Manufacturing Technology: Materials, Processes, and Equipment (Solutions, Instructor Solution Manual) First Edition [1 ed.] 9781439810989, 9781439810859, 1439810850

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SOLUTIONS MANUAL FOR

Manufacturing Technology: Materials, Processes, and Equipment

by Helmi A. Youssef Hassan A. El-Hofy Mahmoud H. Ahmed

SOLUTIONS MANUAL FOR Manufacturing Technology: Materials, Processes, and Equipment

by Helmi A. Youssef Hassan A. El-Hofy Mahmoud H. Ahmed

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2012 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper Version Date: 20110722 International Standard Book Number: 978-1-4398-1098-9 (Paperback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Table of Contents Chapter

Title

Page

2

Properties of Engineering Materials

2

3

Structure of Metals and Alloys

4

7

Casting of Metallic Materials

6

8

Fundamentals of Metal Forming

8

9

Bulk Forming of Metallic Materials

13

10

Sheet Metal Forming Processes

18

11

High Velocity Forming and High Energy Rate Forming

22

13

Processing of Polymeric Materials

24

14

Composite Materials and their Fabrication Processes

25

15

Fundamentals of Traditional Machining Processes

26

16

Machine Tools for Traditional Machining

32

17

Fundamentals Nontraditional Machining

35

18

Numerical Control of Machine Tools

37

19

Industrial Robots and Hexapods

41

20

Surface Technology

42

21

Joining Processes

46

24

Quality Control

50

1

Chapter 2 Properties of Engineering Materials 1- What is the stress in a wire whose diameter is 1 mm, which supports a load of 100 Newton Solution: 

load 100 400 1000 2  127 MPa   2 area   1     4  1000   127 N / mm 2

2- Suppose that the wire in the above problem is cupper and is 30 m long. How much will it stretch? Repeat for a steel and aluminum wires Solution: Cu wire:

 127000000  E 10x1010 Total elongation for copper wire ( l cu ) 

Since



l l

l cu  xl 

127000000 x30  0.038 m =38 mm 10x1010

Steel wire:



 127000000  E 21x1010

Total elongation for copper wire ( l st ) Since



l l

l st  xl 

127000000 x30  0.018 m =18 mm 21x1010

Aluminum wire:

 127000000  E 7 x1010 Total elongation for copper wire ( l al ) 

2

Since



l l

127000000 x30  0.054 m =54 mm 7 x1010 Comment: Steel is stiffer than both cupper and aluminum. It is exactly 3 times more stiff than aluminum l al  xl 

3- Estimate the HB hardness of an alloy steel of ultimate tensile strength TS=1500 MPa using equation 2.9. Use the hardness conversion table to determine the HRC hardness. Solution: TS (MPa) = 3.5 HB l500 HB   428 3.5 Using the conversion table 2.3, the corresponding HRC of that alloy steel = 46 Referring to Figure 2.5 the hardness values are realized.

4- A constructional component made of Al-alloy can be represented as a bar of diameter 20 mm and length 400 mm is subjected to pure tension; calculate: a- Extension of the bar if the imposed load= 8 kN b-The load at which the bar suffers permanent deformation c- Maximum load the bar can withstand fracture

Assume the following mechanical properties: E=70 GPa, YS=496 MPa, TS=560 MPa Solution:

a-

Ao 

x 20 2  314 4

Im posed tensile strength 

mm 2

80000  255 314  255

N/mm 2 MPa

It is thus less than the YS, which means that the deformation will be purly elastic. b-

YS   0.2  496

N/mm 2

Fy   0.2 xA o  496 x 314  156

c-

kN

Fmax  TSxA o  560 x 314  176 3

kN

Chapter 3 Structure of Metals and Alloys 1-Determine the weight density of Cu in g/cm3, provided that: Atomic weight of copper =63.54 0

Atomic radius of Cu Avogadro's number

=1.278 A =6.02x1023

Solution: Cu has fcc lattice form and the number of atoms per unit cell = 4 atoms

acu  4r / 2 0

 3.61 A  3.61x10 -8

cm

a 3 cu  47.21x10 -24 cm 3 4 (atoms) x (63.54/A o ) D cu  a 3 cu 4x63.54   8.94 47.2x10 -24 x 6.02 x10 23

g/cm 3

2- Ti exhibits an hcp crystal structure, and atomic weight of 47.9. if 0

ahcp=2r, r=1.475 A . Calculate the weight density of Ti. Solution: 0

r=1.475 A =ahcp/2 Volume =area of the basal plane x c 3 Volume  3 a 2 hcp x 1.588 a hcp 2  1.5 3 x 1.588 a 3 hcp Number of atoms per unit cell = 6

D Ti 

6 (atoms) x (47.9/6.02x10 23 ) 1.5 3 x1.588x (2x1.475x10 8 ) 3

 4.51

g/cm 3

4

c=1.588 ahcp,

3- The phase diagram of Ag-Cu system is given (CE=71.9 %, TE=780 oC). For an alloy of Co=82 % Ag, determine a-At 900 oC, what are the quantities of Ag and Cu in existence? b-At 800 oC, what phases exist? c-At 400 oC what phases exist? d- At 400 oC what is the amount of eutectic constituent? e- At 400 oC what individual phases coexist? Solution: a- At 900 oC, the alloy (82 % Ag) is in the liquid region. Therefore one-phase exists and consisting of 82 % Ag- 18 % Cu

b- At 800 oC, two phases coexist:  solid solution (silver rich) plus liquid. They are determined by the lever rule



82  78 x100  28.6 92  78

%

Thus the liquid = 71.4 % c- At 400 C, the phases in existence consist of  solid solution and  solid solution , o

regardless of how they are arranged in the microstructure. d- At 400 oC, the eutectic is determined by the lever rule

Eutictic 

98  82 16 x100   61.3 98  71.9 26.1

%

e- At 400 oC, the two phases existing are  and  solid solutions



98  82 x100  16.7 98  2

%

Thus

  83.3

%

5

Chapter 7 Casting of Metals 1. The requirement is to produce a flat CI slab 250 x 200 x 100 mm by sand casting. The sprue is made 400 mm high with conical shape (as shown in Fig. 7.2) with a diameter 40 mm at mid height. Calculate the following: a. The sprue diameter, and the velocity of metal flow at the parting line. b. The rate of metal flow. c. The time for filling the mold cavity Solution:

a Using equation 7.3 : 2 Considering the mid-height to be position 1 and the bottom of the sprue to be position 2:

Using Eq. 7.1

b. c. Using Eq 7.4

v1= √2 9.81 0.2 = 1.98 m/s v2 = √2 9.81 0.4 = 2.8 m/s Q = A1 v1 = A2 v2 d12 v1 = d22v2 (0.04)2 x 1.98 = d22 x 2.8 d2 = 33.64 mm Q = π (0.04)2 x 1.98 / 4 = 0.002488 mm3/s

Volume of mold cavity V = 0.250 x 0.200 x 0.100 = 0.005 mm3 Tmf = V / Q = 0.005 / 0.002488 = 2.01 s

2. If you have three cylindrical castings of identical volume, with height/diameter ratios of 1, 2 & 4 respectively. Calculate which solidifies faster, and comment on the results. Solution:

Applying Chvorinov’s rule: ts = C (V/A)2 Cylinder volume = πd2 h/4 Cylinder surface Area = πdh + 2 πd2/4 Since all three cylinders have the same volume, then we can assume V = 1 For the first cylinder, h = d, then

6

V = πd2 d/4 = π d3/4 = 1 d = (4/π)1/3 A1= π d2+ π d2/2 = 1.5 π d2 = 1.5 π (4/π)2/3 = 5.54 For the second cylinder, h = 2d, then V = πd2 2d/4 = π d3/2 = 1 d = (2/π)1/3 A2= 2π d2+ π d2/2 = 2.5 π d2 =2.5 π (2/π)2/3 = 5.81 For the third cylinder, h = 4d, then V = πd2 4d/4 = π d3 = 1 d = (1/π)1/3 A3= 4π d2+ π d2/2 = 4.5 π d2 = 4.5 π (1/π)2/3 = 6.59 Substituting V = 1 in Chvorinov’s rule, then: ts1 = C / (A1)2 = 0.03248 C ts2 = C / (A2)2 = 0.0296 C ts3 = C / (A3)2 = 0.023 C The first specimen takes longer time than the second to solidify, and the second takes longer time than the third to solidify. Then increasing the surface area of the cylinder for the same volume leads to faster solidification time.

