A Textbook of Production Engineering_P. C. Sharma

Scilab Textbook Companion for A Textbook of Production Engineering by P. C. Sharma1 Created by Mayank Sahu B.E. Mechanical Engineering M. I. T. S Gwalior, M.P College Teacher None Cross-Checked by Bhavani Jalkrish November 3, 2014

1 Funded

by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab codes written in it can be downloaded from the ”Textbook Companion Project” section at the website http://scilab.in

Book Description Title: A Textbook of Production Engineering Author: P. C. Sharma Publisher: S. Chanda & Company, New Delhi Edition: 11 Year: 2008 ISBN: 9788121901116

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Scilab numbering policy used in this document and the relation to the above book. Exa Example (Solved example) Eqn Equation (Particular equation of the above book) AP Appendix to Example(Scilab Code that is an Appednix to a particular Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.

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Contents List of Scilab Codes

4

2 Press Tool Design

9

4 Cost Estimating

21

5 Economics of tooling

34

9 Limits Tolerences and Fits

62

11 Surface finish

69

13 Analysis of metal forming processes

71

14 Theory of metal cutting

80

15 Design and manufacture of cutting tools

94

16 Gear manufacture

101

17 Thread manufacturing

102

21 Statical quality control

103

22 Kinematics of machine tools

114

23 Production planning and control

120

26 Plant layout

125

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List of Scilab Codes Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11

find total pressure and dimensions . . . To find number of draws . . . . . . . . . To calculate bending force . . . . . . . find blanking force and work done . . . To find elastic recovery of material . . . To find cutting forces . . . . . . . . . . To calculate amount of shear . . . . . . To find economy of material . . . . . . . Calculations for designing drawing die . Determine developed length . . . . . . . To calculate bending force . . . . . . . . To calculate bending force . . . . . . . . calculate capacity of double bending die To calculate cutting force . . . . . . . . Determine blank and punch diameter . To find drawing operations and force . . Determine developed length . . . . . . . To calculate total cost and SP . . . . . To find selling price . . . . . . . . . . . To find factory cost . . . . . . . . . . . find production cost and time taken . . To find profit . . . . . . . . . . . . . . . To find lot size and time . . . . . . . . . To find time to change cutter . . . . . . To find tool change time . . . . . . . . . To calculate measuring time allowance . To find direct labour cost . . . . . . . . To find machining time . . . . . . . . . 4

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9 10 10 10 11 11 12 12 13 15 15 16 16 17 17 18 19 21 21 22 23 23 24 24 25 25 25 26

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.9 5.10 5.11 5.13 5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29 5.30

To find time to turn relief . . . . . . . . . . . . calculate time to face on lathe . . . . . . . . . To find time to drill hole . . . . . . . . . . . . To find time to complete cut . . . . . . . . . . To find time to broach . . . . . . . . . . . . . . find feed cutter travel and time . . . . . . . . To find cutting time . . . . . . . . . . . . . . . To find milling time . . . . . . . . . . . . . . . To find time to grind shaft . . . . . . . . . . . To find time to cut threads . . . . . . . . . . . find time to produce one piece . . . . . . . . . To find value of machine tool . . . . . . . . . . To find annual investment . . . . . . . . . . . . find project is economical or not . . . . . . . . selection of economical machine . . . . . . . . . selection of machine . . . . . . . . . . . . . . . selection of economical machine . . . . . . . . . find ERR and economicality of project . . . . . find ERR and economicality of project . . . . To determine acceptance of machine . . . . . . find investment cost and unamortized value . . To make decision of machines replacement . . . Determine economic repair life . . . . . . . . . find time to pay for itself . . . . . . . . . . . . selection of machine for job . . . . . . . . . . . Calculate maximum investment on turret lathe To find years for new machine . . . . . . . . . . To find cost and pieces . . . . . . . . . . . . . . To find number of components . . . . . . . . . To find number of components . . . . . . . . . To find time and profit . . . . . . . . . . . . . To find minimum number of components . . . . To calculate number of pieces . . . . . . . . . . To find cost for new fixture . . . . . . . . . . . find time to amortize fixture . . . . . . . . . . To find profit . . . . . . . . . . . . . . . . . . . To find BEP Cost and Components . . . . . . To find break even point . . . . . . . . . . . . . 5

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27 27 28 28 29 29 30 30 31 31 32 34 35 35 36 37 38 38 39 39 40 40 41 42 43 44 45 46 47 47 48 48 49 49 50 51 51 52

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

5.31 5.32 5.33 5.34 5.35 5.36 5.37 5.38 5.39 5.40 5.41 5.42 5.43 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 11.1 11.2 13.1 13.2 13.3 13.4 13.5 13.7 13.8 13.9 13.10 13.11 13.12 14.1 14.2

To find break even quantity . . . . . . . . . . . . To do break even analysis . . . . . . . . . . . . . To calculate minimum number of pieces . . . . . To determine the point . . . . . . . . . . . . . . To find quantity of pieces . . . . . . . . . . . . . To determine quantity of production . . . . . . . find preference between machines and production To find BEP and various sales . . . . . . . . . . To determine break even point . . . . . . . . . . To calculate economic lot size . . . . . . . . . . . To find EOQ and total cost . . . . . . . . . . . . Determine optimum lot size . . . . . . . . . . . . To find most economical lot size . . . . . . . . . To find allowance and tolerence . . . . . . . . . . Determine dimensions of shaft and hole . . . . . Determine dimensions of hole and shaft . . . . . Calculate fundamental deviations and tolerences Find tolerences limits and clearance . . . . . . . Determine limits of shaft and hole . . . . . . . . Determine dimensions of shaft and hole . . . . . Determine size of bearing and journal . . . . . . Determine size of two mating parts . . . . . . . . Determine size of hole and shaft . . . . . . . . . Calculate CLA value . . . . . . . . . . . . . . . . Calculate average and rms value . . . . . . . . . To find drawing load . . . . . . . . . . . . . . . . Calculate drawing force . . . . . . . . . . . . . . find neutral section slips and pressure . . . . . . To determine maximum force . . . . . . . . . . . Determine sticking radius and total load . . . . . To find drawing load and power . . . . . . . . . . calculate drawing load and power rating . . . . . To calculate forging loads . . . . . . . . . . . . . Determine extrusion load . . . . . . . . . . . . . To find roll pressures . . . . . . . . . . . . . . . . Determine neutral plane . . . . . . . . . . . . . . calculate the tool life . . . . . . . . . . . . . . . . Calculate the optimum cutting speed . . . . . . . 6

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52 54 54 55 55 56 56 57 58 58 59 60 61 62 62 63 63 64 65 66 66 67 68 69 69 71 72 72 73 74 74 75 76 76 77 78 80 80

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10 14.11 14.12 14.13 14.14 14.15 14.16 14.17 14.18 15.1 15.2 15.3 15.4 15.5 15.8 15.9 15.10 16.1 17.1 17.2 21.1 21.2 21.4 21.5 21.6 21.7 21.8 21.9 21.10 22.1 22.2

To find different orthogonal cutting picture . . To find tool life . . . . . . . . . . . . . . . . . . find force and coefficient of friction . . . . . . To find terms of orthogonal cutting . . . . . . . To solve tool life equation . . . . . . . . . . . Determine normal and tangential force . . . . . To find cutting and thrust force . . . . . . . . . find terms of orthogonal rake system . . . . . . Calculate CLA . . . . . . . . . . . . . . . . . . Calculate back and side rake angle . . . . . . . Calculate inclination and rake angle . . . . . . find different powers and resistance . . . . . . . Calculate percentage increase in tool life . . . . To find percentage of total energy . . . . . . . To find power and different energies . . . . . . Determine components of force and power . . . calculate horsepower at cutter and motor . . . Determine broaching power and Design broach Estimate moment thrust force and power . . . Design shell inserted blade reamer . . . . . . . To design single point cutting tool . . . . . . . find various terms for stainless steel . . . . . . To find MRR power and torque . . . . . . . . . find MRR power torque and time . . . . . . . Calculate settings of gear tooth . . . . . . . . . Calculate best wire size . . . . . . . . . . . . . Calculate size and distances over wire . . . . . Construct R and X chart . . . . . . . . . . . . Construct the control charts . . . . . . . . . . . Calculate poisson probabilities . . . . . . . . . Calculate probabilities of defective items . . . Determine producers and consumers risk . . . . Evaluate preliminary and revised control limits Find control limits for c chart . . . . . . . . . find control limits for charts . . . . . . . . . . . Determine producers and consumers risk . . . . Find range of cutting velocity . . . . . . . . . . Determine speed ratios and teeth . . . . . . . . 7

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81 82 83 84 85 85 86 87 88 88 89 89 90 90 91 92 94 94 96 96 97 98 99 99 101 102 102 103 104 106 106 107 107 109 110 112 114 115

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

22.3 22.4 22.5 23.1 23.2 23.3 23.4 23.5 23.6 23.7 23.8 23.9 23.10 26.1

Calculate Calculate Calculate Calculate Calculate Calculate Calculate Calculate Calculate Calculate Calculate Calculate Calculate Calculate

speed and number of teeths common ratio . . . . . . . . gear ratio teeth and speed . forecast . . . . . . . . . . . forecat by SMA method . . forecat by WMA method . forecast for january . . . . . total cost . . . . . . . . . . economical order quantity . economic lot size . . . . . . inventory control terms . . . discount offered . . . . . . . EOQ and reorder point . . . number of machine required

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116 117 118 120 120 121 121 122 122 122 123 123 124 125

Chapter 2 Press Tool Design

Scilab code Exa 2.1 find total pressure and dimensions 1 2 3 4 5 6 7 8 9 10 11 12

clc D = 50 // D i a m e t e r o f w a s h e r i n mm t = 4 // t h i c k n e s s o f m a t e r i a l i n mm d = 24 // d i a m e t e r o f h o l e i n mm p = 360 // s h e a r s t r e n g t h o f m a t e r i a l i n N/mmˆ2 F1 = %pi * D * t * p // b l a n k i n g p r e s s u r e i n N F2 = %pi * d * t * p // p i e r c i n g p r e s s u r e i n N F = F1 + F2 // t o t a l p r e s s u r e i n N d1 = d + 0.4 // p i e r c i n g d i e d i a m e t e r i n mm d2 = D - 0.4 // b l a n k punch d i a m e t e r i n mm c = 0.8* F // p r e s s c a p a c i t y i n N printf ( ” \n B l a n k i n g p r e s s u r e = %d kN\n P i e r c i n g p r e s s u r e = %0 . 3 f KN\n T o t a l p r e s s u r e r e q u i r e d = %0 . 1 f KN” , F1 /1000 , F2 /1000 , F /1000) 13 printf ( ” \n p i e r c i n g punch d i a m e t e r = %0 . 2 f cm\n b l a n k i n g punch d i a m e t r e = %0 . 2 f cm \n p r e s s c a p a c i t y = %0 . 2 f KN\n ” , d1 /10 , d2 /10 , c /1000) 14 // Answers v a r y due t o round o f f e r r o r

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Scilab code Exa 2.2 To find number of draws 1 2 3 4 5 6 7 8 9 10 11 12

clc h = 10 // h e i g h t o f cup i n cm d = 5 // d i a m e t e r o f cup i n cm D = sqrt ( d ^2 + 4* d * h ) // b l a n k d i a m e t e r i n cm // h e i g h t d a i m e t e r r a t i o i s 2 . T h e r e f o r e from t a b l e 2.9 t o t a l reductions are 3 r1 = 0.45* D // f i r s t r e d u c t i o n i s 45% d1 = D - r1 // d i a m e t e r a t f i r s t draw i n cm r2 = d1 *0.25 // s e c o n d r e d u c t i o n i s 25% d2 = d1 - r2 // d i a m e t e r a t s e c o n d draw i n cm r3 = d2 *0.2 // t h i r d r e d u c t i o n i s 20% d3 = d2 - r3 // d i a m e t e r a t t h i r d draw i n cm printf ( ” \n D i a m e t e r a t f i r s t draw = %0 . 2 f cm\n D i a m e t e r a t s e c o n d draw = %0 . 2 f cm\n D i a m e t e r a t t h i r d draw =%0 . 3 f cm” , d1 , d2 , d3 )

Scilab code Exa 2.3 To calculate bending force 1 2 3 4 5 6 7 8

clc K = 1.20 // d i e −o p e n i n g f a c t o r L = 37.5 // Length o f s t r i p i n cm T = 2.5 // t h i c k n e s s o f s t r i p i n mm sigma_ut = 630 // t e n s i l e s t r e n g t h i n N/mmˆ2 W = 16* T // w i d t h o f d i e o p e n i n g i n mm F = ( K * L *10* sigma_ut * T ^2) / W // b e n d i n g f o r c e i n N printf ( ” \n b e n d i n g f o r c e = %0 . 1 f KN” , F /1000)

Scilab code Exa 2.4 find blanking force and work done 1 clc 2 b = 25 // w i d t h o f b l a n k i n mm

10

3 l = 30 // l e n g t h o f b l a n k i n mm 4 tau = 450 // u l t i m a t e s h e a r s t r e s s 5 6 7 8 9 10

o f m a t e r i a l i n N/ mmˆ2 t = 1.5 // t h i c k n e s s o f m e t a l s t r i p i n mm p = 2*( l + b ) // p e r i m e t e r o f b l a n k i n mm f = p * t * tau // b l a n k i n g f o r c e i n N Pt = 0.25* t // punch t r a v e l i n mm w = f * Pt // work done i n Nmm printf ( ” \n b l a n k i n g f o r c e = %0 . 2 f KN\n work done = %0 . 2 f Nm” , f /1000 , w /1000)

Scilab code Exa 2.5 To find elastic recovery of material 1 2 3 4 5 6 7 8 9 10

clc t = 1.5 // t h i c k n e s s i n mm c = 0.05* t // c l e a r a n c e i n mm D = 25.4 // o u t s i d e d i a m e t e r i n mm D_o = D - 0.05 // b l a n k d i e o p e n i n g i n mm B_s = D_o - 2* c // b l a n k i n g punch s i z e i n mm d = 12.7 // i n t e r n a l d i a m e t e r i n mm P_s = d + 0.05 // p i e r c i n g punch s i z e i n mm D_s = P_s + 2* c // p i e r c i n g d i e s i z e i n mm printf ( ” \n c l e a r a n c e = %0 . 3 f mm\n b l a n k d i e o p e n i n g s i z e = %0 . 2 f mm ” ,c , D_o ) 11 printf ( ” \n b l a n k i n g punch s i z e = %0 . 2 f mm\n p i e r c i n g punch s i z e = %0 . 2 f mm\n p i e r c i n g d i e s i z e = %0 . 2 f mm” ,B_s , P_s , D_s )

Scilab code Exa 2.6 To find cutting forces 1 clc 2 D = 25.4 // o u t s i d e d i a m e t e r i n mm 3 d = 12.7 // i n t e r n a l d i a m e t e r i n mm

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4 5 6 7 8 9 10 11

12 13

1.5 // t h i c k n e s s i n mm = 280 // u l t i m a t e s h e a r i n g s t r e n g t h i n N/mmˆ2 %pi *( D + d ) * t * tau // t o t a l c u t t i n g f o r c e i n N = %pi * D * t * tau // c u t t i n g f o r c e when p u n c h e s a r e staggered in N k = 0.6 // p e n e t r a t i o n i = 1 // s h e a r o f punch i n mm F_p = ( t * k * F ) /( k * t + i ) // c u t t i n g f o r c e when b o t h punches act t o g e t h e r in N printf ( ” \n s h e a r f o r c e when b o t h punch a c t a t same t i m e and no s h e a r i s a p p l i e d = %0 . 2 f kN” , F /1000) printf ( ” \n c u t t i n g f o r c e when p u n c h e s a r e s t a g g e r e d = %0 . 1 f kN” , F_s /1000) printf ( ” \n c u t t i n g f o r c e when t h e r e i s p e n e t r a t i o n and s h e a r on punch = %0 . 1 f kN” , F_p /1000)

t = tau F = F_s

Scilab code Exa 2.7 To calculate amount of shear 1 2 3 4 5 6 7 8 9 10

clc D = 60 // h o l e d i a m e t e r i n mm tau = 450 // u l t i m a t e s h e a r s t r e n g t h i n N/mmˆ2 t = 2.5 // t h i c k n e s s i n mm F = %pi * D * t * tau // maximum punch f o r c e i n N w = F *0.4* t // work done i n Nm f = F /2 // p u n c h i n g f o r c e i n N k = 0.4 // p e n e t r a t i o n p e r c e n t a g e i = k * t *( F - f ) / f // s h e a r on punch i n mm printf ( ” \n Amount o f s h e a r on punch = %d mm” , i )

Scilab code Exa 2.8 To find economy of material 1 clc

12

2 3 4 5 6 7 8 9 10 11 12 13 14 15

16 17

// from f i g 2 . 2 7 w = 2.5 // cm t = 3.2 // s t r i p t h i c k n e s s i n mm h = 10 // cm a = t + 0.015* h *10 // back s c r a p and f r o n t s c r a p i n mm b = t // s c r a p b r i d g e i n mm W = h + (2* a ) /10 // w i d t h o f s t r i p i n cm W = ceil ( W ) // cm s = w + b /10 // l e n g t h o f one p i e c e o f s t o c k i n cm L = 240 // l e n g t h o f s t r i p i n cm N = (L - b ) / s // number o f p a r t s y = L - ( N * s + b /10) // s c r a p r e m a i n i n g a t t h e end i n mm printf ( ” \n V a l u e o f back s c r a p = %0 . 1 f mm\n V a l u e o f s c r a p b r i d g e = %0 . 1 f mm ” , a , b ) printf ( ” \n Width o f s t r i p = %0 . 2 f cm\n Length o f one p i e c e o f s t o c k n e e d e d t o p r o d u c e one p a r t = %0 . 2 f cm” , W , s ) printf ( ” \n Number o f p a r t s = %0 . 1 f b l a n k s \n S c r a p r e m a i n i n g a t t h e end = %0 . 2 f mm” , N , y ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 2.9 Calculations for designing drawing die 1 2 3 4 5 6 7 8

clc // from f i g u r e 2 . 7 3 t = 0.8 // t h i c k n e s s i n mm d = 50 // s h e l l d i a m e t e r i n mm r = 1.6 // r a d i u s o f bottom c o r n e r i n mm h = 50 // h e i g h t i n mm D = sqrt ( d ^2 + 4* d * h ) // s h e l l b l a n k s i z e i n mm el = 6.4 // e x t r a l e n g t h r e q u i r e d t o add i n s h e l l blank s i z e 9 D = D + el // mm 13

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

32

pr = 100*(1 -( d / D ) ) // p e r c e n t a g e r e d u c t i o n ratio = h / d n = 2 // number o f draws R1 = 45 // f i r s t r e d u c t i o n D1 = D - R1 * D /100 // d i a m e t e r a t f i r s t r e d u c t i o n i n mm R2 = 100*(1 -( d / D1 ) ) // s e c o n d r e d u c t i o n PR = 4* t // punch r a d i u s i n mm PR = ceil ( PR ) DR = 6 // d i e r a d i u s i n mm DC1 = 0.87 // d i e c l e a r a n c e f o r f i r s t draw i n mm DC2 = 0.88 // d i e c l e a r a n c e f o r s e c o n d draw i n mm PD2 = d - 2* t // punch d i a m e t e r f o r s e c o n d draw i n mm DD2 = PD2 + 2* DC2 // Die o p e n i n g d i a m e t e r f o r s e c o n d draw i n mm PD1 = D1 - 2* t // punch d i a m e t e r f o r f i r s t draw i n mm DD1 = D1 + 2* DC1 // Die o p e n i n g d i a m e t e r f o r f i r s t draw i n mm // Drawing p r e s s u r e c = 0.65 // c o n s t a n t sigma = 427 // N/mmˆ2 F = %pi * d * t * sigma *( D /d - c ) // Drawing p r e s s u r e i n mm Bhp = 30.8 // b l a n k i n g h o l d i n g p r e s s u r e i n kN pc = 150 // p r e s s c a p a c i t y i n kN printf ( ” \n ( i ) s i z e o f b l a n k = %0 . 2 f mm \n ( i i ) P e r c e n t a g e r e d u c t i o n = %0 . 1 f p e r c e n t \n ( i i i ) Number o f draws = %d \n ( i v ) R a d i u s on punch = %d mm \n Die R a d i u s = %d mm \n ( v ) Die c l e a r a n c e f o r f i r s t draw = %0 . 2 f mm \n die clearance f o r s e c o n d draw = %0 . 2 f mm” , D , pr ,n , PR , DR , DC1 , DC2 ) printf ( ” \n Punch d i a m e t e r f o r s e c o n d draw = %0 . 1 f mm \n Die o p e n i n g d i a m e t e r f o r s e c o n d draw = %0 . 2 f mm \n Punch d i a m e t e r f o r f i r s t draw = %0 . 3 f mm \n Die o p e n i n g d i a m e t e r f o r f i r s t draw = %0 . 3 f mm\n ( v i ) Drawing p r e s s u r e = %0 . 2 f mm \n ( 14

v i i ) Blank h o l d i n g p r e s s u r e = %d kN \n ( v i i i ) P r e s s c a p a c i t y = %d kN” , PD2 , DD2 , PD1 , DD1 , F /1000 , Bhp , pc ) 33 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 2.10 Determine developed length 1 2 3 4 5 6 7 8 9 10 11

clc // from f i g u r e 2 . 7 4 l1 = 76 - ( 2.3 + 0.90) // l e n g t h 1 i n mm l2 = 115 - (2.3 + 0.90) // l e n g t h 2 i n mm t = 2.3 // mm r = 0.90 // i n n e r r a d i u s i n mm k = t /3 // mm B = 0.5* %pi *( r + k ) // b e n d i n g a l l o w a n c e i n mm d = l1 + l2 + B // d e v e l o p e d l e n g t h i n mm printf ( ” \n D e v e l o p e d l e n g t h = %0 . 2 f mm” , d ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 2.11 To calculate bending force 1 2 3 4 5 6 7 8 9

clc t = 3.2 // t h i c k n e s s o f b l a n k i n mm l = 900 // b e n d i n g l e n g t h i n mm sigma = 400 // u l t i m a t e t e n s i l e s t r e n g t h i n N/mmˆ2 k = 0.67 // d i e o p e n i n g f a c t o r r1 = 9.5 // punch r a d i u s i n mm r2 = 9.5 // d i e e d g e r a d i u s i n mm c = 3.2 // c l e a r a n c e i n mm w = r1 + r2 + c // w i d t h b e t w e e n c o n t a c t p o i n t s batween d i e and punch i n mm 10 F = ( k * l * sigma * t ^2) / w // b e n d i n g f o r c e i n N 11 F = floor ( F /10) *10 // N 15

12

printf ( ” \n b e n d i n g f o r c e = %0 . 2 f kN” ,F /1000)

Scilab code Exa 2.12 To calculate bending force 1 2 3 4 5 6 7 8

clc k = 1.33 // d i e o p e n i n g f a c t o r l = 1200 // bend l e n g t h i n mm sigma = 455 // u l t i m a t e t e n s i l e s t r e n g t h i n N/mmˆ2 t = 1.6 // b l a n k t h i c k n e s s i n mm w = 8* t // w i d t h o f d i e o p e n i n g i n mm F = k * l * sigma * t ^2/ w // b e n d i n g f o r c e i n N printf ( ” \n b e n d i n g f o r c e = %0 . 2 f kN” , F /1000)

Scilab code Exa 2.13 calculate capacity of double bending die 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc c = 1.25 // c l e a r a n c e i n mm r1 = 3 // d i e r a d i u s i n mm r2 = 1.5 // punch r a d i u s i n mm sigma = 315 // u l t i m a t e t e n s i l e s t r e n g t h i n MPa t = 1 // t h i c k n e s s i n mm l = 50 // w i d t h a t bend i n mm w = r1 + r2 + c // w i d t h b e t w e e n c o n t a c t p o i n t s on d i e and punch i n mm F = 0.67* l * sigma * t ^2/ w // b e n d i n g f o r c e i n N F_p = 0.67* sigma * l * t // pad f o r c e i n N sigma_c = 560 // s e t t i n g p r e s s u r e i n MPa b1 = 2 // b e a d s on punch b = b1 * r1 // mm F_b = sigma_c * l * b // b o t t o m i n g f o r c e i n N F_o = F_p + F_b // F o r c e r e q u i r e d when b o t t o m i n g i s used in N 16

16 F_n = F + F_p // F o r c e r e q u i r e d when b o t t o m i n g

i s not

used in N printf ( ” \n F o r c e r e q u i r e d when b o t t o m i n g i s u s e d = %0 . 1 f t o n n e s ” , F_o /(9.81*1000) ) 18 printf ( ” \n F o r c e r e q u i r e d when b o t t o m i n g i s n o t u s e d = %0 . 3 f t o n n e s ” , F_n /(9.81*1000) ) 17

Scilab code Exa 2.14 To calculate cutting force 1 2 3 4 5 6 7 8 9

clc w = 2 // w i d t h i n mm t = 5 // t h i c k n e s s i n mm theta =6 // s h e a r i n d e g r e e s tau = 382.5 // u l t i m a t e s h e a r s t r e s s i n MPa F = w * t * tau *1000 // c u t t i n g f o r c e i n N l = t / sin ( theta * %pi /180) // l e n g t h t o be c u t i n mm F_i = l * t * tau // c u t t i n g f o r c e i n N printf ( ” \n c u t t i n g f o r c e w i t h p a r a l l e l c u t t i n g e d g e s = %0 . 3 f MN\n c u t t i n g f o r c e when s h e a r i s c o n s i d e r e d = %0 . 2 f kN ” ,F /10^6 , F_i /1000)

