx EI

a x1 1 a b (w ax1)dx1 R EI L0 2

a

+ 2B

¢B =

1 1 w 2 (x + a) B wa(a + x2) x R dx2 R 2EI L0 2 2 2 2

65wa4 48EI

T

Ans.

Ans: w 2 x , M(x1) = wax1, 2 2 x1 1 65wa4 m(x2) = (x2 + a), m(x1) = , ∆B = T 2 2 48EI

M(x2) = wa(a + x2) -

1242

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14–111.  Determine the maximum deflection of the beam caused only by bending, and caused by bending and shear. Take E = 3G.

w

a L

a

Solution For bending and shear, 1#∆ =

L

L fsvV mM dx + dx L0 EI L0 GA

∆ = 2

L0

L>2

1 12 x 2 1 wL 2 x

= ∆ = =

2

EI

4

=

- w x2 2 dx L>2

1 wL 3 wx a x b` EI 6 8 0 5wL4 3wL2 + 384EI 20 GA 5wL4

1 384(3G) 1 12 2 a4

+

+

1 65 2

+ 2

L0

L>2

1 65 2 1 12 2 1 wL 2

- wx 2 dx

GA

wx2 L>2 wL x b` GA 2 2 0 a

3wL2 20(G)a2

20wL4 3wL2 + 4 20Ga2 384Ga

= a

w L 2 5 L 2 3 ba b Ja ba b + R G a 96 a 20

Ans.

For bending only, ∆ =

5w L 4 a b 96G a

Ans.

Ans:

w L 2 5 L 2 3 ba b c a ba b + d, G a 96 a 20 5w L 4 ∆b = a b 96G a

∆ tot = a

1522

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*14–112. The beam is made of oak, for which Eo = 11 GPa. Determine the slope and displacement at point A.

200 mm 4 kN/m

400 mm

A

B 3m

3m

Solution Virtual Work Equation: For the displacement at point A, apply Eq. 14–42. 1#∆ = 1 kN # ∆ A =

L

mM dx L0 EI 1 EI L0 +

∆A = =

3m

2 3 x1 a x1 b dx1 9

1 EI L0

3m

2

(x2 + 3) ( 2.00x2 + 6.00x2 + 6.00 ) dx2

321.3 kN # m3 EI 321.3 ( 103 ) 1 11 ( 109 ) 3 12 (0.2) ( 0.43 ) 4

Ans.

= 0.02738 m = 27.4 m T  For the slope at A, apply Eq. 14–43. 1#u = 1 kN # m # uA =

L

mu M dx L0 EI 1 EI L0

3m

L0

3m

+ uA = =

2 3 1.00 a x1 b dx1 9

1.00 ( 2.00x22 + 6.00x2 + 6.00 ) dx2

67.5 kN # m2 EI 67.5 (1000) 11 ( 10

9

) 3 121 (0.2) ( 0.43 ) 4

= 5.75 ( 10-3 ) rad

Ans.

Ans: ∆ A = 27.4 m T , uA = 5.75 ( 10-3 ) rad 1523

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14–113.  Determine the slope of the shaft at the bearing support A. EI is constant.

w0

A

B

C L – 2

L – 2

Solution 1 # uA =

L

mu M dx L0 EI L

uA =

2 w0L w0 3 1 1 c a1 - x1 b a x x bdx1 EI L0 L 4 1 3L 1

   + =

L0

L 2

a

w0 L w0 3 1 x2 ba x2 x bdx2 L 4 3L 2

5w0 L3 b 192EI

Ans.

Ans: uA = 1524

5w0L3 192 EI

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P

14–114. Determine the vertical displacement of point A on the angle bracket due to the concentrated force P. The bracket is fixed connected to its support. EI is constant. Consider only the effect of bending.

L A

L

Solution 1 # ¢ Av =

L

mM dx L0 EI L

¢ Av =

=

L

1 (x1)(Px1)dx1 + (1L)(PL)dx2 S C EI L0 L0 4PL3 3EI

Ans.

Ans: ¢ Av = 1249

4PL3 3EI

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14–115.  Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A992 steel, determine the vertical displacement of point B due to the loading of 10 kN.

C 10 kN

2m D

A 3m

B 2m

Solution Real Moment Function M(x): As shown in Figure a. Virtual Moment Functions m(x): As shown in Figure b. Virtual Work Equation: For the displacement at point B, 1#∆ =

L0

L

1 kN # ∆ B =

mM nNL dx + EI AE 1 EI L0 +

3m

(0.6667x1)(6.667x1)dx1 1 EI L0 +

∆B = =

2m

(1.00x2)(10.0x2)dx2

1.667(16.667)(2) AE

66.667 kN # m 55.556 kN # m + EI AE 3

66.667(1000) 1 200(109) 3 12 (0.1) 1 0.13 2 4

= 0.04354 m = 43.5 mm

+ T

55.556(1000)

3 p4 1 0.012 2 4 3 200 1 109 2 4

Ans.

Ans: ∆ B = 43.5 mm T 1526

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*14–116. Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A992 steel, determine the slope at A due to the loading of 10 kN.

