Engineering Mechanics 1, Statics Solutions

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os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

D. Gross • W. Hauger J. Schr¨oder • W.A. Wall N. Rajapakse

Engineering Mechanics 1 Statics

Solutions to Supplementary Problems

Gr

The numbers of the problems and the figures correspond to the numbers in the textbook Gross et al., Engineering Mechanics 1, Statics, 2nd Edition, Springer 2012

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 2

Forces with a Common Point of Application

2

11111111 00000000

Example 2.10 A hook is subjec-

ted to three forces (F1 = 180 N, α1 = 45◦ , F2 = 50 N, α2 = 60◦ , F3 = 30 N) as shown in Fig. 2.21. Determine the magnitude and direction of the resultant.

F3

α1

α2

F2

F1

Fig. 2.21

First we choose an x, y-coordinate system; the x-axis coincides with the horizontal. Then we determine the coordinates of the resultant:

Solution

y

αR

F3

α2

F2

x

α1

R

F1

Rx = F1x + F2x + F3x = F1 cos α1 − F2 cos α2 − F3 = 180 cos 45◦ − 50 cos 60◦ − 30 = 72.3 N

and

Ry = F1y + F2y + F3y = −F1 sin α1 −F2 sin α2 = −180 sin 45◦ − 50 sin 60◦ = −170.6 N .

The magnitude and the direction of the resultant follow from   R = Rx2 + Ry2 = 72.32 + 170.62 = 185.3 N , tan αR =

Gr

E2.10

2 Forces with a Common Point of Application

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

2

|Ry | 170.6 = 2.35 = Rx 72.3

→

αR = 67◦ .

3

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

2 Forces with a Common Point of Application

Example 2.11 Determine the magni-

F

tudes F1 and F2 of the components of force F with magnitude F = 5 kN in the directions f1 and f2 (Fig. 2.22).

45◦

30◦

f2

f1

Fig. 2.22

Solution We choose the coordinate system such that the y-axis

coincides with the direction of F . y

F

F1

45◦

30◦

F2

x

f2

f1

The force F is the resultant of the force components F1 and F2 . Hence, we can write Fx = F1x + F2x = 0

→ −F1 sin 30◦ + F2 sin 45◦ = 0 ,

Fy = F1y + F2y = F

→

Thus,

√ 1 2 = 0, −F1 + F2 2 2

F1 cos 30◦ + F2 cos 45◦ = F .

√ √ 3 2 F1 + F2 =F. 2 2

Gr

Solving these two equations for F1 and F2 yields √ 2F 2F √ = 3.7 kN , √ = 2.6 kN . F2 = F1 = 1+ 3 1+ 3

E2.11

A smooth sphere (weight W = 20 N, radius r = 20 cm) is suspended by a wire (length a = 60 cm) as shown in Fig. 2.23. Determine the magnitude of force S in the wire. Example 2.12

1 0 0 1 0 1 0 1 0 1 0 1 0 1 a 0 1 0 1 0 1 0 1 0 1 r 0 1 0 1 W 0 1 0 1 0 1 0 1 0 1 0 1 0 1

Fig. 2.23

Solution The forces acting at the sphere must satisfy the equilibrium conditions. We can make these forces visible if we cut the wire and separate the sphere from the wall. In the free-body diagram we show the force S in the wire (acting in the direction of the wire), the contact force N (acting perpendicularly to the wall) and the weight W . The three forces are concurrent forces. S

N

α

a

α r

W

The equilibrium conditions are given by →:

N − S cos α = 0 ,

↑:

S sin α − W = 0 ,

where the angle α follows from the geometry of the problem:   2 1 20 1 r 1√ = and sin α = 1 − cos α = = = 8. a 60 3 3 3 The second equilibrium condition yields the force S: S=

3 W = √ W = 21.2 N . sin α 8

Gr

E2.12

2 Forces with a Common Point of Application

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

4

Note that the normal force N can be calculated from the first equilibrium condition.

5

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

2 Forces with a Common Point of Application

Example 2.13 Fig. 2.24 shows

a freight elevator. The cable of the winch passes over a smooth pin K. A crate (weight W ) is suspended at the end of the cable. Determine the magnitude of forces S1 and S2 in bars 1 and 2.

Winch

K

1111111 0000000 0000000 1111111 α 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

1

2

W

β

Fig. 2.24

Solution We isolate pin K by passing ima-

ginary sections through the bars and the cable. The internal forces are made visible in the free-body diagram; they are assumed to be tensile forces. The equilibrium condition at the crate is given by ↑:

S−W =0

→

S + S1 sin α + S2 sin β = 0 ,

↓:

S + S1 cos α + S2 cos β = 0 .

Solving for S1 and S2 yields

S2 =

β

α

S2

←:

sin β − cos β W, sin(α − β)

K

S1

S=W

and the equilibrium conditions at the pin are

S1 =

S

S

S

W

cos α − sin α W. sin(α − β)

Gr

If one of the bars forms the angle 45◦ with the vertical, then the other bar has a vanishing internal force (zero-force member).

E2.13

Example 2.14 A smooth circular cy-

linder (weight W , radius r) touches an obstacle (height h) as shown in Fig. 2.25. Find the magnitude of force F necessary to roll the cylinder over the obstacle.

W

r 111 000 000 111 100 00000000 11111111 α 1 00000000 11111111 000 111 h 00000000 11111111 00000000 11111111 10

F

11111111 00000000 00000000 11111111 Fig. 2.25

Solution We isolate the cylinder from the base and the obstacle. The free-body diagram shows the forces acting at the cylinder. The four forces F , W , N1 and N2 are concurrent forces. Thus, the equilibrium conditions are

→:

N2 sin α − F = 0 ,

↑: N1 + N2 cos α − W = 0 .

We obtain the angle α from the geometry of the problem:

F

N1

r

The two equilibrium conditions contain three unknowns: N1 ,

N2

α W

N2

r−h . cos α = r

α

r−h

h

and F .

The force that causes the cylinder to roll over the obstacle also causes the cylinder to lift off the base. Then the contact force N1 vanishes. This leads to N1 = 0

→

N2 =

and

F = N2 sin α = W tan α .

Gr

E2.14

2 Forces with a Common Point of Application

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

6

W cos α

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os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

2 Forces with a Common Point of Application

A large cylinder (weight 4W , radius 2r) lies on top of two small cylinders (each having weight W and radius r) as shown in Fig. 2.26. The small cylinders are connected by a wire S (length 3r). All surfaces are smooth. Determine all contact forces and the magnitude of force S in the wire. Example 2.15

2r

4W

r W

W

S 111111111 000000000 000000000 111111111 000000000 111111111 3r

111111111 000000000

Fig. 2.26

Solution We isolate the three cylinders and introduce the contact

forces in the free-body diagram. The forces acting at each cylinder are concurrent forces. Due to the symmetry of the problem we have only one equilibrium condition at the large cylinder and two equilibrium conditions for one of the small cylinders. These are three equations for the three unknown forces N1 , N2 and S: ① ↑:

2N1 cos α − 4W = 0 ,

② →:

S − N1 sin α = 0 ,

4W

①

N1

W

1 3r/2 = sin α = 3r 2

Gr

This yields √ 4 3 N1 = W, 3

→

◦

α = 30 ,

N2 = 3W ,

N1

W

N1

S

S

N2

N2

3r

↑: N2 − N1 cos α − W = 0 .

The angle α follows from the geometry of the problem:

N1

α

②

3r

α

3r

√ 3 , cos α = 2

√ 3 tan α = . 3

√ 2 3 W. S= 3

Note that we could have determined the contact force N2 from an equilibrium condition for the complete system: ↑: 2N2 − 2W − 4W = 0

→

N2 = 3W .

E2.15

A cable (length l, weight negligible) is attached to two walls at A and B (Fig. 2.27). A cube (weight W ) on a frictionless pulley (radius negligible) is suspended by the cable. Find the distance d of the cube from the left side in the equilibrium position and calculate the force S in the cable.

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1

Example 2.16

11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11

b

B

A

W

d

a

Fig. 2.27

Solution First we introduce the auxiliary quantities α, β, l1 and

l2 where l1 + l2 = l. Then we isolate the pulley by cutting the cable and write down the equilibrium conditions: B

S

α β

S

b

A

α α

l1

d

W

a

→: −S sin α + S sin β = 0 ↑:

l2

S cos α + S cos β − W = 0

→

α =β,

→ S=

W . 2 cos α

The angle α follows from the geometry of the problem:   a 2 a cos α = 1 − . sin α = , l l

The distance d can be calculated from the following geometrical relations: d = l1 sin α,

a = l1 sin α + l2 sin α,

b = −l1 cos α + l2 cos α.

Solving for d yields   a b d= 1− √ . 2 l 2 − a2

Gr

E2.16

2 Forces with a Common Point of Application

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

8

Note: the angle α and thus the force S in the cable are independent of b.

9

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

2 Forces with a Common Point of Application

Example 2.17 A smooth circular cy-

linder (weight W = 500 N) rests on two fixed supports as shown in Fig. 2.28. It is subjected to a force F = 200 N. a) Calculate the contact forces. b) Determine the allowable magnitude of F in order to avoid the cylinder from lifting off.

F

30◦

01W

45◦

0060 11

◦

Fig. 2.28

a) We isolate the cylinder and introduce the contact forces Nl and Nr in the free-body diagram. Solution

F

30◦

Nl

11 00

11 00 00 N 11

45◦ W 60◦

r

These forces are perpendicular to the smooth surface of the cylinder. Thus, Nl , Nr , W and F are a system of concurrent forces. They satisfy the equilibrium conditions →:

↑:

Nl cos 45◦ − Nr cos 60◦ + F cos 30◦ = 0 ,

Nl sin 45◦ + Nr sin 60◦ − F sin 30◦ − W = 0 .

Solving for√Nl , Nr yields with sin 30◦ =√cos 60◦ = 1/2, sin 45◦ = cos 45◦ = 2/2 and sin 60◦ = cos 30◦ = 3/2 the contact forces √ 2 2 √ (W − F ) = 155 N, Nr = F + √ W = 566 N. Nl = 1+ 3 1+ 3

Gr

b) Lift-off takes place at the left support when Nl = 0, i.e. for F ≥ W . Therefore, the allowable F to avoid lift-off is given by Fallow ≤ 500 N .

E2.17

Example 2.18 Two cylinders (weights W1 and W2 ) are pin-con-

nected by a bar (weight negligible). They rest on two smooth inclined planes as shown in Fig. 2.29. Given: W1 = 200 N, W2 = 300 N, α = 60◦ . Calculate the angle ϕ in the equilibrium position and the corresponding force S in the bar.

W1

S

W2

α

ϕ

Fig. 2.29

Solution We isolate the cylinders and introduce the contact forces

N1 , N2 as well as the force S in the bar in the free-body diagram. The forces at each cylinder form a system of concurrent forces. ①

S

α

W1

N1

②

S

ϕ

N2

α

W2

The equilibrium conditions are given by ① →:

S cos ϕ − N1 sin α = 0 ,

↑: −S sin ϕ + N1 cos α − W1 = 0 ,

② →:

−S cos ϕ + N2 cos α = 0 ,

↑:

S sin ϕ + N2 sin α − W2 = 0 .

√ Eliminating N1 and N2 yields with sin α = 3/2 and cos α = 1/2 the following two equations   √ cos ϕ S − sin ϕ + √ = W1 , S sin ϕ + 3 cos ϕ = W2 3 which have the solution ϕ = −19◦ ,

Gr

E2.18

2 Forces with a Common Point of Application

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

10

S = 229 N .

Note that the angle ϕ is negative, i.e., as opposed to Fig. 2.29, in equilibrium cylinder ① is below cylinder ②.

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os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

2 Forces with a Common Point of Application

Example 2.19 A spatial system of concurrent forces consists of the

three forces⎛ ⎞ 1 ⎜ ⎟ F 1 = a1 ⎝2⎠ , 0

⎛

⎞ ⎛ ⎞ −9 8 ⎜ ⎟ ⎜ ⎟ F 2 = a2 ⎝ 6 ⎠ , F 3 = a3 ⎝−7⎠ . 9 1 ⎛ ⎞ 30 ⎜ ⎟ Their resultant is given by R = ⎝−28⎠ kN . 44 Determine the unknown constants a1 , a2 , a3 , the magnitudes of the forces F 1 , F 2 , F 3 and the angle α between F 2 and F 3 .

Solution The coordinates of the resultant are given by

Rx = F1x + F2x + F3x

→

30 kN = a1 − 9a2 + 8a3 ,

Ry = F1y + F2y + F3y

→ −28 kN = 2a1 + 6a2 − 7a3 ,

Rz = F1z + F2z + F3z

→

44 kN = 9a2 + a3 .

From this system of three equations the unknown constants are calculated as a1 = 2 kN ,

a2 = 4 kN ,

a3 = 8 kN .

With the constants ai , the magnitudes of the forces can be determined:  √ 2 + F2 + F2 = a F1 = F1x 1 1 + 4 = 4.5 kN , 1y 1z  √ 2 + F2 + F2 = a F2 = F2x 2 81 + 36 + 81 = 56.3 kN , 2y 2z  √ 2 + F2 + F2 = a F3 = F3x 3 64 + 49 + 1 = 85.4 kN . 3y 3z

The angle α between F 2 and F 3 is calculated using the definition of the dot product:

Gr

F 2 · F 3 = F2 F3 cos α cos α =

→

cos α =

F2 · F3 F2 F3

a2 a3 (−72 − 42 + 9) = −0.699 F2 F3

→

→

α = 134◦ .

E2.19

A smooth sphere R z (weight W , radius R) rests on three F x points A, B and C. These three W R/2 00000000 A points form an equilateral triangle 11111111 C 00000000 11111111 00000000 11111111 in a horizontal plane. The height 00000000 11111111 √ 00000000 11111111 00000000 of the triangle is 3a = 34 3 R (see 11111111 2a a Fig. 2.30). The action line of the applied force F passes through the Fig. 2.30 center of the sphere. Determine the contact forces at A, B and C. Find the force F required to lift the sphere off at C. Example 2.20

Solution The contact forces A, B and C are perpendicular to the

smooth surface of the sphere. Thus, these forces and W and F are a spatial system of concurrent forces. They satisfy the equilibrium condition A + B + C + W + F = 0.

