A Textbook of Production Engineering_P. C. Sharma

7/25/2019 A Textbook of Production Engineering_P. C. Sharma

http://slidepdf.com/reader/full/a-textbook-of-production-engineeringp-c-sharma 1/126

Scilab Textbook Companion for

A Textbook of Production Engineering

by P. C. Sharma1

Created byMayank Sahu

B.E.Mechanical EngineeringM. I. T. S Gwalior, M.P

College Teacher

NoneCross-Checked byBhavani Jalkrish

November 3, 2014

1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilabcodes 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

1



Scilab numbering policy used in this document and the relation to theabove 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 particularExample of the above book)

For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 meansa scilab code whose theory is explained in Section 2.3 of the book.

2



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

3



List of Scilab Codes

Exa 2.1 find total pressure and dimensions . . . . . . . . . . . 9Exa 2.2 To find number of draws . . . . . . . . . . . . . . . . . 10Exa 2.3 To calculate bending force . . . . . . . . . . . . . . . 10Exa 2.4 find blanking force and work done . . . . . . . . . . . 10Exa 2.5 To find elastic recovery of material . . . . . . . . . . . 11Exa 2.6 To find cutting forces . . . . . . . . . . . . . . . . . . 11Exa 2.7 To calculate amount of shear . . . . . . . . . . . . . . 12Exa 2.8 To find economy of material . . . . . . . . . . . . . . . 12Exa 2.9 Calculations for designing drawing die . . . . . . . . . 13Exa 2.10 Determine developed length . . . . . . . . . . . . . . . 15Exa 2.11 To calculate bending force . . . . . . . . . . . . . . . . 15Exa 2.12 To calculate bending force . . . . . . . . . . . . . . . . 16Exa 2.13 calculate capacity of double bending die . . . . . . . . 16

Exa 2.14 To calculate cutting force . . . . . . . . . . . . . . . . 17Exa 2.15 Determine blank and punch diameter . . . . . . . . . 17Exa 2.16 To find drawing operations and force . . . . . . . . . . 18Exa 2.17 Determine developed length . . . . . . . . . . . . . . . 19Exa 4.1 To calculate total cost and SP . . . . . . . . . . . . . 21Exa 4.2 To find selling price . . . . . . . . . . . . . . . . . . . 21Exa 4.3 To find factory cost . . . . . . . . . . . . . . . . . . . 22Exa 4.4 find production cost and time taken . . . . . . . . . . 23Exa 4.5 To find profit . . . . . . . . . . . . . . . . . . . . . . . 23Exa 4.6 To find lot size and time . . . . . . . . . . . . . . . . . 24Exa 4.7 To find time to change cutter . . . . . . . . . . . . . . 24

Exa 4.8 To find tool change time . . . . . . . . . . . . . . . . . 25Exa 4.9 To calculate measuring time allowance . . . . . . . . . 25Exa 4.10 To find direct labour cost . . . . . . . . . . . . . . . . 25Exa 4.11 To find machining time . . . . . . . . . . . . . . . . . 26

4



Exa 4.12 To find time to turn relief . . . . . . . . . . . . . . . . 27

Exa 4.13 calculate time to face on lathe . . . . . . . . . . . . . 27Exa 4.14 To find time to drill hole . . . . . . . . . . . . . . . . 28Exa 4.15 To find time to complete cut . . . . . . . . . . . . . . 28Exa 4.16 To find time to broach . . . . . . . . . . . . . . . . . . 29Exa 4.17 find feed cutter travel and time . . . . . . . . . . . . 29Exa 4.18 To find cutting time . . . . . . . . . . . . . . . . . . . 30Exa 4.19 To find milling time . . . . . . . . . . . . . . . . . . . 30Exa 4.20 To find time to grind shaft . . . . . . . . . . . . . . . 31Exa 4.21 To find time to cut threads . . . . . . . . . . . . . . . 31Exa 4.22 find time to produce one piece . . . . . . . . . . . . . 32Exa 5.1 To find value of machine tool . . . . . . . . . . . . . . 34

Exa 5.2 To find annual investment . . . . . . . . . . . . . . . . 35Exa 5.3 find project is economical or not . . . . . . . . . . . . 35Exa 5.4 selection of economical machine . . . . . . . . . . . . . 36Exa 5.5 selection of machine . . . . . . . . . . . . . . . . . . . 37Exa 5.6 selection of economical machine . . . . . . . . . . . . . 38Exa 5.7 find ERR and economicality of project . . . . . . . . . 38Exa 5.9 find ERR and economicality of project . . . . . . . . 39Exa 5.10 To determine acceptance of machine . . . . . . . . . . 39Exa 5.11 find investment cost and unamortized value . . . . . . 40Exa 5.13 To make decision of machines replacement . . . . . . . 40

Exa 5.15 Determine economic repair life . . . . . . . . . . . . . 41Exa 5.16 find time to pay for itself . . . . . . . . . . . . . . . . 42Exa 5.17 selection of machine for job . . . . . . . . . . . . . . . 43Exa 5.18 Calculate maximum investment on turret lathe . . . . 44Exa 5.19 To find years for new machine . . . . . . . . . . . . . . 45Exa 5.20 To find cost and pieces . . . . . . . . . . . . . . . . . . 46Exa 5.21 To find number of components . . . . . . . . . . . . . 47Exa 5.22 To find number of components . . . . . . . . . . . . . 47Exa 5.23 To find time and profit . . . . . . . . . . . . . . . . . 48Exa 5.24 To find minimum number of components . . . . . . . . 48Exa 5.25 To calculate number of pieces . . . . . . . . . . . . . . 49

Exa 5.26 To find cost for new fixture . . . . . . . . . . . . . . . 49Exa 5.27 find time to amortize fixture . . . . . . . . . . . . . . 50Exa 5.28 To find profit . . . . . . . . . . . . . . . . . . . . . . . 51Exa 5.29 To find BEP Cost and Components . . . . . . . . . . 51Exa 5.30 To find break even point . . . . . . . . . . . . . . . . . 52