C 10 kN

2m D

A 3m

B 2m

Solution Real Moment Function M(x): As shown in Figure a. Virtual Moment Functions mU(x): As shown in Figure b. Virtual Work Equation: For the slope at point A, 1#u =

L

muM nNL dx + AE L0 EI

1 kN # m # uA =

1 EI L0 +

uA = =

3m

(1 - 0.3333x1)(6.667x1)dx1

1 EI L0

2m

0(10.0x2)dx2 +

( -0.3333)(16.667)(2) AE

10.0 kN # m2 11.111 kN EI AE 10.0(1000) 200 1 10

9

23

1 12

(0.1) 1 0.1

3

24

= 0.00529 rad = 0.303°b

-

11.111(1000)

3 1 0.012 2 4 3 200 1 109 2 4 p 4

Ans.

Ans: uA = - 0.303° 1527

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14–117.  Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the vertical displacement of point C due to the loading. Consider only the effect of bending in ABC and axial force in DB.

D

20 kN

300 mm

A

100 mm B 3m

Solution

4m

3m

C

Real Moment Function M(x): As shown in Figure a. Virtual Moment Functions m(x)­: As shown in Figure b. Virtual Work Equation: For the displacement at point C, 1#∆ =

L0

L

mM nNL dx + EI AE

1 kN # ∆ C = 2c ∆C = =

1 EI L0

3m

(1.00x)(20.0x) dx d +

2.50(50.0) (5) AE

360 kN # m3 625 kN # m + EI AE 360(1000) 200 1 10

9

23

1 12

(0.1) 1 0.3

3

24

= 0.017947 m = 17.9 mm T 

+

625(1000)

3 1 0.022 2 4 3 200 1 109 2 4 p 4

Ans.

Ans: ∆ C = 17.9 mmT 1528

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14–118.  Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the slope at A due to the loading. Consider only the effect of bending in ABC and axial force in DB.

D

20 kN

300 mm

A

100 mm B 3m

Solution

4m

3m

C

Real Moment Function M(x): As shown in Figure a. Virtual Moment Functions mU(x): As shown in Figure b. Virtual Work Equation: For the slope at point A, 1#u =

L

muM nNL dx + AE L0 EI

1 kN # m # uA = uA = =

1 EI L0

3m

(1 - 0.3333x)(20.0x)dx +

30.0 kN # m2 104.167 kN EI AE 30.0(1000) 200 1 10

9

23

1 12

(0.1) 1 0.3

3

24

-

( - 0.41667)(50.0)(5) AE

104.167(1000)

3 1 0.022 2 4 3 200 1 109 2 4 p 4

= -0.991 1 10 - 3 2 rad = 0.0568°b

Ans.

Ans: uA = - 0.0568° 1529

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14–119.  The L-shaped frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load, determine the horizontal displacement of point C.

C w L

A B

L

Solution 1 # ∆C =

L

mM dx L0 EI L

∆C = ­­­ =

L

wx21 1 wL2 c (1x1)a bdx1 + (1L)a bdx2 d E I L0 2 2 L0

5wL4 S 8EI

Ans.

Ans: (∆ C)h = 1530

5wL4 S 8EI

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*14–120. The L-shaped frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load, determine the vertical displacement of point B.

C w L

A B

L

Solution 1 # ∆B =

L

mM dx L0 EI

L L wx21 1 wL2 c (0)a bdx1 + (L - x2)a bdx2 d EI L0 2 2 L0 wL4 = c 4EI

∆B =

Ans.

Ans: (∆ B)v = 1531

wL4 c 4EI

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14–121.  Determine the vertical displacement of the ring at point B. EI is constant.

P B

r

Solution Model: The ring can be modeled as a half ring as shown in Figure (a).

A

Real Moment Function M(x): As shown on Figure (a). Virtual Moment Functions m(x) and mU(x): As shown on Figure (b) and (c). Virtual Work Equation: Due to symmetry, the slope at B remains horizontal, i.e., equal to zero. Applying Eq. 14–43, we have 1#u =

L0

L

muM EI

1 # uB = 0 =

Where ds = rdu

ds p

1 Pr 1.00 a sin u - M0 b rdu EI L0 2

Pr p For the vertical displacement at B, apply Eq. 14–42. M0 =

1#∆ = 1 # (∆ B)v = =

L

mM ds L0 EI p

1 Pr Pr (r sin u) a sin u b rdu p EI L0 2 Pr 3 2pEI L0

p

( p sin2 u - 2 sin u ) du

Pr 3 [p(1 - cos 2u) - 4 sin u] du 4pEI L0 p

= (∆ B)v =

Pr3 ( p2 - 8 ) T 4pEI

Ans.

Ans: (∆ B)v = 1532

Pr3 ( p2 - 8 ) T 4pEI

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14–122.  Determine the horizontal displacement at the roller at A due to the loading. EI is constant. r P

A

B

Solution ΣFx = 0;

N - P sin u = 0 N = P sin u

ΣMo = 0;

M - P sin u r = 0 M = Pr sin u

ΣFx = 0;

n - 1 sin u = 0 n = 1 sin u

a+ ΣMo = 0;

m - (1 sin u) r = 0 m = r sin u

1 # (∆ A)h = (∆ A)h =

L

3

p mM Pr 2 Pr 1 1 2 dx = sin u (rdu) = a u - sin 2u b ` EI L0 EI 2 4 0 L0 EI p

pPr 3 d 2EI

Ans.