We now use the coordinate system shown in Fig. 2.30 with its origin at the center of the sphere. In order to obtain the force vectors we first write down their direction vectors with the aid of the coordinates of the points of contact. To this end we introduce the length b which follows from the geometry: √ (2b)2 = b2 + (3a)2 → b = 3 a .

y

B

C

x

b

A

2a

a

Then, for example, the unit vector in the direction from A to the origin of the coordinate system is obtained as ⎛ ⎞ ⎛ √ ⎞ −a − 3 ⎜ √ ⎟ ⎟ 1 1⎜ ⎜ ⎟ ⎜ eA =  3a ⎠ = ⎝ 3 ⎟ ⎝ ⎠. 4 a 1 + 3 + 4/3 √ 2a/ 3 2

Gr

E2.20

2 Forces with a Common Point of Application

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

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2 Forces with a Common Point of Application

13

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Hence, the force vector A (oriented towards the interior of the sphere) is given by ⎛ √ ⎞ − 3 ⎟ A⎜ A= ⎜ 3 ⎟ ⎝ ⎠. 4 2 Similarly, the other force vectors are found to be ⎛ √ ⎞ ⎛ √ ⎞ ⎛ ⎞ − 3 2 3 F ⎟ ⎟ ⎜ ⎟ C⎜ B⎜ ⎜ ⎟ ⎜ ⎟ ⎜ B= −3 ⎠, C = ⎝ 0 ⎠, F = ⎝ 0 ⎟ ⎠, 4 ⎝ 4 2 2 0

⎛

⎞

0

⎜ ⎟ ⎟ W =⎜ ⎝ 0 ⎠. −W

Introduction into the vector equilibrium condition leads to the three equations √ √ √ − 3A − 3B + 2 3C = −4F , 3A − 3B = 0 ,

2A + 2B + 2C = 4W

from which we can calculate the contact forces: 2 1 2 2 A=B= W+√ F , C= W−√ F . 3 3 3 3

When the sphere lifts off at point C, the corresponding contact force vanishes: C = 0. This yields the force which is necessary for the lift-off: √ 3 2 W. W −√ F =0 → F = 2 3

Gr

Note that this force may also be determined from the moment equilibrium condition about the axis passing through the points A and B:  R M (AB) = 0 : aW − F = 0 . 2

Thus, we obtain again √ 2a 3 F = W = W. R 2

Example 2.21 The construction shown in Fig. 2.31 consists of three

bars that are pinconnected at K. A rope attached to a wall is guided without friction through an eye at K. The free end of the rope is loaded with a crate (weight W ). Calculate the forces in the bars.

a

2a

6a

2a

2

1

K

3

4a

3a

W

Fig. 2.31

Solution We isolate pin K by passing imaginary sections through the bars and the rope. The internal forces are made visible in the free-body diagram; they are assumed to be tensile forces. The equilibrium condition is given by

z

y

S2

x

K

S1

W

S3

W

S1 + S2 + S3 + F = 0 ,

where

⎛

⎞

1

S1 ⎜ ⎟ S1 = √ ⎜ −3⎟, 10 ⎝ ⎠ 0

Gr

E2.21

2 Forces with a Common Point of Application

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

14

⎛ ⎞ −1 ⎟ S2 ⎜ ⎜ S 2 = √ ⎝−3⎟ ⎠, 10 0

⎛

⎞

0

S3 ⎜ ⎟ S3 = √ ⎜ −2⎟ 5⎝ ⎠ −1

15

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

2 Forces with a Common Point of Application

and

⎛

⎞

⎛ ⎞ −3 ⎜ ⎟ W ⎜ ⎟ ⎜ ⎟ ⎜ F =W ⎝ 0 ⎠+ 0 ⎟. 5 ⎝ ⎠ −1 −4

0

In coordinates, equilibrium thus requires 1 1 3 √ S1 − √ S2 − W = 0 , 5 10 10 3 2 3 − √ S1 − √ S2 − √ S3 = 0 , 10 10 5 9 1 − √ S3 − W = 0 . 5 5 Solving yields

Gr

9 S1 = √ W , 10

3 S2 = √ W , 10

9 S3 = − √ W . 5

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 3

General Systems of Forces, Equilibrium of a Rigid Body

3

A uniform pole (length l, weight W ) leans against a corner as shown in Fig. 3.28. A rope S prevents the pole from sliding. All surfaces are smooth. Determine the force S in the rope. Example 3.11

W 11111111 00000000 l 00000000 11111111 00000000 11111111 α 00000000 11111111 00000000 11111111 S 00000000 11111111 00000000 11111111

h

Fig. 3.28

Solution We isolate the pole by cutting

the rope and removing the wall and the ground. The contact forces N1 and N2 in the free-body diagram are perpendicular to the respective planes of contact. The forces acting at the pole must satisfy the equilibrium conditions. The force S in the rope can be determined from the two equations  A:

→:

N2

h/ sin α

W

A

α

N1

S l cos α/2

h Wl cos α + N2 = 0 − 2 sin α

S − N2 sin α = 0

→

N2 =

→

S=

Wl sin α cos α , 2h

Wl sin2 α cos α . 2h

Note that this solution is valid only if the contact force N1 is positive. This condition can be checked with the aid of the equilibrium condition of the forces in the vertical direction.

Gr

E3.11

3 General Systems of Forces, Equilibrium of a Rigid Body

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

18

19

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

3 General Systems of Forces, Equilibrium of a Rigid Body

A uniform beam (length l, weight W ) is inserted into an opening (Fig. 3.29). The surfaces are smooth. Calculate the magnitude of the force F required to hold the beam in equilibrium. Is the result valid for an arbitrary ratio a/l? Example 3.12

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 30◦ 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 l, W 11111

→:

F

Fig. 3.29

Solution We isolate the beam

and sketch the free-body diagram. The contact forces N1 and N2 are perpendicular to the respective planes of contact. The equilibrium conditions are

a

N1

A

l−a

l sin 30◦

N2

30◦

W

F

l cos 30◦ /2

N2 sin 30◦ − F = 0 ,

↓: N1 − N2 cos 30◦ + W = 0 ,  l ◦ ◦ A : N2 (l − a) − W cos 30 − F l sin 30 = 0 . 2 Solving for the unknown forces yields √ 3 8a/l − 3 W, W, F = N1 = 6 − 8a/l 6 − 8a/l

N2 = 2 F .

Gr

The solution is valid only if N1 > 0 and N2 > 0. This leads to the condition 3/8 < a/l < 3/4.

E3.12

Example 3.13 Two smooth rollers (each

having weight W and radius r) are connected by a rope (length a) as shown in Fig. 3.30. A lever (length l) subjected to a vertical force F exerts contact forces on the rollers. Determine the contact forces between the rollers and the horizontal plane.

l

F

W W 11111111 00000000 00000000 11111111 00000000 11111111 a Fig. 3.30

Solution We separate the rollers and the lever and draw the free-

body diagrams. ①

D2

S

W

③

②

S

D1

x

W

N2

N1

H

D1

D2

111 000 a cos α N α

O

F

3

The equilibrium conditions for the three bodies are represented by 2 × 2 + 1 × 3 = 7 equations for the 7 unknown forces (D1 , D2 , N1 , N2 , N3 , H and S): ①→:

S − D1 sin α = 0 ,

↑:

N1 − W + D1 cos α = 0 ,

②→:

D2 sin α − S = 0 ,

↑:

N2 − W − D2 cos α = 0 ,

③→:

H + D1 sin α − D2 sin α = 0 ,

↑ : N3 − D1 cos α + D2 cos α − F = 0 ,

 O :

l cos α F − (a cos α + x)D2 + xD1 = 0 .

The angle α follows from the geometry of the problem: r a/2 r , sin α = a/2 α  a/2 cos α = 1 − 4(r/a)2 . r

0000 1111 1111 0000 Addition of the 1st and the 3rd equation yields D = D . Thus,

Gr

E3.13

3 General Systems of Forces, Equilibrium of a Rigid Body

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

20

1

2

H = 0 and N3 = F . Then the unknown distance x cancels in the 7th equation and we finally obtain   l r 2 l r 1 − 4( ) , N2 = W + F 1 − 4( )2 . N1 = W − F a a a a

21

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

3 General Systems of Forces, Equilibrium of a Rigid Body

Example 3.14 Two smooth spheres (each

R

having weight W and radius r) rest in a thin-walled circular cylinder (weight Q, radius R = 4r/3) as shown in Fig. 3.31. Find the magnitude of Q required to prevent the cylinder from falling over.

Q

W

α W r

1111111 0000000 0000000 1111111

Fig. 3.31

Solution We separate the spheres and the cylinder and we draw

the free-body diagrams at the moment when the right-hand side of the cylinder lifts off from the ground. Then the cylinder touches the ground only at point C. The corresponding contact force is denoted by N5 . (Note that the contact force between the cylinder and the ground is distributed over the whole circumference of the cylinder before the lift-off). ①

N1

N1

N2

W

N2

②

Q

N4

W

③

2 r sin α

N4

C

N3

N5

The equilibrium conditions at the sphere and the cylinder are ① ↑:

N2 sin α − W = 0 ,

② ↑:

N3 − N2 sin α − W = 0 ,

→:

N1 − N2 cos α = 0 ,

→:

N2 cos α − N4 = 0 ,

③→:  C :

N4 − N1 = 0 ,

↑:

N5 − Q = 0 ,

(r + 2r sin α)N1 − rN4 − R Q = 0 .

Gr

They lead to N1 = N4 =

W W 3 , N2 = , N3 = 2W, Q = N5 = W cos α. tan α sin α 2

With the geometrical relation cos α = (R − r)/r = 1/3 we obtain the force that causes the cylinder to fall over: Q = W/2 . Thus, falling over is prevented if Q > W/2 .

E3.14

22

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

A rigid body is subjected to three forces: F 1 = F (−2, 3, 1) , F 2 = F (7, 1, −4)T , F 3 = F (3, −1, −3)T . Their points of application are given by the position vectors r 1 = a(4, 3, 2)T , r 2 = a(3, 2, 4)T , r3 = a(3, 5, 0)T . Determine the resultant force R and the resultant moment (A) M R with respect to point A given by r A = a(3, 2, 1)T . Example 3.15 T

Solution The components of the resultant force vector are obtained as the sum of the given force components:  Fix = (−2 + 7 + 3)F = 8F , Rx =  Fiy = (3 + 1 − 1)F = 3F , Ry =  Fiz = (1 − 4 − 3)F = −6F . Rz =

Hence, the resultant can be written as the row vector R = (Rx , Ry , Rz )T = F (8, 3, −6)T .

To determine the resultant moment about point A, we first write down the vectors r i − rA (i = 1, 2, 3) from point A to the points of application of the forces Fi . For example, r 1 − r A = a(1, 1, 1)T .

The moment of force F1 about point A is obtained from (A)

M1

= (r 1 − r A ) × F 1

(A)

→

M1

= aF (−2, −3, 5)T .

Similarly, we find (A)

M2

= aF (−3, 21, 0)T

(A)

(A)

M3

= aF (−10, −3, −9)T . (A)

The moment vector M R is the sum of the vectors M i :  (A) (A) (A) MR = Mi → M R = aF (−15, 15, −4)T .

Gr

E3.15

3 General Systems of Forces, Equilibrium of a Rigid Body

23

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3 General Systems of Forces, Equilibrium of a Rigid Body

A plate in the form of a rectangular triangle (weight negligible) is supported by six bars. It is subjected to the forces F and Q (Fig. 3.32). Calculate the forces in the bars.

Example 3.16

a

a F

Q

6

a

1

2

4

5

3

a

Fig. 3.32

Solution First we sketch the free-body diagram and choose a coordinate system. y

Q

F

z

S6

0

S1

S4

α

S5

S2

45◦

S3

x

Gr

Then we write down the equilibrium conditions (since the geometry of the problem is very simple, we do not resort to the vector formalism): √ √  2 2 S2 + S5 + F = 0, Fx = 0 : 2 2  Fy = 0 : S6 cos α = 0, √ √  2 2 S2 − S3 − S6 sin α − S4 − S5 − Q = 0, Fz = 0 : −S1 − 2 √2  (0) 2 S5 − a Q = 0, Mx = 0 : −2aS4 − 2a 2  (0) a My = 0 : a S3 + Q = 0, 2 √  (0) 2 S5 − aF = 0. Mz = 0 : −2a 2 Solving this system of equations for the forces in the bars yields √ 2 F Q F, S2 = − S3 = − , S1 = , 2 2 2 √ 1 2 F, S4 = (F − Q) , S5 = − S6 = 0 . 2 2

E3.16

A homogeneous rectangular plate (weight W ) is supported by six bars. The plate is subjected to a vertical load F (Fig. 3.33). Calculate the forces in the bars.

Example 3.17

3a 2a2a 2a 2a

3a F

3a

3a

W

1

3 5 B

2

6

4

6a

C

A

Fig. 3.33

Solution We draw the free-body diagram and choose a coordinate

system.

0

F

y

D x

S3 W B

S1

S2

S5

z

S4

S6

C

A

From the geometry of the problem we obtain the relations S1x S1z S1 = = , S2x S2z S2

S3y S3 = , S4y S4

S1x 4a . = S1z 6a

The forces in the bars follow from the equilibrium conditions:   

Fx = 0 : S1x − S2x = 0

→

S1 = S 2 ,

Fy = 0 : S3y + S4y = 0

→

S3 = −S4 ,

(B) Mx

=0:

12aS1z + 12aS2z + 6aW + 9aF = 0 →  (0) Mz = 0 :

−12aS1x + 12aS2x − 8aS4y = 0  (0) My = 0 :

→

Gr

E3.17

3 General Systems of Forces, Equilibrium of a Rigid Body

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24

−8aS2z − 8aS6 − 4aW − 6aF = 0 →  (D) My = 0 : 8aS1z + 8aS5 + 4aW + 2aF = 0

→

√ S2 = − 13(2W + 3F )/24, S4 = 0,

S6 = −(2W + 3F )/8, S5 = −(2W − F )/8.

Bar 5 is under compression for F < 2W and it is under tension for F > 2W .

25

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

3 General Systems of Forces, Equilibrium of a Rigid Body

Example 3.18 A rectangular plate of negligible weight is suspended

by three vertical wires as shown in Fig. 3.34. a) Assume that the plate is subjected to a concentrated vertical force Q. 1 a Determine the location of the point of application of Q so that the forces in the 2 wires are equal. b) Calculate the forces a 4a in the wires if the plate x is subjected to a vertical Fig. 3.34 constant area load p.

3

2a

y

4a

a) We introduce a coordinate system. The unknown coordinates of the point of application of the force Q are denoted by xQ and yQ .

Solution

a)

S x

S

z 0 Q

b)

S

z

S2

y

xQ

yQ

S1

0 F

p

y

3a

S3

2a

x

If the forces in the wires are equal (Si = S, i = 1, 2, 3), the equilibrium conditions (parallel forces) are   

Fz = 0 :

3S − Q = 0 ,

(0) Mx

=0 :

2 aS − yQ Q = 0 ,

(0) My

= 0 : −5 aS − a S − 2aS + xQ Q = 0 .

Gr

This yields S=

Q , 3

yQ =

4 a, 3

xQ =

8 a. 3

E3.18

3 General Systems of Forces, Equilibrium of a Rigid Body

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26

b) Now the plate is subjected to a constant area load p which can be replaced by the resultant force F = 4 · 6a2 p = 24pa2 . The forces in the wires are denoted by S1 , S2 and S3 . The equilibrium conditions  Fz = 0 : S1 + S2 + S3 − 24 pa2 = 0 ,  (0) Mx = 0 : 2 a 24 pa2 − 4 aS3 = 0 ,  (0) My = 0 : 3 a 24 pa2 − 5 aS2 − aS1 − 2 aS3 = 0 now lead to

Gr

S3 = 12 pa2 ,

S1 = 3 pa2 ,

S2 = 9 pa2 .

27

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3 General Systems of Forces, Equilibrium of a Rigid Body

Example 3.19 The circular

arch in Fig. 3.35 is subjected to a uniform tangential line load q0 . Determine the resultant force R and the resul(A) tant moment M R with respect to the center A of the circle. If the load is reduced to a single force alone, find the corresponding line of action.