7

Chapter 8 Fundamentals of Metal Forming 1- A bar of 200 mm initial length is elongated by drawing to a length of 380 mm in three steps. The lengths after each step are 260, 320 and 380 mm respectively: a. Calculate the engineering strain for each step separately, and compare the sum with the overall engineering strain. b. repeat calculations using the true strain and comment on the results Solution: Applying the equations of engineering strain and true strain as given in the table, the following values are obtained: ∆

Change in length

0.30 0.23 0.19 0.72 0.90

200 260 mm 260 320 mm 320 380 mm sum Overall200 380 mm

0.26 0.21 0.17 0.64 0.64

The sum of the true strains is equal to the overall true strain, while the sum of the engineering strains is different from the overall engineering strain. Therefore true strains should be applied for large strains.

2- A mild steel specimen of diameter 12.5 mm and length 50 mm is subjected to a tensile test at room temperature. The loads and lengths are recorded as follows: Load, kN

31

35.2

36.4

37.4

40

44.5

49

Length, mm

50.1

50.4

50.7

52

52.3

53

55

51.9 Necking 64.2

41.8 Fracture 70.3

a. Plot the true stress-strain power relation on a log-log graph, and determine K & n for the material. b. Determine the stress required to deform the material to a plastic strain 0.4. c. On the same graph, plot the relation 315 . representing annealed copper, and the relation 960 . representing stainless steel 410.

8

Solution: .a: Since the volume is constant, then during the uniform deformation AoLo = AL,







The natural strain is calculated from Eq. 8.6. Substituting Lo = 50 mm, and A o= ,



.

122.7

253 0.002

, then the true stresses and true strains are calculated as:

289 0.008

301 0.014

320 0.039

341.1 0.045

384.6 0.058

436.9 0.092

534.8 0.250

Since the power law is reliable for plastic strains appreciably larger than elastic strains, hence the first three points are excluded. The log-log graph is plotted as:

Stainless  steel Mild steel Mild St l

Annealed  Copper

The slope of the line gives n = 0.27, and extrapolating the line to = 1, gives K = 798. b. The power relation for the true stress-strain curve for the M. St. specimen is then written as: σ 798 . Substituting = 0.4 gives 623.1 798 0.4 . c. Plotting the power relations representing annealed copper, and SS 410, gives the lines shown in the graph.

9

3- During a tensile test using a metal that obeys the power law, the tensile strength is found to be 340 MPa. Reaching the maximum load required an elongation of 30 %. Find K and n, and calculate the % elongation when the stress is 300 MPa. Solution: At maximum load n = = 0.3, then 340 0.3 K = 487.9 487.9 . When the stress is 300 MPa, then 300 487.9 ε = 0.198

.

MPa .

4- Consider a stress state where σx = 70 MPa, σy = 35, τxy = 20 MPa, and σz = τyz = τxz = 0 . Calculate the principal stresses. Solution: Given: σx = 70 MPa, σy = 35, τxy = 20 MPa, and σz = τyz = τxz = 0, All stresses in the z direction are zero, then the problem is two dimensional. I1 = σx + σy = 70 + 35 = 105 MPa I2 = σx σy - τxy2 = 70 x 35 – (20)2 = 2450 – 400 = 2050 I3 = 0

The stress equation is 105 2050 0 The roots of the equation are: σ1 = 79.08 MPa, σ2 = 25.92 MPa

The maximum shear stress τmax =





.

.

26.58

The maximum shear stress is in a direction 45º to the principal planes.

5-At a point in a piece of elastic material, direct stresses of 90 MN/m2 tensile, and 50 MN/m2 compressive are applied on mutually perpendicular planes. The planes are also subjected to unknown shear stress. If the maximum principal stress is limited to 100 MN/m2 (tensile), calculate the value of the shear stress, the minimum principal stress, and maximum shear stress. Solution: Given: σx = 90 MPa , σy = -50 MPa ,



From Eq. 8.16: 100





4900

σ1 = 100 MPa







= 100 – 20 = 80

10



τxy = 38.7 MPa













√70

38.7

= -60 MPa τmax = √70

38.7

80

6- A Copper sheet 500 mm long, 100 mm wide and 2 mm thick is pulled in the length direction under tensile force of 200 N, and pressure P across the width on the thickness. The width dimension will not be changed. What value of P is needed to just cause yielding according to both Tresca and Von Mises criteria if the yield stress of the Copper is 150 MPa ? Solution: Since the width does not change under loading, then the problem is plane strain.

σ1 =



1

,

σ3 = P (compressive)

According to Tresca criterion (Eq.8.28): 150 1 - P = 150 , then P = 1 – 150 = -149 MPa Due to plane strain, Eq.8.26 is applied, using ν = 0.5 when the material yields 0.5 According to Von Mises (Eq. 8.29):

√ √













1

173.2







150 √





150

Then P = 172.2 MPa

Notice that P as calculated from Von Mises criterion is higher than P as calculated from Tresca Criterion by 15 % (maximum difference between both theories) due to the plane strain condition.

7- In sheet metal stretching of a steel strip, a grid of circles 5 mm diameter is printed on the sheet surface. One of these circles changed after stretching into an ellipse whose major and minor diameters are 6.5 & 5.5 mm respectively. The stresses in the sheet plane are such that  1   2 , and the final value of σ2 is 300 MPa. Determine the strain components, the maximum principal stress, and the flow stress.

11

Solution: If the original circle diameter = d0, the major and minor diameters are dmaj, and dmin, then the plastic strain increments are calculated as:.

ε1 = ln (dmaj / do) = ln (6.5/5) = 0.26 ε2 = ln (dmin / do) = ln (5.5/5) = 0.095 Due to constant volume in plastic forming: ε1 + ε2 + ε3 = 0 ε3 = - 0.26 – 0.95 = 0.355 The stresses in the sheet plane are: 300 , 0 ,

d 1

From Eq. 8.29:





' 1

d 2



' 2



.









1

From which:





d 3

= =







σ1 = 410









 d

 3'

.







.

.

MPa

From Eq. 8.29 : √

=

√



410

300

300

0

0

410

= 367.6 MPa

8- During the tension test of a Medium Carbon Steel specimen, the strain rate is suddenly increased by a factor of 8. This has led to a corresponding rise of the stress level by 1.8 %. What is the strain rate sensitivity exponent for this material? Solution: Strain rate rises by a factor of 8 , then / 8 Corresponding rise of the stress by 1.8 %, then / From Eq.8.32: / / / m (8) = 1.018 m = 8.58 x 10-3

12



1.018 1.018

Chapter 9 Bulk Forming 1. Plane strain compression is performed on a slab of metal having a yield stress 280 MPa. The width of the slab is 200 mm, while its height is 25 mm. Assuming an average coefficient of friction 0.10, calculate the following:: a) The maximum pressure at the onset of plastic flow, b) The average pressure at the onset of plastic flow, c) Repeat calculation if sticking friction prevailed at each interface, and compare results. Solution: a) At the onset of plastic deformation, σf = 280 MPa ,