Ans: (∆ A)h = 1533

pPr 3 d 2EI

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14–123.

Solve Prob. 14–72 using Castigliano’s theorem.

Solution

333

0.8

kN

333

0.8

2 P+

333

0.8

Member Forces N: Member forces due to external force P and external applied forces are shown on the figure.

AB

0.6667P+3.333

0N 0P 0.6667

BC

0.6667P+3.333

0.6667

3.333 16.6667

96 2.4

26.667

CD

0

0

0

72 1.8

0

DE

0

0

0

96 2.4

0

EF

0

0

0

96 2.4

0

AF

0

0

0

72 1.8

0

AE

–(0.8333P + 4.167)

– 0.8333

–20.8333 – 4.167

120 3.0

52.083

CE

–(0.8333P + 4.167)

– 0.8333

–20.8333 – 4.167

120 3.0

52.083

BE

5.0

0

25.00 5.00

72 1.8

0

N

(C

)

20.

833

1.8 m

3k

N

0.6667P + 16.6667 kN 0.6667P + 16.6667 kN

2.4 m

2.4 m 25 kN

Castigliano’s Second Theorem: Member

P+

N(P==0)0) N(P

L

3.333 16.6667

96 2.4

0N bL 0P 26.667

Na

©157.5 kN # m 0N L ¢ = a Na b 0P AE 157.5 kN ⋅ m (∆ E )v = AE =

157.5(10 3 ) = 0.28125(10 −3 ) m = 0.281 mm T [2.80(10 −3 )][200(109 )]

Ans.

Ans: (∆E)n = 0.281 mm 1258

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*14–124.

Solve Prob. 14–73 using Castigliano’s theorem.

Solution 1.8 m

Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. 2.4 m

2.4 m

Castigliano’s Second Theorem: Member

N

AB

0.6667P

0N 0P 0.6667

BC

0.6667P

CD

N(P N(P= =255kN) kip)

L

3.333 16.6667

96 2.4

0N bL 0P 26.667

0.6667

3.333 16.6667

96 2.4

26.667

0

0

0

72 1.8

0

DE

0

0

0

96 2.4

0

EF

0

0

0

96 2.4

0

AF

0

0

0

72 1.8

0

AE

– 0.8333P

– 0.8333

–20.8333 – 4.167

120 3.0

52.083

CE

– 0.8333P

– 0.8333

–20.8333 – 4.167

120 3.0

52.083

BE

1.00P

1.00

25.00 5.00

72 1.8

45.0 360.00

Na

g202.5 kN · m

0N L b ¢ = a Na 0P AE 202.5 kN ⋅ m (∆ B )v = AE =

202.5(10 3 ) = 0.3616(10 −3 ) m = 0.362 mm T [2.80(10 −3 )][200(109 )]

Ans.

Ans: (∆B)v = 0.362 mm 1257

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14–125.

Solve Prob. 14–75 using Castigliano’s theorem.

Solution ¢ Hv = ©N a

2828.89(10 3 ) 20368 kN · m 20368 0N L 2828.89 b = =  0P AE AE AE 4.5 (29) [2.80(10 −3 )][200(109 )]

= 5.052(10 −3 ) m = 5.05 mm

Ans.

3m

4m 4m 4m 4m 30 kN 40 kN 30 kN 50 kN + P 50 kN + P 2 2

0

333

83.

0.6667P + 66.6667 kN

0.5P + 50

30 kN

P+

333

33.

kN 40 kN

P+

33 .83

0.5P + 50 kN

0.6667P + 66.6667 kN

kN

33 .83

0

1.3333P + 93.333 kN

40 kN

Ans: ∆Hn = 5.05 mm 1261

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14–126.

Solve Prob. 14–76 using Castigliano’s theorem.

Solution ¢ Cv = ©N a

21232 21232 2848.89(10 3 ) 0N L 2948.89 kN · m b = = = 0P AE AE AE [2.80(10 −3 )][200(109 )]

= 5.266(10 −3 ) m = 5.27 mm T

Ans.

3m

4m 4m 30 kN 50 kN + P 2

4m

4m 50 kN + P 2

30 kN

0.

3

3.3

+8

0.

0.6667P + 66.6667 kN

0.5P + 50 kN

3P 833

30 kN

33

3.3

+3

kN

P(T) + 40 kN

3P 833

0.5P + 50 kN

0.6667P + 66.6667 kN

N 3k

1.3333P + 93.333 kN

P + 40 kN

Ans: ∆Cn = 5.27 mm 1260

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14–127.

Solve Prob. 14–77 using Castigliano’s theorem.