A

60

◦

60◦

r

y

q0

x

Fig. 3.35

Solution First we introduce the angle ϕ. Then we consider an element of the line load that is located at ϕ. The x-component of the infinitesimal force dR satisfies the relation dRx (−ϕ) = −dRx (ϕ); the y-component is given by dRy = q0 cos ϕ rdϕ. A

A

x

ϕ

y

r

y

rdϕ

B

dR

x

R

(A)

MR

b = 1.21 r

R

Integration yields the components of the resultant force: ◦

◦

60

Rx =

dRx = 0 ,

−60◦

60

Ry =

−60◦

◦

60 √ dRy = 2q0 r cos ϕdϕ = 3 q0 r . 0

Gr

The moment of the line load with respect to point A is obtained as  2π 2π (A) r·r = q0 r2 . A : MR = q0 3 3

E3.19

3 General Systems of Forces, Equilibrium of a Rigid Body

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28

Thus, the given line load is statically equivalent to   2π 0 (A) q0 r2 . R= √ , MR = 3 3 q0 r

This system is equivalent to the force R alone, if we choose a point B of its line of action according to the condition (see figure)  B:

(A)

M (B) = MR −Ry b = 0

→

(A)

b=

MR Ry

2π r = √ = 1.21 r . 3 3

Thus,



Gr

R=



0

, √ 3 q0 r

xB = 1.21 r ,

yB arbitrary .

3 General Systems of Forces, Equilibrium of a Rigid Body

29

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

1 0 B 0 1 0 1 0 1 0 1 000 111 0 1 000 111 WB 0 1 000 111 00000000 11111111 0 1 00000000 11111111 0 1 00000000 11111111 0 1 WS 11111111 00000000 0 1 00000000 11111111 0 1 00000000 11111111 0 1 00000000 0 11111111 1 l 00000000 11111111 0 1 0 1 00000000 11111111 0 1 0 1 00000000 11111111 000 111 0 1 00000000 11111111 000 0 1 α111 00000000 11111111 b 000 111 0 A1 00000000 11111111 000 111 0 1 00000000 11111111 00 000 111 011 1 00000000 11111111 00 11 000 111 00000000 11111111 00 11

A sphere (weight WS ) is held between a beam (weight WB ) and a wall as shown in Fig. 3.36. The surface of the sphere is smooth. The beam is supported by a hinge at A and a rope at B. Calculate the force S in the rope. Example 3.20

Fig. 3.36

Solution We isolate the sphere and the bar and sketch the free-

body diagram. Since the surface of the sphere is smooth, the contact forces N1 and N2 are perpendicular to the beam and the wall, respectively. S ① N2

00 11 00 11 00 11 00 11 α 00 11

WS

②

N1

N1

AH

W 00 11 111 000 00 B 11 000 111 000 111 b 000 111 0 1 000 111 0 1 0 1 0000 00 11 AV 1111 0000 1111 00 11 l cos α/2 A

1 0 11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 l11 sin α 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11

The equilibrium conditions at the sphere and the bar are ①→:

N2 − N1 sin α = 0 ,

②→:

AH + N1 sin α − S = 0 ,

Gr

↑:  A:

↑:

N1 cos α − WS = 0 ,

AV − N1 cos α − WB = 0 ,

b N1 + l cos α/2 WB − l sin α S = 0 .

These are 5 equations for the 5 unknowns N1 , N2 , AV , AH and S. Solving for S yields S=

b WS WB cot α + . 2 l sin α cos α

E3.20

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 4

Center of Gravity, Center of Mass, Centroids

4

Locate the centroids of the profiles as shown in Fig. 4.16. The measurements are given in mm.

Example 4.9

15

15

8

8

5

20

32

y

45

a

4

x

5

y

x

b

4

45

Fig. 4.16

Solution a) The y-axis coincides with the axis of symmetry. The-

refore, xc = 0. In order to calculate yc , we consider the area to be a)

b) y

y

x

x

composed of three rectangles with known centroids. Then we obtain  yi Ai 5120 2 (4 · 45) + 14(5 · 20) + 28 (8 · 15) = = yc =  Ai 4 · 45 + 5 · 20 + 8 · 15 400 = 12.8 mm .

b) We again consider the area to be composed of three rectangles: xc =

5200 22.5 (4 · 45) + 2.5 (5 · 20) + 7.5 (8 · 15) = = 13 mm , 4 · 45 + 5 · 20 + 8 · 15 400

yc =

2 (4 · 45) + 14 (5 · 20) + 28 (8 · 15) = 12.8 mm . 400

Gr

E4.9

4 Center of Gravity, Center of Mass, Centroids

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32

Note that a displacement of the areas in the x-direction does not change the y-coordinate of the centroid.

33

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

4 Center of Gravity, Center of Mass, Centroids

Locate the centroids of the thin-walled profiles (t a) as shown in Fig. 4.17. Example 4.10

y

t

y

t

a

a

a

x

a

a

x

Fig. 4.17

a

b

a

a

Solution Since t a and constant along the profile, we can apply

the equations for the centroid of a line. We consider the profile to be composed of three parts with known centroids. y ③

y

C3

① C1

C

② C2

③ C3

C

①

x

a

b

C1

②

C2

①

x

a) Individual profile parts: ①:

x1 = 0 ,

y1 = a/2 , l1 = a ,

②:

x2 = a/2 ,

y2 = 0 ,

③:

l2 = a , √ x3 = 3a/2 , y3 = a/2 , l3 = 2 a .

Centroid of the profile:  √ xi li xc =  = (5 2/4 − 1)a , li

 yi li √ yc =  = 2 a/4 . li

b) Individual profile parts:

x1 = 0 , y1 = a/2 ,

l1 = 2a ,

②:

x2 = 0 , y2 = 0 ,

l2 = 2a ,

③:

x3 = 0 , y3 = a + 2a/π ,

l3 = π a .

Gr

①:

Centroid of the profile: xc = 0

(symmetry!) ,

 yi li 3+π a. yc =  = li 4+π

E4.10

Example 4.11 Determine the

y

a

coordinates of the centroid C of the number shown in Fig. 4.18.

3a

a

4a

3a 2a

Fig. 4.18

x

Solution We consider the area as a

y

composite area which consists of five parts. Since the region ④ (semicircle) is void of material, this part has to be considered to be a “ negative ” area. It is practical to perform the calculation in a table:

① C1

②

C2

C ④ C3 C4

⑤

③

x

C5

i

xi /a

yi /a

Ai /a2

xi Ai /a3

yi Ai /a3

1

2.5

8.5

3

7.5

25.5

2

0.5

7

4

2

28

3

9 π 2

3

−2π

0.5

1

3

4

5

4 +1 π 8 +1 3π 0.5



Gr

E4.11

4 Center of Gravity, Center of Mass, Centroids

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

34

8+

This leads to the results  xi Ai = 1.93 a , xc =  Ai

5 π 2

18 +

−

9 π 2

16 − 2π 3 0.5

68 5 + π 3 2

27 π 2

−6π 0.5

54 +

15 π 2

 yi Ai yc =  = = 4.89 a . Ai

35

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

4 Center of Gravity, Center of Mass, Centroids

A circular area is removed from a circle and an ellipse, respectively (Fig. 4.19). Locate the centroids of the remaining areas.

Example 4.12

y

y

r1

r2

x

Fig. 4.19

r1 2

a

b

a

x

b 3

b

Solution We consider each of the remaining areas to be composed of the two parts ① and ②. Since the regions of the small circles are

void of material, the corresponding areas have to be subtracted from the large circle and the ellipse, respectively. y

y

①

①

C1

C

CC

1

x

C2

x

②

②

a

C2

b

a) Individual areas:

y1 = 0 , A1 = π r12 ,

①:

x1 = 0 ,

②:

x2 = r1 /2 , y2 = 0 , A2 = −π r22 .

Centroid of the remaining area:  xi Ai r1 r2 =− 2 2 2 , xc =  Ai 2(r1 − r2 )

yc = 0

(symmetry!) .

Gr

b) Individual areas: ①:

x1 = 0 ,

②:

x2 = b/3 , y2 = −b/3 , A2 = −π b2 /9 .

y1 = 0 ,

Centroid of the remaining area:  xi Ai b2 xc =  =− , Ai 3(9a − b)

A1 = π ab ,

yc = −xc .

Note that the centroid of each of the remaining areas lies on the straight line which connects the individual centroids.

E4.12

A thin metal sheet of even thickness is bent into the shape shown in Fig. 4.20. It consists of a square and two triangles. The measurements are given in cm. Determine the centroid.

Example 4.13

z

2

3

II

3

III

I

x

y

4

4

Fig. 4.20

Solution The metal sheet consists of three parts the centroids of which are known. The unknown position of the centroid (center of mass) follows from    ρi xi Vi ρi yi Vi ρi zi Vi xc =  , yc =  , zc =  . ρi Vi ρi Vi ρi Vi

Since the thickness of the metal sheet and the density of the material are constant these quantities cancel and we can apply the equations    xi Ai yi Ai zi Vi , yc =  , zc =  . xc =  Ai Ai Ai The total area is given by  1 1 A= Ai = 4 · 4 + · 4 · 3 + · 4 · 3 = 28 cm2 . 2 2

With the coordinates zI = 0, xII = 0, yIII = 0 of the individual centroids we obtain 2 2 · 16 + ( · 4) 6 xI AI + xIII AIII 3 = = 1.71 cm , xc = A 28 yc =

2 · 16 + 2 · 6 yI AI + yII AII = = 1.57 cm , A 28

1 1 ( · 3) 6 + ( · 3) 6 zII AII + zIII AIII 3 3 = = 0.43 cm . zc = A 28

Gr

E4.13

4 Center of Gravity, Center of Mass, Centroids

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

36

37

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

4 Center of Gravity, Center of Mass, Centroids

Example 4.14 The area shown in Fig.

4.21 is bounded by the coordinate axes and the quadratic parabola with its apex at x = 0. Determine the coordinates of the centroid.

y 3a/2

a/2

b x

Fig. 4.21

Solution First we write down the equation of the parabola: y = α x2 + β. The constants α and β are determined with the aid of the points x0 = 0, y0 = a/2 and x1 = b, y1 = 3a/2 which yields β = a/2, α = a/b2 . Thus, the parabola is obtained as  x 2 a y=a + . b 2 y

y dA = (b − x)dy

dx

dA = ydx

dy

y/2

x

x

x

Now we choose the infinitesimal area dA = y dx. coordinate of the centroid can be calculated:    b x 2 a   x a + dx b 2 x dA x y dx 0   xc = = = = b    dA y dx x 2 a a + dx b 2 0

b

x

Then the x-

1 2 ab 3 2 = b. 5 5 ab 6

Gr

If we choose the infinitesimal area dA = (b − x)dy in order to determine the y-coordinate, we are led to complicated integrals. For this reason we again use the element dA = y dx. Then we have to take into account that its centroid is located at the height y/2. Hence, we obtain  y  b  y dx 4 a2 2 a2 47 6 2 x 2 a. a 4 + 2x + dx = = yc = 5 10 ab b b 4 100 ab 0 6

E4.14

38

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Locate the 1 2a 2a a a 2a1 0 0 0 0 0 1 11 0 1 0 1 centroid of the area shown 1 0 in Fig. 4.22. 2a

Example 4.15

00 11 a 1 0000 1111 000000 111111 a0 0000 1111 000000 111111 0000 1111 000000 111111 0000 1111 000000 111111 2a 111111 0000 1111 000000 11 00 0000 1111 000000 111111 y 0000 1111 000000 111111 0000 1111 000000 111111 2a 111111 000000 000000 111111 11 00 000000 111111 x 000000 111111

Fig. 4.22

Solution The centroid is located on the x-axis due to the symmetry of the area: yc = 0. We consider the area to be composed of three parts: a square ① and two triangles ②, ③. Since the triangles are void of material, the corresponding areas have to be subtracted from the square. ①

1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 ② 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 C2 0000000000 1111111111 ③ 1111 0000 0000000000 1111111111 C C 1 3 0000 1111 0000000000 1111111111 0000 1111 0000000000 1111111111 0000 1111 0000000000 1111111111 C 1111 0000 0000000000 1111111111 0000 1111 0000000000 1111111111 y 0000 1111 0000000000 1111111111 0000 1111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 x 0000000000 1111111111

The areas and their centroids are given by √ ① : A1 = 64 a2 , x1 = 2 a , √ ② : A2 = −18 a2 , x2 = − 2 a , √ ③ : A3 = −2 a2 , x3 = 2 a/3 .

The centroid of the given area follows from  xi Ai xc =  Ai √ √ √ 2 a 64 a2 + (− 2 a)(−18 a2 ) + 2 a/3(−2 a2 ) = = 2.61 a . (64 − 18 − 2)a2

Gr

E4.15

4 Center of Gravity, Center of Mass, Centroids

39

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

4 Center of Gravity, Center of Mass, Centroids

A thin wire has the shape of the function y = a cosh x/a (Fig. 4.23). Find the centroid.

Example 4.16

y

C

a

a

y = a cosh x a a x

Fig. 4.23

Solution The centroid is located on the y-axis due to the symmetry of the problem: xc = 0. In order to find yc , we first calculate the derivative of the given function: y  = sinh x a . A line element is thus given by    x 2 2 2  ds = (dx) + (dy) = 1 + (y ) dx = 1 + sinh2 dx a x = cosh dx . a

Integration yields the length of the line:   +a x s = ds = cosh dx = 2 a sinh 1 . a −a

The first moment of the line with respect to the x-axis is obtained as 



y ds =

x dx = a a cosh a 2

+a

−a

1 + cosh 2 x a dx = a2 (1 + 1 sinh 2) . 2 2

Gr

Hence, the centroid is located at  y ds a 1 + 12 sinh 2 = = 1.197 a . yc =  ds 2 sinh 2

E4.16

Locate the centroid of a thinwalled spherical shell (radius R, Height H, thickness t R) as shown in Fig. 4.24.

Example 4.17

R

z

H

t

Fig. 4.24

Solution The centroid is located on the z-axis due to the symmetry

of the shell. Its position can be calculated from  z dV . zc =  dV ϕ0 ϕ

R cos ϕ

H

Rdϕ

z

R sin ϕ

In order to evaluate the integrals we choose a ring-shaped volume element: dV = t Rdϕ(2πR sin ϕ) .

Then we get with cos ϕ0 = (R − H)/R and z = R cos ϕ 

dV = 2π tR

2

ϕ0 0



Gr

E4.17

4 Center of Gravity, Center of Mass, Centroids

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

40

zdV = 2π tR

3

sin ϕ dϕ = −2π tR2 (cos ϕ0 − 1) = 2π tRH ,

ϕ0

sin ϕ cos ϕ dϕ = π tR3 sin2 ϕ0

0

= π tR3 (1 − cos2 ϕ0 ) = π tRH(2R − H) . Thus, the z-coordinate of the centroid is obtained as zc = R − H/2 . In the special case of H = R (half-spherical shell, ϕ0 = π/2) we find zc = R/2.

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 5

Support Reactions

5

Example 5.10 The beam

in Fig. 5.26 is supported by three struts and subjected to a triangular line load. Determine the forces in the struts.

q0

3a 4

1

2

a

3

a

a

Fig. 5.26

Solution We isolate the beam by passing cuts through the struts. The free-body diagram shows the forces acting 2a/3 on the beam; the forces in the R struts are assumed to be tenA sile. The line load is replaα α ced by its resultant R = q0 a. S1 In addition, we introduce the S3 S2 angle α. It follows from the a geometry: sin α = 3/5 .