Applying equation 9.2,

p



' f

=

1.15 x 280 MPa

 2  a   exp    x    h 2

The pressure is maximum when x = 0, then

p max



' f

 2   a   exp     h  2 

pmax = 280x 1.15 exp( 2 x 0.1x 0.1/ 2 x 0.025) = 480.37 MPa

b) Pressure is minimum when x = a/2, i.e. pmin = = 322 MPa pav =(pmax + pmin)/2 = 401.19 MPa

a ) 2h = 322 ( 1 + (0.1/2 x 0.025)) = 966 MPa The minimum pressure is the same pmin =322 MPa

c) At sticking, equation 9.7 is applied, p max   'f (1 

Then

pav = (966+322)/2 = 644 MPa

At sticking friction, the average pressure is increased by about 60 % And the maximum pressure is increased by about 100 %

13

2. A cylindrical specimen made of annealed 4135steel (K =1015 MPa, n= 0.17) has a diameter of 150mm, and is 100 mm high. It is upset between flat dies to a height of 50 mm at room temperature. Assuming that the coefficient of friction is 0.2, calculate the force required at the end of the stroke. Solution: The volume is the same before and after upsetting, then A0 h0 = Af hf (150)2 x 100 = (Df)2 x 50 Df = 212.13 mm The flow stress is calculated using the power relation (Eq. 8.8) , where ln / ε = ln (50/100) = 0.693 (negative sign due to compression is ignored) = 1015 (0.693)0.17 = 953.65 MPa

2r ) 3h pav = 953.65 ( 1 + (2x0.2 x 0.106/3x 0.05)) = 1223.2 MPa F = pav x Af = 1223.2 x (0.212)2/4 = 43.18 MN

Applying equation 9.11,

pav   f (1 

3. It is expected that in strip rolling, the rolls will begin to slip if the back tension Pb is too high. Derive an expression for the magnitude of back tension required to make the rolls begin to slip. Solution:

Equation 9.27: p  (

' f

 pb )

h  (H0 H ) e h0

At the exit of the deformation zone: p = σ'f , h = hf , φ = 0, H = 0 h f H 0 e Then:  ' f  ( 'f  Pb ) h0 ′



′

′ 1





4. A 300 mm wide mild steel strip is hot rolled at a temperature 900°C (C = 165 MPa, m = 0.08) from a thickness of 12 mm to 8 mm using 400 diameter rolls rotating at 200 rpm. Assuming a coefficient of friction 0.2, calculate: a) The biting angle, and contact length, b) The forward slip, and maximum possible reduction, c) The approximate roll force and torque, and d) The horsepower for the mill.

14

Solution:

a) Using Eq. 9.19,

h = 1 – (4 / 400) = 0.99 D α = 8.1°

cos   1 

l

Usung Eq. 9.20,

D D sin   h. 2 2

l  4.

b) Applying Eq.7.23,

400  28.28 mm 2

V f  Vr

Forward slip =

Vr

Vr = π D N / 1000 = π x 400 x 200 / 1000 = 251.33 m/min Assume that hn= (h0+hf) / 2 = (12 + 8) / 2 = 10 mm 251.33

314.16

m/min.

Maximum possible reduction is calculated from Eq.9.22, hmax   2 .  hmax  ( 0 .2) 2 x 400 / 2 = 8

D 2

mm

c) Under hot working conditions, Eq. 9.34 is applied: h V  .  r ln( 0 ) = ln (12/8) x 251.33 / 28.28 = 3.6 s-1 l hf Sheet rolling is under plane strain conditions, then: 1.15 1.15 = 210.23 MPa

Applying Eq. 9.30:

. .

1

.

= 1.15 x 165 (3.6)0.08



= 0.028 x 0.3 x 210.23 (1 + (0.2 x 0.028/2 x 0.01)) = 2.26 MN Applying Eq. 9.32,

T = F l/2 = 2.26 x103 x 0.028 / 2 = 31.64 kN.m

d) Applying Eq. 9.33, Power/roll = Then:

 .F .l.N 60,000

kW

mill power = 2 x π x 2.26 x 106x 0.028 x 200 / 60,000 = 1325.3 kW = 1.325 MW

15

5. A copper billet 125 mm diameter and 250 mm long is extruded to a diameter 50 mm at 800 ˚C at a speed of 0.25 m/s. Using square dies and assuming high coefficient of friction 0.24 (dead metal zone at 45˚), estimate the percentage reduction, extrusion ratio, and the force required for this operation. For Copper at the given temperature, C= 130 & m=0.06. Solution

% reduction = (A0-Af) / A0 = (D0)2- (Df)2 / (D0)2 = (1252 - 502) / 502 = 0.84 = 84% From Eq. 9.35,

R = A0/Af = 1252/ 502 = 6.25

Using Eq. 9.40

. 

6V0 ln R D0 s-1

= 6 x 0.25 ln (6.25) / 0.125 = 22

Applying Eq. 9.38,

Then



= 130 (22)0.06 = 156.5

.

ln

F = p A0 = A0

1

1

ln

1

MPa

1

Considering the length to be discarded for a dead zone 45˚, l = (D0 - Df)/ 2 = (125 – 50)/2 = 37.5 mm The length to be extruded L = 250 – 37.5 = 212.5 mm F = 0.125 156.5 ln 6.25 = 10.1

MN

16

1

.

. .

1

6. A round rod of annealed 302 stainless steel (K=1300 MPa & n=0.3) is being drawn from a diameter of 10 mm to a diameter of 8 mm at a speed 0.5 m/s, using a semi die angle 8˚. Calculate the percentage reduction, the applied force due to ideal deformation, friction, and inhomogeneous deformation. Assume coefficient of friction 0.1. Solution:

Applying Eq. 9.41,

 1  p   f .  1   f . ln   1 r  r = (A0-Af) / A0 = (102-82)/102 = 36 %

ln

0.446



Using Eq. 9.40:







. .

.

784.87 MPa

784.87 0.446 350 MPa Fi = 350 x1000 x π x 0.008 /4 = 17.6 kN pf 

f

ln R 1   cot    784 . 87 x 0 . 446 x (1  0 . 1 x cot 8 )

pf = 599 MPa Ff = 599 x1000x π x (0.008)2/4 = 30.1 kN The force including friction is 71 % higher than the ideal force 2 p t   f {ln R 1   cot    tan  } Using Eq. 9.41: 3 pt = 784.87 {0.446(1 + 0.1 cot 8) + 0.667 tan8}= 672.5 MPa Ff = 672.5 x1000x π x (0.008)2/4 = 33.8 kN The total force is 12.3 % higher than the force including friction.

17

Chapter 10 Sheet Metal Forming 1. It is required to cut washers from 2 mm thickness mild steel strips using two punch/die sets on a mechanical press. The outside diameter of the washer is 30 mm, and the inside diameter is 18 mm. If the ultimate tensile strength of this steel is 450 MPa, the shear strength is 70 % of the ultimate strength, and the optimum clearance is 8%. Calculate the required forces, and the diameters of the punches and dies. Solution: σs = 0.7 σu = 0.7 x 450 = 315 MPa The shearing of a washer includes piercing and blanking operations. Using Eq. 10.1: The force required for piercing Fp = t . π D . σs = 0.002 x π x 0.018 x 315 x 103 = 35.63 kN

The force required for blanking Fb = 0.002 x π x 0.03 x 315 x 103 = 59.38 kN Then Ftotal = 35.63 + 59.38 = 95 kN Clearance c = 8 % of t = 2 x 0.08 = 0.16 mm For the piercing operation: Dp = D = 18 mm Dd = D + 2 c = 18.32 mm For the blanking operation: Db = Dd = D =30 mm Dp = D – 2 c = 30 – 2 x 0.16 = 29.68 mm

2. A mild steel strip 0.9 mm thickness, 10 mm width, and length 20 mm is bent to a radius 12.5 mm.. Assuming that the yield stress is 280 MPa, and the modulus of elasticity is 210 GPa, calculate the radius of the part after bending, the spring back factor, and the initial bending angle. Solution: Strip width = 10 mm

Apply Eq. 10.6-b :

10 t , then bending is under plane strain condition  Ri . y Ri  1.69 Rf  E.t

= 1.69

.