Solution Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem:

AB

– 22.5

0N 0P 0

BC

– 22.5

0

– 22.5

1.5

0

AE

37.5

0

37.5

2.5

0

CE

–(1.25P + 37.5)

– 1.25

– 62.5

2.5

195.3125

BE

1.00P

1.00

20.0

2

40.0

DE

0.750P + 45

0.750

60.0

1.5

67.50

Member

N

N(P = 20 kN)

L

– 22.5

1.5

Na

0N bL 0P 0

# a 302.8125 kN m 0N L b ¢ = a Na 0P AE (¢ B)v = =

302.8125 kN # m AE

302.8125 A 103 B

0.400 A 10 - 3 B C 200 A 109 B D

= 3.785 A 10 - 3 B m = 3.79 mm T

Ans.

Ans: (∆B)h = 0.699 (10- 3) in. S 1260

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*14–128.

Solve Prob. 14–78 using Castigliano’s theorem.

Solution Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem:

AB

– 0.750P

0N 0P – 0.750

BC

– 0.750P

– 0.750

–22.5

1.5

25.3125

AE

1.25P

1.25

37.5

2.5

117.1875

CE

–(1.25P + 25.0)

– 1.25

–62.5

2.5

195.3125

BE

20.0

0

20.0

2

0

DE

1.50P+15.0

1.50

60.0

1.5

135.00

Member

N

N(P = 30 kN)

L

–22.5

1.5

0N bL 0P 25.3125

Na

# a 498.125 kN m 0N L b ¢ = a Na 0P AE (¢ A)v = =

498.125 kN # m AE

498.125 A 103 B

0.400 A 10 - 3 B C 200 A 109 B D

= 6.227 A 10 - 3 B m = 6.23 mm T

Ans.

Ans: (¢A)v = 6.23 mm 1259

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14–129.  Solve Prob. 14–81 using Castigliano’s theorem.

Solution Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: Applying Eq. 14–48, we have Member

N 10.0

AB BC

1.00P + 7.50

CD

10.0

AC

0 1.00 0

- 12.5

10.0

2

0N bL 0P

12.5

1.5

18.75

10.0

2

N(P = 5 kN) L Na

0

0

AD

0N 0P

0

0 - 12.5

0

0

1.5

0

2.5

0

Σ 18.75 kN # m

0N L ∆ = a Na b 0P AE (∆ C)h = =

18.75 kN # m AE

18.75 ( 10 3 ) 0.400 ( 10-3 ) [200 ( 109 ) ]

= 0.2344 ( 10-3 ) m = 0.234 mm d

Ans.

Ans: (∆ C)h = 0.234 mm d 1540

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14–130.  Solve Prob. 14–82 using Castigliano’s theorem.

Solution Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: Applying Eq. 14–48, we have Member

0N 0P

N

N(P = 10 kN) L Na

1.00

10.0

2

BC

0.750P + 5.00

0.750

12.5

1.5

CD

1.00P

1.00

10.0

2

1.00P

AB

0

AD AC

0

- 1.25P

- 1.25

0 - 12.5

0N bL 0P

20.00

14.0625 20.00

1.5

0

2.5 39.0625

Σ 93.125 kN # m

0N L ∆ = a Na b 0P AE (∆ D)v = =

93.125 kN # m AE

93.125 ( 103 ) 0.400 ( 10-3 ) [200 ( 109 ) ]

= 1.164 ( 10-3 ) m = 1.16 mm T 

Ans.

Ans: (∆ D)v = 1.16 mm T 1541

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14–131.  Solve Prob. 14–85 using Castigliano’s theorem.

Solution Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: Applying Eq. 14–48, we have Member

0N 0P

N 2.00P

AE

N(P = 4 kN) L

2.00

Na

0N bL 0P

8.00

3

48.00

8.00

3

48.00

ED

2.00P

2.00

CD

- 2.236P

- 2.236

- 8.944 3.354 67.082

- (2.236P + 2.236)

- 2.236

- 11.180 3.354 83.853

BC

CE

- 2.00

0

- 2.00

AC

2.236

0

2.236 3.354

1.5

0 0

Σ 246.935 kN # m

0N L ∆ = a Na b 0P AE (∆ D)h = =

246.935 kN # m AE

246.935 ( 103 ) 0.300 ( 10-3 ) [200 ( 109 ) ]

= 4.116 ( 10-3 ) m = 4.12 mm S

Ans.

Ans: (∆ D)h = 4.12 mm S 1542

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*14–132. Solve Prob. 14–86 using Castigliano’s theorem.

Solution Member Forces N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: Applying Eq. 14–48, we have 0N 0P

Member N 8.00

0

8.00

0

- 8.944

0

AE ED CD BC

- (1.118P + 8.944)

8.00

3

0N bL 0P

8.00

3

0

N(P = 2 kN) L Na

- 1.118

0

- 8.944 3.354

0

- 11.180 3.354 41.926

CE

- 1.00P

- 1.00

- 2.00

AC

1.118P

1.118

2.236 3.354

1.5

3.00 8.385

Σ 53.312 kN # m

0N L ∆ = a Na b 0P AE (∆ E)h = =

53.312 kN # m AE

53.312 ( 103 ) 0.300 ( 10-3 ) [200 ( 109 ) ]

= 0.8885 ( 10-3 ) m = 0.889 mm S

Ans.

Ans: (∆ E)h = 0.889 mm S 1543

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14–133.  Solve Prob. 14–90 using Castigliano’s theorem.