Scilab code Exa 2.15 Determine blank and punch diameter 1 2 3 4 5 6 7 8 9 10

clc d1 = 105 // i n s i d e d i a m e t e r i n mm h = 90 // d e p t h i n mm t = 1 // t h i c k n e s s i n mm D = sqrt ( d1 ^2+4* d1 * h ) // b l a n k d i a m e t e r i n mm tr = t *100/ D // t h i c k n e s s r a t i o // from t a b l e s a f e d r a w i n g r a t i o i s 1 . 8 2 r = 1.82 // draw r a t i o d2 = D / r // d i a m e t e r f o r f i r s t draw i n mm d = 130 // L e t d i a m e t e r f o r f i r s t draw i n mm 17

11 ratio1 = D / d // D/d f o r f i r s t draw 12 ratio2 = d / d1 // D/d f o r s e c o n d draw 13 h1 =(( D ) ^2 -( d ) ^2) /(4* d ) // Depth o f f i r s t draw i n mm 14 sigma = 415 // N/mmˆ2 15 c = 0.65 // c o n s t a n t 16 pc = %pi * d * t * sigma ( D / d - c ) // p r e s s c a p a c i t y i n kN 17 printf ( ” \n Blank d i a m e t e r = %d mm \n T h i c k n e s s r a t i o

= %0 . 3 f \n D i a m e t e r f o r f i r s t draw = %d mm \n Depth o f f i r s t draw = %0 . 2 f mm \n P r e s s c a p a c i t y = %0 . 2 f kN” ,D , tr , d2 , h1 , pc /1000) 18 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 2.16 To find drawing operations and force 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc d = 80 // d i a m e t e r i n mm h = 250 // h e i g h t i n mm D = sqrt (( d ^2+4* d * h ) ) /10 // b l a n k d i a m e t e r i n cm D1 = 0.5* D // d i a m e t e r a f t e r f i r s t draw i n cm // l e t r e d u c t i o n be 40% i n s e c o n d draw D2 = D1 -0.4* D1 // d i a m e t e r a f t e r s c o n d draw i n cm R = (1 - ( d /(10* D2 ) ) ) *100 // p e r c e n t a g e r e d u c t i o n f o r t h i r d draw l1 = (( D ) ^2 -( D1 ) ^2) /(4* D1 ) // h e i g h t o f cup a f t e r f i r s t draw i n cm l2 = (( D ) ^2 -( D2 ) ^2) /(4* D2 ) // h e i g h t o f cup a f t e r f i r s t draw i n cm l3 = (( D ) ^2 -( d /10) ^2) /(4* d /10) // h e i g h t o f cup a f t e r f i r s t draw i n cm t = 3 // mm sigma = 250 // N/mmˆ2 C = 0.66 F = %pi * d /10* t * sigma *(( D *10/ d ) -C ) // d r a w i n g f o r c e i n kN printf ( ” \n D i a m e t e r a f t e r f i r s t draw = %0 . 1 f \n 18

D i a m e t e r a f t e r s e c o n d draw = %0 . 2 f \n P e r c e n t a g e r e d u c t i o n a f t e r t h i r d draw = %d p e r c e n t ” ,D1 , D2 , R ) 17 printf ( ” \n H e i g h t o f cup a f t e r f i r s t draw = %0 . 2 f cm \n H e i g h t o f cup a f t e r s e c o n d draw = %0 . 2 f cm\n H e i g h t o f cup a f t e r t h i r d draw = %0 . 2 f cm” , l1 , l2 , l3 ) 18 printf ( ” \n Drawing f o r c e = %0 . 3 f kN” ,F /1000) 19 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 2.17 Determine developed length 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc // from f i g u r e 2 . 7 5 ( a ) r1 = 30 // r a d i u s i n mm t = 10 // t h i c k n e s s i n mm h1 = 300 // h e i g h t i n mm ir1 = r1 - t // i n n e r r a d i u s o f b e n d s i n mm L1 = h1 -( ir1 + t ) // mm alpha1 = 90 // d e g r e e r2 = 2* t // mm k = 0.33* t // mm L2 = alpha1 *2* %pi *( r2 + k ) /360 // mm w = 200 // mm L3 = w -2*( t + ir1 ) // mm L4 = L2 //mm h2 = 100 // mm L5 = h2 -( t + ir1 ) // mm r3 = 150 //mm ir2 = r3 - t // i n n e r r a d i u s i n mm alpha2 = 180 // d e g r e e L6 = alpha2 *2* %pi *( ir2 + k ) /360 // mm dl = L1 + L2 + L3 + L4 + L5 + L6 // T o t a l d e v e l o p e d l e n g t h i n mm 22 printf ( ” \n T o t a l d e v e l o p e d l e n g t h = %0 . 2 f mm” , dl ) 23 // Answers v a r y due t o round o f f e r r o r 19

20

Chapter 4 Cost Estimating

Scilab code Exa 4.1 To calculate total cost and SP 1 2 3 4 5 6 7 8 9 10 11

clc d_m = 5500 // c o s t o f d i r e c t m a t e r i a l i n Rs d_l = 3000 // m a n u f a c t u r i n g wages i n Rs // f a c t o r y o v e r h e a d i s 100% 0 f m a n u f a c t u r i n g wages f_o = (100* d_l ) /100 // f a c t o r y o v e r h e a d s i n Rs FC = d_m + d_l + f_o // f a c t o r y c o s t i n Rs nm_o = 15* FC /100 // non−m a n u f a c t u r i n g o v e r h e a d s i n Rs tc = FC + nm_o // t o t a l c o s t i n Rs p = 12* tc /100 // p r o f i t i n Rs sp = tc + p // s e l l i n g p r i c e i n Rs printf ( ” \n T o t a l c o s t = Rs %d\n S e l l i n g p r i c e = Rs %d” , tc , sp )

Scilab code Exa 4.2 To find selling price 1 clc 2 // g i v e n

21

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

OS_RM = 20000 // o p e n i n g s t o c k o f raw m a t e r i a l s i n Rs CS_RM = 30000 // c l o s i n g s t o c k o f raw m a t e r i a l s i n Rs TP_RM = 170000 // t o t a l p u r c h a s e i n y e a r i n Rs OS_FG = 10000 // o p e n i n g s t o c k o f f i n i s h e d g o o d s i n Rs CS_FG = 15000 // c l o s i n g s t o c k o f f i n i s h e d g o o d s i n Rs sales = 489500 // s a l e s o f f i n i s h e d g o o d s i n Rs D_W = 120000 // d i r e c t wages i n Rs F_E1 = 120000 // f a c t o r y e x p e n s e s i n Rs NM_E = 50000 // non−m a n u f a c t u r i n g e x p e n s e s i n Rs DMC = OS_RM + TP_RM - CS_RM // d i r e c t m a t e r i a l c o s t FC = DMC + D_W + F_E1 // f a c t o r y c o s t TC = FC + NM_E // t o t a l c o s t FG_S = OS_FG + TC - CS_FG // c o s t o f f i n i s h e d g o o d s s o l d i n Rs P = sales - FG_S // p r o f i t i n Rs F_E2 = ( F_E1 ) / D_W *100 // f a c t o r y e x p e n s e s i n p e r c e n t NM_C = ( NM_E ) / FC *100 // non−m a n u f a c t u r i n g e x p e n s e s to f a c t o r y cost P_C = ( P / FG_S ) *100 // p r o f i t t o c o s t o f s a l e s dm = 20000 // d i r e c t m a t e r i a l i n Rs dw = 30000 // d i r e c t wages i n Rs fe = dw // f a c t o r y e x p e n s e s fc = dm + dw + fe // f a c t o r y c o s t i n Rs nme = NM_C * fc /100 // non−m a n u f a c t u r i n g e x p e n s e s i n Rs tc = fc + nme // t o t a l c o s t i n Rs p = ( P_C * tc ) /100 // p r o f i t i n Rs sp = tc + p // s e l l i n g p r i c e i n Rs printf ( ” \n S e l l i n g p r i c e = Rs %d” , sp )

Scilab code Exa 4.3 To find factory cost 22

1 2 3 4 5 6 7 8 9 10 11 12

clc d = l = p = g = w =

38 // d i a m e t e r o f b a r i n mm 25 // l e n g t h o f b a r i n mm 8.6 // d e n s i t y gm/cmˆ3 9.81 // a c c e l e r a t i o n due t o g r a v i t y i n m/ s ˆ2 ( %pi * d ^2* l * p * g ) /(4*10^6) // w e i g h t o f m a t e r i a l in N mc = w *1.625 // m a t e r i a l c o s t i n Rs lc = (2*90) /60 // l a b o u r c o s t i n Rs fo = 0.5* lc // f a c t o r y o v e r h e a d s i n Rs fc = mc + lc + fo // f a c t o r y c o s t i n Rs printf ( ” \n f a c t o r y c o s t = Rs %0 . 2 f ” , fc ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 4.4 find production cost and time taken 1 2 3 4 5 6 7 8 9 10

clc sp = 65 // s e l l i n g p r i c e i n Rs profit = 0.2* sp // p r o f i t i n Rs tc = sp - profit // t o t a l c o s t i n Rs P = ( sp - profit ) /1.4 // p r o d u c t i o n c o s t i n Rs DM = 15 // c o s t o f d i r e c t m a t e r i a l i n Rs W = ( P - DM ) / 1.4 // d i r e c t l a b o u r c o s t i n Rs tt = W /2 // t i m e t a k e n i n h o u r s printf ( ” \n Time t a k e n = %0 . 3 f Hours ” , tt ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 4.5 To find profit 1 2 3 4

clc mp = 6000 // market p r i c e o f machine i n Rs d = 0.2* mp // d i s c o u n t i n Rs sp = mp - d // s e l l i n g p r i c e o f f a c t o r y i n Rs 23

5 6 7 8 9 10 11

mc = 400 // m a t e r i a l c o s t i n Rs lc = 1600 // l a b o u r c o s t i n Rs fo = 800 // f a c t o r y o v e r h e a d s i n Rs F = mc + lc + fo // f a c t o r y c o s t i n Rs se = 0.5* F // s e l l i n g e x p e n s e s i n Rs profit = sp - ( F + se ) // Rs printf ( ” \n p r o f i t = Rs %d” , profit )

Scilab code Exa 4.6 To find lot size and time 1 2 3 4 5 6 7 8 9 10

clc a = 1500 // r e q u i r e m e n t s o f c o m p o n e n t s s = 30 // c o s t o f e a c h s e t up i n Rs k = 0.2 // c h a r g e f a c t o r c = 5 // c o s t o f e a c h p a r t i n Rs N = 5* sqrt ( a * s ) /( k * c ) // e c o n o m i c l o t s i z e printf ( ” \n Economic l o t s i z e = %d p i e c e s ” , N ) S = ( N * s ) / a // t i m e f o r e a c h s e t up i n h o u r s printf ( ” \n Time f o r e a c h s e t up = %0 . 2 f h o u r s ” , S ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 4.7 To find time to change cutter 1 2 3 4 5 6

clc Tc = 2 // t i m e t a k e n by c u t t e r p e r c y c l e i n m i n u t e s Tk = 10 // t i m e t a k e n t o c h a n g e c u t t e r i n m i n u t e s T = 240 // t o o l l i f e i n m i n u t e s t = ( Tc * Tk ) / T // t i m e t o c h a n g e t h e c u t t e r i n min . printf ( ” \n U n i t t i m e t o c h a n g e t h e c u t t e r = %0 . 3 f min ” , t ) 7 // E r r o r i n t e x t b o o k

24

Scilab code Exa 4.8 To find tool change time 1 clc 2 Tk = 360 // t i m e t a k e n by t o o l t o c u t b e f o r e

s h a r p e n i n g i n min . 3 Tc = 20 // t i m e t a k e n t o c h a n g e t h e t o o l i n min . 4 T = 4320 // t i m e t a k e n b e f o r e i t i s d i s c a r d e d i n min

. 5 t = ( Tc * Tk ) / T // t o o l c h a n g e t i m e p e r c y c l e i n min . 6 printf ( ” \n U n i t t o o l c h a n g e t i m e p e r c y c l e = %0 . 2 f

min ” , t )

Scilab code Exa 4.9 To calculate measuring time allowance 1 2 3 4 5 6 7

clc Tc = 10 // t i m e t a k e n t o c h e c k h o l e i n s e c s F = 2 // f r e q u e n c y o f c h e c k i n g d i m e n s i o n tc = Tc * F // t i m e t a k e n t o c h e c k one p i e c e i n s e c s N = 200 // number o f p i e c e s Tc = tc *( N + 1) // T o t a l t i m e i n s e c printf ( ” \n T o t a l t i m e t a k e n t o c h e c k d i m e n s i o n s = %d min ” , Tc /60)

Scilab code Exa 4.10 To find direct labour cost 1 clc 2 forgings = 40 3 setup = 4 4 Tc = 12 // m a c h i n i n g t i m e i n min .

25

per f o r g i n g

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

nmt = 21 // non−m a c h i n i n g i n min . p e r f o r g i n g st = 45 // s e t up t i m e p e r s e t up ts = 5 // t o t a l s h a r p e n i n g i n min . p e r f o r g i n g f = 20 // f a t i g u e i n p e r c e n t f = f /100 pn = 5 // p e r s o n a l n e e d s i n p e r c e n t pn = pn /100 Tk = 10 // t o o l c h a n h e t i m e i n min . T = 8 // t o o l l i f e i n h o u r s ct = 15 // c h e c k i n g t i m e w i t h 5 c h e c k s i n 15 s e c s R = 1.4 // p e r f o r m a n c e f a c t o r dlc = 5 // d i r e c t l a b o u r c o s t i n Rs p e r h o u r tt = Tc + nmt // m a c h i n i n g and non−m a c h i n i n g t i m e i n min . ft = f * tt // f a t i g u e t i m e i n min . pnt = pn * tt // p e r s o n a l n e e d s i n min . t = ( Tc * Tk ) /( T *60) // t o t a l s h a r p e n i n g t i m e i n min . per f o r g i n g mct = ( ts * ct ) /60 // m e a s u r i n g and c h e c k i n g t i m e i n min . p e r f o r g i n g su = Tc + nmt + pnt + ft + t + mct // sum o f t i m e s i n min . tf = su * forgings // t i m e f o r 40 f o r g i n g s i n min . tst = st * setup // t o t a l s e t up t i m e i n min . Te = tf + tst // t o t a l e s t i m a t e d t i m e i n min . Ta = Te * R // t o t a l a c t u a l t i m e i n min . lc = ( Ta * dlc ) /60 // d i r e c t l a b o u r c o s t i n Rs printf ( ” \n D i r e c t l a b o u r c o s t = Rs %0 . 1 f ” , lc )

Scilab code Exa 4.11 To find machining time 1 clc 2 // from f i g u r e 4 . 4 3 v = 100 // c u t t i n g s p e e d i n m/ min 4 D = 50 // mm

26

5 l1 = 76 // mm 6 f = 7.5 // f e e d i n mm/ r e v . 7 // Case 1 , t i m e t o t u r n 38 mm d i a m e t e r by 76 mm 8 9 10 11 12 13 14 15

length of cut N1 = (1000* v ) /( %pi * D ) // r . p .m tm1 = l1 *10/( f * N1 ) // min . // Case 2 , t i m e t o t u r n 25 mm by 38 mm l e n g t h N2 = (1000* v ) /( %pi *38) // r . p .m l2 = 38 // mm tm2 = l2 *10/( f * N2 ) // min tt = tm1 + tm2 // t o t a l t i m e i n min printf ( ” \n T o t a l t i m e = %0 . 2 f min . ” , tt )

Scilab code Exa 4.12 To find time to turn relief 1 2 3 4 5 6 7 8 9

clc // from f i g u r e 4 . 5 v = 60 // c u t t i n g s p e e d m/ min . f = 0.375 // f e e d i n mm/ r e v D = 38 // mm N = (1000*60) /( %pi * D ) // r e v / min l = 32 // mm Tm = l /( f * N ) // min printf ( ” \n Time t o t u r n e x t e r n a l r e l i e f = %0 . 2 f min . ” , Tm )

Scilab code Exa 4.13 calculate time to face on lathe 1 clc 2 // from f i g u r e 4 . 1 1 3 l = 7.5 // cm 4 Dave = (25+ 10) /2 // a v e r a g e d i a m e t e r i n cm 5 v = 27 // c u t t i n g s p e e d i n m/ min

27

6 f = 0.8 // f e e d i n mm/ r e v 7 N = (1000* v ) /( %pi * Dave *10) // r . p .m. 8 tm = l *10/( f * N ) // min . 9 printf ( ” \n The m a c h i n i n g t i m e t o f a c e on l a t h e = %0

. 2 f min . ” , tm )

Scilab code Exa 4.14 To find time to drill hole 1 2 3 4 5 6 7 8 9

clc D = 12.7 // d i a m e t e r i n mm d = 50 // d e p t h i n mm v = 75 // c u t t i n g s p e e d i n m/ min . f = 0.175 // f e e d i n mm/ r e v l = d + 2*0.29* D // l e m g t h o f d r i l l t r a v e l i n mm N = (1000* v ) /( %pi * D ) // r . p .m. tm = l /( f * N ) // min printf ( ” \n Time t a k e n t o d r i l l h o l e = %0 . 3 f min . ” , tm )

Scilab code Exa 4.15 To find time to complete cut 1 2 3 4 5 6 7 8 9 10

clc k = 1/4 // r e t u r n t i m e t o c u t t i n g r a t i o l = 900 + 2*75 // l e n g t h o f s t r o k e i n mm v = 6 // c u t t i n g s t r o k e i n m/ min f = 2 // f e e d mm/ s t r o k e w = 600 // b r e a d t h i n mm N = ( v *1000) /( l *1.25) // r . p .m N = round ( N ) time = w /( f * N ) // min printf ( ” \n Time r e q u i r e d f o r s h a p e r t o c o m p l e t e one c u t = %d min ” , time ) 28

Scilab code Exa 4.16 To find time to broach 1 2 3 4 5 6

clc l = 70 // l e n g t h o f s t r o k e i n cm cs = 11 // c u t t i n g s p e e d i n m/ min rs = 24 // r e t u r n s p e e d i n m/ min tm = ( l /(100* cs ) ) + ( l /(100* rs ) ) // min printf ( ” \n Time t a k e n t o b r o a c h a f o u r s p l i n e b r a s s = %0 . 4 f min ” , tm ) 7 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 4.17 find feed cutter travel and time 1 2 3 4 5 6 7 8 9 10 11 12 13

clc v = 50 // c u t t i n g s p e e d i n m/ min D = 150 // d i a m e t e r o f f a c e c u t t e r i n mm N = (1000* v ) /( %pi * D ) // r . p .m. f = 0.25 // f e e d mm/ t o o t h n = 10 // number o f t o o t h tf = N * f * n // t a b l e f e e d i n mm/ min l = 200 // l e n g t h o f work p i e c e i n mm d = 25 // d e p t h o f s l o t i n mm tot = sqrt ( D * d - d ^2) // t o t a l o v e r t r a v e l i n mm tct = l + tot // t o t a l c u t t e r t r a v e l i n mm time = tct / tf // min . printf ( ” \n T a b l e f e e d = %d mm/ min . \n T o t a l c u t t e r t r a v e l = %0 . 1 f mm\n Time r e q u i r e d t o machine t h e s l o t = %0 . 3 f min . ” , tf , tct , time )

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Scilab code Exa 4.18 To find cutting time 1 2 3 4 5 6 7 8 9 10 11 12

clc D = w = l = f = n = N = tot

63.5 // d i a m e t e r o f p l a i n m i l l i n g c u t t e r i n mm 30 // w i d t h o f b l o c k i n mm 180 // l e n g t h o f b l o c k i n mm 0.125 // f e e d i n mm/ t o o t h 6 // no . o f t e e t h 1500 // s p i n d l e s p e e d i n r . p .m = ( D - sqrt ( D ^2 - w ^2) ) /2 // t o t a l o v e r t r a v e l i n mm tct = l + tot // t o t a l c u t t e r t r a v e l i n mm Tm = tct /( f * n * N ) // c u t t i n g t i m e i n min printf ( ” C u t t i n g t i m e = %0 . 3 f min . ” , Tm ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 4.19 To find milling time 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc // from f i g u r e 4 . 1 7 d = 19 // d e p t h o f c u t i n mm D1 = 5 // d i a m e t e r o f round b a r i n cm v = 50 // c u t t i n g s p e e d i n m/ min n = 8 // number o f t e e t h f = 0.2 // f e e d i n mm/ t o o t h l = 2* sqrt ( d * D1 *10 - d ^2) // l e n g t h o f c h o r d i n mm D2 = 10 // d a i m e t e r o f c u t t e r i n cm overrun = sqrt ( D2 *10* d + D1 *10* d - d ^2) - sqrt ( D1 *10* d - d ^2) // mm tt = l + overrun // t a b l e t r a v e l i n mm N = (1000* v ) /( %pi * D2 *10) // r . p .m tm = tt /( f * n * N ) // t i m e i n min . printf ( ” \n The m i l l i n g t i m e = %0 . 2 f min . ” , tm )

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Scilab code Exa 4.20 To find time to grind shaft 1 2 3 4 5 6 7 8 9 10 11 12

clc w = 50 // w i d t h o f g r i n d i n g wheep i n mm f = w /2 // f e e d i n mm t = 0.25 // t o a t a l s t o c k i n mm d = 0.025 // d e p t h o f c u t i n mm n = t / d // number o f c u t s v = 15 // c u t t i n g s p e e d i n m/ min D = 38 // d i a m e t e r i n mm N = (1000* v ) /( %pi * D ) // r . p .m. l = 200 // l e n g t h o f p a r t i n mm Tm = ( l *10) /( f * N ) // min . printf ( ” Time r e q u i r e d t o g r i n d t h e s h a f t = %0 . 2 f min . ” , Tm )

Scilab code Exa 4.21 To find time to cut threads 1 2 3 4 5 6 7 8 9

clc v = 6 // c u t t i n g s p e e d i n m/ min n = 5 // number o f c u t s D = 44 // d i a m e t e r i n mm N = (1000* v ) /( %pi * D ) // r . p .m f = 0.5 // f e e d i n cm l = 8.9 // l e n g t h o f c u t i n cm Tm = ( l * n ) /( f * N ) // t i m e i n min printf ( ” \n Time t o c u t t h e t h r e a d s = %0 . 2 f min ” , Tm ) 10 // Answers v a r y due t o round o f f e r r o r

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Scilab code Exa 4.22 find time to produce one piece 1 clc 2 vt = 40 // c u t t i n g s p e e d f o r t u r n i n g i n m/ min 3 vs = 8 // c u t t i n g s p e e d f o r c u t t i n g and k n u r l i n g i n 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

m/ min ft = 0.4 // f e e d f o r t u r n i n g i n mm/ r e v . ff = 0.2 // f e e d f o r f o r m i n g i n mm/ r e v d1 = 25 // d i a m e t e r i n mm l1 = 50 // mm N1 = 1000* vt /( %pi * d1 ) // s p i n d l e s p e e d i n r e v . / min . time1 = l1 /( ft * N1 ) // min . tt = 2* time1 // t o t a l t i m e i n min . d2 = 15 // mm N2 = 1000* vt /( %pi * d2 ) // r e v / min . l2 = 30 // mm time2 = l2 /( ft * N2 ) // min . eft = 0.15 // end f o r m i n g t i m e i n min . d3 = 10 // mm N3 = 1000* vs /( %pi * d3 ) // r e v . / min . l3 = 15 // mm f = 1.5 // f e e d i n min . time3 = l3 /( f * N3 ) // min . N4 = 1000* vs /( %pi * d1 ) // r e v . / min . l4 = 10 // mm time4 = l4 /( ft * N4 ) // min . time5 = 0.15 // t i m e f o r c h a m f e r i n g i n min . Dave = d1 /2 // mm N5 = 1000* vt /( %pi * Dave ) // r . p .m. time6 = Dave /( N5 * ff ) // min , tmt = tt + time2 + time3 + time4 + time5 + time6 + eft // t o t a l m a c h i n i n g t i m e i n min . t = 0.05 // min . ht = time5 +6* time6 +4* t +3* t // h a n d l i n g t i m e i n min . tot = ht + tmt // t o t a l h a n d l i n g t i m e i n min . ct = 15* tot /100 // c o n t i n g e n c y i n min . tct = tot + ct // t o t a l c y c l e t i m e i n min . st = 60 // s e t up t i m e f o r t u r r e t l a t h e 32

35 p = 100 // t o t a l p i e c e s 36 stp = st / p // s e t up t i m e p e r p i e c e i n min . 37 tpt = tct + stp // T o t a l p r o d u c t i o n t i m r p e r p i e c e

in

min . printf ( ” \n T o t a l p r o d u c t i o n t i m r p e r p i e c e = %0 . 2 f min ” , tpt ) 39 // Answers v a r y due t o round o f f e r r o r 38

33

Chapter 5 Economics of tooling

Scilab code Exa 5.1 To find value of machine tool 1 2 3 4 5 6 7 8 9 10 11 12

13 14

clc Co = 250000 // o r i g i n a l v a l u e o f machine t o o l i n Rs Cs = 25000 // s a l v a g e v a l u e i n Rs n = 20 // u s e f u l l i f e i n y e a r s d = ( Co - Cs ) / n // d e p r e c i a t i o n p e r y e a r i n Rs v1 = Co - 10* d // v a l u e o f machine t o o l a t t h e end o f 10 y e a r s i n Rs s = Co - Cs // sum a t t h e end o f u s e f u l l i f e i n Rs i = 8/100 // a n n u a l i n t e r s t r a t e D = ( s * i ) /((1 + i ) ^n -1) // a n n u a l d e p o s i t a = D *((1+ i ) ^10 -1) / i // amount a t t h e end o f 10 y e a r s i n Rs v2 = Co - a // v a l u e a t t h e end o f 10 y e a r s printf ( ” \n V a l u e o f machine a t t h e end o f 10 y e a r s t h r o u g h s t r a i g h t l i n e d e p r e c i a t i o n method = Rs %d ” , v1 ) printf ( ” \n V a l u e o f machine a t t h e end o f 10 y e a r s t h r o u g h s i n k i n g f u n d method = Rs %d” , v2 ) // Answers v a r y due t o round o f f e r r o r