3

  R .   2.25 i y   E.t .

18

   1 

2.25

.

.

1

= 0.958 Rf = 12.5/0.958 = 13.05 mm

From Eq. 10.5: From Eq. 10.2

.







.



.



.

0.96

Lb = α ( R + 0.5 t) α = 20 / ( 12.5 + 0.5 x 0.9) = 1.544 rad. = 88.5°

3. A 1000 x 200 x 1.2 mm annealed mild steel strip is stretched on a tool with circular shape to form an arc with radius 500 mm and angle 120˚. The material 0.26 has strain hardening relation   530  . Calculate the final length, the %

reduction in the cross sectional area, and the maximum permissible stretching length before necking. Solution: The stretched length lf = Radius R . angle α = 500 x 120 x π / 180° lf = 1047.2 mm ε = ln (lf / l0) = ln (1047.2 / 1000) = 0.046 = 4.6 %

% reduction

%r=



To get Af :

A0lo = Af lf Af = (200 x 1.2 ) x 1000 / 1047.2 = 229.18 mm2 % r = (200x1.2 – 229.18) / (200 x 1.2) = 0.0451 = 4.51 % Maximum stretching is up to necking where εmax = n = 0.26 εmax = 0.26 = ln(lmax/l0) = ln (lmax / 1000) lmax = 1000 e0.26 = 1296.9 mm

4. If the material in the previous problem suffers a decrease in thickness 10 % when fully stretched to the maximum length. Calculate the normal anisotropy. Solution: The final thickness tf = 0.9 t0 = 0.9 x 1.2 = 1.08 mm Volume = constant l0 w0 t0 = lmax wf tf wf = (1000 x 200 x 1.2) / (1296.9 x 1.08) = 171.35 mm

Eq. 10.9:



ln

ln



= ln (200 / 171.35) / ln ( 1.2 / 1.08)

R = 1.47 19

5. The table below lists the properties of several sheet materials to be applied in different forming processes: Material A B C D E

E, GPa 207 207 72.4 114 69

σy, MPa 220 241 172 138 7

R0 1.9 1.2 0.7 0.6 1.0

R45 1.2 1.0 0.6 0.9 1.0

R90 2.0 1.2 0.7 0.6 1.0

n 0.25 0.22 0.22 0.5 0.00

You are asked to select the right material for each of the following applications, and state the reason for your selection. a-Which of these materials would have the highest LDR in cupping? b-Which material would shows the greatest amount of earing during cupping? c-Which material would have the greatest uniform elongation in a tension test? d-Excluding material E (because of its low yield strength), which material could be formed into the deepest cup by a hemispherical punch acting on a clamped sheet (no "drawing")? Solution: Using Equations 10.10 & 10.11: R  2 R45  RT R  2 R45  RT R L & R  L 4 2 Anisotropy parameters & ∆ are calculated for each material as shown in the following table: Material E, GPa n σy, MPa ∆R 207 220 0.25 A 1.575 0.75 207 241 0.22 B 1.1 0.2 72.4 172 0.22 C 0.65 0.1 114 138 0.5 D 0.75 -0.3 69 7 0.00 E 1 0

a. The material having highest LDR in cupping is A as it has the maximum b. The material that shows greatest amount of earing during cupping is A as it has maximum ∆R. c. The material that shows highest uniform elongation in a tension test is D as it has maximum strain rate exponent n. d. The material that shows maximum stretching strain is D as it allows maximum uniform deformation n.

20

6. For the materials presented in the FLD of Fig. 10.36, if the major strain in a forming process is equal to 40%, find: a-The maximum positive minor strain for the Aluminum alloy. b-The state of stress at the point of maximum positive minor strain for the alloy. c-The maximum negative minor strain for the same material d-The maximum positive and negative minor strains for high strength steel. Solution: From Figure 10.36: a. Maximum positive minor strain for Al. alloy = 0.4 (40 %) b. State of stress: equal biaxial tension. c. Maximum negative minor strain = - 0.14 (- 14 %) d. For HSLA steel in the same figure: Maximum positive minor strain = 0.1 (10 %) Maximum negative minor strain = - 0.05 (- 5 %)

21

Chapter 11 High Velocity Forming and High Energy Rate Forming 1. A 300 gm RDX explosive charge (1270 kJ/ kg) is used to form a dished end cover for a storage tank from a mild steel plate 1.2 m diameter, and 10 mm thickness. You are asked to: a. b. c. d.

Suggest a suitable standoff distance, Calculate the hydrostatic pressure, Calculate the solid angle subtended by the blank at the charge, Calculate the central height of the formed part if the efficiency is 40 % and the yield strength for the used mild steel is 350 MPa. e. Suggest an alternative explosive for this process. Solution:

a. b. c. d.

For a plate 1.2 m diameter, S = 0.5 D = 0.6 m Hydrostatic head H = 2S = 2 x 0.6 = 1.2 m S = 0.5 D, then subtended angle ө = π/4 = 45° Chemical Energy E = m e = 0.3 x 1270 = 381 kJ Applying equation 11.1

. .



. . 4

350 x 0.01 x π x h2 = (381 x 0.4 x (π/4)2 x 10-3) / 4 h = 46.23 mm e. From Table 11.1, the only explosive providing higher energy than RDX is PENT, then it is a suitable alternative explosive.

2. The charging voltage of an EHF machine is 16 kV, and the bank of capacitors is 120 µF. a-Calculate the machine capacity given that the process efficiency is 20 %. b-Suggest the electrode gap width for this machine. Solution: a. Using equation 11.4: The energy stored in the capacitor E = 0.5 C.V2 E = 0.5 x 120 x 10-6 x (16 x 103)2 = 15,360 J = 15. 36 kJ

22

Since efficiency of the machine is 20 %, Then the machine capacity Em = 15.36 / 0.2 = 76.8 kJ b. Referring to section 11.4.3.3, The electrode gap width is typically 12 mm at 10 kV And is typically 50 mm at 50 kV Using the proportionality principle: If the electrode gap width for the 16 kV machine is w Then (w – 12) /(50 – 12) = (16 – 10) / 50 – 10) w = 17.7 mm 3. A brass tube (σy = 300 MPa) of 150 mm outer diameter, and 2.5 mm thickness is to be formed using EMF, Calculate the magnetic pressure required to deform it, using a multiplying factor of 8 for the high strain rates. What is the hydrostatic pressure inside the tube? Solution: Using equation 11.3:

pm 

2tN y Do

pm = (2 x 2.5 x 10-3 x 8 x 300) / 150 x 10-3 = 80 MPa Using equation 11.2: Hydrostatic pressure

p

2t y Do

= 2 x 2.5 x 10-3 x 300 / 150 x 10-3= 10

23

MPa

Chapter 13 Processing of Polymeric Materials 1. The molecular weights of UHMWPE and LDPE are 4 x 106 and 2 x 104 respectively, and the spacing between carbon atoms in the molecule is 1.26 x 10-4 micrometer. Calculate the degree of polymerization and length of stretched chain in both grades of polyethylene. Solution:

The monomer of PE is C2H4, Atomic wt. for C = 12, Atomic wt. for H = 1 Molecular wt. for monomer = (2 x 12) + (4 x 1) = 28 For UHMWPE, DP = 4,000,000 / 28 = 142,857 For LDPE, DP = 20,000 / 28 = 714 Length of UHMWPE Ch. = 142,857 x 1.26 x 10-4 = 18 µm Length of LDPE Ch. = 714 x 1.26 x 10-4 = 0.09 µm

2. Given that the density of polystyrene is 1.05 g/cm3. Calculate the percentage of gas in expanded polystyrene foam having density 50 kg/m3. Solution: Density of PS = 1.05 g / cm3 = 1050 kg / m3 Percentage of PS in expanded PS = 50 / 1050 = 0.0476 = 4.76 % Then, percentage of air = 100 – 4.76 = 95.24 %

3. In the blow molding process presented in Fig. 13.27, if a HDPE extruded parison is used with outside and inside diameters of 40 and 37 mm respectively. Calculate the thickness of the blown bottle if the mold diameter is 120 mm. Solution: Since the length of the parison will not change during blowing, Then, area before blowing = area after blowing (402 – 372 ) = (1202 – d2 )

Internal diameter of the bottle, d = 119 mm Bottle thickness =

= 0.5 mm

24

Chapter 14 Composite Materials and their Fabrication Processes 1 Given that for a Glass / epoxy composite material, the glass fiber diameter is 10 µm, and its strength is 3500MPa, and the shear strength of the epoxy matrix is 20 MPa. What is the minimum length of fibers that provide good bonding? Solution: For the fibers: fiber diameter d = 10 µm , 3500 For the Epoxy matrix τi = 20 MPa / 2 = 3500 / (2 * 20) = 87.5 Critical l/d =

Critical length = 87.5 * 10 = 875 µm = 0.875 mm Then a length of about 1 mm would be quite sufficient for good bonding.