5



Exa 5.31 To find break even quantity . . . . . . . . . . . . . . . 52

Exa 5.32 To do break even analysis . . . . . . . . . . . . . . . . 54Exa 5.33 To calculate minimum number of pieces . . . . . . . . 54Exa 5.34 To determine the point . . . . . . . . . . . . . . . . . 55Exa 5.35 To find quantity of pieces . . . . . . . . . . . . . . . . 55Exa 5.36 To determine quantity of production . . . . . . . . . . 56Exa 5.37 find preference between machines and production . . . 56Exa 5.38 To find BEP and various sales . . . . . . . . . . . . . 57Exa 5.39 To determine break even point . . . . . . . . . . . . . 58Exa 5.40 To calculate economic lot size . . . . . . . . . . . . . . 58Exa 5.41 To find EOQ and total cost . . . . . . . . . . . . . . . 59Exa 5.42 Determine optimum lot size . . . . . . . . . . . . . . . 60

Exa 5.43 To find most economical lot size . . . . . . . . . . . . 61Exa 9.1 To find allowance and tolerence . . . . . . . . . . . . . 62Exa 9.2 Determine dimensions of shaft and hole . . . . . . . . 62Exa 9.3 Determine dimensions of hole and shaft . . . . . . . . 63Exa 9.4 Calculate fundamental deviations and tolerences . . . 63Exa 9.5 Find tolerences limits and clearance . . . . . . . . . . 64Exa 9.6 Determine limits of shaft and hole . . . . . . . . . . . 65Exa 9.7 Determine dimensions of shaft and hole . . . . . . . . 66Exa 9.8 Determine size of bearing and journal . . . . . . . . . 66Exa 9.9 Determine size of two mating parts . . . . . . . . . . . 67

Exa 9.10 Determine size of hole and shaft . . . . . . . . . . . . 68Exa 11.1 Calculate CLA value . . . . . . . . . . . . . . . . . . . 69Exa 11.2 Calculate average and rms value . . . . . . . . . . . . 69Exa 13.1 To find drawing load . . . . . . . . . . . . . . . . . . . 71Exa 13.2 Calculate drawing force . . . . . . . . . . . . . . . . . 72Exa 13.3 find neutral section slips and pressure . . . . . . . . . 72Exa 13.4 To determine maximum force . . . . . . . . . . . . . . 73Exa 13.5 Determine sticking radius and total load . . . . . . . . 74Exa 13.7 To find drawing load and power . . . . . . . . . . . . . 74Exa 13.8 calculate drawing load and power rating . . . . . . . . 75Exa 13.9 To calculate forging loads . . . . . . . . . . . . . . . . 76

Exa 13.10 Determine extrusion load . . . . . . . . . . . . . . . . 76Exa 13.11 To find roll pressures . . . . . . . . . . . . . . . . . . . 77Exa 13.12 Determine neutral plane . . . . . . . . . . . . . . . . . 78Exa 14.1 calculate the tool life . . . . . . . . . . . . . . . . . . . 80Exa 14.2 Calculate the optimum cutting speed . . . . . . . . . . 80

6



Exa 14.3 To find different orthogonal cutting picture . . . . . . 81

Exa 14.4 To find tool life . . . . . . . . . . . . . . . . . . . . . . 82Exa 14.5 find force and coefficient of friction . . . . . . . . . . 83Exa 14.6 To find terms of orthogonal cutting . . . . . . . . . . . 84Exa 14.7 To solve tool life equation . . . . . . . . . . . . . . . 85Exa 14.8 Determine normal and tangential force . . . . . . . . . 85Exa 14.9 To find cutting and thrust force . . . . . . . . . . . . . 86Exa 14.10 find terms of orthogonal rake system . . . . . . . . . . 87Exa 14.11 Calculate CLA . . . . . . . . . . . . . . . . . . . . . . 88Exa 14.12 Calculate back and side rake angle . . . . . . . . . . . 88Exa 14.13 Calculate inclination and rake angle . . . . . . . . . . 89Exa 14.14 find different powers and resistance . . . . . . . . . . . 89

Exa 14.15 Calculate percentage increase in tool life . . . . . . . . 90Exa 14.16 To find percentage of total energy . . . . . . . . . . . 90Exa 14.17 To find power and different energies . . . . . . . . . . 91Exa 14.18 Determine components of force and power . . . . . . . 92Exa 15.1 calculate horsepower at cutter and motor . . . . . . . 94Exa 15.2 Determine broaching power and Design broach . . . . 94Exa 15.3 Estimate moment thrust force and power . . . . . . . 96Exa 15.4 Design shell inserted blade reamer . . . . . . . . . . . 96Exa 15.5 To design single point cutting tool . . . . . . . . . . . 97Exa 15.8 find various terms for stainless steel . . . . . . . . . . 98

Exa 15.9 To find MRR power and torque . . . . . . . . . . . . . 99Exa 15.10 find MRR power torque and time . . . . . . . . . . . 99Exa 16.1 Calculate settings of gear tooth . . . . . . . . . . . . . 101Exa 17.1 Calculate best wire size . . . . . . . . . . . . . . . . . 102Exa 17.2 Calculate size and distances over wire . . . . . . . . . 102Exa 21.1 Construct R and X chart . . . . . . . . . . . . . . . . 103Exa 21.2 Construct the control charts . . . . . . . . . . . . . . . 104Exa 21.4 Calculate poisson probabilities . . . . . . . . . . . . . 106Exa 21.5 Calculate probabilities of defective items . . . . . . . 106Exa 21.6 Determine producers and consumers risk . . . . . . . . 107Exa 21.7 Evaluate preliminary and revised control limits . . . . 107

Exa 21.8 Find control limits for c chart . . . . . . . . . . . . . 109Exa 21.9 find control limits for charts . . . . . . . . . . . . . . . 110Exa 21.10 Determine producers and consumers risk . . . . . . . . 112Exa 22.1 Find range of cutting velocity . . . . . . . . . . . . . . 114Exa 22.2 Determine speed ratios and teeth . . . . . . . . . . . . 115

7



Exa 22.3 Calculate speed and number of teeths . . . . . . . . . 116

Exa 22.4 Calculate common ratio . . . . . . . . . . . . . . . . . 117Exa 22.5 Calculate gear ratio teeth and speed . . . . . . . . . . 118Exa 23.1 Calculate forecast . . . . . . . . . . . . . . . . . . . . 120Exa 23.2 Calculate forecat by SMA method . . . . . . . . . . . 120Exa 23.3 Calculate forecat by WMA method . . . . . . . . . . 121Exa 23.4 Calculate forecast for january . . . . . . . . . . . . . . 121Exa 23.5 Calculate total cost . . . . . . . . . . . . . . . . . . . 122Exa 23.6 Calculate economical order quantity . . . . . . . . . . 122Exa 23.7 Calculate economic lot size . . . . . . . . . . . . . . . 122Exa 23.8 Calculate inventory control terms . . . . . . . . . . . . 123Exa 23.9 Calculate discount offered . . . . . . . . . . . . . . . . 123