34

Scilab code Exa 5.2 To find annual investment 1 2 3 4 5 6 7 8 9 10

clc p = 200000 // p r e s e n t worth i n Rs i = 10 // a n n u a l i n t e r e s t r a t e i = 10/100 n = 20 // number o f y e a r s a1 = ( p * i ) /((1+ i ) ^n -1) // a n n u a l i n v e s t m e n t u s i n g s i n k i n g f u n d f a c t o r i n Rs a2 = ( p * i *( i +1) ^ n ) /(( i +1) ^n -1) // a n n u a l i n v e s t m e n t u s i n g c a p i t a l r e c o v e r y f a c t o r i n Rs printf ( ” \ nAnnual i n v e s t m e n t u s i n g s i n k i n g f u n d f a c t o r = Rs %d /− p e r y e a r ” , a1 ) printf ( ” \ nAnnual i n v e s t m e n t u s i n g c a p i t a l r e c o v e r y f a c t o r = Rs %d /− p e r y e a r ” , a2 ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.3 find project is economical or not 1 2 3 4 5 6 7 8 9 10 11 12

clc // c a s h i n f l o w s a = 21240 // a n n u a l r e v e n u e i n Rs i = 10 // a n n u a l i n t e r e s t r a t e i = 10/100 n = 5 // p e r o d i n y e a r s f1 = 8000 // s a l v a g e v a l u e i n Rs p1 = ( a *(( i +1) ^n -1) ) /( i *( i +1) ^5) // a n n u a l r e v e n u e i n Rs p2 = f1 /( i +1) ^5 // p r e s e n t worth i n Rs t1 = p1 + p2 // t o t a l c a s h i n f l o w s i n Rs // c a s h o u t f l o w s I = 40000 // i n v e s t m e n t i n Rs 35

13 f2 = 12000 // a n n u a l payment i n Rs 14 p3 = ( f2 *((1+ i ) ^5 -1) ) /( i *(1+ i ) ^5) // a n n u a l payments 15 16 17 18

i n Rs t2 = I + p3 // t o t a l c a s h o u t f l o w s i n Rs printf ( ” \ n T o t a l c a s h i n f l o w s = Rs %0 . 2 f \ n T o t a l c a s h o u t f l o w s = Rs %0 . 2 f ” , t1 , t2 ) disp ( ” S i n c e c a s h o u t f l o w s a r e more t h a n c a s h i n f l o w s t h e r e f o r e p r o j e c t i s not economical ”) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.4 selection of economical machine 1 clc 2 // Machine A 3 f1 = 2000 // a n n u a l b e n e f i t from b e t t e r

production

q u a l i t y i n Rs i = 10 // i n t e r e s t r a t e i = 10/100 f2 = 12000 // s a l v a g e v a l u e i n Rs f3 = 8000 // o p e r a t i n g and m a i n t e n a n c e c o s t i n Rs I1 = 100000 // i n i t i a l c o s t i n Rs n = 5 // y e a r s p1 = ( f1 *((1+ i ) ^n -1) ) /( i *( i +1) ^ n ) p2 = f2 /(1+ i ) ^ n c1 = p1 + p2 // c a s h i n f l o w s i n Rs p3 = ( f3 *((1+ i ) ^n -1) ) /( i *( i +1) ^ n ) c2 = I1 + p3 // c a s h o u t f l o w s i n Rs Pa = c1 - c2 // n e t P .W. i n Rs // Machine B I2 = 60000 // i n i t i a l c o s t i n Rs f4 = 16000 // o p e r a t i n g and m a i n t e n a n c e c o s t i n Rs f5 = 14000 // r e c o n d i t i o n i n g a t t h e end o f t h i r d y e a r i n Rs 20 p4 = (16000*((1+ i ) ^5 -1) ) /( i *(1+ i ) ^5) 21 p5 = f5 /(1+ i ) ^5 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

36

22 Pb = - I2 - p4 - p5 // n e t P .W. i n Rs 23 printf ( ” \n Net P .W. o f Machine A= Rs %0 . 2 f \n Net P .W.

o f Machine B = Rs%0 . 2 f ” , Pa , Pb ) 24 disp ( ” I t i s c l e a r t h a t Net P .W o f Machine A i s n a g a t i v e a s compared t o t h a t o f Machine B , t h e r e f o r e Machine A i s e c o n o m c a l . ” ) 25 // Answers v a r y due t o round o f f e r r o r

less

Scilab code Exa 5.5 selection of machine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc // machine A c1 = 20000 // manual c o s t i n Rs c2 = 40000 // o p e r a t i n g c o s t i n Rs n1 = 2 // machine l i f e i n y e a r s i = 10 // i n t e r e s t r a t e i = 10/100 crf1 = ((1+ i ) ^ n1 -1) /( i *( i +1) ^ n1 ) // c a p i t a l r e c o v e r y factor pw1 = c1 + c2 * crf1 // p r e s e n t worth i n Rs // machine B c3 = 50000 // manual c o s t i n Rs c4 = 30000 // o p e r a t i n g c o s t i n Rs n2 = 4 // machine l i f e i n y e a r s i = 10/100 // i n t e r e s t r a t e crf2 = ((1+ i ) ^ n2 -1) /( i *( i +1) ^ n2 ) // c a p i t a l r e c o v e r y factor pw2 = c3 + c4 * crf2 // p r e s e n t worth i n Rs f o r 4 y e a r s pw3 = ( pw2 * crf1 ) / crf2 // p r e s e n t worth i n Rs f o r 2 years printf ( ” \n P .W. o f e x p e n s e s f o r A = Rs %d\n P .W. o f e x p e n s e s f o r B = Rs %0 . 2 f ” ,pw1 , pw3 ) disp ( ” As t h e e x p e n s e s o f machine B a r e l e s s , s o t h i s i s economical ”) // Answers v a r y due t o round o f f e r r o r 37

Scilab code Exa 5.6 selection of economical machine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc // Machine A i = 8 // // i n t e r e s t r a t e i = i /100 // i n t e r e s t r a t e n1 = 10 // e c o n o m i c l i f e i n y e a r s CRF1 = i *(1+0.08) ^ n1 /((1+ i ) ^ n1 -1) // c a p i t a l recovery factor p1 = 46000 // f i r s t c o s t i n Rs s1 = 8000 // s a l v a g e v a l u e i n Rs o1 = 10000 // o p e r a t i n g c h a r g e s i n Rs AC1 = ( p1 - s1 ) * CRF1 + s1 * i + o1 // a n n u a l c o s t // Machine B n2 = 15 // e c o n o m i c l i f e i n y e a r s CRF2 = i *(1+0.08) ^ n2 /((1+ i ) ^ n2 -1) // c a p i t a l recovery factor p2 = 60000 // f i r s t c o s t i n Rs s2 = 10000 // s a l v a g e v a l u e i n Rs o2 = 9200 // o p e r a t i n g c h a r g e s i n Rs AC2 = ( p2 - s2 ) * CRF2 + s2 * i + o2 // a n n u a l c o s t printf ( ” \n Annual c o s t o f machine A = Rs %0 . 2 Annual c o s t o f Machine B = Rs %0 . 2 f ” ,AC1 , disp ( ” Machine B w i l l be e c o n o m i c a l ” ) // E r r o r i n t e x t b o o k

Scilab code Exa 5.7 find ERR and economicality of project 1 clc 2 a = 100000 // Ej ( p / f , e% , j ) i n Rs 3 n = 5 // l i f e i n y e a r s

38

i n Rs

i n Rs f \n AC2 )

e = 20 // M. A . R . R . e = e /100 // M. A . R . R . i = e A = 32000 // s a v i n g s i n Rs s = 20000 // s a l v a g e v a l u e i n Rs b = (( A *((( i +1) ^ n ) -1) / i ) + s ) / a // ( F/p , I , 5 ) i2 = ( b ) ^(1/ n ) -1 // i n t e r n a l r a t e o f r e t u r n printf ( ” \n ERR = %0 . 4 f \n I n t e r n a l r a t e o f r e t u r n = %0 . 2 f p e r c e n t ” , b , i2 *100) 12 disp ( ” S i n c e I n t e r n a l r a t e o f r e t u r n i s > M. A . R . R , t h e r e f o r e p r o j e c t i s f e a s i b l e ”) 4 5 6 7 8 9 10 11

Scilab code Exa 5.9 find ERR and economicality of project 1 2 3 4 5 6 7 8 9 10 11 12

clc e = 20 // M, A . R . R . i = e // i n t e r e s t r s t e i = i /100 n = 5 // l i f e i n y e a r s s = 32000 // a n n u a l n e t s a v i n g s i n Rs p = 100000 // p r e s e n t worth i n Rs S = 20000 // s a l v a g e v a l u e i n Rs a = (p - S ) *( i /((1+ i ) ^n -1) ) // ( p−s ) (A/F , e% , n ) E = (s - a ) / p // E . R . R . R printf ( ” \n ERRR = %0 . 2 f p e r c e n t ” , E *100) disp ( ” S i n c e E . R . R . R i s > M. A . R . R . t h e r e f o r e p r o j e c t i s f e a s i b l e . ”)

Scilab code Exa 5.10 To determine acceptance of machine 1 clc 2 // machine A 3 r_e1 = 9600 // c a s h f l o w i n Rs

39

p1 = 46000 // i n t i a l c o s t i n Rs s = 0 // s a l v a g e v a l u e e = 8 // M. A . R . R e = e /100 i = 8 // i n v e s t m e n t r a t e i = i /100 n = 6 // l i f e i n y e a r s x = i /((1+ i ) ^n -1) ERRR1 = ( r_e1 - ( p1 - s ) * x ) / p1 // machine B r_e2 = 7200 // c a s h f l o w i n Rs p2 = 32000 // i n t i a l c o s t i n Rs ERRR2 = ( r_e2 - ( p2 - s ) * x ) / p2 printf ( ” \n ERRR1 = %0 . 2 f p e r c e n t \n ERRR2 = %0 . 2 f p e r c e n t ” , ERRR1 *100 , ERRR2 *100) 18 disp ( ” Only machine B i s a c c e p t e b l e ” )

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Scilab code Exa 5.11 find investment cost and unamortized value 1 2 3 4 5 6 7 8

clc pmv = 15000 // p r e s e n t market v a l u e i n Rs ss = 6000 // sum n e e d e d t o make i t s e r v i c e a b l e i n Rs ic = ss + pmv // i n v e s t m e n t c o s t i n Rs pbv = 30000 // p r e s e n t book v a l u e i n Rs sv = 15000 // s a l v a g e v a l u e i n Rs ui = pbv - sv // u n a m o r t i z e d i n v e s t m e n t i n Rs printf ( ” \n I n v e s t m e n t c o s t = Rs %d\n U n a m o r t i z e d i n v e s t m e n t = Rs %d” , ic , ui )

Scilab code Exa 5.13 To make decision of machines replacement 1 clc 2 // E x i s t i n g machine

40

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

22

pmp = 100000 // p r e s e n t market p r i c e i n Rs io = 50000 // i m m e d i a t e o v e r h a u l i n g i n Rs asl = 5 // a d d i t i o n a l s e r v i c e l i f e i n y e a r s aoc = 50000 // a n n u a l o p e r a t i n g c o s t i n Rs svo = 10000 // s a l v a g e v a l u e a f t e r o v e r h a u l i n g i n Rs pc = io + pmp // p r e s e n t c o s t i n Rs i = 10 // i n t e r e s t r a t e i = 10/100 crf1 = ( i *(1+ i ) ^ asl ) /((1+ i ) ^ asl - 1) // c a p i t a l recovery factor AC1 = ( pc - svo ) * crf1 + svo * i + aoc // a v e r a g e c o s t i n Rs // p r o p o s e d machine n = 10 // e x p e c t e d e c o n o m i c l i f e i n y e a r s ic = 300000 // i n i t i a l c o s t i n Rs sv = 100000 // s a l v a g e v a l u e i n Rs o = 30000 // a n n u a l o p e r a t i n g c o s t i n Rs crf2 = ( i *(1+ i ) ^10) /((1+ i ) ^10 - 1) AC2 = ( ic - sv ) * crf2 + sv * i + o // a v e r a g e c o s t i n Rs printf ( ” E x i s t i n g machine = Rs %0 . 3 f \n P r o p o s e d machine = Rs %0 . 2 f ” , AC1 , AC2 ) disp ( ” S i n c e t h e e q u i v a l e n t a n n u a l c o s t o f p r o p o s e d machine i s l e s s t h a n t h a t o f t h e e x i s t i n g machine , t h e r e f o r e , the replacement i s j u s t i f i e d . ”) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.15 Determine economic repair life 1 2 3 4 5 6

clc c = 20000 // f i r s t c o s t o f machine i n Rs s = 1000 // s c r a p v a l u e i n machine i n Rs b = 180 // a n n u a l i n c r e a s e i n c o s t o f r e p a i r s i n Rs n = sqrt (2*( c - s ) / b ) // y e a r s printf ( ” \n Number o f y e a r s o f e c o n o m i c r e p a i r l i f e = 41

%0 . 2 f y e a r s ” , n )

Scilab code Exa 5.16 find time to pay for itself 1 clc 2 Cn = 72000 // c o s t 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

o f new machine i n s t a l l e d and t o o l e d i n Rs Co = 28000 // c o s t o f new machime i n s t a l l e d and t o o l e d i n Rs p = 16 // h o u r l y p i e c e s Nn = 2200* p // e s t i m a t e d a n n u a l p r o d u c t i o n on new machine Ko = 17200 // p r e s e n t book v a l u e o f o l d machine i n Rs So = 6400 // s c r a p v a l u e o f o l d machine i n Rs Sn = 8000 // p r o b a b l e s c r a p v a l u e o f o l d machine i n a t t h e end o f i t s u s e f u l l i f e Rs oco = 2.5 // o p r e a t o r c o s t p e r h o u r mco = 48 // machine c o s t ro = 10 // p r o d u c t i o n r a t e p e r h o u r ocn = 2 // o p r e a t o r c o s t p e r h o u r mcn = 62 // machine c o s t rn = 16 // p r o d u c t i o n r a t e p e r h o u r Po = ( oco + mco ) / ro // l a b o u r and machine c o s t p e r u n i t on o l d machine i n Rs Pn = ( ocn + mcn ) / rn // l a b o u r and machine c o s t p e r u n i t on new machine i n Rs i = 6 // i n t e r e s t on i n v e s t m e n t i = i /100 t = 6 // a n n u a l t a x e s t = t /100 d = 10 // a n n u a l a l l o w a n c e f o r d e p r e c i a t i o n d = d /100 m = 3 // a n n u a l a l l o w a n c e f o r m a i n t e n a n c e m = m /100 42

25 n = (( Cn - Sn ) +( Ko - So ) ) /(( Nn *( Po - Pn ) ) - Cn *( i + t + d + m ) ) 26 printf ( ” \n The number o f y e a r s i n which t h e new

machine w i l l pay f o r

i t s e l f = %0 . 3 f y e a r s ” , n )

Scilab code Exa 5.17 selection of machine for job 1 clc 2 C = 80000 // c o s t 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

o f new machine i n s t a l l e d and t o o l e d i n Rs nel = 2 // number o f e n g i n e l a t h e s c = 32000* nel // f i r s t c o s t o f e n g i n e l a t h e N = 4000 // a n n u a l p r o d u c t i o n o f t u r r e t l a t h e n = 3800 // a n n u a l p r o d u c t i o n i n e n g i n e l a t h e nhp1 = 4 // hp motor L = 2256* nhp1 // a n n u a l l a b o u r c o s t o f t u r r e t l a t h e w = 5 // wage i n p e r h o u r time = 2300 // h o u r s l = time * nel * w // l a b o u r c o s t o f e n g i n e l a t h e nhp2 = 2.5 // hp motor pr = 0.35 // power r a t e i n kwh p = ( nel * nhp2 *746* time * pr ) /1000 // power c o s t P = ( nhp1 *746* time * pr ) /1000 // power c o s t F = 480 // s a v i n g I = 6/100 // i n t e r e s t r a t e T = 4/100 // t a x r a t e D = 10/100 // a l l o w a n c e f o r d e p r e c i a t i o n i n e n g i n e lathe M = 6/100 // a l l o w a n c e f o r m a i n t e n a n c e i n e n g i n e lathe B = 55/100 // l a b o u r b u r d e n i n e n g i n e l a t h e i = 6/100 // i n t e r e s t r a t e t = 4/100 // t a x r a t e d = 10/100 // a l l o w a n c e f o r d e p r e c i a t i o n i n t u r r e t lathe m = 6/100 // a l l o w a n c e f o r m a i n t e n a n c e i n t u r r e t 43

26 27 28

29 30

lathe X = ( L + B * L + P + C *( I + T + D + M ) - F ) / N x = ( l + l * B + p + c *( i + t + d + m ) ) / n printf ( ” \n U n i t p r o d u c t i o n c o s t on t u r r e t l a t h e = Rs %0 . 2 f p e r p i e c e \n U n i t p r o d u c t i o n c o s t on e n g i n e l a t h e = Rs %0 . 2 f p e r p i e c e ” , X , x ) disp ( ” T u r r e t l a t h e w i l l be more e c o n o m i c a l t h a n two engine l a t h e ”) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.18 Calculate maximum investment on turret lathe 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc X = 9.16 // p r o d u c t i o n c o s t on t u r r e t l a t h e N = 4000 // a n n u a l r e q u i r e m e n t c = X * N // c o s t f o r 4 0 0 0 p i e c e s on t u r r e t l a t h e n = 3800 // p r o d u c t i o n o f e n g i n e l a t h e l = 23000 // l a b o u r c o s t p = 3002 // power c o s t i = 6 // i n t e r e s t r a t e i = i /100 t = 4 // t a x r a t e t = t /100 d = 10 // a l l o w a n c e f o r d e p r e c i a t i o n i n t u r r e t l a t h e d = d /100 m = 6 // a l l o w a n c e f o r m a i n t e n a n c e i n t u r r e t l a t h e m = m /100 b = 55/100 // l a b o u r b u r d e n a = i+t+d+m tc = 64000 // f i r s t c o s t o f e n g i n e l a t h e c1 =( N *( l *(1+ b ) + p ) ) / n +( tc * a ) // c o s t f o r e n g i n e lathe 20 s = c1 - c // s a v i n g s 21 amt = s / a // amount i n v e s t e d i n t u r r e t l a t h e o v e r the cost of engine l a t h e 44

printf ( ” \n Amount i n v e s t e d i n t u r r e t l a t h e o v e r t h e c o s t o f e n g i n e l a t h e = Rs %d” , amt ) 23 // Answers v a r y due t o round o f f e r r o r

22

Scilab code Exa 5.19 To find years for new machine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc Cn = 60000 // c o s t o f new machine Sn = 5000 // s c r a p v a l u e o f new machine So = 1000 // s c r a p v a l u e o f o l d machine Nn = 200000 // a n n u a l p r o d u c t i o n I = 10 // i n t e r e s t r a t e I = I /100 M = 7 // a l l o w a n c e f o r m a i n t e n a n e M = M /100 T = 6 // a n n u a l t a x e s T = T /100 D = 1/10 // a l l o w a n c e f o r d e p r e c i a t i o n lco = 300 // l a b o u r c h a r g e s f o r o l d machine m = 12 // months rco = 15000 // r u n n i n g c h a r g e s f o r o l d machine pro = 50000 // p r o d u c t i o n r a t e f o r o l d machine lcn = 500 // l a b o u r c h a r g e s f o r new machine rcn = 10000 // r u n n i n g c h a r g e s f o r o l d machine prn = 200000 // p r o d u c t i o n r a t e f Po = ( lco * m + rco ) / pro // l a b o u r and machine c o s t on o l d machine 21 Pn = ( lcn * m + rcn ) / prn // l a b o u r and machine c o s t on new machine 22 n =(( Cn - Sn ) - So ) /(( Nn *( Po - Pn ) ) - Cn *( I + T + D + M ) ) // y e a r s 23 printf ( ” \n Y e a r s i n which new machine w i l l pay f o r i t s e l f = %0 . 2 f y e a r s ” , n )

45

Scilab code Exa 5.20 To find cost and pieces 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

27

clc a = 1.50 // s a v i n g i n l a b o u r b =55/100 // b u r d e n a p p l i e d on l a b o u r T = 4/100 // a l l o w a n c e f o r t a x e s M = 5/100 // a l l o w a n c e f o r m a i n t e n a n c e I = 8/100 // i n t e r e s t r a t e D = 50/100 // a l l o w a n c e f o r d e p r e c i a t i o n H = 2 // y e a r s t o a m o r t i z e t h e i n v e s t m e n t S = 50 // y e a r l y c o s t f o r s e t up C = 3000 // f i r s t c o s t N1 = ( C *( I + T + M + D ) + S ) /( a *(1+ b ) ) // a n n u a l p r o d u c t i o n when 1 run i s made r = 5 // number o f r u n s N2 = ( C *( I + T + M + D ) + S * r ) /( a *(1+ b ) ) // a n n u a l p r o d u c t i o n when 1 run i s made D1 = 100/100 // a l l o w a n c e f o r d e p r e c i a t i o n N3 = ( C *( I + T + M + D1 ) + S ) /( a *(1+ b ) ) // p r o d u c t i o n when D = 100 n1 = 1530 // p i e c e s C1 = ( n1 *( a *(1+ b ) ) -S ) /( I + T + M + D1 ) // e c o n o m i c a l investment n2 = 950 // p i e c e s a1 = 2 // l a b o u r c o s t r1 = 6 // number o f r u n s S1 = r1 * S // y e a r l y c o s t V = n2 * a1 *(1+ b ) -C *( I + T + M + D ) - S1 // p r o f i t printf ( ” \n Number o f p i e c e s when one run i s made and c o s t i s Rs 3 0 0 0 = %d p i e c e s ” , N1 ) printf ( ” \n Annual p r o d u c t i o n when 5 r u n s a r e made p e r y e a r = %d p i e c e s ” , N2 ) printf ( ” \ nAnnual p r o d u c t i o n when f i x t u r e pay f o r i t s e l f = %d p i e c e s ” , N3 ) printf ( ” \ n E c o n o m i c a l i n v e s t m e n t when 1 5 3 0 p i e c e s f o r s i n g l e run w i t h s a v i n g s Rs 1 . 5 0 p e r p i e c e = Rs %d” , C1 ) printf ( ” \ nAnnual p r o f i t when 950 p i e c e s made p e r 46

y e a r i n 6 r u n s and s a v i n g i n l a b o u r c o s t Rs 2 p e r p i e c e = Rs %d p e r y e a r ” , V ) 28 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 5.21 To find number of components 1 2 3 4 5 6 7 8 9 10 11 12

clc a = 0.125 // s a v i n g i n l a b o u r c o s t p e r u n i t b = 0.4 // o v e r h e a d a p p l i e d on d i r e c t l a b o u r s a v e d D = 1/2 // a l l o w a n c e f o r d e p r e c i a t i o n C = 2400 // f i r s t c o s t I = 6/100 // i n t e r e s t r a t e T = 4/100 // a l l o w a n c e f o r t a x e s M = 10/100 // a l l o w a n c e f o r m a i n t e n a n c e S = 80 // c o s t o f s e t up N = ( C *( I + T + D + M ) + S ) /( a *(1+ b ) ) // p i e c e s p e r y e a r t = N *2 // t o t a l number o f p i e c e s printf ( ” \n T o t a l number o f p i e c e s p r o d u c e d = %d” , t ) 13 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.22 To find number of components 1 2 3 4 5 6 7 8 9 10

clc a = b = D = C = I = T = M = n = S =

0.125 // s a v i n g i n l a b o u r c o s t p e r u n i t 0.4 // o v e r h e a d a p p l i e d on d i r e c t l a b o u r s a v e d 1/2 // a l l o w a n c e f o r d e p r e c i a t i o n 2400 // f i r s t c o s t 6/100 // i n t e r s t r a t e 4/100 // a l l o w a n c e f o r t a x e s 10/100 // a l l o w a n c e f o r m a i n t e n a n c e 6 // number o f b a c h e s 80 // c o s t o f s e t up 47

11 s1 = S * n // t o t a l s e t up c o s t 12 N = ( C *( I + T + D + M ) + s1 ) /( a *(1+ b ) ) // p i e c e s 13 t = N *2 // t o t a l number o f p i e c e s 14 printf ( ” \n T o t a l number o f p i e c e s p r o d u c e d = %d” , t

) 15 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.23 To find time and profit 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc C1 = 2000 // f i r s t c o s t s m a l l t o o l i n Rs N = 5000 // p a r t s p e r y e a r n = 5 // number o f b a t c h e s S = 50* n // c o s t o f s e t up a = 0.15 // s a v i n g i n l a b o u r c o s t p e r u n i t b = 50/100 // b u r d e n a p p l i e d on d i r e c t l a b o u r s a v e d I = 10/100 // i n t e r e s t r a t e T = 5/100 // a l l o w a n c e f o r t a x M = 10/100 // a l l o w a n c e f o r m a i n t e n a n c e H = C1 /(( N * a *(1+ b ) ) -( C1 *( I + T + M ) ) -S ) // y e a r s C2 = 1600 // c o s t o f f i x t u r e D = 1/ H // a l l o w a n c e f o r d e p r e c i a t i o n V = N * a *(1+ b ) - C2 *( I + T + D + M ) -S // p r o f i t printf ( ” \n Number o f y e a r s t a k e n by f i x t u r e o f Rs 2 0 0 0 = %0 . 2 f y e a r s \n p r o f i t made when f i x t u r e o f Rs 1 6 0 0 = Rs %d” , H ,V )

Scilab code Exa 5.24 To find minimum number of components 1 clc 2 c1 = 3 //

machine c o s t p e r component u s i n g e x i s t i n g e u i p m e n t i n Rs 3 c2 = 1 // machine c o s t u s i n g f i x t u r e i n Rs 48

4 s = c1 - c2 // s a v i n g i n machine c o s t p e r p i e c e 5 f = 1000 // c o s t o f f i x t u r e i n Rs 6 N = f /2 // c o m p o n e n ts 7 printf ( ” \n Minimum number o f c o m p o n e n t s t o be

p r o d u c e d i f c o s t o f f i x t u r e t o be r e c o v e r e d = %d” ,N )