2 For a Carbon / Epoxy composite material hand lay-up with a fiber volume fraction 0.68, calculate its stiffness, given that the modulus of elasticity of carbon fibers is 190 GPa, and Epoxy is 4.5 GPa. Compare this stiffness with the separate stiffness of each constituent. Solution: For the composite material: Ec = (Ef - Em) vf + Em , Eq. 14.8 Then Ec = (190 – 4.50 * 0.68 + 4.5 = 130.64 GPa Compared to epoxy: The stiffness of the composite is (130.64 – 4.5)/4.5 = 28 times larger. Compared to carbon fibers: The stiffness is less by (190 – 130.64)/ 190 = 31 %. Then fiber reinforcement increases the stiffness of the matrix material considerably.

25

Chapter 15 Fundamentals of Traditional Machining Processes 1- A turning operation is to be adopted under the following machining conditions: Cutting speed v = 80 m/min Feed rate f = 0.2 mm/rev Depth of cut a = 1.5 mm External workpiece diameter D = 40 mm

Calculate; a) The final workpiece diameter b) The rotational speed of the workpiece c) The tool feed rate expressed in m/min d) The material removal rate Z in mm3/min e) The specific cutting power ks assuming that the measured main force component Fc=750 N. f) The feed power Pf in kW assuming hat the measured feed force component Ff=500 N Solution: a) The final diameter of the workpiece d: Dd a 2 40  d 1.5  2 Then d  37 mm

b) The rotational speed N 1000 v 1000x80 N  D x 40  637 c) The feed rate in m/min

f N 0.2 x637  1000 1000  0.125 d) MRR Z in mm3/min

rev/min



f 

m/min

26

Z  a f v  1.5 x0 .2 x 80000  24

cm 3 /min

 24000

mm3 / min

e) The specific cutting power ks (Fc=750 N) ks 

Pc Fxv F  c  c Z axf xv axf 750   2500 1.5 x 0.2

f) The main cutting power Pc in kW Fxv pC  c 60000 750 x 80   1.0 60000

N/mm 2

kW

g) The feed power Pf in kW (Ff=500 N) Ff x N x f Pf  1000 x 60000 500 x 637x 0.2   0.001 1000 x 60000

kW

2- In an orthogonal cutting test on mild steel, the following results are obtained t1 = 0.25 mm t2 = 0.75 mm b = 2.5 mm Fc = 900 N Fp = 450 N Tool rake  = 0 Calculate: a) Mean angle of friction  on tool face b) Mean shear stress  s

27

Solution: a)  m  tan   

Fp  Fc tan  Fc  Fp tan  450  900 tan0  0.5 900 - 450 tan 0

 =26.6o

b)

s 

Fs , As is the area of shear plane As

Fs  R cos(     ) tan  

r x cos   r  1/ 3 1  r x sin 

  18.4 o 2

2

Fs  Fc  Fp cos(     )  900 2  450 2 cos 45  711

s 

N

Fs F 711  s sin   sin 18.4 As t1 x b 0.25 x 2.5  359

N/mm 2

3-A bar of 100 mm diameter of SAE 1020 steel is to be turned on a lathe using an HSS tool at a speed of 30 m/min. The feed is 0.8 mm/rev and the depth of cut is 3 mm. The main component of the cutting force is measured as 3250 N. What is the power needed to perform this cut ? and what would be the specific power.? Solution: Fc v 60000 3250 x 30   1.625 60000

PC 

ks 

kW

Fc 3250   1354 f x a 0.8 x 3

N/mm 2

28

4- AN HSS tool has been used for a metal cutting operation, shows a T-v Taylor relationship of vT 0.125  44.5 . Originally 15 minutes were required to remove a dull tool. A new tool holder has made it possible to reduce the time to 5 minutes. What increase in cutting speed does this permit to obtain the maximum rate of production from the operation. Solution:

1 T p  (  1) t ch n 1 ( - 1) x 15  105 min 0.125 Therefore Tp = 35 min, if the new holder is used 44.5 v p  0.125  44.5 x (105) -0.125  24.87 T

m/min

44.5  44.5 x (35) -0.125  28.53 0.125 T v p1  v p 2  28.53  24.87  3.66 vp 

m/min m/min

5- For a cemented carbide tool, if vT 0.2  152 , find the economical durability and the durability for maximum productivity and their corresponding speeds assuming the following conditions;  Tool of replaceable clamped insert of 8 cutting edges  Tool insert costs $ 9.6 new  Tip exchange time tch =0.5 min.  Labor cost considering overheads Lc(1+ r)= $ 0.4/min Solution:

  C1 1 Te  (  1)  t ch   n ni ( L(1  r )   (

1 - 1) 0.2

v e  152 x Te

9.6   0.5  8 x0.4   4(0.5  3)  14  

0.2

 152x14 0.2  89.66

1 T p  (  1) t ch n 1 ( - 1) x 0.5  2 0.2

v p  152 x (2) 0.2  132.32 29

m/min

min

m/min

min

6- A shaft made of mild steel ( k s  1990 x h m

-0.26

) with a diameter 55 mm, is ground

under the following conditions; Grinding wheel: dg : 250 mm b : 25 mm ng :2350 rev/min Grain mesh: 60 Workpiece: nw :75 rev/min Table feed ut : 1 m/min Depth of cut e : 0.01 mm It is required to determine: a. Mean chip thickness hm b. Mean force component Fc c. Power of the wheel Pc and work Pf Solution: a)

h m  e

vw vg

e

1 1  dg dw

For  e  33 mm

vw dw x n w 55x75    0.007 vg dg x ng 250x2350 h m  33x 0.007  0.00345

1 1 1  100 250 55 mm

 3.5

m

b) Fc 

vw .e.f tr .f  .k sm vg

f tr 

u t 1000   13.3 nw 75

mm/rev, (b  25 mm)

f   3.5 (Given as based on grain mesh number 60 and chip thickness)

30

k sm  1990 x h m

-0.26

 1990(0.035) -0.26  8650

Therefore

Fc  0.007 x

1 x13.3x3.5 x8650  28 N 100

c) Main power Pc and feed power Pf Pc 

v w .e. f tr . f  .k s 1000



x55 x75 x0.01x13.3 x3.5 x8560 1000 x 60 x1000

 0.86 Pf  Pc

kW

vw  0.86 x 0.007 vg  0.006

kW

31

N/mm2

Chapter 16 Machine Tools for Traditional Machining 1-Calculate suitable gear trains for cutting the following threads: a) 11 tpi on a 4 tpi leadscrew b) 7 threads in 10 mm on 6 mm leadscrew c) 12 tpi on a lathe having 6 mm pitch leadscrew Solution a) 11 tpi on a 4 tpi leadscrew

t u c e b o t h c t i P

1 4 20 40  11    1 11 55 110 4

w r c s d a e l f o h c t i P

Drivers  Driven

This gives either 20/55 or 40/110 in a simple train

b) 7 threads in 10 mm in 6 mm leadscrew The pitch of the thread =10/7 10 10 5x2 50 20 Drivers x  7    6 42 7x6 70 60 Driven