Exa 23.10 Calculate EOQ and reorder point . . . . . . . . . . . . 124Exa 26.1 Calculate number of machine required . . . . . . . . . 125

8



Chapter 2

Press Tool Design

Scilab code Exa 2.1 find total pressure and dimensions

1 clc

2 D = 5 0 // D ia me te r o f w as he r i n mm3 t = 4 // t h i c k n e s s o f m a t e r i a l i n mm4 d = 2 4 // d ia me te r o f h o l e i n mm5 p = 360 // s h e ar s t r e n g t h o f m a t e r i a l i n N/mmˆ26 F 1 = % pi * D* t *p // b l a nk in g p r e s s u r e i n N

7 F 2 = % pi * d* t *p // p i e r c i n g p r e s s u r e i n N8 F = F 1 + F 2 // t o t a l p r e s s u r e i n N9 d1 = d + 0.4 // p i e r c i n g d i e d ia me te r i n mm

10 d2 = D - 0.4 / / b l an k punch d i a m et e r i n mm11 c = 0.8* F // p r e s s c ap a c i t y i n N12 printf ( ” \n B l an k in 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 ta l p r e s s u re r e q u i r e d =%0 . 1 f KN” , F 1 / 1 0 0 0 , F 2 / 1 0 0 0 , F / 1 0 0 0 )

13 printf ( ” \n p i e r c i n g punch d i a me t er = %0 . 2 f cm\nb l a n k i n g p un ch 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 / 10 00 )14 / / A nswers v ar y d ue t o ro und o f f e r r o r

9



Scilab code Exa 2.2 To find number of draws

1 clc

2 h = 1 0 // h e i g h t o f cup i n cm3 d = 5 // d ia me te r o f cup i n cm4 D = sqrt ( d ^2 + 4 *d * h) // b la nk d i am e te r i n cm5 // h e i g h t d ai me te r r a t i o i s 2 . T he re f o r e from t a b l e

2 . 9 t o t a l r e d u c t i o n s a re 36 r 1 = 0 .4 5* D // f i r s t r e d u c t io n i s 45%7 d1 = D - r 1 // d i a me t e r a t f i r s t draw i n cm8 r 2 = d 1 *0 .2 5 // s ec on d r e du c t io n i s 25%9 d2 = d1 - r 2 // d i am e te r a t s ec on d draw i n cm

10 r3 = d2 * 0.2 // t h i r d r e d u ct i o n i s 20%11 d3 = d2 - r 3 // d i a me te r a t t h i r d draw i n cm12 printf ( ” \n D i am e te r a t f i r s t draw = %0 . 2 f cm\n

D ia me te r a t s e co n d draw = %0 . 2 f cm\n D ia me te r a tt h i r d d ra w =%0 . 3 f cm” , d1 , d2 , d3 )

Scilab code Exa 2.3 To calculate bending force

1 clc2 K = 1.20 // d ie −o pe ni ng f a c t o r3 L = 37.5 // Length o f s t r i p i n cm4 T = 2.5 // t h i c k n e s s o f s t r i p i n mm5 s i g ma _ ut = 6 30 // t e n s i l e s t r e n g t h i n N/mmˆ26 W = 16* T // wi dt h o f d i e o pe ni ng i n mm7 F = ( K * L *1 0* s i g ma _ ut * T ^ 2) / W // b en d i ng f o r c e i n N8 printf ( ” \n b e n di n g f o r c e = %0 . 1 f KN” , F / 1 00 0)

Scilab code Exa 2.4 find blanking force and work done

1 clc

2 b = 2 5 // w id th o f b la nk i n mm

10



3 l = 3 0 // l e ng t h o f b la nk i n mm

4 t au = 450 // u l t i m a t e s he ar s t r e s s o f m a t e r i a l i n N/mmˆ25 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 mm6 p = 2*( l + b ) // p e r i me t e r o f b la nk i n mm7 f = p *t * tau // b la nk i ng f o r c e i n N8 P t = 0 .2 5* t // punch t r a v e l i n mm9 w = f * Pt / / w or k d o ne i n Nmm

10 printf ( ” \n b l an k i ng f o r c e = %0 . 2 f KN\n work d one =%0 . 2 f Nm” , f / 1 00 0 , w / 1 0 0 0 )

Scilab code Exa 2.5 To find elastic recovery of material

1 clc

2 t = 1.5 // t h i c k n e s s i n mm3 c = 0. 05 * t // c l e a r a n c e i n mm4 D = 25.4 // o u t s i d e d i am et e r i n mm5 D_o = D - 0.05 // b la nk d i e o pe ni ng i n mm6 B_s = D_o - 2* c // b l a nk i n g punch s i z e i n mm7 d = 12.7 // i n t e r n a l d i a me te r i n mm

8 P_s = d + 0.05 // p i e r c i n g punch s i z e i n mm9 D_s = P_s + 2* c // p i e r c i n g d i e s i z e i n mm

10 printf ( ” \n c l e a r a n c e = %0 . 3 f mm\n b la nk d i e o pe ni ngs 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 gp un ch 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 am et e r i n mm3 d = 12.7 // i n t e r n a l d i a me te r i n mm

11



4 t = 1.5 // t h i c k n e s s i n mm

5 t au = 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/mmˆ 26 F = %pi *( D + d )* t* tau // t o t a l c ut t i n g f o r c e i n N7 F _ s = % pi * D * t* t au // c u t t i n g f o r c e when p un ch es a r e

s t a g ge r e d i n N8 k = 0.6 // p e n e t r a t i o n9 i = 1 // s h ea r o f punch i n mm

10 F _p = ( t *k *F ) /( k *t + i) // c u t t i n g f o r c e when b o thp un ch es a ct t o g e t h e r i n N

11 printf ( ” \n s h e ar f o r c e when b ot h punch a c t a t samet im e and no s h ea r i s a p p l i e d = %0 . 2 f kN” , F

/1000)

12 printf ( ” \n c u t t i n g f o r c e when p unc he s a r e s t a gg e r ed= %0. 1 f kN ”, F _ s / 1 00 0 )