Scilab code Exa 5.25 To calculate number of pieces 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc C = 1000 // c o s t o f f i x t u r e Co = 700 // c o s t o f o l d f i x t u r e Cs = 250 // s c r a p v a l u e a = 10 // s a v i n g p e r p i e c e i n p a i s a a = a /100 b = 30 // o v e r h e a d a p p l i e d on d i r e c t l a b o u r s a v e d b = b /100 I = 8 // i n t e r e s t r a t e I = I /100 M = 3 // a l l o w a n c e f o r m a i n t e n a n c e M = M /100 T = 12 // a l l o w a n c e f o r t a x T = T /100 H = 3/2 // a m o r t i z a t i o n D = 1/ H // a l l o w a n c e f o r d e p r e c i a t i o n N = ( C *( I + T + D + M ) +( Co - Cs ) * I ) /( a *(1+ b ) ) // p i e c e s p e r year 18 printf ( ” \n Number o f p i e c e s which must be p r o d u c e d t o b r e a k e v e n s o t h a t f i x t u r e may pay f o r i t s e l f = %d p i e c e s p e r y e a r ” , N ) 19 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.26 To find cost for new fixture 49

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc N = 9000 // number o f p i e c e s Co = 700 // c o s t o f o l d f i x t u r e Cs = 250 // s c r a p v a l u e a = 10 // s a v i n g p e r p i e c e i n p a i s a a = a /100 b = 30 // o v e r h e a d a p p l i e d on d i r e c t l a b o u r s a v e d b = b /100 I = 8 // i n t e r e s t r a t e I = I /100 M = 3 // a l l o w a n c e f o r m a i n t e n a n c e M = M /100 T = 12 // a l l o w a n c e f o r t a x T = T /100 H = 3/2 // a m o r t i z a t i o n D = 1/ H // a l l o w a n c e f o r d e p r e c i a t i o n C = ( N * a *(1+ b ) -( Co - Cs ) * I ) /( I + T + D + M ) // c o s t i n Rs printf ( ” \n C o s t f o r new f i x t u r e = Rs %d” , C ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.27 find time to amortize fixture 1 2 3 4 5 6 7 8 9 10 11 12 13

clc n = c = a = a = b = b = I = I = M = M = T = T =

6500 // y e a r l y p r o d u c t i o n 1350 // c o s t o f f i x t u r e 10 // s a v i n g p e r p i e c e i n p a i s a a /100 30 // o v e r h e a d a p p l i e d on d i r e c t l a b o u r s a v e d b /100 8 // i n t e r e s t r a t e I /100 3 // a l l o w a n c e f o r m a i n t e n a n c e M /100 12 // a l l o w a n c e f o r t a x T /100 50

14 co = 700 // c o s t o f o l d f i x t u r e 15 cs = 250 // s c r a p v a l u e 16 H = ( c ) /(( n * a *(1+ b ) ) -I *( co - cs ) -c *( I + T + M ) ) // 17

amotization in years printf ( ” \n Time t a k e n t o a m o r t i z e t h e f i x t u r e = %0 . 1 f y e a r s ” , H)

Scilab code Exa 5.28 To find profit 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc n = 9000 // p r o d u c t i o n o f p i e c e s p e r y e a r c = 1000 // f i x t u r e c o s t s Co = 700 // c o s t o f o l d f i x t u r e Cs = 250 // s c r a p v a l u e a = 10 // s a v i n g p e r p i e c e i n p a i s a a = a /100 b = 30 // o v e r h e a d a p p l i e d on d i r e c t l a b o u r s a v e d b = b /100 I = 8 // i n t e r e s t r a t e I = I /100 M = 3 // a l l o w a n c e f o r m a i n t e n a n c e M = M /100 T = 12 // a l l o w a n c e f o r t a x T = T /100 h = 1.5 // a m o r t i z a t i o n D = 1/ h // a l l o w a n c e f o r d e p r e c i a t i o n V = ( n * a *(1+ b ) ) -( c *( I + T + D + M ) ) -(( Co - Cs ) * I ) // p r o f i t printf ( ” \n p r o f i t = Rs %d ” , V ) // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.29 To find BEP Cost and Components 1 clc

51

2 3 4 5 6 7 8 9 10 11

fc1 = 100000 // f i x e d c o s t i n Rs vc1 = 100 // v a r i a b l e c o s t i n Rs p e r u n i t sp = 200 // s e l l i n g p r i c e i n Rs p e r u n i t q1 = fc1 /( sp - vc1 ) // q u a n t i t y o f p r o d u c t i o n a t b r e a k even p o i n t fc2 = 125000 // f i x e d c o s t i n Rs vc2 = 90 // v a r i a b l e c o s t i n Rs p e r u n i t q2 = fc2 /( sp - vc2 ) // q u a n t i t y o f p r o d u c t i o n a t b r e a k even p o i n t p = 20000 // p r o f i t i n Rs q3 = ( fc1 + p ) /( sp - vc1 ) // q u a n t i t y o f p r o d u c t i o n a t p r o f i t o f Rs 2 0 0 0 0 printf ( ” \n Break e v e n p o i n t = %d p i e c e s \n I f f i x e d c o s t i s 1 2 5 0 0 0 and v a r i a b l e c o s t i s Rs 90 p e r u n i t t h e n b r e a k e v e n p o i n t = %d p i e c e s \n Number o f c o m p o n e nt s t o g e t p r o f i t o f Rs 2 0 0 0 0 = %d p i e c e s ” , q1 , q2 , q3 )

Scilab code Exa 5.30 To find break even point 1 2 3 4

clc fc1 = 12000 // f i x e d c o s t f o r machine A i n Rs fc2 = 48000 // f i x e d c o s t f o r machine B i n Rs n1 = 6 // u n i t p r o d u c t i o n c o s t i n Rs p e r p i e c e f o r machine A 5 n2 = 1.2 // u n i t p r o d u c t i o n c o s t i n Rs p e r p i e c e f o r machine B 6 q = ( fc2 - fc1 ) /( n1 - n2 ) // b r e a k e v e n p o i n t 7 printf ( ” \n Break e v e n p o i n t = %d p i e c e s ” , q )

Scilab code Exa 5.31 To find break even quantity 1 clc

52

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

// c a p s t a n l a t h e tc1 = 300 // t o t a l c o s t i n Rs mc1 = 2.5 // m a t e r i a l c o s t p e r p i e c e i n Rs olc1 = 5 // o p e r a t i o n l a b o u r c o s t p e r h o u r i n Rs ct1 = 5 // c y c l e t i m e p e r p i e c e i n min . slc1 = 20 // s e t t i n g up l a b o u r c o s t i n Rs p e r h o u r st1 = 1 // s e t t i n g up t i m e i n h o u r mo1 = 300/100 // machine o v e r h e a d s o f o p e r a t i o n labour cost o1 = mo1 * olc1 // o v e r h e a d s o f c a p s t a n l a t h e i n Rs per hour fc1 = tc1 + slc1 * st1 + o1 * st1 // f i x e d c o s t o f c a p s t a n l a t h e i n Rs vc1 = mc1 + ( olc1 * ct1 ) /60 + ( o1 * ct1 ) /60 // v a r i a b l e c o s t i n Rs // A u t o m a t i c ( s i n g l e s p i n d l e ) tc2 = 300 // t o t a l c o s t i n Rs cc2 = 1500 // c o s t o f cams i n Rs mc2 = 2.5 // m a t e r i a l c o s t p e r p i e c e i n Rs olc2 = 2 // o p e r a t i o n l a b o u r c o s t p e r h o u r i n Rs ct2 = 1 // c y c l e t i m e p e r p i e c e i n min . slc2 = 20 // s e t t i n g up l a b o u r c o s t i n Rs p e r h o u r st2 = 8 // s e t t i n g up t i m e i n h o u r mo2 = 1000/100 // machine o v e r h e a d s o f o p e r a t i o n labour cost o2 = mo2 * olc2 // o v e r h e a d s o f s i n g l e s p i n d l e i n Rs per hour fc2 = tc2 + cc2 + slc2 * st2 + o2 * st2 // f i x e d c o s t o f s i n g l e s p i n d l e i n Rs vc2 = mc2 + ( olc2 * ct2 ) /60 + ( slc2 ) /60 // v a r i a b l e c o s t i n Rs q = ( fc2 - fc1 ) /( vc1 - vc2 ) // b r e a k e v e n q u a n t i t y printf ( ” \n Break e v e n q u a n t i t y f o r a component which can be p r o d u c e d on e i t h e r t h e c a p s t a n l a t h e o r s i n g l e s p i n d l e a u t o m a t i c = %d p i e c e s ” , q )

53

Scilab code Exa 5.32 To do break even analysis 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc // E n g i n e l a t h e t = 12 // t i m e / p i e c e i n min . l = 7 // o v e r h e a d c o s t / h r o = 4 // d i r e c t l a b o u r c o s t / h r s = 2 // s e t up t i m e i n h o u r sr = 8 // s e t up r a t e p e r // t u r r e t l a t h e T = 5 // t i m e / p i e c e i n min . L = 5 // o v e r h e a d c o s t / h r O = 8 // d i r e c t l a b o u r c o s t / h r S = 8 // s e t up t i m e i n h o u r SR = 8 // s e t up r a t e p e r q = 60*( S * SR - s * sr ) /( t *( l + o ) -T *( L + O ) ) // b r e a k e v e n point 15 q = round ( q ) 16 printf ( ” \n Break e v e n p o i n t = %d p i e c e s ” , q )

Scilab code Exa 5.33 To calculate minimum number of pieces 1 2 3 4 5 6 7 8

clc fc1 = 80000 // f i x e d c o s t f o r t u r r e t l a t h e i n Rs fc2 = 32000 // f i x e d c o s t f o r e n g i n e l a t h e i n Rs n1 = 16 // p r o d u c t i o n o f p i e c e s p e r y e a r i n t u r r e t lathe n2 = 10 // p r o d u c t i o n o f p i e c e s p e r y e a r i n e n g i n e lathe vc1 = 2 // o p e r a t o r s c o s t i n t u r r e t l a t h e vc2 = 2.5 // o p e r a t o r s c o s t i n e n g i n e l a t h e Q = poly (0 , ’Q ’ ) 54

9 Q = roots (( fc1 +1/ n1 * vc1 * Q ) -( fc2 +2.5* Q /10) ) 10 printf ( ” \n Break e v e n p o i n t = %d p i e c e s ” , Q )

Scilab code Exa 5.34 To determine the point 1 2 3 4 5 6 7

clc st1 = 15 // s e t up t i m e f o r e n g i n e l a t h e i n min . ut1 = 15 // u n i t t i m e f o r e n g i n e l a t h e i n min . st2 = 90 // s e t up t i m e f o r a u t o m a t i c l a t h e i n min . ut2 = 1.5 // u n i t t i m e f o r e n g i n e l a t h e i n min . q = ( st2 - st1 ) /( ut1 - ut2 ) // q u a n t i t y o f p r o d u c t i o n printf ( ” \n The p o i n t a t which t h e a u t o m a t i c l a t h e w i l l be j u s t i f i e d = %0 . 2 f ” , q ) 8 // Answers v a r y due t o round o f f e r r o r

Scilab code Exa 5.35 To find quantity of pieces 1 2 3 4 5 6 7 8 9 10 11 12

clc // A u t o m a t i c l a t h e p = 30 // number o f p i e c e s p r o d u c e d p e r h o u r l = 4 // l a b o u r r a t e p e r h o u r i n Rs d = 4.50 // h o u r l y d e p r e c i a t i o n r a t e p e r machine i n hour s = 4 // s e t up t i m e i n h o u r // t u r r e t l a t h e P = 10 // number o f p i e c e s p r o d u c e d p e r h o u r L = 4 // l a b o u r r a t e p e r h o u r i n Rs D = 1.50 // h o u r l y d e p r e c i a t i o n r a t e p e r machine i n hour S = 2 // s e t up t i m e i n h o u r q = ( P * p *( S * L + S *D - s *l - s * d ) ) /( P *( l + d ) -p *( L + D ) ) // q u a n t i t y o f p i e c e s at break even p o i n t 55

13

printf ( ” \n Q u a n t i t y o f p i e c e s a t Break e v e n p o i n t = %d p i e c e s ” , q )

Scilab code Exa 5.36 To determine quantity of production 1 clc 2 Pa = 8.4 // u n i t 3 4 5 6 7

t o o l p r o c e s s c o s t f o r method A i n Rs Pb = 14.8 // u n i t t o o l p r o c e s s c o s t f o r method B i n Rs Ta = 6480 // t o t a l t o o l c o s t f o r method A i n Rs Tb = 1616 // t o t a l t o o l c o s t f o r method B i n Rs q = ( Ta - Tb ) /( Pb - Pa ) // b r e a k e v e n p o i n t printf ( ” \n Q u a n t i t y o f p r o d u c t i o n a t b r e a k e v e n p o i n t = %d p i e c e s ” , q )

Scilab code Exa 5.37 find preference between machines and production 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc // machine A ic1 = 50000 // i n i t i a l c o s t hoc1 = 10 // h o u r l y o p e r a t i n g c h a r g e s pp1 = 5 // p i e c e s p r o d u c e d p e r h o u r i = 15 // i n t e r e s t r a t e i = i /100 oh = 2000 // o p e r a t i n g h o u r s fc1 = ic1 * i // f i x e d c o s t vc1 = oh * hoc1 // v a r i a b l e c o s t tc1 = fc1 + vc1 // t o t a l c h a r g e s ao1 = oh * pp1 // a n n u a l o u t p u t c1 = tc1 / ao1 // c o s t p e r u n i t // machine B ic2 = 80000 // i n i t i a l c o s t 56

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

hoc2 = 8 // h o u r l y o p e r a t i n g c h a r g e s pp2 = 8 // p i e c e s p r o d u c e d p e r h o u r fc2 = ic2 * i // f i x e d c o s t vc2 = oh * hoc2 // v a r i a b l e c o s t tc2 = fc2 + vc2 // t o t a l c h a r g e s ao2 = oh * pp2 // a n n u a l o u t p u t c2 = tc2 / ao2 // c o s t p e r u n i t printf ( ” \n ( i ) C o s t p e r u n i t f o r machine A = Rs %0 . 2 f \n C o s t p e r u n i t machine B = Rs %0 . 2 f ” ,c1 , c2 ) disp ( ” machine B w i l l be p r e f e r r e d ” ) // machine A ao3 = 4000 // a n n u a l o u t p u t oc3 = ao3 * hoc1 / pp1 // o p e r a t i n g c h a r g e s tc3 = oc3 + fc1 // t o t a l a n n u a l c h a r g e c3 = tc3 / ao3 // c o s t / p i e c e // machine B ao4 = 4000 // a n n u a l o u t p u t oc4 = ao4 * hoc2 / pp2 // o p e r a t i n g c h a r g e s tc4 = oc4 + fc2 // t o t a l a n n u a l c h a r g e c4 = tc4 / ao4 // c o s t / p i e c e printf ( ” \n ( i i ) C o s t p e r u n i t f o r machine A = Rs %0 . 2 f \n C o s t p e r u n i t machine B = Rs %0 . 2 f ” ,c3 , c4 ) disp ( ” machine A w i l l be p r e f e r r e d ” ) A = hoc1 / pp1 // o p e r a t i n g c o s t p e r p i e c e on machine A B = hoc2 / pp2 // o p e r a t i n g c o s t p e r p i e c e on machine B Q = fc2 - fc1 // a n n u a l p r o d u c t i o n printf ( ” \n ( i i i ) Annual p r o d u c t i o n t o make c o s t p e r p i e c e e q u a l f o r two m a c h i n e s = %d p i e c e s ” , Q )

Scilab code Exa 5.38 To find BEP and various sales 1 clc 2 as = 80000 // a n n u a l

s a l e s i n Rs 57

3 4 5 6 7 8 9 10 11 12 13

vc = 64000 // v a r i a b l e e x p e n s e s i n Rs c = 16000 // c o n t r i b u t i o n i n Rs fc = 24000 // f i x e d e x p e n s e s i n Rs l = 8000 // l o s s e s i n Rs p = 9000 // p r o f i t i n Rs s1 = fc + vc // s a l e s a t B . E . P i n Rs s2 = ( fc + vc + p ) /0.945 // s a l e s a t n e t i n c o m e o f Rs9000 and c o r p o r a t e t a x r a t e b e i n g 5 . 5% q = 10000 // q u a n t i t y o f u n i t s sp = ( fc + vc ) / q // s e l l i n g p r i c e p e r u n i t i n Rs printf ( ” \n S a l e s a t b r e a k e v e n p o i n t = %d u n i t s ” , s1 ) printf ( ” \n S a l e s a t n e t i n c o m e o f Rs9000 and c o r p o r a t e t a x r a t e b e i n g 5 . 5 = Rs %0 . 2 f \n S a l e s p e r u n i t i f B . E . P b r o u g h t down t o 1 0 0 0 0 u n i t s = Rs %0 . 2 f p e r u n i t ” , s2 , sp )

Scilab code Exa 5.39 To determine break even point 1 2 3 4 5 6 7 8

clc fc = 55000 // f i x e d c o s t i n Rs vc = 45 // v a r i a b l e c o s t p e r p i e c e i n Rs sp = 100 // s e l l i n g p r i c e p e r p i e c e i n Rs p = ( vc / sp ) *100 // p e r c e n t a g e o f v a r i a b l e c o s t t o pm = 100 - p // p r o f i t m a r g i n bep = ((55000/55) *100) /100 // Break e v e n p o i n t printf ( ” \n Break e v e n p o i n t = %d p i e c e s ” , bep )

Scilab code Exa 5.40 To calculate economic lot size 1 clc 2 f1 = 335 // f i x e d 3 k = 0.25 // s t o c k

c o s t i n Rs f o r c a p s t a n l a t h e carrying f a c t o r in p a i s e per p i e c e 58

4 k = k /100 5 N1 = sqrt ( f1 / k ) // p i e c e s f o r c a p s t a n l a t h e 6 a1 = 4.16 // v a r i a b l e c o s t p e r p i e c e f o r c a p s t a n

lathe tc1 = a1 + f1 / N1 + k * N1 // t o t a l c o s t f o r c a p s t a n l a t h e f2 = 2120 // f i x e d c o s t i n Rs f o r t u r r e t l a t h e N2 = sqrt ( f2 / k ) // p i e c e s f o r t u r r e t l a t h e a2 = 2.863 // v a r i a b l e c o s t p e r p i e c e f o r t u r r e t lathe 11 tc2 = a2 + f2 / N2 + k * N2 // t o t a l c o s t f o r t u r r e t l a t h e 12 printf ( ” \n T o t a l c o s t p e r p i e c e f o r c a p s t a n l a t h e = Rs %0 . 2 f \n T o t a l c o s t p e r p i e c e f o r t u r r e t l a t h e = Rs %0 . 2 f ” , tc1 , tc2 ) 13 // Answers v a r y due t o round o f f e r r o r 7 8 9 10

Scilab code Exa 5.41 To find EOQ and total cost 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc R =500 // c o s t o f o r d e r i n g i n Rs p e r o r d e r A =12000 // a n n u a l c o n s u m p t i o n u n i t s C =3.00 // u n i t c o s t o f i t e m K =3 // u n i t s t o r a g e c o s t I1 =0.2 // i n t e r e s t r a t e function y = f ( N ) function G = f2 ( N ) G = C * A + I1 * C * N /2+ K * N /2+ A * R / N // t o t a l c o s t p e r y e a r endfunction y = derivative ( f2 , N ) endfunction funcprot (0) N = fsolve (2000 , f ) O = A / N // number o f o r d e r s N1 = 2400 // u n i t s tc = C * A + I1 * C * N1 /2 + K * N1 /2 + A * R / N1 // t o t a l c o s t i n Rs 59

18 I2 = (2* R * A ) /( C * N1 ^2) 19 printf ( ” \n Economic o r d e r q u a n t i t y = %d u n i t s \n T o t l

c o s t = Rs %d p e r y e a r \n I = %0 . 4 f ” ,N1 , tc , I2 ) 20 disp ( ” I t i s c l e a r t h a t i n v e n t o r y c o s t w i l l g e t i n c r e a s e d very g r e a t l y ”)

Scilab code Exa 5.42 Determine optimum lot size 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

17 18

clc A = 40000 // number o f u n i t s p e r y e a r I = 25 // c a r r y i n g c o s t i n p e r c e n t I = I /100 C1 = 8 // c o s t f o r 0 < N < 1 0 0 0 p e r u n i t i n Rs C2 = 7.5 // c o s t f o r 1 0 0 0 < N < 1 0 0 0 0 p e r u n i t i n Rs C3 = 7.25 // c o s t f o r N >= 1 0 0 0 0 p e r u n i t i n Rs R = 250 // o r d e r i n g c o s t p e r o r d e r i n Rs N = 10000 // u n i t s N1 = sqrt (2* R * A /( I * C3 ) ) // o p t i m a l q u a n t i t y f o r lowest curve G1 = C3 * A +( A * R ) / N + I * C3 * N /2 // t o t a l c o s t i n Rs N2 = sqrt (2* R * A /( I * C2 ) ) // o p t i m a l q u a n t i t y f o r higher curve G2 = C2 * A +( A * R ) / N2 + I * C2 * N2 /2 // t o t a l c o s t i n Rs N3 = sqrt (2* R * A /( I * C1 ) ) // o p t i m a l q u a n t i t y f o r highest curve G3 = C1 * A +( A * R ) +1 // t o t a l c o s t i n Rs printf ( ” \n T o t a l c o s t f o r l o w e s t c o s t c u r v e = Rs %0 . 2 f \n T o t a l c o s t f o r n e x t h i g h e r c u r v e = Rs %0 . 2 f \n T o t a l c o s t f o r h i g h e s t c u r v e = Rs %0 . 2 f ” , G1 , G2 , G3 ) disp ( ” Comparing a l l t o t a l c o s t l o w e s t i s Rs 3 0 0 , 0 6 2 . 5 0 f o r an o r d e r q u a n t i t y o f 1 0 , 0 0 0 . ” ) disp ( ”N = 1 0 , 0 0 0 and No . o f o r d e r s = 4 ” )

60

Scilab code Exa 5.43 To find most economical lot size 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc c = 50000 // c o m p o n e n ts R =500 // c o s t o f o r d e r i n g i n Rs p e r o r d e r A =12000 // a n n u a l c o n s u m p t i o n u n i t s C =3.00 // u n i t c o s t o f i t e m K =1.50 // u n i t s t o r a g e c o s t I =0.2 // i n t e r e s t r a t e function y = f ( N ) function G = f2 ( N ) G =0.02* N +1500000/ N endfunction y = derivative ( f2 , N ) endfunction funcprot (0) N = fsolve (2000 , f ) l = c / N // number o f l o t s l = ceil ( l ) ls = c / l // l o t s i z e printf ( ” \n The l o t s i z e = %d c o m p o n e n t s ” , ls )

61

Chapter 9 Limits Tolerences and Fits

Scilab code Exa 9.1 To find allowance and tolerence 1 2 3 4 5 6 7 8 9

clc h1 = 37.52 // h i g h l i m i t o f h o l e i n mm h2 = 37.50 // low l i m i t o f h o l e i n mm s1 = 37.47 // h i g h l i m i t o f s h a f t i n mm s2 = 37.45 // low l i m i t o f s h a f t i n mm ht = h1 - h2 // h o l e t o l e r e n c e i n mm st = s1 - s2 // s h a f t t o l e r e n c e i n mm a = h2 - s1 // a l l o w a n c e i n mm printf ( ” \n H o l e t o l e r e n c e = %0 . 2 f mm\n S h a f t t o l e r e n c e = %0 . 2 f mm\n A l l o w a n c e = %0 . 2 f mm” , ht , st ,a )

Scilab code Exa 9.2 Determine dimensions of shaft and hole 1 2 3 4

clc t = 0.075 // t o l e r e n c e i n mm h2 = 75 // low l i m i t o f h o l e i n mm a = 0.10 // a l l o w a n c e i n mm 62

5 h1 = h2 + t // h i g h l i m i t o f h o l e i n mm 6 s1 = h2 - a // h i g h l i m i t o f s h a f t i n mm 7 s2 = s1 - t // low l i m i t o f s h a f t i n mm 8 printf ( ” \n High l i m i t o f h o l e = %0 . 3 f mm\n High

l i m i t o f s h a f t = %0 . 2 f mm\n Low l i m i t o f s h a f t = %0 . 3 f mm” , h1 , s1 , s2 )

Scilab code Exa 9.3 Determine dimensions of hole and shaft 1 2 3 4 5 6 7 8

clc t = 0.225 // t o l e r e n c e i n mm h2 = 75 // low l i m i t o f h o l e i n mm a = 0.0375 // i n t e r f e r e n c e i n mm h1 = h2 + t // h i g h l i m i t o f h o l e i n mm s2 = h1 + a // low l i m i t o f s h a f t i n mm s1 = s2 + t // h i g h l i m i t o f s h a f t i n mm printf ( ” \n High l i m i t o f h o l e = %0 . 3 f mm\n Low l i m i t o f s h a f t = %0 . 4 f mm\n High l i m i t o f s h a f t = %0 . 4 f mm” , h1 , s2 , s1 )