Solution Set M′ = 0 L0

L

L0

a

= =

Pa2 Pa2 5Pa2 + = 3EI 2EI 6EI

uC =

Ma

0M dx b 0M′ EI

(Px1)(a1x1)dx1 EI

+

L0

a

(Px2)(1)dx2 EI Ans.

Ans: uC = 1544

5Pa 2 6EI

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14–134.  Solve Prob. 14–91 using Castigliano’s theorem.

Solution 0M1 x1 = 1 0M′ a

0M2 = 0 0M′

Set M′ = 0 M1 = -Px1 uA = =

L0

L

Pa2 6EI

Ma

M2 = Px2 a a x1 0M dx 1 - Pa2 b = c ( -Px1) a1 - bdx1 + (Px2)(0)dx2 d = 0M′ EI EI L0 a 6EI L0

Ans.

Ans: uA = 1545

Pa 2 6EI

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14–135.

Solve Prob. 14–106 using Castigliano’s theorem.

Solution ∂M1 ∂M2 = = 1.5 − 0.5 x2 0 ∂p ∂p

∂M3 = x3 ∂p

Set p = 0 = = = M1 40 x1 M2 20 x2 M 3 0 L  ∂M  dx ∆C = ∫0 M  ∂p  EI

=

3m 3m 1  1.5m 40x1(0)dx1 + ∫ 20x2 (1.5 − 0.5x2 )dx2 + ∫ (0)(x3 )dx3  ∫   0 0 0 EI 

45 kN ⋅ m 3 45(10 3 ) = EI [200(109 )][22.3(10 −6 )]

=

= 0.01009 m = 10.1 mm ↓

Ans.

Ans: ¢C = 10.1 mm 1573

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*14–136.

Solve Prob. 14–107 using Castigliano’s theorem.

Solution ∂M1 ∂M2 ∂M3 = = = 0 1 − 0.3333 x2 0 ∂M ′ ∂M ′ ∂M ′ Set M ′ = 0

= = = M1 40 x1 M2 20 x2 M 3 0 L

 ∂M  dx   ∂M ′  EI

θB = ∫ M  0 = =

3m 1  1.5m 40x1(0)dx1 + ∫ 20x2 (1 − 0.3333 x2 )dx2 + 0   0 EI  ∫0

30 kN ⋅ m 2 30(10 3 ) = EI [200(109 )][22.3(10 −6 )] 6.7265(10 −3 ) πad = 0.385°

Ans.

uB

Ans: uB = 0.385° 1574

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14–137.  Solve Prob. 14–95 using Castigliano’s theorem.

Solution 0M1 = 0.4286x1 0P

0M2 = 0.4286x2 + 0.17144 0P

0M3 = 0.5714x3 0P

0M4 = 0.5714x4 + 0.17144 0P

Set P = 2 kN M1 = 4.3572x1

M2 = 0.3572x2 + 1.7429

M3 = 4.6428x3

M4 = 1.6428x4 + 1.3929

∆B =

L0

L

Ma

0M dx b 0P EI

0.4

=

=

1 c (4.3572x1) (0.4286x1) dx1 + EI L0 L0

0.4

L0

0.3

L0

0.3

(0.3572x2 + 1.7429)(0.4286x2 + 0.17144)dx2 +

(4.6428x3)(0.5714x3)dx3 +

(1.6428x4 + 1.3929)(0.5714x4 + 0.17144)dx4 d

0.37944(103) 0.37944 kN # m3 = = 0.0478 m = 47.8 mm EI 200(109)p4 (0.015)4

Ans.

Ans: ∆ B = 47.8 mm 1548

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14–138.  Solve Prob. 14–96 using Castigliano’s theorem.

Solution 0M1 = 0 0P

0M2 = - 0.5x2 0P

Set P = 0 M1 = 18 ∆D =

L0

M2 = 18 L

= (2)

=

Ma

0M dx b 0P EI 4

3

1 c (18)(0)dx1 + (18)( - 0.5x2)dx2 d EI L0 L0

81(10)3 81 kN # m3 = = 3.24(10-3) m = 3.24 mm EI 200(109)(125)(10-6)

Ans.

Ans: ∆ D = 3.24 mm 1549

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14–139.  Solve Prob. 14–97 using Castigliano’s theorem.

Solution 0M2 = 1 - 0.1667x2 0M′

0M1 = 1 0M′

0M3 = 0 0M′

­

Set M′ = 18 kN # m M1 = 18 kN # m uA =

=

L0

L

Ma

M2 = 18 kN # m 4

M3 = 18 kN # m 6

4

0M dx 1 b = c (18)(1)dx1 + 18(1 - 0.1667x2)dx2 + (18)(0)dx3 d 0M′ EI EI L0 L0 L0

126(103) 126 kN # m2 = = 5.04(10 - 3) rad = 0.289° EI 200(109)(125)(10 - 6) ­

Ans.

Ans: uA = 0.289° 1550

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*14–140. Solve Prob. 14–98 using Castigliano’s theorem.

Solution uB =

=

L0

L

Ma

6 ( -18) 1 - 16x 2 dx(103) 0M dx b = 0M′ E I EI L0

18(62)(103)

6(2)(200)(109)(125)(10 - 6)

Ans.