The equilibrium conditions lead to the unknown forces:  A:

aS3 sin α + 2aR/3 = 0

→ S3 = −10 q0a/9 ,

←:

S1 cos α + S3 cos α = 0

→ S1 = 10 q0 a/9 ,

↓ : S1 sin α + S2 + S3 sin α + R = 0

→ S2 = −q0 a .

The negative signs of the forces S2 and S3 indicate that their directions in reality are opposite to those assumed in the freebody diagram. Hence, these forces are compressive forces.

Gr

E5.10

5 Support Reactions

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

42

43

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

5 Support Reactions

Example 5.11 The struc-

ture shown in Fig. 5.27 consists of a beam and three bars. It carries a concentrated force F . Determine the support reaction at A and the forces in the bars.

l

11 00 00 11 00 11 00 11 00 11 00 11 00 11

F

A

h

Fig. 5.27

2

K

a

Solution We remove the

l

support at A and we pass a cut through bar 1. AH The free-body diagram shows the corresponding h forces AH , AV (their senses of direction have been chosen arbitrarily), S1 (which is assumed to be a tensile force) and the given load F . The equilibrium conditions yield ↑ : AV − F = 0

3 α

1

F

A

AV

S1

K

S2

S1

S3

α

K

→ AV = F ,

 A:

l F + h S1 = 0 →

→:

AH − S1 = 0

l S1 = − F , h l → AH = − F . h

Now we pass cuts through the bars 2 and 3 and write down the equilibrium conditions at joint K: l F, h sin α l F. S2 + S3 cos α = 0 → S2 = h tan α

→ : −S1 + S3 sin α = 0

Gr

↑:

→ S3 = −

Note that the support reaction at A and the forces in the bars are independent of a.

E5.11

Example 5.12 The simply supported beam (length a = 1 m) shown

in Fig. 5.28 is subjected to the three concentrated forces F1 = 4 kN, F2 = 2 kN, F3 = 3 kN, the line load q0 = 5 kN/m and the moment M0 = 4 kNm. q◦ F1 F3 F2 M◦ Calculate the support ◦ 45 reactions.

10 0 10 00 11 1a

A

10 0 10 00 11 1

B

a

a

Fig. 5.28

Solution We free the beam from the supports and make the reactions A, BH and BV visible in the free-body diagram. The line load is replaced by its resultant q◦ a. 3a/2

F1

45

A

◦

F2

q◦ a

F3

B

BV

A

BH M◦

Then we write down the equilibrium conditions:  A:  B :

3 3a BV − M0 − 2a F3 − a (q◦ a) − a F2 sin α = 0 , 2 3 −3a A + 3a F1 + 2a F2 sin α + a (q◦ a) + a F3 − M◦ = 0 , 2 → : F2 cos α − BH = 0 .

They lead to

√ · 5 + 2 · 12 2 = 6.30 kN , BV = 3 √ 12 + 2 · 2 · 12 2 + 32 · 5 + 3 − 4 A = = 7.11 kN , 3 1√ 2 = 1.41 kN . BH = 2 · 2 4+6+

3 2

Gr

E5.12

5 Support Reactions

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

44

As a check we use the force equilibrium in the vertical direction: ↑:

A + BV − F1 − F2 sin α − q◦ a − F3 = 0 , →

6.30 + 7.11 − 4 − 2 · 0.71 − 5 − 3 = 0 .

Note that the support reactions are given with an accuracy of only two digits after the decimal point. Therefore, this equation is not satisfied exactly.

5 Support Reactions

45

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

q◦

Example 5.13 Find the support

reactions for the hinged beam shown in Fig. 5.29.

A

C

B

P

3a

00 11 0D 1 11 00 00 11 0 1

a

3a

Fig. 5.29

Solution The hinged beam is statically determinate supported.

The fixed support A transmits three reactions: two force components AH , AV and the moment MA . They are made visible in the free-body diagram where their senses of direction have been chosen arbitrarily. The force D in the bar DC is assumed here to be a compressive force. 2a

AH

a/2

4q◦ a

②

MA A

BH

α

①

AV

D

q◦ a ②

BV

P

B

D

P

The equilibrium conditions for the complete system (the hinge is regarded as being frozen) are ↓:

−D sin α − AV + P + q◦ 4a = 0 ,

→ : AH + D cos α = 0 ,  A:

−MA + 4a D sin α − 2a 4q◦ a − 4aP = 0 .

In addition, we write down a moment equilibrium condition for the right-hand part ②: a D sin α − P a −

Gr

 B :

1 aq◦ a = 0 . 2

With sin α = 3/5 and cos α = 4/5 these four equations yield the four unknown reactions: D=

5 2 5 7 4 P + q◦ a, AV = q◦ a, AH = − P − q◦ a, MA = −3q◦ a2 . 3 6 2 3 3

E5.13

Example 5.14 The hinged beam in Fig. 5.30 carries a concentrated

force and a triangular line load. Determine the support α reactions and the force in A the hinge.

q0

F

G

B

a

a

a a

C

3a

Fig. 5.30

Solution The hinged beam is statically determinate supported.

We draw the free-body diagram for the complete system, for the left part ① and for the right part ②, respectively. The triangular line load is replaced by its resultant force R = 3q0 a/2.

a

R

F

G

AH

②

① AV

B

C

a

R

F

G GH

A

AH

① AV

GV B

GH

C

G

②

GV

B

C

It is advantageous to use the following 6 equilibrium conditions to determine the 6 unknowns AH , AV , B, C, GH and GV . The force equilibrium in the horizontal direction for the complete system requires →:

−AH + F cos α = 0

→

AH = F cos α .

The equilibrium conditions for the left part ① are  A:  G:

Gr

E5.14

5 Support Reactions

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

46

→:

−aF sin α + 2aGV = 0

→

GV = F/2 sin α ,

−2aAV + aF sin α = 0

→

AV = F/2 sin α ,

−AH + F cos α − GH = 0

→ GH = 0

and moment equilibrium for the right part ② requires  B :  C :

aGV − 3aR + 4aC = 0

→

C = (9q0 a − F sin α)/8 ,

5aGV − 4aB + aR = 0 → B = (3q0 a + 5F sin α)/8 .

47

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

5 Support Reactions

Example 5.15 Determine the

01 100 11 00 1 2R 10A 10 11 00 B

support reactions for the structure shown in Fig. 5.31. The pulley is frictionless.

R

C

R

D

F

2R

Fig. 5.31

3R

Solution First we verify that the structure is statically determina-

te. It consists of n = 3 bodies; there exist r = 4 support reactions (2 force components at A and 2 components at B). The number v of the transferred joint reactions is 5, i.e. 2 reactions at C, 2 reactions at D and the force in the rope). Thus, the condition f = 3n − (r + v) = 3 · 3 − (4 + 5) = 0

is satisfied. (Note that the two beams can be regarded as a threehinged arch.) S Cx

Cx

Bx

Cy

By

①

②

Dx

Cy

Dy

Ax

S

y

Ay

x

Dx

③

Dy

F

We isolate the three bodies and write down the equilibrium conditions for the disk ③  D : RS − RF = 0 → S = F , ↑ : −Dy − F = 0

Gr

→ : −Dx − S = 0

→ Dy = −F , →

Dx = −S = −F ,

E5.15

5 Support Reactions

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

48

for the angled member ①  A:

↑:

→:

2R Cx − 2R Cy − 3R S = 0 , Ay − Cy = 0 ,

Ax − Cx + S = 0

and for the beam ② (using the results for the disk):  D:

−5R By − 3R Cy = 0 ,

↑:

By + Cy − F = 0 ,

→:

Bx + Cx − F = 0 .

The four support reactions and the two joint reactions at C can be calculated from the last six equations: By = −3 F/2 , Cx = 4 F ,

Cy = Ay = 5 F/2 ,

Bx = −3 F ,

Ax = 3 F .

Note that the support reactions in the horizontal direction can also be determined from the equilibrium condition for the complete system: Bx

F

By

Ax

Ay

 A:

6R F + 2R Bx = 0

→ : Ax + Bx = 0

→

Bx = −3F ,

→

Ax = 3F ,

Gr

In order to find Ay and By we have to pass a cut through the structure.

49

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

5 Support Reactions

Example 5.16 A homogeneous beam (weight W ) hangs on a crane

(Fig. 5.32). Determine the support reactions at A and B and the force at hinge C. 4a

4a

C

3a

B

A

111111111111111 000000000000000 000000000000000 111111111111111 W 000000000000000 111111111111111 000000000000000 111111111111111 3a

5a 2

Fig. 5.32

3a

a

a 2

Solution The structure is a three-hinged arch. In order to determi-

ne the unknown forces we isolate the beam, remove the supports A and B and we pass a cut through hinge C. C

S1

S2

A

AH

E

D

2a

BV

AV

W

4a

W

CV

CV

CH

CH

AH

Gr

A

AV

S2

Equilibrium at the beam yields  D:  E :

BH

B

S1

BH

B

4a W − 6a S1 = 0 →

S1 = 2W/3 ,

2a W − 6a S2 = 0 →

S2 = W/3 .

BV

E5.16

5 Support Reactions

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

50

Now we apply moment equilibrium conditions to the complete crane:  B : 14a AV − 8a W = 0 → AV = 4W/7 ,  A : 14a BV − 6a W = 0 → BV = 3W/7 . Subsequently, we write down the equilibrium conditions for the left-hand part: ↑:

AV − S1 + CV = 0

→

 C :

7a AV − 4a AH − 3a S1 = 0 →

→:

AH + CH = 0

→

CV = 2W/21 , AH = W/2 ,

CH = −W/2 .

Finally, force equilibrium in the horizontal direction for the complete system leads to

Gr

→:

AH + BH = 0

→

BH = −W/2 .

51

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

5 Support Reactions

Example 5.17 A mast (weight W1 ) has a hinged support (ball-and-

socket connection) at A. In addition it is supported by two struts. Its upper end carries a weight W2 (Fig. 5.33). Determine the reaction force at A and the forces in the struts.

W1

2a

E5.17

W2

z

2

2a

1

y

A

a

x

a

a

a

Fig. 5.33

Solution First we isolate the mast by removing support A and

cutting the struts.

Then we use the geometrical relations

W1

z

S1x : S1y : S1z = S2x : S2y : S2z = 1 : 1 : 2 . Due to the symmetry we obtain

W2

S2

Az

S1

A

S1 = S2

Ax x

Ay

B y

and the equilibrium conditions yield  Fix = 0 : Ax + S1x − S2x = 0 → Ax = 0 ,  

→ Az = W1 /2 ,

(A)

Mix = 0 : −aS1z − aS2z − 2aS1y − 2aS2y − aW1 − 2aW2 = 0 √ → S1 = − 6 (W1 + 2W2 )/8 , Fiy = 0 : Ay + S1y + S2y = 0

Gr



(B)

Mix = 0 : a W1 − 2a Az = 0

→

Ay = (W1 + 2W2 )/4 .

Example 5.18 Determine the

support reactions for the frame shown in Fig. 5.34.

11 00 000 11 101A

F

l/2

l

00 11 010 11 00 1

B

l

l/2

C 11 00 00 11 001 11 0 00 11 00 11 00 11 0 1 l/2

Fig. 5.34

Solution The free-body diagram shows the 5 unknown support

reactions: 2 force components in A and in C, respectively, and the force B in the bar (here assumed to be a compressive force). First we write down the equilibrium conditions for the complete system (the hinges are assumed to be frozen):

AH

A

AV

①

F

D

B

②

E

③

CH

CV

3 l 1√ 3 1√ 2+ B 2 − l F − l CH + 2l CV = 0 , 2 2 2 2 2 1√ 2 + CV + AV − F = 0 , ↑: B 2 1√ 2 − CH = 0 . → : AH + B 2  A:

lB

Then we use moment equilibrium conditions for the part ① to the left of hinge D and the part ③ to the right of hinge E, respectively:  D:  E :

−l AV −

l AH = 0 , 2

l CV − l CH = 0 . 2

Solving these 5 equations for the 5 unknowns yields √ F F F 2 F , CH = , AH = , AV = − , B = 3 6 6 2

Gr

E5.18

5 Support Reactions

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

52

CV = F .

Note that the support reactions could be determined without calculating the forces in the hinges.

53

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

5 Support Reactions

Calculate the support reactions for the spatial structure in Fig. 5.35.

Example 5.19

q◦

y

A

z

C

x

D

a

B

q◦

2a

Fig. 5.35

Solution We isolate the structure in the free-body diagram. The

directions of the forces in the struts B, C and D are known. Support A transfers three force components. q◦

Ax

Cy

A

Ay

Az

Dz

q◦

Bz

The equilibrium conditions yield  Fx = 0 : Ax − 2q◦ a = 0   

→

Ax = 2q◦ a ,

→

Dz = −q◦ a ,

→

Bz = −

→

Cy = 2q◦ a ,

Fy = 0 : Ay + Cy = 0

→

Ay = −2q◦ a ,

Fz = 0 : Az + Bz + Dz + q◦ a = 0

→

Az = q◦

(A)

= 0 : −Dz 2a − q◦ a 2a = 0

(A)

= 0 : −Bz a − q◦ a

(A)

= 0 : Cy a − 2q◦ a a = 0

Mx My Mz 



a =0 2

q◦ a , 2

a . 2

Gr

Note that an appropriate choice of the reference points for the moment equations simplifies the calculation.

E5.19

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 6

Trusses

6

The truss shown in Fig. 6.13 carries the two forces F1 = F and F2 = 3F . Calculate the forces in the members 1, 2 and 3.

F1

Example 6.4

F2

1

2

A

B

a

a

a a

3

a

a

a

a

Fig. 6.13

Solution First we determine the force (directed vertically) at the

support A. Applying the moment equilibrium condition about F1

F2

F1

S1

B

A

C

S2

A

B

D

S3

point B of the free-body diagram for the whole truss yields  B :

6a A − 5a F1 − a F2 = 0

→

A=

4 F. 3

Then we pass an imaginary section through the members 1, 2, 3. We assume the forces in the bars to be tensile and write down the equilibrium conditions for the left-hand part of the truss:  C :

 D:

3a A − 2a F1 − 2a S3 = 0

a A + 2a S1 = 0 √ 2 S2 = 0 ↑ : A − F1 + 2

Gr

E6.4

6 Trusses

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

56

→ S3 = F ,

2 S1 = − F , 3 √ 2 F. → S2 = − 3

→

57

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

6 Trusses

Determine the forces in the members 1, 2 and 3 of the truss shown in Fig. 6.14. Example 6.5

F

F

1

2

a

3

B

a

2F

A

a

a

Fig. 6.14

a

a

Solution First we draw the free-body diagram for the complete F

F

F

S1

S2

I

S3

2F

AH

2F

B

B

A

AV

truss and determine the support reactions: →:

F − AH = 0

 A:

2a F + 6a F + 2a F − 4a B = 0 → B =

↑:

AV + B − F − 2F = 0

→

→

AH = F ,

5 F, 2 1 AV = F . 2

Gr

In order to obtain the forces in the members 1, 2, 3 we pass an imaginary cut through these members. The forces are assumed to be tensile forces. Equilibrium for the right-hand part of the truss leads to √ √ 2 2 S2 + B − 2F = 0 F, → S2 = − ↑: 2 2  3 → S1 = − F , I : a F − a S1 − a B = 0 2 √ 2 S2 − F = 0 → S3 = 3F . ← : S3 + S 1 + 2

E6.5

6 Trusses

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

58

Gr

As a check we apply the force equilibrium condition to the lefthand part: √ 2 1 1 S2 = F − F + F = 0 . ↑ : AV − F − 2 2 2

59

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

6 Trusses

Example 6.6 The structure

q0

in Fig. 6.15 consists of a hinged beam AB and five bars. It is subjected to a line load q0 . Determine the forces in the bars.