Scilab code Exa 9.4 Calculate fundamental deviations and tolerences 1 2 3 4 5 6 7 8 9

clc s1 = 50 // d i a m e t e r o f s t e p 1 i n mm s2 = 80 // d i a m e t e r o f s t e p 2 i n mm d = ( s1 * s2 ) ^(1/2) // mm i = (0.45*( d ) ^(1/3) +0.001* d ) /10^3 // mm t1 = 25* i // t o l e r e n c e f o r h o l e i n mm t2 = 16* i // t o l e r e n c e f o r s h a f t i n mm a1 = 0 // f u n d a m e n t a l d e v i a t i o n f o r h o l e i n mm a2 = 5.5*( d ) ^0.41 // f u n d a m e n t a l d e v i a t i o n f o r s h a f t in microns 10 a2 = a2 /10^4 // mm 63

h1 = 60 // low l i m i t o f h o l e i n mm h2 = h1 + t1 // h i g h l i m i t o f t o l e r e n c e i n mm s1 = h1 - t2 // h i g h l i m i t o f s h a f t i n mm s2 = s1 - t2 // low l i m i t o f s h a f t i n mm printf ( ” \n T o l e r e n c e f o r h o l e = %0 . 3 f mm\n T o l e r e n c e f o r s h a f t = %0 . 3 f mm” , t1 , t2 ) 16 printf ( ” \n Fundamental d e v i a t i o n f o r h o l e = %0 . 2 f mm \n Fundamental d e v i a t i o n f o r s h a f t = %0 . 3 f mm” , a1 , a2 ) 17 printf ( ” \n Low l i m i t o f h o l e = %d mm\n High l i m i t o f h o l e = %0 . 3 f mm\n High l i m i t o f s h a f t = %0 . 2 f mm \n Low l i m i t o f h o l e = %0 . 2 f mm” , h1 , h2 , s1 , s2 ) 18 // Answers v a r y due t o round o f f e r r o r 11 12 13 14 15

Scilab code Exa 9.5 Find tolerences limits and clearance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc b = 30 // b a s i c s i z e i n mm s1 = 0.005 // maximum l i m i t o f s h a f t i n mm s2 = 0.018 // minimum l i m i t o f s h a f t i n mm h1 = 0.020 // maximum l i m i t o f h o l e i n mm h2 = 0.0 // minimum l i m i t o f h o l e i n mm t1 = s2 - s1 // s h a f t t o l e r e n c e i n mm t2 = h1 - h2 // h o l e t o l e r e n c e i n mm Sh = b - s1 // h i g h l i m i t o f s h a f t i n mm Sl = b - s2 // low l i m i t o f s h a f t i n mm Hh = b + h1 // h i g h l i m i t o f h o l e i n mm Hl = b + h2 // low l i m i t o f h o l e i n mm c1 = Hh - Sl // maximum c l e a r a n c e i n mm c2 = Hl - Sh // minimum c l e a r a n c e i n mm printf ( ” \n B a s i c s i z e = %d mm\n S h a f t t o l e r e n c e = %0 . 3 f mm\n H o l e t o l e r e n c e = %0 . 3 f mm” ,b , t1 , t2 ) 16 printf ( ” \n High l i m i t o f s h a f t = %0 . 3 f mm\n Low l i m i t o f s h a f t = %0 . 3 f mm\n High l i m i t o f h o l e = 64

%0 . 3 f mm \n Low l i m i t o f h o l e = %0 . 3 f mm” ,Sh , Sl , Hh , Hl ) 17 printf ( ” \n Maximum c l e a r a n c e = %0 . 3 f mm\n Minimum c l e a r a n c e = %0 . 3 f mm” ,c1 , c2 )

Scilab code Exa 9.6 Determine limits of shaft and hole 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc minc = 0.01 // minimum c l e a r a n c e i n mm bs = 25 // b a s i c s i z e i n mm maxc = 0.02 // maximum c l e a r a n c e i n mm x = poly (0 , ’ x ’ ) y =1.5* x x = roots ( y +0.01+ x -0.02) y = horner (y , x ) // h o l e b a s i s s y s t e m low_h1 = bs // low l i m i t o f h o l e i n mm high_h1 = bs + y // h i g h l i m i t o f h o l e i n mm u_s = low_h1 - minc // u p p e r l i m i t o f s h a f t i n mm low_s1 = u_s - x // l o w e r l i m i t o f s h a f t i n mm // s h a f t b a s i s s y s t e m high_s = bs // h i g h l i m i t o f s h a f t i n mm low_s2 = bs - x // low l i m i t o f s h a f t i n mm low_h2 = bs + minc // low l i m i t o f h o l e i n mm high_h2 = low_h2 + y // h i g h l i m i t o f h o l e i n mm printf ( ” H o l e b a s i s s y s t e m \n Lower l i m i t o f h o l e = %d mm\n H i g h e r l i m i t o f h o l e = %0 . 3 f mm\n H i g h e r l i m i t o f s h a f t = %0 . 3 f mm \n Lower l i m i t o f s h a f t = %0 . 3 f mm” , low_h1 , high_h1 , u_s , low_s1 ) 20 printf ( ” \n S h a f t b a s i s s y s t e m \n h i g h l i m i t o f s h a f t = %0 . 3 f mm\n l o w e r l i m i t o f s h a f t = %0 . 3 f mm\n l o w e r l i m i t o f h o l e = %0 . 3 f mm\n u p p e r l i m i t o f h o l e = %0 . 3 f mm” , high_s , low_s2 , low_h2 , high_h2 )

65

Scilab code Exa 9.7 Determine dimensions of shaft and hole 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

19

clc bs = 100 // b a s i c s i z e i n mm s1 = 120 // d i a m e t e r o f s t e p 1 i n mm s2 = 80 // d i a m e t e r o f s t e p 2 i n mm d = ( s1 * s2 ) ^(1/2) // mm d = ceil ( d ) i = (0.45*( d ) ^(1/3) +0.001* d ) /10^3 // mm t1 = 16* i // t o l e r e n c e f o r h o l e i n mm t2 = 25* i // t o l e r e n c e f o r s h a f t i n mm G = (2.5*( d ) ^0.34) /10^3 // f u n d a m e n t a l d e v i a t i o n f o r h o l e i n mm e = (11*( d ) ^0.11) /10^3 // f u n d a m e n t a l d e v i a t i o n f o r s h a f t in microns // H o l e LLh = bs + G // l o w e r l i m i t o f h o l e i n mm HLh = LLh + t1 // h i g h e r l i m i t o f h o l e i n mm // s h a f t ULs = bs - e // u p p e r l i m i t o f s h a f t i n mm LLs = ULs - t2 // l o w e r l i m i t o f s h a f t i n mm printf ( ” \n l o w e r l i m i t o f h o l e = %0 . 3 f mm\n h i g h e r l i m i t o f h o l e = %0 . 3 f mm\n u p p e r l i m i t o f s h a f t = %0 . 3 f mm\n l o w e r l i m i t o f s h a f t = %0 . 3 f mm” , LLh , HLh , ULs , LLs ) // E r r o r i n t e x t b o o k

Scilab code Exa 9.8 Determine size of bearing and journal 1 clc 2 tb = 0.005 // t o l e r e n c e on b e a r i n g i n mm 3 tj = 0.004 // t o l e r e n c e on j o u r n a l i n mm

66

a = 0.002 // a l l o w a n c e i n mm // h o l e −b a s i s s y s t e m b = 100 // b a s i c s i z e i n mm Bl = b // l o w e r l i m i t o f b e a r i n g i n mm Bh = Bl + tb // h i g h e r l i m i t o f b e a r i n g i n mm Jh = Bl - a // h i g h e r l i m i t o f j o u r n a l i n mm Jl1 = Jh - tj // l o w e r l i m i t o f j o u r n a l i n // s h a f t −b a s i s s y s t e m Ju = b // u p p e r l i m i t o f j o u r n a l i n mm Jl2 = Ju - tj // l o w e r l i m i t o f j o u r n a l i n mm Bl = Ju + a // l o w e r l i m i t o f b e a r i n g i n mm Bu = Bl + tb // u p p e r l i m i t o f b e a r i n g i n mm printf ( ” \n H o l e b a s i s s y s t e m \n Lower l i m i t o f j o u r n a l = %d mm\n H i g h e r l i m i t o f b e a r i n g = %0 . 3 f mm\n H i g h e r l i m i t o f j o u r n a l = %0 . 3 f mm \n Lower l i m i t o f j o u r n a l = %0 . 3 f mm” , Bl , Bh , Jh , Jl1 ) 17 printf ( ” \n s h a f t b a s i s s y s t e m \n u p p e r l i m i t o f j o u r n a l = %0 . 3 f mm\n l o w e r l i m i t o f j o u r n a l = %0 . 3 f mm\n l o w e r l i m i t o f b e a r i n g = %0 . 3 f mm\n u p p e r l i m i t o f b e a r i n g = %0 . 3 f mm” , Ju , Jl2 , Bl , Bu )

4 5 6 7 8 9 10 11 12 13 14 15 16

Scilab code Exa 9.9 Determine size of two mating parts 1 2 3 4 5 6 7 8 9 10 11

clc // Hole−b a s i s s y s t e m b = 100 // b a s i c s i z e i n mm i1 = 0.12 // maximum i n t e r f e r e n c e i n mm i2 = 0.05 // minimum i n t e r f e r n c e i n mm t = ( i1 - i2 ) /2 // t o l e r e n c e i n mm Sh = b + i1 // u p p e r l i m i t o f s h a f t i n mm Hl = b // l o w e r l i m i t o f h o l e i n mm Hh = b + t // h i g h e r l i m i t o f h o l e i n mm Sl1 = Sh - t // l o w e r l i m i t o f s h a f t i n mm // s h a f t −b a s i s s y s t e m 67

Su = b // u p p e r l i m i t o f s h a f t i n mm Sl2 = b - t // l o w e r l i m i t o f s h a f t i n mm Hl1 = b - i1 // l o w e r l i m i t o f h o l e i n mm Hu = Hl1 + t // h i g h e r l i m i t o f h o l e i n mm printf ( ” \n H o l e b a s i s s y s t e m \n u p p e r l i m i t o f s h a f t = %0 . 3 f mm\n l o w e r l i m i t o f h o l e = %0 . 3 f mm\n h i g h e r l i m i t o f h o l e = %0 . 3 f mm\n l o w e r l i m i t o f s h a f t = %0 . 3 f mm” , Sh , Hl , Hh , Sl1 ) 17 printf ( ” \n S h a f t b a s i s s y s t e m \n u p p e r l i m i t o f s h a f t = %0 . 3 f mm\n l o w e r l i m i t o f s h a f t = %0 . 3 f mm\n l o w e r l i m i t o f h o l e = %0 . 3 f mm\n u p p e r l i m i t o f h o l e = %0 . 3 f mm” , Su , Sl2 , Hl1 , Hu )

12 13 14 15 16

Scilab code Exa 9.10 Determine size of hole and shaft 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc aa = 0.04 // a v e r a g e a l l o w a n c e i n mm a = 0.012 // a l l o w a n c e i n mm Max = aa + a // maximum a l l o w a n c e i n mm Min = aa - a // minimum a l l o w a n c e i n mm t = ( Max - Min ) /3 // t o l e r e n c e i n mm ts = t // t o l e r e n c e i n s h a t i n mm th = 2* t // t o l e r e n c e i n h o l e i n mm b = 100 // b a s i c s i z e i n mm Hl = b // l o w e r l i m i t o f h o l e i n mm Hu = b + th // u p p e r l i m i t o f h o l e i n mm Su = b -0.028 // u p p e r l i m i t o f s h a f t i n mm Sl = Su - ts // l o w e r l i m i t o f s h a f t i n mm printf ( ” \n l o w e r l i m i t o f h o l e = %d mm\n u p p e r l i m i t o f h o l e = %0 . 3 f mm\n u p p e r l i m i t o f s h a f t = %0 . 3 f mm\n l o w e r l i m i t o f s h a f t = %0 . 3 f mm” ,Hl , Hu , Su , Sl )

68

Chapter 11 Surface finish

Scilab code Exa 11.1 Calculate CLA value 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc v = 15000 // v e r t i c a l m a g n i f i c a t i o n h = 100 // h o r i z o n t a l m a g n i f i c a t i o n l = 0.8 // s a m p l i n g l e n g t h i n mm a1 = 160 // a r e a a b o v e datum l i n e i n mmˆ2 a2 = 90 // a r e a a b o v e datum l i n e i n mmˆ2 a3 = 180 // a r e a a b o v e datum l i n e i n mmˆ2 a4 = 50 // a r e a a b o v e datum l i n e i n mmˆ2 a5 = 95 // a r e a b e l o w datum l i n e i n mmˆ2 a6 = 65 // a r e a b e l o w datum l i n e i n mmˆ2 a7 = 170 // a r e a b e l o w datum l i n e i n mmˆ2 a8 = 150 // a r e a b e l o w datum l i n e i n mmˆ2 a = ( a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 ) /( v * h ) CLA = a / l printf ( ” \n C . L . A v a l u e = %0 . 2 f ∗10ˆ −6 m ” , CLA *1000)

Scilab code Exa 11.2 Calculate average and rms value

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

clc // from f i g u r e 1 1 . 2 3 y1 = 0.15 // mu m y2 = 0.25 // mu m y3 = 0.35 // mu m y4 = 0.25 // mu m y5 = 0.30 // mu m y6 = 0.15 // mu m y7 = 0.10 // mu m y8 = 0.30 // mu m y9 = 0.35 // mu m y10 = 0.10 // mu m y1sqr = y1 ^2 // mu m y2sqr = y2 ^2 // mu m y3sqr = y3 ^2 // mu m y4sqr = y4 ^2 // mu m y5sqr = y5 ^2 // mu m y6sqr = y6 ^2 // mu m y7sqr = y7 ^2 // mu m y8sqr = y8 ^2 // mu m y9sqr = y9 ^2 // mu m y10sqr = y10 ^2 // mu m n = 10 yn = ( y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 + y9 + y10 ) / n // a r i t h m e t i c a v e r a g e i n mu m 25 rms = sqrt (( y1sqr + y2sqr + y3sqr + y4sqr + y5sqr + y6sqr + y7sqr + y8sqr + y9sqr + y10sqr ) / n ) // r .m. s v a l u e i n mu m 26 printf ( ” \n The a r i t h m e t i c a v e r a g e = %0 . 2 f ∗10ˆ −6 m \n The r .m. s . v a l u e = %0 . 3 f ∗10ˆ −6 m” ,yn , rms )

70

Chapter 13 Analysis of metal forming processes

Scilab code Exa 13.1 To find drawing load 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc sigma_0 = 240 // N/mmˆ2 d1 = 5 // i n i t i a l w i r e d i a m e t e r i n mm d0 = 5.5 // f i n a l w i r e d i a m e t e r i n mm x = d1 / d0 // mm alpha = 8 // a n g l e o f c o n t a c t alpha = alpha * %pi /180 mu = 0.1 // c o e f f i c i e n t o f f r i c t i o n B = mu * cotg ( alpha ) sigma_d = ( sigma_0 *(1+ B ) *(1 -( x ) ^(2* B ) ) ) / B // N/mmˆ2 l = 3 // d i e l a n d i n mm mu = 0.1 // c o e f f i c i e n t o f f r i c t i o n r1 = d1 /2 // mm sigma_t = sigma_0 - ( sigma_0 - sigma_d ) / exp ((2* mu * l ) / r1 ) // N/mmˆ2 15 dl = sigma_t * %pi *( r1 ) ^2 // d r a w i n g l o a d i n N 16 printf ( ” \n T o t a l d r a w i n g l o a d = %0 . 1 f N” , dl ) 17 // Answers v a r y due t o round o f f e r r o r

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Scilab code Exa 13.2 Calculate drawing force 1 2 3 4 5 6 7 8 9 10 11 12 13 14

15

clc alpha = 15 // a n g l e o f c o n t a c t alpha = alpha * %pi /180 bita = 0 // d e g r e e mu = 0.1 // c o e f f i c i e n t o f f r i c t i o n mu1 = mu mu2 = mu h1 = 1.75 // mm h0 = 2.5 // mm B = ( mu1 + mu2 ) /( tan ( alpha ) - tan ( bita ) ) y1 = (1+ B ) *(1 -( h1 / h0 ) ^ B ) / B // s i g m a d / s i g m a 0 f o r p l u g m a n d r e l s i n N/mmˆ2 z = 1/(( h0 / h1 ) -1) y2 = log10 ( z ) // s i g m a d / s i g m a 0 f o r movable m a n d r e l s i n N/mmˆ2 printf ( ” \n The p i p e d r a w i n g f o r c e f o r c e on p l u g m a n d r e l s = %0 . 3 f \n The p i p e d r a w i n g f o r c w on m a n d r e l s = %0 . 3 f ” ,y1 , y2 ) disp ( ” Use o f movable m a n d r e l s u b s t a n t i a l l y r e d u c e s drawing f o r c e ”)

Scilab code Exa 13.3 find neutral section slips and pressure 1 2 3 4 5 6 7

clc h0 = 25 h1 = 20 delta_h sigma = D = 500 r = D /2

// t h i c k n e s s o f p l a t e i n mm // mm = h0 - h1 // mm 100 // maximum p r e s s u r e i n N/mmˆ2 // r o l l e d d i a m e t e r i n mm // r o l l e d r a d i u s i n mm 72

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

alpha = acos (1 -( delta_h / D ) ) // a n g l e o f c o n t a c t i n radians mu = tan ( alpha ) // c o e f f i c i e n t o f f r i c t i o n Ho = 2* sqrt ( r / h1 ) * atan ( sqrt ( r / h1 ) * mu ) Hn = ( Ho - ( log ( h0 / h1 ) ) / mu ) /2 theta = sqrt ( h1 / r ) * tan ( sqrt ( h1 / r ) *( Hn /2) ) // r a d i a n hn = h1 + r * theta ^2 // n e u t r a l s e c t i o n i n mm x = hn / h0 bs = (1 - x ) *100 // backward s l i p y = hn / h1 fs = (y -1) *100 // f o r w a r d s l i p sigma0 = 2* sigma / sqrt (3) pn = sigma0 * hn * exp ( mu * Hn ) / h1 //N/mmˆ2 printf ( ” \n N e u t r a l s e c t i o n = %0 . 1 f mm” , hn ) printf ( ” \n Backward s l i p = %0 . 1 f p e r c e n t \n Forward s l i p = %0 . 1 f p e r c e n t ” , bs , fs ) printf ( ” \n Maximum p r e s s u r e = %0 . 1 f N/mmˆ2 ” , pn ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 13.4 To determine maximum force 1 2 3 4 5 6 7 8 9 10

clc Do = 250 // d i a m e t e r i n mm ho = 250 // h i e g h t i n mm delta_h = 100 // mm h = 150 // mm sigma0 = 55 // N/mmˆ2 d = Do * sqrt ( ho /( ho - delta_h ) ) // d i a m e t e r i n mm mu = 0.42 // c o e f f i c i e n t o f f r i c t i o n R = 162.5 // mm pa = sigma0 /2*( h /( mu * R ) ) ^2*( %e ^(2* mu * R / h ) -2* mu * R /h -1) // N/mmˆ2 11 p = pa * %pi *( R ) ^2 // f o r c e i n kN 12 printf ( ” \n F o r c e = %d kN” ,p /1000)

73

Scilab code Exa 13.5 Determine sticking radius and total load 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc d = 150 // d i a m e t e r i n mm h = 10 // t h i c k n e s s i n mm R = d /2 // r a d i u s i n mm mu = 0.2 // c o e f f i c i e n t o f f r i c t i o n sigma_0 = 200 // N/mmˆ2 Rs = R - ( h /(2* mu ) ) * log (1/( sqrt (3) * mu ) ) // s t i c k i n g r a d i u s i n mm Ps = sigma_0 * exp (2* mu *( R - Rs ) / h ) // p r e s s u r e a t s t i c k i n g r a d i u s i n N/mmˆ2 function y = f ( r ) y =2* %pi * r * sigma_0 * exp (2* mu / h *( R - r ) ) endfunction L_sld = intg (48.5 ,75 , f ) L_sld = L_sld /1000 // l o a d on s l i d i n g p o r t i o n i n kN Pc = Ps + (2* sigma_0 * Rs ) /( h * sqrt (3) ) // p r e s s u r e a t c e n t r e i n N/mmˆ2 L_sp = ( Pc + Ps ) * %pi *( Rs ) ^2/(2*1000) // l o a d on s t i c k i n g p o r t i o n i n kN F_l = L_sld + L_sp // t o t a l f o r g i n g l o a d i n kN printf ( ” \n S t i c k i n g r a d i u s = %0 . 1 f mm \n T o t a l f o r g i n g l o a d = %0 . 3 f MN” , Rs , F_l /1000) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 13.7 To find drawing load and power 1 clc 2 RA = 0.30 3 d = 12 // d i a m e t e r i n mm 4 alpha = 6 // a n g l e o f c o n t a c t i n d e g r e e

74

alpha = 6* %pi /180 // a n g l e o f c o n t a c t i n r a d i a n mu = 0.10 // c o e f f i c i e n t o f f r i c t i o n sigma_0 = 240 // N/mmˆ2 B = mu * cotg ( alpha ) x = 1 - RA sigma_d = ( sigma_0 *(1+ B ) *(1 -( x ) ^ B ) ) / B // N/mmˆ2 r1 = sqrt ( x ) *( d /2) // mm l = sigma_d * %pi *( r1 ) ^2 // l o a d i n kN ita = 98 // e f f i c i e n c y ita = ita /100 s = 2.3 // d r a w i n g s p e e d i n m/ s P = ( l * s ) / ita // kW printf ( ” \n Drawing l o a d = %0 . 2 f kN\n Power o f motor = %0 . 2 f kW” , l /1000 ,P /1000 ) 18 // ’ Answers v a r y due t o round o f f e r r o r ’

5 6 7 8 9 10 11 12 13 14 15 16 17

Scilab code Exa 13.8 calculate drawing load and power rating 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc mu1 = 0.15 // c o e f f i c i e n t o f f r i c t i o n mu2 = 0.18 // c o e f f i c i e n t o f f r i c t o n alpha = 14 // a n g l e o f c o n t a c t i n d e g r e e alpha = alpha * %pi /180 bita = 10 // semi −c o n e a n g l e i n d e g r e e bita = bita * %pi /180 sigma_0 = 1.40 // kN/mmˆ2 h0 = 1.5 //mm h1 = 1 // mm B = ( mu1 + mu2 ) /( tan ( alpha ) + tan ( bita ) ) sigmad = ( sigma_0 *(1+ B ) *(1 -( h1 / h0 ) ^ B ) ) / B // d r a w i n g s t r e s s i n kN/mmˆ2 d1 = 11 // o u t s i d e d i a m e t e r i n mm t = 1 // t h i c k n e s s i n mm d2 = d1 -2* t // mm a = ( %pi *(( d1 ) ^2 -( d2 ) ^2) ) /4 // a r e a i n mmˆ2 75

l = sigmad * a // l o a d i n kN s = 0.65 // d r a w i n g s p e e d i n m/ s w = l * s // work i n kJ / s p = w // power i n kW printf ( ” \n Drawing l o a d = %0 . 3 f kN\n Power r a t i n g o f motor = %0 . 2 f kW” , l , p ) 22 // ’ Answers v a r y due t o round o f f e r r o r ’

17 18 19 20 21

Scilab code Exa 13.9 To calculate forging loads 1 2 3 4 5 6 7 8 9 10 11 12

clc sigma_0 = 50 // p r e s s u r e a t s t a r t i n MPa B = 0.9 // w i d t h i n m h1 = 0.2 // t h i c k n e s s i n m b = 0.3 // t o o l b i t e i n m // At commencement o f f o r g i n g FL = sigma_0 * B * b *(1+( b /(4* h1 ) ) ) // f o r g i n g l o a d i n MN // At c o m p l e t i o n o f f o r g i n g h2 = 0.1 // t h i c k n e s s i n m sigma_0c = 150 // p r e s s u r e a t c o m p l e t i o n i n MPa FL2 = sigma_0c * B * b *(1+( b /(4* h2 ) ) ) // f o r g i n g l o a d i n MN printf ( ” \n F o r g i n g l o a d a t s t a r t o f f o r g i n g = %0 . 4 f MN\n F o r g i n g l o a d a t c o m p l e t i o n o f f o r g i n g = %0 . 3 f MN” , FL , FL2 )

Scilab code Exa 13.10 Determine extrusion load 1 clc 2 sigma_0 = 250 // N/mmˆ2 3 d1 = 5 // i n i t i a l w i r e d i a m e t e r i n mm 4 d0 = 15 // f i n a l w i r e d i a m e t e r i n mm

76

5 6 7 8 9 10 11 12 13 14 15 16

r0 = d0 /2 r1 = d1 /2 x = ( r0 / r1 ) ^2 // mm alpha = 45 // a n g l e o f c o n t a c t alpha = alpha * %pi /180 mu = 0.1 // c o e f f i c i e n t o f f r i c t i o n B = mu * cotg ( alpha ) sigma_x0 = ( sigma_0 *(1+ B ) *(1 -( x ) ^ B ) ) / B // N/mmˆ2 sigma_x0 = - sigma_x0 l = 37.5 // l e n g t h 0 f b i l l e t i n mm tau1 = sigma_0 /2 // Mpa Pe = sigma_x0 + (4* tau1 * l ) / d0 // e x t r u s i o n p r e s s u r e i n Mpa 17 el = Pe * %pi *( r0 ) ^2 // e x t r u s i o n l o a d i n MN 18 printf ( ” \n E x t r u s i o n l o a d = %d MN” , el /10000)

Scilab code Exa 13.11 To find roll pressures 1 2 3 4 5 6 7 8 9 10 11 12 13

clc h0 = 4.05 // t h i c k n e s s o f p l a t e i n mm h1 = 3.55 // mm D = 500 // r o l l e d d i a m e t e r i n mm r = D /2 // r o l l e d r a d i u s i n mm mu = 0.04 // c o e f f i c i e n t o f f r i c t i o n sigma = 210 // N/mmˆ2 delta_h = h0 - h1 // mm p = 2* sigma / sqrt (3) // N/mmˆ2 alpha = acos (1 -( delta_h / D ) ) // a n g l e o f c o n t a c t Ho = 2* sqrt ( r / h1 ) * atan ( sqrt ( r / h1 ) * alpha ) Hn1 = ( Ho - ( log ( h0 / h1 ) ) / mu ) /2 theta = sqrt ( h1 / r ) * tan ( sqrt ( h1 / r ) *( Hn1 /2) ) // radians 14 hn = h1 + 2* r *(1 - cos ( theta ) ) // mm 15 pn1 = p * hn * exp ( mu * Hn1 ) / h1 // r o l l p r e s s u r e i n N/mmˆ2 16 // b ) r o l l p r e s s u r e when c o e f f i c i e n t o f f r i c t i o n i s 77