= 0.00216 rad = 0.124°

Ans: uB = 0.124° 1551

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14–141.  Solve Prob. 14–108 using Castigliano’s theorem.

Solution M′ does not influence the moment within the overhang. M =

M′ wx2 x - M′ L 2

0M x = - 1 0M′ L Setting M′ = 0, uA =

=

L0

L

Ma

wL3 24EI

L

1 wx2 - w L3 L3 0M dx x b = ab a - 1b dx = c d 0M′ EI EI L0 2 L 2EI 4 3 Ans.

Ans: uA = 1552

wL3 24EI

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14–142.  Solve Prob. 14–119 using Castigliano’s theorem.

Solution wx12 2

M1 = Px1 +

0M1 0M2 = x1 = L 0P      0P Setting P = 0 M1 = ∆C =

wx12 wL2 M2 = 2      2 L0

L

Ma

L L wx21 0M dx 1 wL2 5wL4 b = c (x1)dx1 + Ldx2 d = 0P EI EI L0 2 8EI L0 2

Ans.

Ans: ∆C = 1553

5wL4 8EI

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14–143.  Solve Prob. 14–120 using Castigliano’s theorem.

Solution P does not influence moment within segment. M = Px =

wL2 2

0M = x 0P Set P = 0 ∆B =

L0

L

Ma

L

0M dx wL2 dx wL4 b = ab (x) = 0P EI L0 2 EI 4EI

Ans.

Ans: ∆B = 1554

wL4 4EI

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*14–144. Solve Prob. 14–105 using Castigliano’s theorem.

Solution Internal Moment Function M(x): The internal moment function in terms of the load P′ and couple of moment M′ and externally applied load are shown on figures (a) and (b), respectively. Castigliano’s Second Theorem: The displacement at C can be determined using 0M(x) x Eq. 14–49 with = and set P′ = P. 0P′ 2 ∆ =

L0

L

Ma

0M dx b 0P′ EI L

∆ C = 2c =

2 1 P x a xba bdx d EI L0 2 2

PL3 T 48EI

To determine the slope at B, we apply Eq. 14–50 with and setting M′ = 0. u =

L0

L

Ma L

uB =

Ans. 0M(x1) 0M′

=

x1 0M(x2) x2 , = 1 L 0M′ L

0M dx b 0M′ EI

2 x1 1 P a x ba bdx1 EI L0 2 1 L L

2 x2 1 P + a x ba1 - bdx2 EI L0 2 2 L

=

PL2 16EI

Ans.

Ans:

PL3 T 48EI 2 PL uB = 16EI ∆C =

1555

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R14–1.  Determine the total axial and bending strain energy in the A992 steel beam. A = 2300 mm2, I = 9.5(106) mm4.

1.5 kN/m

15 kN 10 m

Solution Axial Load: L

(Ue)i = (Ue)i =

N2dx N2L = 2EA L0 2EA ((15)(103))2(10) 2(200)(109)(2.3)(10 - 3)

= 2.4456 J

Bending: 10

L

(Ub)i =

M2dx 1 = [ (7.5) (103) x - 0.75 (103) x2 ] 2dx 2EI 2EI L0 L0 10

= (Ub)i =

1 [56.25 (106) x2 + 562.5 (103) x4 - 11.25 (106) x3 ] dx 2EI L0 0.9375(109) 200(109)(9.5)(10 - 6)

= 493.4210 J

Ui = (Ua)i + (Ub)i = 2.4456 + 493.4210 = 496 J

Ans.

Ans: Ui = 496 J 1556

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R14–2.  The 200-kg block D is dropped from a height h = 1 m onto end C of the A992 steel W200 * 36 overhang beam. If the spring at B has a stiffness k = 200 kN>m, determine the maximum bending stress developed in the beam.

D h A B 4m

Solution Equilibrium: The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a. Ue = Ui L

1 M2dx P∆ st = Σ 2 L0 2EI 1 1 P∆ st = J 2 2EI L0 ∆ st =

4m

8P EI

a

2 P x2 b dx + 2 L0

2m

(Px1)2 dx R

Here, I = 34.4 1 106 2 mm4 = 34.4 1 10 - 6 2 m4 (see the appendix) and E = Est = 200 GPa. Then, the equivalent spring constant can be determined from P = kb ∆ st P = kb a

8P b EI

EI kb = = 8

200 1 109 2 c 34.4 1 10 - 6 2 d 8

From the free-body diagram, Fsp =

3 P 2

ksp ∆ sp =

∆ sp =

= 860 1 103 2 N>m

3 (k ∆ ) 2 b b

3 3 kb 3 860 1 10 2 ¢ ≤∆ b = £ ≥∆ b = 6.45 ∆ b 2 ksp 2 200 1 103 2

(1)