1

4a

B

G

2

4

5

2a

3a

3

2a

4a

Fig. 6.15

The support reactions follow from the equilibrium

Solution 3a

A

R = 6q0 a

GV

A A

B

1

G

2

G

3

4

5

I

GH

B

α

S3

S4

B

II

S5

S5

II

conditions for the complete truss:  B :  A:

12a A − 9a R = 0

→

A = 9q0 a/2 ,

12a B − 3a R = 0

→

B = 3q0 a/2 .

Now we pass an imaginary section through hinge G and bar 5. The senses of direction of the forces GH and GV are chosen arbitrarily; the force in bar 5 is assumed to be tensile. Moment equilibrium for the right-hand part yields  G:

6a B − 3a S5 = 0

→

S5 = 3q0 a .

Then we isolate joint II, introduce the angle α and write down the equilibrium conditions (sin α = 4/5, cos α = 3/5): →:

S4 = 15q0 a/4 ,

→

S3 = −9q0 a/4 .

S3 + S4 cos α = 0

Gr

↑:

−S5 + S4 sin α = 0 →

Finally, from symmetry: S1 = S4 ,

S2 = S3 .

E6.6

Example 6.7 Determine

4

1

the forces in the members 1-7 of the truss shown in Fig. 6.16.

7

a

2

A

3D

a

a

5

6

2F a

B

a

a

30◦

F

Fig. 6.16

Solution

The support reactions follow from the equilibrium C

S1

S2

AH

AH

AV

2F

B

F

AV

S3

S2

S3

D

S5 S6

2F

conditions for the complete truss:  A:

1 2a 2F − 4a B − 5a F = 0 2 1 ↑ : AV + B − 2F + F = 0 2 √ 3 → : AH − F =0 2

3 F, 8 9 → AV = F , 8 √ 3 → AH = F. 2 →

B=

In order to determine the forces in the members 1, 2, 3 we pass an imaginary section through these members. The equilibrium conditions for the left-hand part of the truss yield √ 9  3 − − a A − a S = 0 → S = F, : a A C V H 3 3 8 2 √ 2 9√ S2 = 0 → S2 = 2F , ↑ : AV − 2 8 √ 9 2 S 2 = 0 → S1 = − F . → : AH + S1 + S3 + 2 4 By inspection we see that bar 7 is a zero-force member and that S4 = S1 . Finally, equilibrium at joint D leads to √ √ 7√ 2 2 S2 + S5 − 2F = 0 → S5 = 2F , ↑: 2 2 8 √ √  11 √3 2 2 →: S5 − S 2 + S 6 − S 3 = 0 → S6 = − F. 2 2 8 2

Gr

E6.7

6 Trusses

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

60

61

1 0 0 1 0 1 0 1 0 1 0 V 1 0 1 0 1

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

6 Trusses

Example 6.8 The truss shown in

II

Fig. 6.17 carries the forces F1 = 10 kN and F2 = 20 kN. Calculate the forces in all the members.

1

60◦

I

IV 8

4

3

5

60◦ 6

2 III

F1

7

F2

Fig. 6.17

Solution In this example the method of joints can be applied wi-

thout having to determine the support reactions. We draw the free-body diagrams for each joint, assuming that all the forces in the members are tensile forces. Beginning with the loaded joint I we write down the equilibrium conditions: I

S2 + S1 cos 60 = 0 ,

2 S1 = − √ F1 = 11.6 kN , 3

→

↓:

1 S2 = − S1 = −5.8 kN . 2 II

→ : S4 − S1 cos 60◦ + S3 cos 60◦ = 0 ,

S4 = S1 = 11.6 kN .

→ : −S2 + (S5 − S3 ) cos 60◦ + S6 = 0 , →

S4

S3

S1

S3

↑ : (S3 + S5 ) sin 60◦ − F2 = 0 ,

Gr

III

S3 = −S1 = −11.6 kN ,

S2

F1

S1 sin 60◦ + S3 sin 60◦ = 0 ,

→

60◦

I

◦

→:

II

S1

↑ : S1 sin 60◦ − F1 = 0 ,

2 S5 = √ (F1 + F2 ) = 34.6 kN , 3

S2

S5

S6

III

F2

S6 = −28.9 kN .

E6.8

6 Trusses

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

62

IV

↓ : S5 sin 60◦ + S7 sin 60◦ = 0 ,

→ : −S4 + (S7 − S5 ) cos 60◦ + S8 = 0 , →

IV

S4

S7 = −S5 = −34.6 kN ,

S5

S8

S7

S8 = 46.2 kN .

It is useful to present the results in dimensionless form in a table, including the negative signs: i

1

2

3

4

5

6

7

8

Si /kN

11.6

-5.8

-11.6

11.6

34.6

-28.9

-34.6

46.2

As a check we determine the forces in the members 6, 7, 8 with the method of sections:

Gr

↓:

S7

a sin 60◦

F2

a

 V:

S8

S6 V

F1

 IV :

IV

a/2 a/2

3 a aF1 + F2 + a sin 60◦ S6 = 0 → S6 = −28.9 kN , 2 2

2a F1 + a F2 − a sin 60◦ S8 = 0

→

S8 = 46.2 kN ,

F1 + F2 + S7 cos 30◦ = 0

→

S7 = −34.6 kN .

63

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6 Trusses

Example 6.9 A rope is

4a

guided around the smooth pin C of the truss shown in Fig. 6.18. One end of the rope is connected to a rigid wall; the other end carries a box of weight W . Calculate the force S11 in member 11.

a

2a

a

2a a

11

a

45◦

C

W

Fig. 6.18

Solution By inspection we see that bar 8 is a zero-force member.

This is why, applying the method of sections, we may divide the truss by a cut through the four members 13, 11, 8 and 9 into 9

cut

7

5 3

S9

10

12

8

W

6

13

A

4

11

W

W

S11

1

2

B

BH

S13

B

W

BV

Gr

two parts. We also cut the rope; the force in the rope is equal to the weight W of the box (smooth pin C!). Now we consider the right-hand part of the system where the zero-force member is omitted. Moment equilibrium about point B immediately yields the force in member 11: √ √  B : 4a(S11 + 2/2 W −W ) = 0 → S11 = (1 − 2/2) W .

E6.9

Example 6.10 The truss

F1

shown in Fig. 6.19 carries the forces F1 = 10 kN and F2 = 20 kN (given: a = 3 m). Calculate the forces in the members 6 and 8.

F2

4

2

5

G

1

7

6

3

8

9

a

12

a

10

A

11

a

a

a

B

Fig. 6.19

Solution The truss represents a three-hinged arch. In a first step

we calculate the support reactions at B. We isolate the two parts F1

F2

F2

4

II

5

I

S4

I

II

5

G

7

6

8

7

9

8

S6

9

A

AH

AV

F1

GH

F2

4

I

BH

II

5

BH

BV

BV

G

7

GV

8

F2

9

6

S5

I

S5

II

S4

S7

S9

S8

Gr

E6.10

6 Trusses

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

64

BH BV

65

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6 Trusses

of the truss and write down the moment equilibrium conditions about point A of the complete (frozen) system and point G of the right-hand part:  A:  G:

aF1 + 2aF2 − 3aBV + aBH = 0 ,

aF2 − 2aBV + 2aBH = 0

→

BH = BV − F2 /2 .

Solving these two equations yields BH = 10 kN ,

BV = 20 kN .

Now we pass a cut through the members 4 and 6 and apply the equilibrium conditions in the vertical and horizontal directions to the free-body diagram: √ 2/2 S6 − F2 + BV = 0 → S6 = 0 , ↑: ←:

S 4 + BH = 0

→ S4 = −10 kN .

Equilibrium in the horizontal direction at the isolated joint I leads to →:

S5 = S4 = −10 kN .

Finally, we isolate joint II and write down the equilibrium condition in the horizontal direction: √ √ ← : S5 + 2/2 S8 = 0 → S8 = − 2 S5 = 14.1 kN .

Gr

Note that the forces in the members are independent of the value of a.

Example 6.11 Determine the

z

F

support reactions and the forces in the members of the space truss shown in Fig. 6.20.

a

3

C

2

6

1

a

B

4

y

5

a

a

A

x

Fig. 6.20

Solution The truss consists of

j = 4 joints, m = 6 members and there are r = 6 support reactions. Therefore, the necessary condition for statical determinacy is satisfied:

I

F

3 IV

2

6

1

4

Cz 5

II

Ax

f = 3j − (m + r)

Ay

Az

= 12 − (6 + 6) = 0 .

The equilibrium  Fx = 0 :  Fy = 0 :  Fz = 0 :  Mx = 0 :  My = 0 :  Mz = 0 :

conditions for the complete truss Ax = 0 ,

Ay + By + F = 0 ,

Az + Bz + Cz = 0 , −a F + a Bz = 0 , a Cz − a Az = 0 , a Ay = 0

yield the support reactions: Ax = 0 ,

Gr

E6.11

6 Trusses

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66

By = −F ,

Ay = 0 ,

1 Az = − F , 2

Bz = F ,

1 Cz = − F . 2

III By

Bz

67

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6 Trusses

In order to determine the forces in the members we apply the method of joints. Note that (with the exception of bar 4) all the members have the angle 45◦ with respect to the corresponding coordinate axes. Thus, equilibrium at the joints I and II leads to I





II

Fx = 0 :

1 1 √ S1 − √ S3 = 0 , 2 2

Fy = 0 :

1 √ S2 + F = 0 , 2



1 1 1 Fz = 0 : − √ S1 − √ S2 − √ S3 = 0 , 2 2 2 √ √ 2 F , S2 = − 2 F . → S1 = S3 = 2



1 1 Fx = 0 : Ax − √ S1 − S4 − √ S5 = 0 , 2 2



1 Fy = 0 : Ay + √ S5 = 0 , 2 →

1 S4 = − F , 2

S5 = 0 .

From symmetry we find S6 = S5 = 0 .

As a check we write down the equilibrium conditions for joint IV: 



1 1 √ S 6 + S4 + √ S 3 = 0 2 2

Fy = 0 :

1 √ S6 = 0 , 2

Fz = 0 :

1 Cz + √ S3 = 0 2

Gr



Fx = 0 :

→

0−

→ −

F F + = 0, 2 2

F F + = 0. 2 2

Example 6.12 Calculate the forces in all the members of the space

truss in Fig. 6.21. C

G

11

5

10

B

9

6

1

7

F a

3a

E

a

2

3

4

8

12

A

D

a

a

x

y z

P

Fig. 6.21

Solution The truss has j = 7 joints, m = 12 members and r = 9

reaction forces. It is statically determinate: f = 3j − (r + m)

→

f = 21 − (9 + 12) = 0 .

We calculate the forces in the members with the aid of the method of joints and write down the corresponding equilibrium conditions. S3

S5

S2

45◦

45◦

45◦

S1

45◦

D P

E

S9

S2

S4

Joint D  Fx = 0 : −S1 cos 45◦ − S2 cos 45◦ − S3 cos 45◦ = 0 ,  Fy = 0 : S1 sin 45◦ − S2 sin 45◦ = 0 ,  Fz = 0 : P − S3 sin 45◦ = 0 √ √ → S3 = 2 P , S1 = S2 = − 2/2 P . Joint E  Fx = 0 : −S9 + S2 sin 45◦ = 0 ,  Fy = 0 : S4 + S5 cos 45◦ + S2 cos 45◦ = 0 ,  Fz = 0 : S5 sin 45◦ = 0

Gr

E6.12

6 Trusses

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68

→

S9 = −P/2 ,

S5 = 0 ,

S4 = P/2 .

6 Trusses

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

S11

69

S6

S8

S4

γ

Joint F  Fz = 0 :  Fx = 0 :  Fy = 0 :

α

S10

S12

S1

S7

G

F

S3

β

S6 sin 45◦ = 0

S1 sin 45◦ − S7 − S8 cos γ = 0 ,

−S1 cos 45◦ − S6 cos 45◦ − S8 sin γ − S4 = 0

→

S6 = 0 ,

S7 = −P/2 ,

S8 = 0 .

Note that these results also follow from the symmetry of the loading: S6 = S5 , S7 = S9 , S8 = 0.

Joint G We now introduce the auxiliary angles α (between member 12 and the vertical line passing through G) and β (between the projection of 12 onto the x, y-plane and the x-axis). Thus, from geometry √ √ √ √ cos α = 1/ 11 , sin α = 10/ 11 , cos β = 3/ 10 .  Using S5 = S6 = 0, the equilibrium condition Fy = 0 yields S10 = S12 . The other equilibrium conditions lead to  Fz = 0 : S3 cos 45◦ + 2 S12 cos α = 0 ,  Fx = 0 : −S11 − 2 S12 sin α cos β + S3 sin 45◦ = 0 √ → S10 = S12 = − 11/2 P , S11 = 4 P .

Gr

As a check we determine S11 from the equilibrium of the complete truss. Moment equilibrium about the axis passing through A and B yields  My = 0 : 4a P − a S11 = 0 → S11 = 4 P .

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 7

Beams, Frames, Arches

7

Example 7.13 A crab on two

W ξ wheels can move on a beam (weight negligible). Its weight B A a W is linearly distributed as indicated in Fig. 7.27. l Determine the value Fig. 7.27 ξ = ξ ∗ for which the bending moment attains its maximum value Mmax . Calculate Mmax .

Solution First we assume the crab to be at an arbitrary position.

The beam is subjected to the concentrated forces 2W/3 at x = ξ and W/3 at x = ξ + a. Equilibrium for the beam as a whole yields the support reactions: A=

W W (3l−3ξ−a), B = (3ξ+a). 3l 3l

The bending moment is a piecewise linear function of the coordinate x. The slope discontinuities and thus the maximum values of the bending moment are found at the positions of the wheels. Therefore we now imagine the beam cut at x = ξ. The stress resultants M1 and V1 are shown with their positive directions in the freebody diagram (Note that N = 0). The bending moment at x = ξ follows from the equilibrium of moments of the left-hand part: M1 (ξ) = Aξ =

W [(3l − a)ξ − 3ξ 2 ]. 3l

Similarly, the bending moment M2 at x = ξ + a is obtained as

Gr

E7.13

7 Beams, Frames, Arches

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72

W

2W/3

W/3

2W/3

W/3

A

x

z

M

M1

M2

M1

C

A

B

V1

ξ

M2

C

V2

ξ+a

B

l−ξ−a

73

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7 Beams, Frames, Arches

M2 (ξ) = B (l − ξ − a) =

W [(3l − 4a)ξ − 3ξ 2 + al − a2 ]. 3l

In order to find the maximum values of M1 and M2 as functions of ξ, we determine the derivatives: dM1 1 1  a 2 = 0 → ξ1 = (3l − a) → M1max = 3− W l, dξ 6 36 l dM2 1 1  a 2 = 0 → ξ2 = (3l − 4a) → M2max = 3−2 W l. dξ 6 36 l By inspection we see M1max > M2max

→

Mmax = M1max ,

ξ ∗ = ξ1 ,

Gr

The solution is valid only for ξ1 + a < l, that is for a < 3l/5.