13 printf ( ” \n c u t t i ng f o r c e when t h e r e i s p e ne t ra t i o na nd s h e a r on p un ch = %0 . 1 f kN” , F _ p / 1 0 0 0 )

Scilab code Exa 2.7 To calculate amount of shear

1 clc

2 D = 6 0 // h o l e d i am et er i n mm3 t au = 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/mmˆ24 t = 2.5 // t h i c k n e s s i n mm5 F = % pi * D *t * ta u / / maximum p un ch f o r c e i n N6 w = F *0 .4 *t / / wo rk d on e i n Nm7 f = F /2 // p un ch in g f o r c e i n N8 k = 0.4 // p e n e t r a t i o n p e r ce n t a ge9 i = k *t *( F -f ) /f / / s h e a r on punch i n mm

10 printf ( ” \n Amount o f s h e a r o n p un ch = %d mm” , i )

Scilab code Exa 2.8 To find economy of material

1 clc

12



2 // f rom f i g 2 . 2 7

3 w = 2.5 / / cm4 t = 3.2 // s t r i p t h i c k n e ss i n mm5 h = 1 0 // cm6 a = t + 0 .015* h *10 // back s c r a p and f r o n t s c ra p i n

mm7 b = t // s c ra p b r i d ge i n mm8 W = h + ( 2 * a ) / 1 0 // w i dt h o f s t r i p i n cm9 W = ceil ( W ) // cm

10 s = w + b /10 // l en g t h o f one p i e c e o f s to c k i n cm11 L = 240 // l en g t h o f s t r i p i n cm12 N = (L - b) /s // number o f p a r t s

13 y = L - ( N * s + b /10) // s c r a p r em ai ni ng a t t he endi n mm

14 printf ( ” \n V al ue o f b ac k s c r a p = %0 . 1 f mm\n V al ue o f s c r a p b r i d g e = %0 . 1 f mm ” , a , b )

15 printf ( ” \n Width o f s t r i p = %0 . 2 f cm\n L en gt h o f on ep i e c e o f s t oc k n ee d ed t o p ro du ce one p ar t = %0 . 2

f cm” , W , s )

16 printf ( ” \n Number o f p a r t s = %0 . 1 f b l a n k s \n S cr apr e m a i n i n g a t t h e e nd = %0 . 2 f mm” , N , y )

17 / / A nswers v ar y d ue t o ro und o f f e r r o r

Scilab code Exa 2.9 Calculations for designing drawing die

1 clc

2 // from f i g u r e 2 . 7 33 t = 0.8 // t h i c k n e s s i n mm4 d = 5 0 // s h e l l d ia me te r i n mm5 r = 1.6 // r a d i u s o f bottom c o r n er i n mm

6 h = 5 0 // h e i g ht i n mm7 D = sqrt ( d ^2 + 4 *d * h) // s h e l l b l an k s i z e i n mm8 e l = 6.4 // e xt ra l en gt h r e q u i r e d t o add i n s h e l l

b la nk s i z e9 D = D + el // mm

13



10 p r = 1 00 *( 1 - ( d / D) ) // p e r ce n t ag e r e d u c t io n

11 r at io = h /d12 n = 2 / / number o f d ra ws13 R 1 = 45 / / f i r s t r e d u c t i o n14 D1 = D - R1 *D /100 // d i am e te r a t f i r s t r e d u c ti o n i n

mm15 R 2 = 1 00 *( 1 - ( d / D1 ) ) // s ec on d r e d u c t io n16 P R = 4* t // punch r a d i u s i n mm17 PR = ceil ( P R )

18 D R = 6 // d i e r a d i u s i n mm19 DC1 = 0 .8 7 // d i e c l e a r a n c e f o r f i r s t draw i n mm20 DC2 = 0 .8 8 // d i e c l e a r a n c e f o r s ec on d draw i n mm

21 PD2 = d - 2* t // punch d i am e te r f o r s ec on d draw i nmm

22 DD2 = PD2 + 2* DC2 // Di e o pe ni ng d i am e te r f o r s ec on ddraw i n mm

23 PD1 = D1 - 2* t / / punch d i a me t er f o r f i r s t draw i nmm

24 DD1 = D1 + 2* DC1 // D ie o p en i ng d i a me t er f o r f i r s tdraw i n mm

25 / / D ra win g p r e s s u r e26 c = 0.65 // c o n st a n t27 s i gm a = 4 27

/ / N/mmˆ228 F = % pi * d * t* s i gm a * ( D/ d - c) / / Dra wi ng p r e s s u r e i n mm29 Bhp = 3 0. 8 // b l an k i n g h o ld i ng p r e s s ur e i n kN30 p c = 150 // p r e s s c a pa c i t y i n kN31 printf ( ” \n ( i ) s i z e o f b la nk = %0 . 2 f mm \n ( i i )

P e r ce n t ag 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 d r a ws = %d \n ( i v ) R a d iu s o n p un ch = %d

mm \n Die R adi us = %d mm \n ( v ) Di e c l e a r a nc ef o r f i r s t d ra w = %0 . 2 f mm \n d i e c l e a r a n c e

f o r s e c o n d d ra w = %0 . 2 f mm” , D , pr , n , PR , DR , DC1 ,

D C 2 )

32 printf ( ” \n Punch d i am et er f o r s ec on d draw = %0 . 1 f mm \n Di e o pe ni ng d ia me te r f o r s ec on d draw = %0. 2 f mm \n Punch d ia me te r f o r f i r s t draw = %0 . 3f mm \n Di e o pe n in g d ia me te r f o r f i r s t draw =%0 . 3 f mm\n ( v i ) D ra wi ng p r e s s u r e = %0 . 2 f mm \n (

14



v i i ) Bl ank h o l d in 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/ 100 0 , Bhp , pc )


Scilab code Exa 2.10 Determine developed length

1 clc

2 // from f i g u r e 2 . 7 4

3 l1 = 76 - ( 2.3 + 0.90) / / l e n g th 1 i n mm4 l 2 = 115 - (2.3 + 0.90) / / l e n g th 2 i n mm5 t = 2.3 / / mm6 r = 0.90 // i n n er r a d i us i n mm7 k = t /3 / / mm8 B = 0 .5 * %pi *( r + k ) // b en d in g a l l o w a n c e i n mm9 d = l1 + l2 + B // d e ve l op e d l e n g t h i n mm

10 printf ( ” \n D e v el o p e d l e n g t h = %0 . 2 f mm” , d )