A compound train with 50 teeth and 20 teeth as the drivers and 70 teeth and 60 teeth as the driven c) 12 tpi on a lathe having 6 mm pitch leadscrew Now dealing with English threads (inch) on lathes with metric leadscrew (mm) Drivers 127  Driven 5pN Drivers 127 127 20 127    x Driven 5x6x12 6x60 60 120

A compound train with 20 teeth and 127 teeth as the drivers and 60 teeth and 120 teeth as the driven

32

2- Calculate the broach length required for machining a key way of 8 mm depth and 20 mm wide for 100 mm length in a steel workpiece, Sz is taken as 0.08 and chip space number is 8. Solution

P  3 S z .l.x

Since the pitch

P  3 0.08 x100 x 8  24

mm

l P  (  50) 2

mm

Zc 

t 8   100 S z 0.08

teeth

t is the depth o be removed l  Z c .P  2400 b

mm

3-Calculate the indexing particulars for cutting eight flutes in a reamer blank Solution The indexing required is;

40 40  5 z 8 Turn the index 5 complete turns n

4-Calculate the indexing movements and gear ratio required for cutting 57 gear teeth. Solution The indexing required is;

n

40 z

Select Z'=58

40 5  56 7 Move the crank 15 holes on 21 holes circle The change gears supplied to match the three plate system (Brown and Sharp) are: 24(2), 28, 32, 40,44,48,56,64,72,86 and 100 teeth The number of teeth of the change gears, a, b, c, d are determined from; n



) Z ` Z ( 0` 4Z

cd . ab

Gear ratio

33

) 7 5 6 5 ( 06 45

cd . ab 

5 40 40 24   x 7 56 24 56

Gear train is a=40, b=24, c=24, and d The negative sign means that the index plate rotates counter clocwise, and an ideler gear must be used 5- Calculate the time required to achieve the hobbing operation using HSS-hob of pitch diameter 80 mm to cut a spur gear of 48 teeth at a cutting speed of 40 m/min. the gear toot has a face width of 94 mm. The hob is fed at a rate of 2 mm/rev of the blank, and assuming the approach and over travel of 36 mm. Solution Hob rotational speed is N

N

1000V 1000x 40   159.154 D x80

rev/min

Axial feed ua ua  fxN  2 x159.154  318.3

mm/min

Time to achieve the cut t;

t

94  36  0.41 318.309.

min

6-Calculate the angle of tilting the compound slide when turning a short taper of steel bar having D=80 mm, d=60 mm, and length of taper l=40 mm, see Figure 16.2-.a. Solution According to Figure 16.2-a

D  d 100  60   0.25 2l 2 x 40   sin -1 0.25  14 o 28 '

Sin  Hence the tilting angle

7-Find the tailstock setting required for turning a taper of 85 mm diameter to 75 mm diameter over a length of 200 mm. The total length of the job is 300 mm. Solution According to Figure 16.2-c Dd (85  75) hL  300  15 2l 200

34

mm

Chapter 17 Fundamentals of Nontraditional Machining 1-Calculate the traverse speed vf in mm/s for cutting tungsten carbide sheet of 2.5 mm thick, if EB equipment of 8 kW is used. The equipment is capable of focusing the beam to a diameter of 0.4 mm. The following workpiece data are given:  m  3400

o

C,

C d  8.1x10 5 m 2 /s, k t  214

N/s. o C

Solution:

Referring to equation 17.8 C vf  d df

 Pe  0.1    m .t 1 .k t 

2

m/s

Therefore

8.1x10 -5 vf  0.0005  0.0313  31.3

8x1000   0.1 3400x 0.0025x 214  m/s mm/s

2

2-In an EDM operation using Lasarenko's relaxation generator, Vo=250 volt, R=10  , C=3 F . If the cut is required to be performed under maximum removal rate condition, calculate: a) Discharge (breakdown) voltage Vs b) Charging time tc c) Cycle frequency fr d) Energy per individual discharge of capacitor, Ed Solution:

The cut is performed under maximum removal rate condition, therefore: a) Discharge (breakdown) voltage Vs

V s  0.73 VO  0.73x250  182.5

V

35

b) Charging time tc Vs  Vo (1  e  t c / RC ) Vs  0.73  1  e  t c / 30 Vo

(C in F and t c in s)

From which tc=39.2 s c) Cycle frequency fr t d  0 .1 t c  3.9 s t c  t d  39.2  3.9  43.1  43.x10

s -6

s

Therefore fr 

1 10 6  t c  t d 43.1  23.2

Hz kHz

d) Energy per individual discharge of capacitor, td

1 2 CVs 2 1  x3x10 -6 x(182.5) 2 2  0.05

Ed 

J

36

Chapter 18 Numerical Control of Machine Tools 1. In which year is the first NC machine tool developed? A-1700 B-1860 C-1952 Solution: C

D-1977

2. The binary number 10010101 is equivalent to what number in the decimal system? A-159 Solution: B

B-149

C-139

D-129

3. The magic three method S675 would represent a spindle speed of A-750 B-368 rpm C-967 rpm D-975 rpm rpm Solution: A 4. The magic three method for feed rate 40 inch/min is A-f545 B- f740 C-f 745 Solution: D

D-f540

5. The term applied to the total number of holes in each row of a tape perpendicular to direction of feed is A- EIA

B- binary -coded decimal

C- even or odd parity

D- ASCII subset

Solution: C 6. The decimal number 45 is equal to A-110111 B-101101 Solution: B

C-101001

D-111011

7. Write down a part program for milling, slotting, and drilling the part shown in Figure 18.38  Preset the absolute tool reference at X= -2 inch ; and Y=-2 inch  Use letters A, B, C, D, to describe the tool paths

37

Solution: Workpiece coordinates

Point location Tool change A B C D E F G H I J K

X coordinate (Inches) -2 0 3.5 3.5 2.75 2 1.5 1.5 0.0 1.0 1.0 3.0

Y coordinate (Inches) -2 3.5 3.5 1.5 0.75 0.75 0.25 0.0 0.0 1.75 2.75 3.0

Part program N10 N20 N30 N40 N50 N60 N70 N80 N90 N100

G90 G20 G40 M06 T01 G54X-2Y-2 M03 S300 G00 X-1 Y-1 Z0.1 Z-0.5 G41 X0.0Y0.0 G01X0.0 Y3.5 F3

Absolute, inch programming Cutter diameter cancel Change to tool 1 Workpiece preset position Rotate spindle cw at 300 rpm Rapid positioning

Cutter diameter compensation Linear interpolation to point A

38

N110 N120 N130 N140 N150 N160 N170 N180 N190 N200 N210 N220 N230 N240 N250 N260 N270 N280

X3.5

Linear interpolation to point B Linear interpolation to point C Circular interpolation cw to point D Linear interpolation to point E Circular interpolation ccw to point F Linear interpolation to point G Linear interpolation to point H Cutter diameter cancel Rapid positioning Linear interpolation to depth Mill slot

Y1.5 G02 X2.75Y0.75I-0.75J0 G01 X2.0 G03 X1.5 Y0.25 I0 J-0.5 G01 X1.5 Y0.0 X0.0 Y0.0 G40 Y-1.0 G00 Z0.1 G01 Z-0.5 X1.0 Y1.75 Z0.1 G00 X3.0Y3.0 G83 Z-0.5R 0.1Q2 G80 G28 M05 M30

Rapid positioning Drilling cycle Cancel drilling cycle Home position Spindle stop Rewind the program

8. Write down a part program to finish turn the part shown in Figure 18.39. Preset the tool at X=6" and Z=10".

Solution: Workpiece coordinates

Point location Tool change A B C D E F G

X coordinate (Inches) 6 0 0.65 0.75 0.75 1.0 1.0 1.25

Z coordinate (Inches) 10 0 0 -0.1 -1.0 -1.5 -1.625 -1.75 39

Part program N010 G90 G20

Absolute and inch programming

N020 G40

Cancel tool nose radius (TNR)