Example 7.14 Determine the

x

q0

bending moment for a cantilever subjected to a sinusoidal line load (Fig. 7.28).

l

Fig. 7.28

Solution Since the shear force and the bending moment are zero

at the free end, it is convenient to use a coordinate system where the x-axis points to the left. The load is given by q(x) πx . q(x) = q0 sin l x Integration leads to z  l πx πx V (x) = − q0 sin dx = q0 cos + C1 , l π l  l 2 πx + C1 x + C2 . M (x) = q0 sin π l The constants of integration follow from the boundary conditions: V (0) = 0

→

C1 = −

q0 l , π

M (0) = 0

→

C2 = 0 .

Thus, the stress resultants are obtained as πx πx q0 l 2  x q0 l  −1 , − sin cos M (x) = − . V (x) = π l π l l V

−2q0

M

l π

−q0

l2 π

Their maximum values are found at the clamping (x = l): V (l) = −

2 q0 l , π

M (l) = −

1 q0 l 2 . π

Note that the shear force is negative since the x-axis points to the left (positive face!).

Gr

E7.14

7 Beams, Frames, Arches

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

74

M

V

x

z

75

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7 Beams, Frames, Arches

The structure in Fig. 7.29 consists of a hinged beam and five bars. It is subjected to two concentrated forces. Determine the forces in the bars and the bending moment in the beam.

Example 7.15

2F

F

2

1

a

4

3

a

a

5

a

a

Fig. 7.29

Solution First we determine the support reactions from the equi-

librium conditions for the complete structure: 5 F, 4 7 B= F, 4 AH = 0 . AV =

F

AH A AV

1

2

GH

E

C

GV

S3

2F

GV

G GH

G

S3

H

4

B

5

B

D

Then we pass an imaginary section through hinge G and bar 3. Equilibrium at the right-hand part yields the forces GH , GV and S3 : → GV = F/4 , ↑ : GV + B − 2F = 0  G : a S3 + 2a F − 2a B = 0 → S3 = 3F/2 , →:

−GH − S3 = 0

→

GH = −3F/2 .

Now we free joint C (or joint D) and write down the corresponding equilibrium conditions: √ 2 3√ S 1 + S 3 = 0 → S1 = 2F , →: − 45◦ S 2 2 2 √ S 1 2 3 S3 S1 = 0 → S2 = − F . ↑ : S2 + C 2 2 Note that we can see by inspection that S5 = S1 and S4 = S2 .

Gr

Since the beam is loaded by concentrated forces, the bending moment is piecewise linear along the axis. In order to calculate the bending moment at the point E of application of the force F , we

E7.15

7 Beams, Frames, Arches

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

76

section the beam at this point and at the hinge G. The moment equilibrium condition for the portion VE EG of the beam yields G GH ME = −a GV = −

aF . 4

ME

E

GV

Similarly we obtain the moment at the point H of application of the force 2F : MH =

aF . 4

This determines the following bending-moment diagram. aF/4

M

Gr

−aF/4

77

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

7 Beams, Frames, Arches

Example 7.16 A simply supported beam carries a linearly varying

line load as shown in Fig. 7.30. Calculate the location and the magnitude of the maximum bending moment for q1 = 2q0 .

x

q1

q0

l

Fig. 7.30

Solution The linearly varying load is described by

q(x) = a + b x .

The constants a and b follow from the values of q(x) at the boundaries: q(0) = q0

→ a = q0 ,

q(l) = q1

→ q1 = a + b l

Thus,

q(x) = q0 +

→

b=

q1 − q0 . l

q1 − q0 x. l

Integration yields

q1 − q0 x2 + C1 , l 2 x2 q1 − q0 x3 − + C1 x + C2 . M (x) = −q0 2 l 6 V (x) = −q0 x −

The constants of integration are obtained from the boundary conditions: M (0) = 0 →

C2 = 0 , q0 l q1 − q0 l2 + . M (l) = 0 → C1 = 2 l 6

Gr

Hence, with q1 = 2q0 , the stress resultants are x2 q0 x2  q0 l q0 l 2 + + − q0 x + q0 l , = −q0 V (x) = −q0 x − l 2 2 6 2l 3 x3 x2 2 − q0 + q0 lx . M (x) = −q0 6l 2 3

E7.16

7 Beams, Frames, Arches

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78

The maximum bending moment follows from the condition of a vanishing shear force (M  = V ):   7 4 − 1 = 0.53 l . V = 0 → x∗ = −l ± l2 + l2 = l 3 3 Introduction into M (x) finally yields

Gr

Mmax = M (x∗ ) = 0.19 q0 l2 .

79

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

7 Beams, Frames, Arches

Example 7.17 Draw the shear-

1 0 0 1 C

q0

force and bending-moment diagrams for the hinged beam shown in Fig. 7.31.

G1

A

x 2a

G2 B

a

a

a

Fig. 7.31

Solution First we calculate the support reactions. To this end we isolate the three parts ①, ② and ③ of the beam. The equilibrium

A z1

x1 ①

B

q0 ② x2

MC

③

R = q0 x1

C

M

z2

D

A

V q0 a

V

x1

3 qa 2 0

M

V

MC

E

M

−2q0 a

quadratic parabola

C

x2

3 q a2 2 0

2a

1 q a2 2 0

− 32 q0 a2

conditions for the portions lead to A = q0 a ,

B=

7 q0 a , 2

3 C = − q0 a , 2

MC =

3 q0 a2 . 2

Gr

Then we divide the structure into the two regions AB and BC and use the coordinates x1 and x2 , respectively. An imaginary section at an arbitrary position x1 and the equilibrium conditions for the free-body diagram yield ↓ : V (x1 ) − A + R = 0

 D:

M (x1 ) − Ax1 + R

x1 =0 2

→ V (x1 ) = q0 (a − x1 ) ,  x x2 1 1 → M (x1 ) = q0 a2 2 − 21 . 2 a a

E7.17

7 Beams, Frames, Arches

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

80

Similarly, a cut at x2 leads to

→

↑ : V (x2 ) + C = 0

 E :

M (x2 ) − MC − C(2a − x2 ) = 0 →

M (x2 ) =

V (x2 ) =

3 q0 a , 2

 3 x2 q0 a2 −1 + . 2 a

Gr

The shear force and the bending moment are displayed in diagrams.

81

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

7 Beams, Frames, Arches

The beam shown in Fig. 7.32 carries a uniformly distributed line load q0 and a couple M0 = 4q0 a2 . Draw the shear-force and bending-moment diagrams. Example 7.18

q0

M0

A

2a

S 11 00 004a 11

α

x

2a

Fig. 7.32

Solution First we calculate the support reaction AV and the ver-

tical component SV of the force S in the bar (the horizontal components AH and SH need not be known in order to determine the shear force and the benq0 M0 ding moment). The equilibriAH um conditions for the complex te beam yield AV S AV = −q0 a ,

SV = −5q0 a .

z

4q0 a

Gr

Then we introduce a coordiV nate system. The shear force is constant in the region −q0 a 0 < x < 4a; it is equal to M AV : V = −q0 a. In the region 4a < x < 8a, the shear force −2q0 a2 is a linear function of x. The quadratic parabola −6q0 a2 jump at x = 4a is equal to SV , −8q0 a2 and V = 0 at the free end. The bending moment is linear with equal slopes for 0 < x < 2a and for 2a < x < 4a; the jump at x = 2a is equal to M0 . In the region 4a < x < 8a, the bending moment is represented by a quadratic parabola. The value M (4a) = −8q0 a2 can easily be calculated from equilibrium of the corresponding right-hand portion of the beam. Finally, M (8a) = 0 and M  (8a) = 0.

E7.18

Example 7.19 Determine the distance a of hinge G from the support B (Fig. 7.33) so that the q0

magnitude of the maximum bending moment becomes minimal.

B

A

a G

l

C

l

Fig. 7.33

Solution First we calculate the support reactions and the force

in hinge G. Separating the two beams at G and employing the q0

q0

②

①

A

B

G

C

G

equilibrium conditions leads to ① ↑:  G:

A + B − G − q0 (l + a) = 0 , q0 (l + a)2 = 0, (l + a) A + a B − 2 ② ↑ : G + C − q0 (l − a) = 0 ,  q0 (l − a)2 − (l − a) C = 0 G: 2 q0 (l − a) → A=G=C= , B = q0 (l + a) . 2

Then we determine the bending moment at B. Equilibrium at the free-body diagram yields 1 q0 l2 MB = l A − = − q0 l a . 2 2

q0

A

l

MB

VB

This enables us to draw the diagrams of the stress resultants. The shear-force diagram is antisymmetrical, the bending moment is symmetrical with respect to B. The relative maxima of the bending moment are located at the distance b = (l − a)/2 from the supports A and C, respectively (shear force equal to zero!).

Gr

E7.19

7 Beams, Frames, Arches

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

82

83

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7 Beams, Frames, Arches

V

1 q (l 2 0

1 q (l 2 0

− a)

+ a)

− 12 q0 (l + a)

b

M

− 12 q0 (l − a)

1 q (l 8 0

b

− a)2

− 12 q0 la

They have the values

q0 q0 b2 = (l − a)2 . M∗ = b A − 2 8

M∗

q0

A

The magnitude of the maximum bending moment becomes minimal if the condition

b

|MB | = |M ∗ |

is satisfied. This yields the result

Gr

1 1 q0 l a = q0 (l − a)2 2 8

→

a = (3 −

√ 8)l = 0.172 l .

Example 7.20 Determine the

stress resultants for the clamped angled member shown in Fig. 7.34.

q0

A

C

y1

z1

x1

a

B

x2

b

y2

z2

Fig. 7.34

Solution First we separate the two parts of the angled member by a cut at B. In order to define the algebraic signs of the stress resultants we introduce two coordinate systems. Then q0 we pass imaginary cuts at the MB arbitrary positions x1 and x2 , y Mx 1 x1 respectively. In region ① we z1 ① use the relationships between My VB the loading and the stress reVz VB sultants. With the bounda② ry conditions Vz (0) = 0 and x2 My (0) = 0 the integration of MB y2 q = q0 yields z2

Vz = −q0 x1 ,

1 My = − q0 x21 . 2

Thus, at the corner B we obtain the values VB = Vz (a) = −q0 a ,

1 MB = My (a) = − q0 a2 . 2

In region ② we apply the equilibrium conditions to the free-body diagram:  Fz = 0 : Vz = VB = −q0 a ,  

1 q0 a2 , 2

Mx = 0 :

Mx = −MB =

My = 0 :

My = x2 VB = −q0 ax2 .

Gr

E7.20

7 Beams, Frames, Arches

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84

The other stress resultants are zero.

85

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7 Beams, Frames, Arches

It shall be noted that the support reactions can be obtained from these results: they are equal to the values of the stress resultants at x2 = b in region ②: A = −Vz (b) = q0 a ,

MxA = Mx (b) =

MyA = My (b) = −q0 ab .

q0 a2 , 2

Gr

Note that the bending moment My in region ① is transferred at the corner B to the torque Mx in region ②.

Determine the distance a of the hinge G (Fig. 7.35) so that the magnitude of the maximum bending moment becomes minimal.

Example 7.21

l

q0

a

G

l

Fig. 7.35

Solution The structure is a three-hinged arch. In order to cal-

culate the support reactions we separate the two bodies. The equilibrium conditions for GV q0 C ② GH D the structure as a whoGH le and for the right-hand ② portion GV ↑ : AV + BV = 0 ,

①

→ : q0 l − AH − BH = 0 ,  A : 12 q0 l2 − l BV = 0 ,  G : l BH − (l − a)BV = 0

③

x

AH

BH

AV

BV

lead to

BV = −AV =

1 2

q0 l ,

BH =

1 2

q0 (l − a) ,

AH =

1 2

q0 (l + a) .

The bending moment is linear in the regions ② and ③. At the points C and D it has the values MC = l AH −

1 2

q0 l2 =

1 2

MD = − 12 q0 l(l − a) .

q0 la ,

In region ① the bending moment is given by M (x) = x AH −

Gr

E7.21

7 Beams, Frames, Arches

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86

1 2

q0 x2 =

1 2

q0 [(l + a)x − x2 ] .

Its maximum value follows through differentiation: dM =0 : dx

l + a − 2x = 0

→

x∗ =

l+a 2

87

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7 Beams, Frames, Arches

→

M ∗ = M (x∗ ) =

1 8

q0 (l + a)2 .

The largest bending moments are found at C, D and x∗ . If, in turn, we equate the magnitudes of two moments, we obtain |MC | = |MD | : →

a = l/2

|MC | = |MD | = q0 l2 /4 = 0.25 q0 l2 ,

M ∗ = 9q0 l2 /32 = 0.28 q0l2 ,

|MC | = |M ∗ | : →

4la = (l + a)2 ∗

→

a=l

2

|MC | = |M | = 0.5 q0 l ,

MD = 0 ,

∗

√ (l + a)2 = 4l(l − a) → a = l( 12 − 3) √ → |M ∗ | = |MD | = (4 − 12)q0 l2 /2 = 0.268 q0l2 , √ MC = (−3 + 12)q0 l2 /2 = 0.232 q0l2 .

|M | = |MD | :

We see that the magnitude of the largest moment is a minimum if √ a = l ( 12 − 3) = 0.464 l .

Then the support reactions BH and AH are √ √ 12 − 2 4 − 12 q0 l = 0.268 q0 l , AH = q0 l = 0.732 q0 l BH = 2 2 and the corresponding M -line can be drawn: 0.232 q0l2

−0.268 q0 l2

0.268 q0l2

Gr

x∗

M - line

Example 7.22 Draw the shear-

q0

force and bending-moment diagrams for the frame shown in Fig. 7.36.

2a

A

B

a

2a

a

Fig. 7.36

Solution Due to the symmetry of the problem the support reac-

tions are

A = B = q0 a .

To obtain the stress resultants in the parts ① and ② we pass imaginary √ sections at arbitrary √ positions of these parts. Using cos α = 1/ 5 and sin α = 2/ 5, the equilibrium conditions for the free-body diagrams yield ①:

N1 = −A sin α = − √25 q0 a ,

M1

C

: V1 = A cos α = √15 q0 a ,  C : M1 = x1 A = x1 q0 a ,

②→:

A

N1

① V1 x1 q0

N2 = 0 ,

M2

C V2 ②

↑ : V2 = A − q0 x2 = q0 (a − x2 ) ,  C : M2 = (a + x2 )A − 12 q0 x22

A

= q0 (a2 + ax2 − 12 x22 ) .

a

N2

x2

The stress resultants are displayed in the following diagrams. √ q0 a/ 5

q0 a

3q0 a2 /2

2

q0 a

Gr

E7.22

7 Beams, Frames, Arches

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

88

V - line

M - line

89

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7 Beams, Frames, Arches

Example 7.23 The arch shown in

Fig. 7.37 carries a constant line load q0 . Calculate the maximum values of the normal force and the bending moment.

q0

r

Fig. 7.37

Solution First we calculate the support reactions from the equili-

brium conditions for the complete arch: AV = q0 r ,

AH = B =

1 2

q0

q0 r .