1 clc

2 t = 3.2 // t h i c k n e s s o f b la nk i n mm3 l = 900 // b en d in g l e n g t h i n mm4 s i gm a = 4 00 // u l t i m at e t e n s i l e s t r e n g t h i n N/mmˆ25 k = 0.67 // d i e o pe ni ng f a c t o r6 r 1 = 9.5 / / punch r a d i u s i n mm7 r 2 = 9.5 // d i e e d ge r a d i u s i n mm

8 c = 3.2 // c l e a r a n c e i n mm9 w = r1 + r2 + c // w id th b etw een c o n ta c t p o i n t s

b at we en d i e and p unc h i n mm10 F = ( k * l * s i gm a * t ^ 2 ) / w // b e n d in g f o r c e i n N11 F = floor ( F / 1 0 ) * 1 0 // N

15



12 printf ( ” \n b en d in g f o r c e = %0 . 2 f kN” , F / 1 0 0 0 )


1 clc

2 k = 1.33 // d i e o pe ni ng f a c t o r3 l = 1200 // bend l e n g t h i n mm4 s i gm a = 4 55 // u l t i m at e t e n s i l e s t r e n g t h i n N/mmˆ25 t = 1.6 // b la nk t h i c k n e s s i n mm

6 w = 8* t // w id th o f d i e o pe ni ng i n mm7 F = k * l * si gm a * t ^2 / w // b en di n g f o r c e i n N8 printf ( ” \n b en di ng f o r c e = %0 . 2 f kN” , F / 1 00 0)

Scilab code Exa 2.13 calculate capacity of double bending die

1 clc

2 c = 1.25 // c l e a r a n c e i n mm

3 r 1 = 3 // d i e r a d i u s i n mm4 r 2 = 1.5 / / punch r a d i u s i n mm5 s i gm a = 3 15 // u l t i m at e t e n s i l e s t r en g t h i n MPa6 t = 1 // t h i c k n e s s i n mm7 l = 5 0 / / w id th a t bend i n mm8 w = r1 + r2 + c // w id th b et wee n c o n t ac t p o i n t s on

d i e and p un ch i n mm9 F = 0 .6 7* l * s i gm a * t ^2 / w // b en d i ng f o r c e i n N

10 F _p = 0 .6 7* s i g ma * l * t // pad f o r c e i n N11 s ig ma _c = 5 60 // s e t t i n g p r e s su r e i n MPa12 b 1 = 2 // b e ad s on p un ch13 b = b1 *r1 / / mm14 F _ b = s ig m a_ c * l *b // b ot to mi ng f o r c e i n N15 F _o = F_p + F_b // F or ce r e q u i r e d when b ot to mi ng i s

u se d i n N

16



16 F_n = F + F_p // F or ce r e q u i r e d when b ot to mi ng i s n ot

us ed i n N17 printf ( ” \n F o rc e r e q u i r e d when b ot to mi ng i s u se d =%0 . 1 f t o n n e s ” , F _ o / ( 9 . 8 1 * 1 0 0 0 ) )

18 printf ( ” \n F or ce r e q u i r e d when b ot to mi ng i s n ot u se d= %0 . 3 f t o n n e s ” , F _n / ( 9 . 81 * 1 0 00 ) )

Scilab code Exa 2.14 To calculate cutting force

1 clc2 w = 2 / / w id t h i n mm3 t = 5 // t h i c k n e s s i n mm4 t h e t a = 6 // s he ar i n d e g r e e s5 t a u = 3 82 .5 // u l t i m at e s h ea r s t r e s s i n MPa6 F = w * t * ta u * 10 00 // c u t t i n g f o r c e i n N7 l = t / sin ( t h e t a * % p i / 1 8 0 ) // l e ng t h t o be c ut i n mm8 F _i = l * t* ta u // c u t t i n g f o r c e i n N9 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 ge s

= %0. 3 f MN\n c u t t i n g f o r c e when s he ar i sc o n s i d e r e d = %0 . 2 f kN ” , F / 1 0^ 6 , F _i / 1 0 0 0 )

Scilab code Exa 2.15 Determine blank and punch diameter

1 clc

2 d 1 = 105 // i n s i d e d i am et er i n mm3 h = 9 0 / / d ep th i n mm4 t = 1 // t h i c k n e s s i n mm5 D = sqrt ( d 1 ^ 2 + 4 * d 1 * h ) // b l an k d i a me t er i n mm

6 t r = t * 10 0/ D // t h i c k n e s s r a t i o7 // f rom 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 28 r = 1.82 // draw r a t i o9 d 2 = D / r / / d i a m e t e r f o r f i r s t draw i n mm

10 d = 130 / / L et d i a me t e r f o r f i r s t draw i n mm

17



11 r at io 1 = D /d / / D/ d f o r f i r s t d ra w

12 r at io 2 = d / d1 // D/ d f o r s e co n d draw13 h 1 = ( ( D ) ^2 - ( d ) ^ 2 ) / ( 4* d ) / / D ep th o f f i r s t d ra w i n mm14 s i gm a = 4 15 / / N/mmˆ215 c = 0.65 // c o n st a n t16 p c = % pi * d* t * si gm a (D /d - c ) // p r e s s c a p a ci t y i n kN17 printf ( ” \n B l an k d i a m e t e r = %d mm \n T hi ck ne ss 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 d ra w = %d mm \nDepth of f i r s t draw = %0. 2 f mm \n P r e ss c a p ac i t y= %0. 2 f kN ” , D , t r , d 2 , h 1 , p c / 1 0 0 0 )


Scilab code Exa 2.16 To find drawing operations and force

1 clc

2 d = 8 0 / / d i a me t e r i n mm3 h = 250 / / h e i g h t i n mm4 D = sqrt ( ( d ^ 2 + 4 * d * h ) ) / 1 0 // b la nk d i am et er i n cm5 D 1 = 0. 5* D / / d i a me t er a f t e r f i r s t draw i n cm6 / / l e t r e du c t i o n be 40% i n s ec on d draw