17 18 19 20 21 22 23 24 25

26

27

28

0.4 mu2 = 0.4 // c o e f f i c i e n t o f f r i c t i o n Hn2 = ( Ho - ( log ( h0 / h1 ) ) / mu2 ) /2 theta = sqrt ( h1 / r ) * tan ( sqrt ( h1 / r ) *( Hn2 /2) ) // radians hn2 = h1 + r * theta ^2 // mm pn2 = ( p * hn2 * exp ( mu2 * Hn2 ) ) / h1 // r o l l p r e s s u r e i n N/ mmˆ2 // c ) i f t e n s i o n i s a p p l i e d o f 35 N/mmˆ2 sigma_f = 35 // f r o n t t e n s i o n i n N/mmˆ2 pn3 = (p - sigma_f ) * hn * exp ( mu * Hn1 ) / h1 // r o l l r e s s u r e i n N/mmˆ2 printf ( ” \n ( a ) R o l l p r e s s u r e a t e n t e r and e x i t = %0 . 1 f N/mmˆ2\ n Roll p r e s s u r e at n e u t r a l plane = %0 . 2 f N/mmˆ2 ” ,p , pn1 ) printf ( ” \n ( b ) R o l l p r e s s u r e a t n e u t r a l p o i n t when co− e f f i c i e n t o f f r i c t i o n i s 0 . 4 0 = %0 . 2 f N/mmˆ2 ” , pn2 ) printf ( ” \n ( c ) R o l l p r e s s u r e when 35 N/mmˆ2 t e n s i o n i s a p p l i e d a t n e u t r a l p o i n t = %0 . 2 f N/mmˆ2 ” , pn3 ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 13.12 Determine neutral plane 1 2 3 4 5 6 7 8 9 10

clc h1 = 6.35 // t h i c k n e s s i n mm mu = 0.2 // c o e f f i c i e n t o f f r i c t i o n r = 50 // r o l l e d r a d i u s i n cm r = r *10 // mm R = 30 // r e d u c t i o n i n p e r c e n t h0 = h1 *100/(100 - R ) // mm delta_h = h0 - h1 // mm alpha = acos (1 -( delta_h /(2* r ) ) ) // a n g l e o f c o n t a c t Ho = 2* sqrt ( r / h1 ) * atan ( sqrt ( r / h1 ) * alpha ) 78

11 Hn = ( Ho - ( log ( h0 / h1 ) ) / mu ) /2 12 theta = sqrt ( h1 / r ) * tan ( sqrt ( h1 / r ) *( Hn /2) ) // n e u t r a l

plane in radians 13 theta = theta *180/ %pi // n e u t r a l p l a n e i n d e g r e e s 14 printf ( ” \n N e u t r a l p l a n e = %0 . 2 f d e g r e e ” , theta ) 15 // ’ Answers v a r y due t o round o f f e r r o r ’

79

Chapter 14 Theory of metal cutting

Scilab code Exa 14.1 calculate the tool life 1 2 3 4 5 6 7 8

clc v1 = 18 // c u t t i n g s p e e d i n m/ min t1 = 3 // t o o l l i f e i n h o u r s n = 0.125 // e x p o n e n t c = v1 *( t1 *60) ^ n // c o n s t a n t v2 = 24 // c u t t i n g s p e e d i n m/ min t = ( c / v2 ) ^(1/0.125) // t o o l l i f e i n min . printf ( ” T o o l l i f e = %d min . ” , t )

Scilab code Exa 14.2 Calculate the optimum cutting speed 1 2 3 4 5 6 7

clc c_t = 8 // t o o l c h a n g e t i m e i n min . r_t = 5 // t o o l r e −g r i n d t i m e i n min . mr_c = 5 // machine r u n n i n g c o s t p e r h o u r d = 30 // t o t a l d e p r e c i a t i o n p e r r e −g r i n d i n p a i s a n = 0.25 // e x p o n e n t c = 150 // c o n s t a n t 80

c_c = mr_c * c_t /60 // t o t a l c h a n g e c o s t i n Rs r_c = mr_c * r_t /60 // r e g r i n d c o s t i n Rs ct = c_c + r_c + d /100 // t o o l i n g c o s t i n Rs cm = mr_c /60 // m a c h i n i n g c o s t i n Rs v = c *(( cm * n ) /( ct *(1 - n ) ) ) ^ n // c u t t i n g s p e e d i n m/ min . 13 printf ( ” \n C u t t i n g s p e e d = %0 . 1 f m/ min . ” , v ) 8 9 10 11 12

Scilab code Exa 14.3 To find different orthogonal cutting picture 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc mu1 = 0.15 // c o e f f i c i e n t o f f r i c t i o n mu2 = 0.18 // c o e f f i c i e n t o f f r i c t o n alpha = 14 // a n g l e o f c o n t a c t i n d e g r e e alpha = alpha * %pi /180 bita = 10 // semi −c o n e a n g l e i n d e g r e e bita = bita * %pi /180 sigma_0 = 1.40 // kN/mmˆ2 h0 = 1.5 //mm h1 = 1 // mm B = ( mu1 + mu2 ) /( tan ( alpha ) + tan ( bita ) ) sigmad = ( sigma_0 *(1+ B ) *(1 -( h1 / h0 ) ^ B ) ) / B // d r a w i n g s t r e s s i n kN/mmˆ2 d1 = 11 // o u t s i d e d i a m e t e r i n mm t = 1 // t h i c k n e s s i n mm d2 = d1 - t // mm a = ( %pi *(( d1 ) ^2 -( d2 ) ^2) ) /4 // a r e a i n mmˆ2 l = sigmad * a // l o a d i n kN s = 0.65 // d r a w i n g s p e e d i n m/ s w = l * s // work i n kJ / s p = w // power i n kW printf ( ” \n Drawing l o a d = %0 . 3 f kN\n Power r a t i n g o f motor = %0 . 2 f kW” , l , p ) clc t = 0.127 // u n c u t c h i p t h i c k n e s s i n mm 81

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

b = 6.35 // w i d t h o f c u t i n mm v = 2 // c u t t i n g s p e e d i n m/ s alpha = 10 // r a k e a n g l e i n d e g r e e s fc = 567 // c u t t i n g f o r c e i n N ft = 227 // t h r u s t f o r c e i n N tc = 0.228 // c h i p t h i c k n e s s i n mm r = t / tc // c h i p t h i c k n e s s r a t i o alpha = alpha * %pi /180 // r a k e a n g l e i n r a d i a n s phi = atan ( r * cos ( alpha ) /(1 -( r * sin ( alpha ) ) ) ) // s h e a r angle phi1 = phi *180/ %pi // s h e a r a n g l e printf ( ” \n S h e a r a n g l e = %0 . 2 f d e g r e e ” , phi1 ) mu =(( fc * sin ( alpha ) + ft * cos ( alpha ) ) /( fc * cos ( alpha ) - ft * sin ( alpha ) ) ) // c o e f f i c i e n t o f f r i c t i o n bita = atan ( mu ) // f r i c t i o n a n g l e bita = bita *180/( %pi ) printf ( ” \n F r i c t i o n a n g l e = %0 . 2 f d e g r e e ” , bita ) fs = fc * cos ( phi ) - ft * sin ( phi ) // s h e a r f o r c e i n N taus = ( fs * sin ( phi ) ) /( b * t ) // s h e a r s t r e s s printf ( ” \n S h e a r s t r e s s = %0 . 1 f N/mmˆ2 ” , taus ) cp = fc * v /1000 // c u t t i n g power i n kw printf ( ” \n C u t t i n g power = %0 . 3 f kw ” , cp ) vc = v * r // c h i p v e l o c i t y i n m/ s printf ( ” \n Chip v e l o c i t y = %0 . 3 f m/ s ” , vc ) ss = cotg ( phi ) + tan ( phi - alpha ) // s h e a r s t r a i n printf ( ” \n s h e a r s t r a i n = %0 . 3 f ” , ss ) spl = t / sin ( phi ) // s h e a r p l a n e l e n g t h vs = v * cos ( alpha ) / cos ( phi - alpha ) // s h e a r v e l o c i t y S = vs *10/ spl // s h e a r s t r a i n r a t e S = S *10^3 // s h e a r s t r a i n r a t e printf ( ” \n S h e a r s t r a i n r a t e = %. 3 f s ˆ−1” , S ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.4 To find tool life

82

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc v = 30 // c u t t i n g s p e e d i n m/ min feed = 0.3 // f e e d r a t e i n mm/ r e v . d = 2.5 // d e p t h o f c u t i n mm t = 60 // t o o l l i f e i n min . c = v * t ^0.13* feed ^0.77* d ^0.37 // c o n s t a n t printf ( ” \n c o n s t a n t = %0 . 2 f ” , c ) v2 = v *1.2 // c u t t i n g s p e e d i n m/ min t2 = ( c /( v2 * feed ^0.77* d ^0.37) ) // t o o l l i f e when c u t t i n g s p e e d i n c r e a s e d by 20% i n min . t2 = t2 ^(1/0.13) f2 = feed *1.2 // f e e d r a t e i n mm/ r e v . t3 = ( c /( v * d ^0.37* f2 ^0.77) ) // t o o l l i f e when f e e d r a t e i n c r e a s e d by 20% i n min . t3 = t3 ^(1/0.13) d2 = d *1.2 // d e p t h o f c u t i n mm t4 = ( c /( v * feed ^0.77* d2 ^0.37) ) // t o o l l i f e when d e p t h o f c u t i n c r e a s e d by 20% i n min . t4 = t4 ^(1/0.13) t5 = ( c /( v2 * d2 ^0.37* f2 ^0.77) ) // t o o l l f e i n min . t5 = t5 ^(1/0.13) printf ( ” \n T o o l l i f e when c u t t i n g s p e e d i n c r e a s e d by 20 = %0 . 2 f min . ” , t2 ) printf ( ” \n T o o l l i f e when f e e d r a t e i n c r e a s e d by 20 = %0 . 2 f min . ” , t3 ) printf ( ” \n T o o l l i f e when d e p t h o f c u t i n c r e a s e d by 20 = %0 . 2 f min . ” , t4 ) printf ( ” \n T o o l l i f e when a l l t a k e n t o g e t h e r a f t e r i n c r e a s i n g by 20 = %0 . 2 f min . ” , t5 ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.5 find force and coefficient of friction 1 clc 2 t = 0.25 // u n c u t c h i p

t h i c k n e s s i n mm 83

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

b = 2.5 // w i d t h o f c u t i n mm v = 2.5 // c u t t i n g s p e e d i n m/ s alpha = 10 // r a k e a n g l e i n d e g r e e s fc = 1130 // c u t t i n g f o r c e i n N ft = 295 // t h r u s t f o r c e i n N tc = 0.45 // c h i p t h i c k n e s s i n mm r = t / tc // c h i p t h i c k n e s s r a t i o alpha = alpha * %pi /180 // r a k e a n g l e i n r a d i a n s phi = atan (( r * cos ( alpha ) ) /(1 - r * sin ( alpha ) ) ) // s h e a r angle phi2 = phi *180/ %pi // s h e a r a n g l e fs = fc * cos ( phi ) - ft * sin ( phi ) // s h e a r f o r c e i n N printf ( ” \n F o r c e o f s h e a r a t s h e a r p l a n e = %0 . 2 f N” , fs ) mu = atan (( fc * sin ( alpha ) + ft * cos ( alpha ) ) /( fc * cos ( alpha ) - ft * sin ( alpha ) ) ) // f r i c t i o n a n g l e l e printf ( ” \n F r i c t i o n a n g l e = %0 . 3 f d e g r e e ” , mu ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.6 To find terms of orthogonal cutting 1 2 3 4 5 6 7 8 9 10 11 12

clc t = 0.2 // u n c u t c h i p t h i c k n e s s i n mm alpha = 15 // r a k e a n g l e i n d e g r e e s tc = 0.62 // c h i p t h i c k n e s s i n mm r = t / tc // c h i p t h i c k n e s s r a t i o crc = 1/ r // c h i p r e d u c t i o n c o e f f i c i e n t printf ( ” \n C u t t i n g r a t i o = %0 . 3 f \n Chip r e d u c t i o n co − e f f i c i e n t = %0 . 1 f ” , r , crc ) alpha = alpha * %pi /180 // r a k e a n g l e i n r a d i a n s phi = atan ( r * cos ( alpha ) /(1 - r * sin ( alpha ) ) ) // s h e a r angle phi = phi *180/ %pi // s h e a r a n g l e printf ( ” \n S h e a r a n g l e = %0 . 2 f d e g r e e ” , phi ) ss = cotg ( phi * %pi /180) + tan (( phi * %pi ) /180 -( alpha * 84

%pi ) /180) // s h e a r s t r a i n 13 printf ( ” \n s h e a r s t r a i n = %0 . 3 f ” , ss ) 14 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.7 To solve tool life equation 1 2 3 4 5 6 7 8 9 10

clc v1 = 25 // c u t t i n g s p e e d i n m/ min t1 = 90 // t o o l l i f e i n min . v2 = 35 // c u t t i n g s p e e d i n m/ min t2 = 20 // t o o l l i f e i n min n = log ( v2 / v1 ) / log ( t1 / t2 ) // e x p o n e n t C = v1 *( t1 ) ^ n // c o n s t a n t t = 60 // t o o l l i f e i n min . v = C /( t ) ^ n // c u t t i n g s p e e d i n m/ min . printf ( ” \n n = %0 . 3 f \n C = %0 . 1 f \n C u t t i n g s p e e d = %0 . 2 f m/ min . ” , n , C , v ) 11 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.8 Determine normal and tangential force 1 2 3 4 5 6 7

clc t = 0.5 // u n c u t c h i p t h i c k n e s s i n mm b = 3 // w i d t h o f c u t i n mm alpha = 15 // r a k e a n g l e i n d e g r e e s alpha = alpha * %pi /180 // r a k e a n g l e i n r a d i a n s r = 0.383 // c h i p t h i c k n e s s r a t i o mu = 0.7 // a v e r a g e c o e f f i c i e n t o f f r i c t i o n on t o o l face 8 bita = atan ( mu ) // f r i c t i o n a n g l e 9 tau = 280 // y i e l d s t r e s s i n N/mmˆ2 10 phi = atan (( r * cos ( alpha ) ) /(1 - r * sin ( alpha ) ) ) // s h e a r angle 85

11 fc = ( tau * b * t ) /( sec ( bita - alpha ) * cos ( phi + bita - alpha ) *

sin ( phi ) ) // c u t t i n g f o r c e i n N 12 ft = ( fc *( mu - tan ( alpha ) ) ) /(1+ mu * tan ( alpha ) ) //

thrust force in N 13 F = fc * sin ( alpha ) + ft * cos ( alpha ) // t a n g e n t i a l 14 15 16 17

force

on t o o l f a c e i n N F = ceil ( F ) N = fc * cos ( alpha ) - ft * sin ( alpha ) // n o r m a l f o r c e on tool face in N printf ( ” T a n g e n t i a l f o r c e on t o o l f a c e = %d N\n n o r m a l f o r c e on t o o l f a c e = %0 . 1 f N” , F , N ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.9 To find cutting and thrust force 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc t = 0.25 // u n c u t c h i p t h i c k n e s s i n mm b = 0.5 // w i d t h o f c u t i n cm v = 8.2 // c u t t i n g s p e e d i n m/ min . alpha = 20 // r a k e a n g l e i n d e g r e e s alpha2 = alpha * %pi /180 // r a k e a n g l e i n r a d i a n s r = 0.351 // c u t t i n g r a t i o phi = atan ( r * cos ( alpha2 ) /(1 - r * sin ( alpha2 ) ) ) // s h e a r angle in radians phi2 = phi *180/ %pi // s h e a r a n g l e i n d e g r e e s alpha2 = alpha * %pi /180 // r a k e a n g l e i n r a d i a n s bita = 35+ alpha - phi2 // d e g r e e s s = cotg ( phi ) + tan ( phi - alpha2 ) // s h e a r s t r a i n e = s / sqrt (3) // n a t u r a l s t r a i n sigma = 784*( e ) ^0.15 // t e n s i l e p r o p e r t y i n N/mmˆ2 tau = sigma / sqrt (3) // y i e l d s h e a r s t r e s s i n N/mmˆ2 As = ( b *10* t ) / sin ( phi ) // s h e a r p l a n e a r e a i n mmˆ2 Fs = tau * As // s h e a r g o r c e i n N R = Fs / cos ( phi +( bita * %pi /180) - alpha2 ) Fc = R * cos (( bita * %pi /180) - alpha2 ) // c u t t i n g f o r c e 86

in N 20 Ft = R * sin (( bita * %pi /180) - alpha2 ) // t h r u s t f o r c e i n N 21 printf ( ” \n C u t t i n g f o r c e = %0 . 1 f N\n T h r u s t f o r c e = %0 . 1 f N” , Fc , Ft ) 22 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.10 find terms of orthogonal rake system 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

20

clc f = 0.2 // f e e d i n mm/ r e v . t = 0.2 // u n c u t c h i p t h i c k n e s s i n mm alpha = 10 // r a k e a n g l e i n d e g r e e s fc = 1600 // c u t t i n g f o r c e i n N ft = 850 // t h r u s t f o r c e i n N tc = 0.39 // c h i p t h i c k n e s s i n mm r = t / tc // c h i p t h i c k n e s s r a t i o d = 2 // d e p t h o f c u t i n mm b = 2 // mm alpha2 = alpha * %pi /180 // r a k e a n g l e i n r a d i a n s phi = atan ( r * cos ( alpha2 ) /(1 - r * sin ( alpha2 ) ) ) // s h e a r angle in radians phi2 = phi *180/ %pi // s h e a r a n g l e i n d e g r e e fs = fc * cos ( phi ) - ft * sin ( phi ) // s h e a r f o r c e i n N fn = fc * sin ( phi ) + ft * cos ( phi ) // n o r m a l f o r c e i n N f = fc * sin ( alpha2 ) + ft * cos ( alpha2 ) // f r i c t i o n f o r c e in N mu =(( fc * tan ( alpha2 ) + ft ) /( fc - ft * tan ( alpha2 ) ) ) // kinetic c o e f f i c i e n t of f r i c t i o n s = fc /( b * t ) // s p e c i f i c c u t t i n g e n e r g y i n N/mmˆ2 printf ( ” \n S h e a r f o r c e = %d N\n Normal f o r c e = %0 . 1 f N\n F r i c t i o n f o r c e = %0 . 1 f N\n K i n e t i c c o e f f i c i e n t o f f r i c t i o n = %0 . 3 f ” , fs , fn ,f , mu ) printf ( ” \n S p e c i f i c c u t t i n g e n e r g y = %d N/mmˆ2 ” , s ) 87

21

// ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.11 Calculate CLA 1 2 3 4 5 6 7 8 9 10 11

clc cs = 20 // s i d e c u t t i n g e d g e a n g l e i n d e g r e e ce = 30 // end c u t t i n g e d g e a n g l e i n d e g r e e f = 0.1 // f e e d i n mm/ r e v . r = 3 // n o s e r a d i u s i n mm cs2 = cs * %pi /180 // s i d e c u t t i n g e d g e a n g l e i n radians ce2 = ce * %pi /180 // end c u t t i n g e d g e a n g l e i n radians h = (1 - cos ( ce2 ) ) * r + f * sin ( ce2 ) * cos ( ce2 ) - sqrt ((2* f * r *( sin ( ce2 ) ) ^3) -(( f ^2) *( sin ( ce2 ) ) ^4) ) Ra = h /4 // C e n t r e l i n e a v e r a g e r o u g h n e s s i n mm printf ( ” \n C e n t r e l i n e a v e r a g e r o u g h n e s s = %0 . 2 f ∗10ˆ −6m” , Ra *10^3) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.12 Calculate back and side rake angle 1 clc 2 i = 0 // i n c l i n a t i o n a n g l e i n d e g r e e 3 alpha = 10 // o r t h o g o n a l r a k e a n g l e i n d e g r e e 4 lemda = 75 // p r i n c i p a l c u t t i n g e d g e a n g l e i n d e g r e e 5 alpha = alpha * %pi /180 // o r t h o g o n a l r a k e a n g l e i n

radian lemda = lemda * %pi /180 // p r i n c i p a l c u t t i n g e d g e angle in radian 7 alpha_b = atan ( cos ( lemda ) * tan ( alpha ) + sin ( lemda ) * tan ( i ) ) // back r a k e a n g l e i n r a d i a n s

6

88

alpha_b = alpha_b *180/ %pi // back r a k e a n g l e i n degree 9 alpha_s = atan ( sin ( lemda ) * tan ( alpha ) - cos ( lemda ) * tan ( i ) ) // s i d e r a k e a n g l e i n r a d i a n s 10 alpha_s = alpha_s *180/ %pi // s i d e r a k e a n g l e i n degree 11 printf ( ” \n Back r a k e a n g l e = %0 . 2 f d e g r e e \n S i d e r a k e a n g l e = %0 . 2 f d e g r e e ” , alpha_b , alpha_s ) 8

Scilab code Exa 14.13 Calculate inclination and rake angle 1 clc 2 alphab = 8 // back r a k e i n d e g r e e 3 alphas = 4 // s i d e r a k e i n d e g r e e 4 cs = 15 // s i d e c u t t i n g e d g e a n g l e i n d e g r e e 5 lemda = 90 - cs // a p p r o a c h a n g l e i n d e g r e e 6 alphab = alphab * %pi /180 // back r a k e i n r a d i a n 7 alphas = alphas * %pi /180 // s i d e r a k e i n r a d i a n 8 cs = cs * %pi /180 // s i d e c u t t i n g e d g e a n g l e i n r a d i a n 9 lemda = lemda * %pi /180 // a p p r o a c h a n g l e i n r a d i a n 10 alpha = atan ( tan ( alphas ) * sin ( lemda ) + tan ( alphab ) * cos (

lemda ) ) // o r t h o g o n a l r a k e a n g l e i n r a d i a n alpha = alpha *180/ %pi // o r t h o g o n a l r a k e a n g l e i n degree 12 i = atan ( sin ( lemda ) * tan ( alphab ) - cos ( lemda ) * tan ( alphas ) ) // i n c l n a t i o n a n g l e i n r a d i a n 13 i = i *180/ %pi // i n c l n a t i o n a n g l e i n d e g r e e 14 printf ( ” \n O t h o g o n a l r a k e a n g l e = %0 . 2 f d e g r e e \n I n c l i n a t i o n a n g l e = %0 . 1 f d e g r e e ” , alpha , i ) 11

Scilab code Exa 14.14 find different powers and resistance 1 clc

89

cs = 15 // s i d e c u t t i n g e d g e a n g l e i n d e g r e e v = 0.2 // c u t t i n g s p e e d i n m/ s f = 0.5 // f e e d r a t e i n mm/ r e v . d = 3.2 // d e p t h o f c u t i n mm fc = 1593*( f ) ^0.85*( d ) ^0.98 // c u t t i n g f o r c e i n N pc = fc * v /1000 // c u t t i n g power i n kw ita_mt = 0.85 // e f f i c i e n c y o f l a t h e pm = pc / ita_mt // motor power i n kw a = f * d // a r e a o f u n c u t c h i o i n mmˆ2 r = fc / a // s p e c i f i c c u t t i n g r e s i s t a n c e i n N/mmˆ2 p = pc /( a * v ) // u n i t power i n W/ (mmˆ 3 ) ∗ s printf ( ” \n C u t t i n g power = %0 . 3 f kw\n Motor power = %0 . 2 f kw\n S p e c i f i c c u t t i n g r e s i s t a n c e = %0 . 2 f N/ mmˆ2\ n U n i t power = %0 . 3 f W/ (mmˆ 3 ) ∗ s ” , pc , pm ,r , p ) 14 // ’ Answers v a r y due t o round o f f e r r o r ’

2 3 4 5 6 7 8 9 10 11 12 13

Scilab code Exa 14.15 Calculate percentage increase in tool life 1 2 3 4 5 6 7

clc C = 400 n =0.5 a =2 // ( T1/T2 ) ˆn b =2^(1/ n ) // T2 i = (b -1) *100 // p e r c e n t a g e i n c r e a s e printf ( ” \n P e r c e n t a g e i n c r e a s e = %d p e r c e n t ” , i )

Scilab code Exa 14.16 To find percentage of total energy 1 2 3 4

clc t = 0.127 // u n c u t c h i p t h i c k n e s s i n mm b = 6.35 // w i d t h o f c u t i n mm v = 1.20 // c u t t i n g s p e e d i n m/ min . 90

alpha = 10 // r a k e a n g l e i n d e g r e e s fc = 556.25 // c u t t i n g f o r c e i n N ft = 222.50 // t h r u s t f o r c e i n N tc = 0.229 // c h i p t h i c k n e s s i n mm r = t / tc // c h i p t h i c k n e s s r a t i o R = sqrt (( fc ^2) +( ft ^2) ) bita = ( acos ( fc / R ) ) + alpha * %pi /180 // f = R * sin ( bita ) // fe = f * r // f r i c t i o n e n e r g y p = ( f * r *100) / fc // p e r c e n t a g e o f f r i c t o n e n r g y and t o t a l energy 15 printf ( ” \n The p e r c e n t a g e o f t o t a l e n e r g y t h a t g o e s i n t o overcoming f r i c t i o n at t o o l chip i n t e r f a c e = %0 . 2 f p e r c e n t ” , p ) 16 // ’ Answers v a r y due t o round o f f e r r o r ’ 5 6 7 8 9 10 11 12 13 14