N030 T0101

Finish turning tool

N040 G92 X6.0 Z 10.0

Workpiece coordinate setting

N050 G96 S400 M03

Constant surface speed

N060 G00 G42 X0.0 Z0.1

Rapid positioning, TNR compensation

N070 G01 X0 Z0

Linear interpolation to point A

N080 X0.65

Linear interpolation to point B

N090 G01 X.75 Z-.1

Linear interpolation to point C

N100

Z-1.0

Linear interpolation to point D

N110

X1.0 Z-1.5

Linear interpolation to point E

N120

Z-1.625

Linear interpolation to point F

N130 G02 X1.25 Z-1.75 I0.125 K0

Circular interpolation cw to point G

N140 G28

Home position

N150 G40

Cancel tool nose radius (TNR)

N160 T0404

Drilling tool

N170 S800 M03

Spindle speed at 800 rpm

N180 G00 X0 Z.2

Rapid positioning near point A

N190 G74 X0 Z-.75 F0.01 K1.125

Peck drilling cycle

N200 G28

Home position

N210 M30

Rewind the program

40

Chapter 19 Industrial Robots and Hexapods 1- For a new robot installation, the total cost of an IR including tooling is $ 80,000. The estimated annual maintenance and programming cost is $ 10,000 for one shift, and $12,000 for two shifts. The robot replaces one worker whose annual salary including overheads is $ 30,000. What would be the payback period P for one- and two-shift use? Solution:

Referring to equation 19.1 in the textbook P

CR LM 80,000  4 30,000  10,000 80,000   1.67 60,000  12,000

years for one shift years for two hift

2- Five IRs are replacing 14 workers in a production line. Each will cost $ 70,000, including accessories. The annual maintenance is estimated as $ 4000/robot, and the programming cost for the whole system is $ 12,000.What is the payback period P, as based on one shift if the annual salary of each worker is $ 25,000 including overheads. Solution:

Maintenance and programming cost=M + P =5x4,000 + 12,000= 32,000 $ CR=5x70,000= 350,000 $ L=12x25,000= 300,000 $

P

CR LM 

350,000  1 .3 300,000  32,000

41

years for one shift

Chapter 20 Surface Technology

1- An automobile bumper has an area of 1.4 m2. It receives one of the following listed electroplates and a cell voltage of 6 V. Material

Allow. current density (A/m2)

Thickness ( m )

1-Cu plate 2-Ni plate 3-Cr plate 4-Cd plate

12 12.5 0.5 12.5

Anode material properties Atomic Valence Specific weight n weight  (Aw) ( g/cm3)

130 270 1600 215

64 59 52 112

1 2 6 2

8.9 8.8 6.9 8.6

Current efficiency

 % 50 95 5 90

How much plate material and electrical energy are consumed for each bumper, and how long should be the plating time for the used allowable current densities, Solution: 1- Cu plate

Cu quantity 

12 x 1.4 x10 6  17 6 10  17x8.9  151

Faraday's law of electrolysis 1 A I ..t 96500 n 0.5 64 151  x x130x1.4 t 96500 1 mp  

Plating time, t

t

151x96500  2502 0.5x64x130x1.4  41.8

sec min

Electrical energy consumed:

E  V .I .t 

6 x130 x1.4 x 41.8  0.76 1000 x 60 42

kWh

cm 3 g

2- Ni plate

Ni quantity 

12.5 x 1.4 x10 6  17.6 10 6  17.5x8.8  154

cm 3 g

Faraday's law of electrolysis 1 A I. t 96500 n 0.95 59 x270x1.4 t x 154  96500 2 mp  

Plating time, t:

t

154x96500x2  1403 0.95x59x270x1.4  23.4

sec min

Electrical energy consumed: E  V .I .t 6 x 270 x1.4 x 23.4   0.88 1000 x 60

kWh

3- Cr plate

Cr  quantity 

0.5 x 1.4 x10 6  0.7 6 10  0.7x6.9  4.8

Faraday's law of electrolysis 1 A I ..t 96500 n 0.15 52 x1600x1.4 t x 4.8  96500 6 mp  

Plating time, t

t

4.8x96500x6  160 0.15x52x1600x1.4  2.7

sec min

Electrical energy consumed: E  V.I..t 

6 x1600 x1.4 x 2.7  0.42 1000 x 60

43

kWh

cm 3 g

4- Cd plate Cd  quantity 

12.5 x.1.4 x10 6  17.5 10 6  17.5 x 8.6  151

cm 3 g

Faraday's law of electrolysis 1 A I ..t 96500 n 0.9 112 x215 x 1.4 t x 151  96500 2 mp  

Plating time, t

t

151 x 96500 x 2  956 0.9x112x215x1.4  16

sec min

Electrical energy consumed: E  V .I .t 

6 x 215 x1.4 x16  0.48 1000 x 60

kWh

2- Instead of plating Cu and Ni on a bumper, the same requirements can be met by a single Ni-plate of 25 m thick. In either case, a final Cr plate of 0.5 m must be applied. Cu costs $ 1.7/kg, and Ni costs $ 5/kg. Electricity costs 30 cent/kWh. Which of the two alternatives do you recommend from the economical point of view? Solution:

Referring to the above problem First alternative (12.0 m Cu, 12.5 m Ni, and 0.5 m Cr) Cost of materials: 12 m Cu-layer cost=1.7x0.151 = 0.257 12.5 m Ni-layer cost = 5 x 0.154=0.77

$ $

Total cost of Cu and Ni layers = 0.257 + 0.770 = 1.027 Cost of electricity: 44

$

Electricity cost (Cu-layer)=0.3 x 0.76=0.23 $ Electricity cost (Ni-layer) =0.3x 0.88=0.26

$

Total electricity cost = 0.23 + 0.26 = 0.49

$

Total material and electricity cost=1.027 + 0.49 = 1.517

$

Cr-layer is the same as in the second alternative Second alternative (25 m Ni, and 0.5 m Cr) 25 m Ni-layer cost = 5 x 0.154x2=1.54

$

Electricity cost (Ni-layer) =0.3x 0.88x2=0.53

$

Total material and electricity cost=1.54 + 0.53 =2.07 Cr-layer is the same as in the first alternative Therefore the first alternative is economically feasible

45

$

Chapter 21 Joining Processes

1. During arc welding, if the current used is 80 A, arc voltage is 20 V, and the welding speed is 96 mm/min, calculate:  Energy input to the arc  Heat input in J/mm Solution: Power inpu t to the a rc ( W )  volt (V) xcurrent (I)

Heat input to the ar c  20 X80  1600

J/s

︵ ︶

m m / J

c r a e h t o t t u p n i t a e H 

0 6 x w Iv d e e p s g n i d l e w

W

t n e r r u c x V t l o v

Power inpu t to the a rc ( W )  20 X80  1600 ︵︶

For a welding speed 960 mm/min

︵ ︶ 80 x 20 x 60  1000 Heat input to the arc  96

J/mm

2. During arc welding using a current of 70 A, voltage 18 V, calculate the welding speed if the heat input to the arc is 400 J/mm. How much heat in kcaloris. is required for seam welding of 100 mm length? Solution: Power inpu t to the a rc (W )  18 X 70  1260

W

Heat input to the ar c  20 X 80  1260

J/s

Since Heat input to the arc 

volt (V) x current (I) x 60 welding speed (V w )

Welding speed (Vw ) 

volt (V) xcurrent (I) x 60 1260 x60   189 heat to the arc ( J / mm) 400

Since 1J=2.3901x10-4

kcal 46

J/mm

mm/min

Heat input  1260 x 2.3901x10 4  0.3012

kcal/s

The time required for 100 mm weld=100x60=600

s

The total heat required=0.3012x600=180.692

kcal

3. Calculate the welding speed during arc welding using 100 A, and 20 volt if the heat input is 600 J/mm. If the thickness is doubled, the welding speed is reduced to 50 %, calculate the heat input required in the new case. Solution: Welding speed (V w ) 

volt (V) xcurrent (I) x 60 heat to the arc (J/mm)