Then we introduce the dashed line and cut the arch at an arbitrary position α. The equilibrium conditions for the lefthand part yield the stress resultants: :

AH

AV

N (α) = −[AV − q0 r(1 − cos α)] cos α − AH sin α = − 12 q0 r(2 cos2 α + sin α) ,

:  C :

B

N

M

q0

V (α) = [AV − q0 r(1 − cos α)] sin α − AH cos α

C

AH

= 12 q0 r(2 cos α sin α − cos α) ,

V

α

AV M (α) = AV r(1 − cos α) − AH r sin α − 12 q0 r2 (1 − cos α)2

= 12 q0 r2 (1 − sin α − cos2 α) .

The maximum values of the normal force and the bending moment, respectively, follow from dM = 0 : (−1 + 2 sin α) cos α = 0 , dα cos α1 = 0 → α1 = π/2 sin α2 =

1 2

→

→

M (α1 ) = 0 ,

→

M (α2 ) = − 18 q0 r2 ,

α2 = π/6

Gr

dN = 0 : (−4 sin α + 1) cos α = 0 , dα cos α3 = 0 → α3 = π/2 sin α4 =

1 4

→

cos2 α4 =

→ N (α3 ) = − 12 q0 r , 15 16

→ N (α4 ) =

17 16

q0 r .

E7.23

A clamped arch in the form of a quarter-circle (weight negligible) supports a line load q0 (Fig. 7.38). Determine the stress resultants as functions of the coordinate ϕ. Example 7.24

111 000 000 111 A 000 111 000 111 000 111

r

q0

B ϕ

Fig. 7.38

Solution We section the arch at an arbitrary position ϕ and con-

sider the cut-off portion of the arch. To define the algebraic signs of the stress resultants, a local x, y, z-coordinate system is used. The shear force Vz , the torque MT , the bending moment My and the resultant R = q0 rϕ of the line load are drawn into the free-body diagram with r their positive directions. The ϕ/2 ϕ other stress resultants are zea ro; they are omitted in the R = q0 rϕ free-body diagram. The posiy M y tion of the action line of the resultant of the line load acVz ting on the cut-off portion is a sin(ϕ/2) MT given by x r − a cos(ϕ/2) ϕ 2 a = r sin . ϕ 2 The equilibrium conditions yield  Fz = 0 : Vz + q0 rϕ = 0  (0) Mx = 0 :  ϕ =0 MT + q0 rϕ r − a cos 2  (0) ϕ My = 0 : My + q0 rϕa sin = 0 2

Gr

E7.24

7 Beams, Frames, Arches

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

90

→ Vz = −q0 rϕ,

→ MT = −q0 r2 (ϕ − sin ϕ), → My = −q0 r2 (1 − cos ϕ).

7 Beams, Frames, Arches

91

x

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Draw the shear-force and bendingmoment diagrams for the hinged beam shown in Fig. 7.39. Example 7.25

q0

A

1

5

4a

B

G

2

2a

3

2a

3a

4

4a

Fig. 7.39

Solution The support reactions follow from the equilibrium for

the structure as a whole: A=

9 2

q0 a ,

B=

3 2

q0 a .

The forces in the bars connected with the beam can be taken from Example 6.6: S 1 = S4 =

15 4

q0 a ,

S2 = S3 = − 94 q0 a .

q0

A

S1

S2

S3

S4

B

V

9 qa 4 0

2 qa 4 0

3 qa 4 0

Gr

The shear force is linear with 9 a equal slopes for 0 < x < 4a 4 − 64 q0 a 3 − 74 q0 a a and 4a < x < 6a, respec2 81 q a2 tively. The value at x = 0 M 32 0 9 q a2 is given by the difference 8 0 of the support reaction A and the vertical component q0 a2 S1V = 9 q0 a/4 of the force S1 : V (0) = A − S1V = 9 q0 a/4. The jump at x = 4a is −3q0 a2 equal to the force S2 . In the regions 6a < x < 8a and 8a < x < 12a, respectively, the shear force is constant. The jump at x = 8a is equal to S3 . The bending moment is described by quadratic parabolas for 0 < x < 4a and 4a < x < 6a, respectively. The value M (4a) = q0 a2 can be calculated from equilibrium at a cut-off portion. The maximum values of M are located at the points of vanishing shear force. In the regions 6a < x < 8a and 8a < x < 12a the bending moment is represented by two straight lines. The value M (8a) = −3 q0 a2 again follows from equilibrium of a cut-off portion.

E7.25

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Gr

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 8

Work and Potential Energy

8

Example 8.11 The mechanism

shown in Fig. 8.19 is subjected to a force F and a moment M0 . The weights of the three links may be neglected. Apply the principle of virtual work to find the equilibrium position ϕ = ϕ∗ .

M0

l

ϕ

ϕ

F

l

Fig. 8.19

Solution The system has one degree of freedom. If the arbitrary position ϕ is changed by a virtual displacement δϕ, the point of application of the force F is disM0 placed. In order to determine the virtual work of F we have to δϕ consider only the vertical component δv of the displacement. With δv = lδϕ sin ϕ and the fact ϕ δϕ that F acts in the opposite diδv rection of the displacement, the lδϕ principle of virtual work δU = 0 ϕ F yields

M0 δϕ − F lδϕ sin ϕ = 0

→

(M0 − F l sin ϕ)δϕ = 0 .

Since δϕ = 0, the equilibrium position follows from M0 − F l sin ϕ = 0

→

ϕ∗ = arcsin

M0 . Fl

Equilibrium is possible only for |M0 /F l| ≤ 1.

Gr

E8.11

8 Work and Potential Energy

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94

95

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8 Work and Potential Energy

A hinged beam is subjected to a line load q0 and a concentrated force F (Fig. 8.20). Determine the support reaction B with the aid of the principle of virtual work. Example 8.12

F

q0

α

A

G

a

C

B

a

a

a

Fig. 8.20

Solution We remove the support B and replace it by the support reaction. Then the point of application of the support reaction is able to move in the vertical direction and the support reaction has to be treated as an external load. The line load is replaced by the resultant forces R1 = q0 a and R2 = q0 a, respectively, R1 = q0 a

R2 = q0 a

B

δϕ

δw1 δw2δwB δwG

F

δψ

δwF

to the left and to the right of the hinge G. We now impose a virtual displacement δwG on the system. Since only the component F sin α of the force F in the direction of the displacement has to be taken into account, the principle of virtual work reads δU = 0 :

R1 δw1 + R2 δw2 − BδwB + F sin αδwF = 0 .

The following geometrical relations between the various virtual displacements can be taken from the figure: δwG = aδϕ = 3aδψ δw1 = 12 aδϕ ,

→

δϕ = 3δψ ,

δw2 = 52 aδψ ,

δwB = 2aδψ ,

δwF = aδψ .

Gr

Introduction into the principle of virtual work yields 3 5 q0 a + q0 a − 2B + F sin α aδψ = 0 2 2 1 → B = 2q0 a + F sin α . 2

E8.12

The system in Fig. 8.21 is held by a spring (stiffness k) and a torsion spring (stiffness kT ). The forF ce in the spring and the moment in the torsion spring are k zero in the equilibrium posil tion shown in the figure. This equilibrium position is unstakT ble if the applied force F exceeds a critical value Fcrit . l l Find Fcrit . Example 8.13

Fig. 8.21

Solution We introduce the angle ϕ and consider an arbitrary position of the system. The total potential energy V is the sum of the potential energy VF of the force F , the potential energy y

F

ϕ

h

ϕ

x

ϕ

xF

Vk of the force in the spring and the potential energy VkT of the moment in the torsion spring: V = VF + Vk + VkT = F h +

1 2

k x2F +

1 2

2

kT (2ϕ)2 2

= F l cos ϕ +

1 2

k (l sin ϕ) + 12 kT (2 ϕ)

= F l cos ϕ +

1 2

k l2 sin2 ϕ + 2 kT ϕ2 .

The equilibrium positions follow from

dV = −F l sin ϕ + k l sin ϕ cos ϕ + 4 kT ϕ = 0 . dϕ

Gr

E8.13

8 Work and Potential Energy

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96

97

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8 Work and Potential Energy

In addition to the trivial position ϕ = 0 there exist further equilibrium positions which can be determined from this transcendental equation. Since we are interested only in the stability of the position ϕ = 0, we now insert this value into the second derivative of the total potential: d2 V = −F l cos ϕ + k l2 cos 2ϕ + 4 kT dϕ2  d2 V  → = −F l + k l2 + 4 kT . dϕ2 ϕ=0

The position is unstable if the second derivative is negative. Thus, the critical load follows from V  = 0:

Gr

Fcrit = k l + 4

kT . l

A hydraulic ramp is schematically depicted in Fig. 8.22. The two beams (each length l) are pinW connected at their centers M . A car (weight W ) B 30◦ stands on the ramp. M Determine the force F l ◦ 30 which has to be generaa F 30◦ ted in the hydraulic piston A and applied to the lever of length a in order to keep the 2a system in equilibrium.

Example 8.14

Fig. 8.22

Solution We impose a virtual displacement δα on the beam AB.

Due to the symmetry of the system, the ramp stays horizontal during the displacement. Thus, the vertical displaceW ments δwW of the point of δwW application of the force W δwB lδα and of the point B are equal: ◦ 30 B δwW = δwB . According to the principle of virtual work δf we have (the weight acts in δα δα the opposite direction of the F A displacement) δU = 0 :

F δf − W δwW = 0 .

The virtual displacements δwW and δf can be expressed by δα: √ 3 ◦ l δα , δf = a δα . δwW = δwB = l δα cos 30 = 2 Introduction into δU = 0 yields √ √  3 3l W l δα = 0 → F = W. Fa − 2 2a

Gr

E8.14

8 Work and Potential Energy

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

98

99

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8 Work and Potential Energy

Example 8.15 A wheel (weight W , radius r) rolls on a circular

cylinder (radius R) without sliding. It is connected to a wall by a spring (stiffness k). The spring is kept in a horizontal position by the support; the force in the spring is zero in the position shown in Fig. 8.23. Determine the equilibrium positions and investigate their stability.

ϕ

k

11111111 00000000 00000000 11111111 W

R

Fig. 8.23

Solution The system has one

degree of freedom: the position of the wheel can be described by the angle ϕ. The potential energy is the sum of the potential energies VW of the weight and Vk of the spring force:

r

(R + r) sin ϕ Fk

111111111 000000000 W 000000000 111111111 (R + r) cos ϕ ϕ

V = W (R + r) cos ϕ + 12 k(R + r)2 sin2 ϕ .

We now determine the first and the second derivative of the potential energy: V  = −W (R + r) sin ϕ + k(R + r)2 sin ϕ cos ϕ ,

V  = −W (R + r) cos ϕ + k(R + r)2 (2 cos2 ϕ − 1) . The equilibrium positions follow from V =0

→

→

sin ϕ[−W + k(R + r) cos ϕ] = 0 W ϕ1 = 0 , ϕ2,3 = ± arccos . k(R + r)

Gr

Note that the positions ϕ = ϕ2,3 = 0 exist only for W < k(R + r).

E8.15

8 Work and Potential Energy

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100

In order to investigate the type of stability of these positions, we discuss several cases: a) W < k(R + r):

V  (ϕ1 ) = (R + r)[−W + k(R + r)] > 0

→

stable ,

V  (ϕ2,3 ) = (R + r)[−W cos ϕ2,3 + k(R + r)(2 cos2 ϕ2,3 − 1)] 1 = [W 2 − k 2 (R + r)2 ] < 0 → unstable . k

b)

W > k(R + r):

V  (ϕ1 ) < 0

c)

→

unstable .

→

unstable .

W = k(R + r):

V  (ϕ1 ) = 0 ,

V  (ϕ1 ) = 0 ,

Gr

V IV (ϕ1 ) = −3k(R + r)2 < 0

8 Work and Potential Energy

101

1 0 0 1 0 01 01 1 0 1 0 1 B 0 1 0 1 0 1 0M 1 α 1 0 0A 1

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Example 8.16 The slider crank mecha-

F

nism shown in Fig. 8.24 consists of the crank AC and the connecting rod BC. Their weights can be neglected in comparison with the force F acting at B. Determine the moment M (α) which is necessary to keep the system in equilibrium at an arbitrary angle α.

l

C

r

Fig. 8.24

Solution We introduce the x, y-coordinate

system and the position xB (yB = 0) of the piston. Then we impose a virtual displacement δα on the crank AC which causes a virtual displacement δxB of the piston. The moment M acts in the opposite direction of δα and the force F is directed against the positive δxB . Thus, the principle of virtual work reads δU = −M δα − F δxB = 0 .

1 00 0 11 0 0 1 B1 01 1 0 0 1 0 1 β 1 l 0 x 0 1 0x 1 0 1 M α 1 r y 0000 11110 000 1 11 F

B

To express xB in terms of α we introduce the auxiliary angle β and use the geometric relations xB = r cos α + l cos β ,

r yB = 0 = r sin α − l sin β → sin β = sin α ,   l 2 cos β = 1 − sin β = 1 − (r/l)2 sin2 α .

The virtual displacements are obtained through differentiation: δxB = −r sin α δα − l sin β δβ ,

Gr

δyB = 0 = r cos α δα − l cos β δβ

→

δβ =

r cos α δα . l cos β

The negative signs at δxB show that it is directed downwards for positive δα and δβ.

E8.16

8 Work and Potential Energy

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102

Introducing δxB into δU = 0 and eliminating the auxiliary angle β and δβ yields  r cos α r −M δα + F r sin α δα + l sin α  δα = 0 l l 1 − (r/l)2 sin2 α  r cos α M = F r sin α 1 +  . l2 − r2 sin2 α

Gr

or

103

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

8 Work and Potential Energy

Example 8.17 Calcu-

q0

late the force S1 in member 1 of the structure in Fig. 8.25.

1

a

a

1 0 0 1 0 1 0 1

2q0

2

a

a

Fig. 8.25

First we replace the line loads by three concentrated forces as shown in 2q0 a the figure. Then we δϕ cut member 1 and subject the strucS1 2q0 a 2q0 a ture to a virtual S1 displacement. The principle of virtual δψ work reads

Solution

1 0 0 1 0 1 0 1

δU = −2 q0 a · a δϕ − S1 · a δϕ + S1 · 2 a δψ − 2 q0 a ·

a δψ = 0 2

Now we insert the geometrical relation 2a δϕ = a δψ

→

δψ = 2 δϕ

and obtain

−2 q0 a2 δϕ − S1 a δϕ + 2 a S1 2 δϕ − q0 a2 2 δϕ = 0 . This yields S1 =

4 q0 a . 3

Gr

Note that the line load acting at the lower beam may not be replaced by the resultant force of the complete line load which acts at the hinge since this force does no work under a virtual displacement.