Scilab code Exa 14.17 To find power and different energies 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc D = 300 // d i a m e t e r i n mm r = 45 // r e v / min . d = 2 // d e p t h o f c u t i n mm f = 0.3 // f e e d i n mm/ r e v fc = 1850 // c u t t i n g f o r c e i n N ff = 450 // f e e d f o r c e i n N V = 2.5*10^6 // m e t a l removed i n mm v = ( %pi * D * r ) /(60*1000) // c u t t i n g v e l o c i t y i n m/ s pc = fc * v /1000 // c u t t i n g power i n kW fv = f * r /60*1000 // f e e d v e l o c i t y i n m/ s fp = fv * ff // f e e d power i n W mrr = d * f * v *60*1000 // mmˆ3/ min . ce = pc *1000*60/ mrr // s p e c i f i c c u t t i n g e n e r g y i n W∗ s /mmˆ2 15 E = ce * V /(3600*1000) // e n e r g y consumed 16 printf ( ” \n Power c o n s u m p t i o n = %0 . 2 f W\n S p e c i f i c 91

c u t t i n g e n e r g y = %0 . 2 f W∗ s /mmˆ3\ n Energy consumed = %0 . 3 f kWh” ,pc , ce , E ) 17 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 14.18 Determine components of force and power 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

22

clc D = 100 // d i a m e t e r i n mm cs = 30 // s i d e c u t t i n g e d g e a n g l e i n d e g r e e lemda = 90 - cs // a p p r o a c h a n g l e i n d e g r e e d = 2.5 // d e p t h o f c u t i n mm f = 0.125 // f e e d i n mm/ r e v . N = 300 // t u r n i n g s p e e d o f j o b i n r e v . / min . mu = 0.6 // c o e f f i c i e n t o f f r i c t i o n tau = 400 // u l t i m a t e s h e a r s t r e s s i n Mpa bita = atand ( mu ) // f r i c t i o n a n g l e i n r a d i a n alphas = 10 // s i d e r a k e a n g l e alphab = 6 // back r a k e a n g l e alpha = atand ( tand ( alphas ) * sind ( lemda ) + tand ( alphab ) * cosd ( lemda ) ) // o r t h o g o n a l r a k e a n g l e i n d e g r e e phi = 45 - ( bita - alpha ) // s h e a r a n g l e Fc = tau * d * f /( secd ( bita - alpha ) * cosd ( phi + bita - alpha ) * sind ( phi ) ) // c u t t i n g f o r c e i n N Ft = Fc * tand ( bita - alpha ) // t h r u s t component i n N Ff = Ft * sind ( lemda ) // f e e d f o r c e a l o n g a x i s o f j o b in N Rf = Ft * cosd ( lemda ) // r a d i a l f o r c e n o r m a l t o a x i s of job in N v = %pi * D * N /(1000*60) // v e l o c i t y i n m/ s p = Fc * v // power i n w a t t s printf ( ” \n C u t t i n g f o r c e = %d N\n T h r u s t f o r c e = %0 . 3 f N\n Feed f o r c e = %0 . 1 f N\n R a d i a l f o r c e = %0 . 3 f N\n C u t t i n g power = %d w a t t s ” , Fc , Ft , Ff , Rf , p ) // ’ Answers v a r y due t o round o f f e r r o r ’ 92

93

Chapter 15 Design and manufacture of cutting tools

Scilab code Exa 15.1 calculate horsepower at cutter and motor 1 2 3 4 5 6 7 8 9 10 11 12 13

clc w = 10 // w i d t h o f c u t i n cm h = 0.32 // d e p t h o f c u t i n cm n = 8 // number o f t e e t h i n c u t t e r ft = 0.033 // f e e d r a t e p e r t o o t h N = 200 // c u t t e r s p e e d i n rpm ita = 60/100 // e f f i c i e n c y f = n * ft * N // f e e d r a t e i n cm/ min . mrr = w * h * f // m e t a l r e m o v a l r a t e i n cmˆ3/ min . k = 8.2 // m a c h i n i b i l i t y f a c t o r from t a b l e 1 5 . 3 hpc = mrr / k // h o r s e p o w e r a t c u t t e r hpm = hpc / ita // h o r s e p o w e r a t motor printf ( ” \n H o r s e p o w e r a t c u t t e r = %0 . 2 f \n H o r s e p o w e r a t motor = %0 . 2 f ” , hpc , hpm ) 14 // ’ Answers v a r y due t o round o f f e r r o r ’

94

Scilab code Exa 15.2 Determine broaching power and Design broach 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

31

clc l = 35 // l e n g t h o f b o r e i n mm v = 0.15 // c u t t i n g s p e e d i n m/ s t1 = 0.01 // u p p e r l i m i t i n mm t2 = 0.05 // u p p e r l i m i t i n mm D = 32.25 // f i n i s h e d b r o a c h i n mm D1 = 32.25+ t2 // mm d = 32.75 // f i n i s h d i a m e t e r i n mm d1 = 32.75 + t1 // f i n i s h d i a m e t e r o f h o l e i n mm s = 0.05 // mm B = 1.30 // b l u n t b r o a c h f a c t o r c = 45 // s p e c i f i c c u t t i n g f o r c e i n N/mmˆ2 n = 3 // number o f t e e t h c u t t i n g a t a t i m e F = n * %pi * d1 * s * c * B // f o r c e n e e d e d f o r b r o a c h i n g i n N bp = F * v /1000 // B r o a c h i n g power i n kw // b r o a c h d e s i g n p = 1.75* sqrt ( l ) // p i t c h i n mm theta = 10 // f a c e a n g l e i n d e g r e e alha1 = 1.5 // r e l i e f a n g l e f o r r o u g h i n g i n d e g r e e alha2 = 1.0 // r e l i e f a n g l e f o r f i n i s h i n g i n d e g r e e w = 0.3* p // w i d t h o f l a n d i n mm h = 0.4* p // d e p t h o f c u t t i n g t e e t h i n mm r = 0.3* p // t o o t h f i l l e t r a d i u s i n mm T = ( d1 - D1 ) /2 // mm n = T / s // number o f c u t t i n g t e e t h n = round ( n ) l = ( n +7) * p // l e n g t h o f t o o t h e d p o r t i o n o f b r o a c h i n mm printf ( ” \n ( i ) B r o a c h i n g power = %0 . 4 f kW” , bp ) disp ( ” ( i i ) Broach D e s i g n ” ) printf ( ” ( a ) P i t c h d i a m e t e r = %0 . 2 fmm\n ( b ) w i d t h o f l a n d = %0 . 2 f mm \n depth o f c u t t i n g t e e t h = %0 . 2 f mm\n Tooth f i l l e t r a d i u s= %0 . 2 f mm” , p ,w ,h , r ) printf ( ” \n ( c ) Length o f t o o t h e d p o r t i o n o f b r o a c h = 95

%d mm” , l ) 32 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 15.3 Estimate moment thrust force and power 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

19

clc Hb = 200 // b r i n e l l h a r d n e s s d = 12.7 // d i a m e t e r i n mm f = 0.254 // f e e d i n mm/ r e v . N = 100 // rpm M = ( Hb *( d ) ^2* f ) /8 // moment i n k g f −mm k = 1.1 // m a t e r i a l f a c t o r p = (1.25*( d ) ^2* k * N *(0.056+1.5* f ) ) /(10) ^5 // power i n kW T1a = (1.7* M ) / d // t h r u s t f o r c e k g f T1b = (3.5* M ) / d // k g f T1 = ( T1a + T1b ) /2 // a v e r a g e w = 0.14* d // t h i c k n e s s i n mm T2a = (0.1* %pi *( w ) ^2* Hb ) /4 // t h r u s t f o r c e k g f T2b = (0.2* %pi *( w ) ^2* Hb ) /4 // t h r u s t f o r c e k g f T2 = ( T2a + T2b ) /2 // a v e r a g e avg = T1 + T2 // k g f thrust = 1.16* k * d *(100* f ) ^0.85 // k g f printf ( ” \n Moment = %0 . 1 f k g f −mm\n Power = %0 . 3 f hp \ n A v e r a g e f o r c e = %d k g f \n T h r u s t f o r c e = %0 . 1 f k g f ” ,M , p , avg , thrust ) // E r r o r i n t e x t b o o k

Scilab code Exa 15.4 Design shell inserted blade reamer 1 clc 2 d = 55 // d i a m e t e r i n mm 3 ul = 0.035 // u p p e r l i m i t

i n mm 96

4 ll = 0.000 // l o w e r l i m i t i n mm 5 Dmax = d + ul // maximum d i a m e t e r o f h o l e i n mm 6 Dmin = d + ll // minimum d i a m e t e r o f h o l e i n mm 7 IT = 0.035 // h o l e t o l e r e n c e i n mm 8 dmax = Dmax -0.15* IT // maximum d i a m e t e r o f r e a m e r i n 9 10 11 12 13

14

mm dmin = dmax -0.35* IT // minimum d i a m e t e r o f r e a m e r i n mm l = (( d /4) +( d /3) ) /2 // l e n g t h o f g u i d i n g s e c t i o n i n mm Z = 1.5* sqrt ( d ) +2 // number o f t e e t h Z = ceil ( Z ) printf ( ” \n 1 D i a m e t e r o f r e a m e r \n Maximum d i a m e t e r o f r e a m e r = %0 . 3 f mm \n Minimum d i a m e t e r o f r e a m e r = %0 . 3 f mm \n 2 Back t a p e r = 0 . 0 5 mm \n 3 V a l u e s o f v a r i o u s a n g l e s \n Rake a n g l e = 5 d e g r e e \n Plan a p p r o a c h a n g l e = 45 d e g r e e \n C i r c u l a r l a n d = 0 . 2 5 t o 0 . 5 0 mm \n S e c o n d a r y c l e a r a n c e a n g l e = 10 d e g r e e \n 4 Length o f r e a m e r \n Length o f f l u t e d p o r t i o n = 8 2 . 5 mm \n Length o f r e a m i n g a l l o w a n c e = 0 . 1 8 mm \n Length o f c u t t i n g s e c t i o n = 2 . 2 5 mm \n Length o f g u i d i n g s e c t i o n = %d mm \ n 5 Number o f t e e t h = %d” , dmax , dmin ,l , Z ) // Answer v a r y due t o round o f f e r r o r

Scilab code Exa 15.5 To design single point cutting tool 1 2 3 4 5 6 7 8

clc Pm = 10 // power o f motor i n kw v = 40 // c u t t i n g s p e e d i n m/ min . ita = 70 // e f f i c i e n c y ita = ita /100 Pc = Pm * ita Fc = ( Pc *1000*60) / v // c u t t i n g f o r c e sigmab = 250 // s t r e s s i n Mpa 97

9 B = sqrt (( Fc *1.25*6) /( sigmab *1.6) ) // w i d t h o f s h a n k

i n mm 10 h = 1.6* B // h i e g h t o f s h a n k i n mm 11 l = 1.25* h // s h a n k o v e r a n g i n mm 12 printf ( ” \n The w i d t h o f s h a n k = %0 . 1 f mm\n H e i g h t o f

s h a n k = %0 . 2 f mm\n Shank o v e r h a n g = %0 . 1 f mm” , B ,h , l ) 13 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 15.8 find various terms for stainless steel 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc l = 150 // l e n g t h i n mm D = 12.70 // d i a m e t e r i n mm dia = 12.19 // d i a m e t e r on c e n t r e l a t h e i n mm N = 400 // s p i n d l e s p e e d i n r e v . / min s = 203.20 // a x i a l s p e e d i n mm/ min.∗#### v = ( %pi * D * N ) /(1000*60) // c u t t i n g v e l o c i t y i n m/ s d = (D - dia ) /2 // d e p t h o f c u t i n mm f = s / N // f e e d i n mm/ r e v . Dave = ( D + dia ) /2 // a v e r a g e d i a m e t e r i n mm V = %pi * Dave * N a = d * f // a r e a o f c u t i n mmˆ2 mrr = a * V // m e t a l r e m o v a l r a t e i n mmˆ3/ min . T = l /( f * N ) // machine t i m i n g i n min . c = 56 // c o n s t a n t from t a b l e p = d * f * v *60* c // power i n w a t t s omega = (2* %pi * N ) /60 // rpm t = p / omega // t o r q u e i n Nm Fc = (2* t *1000) / Dave // c u t t i n g f o r c e i n N printf ( ” \n C u t t i n g s p e e d = %0 . 2 f m/ s \n MRR = %d mm ˆ3/ min . \ n Time t o c u t = %0 . 2 f min . \ n Power = %0 . 1 f w a t t s \ n C u t t i n g f o r c e = %d N” , v , mrr , T ,p , Fc ) 21 // Answers a r e g i v e n wrong i n book 98

Scilab code Exa 15.9 To find MRR power and torque 1 2 3 4 5 6 7 8 9 10

clc f = 0.2 // f e e d i n mm/ r e v . N = 800 // s p i n d l e s p e e d i n r e v . / min . d = 10 // d o a m e t e r o f h o l e i n mm mrr = %pi *( d ^2) * f * N /4 // m e t a l r e m o v a l r a t e i n mmˆ3/ min . mrr = mrr /60 // mmˆ3/ s p = 0.5* mrr // c u t t i n g power from t a b l e 1 4 . 2 i n watts omega = 2* %pi * N /60 // rpm T = p / omega // t o r q u e i n N .m printf ( ” \n MRR = %0 . 2 f mmˆ3/ s \n C u t t i n g power = %0 . 3 f w a t t s \n Torque = %0 . 2 f N .m” , mrr ,p , T )

Scilab code Exa 15.10 find MRR power torque and time 1 2 3 4 5 6 7 8 9 10 11 12 13

clc l = 300 // l e n g t h i n mm w = 100 // w i d t h i n mm f = 0.25 // f e e d i n mm/ t o o t h d = 3.2 // d e p t h o f c u t i n mm D = 50 // c u t t e r d i a m e t e r i n mm n = 20 // number o f c u t t e r t e e t h N = 100 // c u t t e r s p e e d i n r e v . / min . tf = f * n * N // t a b l e f e e d i n mm/ min . mrr = w * d * tf // m e t a l r e m o v a l r a t e i n mmˆ3/ min . mrr = mrr /60 // mmˆ3/ s p = 6* mrr // c u t t i n g power from t a b l e 1 4 . 2 i n w a t t s omega = 2* %pi * N /60 // rpm 99

14 15 16 17 18

T = p / omega // t o r q u e i n N .m att = sqrt (( D * d ) -( d ^2) ) // added t a b l e t r a v e l i n mm t = ( l + att ) / tf // c u t t i n g t i m e i n min . t = t *60 // s printf ( ” \n MRR = %0 . 2 f mmˆ3/ s \n C u t t i n g power = %d w a t t s \n Torque = %0 . 2 f N .m\n C u t t i n g t i m e = %0 . 1 f s ” , mrr ,p ,T , t )

100

Chapter 16 Gear manufacture

Scilab code Exa 16.1 Calculate settings of gear tooth 1 2 3 4 5

clc n = 34 // number o f t e e t h s m = 5 // module i n mm w = m * n * sin ( %pi /( n *2) ) // t o o t h t h i c k n e s s i n mm h = m *(1+( n *(1 - cos ( %pi /( n *2) ) ) /2) ) // c h o r d a l addendum i n mm 6 printf ( ” \n Tooth t h i c k n e s s = %0 . 3 f mm\n C h o r d a l addendum = %0 . 3 f mm” ,w ,h ) 7 // ’ Answers v a r y due t o round o f f e r r o r ’

101

Chapter 17 Thread manufacturing

Scilab code Exa 17.1 Calculate best wire size 1 2 3 4 5

clc d = 80 // o u t s i d e d i a m e t e r i n mm p = 6 // p i t c h d i a m e t e r i n mm d = 0.5774* p // b e s t w i r e s i z e i n mm printf ( ” \n B e s t w i r e s i z e = %0 . 3 f mm” , d )

Scilab code Exa 17.2 Calculate size and distances over wire 1 2 3 4 5 6

clc D = 20 // d i a m e t e r i n mm p = 2.5 // p i t c h d i a m e t e r i n mm d = 0.5774* p // mm W = D +3* d -1.5156* p // b e s t w i r e s i z e i n mm printf ( ” \n B e s t w i r e s i z e = %0 . 3 f mm\n D i s t a n c e o v e r w i r e s = %0 . 3 f mm” ,d , W ) 7 // Answer v a r y due t o round o f f e r r o r

102

Chapter 21 Statical quality control

Scilab code Exa 21.1 Construct R and X chart 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc clf () n = 10 // number o f s a m p l e s A2 = 0.577 D3 = 0 D4 = 2.115 // number o f d e f e c t i v e s x1 = 11.274 x2 = 11.246 x3 = 11.204 x4 = 11.294 x5 = 11.252 x6 = 11.238 x7 = 11.230 x8 = 11.276 x9 = 11.208 x10 = 11.266 r1 = 0.15 r2 = 0.20 r3 = 0.33 r4 = 0.46 103

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

r5 = 0.10 r6 = 0.15 r7 = 0.20 r8 = 0.23 r9 = 0.50 r10 = 0.30 x = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 r = r1 + r2 + r3 + r4 + r5 + r6 + r7 + r8 + r9 + r10 Xavg = x / n Ravg = r / n // f o r X c h a r t ucl1 = Xavg + A2 * Ravg lcl1 = Xavg - A2 * Ravg // f o r R c h a r t ucl2 = D4 * Ravg lcl2 = D3 * Ravg printf ( ” \n c o n t r o l l i m i t s \n For X c h a r t s \n UCL = %0 . 2 f cm \n LCL = %0 . 2 f cm\n For R c h a r t s \n UCl = %0 . 3 f \n LCL = %0 . 3 f ” , ucl1 , lcl1 , ucl2 , lcl2 ) 39 // X c h a r t 40 x =[1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 ,10]; 41 y =[11.274 ,11.246 ,11.204 ,11.294 ,11.252 ,11.238 ,11.230 ,11.276 ,11.208 ,1 42 plot (x , y ) 43 xtitle ( ”X c h a r t ” ,” Sample No . ” ,”X” ) 44 // R c h a r t 45 xset ( ” window ” ,1) 46 z =

[0.15 ,0.20 ,0.33 ,0.46 ,0.10 ,0.15 ,0.20 ,0.23 ,0.50 ,0.30] 47 48

plot (x , z ) xtitle ( ”R c h a r t ” ,” Sample no . ” , ”R” )

Scilab code Exa 21.2 Construct the control charts 104

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

32 33

clc clf () n = 100 // t o t a l number o f sub g r o u p s s = 10 // number o f s a m p l e s // number o f d e f e c t i v e s d1 = 3 d2 = 2 d3 = 3 d4 = 5 d5 = 3 d6 = 3 d7 = 2 d8 = 4 d9 = 3 d10 = 2 d = d1 + d2 + d3 + d4 + d5 + d6 + d7 + d8 + d9 + d10 // t o t a l number of defectives p1 = d /( n * s ) // a v e r a g e f r a c t i o n o f d e f e c t i v e s sigmap1 = sqrt ( p1 *(1 - p1 ) / n ) ucl1 = p1 + 3* sigmap1 lcl1 = p1 - 3* sigmap1 // c o n t r o l c h a r t f o r f r a c t i o n d e f e c t i v e s x = linspace (0 ,10 ,10) y = linspace (0 ,0.081 ,10) plot (x , y ) xtitle ( ” C o n t r o l c h a r t f o r f r a c t i o n d e f e c t i v e s ” , ” S a m p l e s ” ,” F r a c t i o n d e f e c t i v e s ” ) // p e r c e n t d e f e c t i v e ( mean ) p1 = p1 *100 sigmap2 = sqrt ( p1 *(100 - p1 ) / n ) ucl2 = p1 + 3* sigmap2 lcl2 = p1 - 3* sigmap2 printf ( ” \n C o n t r o l l i m i t s \n F r a c t i o n d e f e c t i v e s \n UCL = %0 . 3 f \n LCL = %0 . 4 f \n P e r c e n t d e f e c t i v e s \n UCL = %0 . 1 f \n LCL = %0 . 1 f ” , ucl1 , lcl1 , ucl2 , lcl2 ) // c o n t r o l c h a r t f o r p e r c e n t d e f e c t xset ( ” window ” ,1) 105

34 z = linspace (0 ,8.1 ,10) 35 plot (x , z ) 36 xtitle ( ” C o n t r o l c h a r t f o r 37

//

p e r c e n t d e f e c t s ” , ” Sample no . ” , ” p e r c e n t d e f e c t s ” ) ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 21.4 Calculate poisson probabilities 1 2 3 4 5 6 7 8 9 10

clc n = s = d = z = pp0

1000 // number o f u n i t s 4 // random s a m p l e 50 // d e f e c t i v e s d*s/n = exp ( -0.2) *1 // p o i s s o n p r o b a b i l i t i e s f o r 0 defectives pp1 = exp ( -0.2) *( z ) // p o i s s o n p r o b a b i l i t i e s f o r 1 defectives pp2 = exp ( -0.2) *( z ^2/ factorial (2) ) // p o i s s o n probabilities for 2 defectives pp3 = exp ( -0.2) *( z ^3/ factorial (3) ) // p o i s s o n probabilities for 3 defectives printf ( ” \n P r o a b i l i t i e s f o r 0 , 1 , 2 and 3 d e f e c t i v e s a r e : %0 . 3 f ,%0 . 4 f , %0 . 4 f , %0 . 5 f ” , pp0 , pp1 , pp2 , pp3 )

Scilab code Exa 21.5 Calculate probabilities of defective items 1 2 3 4 5 6

clc d = l = p = q = n =

50 // d e f e c t i v e s 1000 // l o t o f p i e c e s d / l // p r o a b i l i t y o f an e v e n t h a p p e n i n g 1 - p // p r o a b i l i t y o f an e v e n t n o t h a p p e n i n g 4 // s a m p l e s i z e 106

7 8 9 10 11

p0 = q ^ n // p r o b a b i l i t i e s f o r 0 d e f e c t i v e s p1 = 4*( q ) ^3* p // p r o b a b i l i t i e s f o r 1 d e f e c t i v e s p2 = 6*( q ) ^2* p ^2 // p r o b a b i l i t i e s f o r 2 d e f e c t i v e s p3 = 4* q *( p ) ^3 // p r o b a b i l i t i e s f o r 3 d e f e c t i v e s printf ( ” \n P r o a b i l i t i e s f o r 0 , 1 , 2 and 3 d e f e c t i v e s a r e : %0 . 4 f %0 . 4 f %0 . 4 f %0 . 6 f ” , p0 , p1 , p2 , p3 )

Scilab code Exa 21.6 Determine producers and consumers risk 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc // p r o d u c e r ’ s r i s k n = 71 // s a m p l e s i z e AQL = 0.005 LTPD = 0.05 l_s = 500 // l o t s i z e z1 = n * AQL // mean number o f d e f e c t s pp1 = exp ( - z1 ) + z1 * exp ( - z1 ) // p o i s s o n p r o a b i l i t y f o r 1 or l e s s d e f e c t i v e alpha = (1 - pp1 ) *100 // p r o d u c e r ’ s r i s k alpha = ceil ( alpha ) // consumer ’ s r i s k z2 = n * LTPD // mean number o f d e f e c t s pp2 = exp ( - z2 ) + z2 * exp ( - z2 ) // p o i s s o n p r o a b i l i t y f o r 1 or l e s s d e f e c t i v e bita = pp2 *100 // consumer ’ s r i s k printf ( ” \n P r o d u c e r s r i s k = %d p e r c e n t \n Consumers r i s k = %0 . 2 f p e r c e n t ” , alpha , bita )

Scilab code Exa 21.7 Evaluate preliminary and revised control limits 1 clc 2 td1 = 20 // t o t a l number o f d a y s 3 n1 = 200 // s a m p l e s i z e

107

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

26 27 28 29 30 31 32 33 34 35 36 37

// number o f d e f e c t i v e s d1 = 10 d2 = 15 d3 = 10 d4 = 12 d5 = 11 d6 = 9 d7 = 22 d8 = 4 d9 = 12 d10 = 24 d11 = 21 d12 = 15 d13 = 8 d14 = 14 d15 = 4 d16 = 10 d17 = 11 d18 = 11 d19 = 26 d20 = 13 d = d1 + d2 + d3 + d4 + d5 + d6 + d7 + d8 + d9 + d10 + d11 + d12 + d13 + d14 + d15 + d16 + d17 + d18 + d19 + d20 // t o t a l number o f defectives p1 = d /( n1 * td1 ) // a v e r a g e f r a c t i o n o f d e f e c t i v e s sigmap1 = sqrt ( p1 *(1 - p1 ) / n1 ) ucl1 = p1 + 3* sigmap1 lcl1 = p1 - 3* sigmap1 // r e v i s e d c o n t r o l l i m i t s td2 = 18 // t o t a l number o f d a y s D = d - ( d10 + d19 ) // number o f d e f e c t s p2 = D /( n1 * td2 ) sigmap2 = sqrt ( p2 *(1 - p2 ) / n1 ) ucl2 = p2 + 3* sigmap2 lcl2 = p2 - 3* sigmap2 printf ( ” \n P r e l i m i n a r y c o n t r o l l i m i t s \n UCL = %0 . 3 f \n LCL = %0 . 3 f \n R e v i s e d c o n t r o l l i m i t s \n UCL = %0 . 3 f \n LCL = %0 . 3 f ” , ucl1 , lcl1 , ucl2 , lcl2 ) 108

Scilab code Exa 21.8 Find control limits for c chart 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

clc n1 = 15 // t o t a l number o f sub g r o u p s // number o f d e f e c t i v e s d1 = 77 d2 = 64 d3 = 75 d4 = 93 d5 = 45 d6 = 61 d7 = 49 d8 = 65 d9 = 45 d10 = 77 d11 = 59 d12 = 54 d13 = 84 d14 = 40 d15 = 92 d = d1 + d2 + d3 + d4 + d5 + d6 + d7 + d8 + d9 + d10 + d11 + d12 + d13 + d14 + d15 // t o t a l number o f d e f e c t i v e s c1 = d / n1 ucl1 = c1 + 3* sqrt ( c1 ) lcl1 = c1 - 3* sqrt ( c1 ) // r e v i s e d c o n t r o l l i m i t s n2 = 12 // t o t a l number o f sub g r o u p s D = d - ( d4 + d14 + d15 ) // number o f d e f e c t s c2 = D / n2 ucl2 = c2 + 3* sqrt ( c2 ) lcl2 = c2 - 3* sqrt ( c2 ) printf ( ” \n P r e l i m i n a r y c o n t r o l l i m i t s \n UCL = %0 . 2 f \n LCL = %0 . 2 f \n R e v i s e d c o n t r o l l i m i t s \n UCL = %0 . 3 f \n LCL = %0 . 3 f ” , ucl1 , lcl1 , ucl2 , lcl2 ) 109