Conservation of Energy: mg ah + ∆ b +

3 1 1 ∆ b = ksp ∆ sp2 + kb ∆ b2 2 sp 2 2

1557

C

k 2m

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R14–2.  Continued

Solution Substiuting Eq. (1) into this equation, 200(9.81) c 1 + ∆ b +

3 1 1 (6.45∆ b) d = c 200 1 103 2 d (6.45∆ b)2 + c 860 1 103 2 d ∆ b 2 2 2 2

4590.25 1 103 2 ∆ b 2 - 20944.35∆ b - 1962 = 0

Solving for the positive root ∆ b = 0.02308 m

Maximum Stress: The maximum force on the beam is Pmax = kb ∆ b =

860 1 103 2 (0.02308) = 19.85 1 103 2 N. The maximum moment occurs at the

supporting spring, where Mmax = Pmax L = 19.85 1 103 2 (2) = 39.70 1 103 2 N # m. d 0.201 Applying the flexure formula with c = = = 0.1005 m, 2 2 smax =

39.70 1 103 2 (0.1005) Mmax c = = 115.98 MPa = 116 MPa I 34.4 1 10 - 6 2

Ans.

Since smax 6 sY = 345 MPa, this result is valid.

­­­­Ans: smax = 116 MPa 1558

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R14–3.  Determine the maximum height h from which the 200-kg block D can be dropped without causing the A992 steel W200 * 36 overhang beam to yield. The spring at B has a stiffness k = 200 kN>m.

D h A B 4m

Solution Equilibrium: The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig. a. Ue = Ui L

1 M2dx P∆ st = Σ 2 L0 2EI 1 1 P∆ st = J 2 2EI L0 ∆ st =

8P EI

4m

a

2 P x2 b dx + 2 L0

2m

(Px1)2 dx R

Here, I = 34.4 1 106 2 mm4 = 34.4 1 10 - 6 2 m4 (see the appendix) and E = Est = 200 GPa. Then, the equivalent spring constant can be determined from P = kb ∆ st

8P b EI 200 1 109 2 c 34.4 1 10 - 6 2 d EI kb = = = 860 1 103 2 N>m 8 8 P = kb a

From the free-body diagram, Fsp =

3 P 2

ksp ∆ sp = ∆ sp =

3 (k ∆ ) 2 b b

3 3 kb 3 860 1 10 2 ¢ ≤∆ b = £ ≥∆ b = 6.45∆ b 2 ksp 2 200 1 103 2

(1)

1559

C

k 2m

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R14–3.  Continued

Solution Maximum Stress: The maximum force on the beam is Pmax = kb ∆ b = 860 1 103 2 ∆ b. The maximum moment occurs at the supporting spring, where Mmax = Pmax L = d 860 1 103 2 ∆ b(2) = 1720 1 103 2 ∆ b. Applying the flexure formula with c = = 2 0.201 = 0.1005 m, 2 smax =



Mmaxc I

345 1 106 2 =

1720 1 103 2 ∆ b(0.1005)

∆ b = 0.06866 m

34.4 1 10 - 6 2

Substituting this result into Eq. (1),

∆ sp = 0.44284 m

Conservation of Energy: mg ah + ∆ b +

3 1 1 ∆ sp b = ksp ∆ sp2 + kb ∆ b2 2 2 2

200(9.81) c h + 0.06866 +

3 1 (0.44284) d = c 200 1 103 2 d (0.44284)2 2 2 +

h = 10.3 m

1 c 860 1 103 2 d (0.06866)2 2

Ans.

­­­­Ans: h = 10.3 m 1560

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*R14–4. The A992 steel bars are pin connected at B and C. If they each have a diameter of 30 mm, determine the slope at E.

B

A 3m

300 Nm E 2m

2m

C

D 3m

Solution L

2

3

2

Ui =

M dx 1 1 65625 = (2) (75x1)2dx1 + (2) (-75x2)2dx2 = 2EI 2EI 2EI EI L0 L0 L0

Ue =

1 1 (M′)u = (300)uE = 150uE 2 2

Conservation of Energy: Ue = Ui 150uE = uE =

65625 EI

473.5 473.5 = 0.0550 rad = 3.15°b = EI (200)(109)(p4 )(0.0154)

Ans.

Ans: uE = - 3.15° 1561

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R14–5. The steel chisel has a diameter of 12 mm and a length of 250 mm. It is struck by a hammer of mass 1.5 kg, and at the instant of impact it is moving at 3.6 m/s. Determine the maximum compressive stress in the chisel, assuming that 80% of the impacting energy goes into the chisel. Est = 200 GPa, sY = 700 MPa. 250 mm

Solution 2 9 π AE  4 (0.012 ) [200(10 )] = = 90.4779(106 ) N−m k = 0.25 L

0.8U e = U i 1  1 2 0.8  (1.5)(3.6 2 ) + 1.5(9.81) [90.4779(106 )]∆ max = ∆ max  2   2 2 45.2389(106 )∆ max 0 − 11.772 ∆ max − 7.776 =

0.41472(10 −3 ) m ∆ max = [90.4779(106 )][0.41472(10 −3 )] = 37.5233(10 3 ) N Pmax = k ∆ max =

σ= max

Pmax 37.5233(10 3 ) 2 = 332 MPa < σ Y = = 331.78(106 ) N−m π (0.012 2 ) A 4

O.K.

Ans.