E8.17

Example 8.18 A concentrated mass

m is attached to a circular disk (radius R, mass M ) as shown in Fig. 8.26. The disk can roll on an inclined plane (no sliding!). Determine the positions of equilibrium and investigate their stability.

r

M

m

ϕ

R

α

Fig. 8.26

Solution We choose the angle ϕ such that it is zero in the po-

sition of the system where the concentrated mass is located at the same height as the center of α x sin α the disk. Since the disk rolls without slipping, a displacement x R cos α r sin ϕ α along the inclined plane and the ϕ C angle ϕ are not independent: x

α

x = Rϕ.

The zero-level of the potential energy is chosen to be the horizontal plane passing through the point C of contact in the position ϕ = 0. Then the potential energy in an arbitrary position is given by V = M g(R cos α − x sin α) + mg(R cos α − x sin α + r sin ϕ) = (M + m)gR(cos α − ϕ sin α) + mgr sin ϕ .

The first two derivatives of the potential energy are V  = −(M + m)gR sin α + mgr cos ϕ ,

V  = −mgr sin ϕ .

The equilibrium positions follow from V =0

→

cos ϕ =

M +m R sin α =: k m r

Note that equilibrium exists only for k ≤ 1.

Gr

E8.18

8 Work and Potential Energy

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104

→

ϕ = ± arccos k .

105

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8 Work and Potential Energy

In order to investigate the type of stability of the equilibrium positions, we discuss several cases: a) k < 1:

ϕ1,2 = ± arccos k ,

 V  (ϕ1 ) = −mgr 1 − k 2 < 0  V  (ϕ2 ) = mgr 1 − k 2 > 0

→

unstable ,

→

stable .

→

point of inflection

b) k = 1:

ϕ1 = 0 ,

V  (ϕ1 ) = 0 ,

V  (ϕ1 ) = −mgr = 0

Gr

with a horizontal tangent: unstable .

Figure 8.27 shows schematically the door CD (weight W , height 2r) of a garage. It is supported by a lever BC and a spring AB (stiffness k). The spring is unstretched for α = π. The distance between the points B and M is denoted by a. Investigate the stability of the equilibrium configurations for the case a r and W r/ka2 = 3. Example 8.19

β

D

r

r

B

a

W

M

r

r

α

r

C

A

Fig. 8.27

Solution We first determine the elongation f of the spring during

a displacement from the position α = π. With the length r − a of the unstretched spring and the aid of the law B of cosines we can write down the relation a (r − a + f )2 = r2 + a2 + 2ar cos α .

Since a r and thus also f r this relation is approximated by

M π−α

r−a+f

r

f = a (1 + cos α) .

A

We now choose the zero-level of the potential energy of the force W at the level of point M . Then the total potential energy is given by V = W (r sin β − r cos α) +

1 2

k f2 .

Using the geometrical relation 2r sin β − r cos α = r

→

sin β = 12 (1 + cos α)

the potential energy can be written as

Gr

E8.19

8 Work and Potential Energy

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106

V = 12 W r(1 − cos α) + 12 ka2 (1 + cos α)2 = V (α) .

8 Work and Potential Energy

107

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Equilibrium positions follow from (note: W r/ka2 = 3)

V  = 12 W r sin α − ka2 (1 + cos α) sin α = ka2 sin α ( 12 − cos α) = 0 . Hence,

sin α = 0

1 2

cos α =

→ α1 = 0 ,

→

α2 = π ,

α3 = π/3 .

To apply the stability criterion, we consider the second derivative of the potential: V  =

1 2

W r cos α − k a2 (cos α + cos 2α) .

This yields

→

unstable ,

V  (α2 ) < 0

→

unstable ,

V  (α3 ) > 0

→

stable .

Gr

V  (α1 ) < 0

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

Gr Chapter 9

Static and Kinetic Friction

9

Example 9.6 A sphere (weight W1 ) and a wedge (weight W2 ) are

jammed between two vertical walls with rough surfaces (Fig. 9.11). The coefficient of static friction between the sphere and the left wall and between the wedge and the right wall, respectively, is μ0 . The inclined surface O of the wedge is smooth. Determine the required value of μ0 in order to keep the system in equilibrium.

1 0 0 1 0 1 O 0 1 0 1 0 1 0 1 W1 0 1 0 1 0 1 W2 0 1 0 1 0μ 1

μ0

0

1 0 0 1 0 1 0 1 0 1 0 1 0 1 α 0 1 0 1 0 1 0 1 0 1

Fig. 9.11

First we draw the free-body diagrams of the isolated

Solution

Nl

Hr

Nl

C

α

Hl

W1

Nr

N

Hl

W1

W2

sphere and of the complete system. The force Hr of static friction is directed such that it opposes the direction of motion that would occur in the absence of friction. The force Hl is zero as can be seen from the equilibrium conditions at the sphere:  C:

Hl = 0 ,

↑ : N sin α − W1 = 0 ,

→:

Nl − N cos α = 0

→

Nl = W1 cot α .

The normal force Nr and the friction force Hr follow from equilibrium of the complete system: → : Nl − Nr = 0 ↑:

Hr − W1 − W2 = 0

→

Nr = Nl = W1 cot α ,

→

Hr = W1 + W2 .

They have to satisfy the condition of static friction (the corresponding condition at the left-hand side is automatically satisfied since Hl = 0):

Gr

E9.6

9 Static and Kinetic Friction

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110

Hr ≤ μ0 Nr

→

μ0 ≥ (1 + W2 /W1 ) tan α .

111

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9 Static and Kinetic Friction

Example 9.7 The excentric device in Fig. 9.12 is used to exert a

large normal force onto the base. The applied force F , desired normal force N , coefficient of static friction μ0 , length l, radius r and angle α are given. Calculate the required eccentricity e.

F

l

1 0

α

r

e μ0 111111111 000000000 000000000 111111111 000000000 111111111

Fig. 9.12

Solution First we isolate the device and draw the free-body diagram. The static friction force H is directed such that it prevents F

AV

AH

A

C

B

H

A

e α e cos α

C

e sin α

N

a motion. Then we write down the equilibrium conditions: →:

AH + H + F sin α = 0 ,

↑:  C:

−AV + N − F cos α = 0 ,

F (l − e) − AH e sin α − AV e cos α − H r = 0 .

If we eliminate the support reactions AH and AV we obtain the friction force: H=

F l − N e cos α . r − e sin α

Gr

Introduction into the condition of static friction (the normal force N is given) |H| < μ0 N

E9.7

9 Static and Kinetic Friction

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112

yields

F l − N e cos α < μ0 N (r − e sin α) .

This can be solved for the required eccentricity: F − μ0 r N e> . cos α − μ0 sin α

Gr

l

113

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9 Static and Kinetic Friction

Example 9.8 A horizontal force F

F

is exerted on a vertical lever to prevent a load (weight W ) from falling downwards (Fig. 9.13). The drum can rotate without friction about point B; the coefficient of static friction between the drum and the block is μ0 . Determine the magnitude of the force F needed to prevent the drum from rotating.

μ0

a

B

b

A

c

Fig. 9.13

r

W

Solution We separate the lever and the drum. The static fric-

tion force H opposes the direction of the motion that would occur in the absence of friction (here: a clockwise rotation of the drum). The friction force and the normal force N can be calculated from moment equations at the drum ① and at the lever ②, respectively:

F

②

①

H

N

B

N

H

A

 ①B :

rH −rW = 0

→

H =W,

 ② A:

b N + c H − (a + b)F = 0

→

N=

W

1 [(a + b)F − c W ] . b

A rotation of the drum is prevented as long as the condition of static friction is fulfilled:

Gr

H ≤ μ0 N

→

F ≥

b + μ0 c W. μ0 (a + b)

E9.8

Example 9.9 A wall and a beam (weight W2 = W ) keep a roller

(weight W1 = 3W ) in the position as shown in Fig. 9.14. The beam adheres to the rough base; all the other areas of contact are smooth. Determine the minimum value of the coefficient of static friction μ0 between the base and the beam in order to prevent slipping.

11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 5a 4

W1

We isolate the roller and the beam. Since the surface of the roller is smooth, the forces acting at the roller are a system of concurrent forces. Equilibrium yields

μ0

Fig. 9.14

Solution

↑:

D cos 60◦ − W1 = 0

30◦

W2

a

B

60◦

C

W1

B

D

W2

D

H

→

D = 6W .

N

The normal force N at the base of the beam and the friction force H follow from the equilibrium conditions at the beam: N − D cos 60◦ − W2 = 0 →

↑:

 B :

aN +

√ a 5 3 a H − a D − W2 = 0 → 4 2

N = 4W , H=

4√ 3W . 3

They are inserted into the condition of static friction in order to determine the required coefficient of static friction: √ 3 H ≤ μ0 N → μ0 ≥ . 3

Gr

E9.9

9 Static and Kinetic Friction

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

114

115

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9 Static and Kinetic Friction

A block (weight W2 ) is clamped between two cylinders (each weight W1 ) as μ0 shown in Fig. 9.15. All the surμ0 faces are rough (coefficient of W1 static friction μ0 ). α α Find the maximum value of W2 W2 in order to prevent slipping. Fig. 9.15 Example 9.10

Solution We draw the free-body diagram of the block and of one

of the cylinders. The directions of the forces H1 and H2 are chosen such that they oppose the direction of motion that would occur in the absence of friction. The equilibrium conditions for the block ↑:

2H2 − W2 = 0

and for the cylinder ↑:

N2

N2

H2

N1

→ : N1 sin α + H1 cos α − N2 = 0 ,  A : H2 r − H1 r = 0

yield the normal forces and the friction forces: W2 , 2 W2 (1 + sin α) + 2W1 N1 = , 2 cos α W2 (1 + sin α) + 2W1 sin α N2 = . 2 cos α

H 1 = H2 =

Gr

Introduction into the conditions of static friction H2 < μ 0 N 2

H2

W1

H1 α

N1 cos α − H2 − H1 sin α − W1 = 0 ,

H 1 < μ0 N 1 ,

W2

H2

A r

N2

E9.10

9 Static and Kinetic Friction

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116

leads to W2


H 1 √ = 0.268 = NA 2+ 3

and between the lower cylinders and the base μ0 >

H 1 √ = 0.089 = NC 3(2 + 3)

have to be satisfied in order to avoid slipping. Thus, the minimum value of μ0 is obtained as

Gr

μ0 = 0.268 .

Example 9.14 A rotating drum (weight W1 ) exerts a normal force

and a kinetic friction force on a wedge (Fig. 9.19). The wedge lies on a rough base (coefficient of static friction μ0 ). Find the value of the coefficient of kinetic friction μ between the drum and the wedge that is required to move the wedge to the right.

W1

W

μ

α

μ0

Fig. 9.19

Solution The center of gravity of the

drum is not moving: it is in equilibrium. Therefore we can apply the equilibrium conditions →:

N1 sin α − R1 cos α − A = 0 ,

↑:

N1 cos α + R1 sin α − W1 = 0

A

W1

R1

to the drum. The normal force N1 and the friction force R1 follow with the law of friction

N1

W

H2

R1 = μN1

α

N1

R1

N2

as

N1 =

W1 , cos α + μ sin α

R1 = μ

W1 . cos α + μ sin α

Introduction into the equilibrium conditions for the wedge → : R1 cos α − N1 sin α − H2 = 0 , ↑:

N2 − N1 cos α − R1 sin α − W = 0

yields

Gr

E9.14

9 Static and Kinetic Friction

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122

H2 = W1

μ cos α − sin α , cos α + μ sin α

N2 = W1 + W .

123

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9 Static and Kinetic Friction

The wedge is on the verge of moving if the condition H 2 = μ0 N 2

is satisfied. Solving for μ yields μ=

μ0 (1 + W/W1 ) + tan α . 1 − μ0 (1 + W/W1 ) tan α

Gr

Note that the system is self-locking for μ0 > cot α/(1 + W/W1 ) . In this case the wedge will not move. If α = 0, the result simplifies to μ = μ0 (1 + W/W1 ).

Example 9.15 A beam (length 2a, weight W ) rests on support

A. The triangle attached to its right end touches a rotating drum (Fig. 9.20). The coefficient of static friction μ0 at A and the coefficient of kinetic friction μ at B are given. a) Calculate the maximum allowable value of x in order to prevent slipping at A. b) Determine the necessary value of μ0 so that the beam does not slip for arbitrary values of x (0 ≤ x ≤ a).

1 0 0 1 0 1 0 1

0 1 1 0 0 1 a 1 0 0 1 0 1 0 1 0 1 0 1 0 1 W 0 1 0 1 0 1 0μ 0 B1 1 11 μ0 00 b x

2a

A

11 00 00 11

Fig. 9.20

Solution a) First we draw the free-body diagram of the isolated

beam. Due to the rotating drum we have a kinetic friction force RB at point B, whereas a static friction force HA acts at point A. The three equilibrium conditions 0 1 ↑:

NA + N B − W = 0 ,

→ : −HA + RB = 0 ,  A : −(a + x)NB − b RB + x W = 0

0 1 000 111 A 0 1 0 1 N 0 1 0A 1

HA

1 0 0 1 0 1 W 0 1

B 111 000 0 RB 1 0 1 0 1 N 0 1 0B 1

and the law of friction RB = μNB

yield the four unknowns NA , NB , HA and RB for a given value of x: NA =

a + μb W, a + x + μb

NB =

x W, a + x + μb

RB = HA = μ NB .

In order to prevent slipping at point A the condition of static friction

Gr

E9.15

9 Static and Kinetic Friction

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

124

H A ≤ μ0 N A

125

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

9 Static and Kinetic Friction

has to be satisfied. This leads to the maximum allowable value of x: a +b . → xmax = μ0 μ x ≤ μ0 (a + μ b) μ

Gr

b) The beam does not slip if the inequality above is satisfied for any x ≤ a. The left-hand side becomes a maximum for x = a. Hence, the minimum necessary value of μ0 is given by μa μ a ≤ μ0 (a + μ b) → μ0min = . a+ μb

The rotating drum in Fig. 9.21 is encircled by a break band that is tightened by the applied force F . The coeffiμ cient of kinetic friction between the drum and the band is μ. r Calculate the magnitude of F F that is necessary to induce a given breaking moment MB if the rotation of the drum is A l clockwise (c) and if it is counterclockwise (cc). Fig. 9.21

Example 9.16

First we draw the free-body diagram. Then we write down the moment equilibrium condition for the lever:

Solution

 A:

−S2 2 r + F l = 0 .

This yields S2 = F

S2

S1

l . 2r

If the rotation of the drum is clockwise, S1 > S2 is valid and the formula for belt friction is given by

S1

S2

A

A

S1 = S2 e μπ .

Thus, the breaking moment becomes

MB = S1 r − S2 r = S2 r (e μπ − 1) .

Introduction of S2 leads to the necessary breaking force:

Gr

E9.16

9 Static and Kinetic Friction

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

126

Fc =

2MB . l (e μπ − 1)

F

127

os s, H En aug gin er ee , Sc rin g M hröd Sp ec er, rin han Wa ge r 2 ics 1 ll, Ra 01 2 , Sta japak tics se

9 Static and Kinetic Friction

If the rotation of the drum is counterclockwise we have S2 > S1 and S2 = S1 e μπ .

Then the breaking moment is

MB = S2 r − S1 r = S2 r (1 − e−μπ )

and the necessary breaking force becomes Fcc =

2MB e μπ . l (e μπ − 1)

Gr

Note that e μπ > 1. Therefore, in order to generate the same breaking moment, Fcc > Fc .