Scilab code Exa 21.9 find control limits for charts 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

clc n = 20 // number o f s a m p l e s A = 1.342 A1 = 1.596 A2 = 0.577 d2 = 2.326 d3 = 0.864 D1 = 0 D2 = 4.918 D3 = 0 D4 = 2.115 // number o f d e f e c t i v e s x1 = 3290 x2 = 3180 x3 = 3350 x4 = 3470 x5 = 3080 x6 = 3240 x7 = 3260 x8 = 3310 x9 = 3640 x10 = 4110 x11 = 3220 x12 = 3590 x13 = 4270 x14 = 4040 x15 = 3580 x16 = 3500 x17 = 3570 x18 = 3560 x19 = 2740 x20 = 3200 110

33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

r1 = 560 r2 = 410 r3 = 200 r4 = 300 r5 = 90 r6 = 650 r7 = 890 r8 = 410 r9 = 1120 r10 = 520 r11 = 580 r12 = 670 r13 = 480 r14 = 250 r15 = 170 r16 = 670 r17 = 440 r18 = 660 r19 = 560 r20 = 590 x = x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20 r = r1 + r2 + r3 + r4 + r5 + r6 + r7 + r8 + r9 + r10 + r11 + r12 + r13 + r14 + r15 + r16 + r17 + r18 + r19 + r20 Xavg = x / n Ravg = r / n // f o r X c h a r t ucl1 = Xavg + A2 * Ravg lcl1 = Xavg - A2 * Ravg // f o r R c h a r t ucl2 = D4 * Ravg lcl2 = D3 * Ravg // R e v i s e d c o n t r o l l i m i t s n1 = 15 n2 = 19 X = ( x - ( x5 + x10 + x13 + x14 + x19 ) ) / n1 R = ( r - ( r9 ) ) / n2 // f o r X c h a r t 111

69 70 71 72 73 74

ucl3 = X + A2 * R lcl3 = X - A2 * R // f o r R c h a r t ucl4 = D4 * R lcl4 = D3 * R printf ( ” \n P r e l i m i n a r y c o n t r o l l i m i t s \n For X c h a r t s \n UCL = %0 . 2 f \n LCL = %0 . 2 f \n For R c h a r t s \n UCl = %0 . 3 f \n LCL = %0 . 3 f \n R e v i s e d c o n t r o l l i m i t s \n For X c h a r t \n UCL = %0 . 3 f \n LCL = %0 . 3 f \n For R c h a r t s \n UCl = %0 . 3 f \n LCL = %0 . 3 f ” , ucl1 , lcl1 , ucl2 , lcl2 , ucl3 , ucl3 , ucl4 , lcl4 ) 75 // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 21.10 Determine producers and consumers risk 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc clf () n = 50 // s a m p l e s i z e rn = 2 // r e j e c t i o n number AQL = 0.02 LTPD = 0.08 // P r o d u c e r ’ s r i s k z1 = n * AQL // mean number o f d e f e c t i v e s pp1 = exp ( - z1 ) + z1 * exp ( - z1 ) // p o i s s o n p r o a b i l i t y f o r 1 or l e s s d e f e c t i v e alpha = (1 - pp1 ) *100 // p r o d u c e r ’ s r i s k // consumer ’ s r i s k z2 = n * LTPD // mean number o f d e f e c t i v e s bita = ( exp ( - z2 ) + z2 * exp ( - z2 ) ) *100 // consumer ’ s r i s k d1 = 1 // i n c o m i n g d e f e c t i v e i n p e r c e n t z3 = n * d1 /100 // a v e r a g e number o f d e f e c t i v e ppa1 = exp ( - z3 ) + z3 * exp ( - z3 ) // p r o a b i l i t y o f acceptance ppa1 = ppa1 *100 112

18 ppr1 = 100 - ppa1 // p r o a b i l i t y o f r e j e c t i o n 19 AOQ1 = ppr1 *0 + ppa1 * d1 /100 20 d2 = 2 // i n c o m i n g d e f e c t i v e i n p e r c e n t 21 z4 = n * d2 /100 // a v e r a g e number o f d e f e c t i v e 22 ppa2 = exp ( - z4 ) + z4 * exp ( - z4 ) // p r o a b i l i t y o f 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

acceptance ppa2 = ppa2 *100 ppr2 = 100 - ppa2 // p r o a b i l i t y o f r e j e c t i o n AOQ2 = ppr2 *0 + ppa2 * d2 /100 d3 = 4 // i n c o m i n g d e f e c t i v e i n p e r c e n t z5 = n * d3 /100 // a v e r a g e number o f d e f e c t i v e ppa3 = exp ( - z5 ) + z5 * exp ( - z5 ) // p r o a b i l i t y o f acceptance ppa3 = ppa3 *100 ppr3 = 100 - ppa3 // p r o a b i l i t y o f r e j e c t i o n AOQ3 = ppr3 *0 + ppa3 * d3 /100 d4 = 6 // i n c o m i n g d e f e c t i v e i n p e r c e n t z6 = n * d4 /100 // a v e r a g e number o f d e f e c t i v e ppa4 = exp ( - z6 ) + z6 * exp ( - z6 ) // p r o a b i l i t y o f acceptance ppa4 = ppa4 *100 ppr4 = 100 - ppa4 // p r o a b i l i t y o f r e j e c t i o n AOQ4 = ppr4 *0 + ppa4 * d4 /100 d5 = 8 // i n c o m i n g d e f e c t i v e i n p e r c e n t z7 = n * d5 /100 // a v e r a g e number o f d e f e c t i v e ppa5 = exp ( - z7 ) + z7 * exp ( - z7 ) // p r o a b i l i t y o f acceptance ppa5 = ppa5 *100 ppr5 = 100 - ppa5 // p r o a b i l i t y o f r e j e c t i o n AOQ5 = ppr5 *0 + ppa5 * d5 /100 printf ( ” \n P r o d u c e r s r i s k = %0 . 2 f p e r c e n t \n Consumers r i s k = %0 . 3 f p e r c e n t ” , alpha , bita ) x = [1 ,2 ,4 ,6 ,8] y = [0.91 ,1.4716 ,1.624 ,1.194 ,0.733] plot (x , y ) xtitle ( ”AOQ c u r v e ” ,” P e r c e n t d e c t i v e s ” , ”AOQ o f l o t ” )

113

Chapter 22 Kinematics of machine tools

Scilab code Exa 22.1 Find range of cutting velocity 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

clc d1 = 10 // min . d i a o f c u t t e r i n mm d2 = 60 // max . d i a o f c u t t e r i n mm v = 30 e3 // o p e r a t i n g s p e e d i n m/ min n1 = v / ( %pi * d2 ) // n min i n rpm n2 = v / ( %pi * d1 ) // n max i n rpm phi = ( n2 / n1 ) ^(1/5) spindle_speeds = zeros () for i =0:5 spindle_speeds ( i +1) = phi ^ i * n1 end cutter_dia = v ./ ( %pi * spindle_speeds ) clf () y = [0; v ] plot ([0; cutter_dia (1) ] , y , [0; cutter_dia (2) ] , y , [0; cutter_dia (3) ] , y , [0; cutter_dia (4) ] , y , [0; cutter_dia (5) ] , y , [0; cutter_dia (6) ] , y ) 16 xtitle ( ” ” ,” c u t t e r d i a m e t e r mm” ,” c u t t i n g v e l o c i t y , m/ min ” ) 17 // from g r a p h 18 vmax1 = 36 // m/ min 114

19 vmin1 = 24.5 // m/ min 20 r1 = vmax1 - vmin1 // Range o f 21 22 23 24

c u t t i n g s p e e d f o r 12 mm d i a m e t e r i n m/ min vmax2 = 36.5 // m/ min . vmin2 = 26 // m/ min . r2 = vmax2 - vmin2 // Range o f c u t t i n g s p e e d f o r 36 mm d i a m e t e r i n m/ min printf ( ” \n Range o f c u t t i n g s p e e d f o r 12 mm d i a m e t e r = %0 . 1 f m/ min . \ n Range o f c u t t i n g s p e e d f o r 36 mm d i a m e t e r = %0 . 1 f m/ min . ” , r1 , r2 )

Scilab code Exa 22.2 Determine speed ratios and teeth 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc m = 2.5 // module i n mm phi = 1.2 // common r a t i o n = 150 // s p e e d i n r e v / min . n1 = 70 // s p e e d i n r e v / min . n2 = ( phi ) ^1* n1 // s p e e d i n n3 = ( phi ) ^2* n1 // s p e e d i n n4 = ( phi ) ^3* n1 // s p e e d i n T1 = poly (0 , ’ T1 ’ ) t1 = n1 / n * T1 T1 = roots ( t1 + T1 -80) t1 = horner ( t1 , T1 ) T2 = poly (0 , ’ T2 ’ ) t2 = n2 / n * T2 T2 = roots ( t2 + T2 -80) t2 = horner ( t2 , T2 ) T3 = poly (0 , ’ T3 ’ ) t3 = n3 / n * T3 T3 = roots ( t3 + T3 -80) t3 = horner ( t3 , T3 ) T4 = poly (0 , ’ T4 ’ ) t4 = n4 / n * T4 115

of driving of driven r e v / min . o f r e v / min . o f r e v / min . o f

shaft shaft driven shaft driven shaft driven shaft

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

T4 = roots ( t4 + T4 -80) t4 = horner ( t4 , T4 ) t1 = floor ( t1 ) // number o f t e e t h on d r i v i n g s h a f t T1 = ceil ( T1 ) // number o f t e e t h on d r i v e n s h a f t t2 = ceil ( t2 ) // number o f t e e t h on d r i v i n g s h a f t T2 = floor ( T2 ) // number o f t e e t h on d r i v e n s h a f t t3 = floor ( t3 ) // number o f t e e t h on d r i v i n g s h a f t T3 = ceil ( T3 ) // number o f t e e t h on d r i v e n s h a f t t4 = ceil ( t4 ) // number o f t e e t h on d r i v i n g s h a f t T4 = floor ( T4 ) // number o f t e e t h on d r i v e n s h a f t // r u n n i n g s p e e d s n1 = n * t1 / T1 n2 = n * t2 / T2 n3 = n * t3 / T3 n4 = n * t4 / T4 printf ( ” \n Number o f t e e t h on d r i v e r and d r i v e n a r e :− \n t 1 = %d , T1 = %d\n t 2 = %d , T2 = %d \n t 3 = %d , T3 = %d \n t 4 = %d , T4 = %d ” ,t1 , T1 , t2 , T2 , t3 , T3 , t4 , T4 ) 39 printf ( ” \n The a c t u a l r u n n i n g s p e e d o f d r i v e n s h a f t w i l l be : \n n1 = %0 . 2 f r e v / min \n n2 = %0 . 2 f r e v / min \n n3 = %0 . 2 f r e v / min \n n4 = %0 . 2 f r e v / min ” , n1 , n2 , n3 , n4 ) 40 // Answer o f n3 i s g i v e n wrong i n book 41 // Answer v a r y due t o round o f f e r r o r

Scilab code Exa 22.3 Calculate speed and number of teeths 1 2 3 4 5 6 7

clc z = 6 // number o f s t e p s n1 = 180 // r e v / min n2 = 100 // r e v / min Rn = n1 / n2 phi = ( Rn ) ^(1/( z -1) ) // common r a t i o n3 = phi * n2 // r e v / min 116

n4 = ( phi ) ^2* n2 // r e v / min n5 = ( phi ) ^3* n2 // r e v / min n6 = ( phi ) ^4* n2 // r e v / min n7 = 225 // s p e e d o f i n p u t s h a f t i n r e v / min Ta = poly (0 , ’ Ta ’ ) tb = n7 / n5 * Ta Ta = roots ( tb + Ta -52) tb = horner ( tb , Ta ) tb = ceil ( tb ) Tc = poly (0 , ’ Tc ’ ) td = n7 / n6 * Tc Tc = roots ( td + Tc -52) td = horner ( td , Tc ) Tc = ceil ( Tc ) Te = poly (0 , ’ Te ’ ) tf = n7 / n1 * Te Te = roots ( tf + Te -52) tf = horner ( tf , Te ) tf = ceil ( tf ) Th = poly (0 , ’ Th ’ ) tj = n2 / n5 * Th Th = roots ( tj + Th -46) Th = ceil ( Th ) tj = horner ( tj , Th ) tj = floor ( tj ) Ti = poly (0 , ’ Ti ’ ) tg = n5 / n5 * Ti Ti = roots ( tg + Ti -46) tg = horner ( tg , Ti ) printf ( ” \n Ta = %d Tb = %d \n Tc = %d Td = %d \n Te = %d t f = %d \n Th = %d Tj = %d \n Ti = %d Tg = %d” , Ta , tb , Tc , td , Te , tf , tj , Th , Ti , tg ) 38 // ’ Answers v a r y due t o round o f f e r r o r ’

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

Scilab code Exa 22.4 Calculate common ratio 117

1 clc 2 v = 21 // c u t t i n g s p e e d i n r e v / min . 3 z = 6 4 dmin = 5 // d a i m e t e r i n mm 5 dmax = 20 // d a i m e t e r i n mm 6 nmax = 1000* v /( %pi * dmin ) // s p i n d l e s p e e d i n r e v / min 7 8 9 10 11 12 13 14 15

16

. nmin = 1000* v /( %pi * dmax ) // s p i n d l e s p e e d i n r e v / min . phi = ( nmax / nmin ) ^(1/( z -1) ) // common r a t i o n1 = nmin // r e v / min . n2 = phi * n1 // r e v / min . n3 = ( phi ) ^2* n1 // r e v / min . n4 = ( phi ) ^3* n1 // r e v / min . n5 = ( phi ) ^4* n1 // r e v / min . n6 = ( phi ) ^5* n1 // r e v / min . printf ( ” \n Common r a t i o = %0 . 2 f \n S p i n d l e s p e e d s a r e : %0 . 2 f , %0 . 1 f , %0 . 2 f , %0 . 2 f ,%0 . 2 f and %0 . 1 f r e v / min . ” ,phi , n1 , n2 , n3 , n4 , n5 , n6 ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 22.5 Calculate gear ratio teeth and speed 1 // from f i g . 2 2 . 1 8A 2 clc 3 // Three g e a r r a t i o s b e t w e e n i n p u t and i n t e r m e d i a t e 4 5 6 7 8 9 10

shaft nmax = 1400 // maximum s p e e d i n r e v / min . i1 = 1/1 i2 = 1/1.26 i3 = 1/(1.26) ^2 // The two r a t i o s b e t w e e n i n t e r m e d i a t e and o u t p u t shaft i4 = 1/1 i5 = 1/(1.26) ^3 118

11 // number o f t e e t h f o r i n p u t and i n t e r m e d i a t e s h a f t 12 t1 = 27/27 13 t2 = 24/30 14 t3 = 21/33 15 // number o f t e e t h f o r o u t p u t and i n t e r m e d i a t e 16 17 18 19 20 21 22 23 24 25

26

shaft t4 = 34/34 t5 = 20/48 // o u t p u t s p e e d s i n r e v . / min n1 = t3 * t5 * nmax n2 = t2 * t5 * nmax n3 = t1 * t5 * nmax n4 = t3 * t4 * nmax n5 = t2 * t4 * nmax n6 = t1 * t4 * nmax printf ( ” \n Three g e a r r a t i o s b e t w e e n i n p u t and i n t e r m e d i a t e s h a f t i 1 = %d i 2 = %0 . 2 f i 3 = %0 . 3 f \n The two r a t i o s b e t w e e n i n t e r m e d i a t e and o u t p u t s h a f t i 4 = %d i 5 = %0 . 3 f \n number o f t e e t h f o r e a c h p a i r b e t w e e n i n p u t and i n t e r m e d i a t e s h a f t t1 = 27/27 , t2 = 24/30 , t3 = 2 1 / 3 3 \n number o f t e e t h f o r e a c h p a i r b e t w e e n o u t p u t and i n t e r m e d i a t e s h a f t = t 4 = 3 4 / 3 4 , t 5 = 2 0 / 4 8 \n Output s p e e d s \n n1 = %d r e v / min , n2 = %d r e v / min , n3 = %d r e v / min \n n4 = %d r e v / min , n5 = %d r e v / min , n6 = %d r e v / min ” , i1 , i2 , i3 , i4 , i5 , n1 , n2 , n3 , n4 , n5 , n6 ) // Answer v a r y due t o round o f f e r r o r

119

Chapter 23 Production planning and control

Scilab code Exa 23.1 Calculate forecast 1 2 3 4 5 6

clc d1 = 90 // demand f o r f i r s t q u a r t e r d2 = 100 // demand f o r s e c o n d q u a r t e r d3 = 80 // demand f o r t h i r d q u a r t e r sa = ( d1 + d2 + d3 ) /3 // s i m p l e a v e r a g e printf ( ” \n F o r e c a s t = %d” , sa )

Scilab code Exa 23.2 Calculate forecat by SMA method 1 2 3 4 5 6 7

clc d1 = d2 = d3 = d4 = d5 = d6 =

300 350 400 500 600 700

// // // // // //

demand demand demand demand demand demand

for for for for for for

july august september october november december 120

8 9 10

// a s s u m i n g n = 3 , where n i s number o f t i m e p e r i o d forecast = ( d6 + d5 + d4 ) /3 // f o r e c a s t printf ( ” \n F o r e c a s t = %d” , forecast )

Scilab code Exa 23.3 Calculate forecat by WMA method 1 2 3 4 5 6 7 8 9

clc d1 = 500 // demand f o r o c t o b e r d2 = 600 // demand f o r november d3 = 700 // demand f o r d e c e m b e r w1 = 0.25 // r e l a t i v e w e i g h t w i t h d e c e m b e r w2 = 0.25 // r e l a t i v e w e i g h t w i t h november w3 = 0.5 // r e l a t i v e w e i g h t w i t h o c t o b e r f = w1 * d1 + w2 * d2 + w3 * d3 // f o r e c a s t printf ( ” \n F o r e c a s t by w e i g h t e d moving a v e r a g e = %d” , f)

Scilab code Exa 23.4 Calculate forecast for january 1 2 3 4 5 6

clc alpha = 0.7 // s m o o t h i n g c o e f f i c i e n t d1 = 250 // demand f o r november d2 = 300 // demand f o r d e c e m b e r f1 = 200 // f o r e c a s t f o r november f2 = alpha * d1 + (1 - alpha ) * f1 // f o r e c a s t f o r december 7 f3 = alpha * d2 + (1 - alpha ) * f2 // f o r e c a s t f o r j a n u a r y 8 f3 = ceil ( f3 ) 9 printf ( ” \n F o r e c a s t f o r j a n u a r y = %d u n i t s ” , f3 )

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Scilab code Exa 23.5 Calculate total cost 1 2 3 4 5 6 7 8 9 10 11

clc s = c = a = i =

600 // s e t up c o s t p e r l o t i n Rs 6 // u n i t c o s t o f i t e m i n Rs 100000 // a n n u a l demand f o r i t e m 25 // a n n u a l c a r r y i n g c h a r g e s o f a v e r a g e inventory i = 25/100 k = c * i // c a r r y i n g c o s t f a c t o r i n u n i t / y e a r n = sqrt (2* s * a / k ) // most e c o n o m i c l o t s i z e tc = a * c + s * a / n + k * n /2 // t o t a l c o s t i n Rs printf ( ” \n T o t a l c o s t = Rs %0 . 2 f ” , tc ) // ’ Answers v a r y due t o round o f f e r r o r ’

Scilab code Exa 23.6 Calculate economical order quantity 1 2 3 4 5 6 7 8 9

clc a = c = r = i =

8000 // a n n u a l r e q u i r e m e n t o f p a r t s 60 // u n i t c o s t o f p a r t i n Rs 150 // o r d e r i n g c o s t p e r l o t i n Rs 30 // a n n u a l c a r r y i n g c h a r g e s o f a v e r a g e inventory i = 30/100 k = i * c // c a r r y i n g c o s t p e r u n i t p e r y e a r n = sqrt (2* r * a / k ) // most e c o n o m i c a l o r d e r q u a n t i t y printf ( ” \n Most e c o n o m i c a l o r d e r i n g q u a n t i t y = %d u n i t s ” , n)

Scilab code Exa 23.7 Calculate economic lot size 1 clc 2 a = 12000 // a n n u a l r e q u i r e m e n t

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3 4 5 6 7 8 9 10

c = 5 // u n i t c o s t o f p a r t s = 60 // s e t up c o s t p e r l o t p = 18750 // p r o d u c t i o n r a t e p e r y e a r i = 20 // i n v e n t o r y c a r r y i n g c o s t i = 20/100 k = i * c // c a r r y i n g c o s t p e r u n i t p e r y e a r n = sqrt (2* s /(1/ a -1/ p ) * k ) // Most e c o n o m i c l o t s i z e printf ( ” \n Most e c o n o m i c l o t s i z e = %d p a r t s ” , n )

Scilab code Exa 23.8 Calculate inventory control terms 1 2 3 4 5 6 7 8 9 10

clc a = 15625 // a n n u a l r e q u i r e m e n t o f p a r t s c = 12 // u n i t c o s t o f p a r t i n Rs r = 60 // o r d e r i n g c o s t p e r l o t i n Rs k = 1.2 // i n v e n t o r y c a r r y i n g c o s t p e r u n i t n = sqrt (2* r * a / k ) // e c o n o m i c a l o r d e r q u a n t i t y oc = r * a / n // o r d e r i n g c o s t i n Rs cc = k * n /2 // c a r r y i n g c o s t i n Rs tc = oc + cc // t o t a l i n v e n t o r y c o s t i n Rs printf ( ” \n E c o n o m i c a l o r d e r q u a n t i t y = %d u n i t s \n o r d e r c o s t = Rs %d\n c a r r y i n g c o s t = Rs %d\n T o t a l i n v e n t o r y c o s t = Rs %d” , n , oc , cc , tc )

Scilab code Exa 23.9 Calculate discount offered 1 2 3 4 5 6 7

clc // c a s e a a = 50 // a n n u a l r e q u i r e m e n t c = 500 // u n i t c o s t o f p a r t r = 100 // o r d e r i n g c o s t p e r i = 20 // i n v e n t o r y c a r r y i n g i = i /100 123

of parts in tonnes i n Rs o r d e r i n Rs cost

d = 2 // d i s c o u n t o f p u r c h a s e c o s t i n p e r c e n t k = i * c // i n v e n t o r y c a r r y i n g c o s t p e r u n i t n1 = sqrt (2* r * a / k ) // e c o n o m i c a l o r d e r q u a n t i t y oc1 = r * a / n1 // o r d e r i n g c o s t i n Rs cc1 = k * n1 /2 // c a r r y i n g c o s t i n Rs tc1 = oc1 + cc1 // t o t a l i n v e n t o r y c o s t i n Rs // c a s e b n2 = 25 // o r d e r p e r l o t oc2 = r * a / n2 // o r d e r i n g c o s t i n Rs cc2 = k * n2 /2 // c a r r y i n g c o s t i n Rs tc2 = oc2 + cc2 // t o t a l i n v e n t o r y c o s t i n Rs i = tc2 - tc1 // i n c r e a s e i n c o s t i n Rs d_o = d * c * a /100 // d i s c o u n t o f f e r e d printf ( ” \n I n c r e a s e i n i n v e n t o r y c o s t = Rs %d\n D i s c o u n t o f f e r e d = Rs%d” ,i , d_o ) 22 disp ( ” o f f e r i s worth a c c e p t i n g ” )

8 9 10 11 12 13 14 15 16 17 18 19 20 21

Scilab code Exa 23.10 Calculate EOQ and reorder point 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc a = 1000000 // a n n u a l r e q u i r e m e n t o f p a r t s r = 32 // o r d e r i n g c o s t p e r l o t i n Rs k = 4 // i n v e n t o r y c a r r y i n g c o s t p e r u n i t d1 = 250 // number o f w o r k i n g d a y s d2 = 2 // d a y s f o r s a f e t y s t o c k d3 = 4 // l e a d t i m e i n d a y s eoq = sqrt (2* r * a / k ) // e c o n o m i c a l o r d e r q u a n t i t y oc = r * a / eoq // o r d e r i n g c o s t i n Rs cc = k * eoq /2 // c a r r y i n g c o s t i n Rs tc = oc + cc // t o t a l i n v e n t o r y c o s t i n Rs ss = a * d2 / d1 // s a f e t y s t o c k ro_p = ss + eoq * d3 // r e o r d e r p o i n t printf ( ” \n Economic o r d e r q u n a n t i t y = %d c o m po n e n t s \ n Re−o r d e r p o i n t = %d c o m p o n e n t s ” , eoq , ro_p )

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Chapter 26 Plant layout

Scilab code Exa 26.1 Calculate number of machine required 1 2 3 4 5 6 7 8 9 10 11

clc N = 100000 // a n n u a l o u t p u t o f p a r t s s = 2 // e x p e c t e d s c r a p t = 105 // e s t i m a t e d t i m e p e r p a r t i n s ita = 80 // p r o d u c t i o n e f f i c i e n c y o f machine a = 2300 // number o f w o r k i n g h o u r s output = (3600* ita ) /( t *100) // p a r t s r e q u i r e d p e r hour pr = N *(100+ s ) /( a *100) // o u t p u t from one machine per hour mr = pr / output // m a c h i n e s r e q u i r e d printf ( ” \n Number o f m a c h i n e s r e q u i r e d = %0 . 2 f ” , mr ) disp ( ” I f machine i s t o be u s e d e x c l u s i v e l y f o r p a r t c o n s i d e r e d two m a c h i n e s r e q u i r e d ” )

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