7 D 2 = D1 - 0 .4 * D1 // d i am e te r a f t e r s co nd draw i n cm8 R = (1 - ( d /( 10 * D2 ) )) * 10 0 // p e r ce n t ag e r e d u c t io n

f o r t h i r d draw9 l 1 = ( ( D) ^ 2 -( D 1 ) ^2 ) / (4 * D1 ) // h e i g h t o f cup a f t e r

f i r s t d ra w i n cm10 l 2 = ( ( D) ^ 2 -( D 2 ) ^2 ) / (4 * D2 ) // h e i g h t o f cup a f t e r

f i r s t d ra w i n cm11 l 3 = ( ( D ) ^2 - ( d / 1 0 ) ^ 2) / ( 4 * d / 1 0) // h e i g h t o f cup

a f t e r f i r s t draw i n cm12 t = 3 // mm

13 s i gm a = 2 50 / / N/mmˆ214 C = 0.66

15 F = % pi * d / 1 0 * t * s i gm a * ( ( D * 1 0/ d ) - C ) // d ra wi ng f o r c ei n kN

16 printf ( ” \n D i am e te r a f t e r f i r s t dra w = %0 . 1 f \n

18



D ia me te r a f t e r s e co n d draw = %0 . 2 f \n P e r ce n t ag e

r e d u c ti o n a f t e r t h i r d draw = %d p e r ce n t ” , D 1 , D 2 , R )17 printf ( ” \n H ei g ht o f cup a f t e r f i r s t draw = %0 . 2 f cm\n H ei gh t o f cup a f t e r s ec on d draw = %0 . 2 f cm\nH ei gh t o f cup a f t e r t h i r d draw = %0 . 2 f cm” , l1 , l 2

, l 3 )

18 printf ( ” \n D ra wi ng f o r c e = %0 . 3 f kN” , F / 1 0 0 0 )


Scilab code Exa 2.17 Determine developed length

1 clc

2 / / from f i g u r e 2 . 7 5 ( a )3 r 1 = 30 // r a d i u s i n mm4 t = 1 0 // t h i c k n e s s i n mm5 h 1 = 300 // h e i g h t i n mm6 i r1 = r1 - t // i n n er r a d i u s o f be nds i n mm7 L 1 = h1 - ( ir 1 +t ) // mm8 a l ph a1 = 90 // d e gr e e9 r 2 = 2* t / / mm

10 k = 0. 33 * t // mm11 L 2 = a l p ha 1 * 2 * % p i * ( r2 + k ) / 3 6 0 / / mm12 w = 200 / / mm13 L 3 = w - 2* ( t+ i r1 ) // mm14 L 4 = L2 //mm15 h 2 = 100 / / mm16 L5 = h2 -( t+ ir1 ) / / mm17 r 3 = 150 //mm18 ir2 = r3 - t // i n n er r a d i u s i n mm19 a lp ha 2 = 1 80 // d e g re e

20 L 6 = a l p ha 2 * 2 * % p i * ( i r2 + k ) / 3 6 0 / / mm21 d l = L 1 + L2 + L 3 + L4 + L 5 + L6 // T ot al d ev el op ed l e ng t h i nmm

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” , d l )


19



20



Chapter 4

Cost Estimating

Scilab code Exa 4.1 To calculate total cost and SP

1 clc

2 d _m = 5 50 0 // c o s t o f d i r e c t m a t e r i a l i n Rs3 d _l = 3 00 0 // m a n uf a ct u ri n g w ag es i n Rs4 // f a c t o r y o ve rh ea d i s 1 00% 0 f m a nu fa ct ur in g wa ges5 f _o = ( 10 0* d _ l ) /1 00 // f a c t o r y o ve rh ea ds i n Rs6 F C = d_m + d_l + f_o // f a c t o r y c o s t i n Rs

7 n m_ o = 1 5* F C / 10 0 / / non−m a n uf a ct u ri n g o v e rh e a d s i nRs

8 t c = FC + nm _o // t o t a l c os t i n Rs9 p = 1 2* t c /1 00 // p r o f i t i n Rs

10 s p = tc +p / / s e l l i n g p r i c e i n Rs11 printf ( ” \n T ot al c o s t = Rs %d\n S e l l i n g p r i c e = Rs

%d” , tc , s p)

Scilab code Exa 4.2 To find selling price

1 clc

2 / / g i ve n

21



3 O S_ RM = 2 00 00 // o pe n in g s t o c k o f raw m a t e r i a l s i n

Rs4 C S_ RM = 3 00 00 // c l o s i n g s t o ck o f raw m a t e r i a l s i nRs

5 T P _ RM = 1 7 00 00 // t o t a l p ur ch as e i n y ea r i n Rs6 O S_ FG = 1 00 00 // o pe n in g s t o c k o f f i n i s h e d g o o ds i n

Rs7 C S_ FG = 1 50 00 // c l o s i n g s t o ck o f f i n i s h e d g o o d s i n

Rs8 s a l es = 4 8 95 00 // s a l e s o f f i n i s h e d g o o d s i n Rs9 D _W = 1 20 00 0 // d i r e c t wages i n Rs

10 F _ E1 = 1 20 00 0 // f a c t o r y e xp en se s i n Rs

11 N M_ E = 5 00 00 // non−m a n uf a ct u ri n g e x p e ns e s i n Rs12 DMC = O S_ RM + TP _R M - C S_R M // d i r e c t m a t e r i a l c o s t13 FC = DMC + D_W + F_E1 // f a c t o r y c o st14 T C = FC + NM_E // t o t a l c o s t15 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 Rs16 P = sales - FG_S // p r o f i t i n Rs17 F _E 2 = ( F _E 1 ) / D_ W * 10 0 // f a c t o r y e xp e n s es i n p e rc e n t18 N M_ C = ( N M_ E ) / FC * 1 00 / / non−m a n u f a ct u r i n g e x p e n s e s

t o f a c t o r y c o s t19 P _C = ( P / F G_ S ) * 10 0

// p r o f i t t o c o st o f s a l e s20 d m = 2 00 00 // d i r e c t m a t e r i a l i n Rs21 d w = 3 00 00 // d i r e c t wages i n Rs22 f e = dw // f a c t o r y e x p e n se s23 f c = dm + dw + fe // f a c t o r y c o s t i n Rs24 n m e = N M_ C * fc / 1 00 / / non−m a n uf a ct u ri n g e x p e n se s i n

Rs25 tc = fc + nme // t o t a l c os t i n Rs26 p = ( P _C * t c ) /1 00 // p r o f i t i n Rs27 s p = tc +p / / s e l l i n g p r i c e i n Rs28 printf ( ” \n S e l l i n g p r i c e = Rs %d” , s p )