Ans: smax = 332 MPa 1590

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R14–6. Determine the total strain energy in the A-36 steel assembly. Consider the axial strain energy in the two 12mm-diameter rods and the bending strain energy in the beam for which I = 17.0(106) mm4.

3 ftm 0.9

500kN lb 2.5

Solution

4 ft 1.2 m

4 ft 1.2 m

Support Reactions: As shown FBD(a). 1.25 kN

1.25 kN

2.5 kN

Internal Moment Function: As shown on FBD(b). Total Strain Energy: L

(Ui)T =

M 2 dx N 2L + 2EI 2AE L0

= 2B

=

1 2EI L0

1.2 m 1.25 kN

1.25 250(0.9) (3) 22 (250x) dx (1.25x) dxRd ++22c B Rd 2AE 2AE 22

4 ftm 1.2

1.25x

0.9 kN 2 ⋅ m 3 1.40625 kN 2 ⋅ m 3 + EI AE 0.9 kN 2 ⋅ m 3 9

2

−6

4

[200(10 ) N−m ][17.0(10 ) m ] =

1.2 m

+

1.40625(10 3 N)2 ⋅ m 3  π (0.012 m)2  [200(109 ) N−m 2 ] 4 

0.3269 = N ⋅ m 0.327 J

Ans.

Ans:

1272

1Ui 2T = 0.327 J

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R14–7. Determine the vertical displacement of joint E. For each member A = 400 mm2, E = 200 GPa. Use the method of virtual work.

F

E

D

1.5 m

A

Solution Member

n

N

L

2m

nNL

AF

0

0

1.5

0

AE

– 0.8333

– 37.5

2.5

78.125

AB

0.6667

30.0

2.0

40.00

EF

0

0

2.0

0

EB

– 0.50

22.5

1.5

– 16.875

ED

– 0.6667

– 30.0

2.0

40.00

BC

0

0

2.0

0

BD

0.8333

37.5

2.5

78.125

CD

– 0.5

– 22.5

1.5

16.875

B 45 kN

C 2m

© = 236.25 1 # ¢ Bv = © ¢ Bv =

nNL AE

236.25 A 103 B

400 A 10 - 6 B (200) A 109 B

= 2.95 A 10 - 3 B = 2.95 mm

Ans.

Ans: ∆Bn = 2.95 mm 1272

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*R14–8.

F

Solve Prob. 14–152 using Castigliano’s theorem.

E

D

1.5 m

A

Solution Member

2m

0N>0P

N

N(P = 45)

L

B 45 kN

C 2m

N(0N>0P)L

AF

0

0

0

1.5

0

AE

–(0.8333P + 37.5)

– 0.8333

–37.5

2.5

78.125

AB

0.6667P + 30

0.6667

30.0

2.0

40.00

BE

22.5–0.5P

– 0.5

22.5

1.5

–16.875

BD

0.8333P + 37.5

0.8333

37.5

2.5

78.125

BC

0

0

0

2.0

0

CD

–(0.5P + 22.5)

– 0.5

–22.5

1.5

16.875

DE

–(0.6667P + 30)

– 0.6667

–30.0

2.0

40.00

EF

0

0

0

2.0

0 © = 236.25

¢ Bv = ©N =

236.25 0N L = 0P AE AE

236.25 A 103 B

400 A 10 - 6 B (200) A 109 B

= 2.95 A 10 - 3 B m = 2.95 mm

Ans.

Ans: ¢Bv = 2.95 mm 1273

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R14–9.  The cantilevered beam is subjected to a couple moment M0 applied at its end. Determine the slope of the beam at B. EI is constant. Use the method of virtual work.

M0

A B L

Solution L

uB = =

L (1) M0 muM dx = dx EI EI L0 L0

M0L EI

Ans.

Ans: uB = 1566

M0 L EI

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R14–10.  Solve Prob. R14–9 using Castigliano’s theorem. M0

A B L

Solution uB = =

L0

L

ma

M0L EI

L M0(1) dm dy b = dx dm′ EI EI L0

Ans.

Ans: uB = 1567

M0 L EI

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R14–11.  w

Determine the slope and displacement at point C. EI is constant.

w

C A a

B 2a

a

Solution L

uC = =

mu M dx L0 E I a a wx21 wx22 1 c (0)a bdx1 + (1)a bdx2 EI L0 2 2 L0

+ = -

L0

2a

a1 -

x3

2a

2 w a3 b 3EI

2

ba

wa bdx3 d 2

Ans.

L

∆C =

mM dx L0 E I a

=

+ =

a

wx21 wx22 1 c (0) a bdx1 + (x2) a bdx2 EI L0 2 2 L0 L0

5 w a4 T 8EI

2a

aa -

x3 2

ba

wa2 bdx3 d 2

Ans.

Ans:

2wa3 3EI 5wa4 = T 8EI

uC = ∆C 1568

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*R14–12. Determine the displacement at B. EI is constant.

w

C

A L — 2

B

L — 2

Solution 1 # ∆B =

∆B = =

L0

L

mM dx L0 E I L 2

( - 1 x) 1 - w2 x

w L4 T 128 EI

EI

2

2

dx =

w 1 L2 2

4

8E I

Ans.

Ans: ∆B = 1569

wL4 T 128 EI