Welding speed (Vw ) 

20 x100 x 60  200 600

mm / min

The new welding speed = 100 mm/min Heat input to the arc 

volt (V) xcurrent (I) x 60 welding speed (Vw )

Heat to the arc (J / mm) 

20 x100 x 60  1200 100

j / mm

Doubling the thickness reduces the welding speed to 50 % and raises the heat input to 200 % 4. During arc welding a seam deposited during 10 minutes using an arc current of 200 A and voltage of 20 V, calculate:  The amount of kW hours consumed  The amount of heat used in kcal Solution: Power inpu t to the a rc  200x20  4000

W

The amount of kW hour can be calculated from

Power concsumed 

4000 60 x  24 1000 10

kW.hr

Heat input to the ar c  4000

J/s

Heat input  4000 x 2.3901x10 4  0.956 The time taken for welding= 10x60=600

kcal/s

s

Heat input  0.956 x 600  573.624

47

kcal

5. Calculate the amount of energy (J) and the heat required (kcal) during resistance spot welding under the following conditions: 1. Welding cycle: 3 on, 2 off 2. Power supply: 50 Hz 3. Welding current : 12,000 A 4. Resistance between electrodes: 100 µΩ Solution:

During resistance spot welding The energy required  I 2 xRxt x K 1

J

K1=1

The energy required  12000 2 x100 x10 6 x

3  864 50

Heat required  864 x 2.3901x10 4  0.207

J kcal

6. Calculate the number of welds per minute, welding speed, electrode rotational speed, and the amount of energy required during resistance seam welding under the following conditions: 1. Electrode diameter: 250 mm 2. Welding current : 10000 A : 3. Welding rate: 4 welds/cm 4. Welding cycle 3 on, 2 off 5. Power supply: 50 Hz 6. Effective resistance between electrodes 100 µΩ Solution:

Welds / min) 

Frequency x 60 Cycles of weld  int erval cycles

Welds / min) 

50 x 60  600 3 2

Welding speed (mm / min) 

weld/min weld / min 600   1500 weld require / mm 0.4

Electrode rotational speed 

weling speed (mm/min)  x electrode diameter (mm )

48

mm / min

Electrode rotational speed (rpm) 

1500  1.91  x 250

The energy required  I 2 xRxt

The energy required  10,000 2 x100x10 6 x

49

rpm J

3  600 50

J

Chapter 24 Quality Control 



1. A control chart for X and R is used to control an important part dimension. The 



subgroup size is 5. The values of X and R are computed for each subgroup and the values of



X

and



R

after 25 subgroup are 614.8 mm and 120.0 mm 

respectively. Determine the control limits for X Solution mc 



X

X

j

j1



mc

614.8  24.952 mm 25

mc

R



j1

Rm 

mc

j



120  4.8 25









mm

UCL   X  A 2 R m  24.952  0.577 x 4.8  27.722 X

LCL   X  A 2 R m  24.952  0.577 x 4.8  22.182 X

A2 is taken from Table 24.3 for n=5

mm mm

as 0.577

2. During a quality control inspection of manufactured, parts, the following data are available: Number of subgroups=25 Subgroup size=25 

X  394.536 

R m  21.52

mm mm 



- Determine the control limits for X and R - If the specification limits for each item are 400±30 capability with the specifications

50

mm, compare the process

Solution For X- bar chart

For n=5 at Table 24.3 A2=0.577 



UCL   X  A 2 R m  394.536  0.577 x 21.52  406.953 mm X





LCL   X  A 2 R m  394.536  0.577 x 21.52  382.119 mm X

For R- bar chart

Table 24.3 for n=5 D3=0 D4=2.115 

UCL R  D 4 R m  2.115x 21.52  45.51 mm 

LCL R  D 3 R m  0 

R 21.52   m   9.11 d2 2.326

Hence, from Table 24.3

d2=2,326

6  6 x9.11  54.66 From the specification limits, USL-LSL=30+30=60

mm



Process spread = X ±3  

X +3  =394.536+3x9.11= 421.866 mm 

X -3  =394.536-3x9.11=367.206 mm For the specification limits; USL=400+30=430

mm

LSL=400-30=370

mm 

Since 6  USL  LSL the process can be kept in control by adjusting the center or X .

51

3. For the hole basis system of 38 mm diameter, specify the tolerance for shaft and hole for a close running fit. Solution

For the required type of fit, select H8f7 For the hole H8 The fundamental tolerance for size 38 mm, i , becomes:

i  0.453 D  0.001D ,

(D in mm)

i  0.453 38  0.001x38  1.53

µm

For tolerance grade 8, the amount of tolerance, Table 24.5=25 i=25x1.53=39 µm The fundamental deviation for H=0 µm The minimum hole diameter = 38.000 The minimum clearance=38.039 mm For the shaft f7 For tolerance grade 7, the amount of tolerance, Table 24.5=25 i=16x1.53=25 µm The fundamental deviation for shaft f, Table 24.6, =-5.5D0.41=-5.53D0.41=-25 µm

| As shown above; The maximum clearance = 38.039-37.950=0.089

mm

The minimum clearance=38.000-37.975=0.050

mm

52

4. Specify the type of fit for 25H7k6 Solution

For the hole H7 The fundamental tolerance for size D=25 mm, i , becomes:

i  0.453 D  0.001D i  0.453 25  0.001x 25  1.33

µm

For tolerance grade 7, the amount of tolerance, Table 24.5=16 i=16x1.33=22 µm The fundamental deviation for H=0 µm Minimum hole diameter=25.000 mm Maximum hole diameter = 25.022 mm For the shaft k6 For tolerance grade 6, the amount of tolerance, Table 21.5=10 i=16x1.33=14 µm The fundamental deviation for shaft f, Table 24.6,  0.63 25 =2 µm The maximum shaft diameter =25.000+0.002+0.014=25.016 mm The minimum shaft diameter= 25.002 mm The amount of clearance=25.002-25.022. =0.020 mm The amount of interference=25.016-25.000=0.016 mm These results are shown in the following sketch;

53

5. Choose the suitable block gauges to assemble the following dimensions:; a-62.31 mm b-36.685 mm Solution:

a-62.31 mm

b-36.685 mm

1.010

1.005

+ 1.300

1.080

+60.000

1.600

62.310

8.000 25.000 36.685

6. Using the set of block gauges shown in Table 24.9, make up a set of slip gauges to check a gauge measuring of 2.1758 inch Solution:

The metric equivalent of 2.1758 inch =2.1758 x 25.4=55.26532 mm The nearest size to this is 55.265 mm. gauges required are as follows; 1.005 1.060 1.200 2.000 50.000 55.265 7. Calculate the setting of a 200 mm sine bar to measure an angle of 36o 45'

Solution

Since

sine 36o 45'= 0.598

sine θ 

H Height of slip gauges  s Center distance of roller axes L

H s  L sin e Hs=200x0.598=119.665 54

mm

8. Calculate the setting of a sine bar 250 mm to check the angle of a taper 1:20 Solution:

tan

 2

 2



1/2 1   0.025 20 40

 1.432

  1.432 x2  2.874 H s  L sin   250 sin 2.870  12.491 mm 9. Draw the vernier reading for the following values a- 22.32 mm

b- 1.859 inch

Solution

a- 22+16x0.02=22.32 mm

b- =1.8+ 2/40 + 9/1000= 1.859 inch

10. Draw the external micrometer reading for the following values a- 22.32 mm b- 0.272 inch Solution

a) 7 + 0.5 + 0.29 =7.79 mm

b) 0.4 + 2/40 + 22/1000 = 0.472 inch 55

K10546

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