Scilab code Exa 4.3 To find factory cost

22



1 clc

2 d = 3 8 // d i am et e r o f b ar i n mm3 l = 2 5 // l e ng t h o f b ar i n mm4 p = 8.6 / / d e n s i t y gm/ cm ˆ35 g = 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 ˆ26 w = ( % p i * d ^ 2* l * p * g ) / ( 4 * 1 0^ 6 ) // w ei gh t o f m a t e r i a l

i n N7 m c = w * 1. 62 5 // m a te r ia l c o st i n Rs8 l c = ( 2* 90 ) /6 0 // l ab ou r c o st i n Rs9 f o = 0 .5 * lc // f a c t o r y o v e rh ea ds i n Rs

10 fc = mc + lc + fo // f a c t or y c o s t i n Rs11 printf ( ” \n f a c t o r y c o s t = Rs %0 . 2 f ” , f c )


Scilab code Exa 4.4 find production cost and time taken

1 clc

2 s p = 65 // s e l l i n g p r i c e i n Rs3 p r o fi t = 0 .2 * s p // p r o f i t i n Rs4 t c = sp - pro fit // t o t a l c os t i n Rs

5 P = ( sp - p ro fi t) /1 .4 // p r o d uc t i o n c o s t i n Rs6 D M = 15 // c os t o f d i r e c t m a t e r i a l i n Rs7 W = ( P - DM )/ 1.4 // d i r e c t l ab ou r c o s t i n Rs8 tt = W /2 // t im e t ak en i n h ou rs9 printf ( ” \n Time t a k e n = %0 . 3 f H ou rs ” , t t )


Scilab code Exa 4.5 To find profit

1 clc

2 m p = 60 00 // m arket p r i c e o f ma ch in e i n Rs3 d = 0 .2 * mp // d i s co u n t i n Rs4 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 m c = 400 // m a t e r i a l c o s t i n Rs

6 l c = 16 00 // l ab ou r c o st i n Rs7 f o = 800 // f a c t o r y o ve rh ea ds i n R s8 F = m c + l c + f o // f a c t o r y c o s t i n Rs9 s e = 0. 5* F // s e l l i n g e x pe n s es i n Rs

10 pro fit = sp - ( F + se ) / / Rs11 printf ( ” \n p r o f i t = Rs %d” , p ro f it )

Scilab code Exa 4.6 To find lot size and time

1 clc

2 a = 1500 // r e q u i r e m en t s o f c om po ne nt s3 s = 3 0 // c o s t o f e a c h s e t up i n Rs4 k = 0.2 // c ha rg e f a c t o r5 c = 5 // c o s t o f e a c h p ar t i n Rs6 N = 5 * sqrt ( a * s ) / ( k * c ) // e c o n o m i c l o t s i z e7 printf ( ” \n Economic l o t s i z e = %d p i e c e s ” , N )

8 S = ( N* s) /a // t i m e f o r e a ch s e t up i n h ou rs9 printf ( ” \n Time f o r e ac h s e t up = %0 . 2 f h ou rs ” , S )


Scilab code Exa 4.7 To find time to change cutter

1 clc

2 T c = 2 // t im e t ak en by c u t t e r p er c y c l e i n m in ut es3 T k = 10 // t i me t ak en t o c h ang e c u t t e r i n m in ut es4 T = 240 / / t o o l l i f e i n m in ut es5 t = ( Tc * Tk ) /T // t im e t o c h ang e t he c u t t e r i n min .6 printf ( ” \n U ni t t im e t o c ha ng e t he c u t t e r = %0 . 3 f

min” , t )

7 / / E rr or i n t ex tb oo k

24



Scilab code Exa 4.8 To find tool change time

1 clc

2 T k = 360 // t i m e t a ke n by t o o l t o c u t b e f o r es h a r p e ni n g i n min .

3 T c = 20 // t i me t ak en t o c h ang e t he t o o l i n min .4 T = 4320 // t i m e t a ke n b e f o r e i t i s d i sc a r d e d i n min

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

min” , t )

Scilab code Exa 4.9 To calculate measuring time allowance

1 clc

2 T c = 10 // t i m e t ak e n t o c h ec k h ol e i n s e c s

3 F = 2 // f r eq u e nc y o f c h ec k in g d im en si o n4 t c = Tc *F // t i m e t a ke n t o c h e c k one p i e c e i n s e c s5 N = 200 // number o f p i e c e s6 Tc = tc *( N + 1) // T o ta l t i m e i n s e c7 printf ( ” \n T ot al t im e t ak en t o c he ck d i me n si o ns = %d

min” , T c / 60 )

Scilab code Exa 4.10 To find direct labour cost

1 clc

2 f or gi ng s = 40

3 setup = 4

4 T c = 12 // m ac hi ni ng t im e i n min . p er f o r g i n g

25



5 n mt = 21 / / non−m ac hi ni ng i n min . p er f o r g i n g

6 s t = 45 // s e t up t i m e p e r s e t up7 t s = 5 // t o t a l s ha rp en i n g i n min . p er f o r g i n g8 f = 2 0 // f a t i g u e i n p e r ce nt9 f = f / 1 0 0

10 p n = 5 // p e r so n a l n ee ds i n p e rc e n t11 p n = pn / 10 0

12 T k = 10 / / t o o l ch an he t im e i n min .13 T = 8 / / t o o l l i f e i n h ou rs14 c t = 15 // c he ck in g t i m e w i t h 5 c he ck s i n 15 s e c s15 R = 1.4 // p er fo rm a nc e f a c t o r16 dlc = 5 // d i r e c t l ab ou r c o st i n Rs p er hour

17 tt = Tc + nmt / / m a ch i n in g and non−m ac hi ni ng t im e i nmin .

18 f t = f *tt / / f a t i g u e t i me i n min .19 p nt = pn * tt // p e r s o n a l n ee ds i n min .20 t = ( T c * Tk ) / ( T *6 0) / / t o t a l s h ar p en i ng t ime i n min .

p er f o r g i n g21 m c t = ( t s * ct ) / 60 // m ea su ri ng and c h ec k in g t im e i n

min . p e r f o r g i n g22 su = Tc + nmt + p nt + ft + t + mct // sum o f t im es i n

min .23 t f = s u * fo r gi n gs

// t i m e f o r 40 f o r g i n g s i n min .24 t s t = s t * se tu p / / t o t a l s e t up t i m e i n min .25 Te = tf + tst // t o t a l e s ti m at e d t i me i n min .26 T a = Te *R // t o t a l a c t ua l t i me i n min .27 l c = ( T a * dl c ) /6 0 // d i r e c t l ab ou r c o s t i n Rs28 printf ( ” \n D i r e ct l a b ou r c o s t = Rs %0 . 1 f ” , l c )

Scilab code Exa 4.11 To find machining time

1 clc

2 // f rom f i g u r e 4 . 43 v = 100 // c u t t i n g s pe ed i n m/ min4 D = 5 0 / / mm

26



5 l 1 = 76 / / mm

6 f = 7.5 / / f e e d i n mm/ r e v .7 // Case 1 , t im e t o t ur n 38 mm d i am e te r by 76 mml e n g t h o f c u t

8 N 1 = ( 10 0 0* v ) / ( %p i * D) // r . p .m9 t m 1 = l 1 * 10 /( f * N 1 ) // min .

10 // Case 2 , t im e t o t ur n 25 mm by 38 mm l e n g t h11 N 2 = ( 10 0 0* v ) / ( %p i * 38 ) // r . p .m12 l 2 = 38 / / mm13 t m 2 = l 2 * 10 /( f * N 2 ) / / min14 t t = tm1 + tm2 // t o t a l t i me i n min15 printf ( ” \n T o t a l t i m e = %0 . 2 f min . ” , t t )

Scilab code Exa 4.12 To find time to turn relief

1 clc

2 // f rom f i g u r e 4 . 53 v = 6 0 // c u t t i n g s pe ed m/ min .4 f = 0. 37 5 / / f e e d i n mm/ r e v5 D = 3 8 / / mm

6 N = ( 1 00 0 *6 0 ) /( % p i *D ) // r e v / min7 l = 3 2 / / mm8 T m = l /( f *N ) / / mi n9 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 .

” , T m )

Scilab code Exa 4.13 calculate time to face on lathe

1 clc2 // from f i g u r e 4 . 1 13 l = 7.5 / / cm4 D av e = ( 25 + 1 0) / 2 // a v er a ge d i am et er i n cm5 v = 2 7 // c u t t i n g s pe ed i n m/ min

27



6 f = 0.8 / / f e e d i n mm/ r e v

7 N = ( 1 00 0 * v ) / ( % pi * D a v e * 1 0) // r . p .m.8 t m = l * 10 /( f * N) // min .9 printf ( ” \n The m ac hi ni ng t im e t o f a c e on l a t h e = %0

. 2 f min . ” , t m )

Scilab code Exa 4.14 To find time to drill hole

1 clc

2 D = 12.7 / / d i a me t er i n mm3 d = 5 0 / / d ep th i n mm4 v = 7 5 / / c u t t i n g s p ee d i n m/ min .5 f = 0. 17 5 / / f e e d i n mm/ r e v6 l = d + 2*0. 29* D // l em gt h o f d r i l l t r a v e l i n mm7 N = ( 10 00 * v ) /( % pi * D ) // r . p .m.8 t m = l /( f *N ) / / mi n9 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 clc

2 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 o3 l = 900 + 2*75 // l e ng t h o f s t r o k e i n mm4 v = 6 // c u t t i n g s t r o k e i n m/ min5 f = 2 / / f e e d mm/ s t r o k e6 w = 600 / / b r ea d th i n mm7 N = ( v * 10 00 ) / ( l *1 . 25 ) // r . p .m

8 N = round ( N )9 time = w /( f* N) / / min

10 printf ( ” \n Time r e q ui r e d f o r s ha pe r t o c om pl et e onec u t = % d m i n ” , t im e )

28



Scilab code Exa 4.16 To find time to broach

1 clc

2 l = 7 0 // l e n g t h o f s t r o k e i n cm3 c s = 11 // c u t t i n g s pe ed i n m/ min4 r s = 24 // r e t u r n s p ee d i n m/ min5 t m = ( l / (1 00 * c s )) + ( l / (1 0 0* r s ) ) / / min6 printf ( ” \n Time t ak en t o b ro ac h a f o u r s p l i n e b r as s

= %0 . 4 f m in ” , t m )


Scilab code Exa 4.17 find feed cutter travel and time

1 clc

2 v = 5 0 // c u t t i n g s pe ed i n m/ min3 D = 150 // d i a me te r o f f a c e c u t t e r i n mm

4 N = ( 10 00 * v ) /( % pi * D ) // r . p .m.5 f = 0.25 / / f e e d mm/ t o o t h6 n = 1 0 // n umber o f t o o t h7 t f = N *f *n // t a b l e f e e d i n mm/ min8 l = 200 // l e ng t h o f work p i e c e i n mm9 d = 2 5 // de pt h o f s l o t i n mm

10 t ot = sqrt ( D* d - d ^2) // t o t a l o v e r t r a v e l i n mm11 tct = l + tot // t o t a l c u t t e r t r a v e l i n mm12 t im e = t ct / tf // min .13 printf ( ” \n T abl e f e e d = %d mm/min . \n T ot al 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 m ac hi ne t h e

s l o t = %0 . 3 f min . ” , tf , tct , time )

29



Scilab code Exa 4.18 To find cutting time

1 clc

2 D = 63.5 // d ia me te r o f p l a i n m i l l i n g c u t t e r i n mm3 w = 3 0 // w id th o f b l o ck i n mm4 l = 180 // l e ng t h o f b l o c k i n mm5 f = 0. 12 5 / / f e e d i n mm/ t o o t h6 n = 6 // no . o f t e e t h7 N = 1500 // s p i n d l e s pe ed i n r . p .m8 tot = ( D - sqrt ( D ^2 - w ^ 2) ) /2 // t o t a l o v e r t r a v e l

i n mm9 tct = l + tot // t o t a l c u t t e r t r a v e l i n mm

10 T m = t ct / ( f *n * N ) // c u t t i n g t i me i n min11 printf ( ” C u t t i n g t i m e = %0 . 3 f min . ” , T m )


Scilab code Exa 4.19 To find milling time

1 clc

2 // from f i g u r e 4 . 1 7

3 d = 1 9 // d ep th o f c ut i n mm4 D 1 = 5 // d i am et e r o f ro un d b ar i n cm5 v = 5 0 // c u t t i n g s pe ed i n m/ min6 n = 8 // number o f t e e t h7 f = 0.2 / / f e e d i n mm/ t o o t h8 l = 2 * sqrt ( d *D1 * 10 - d ^ 2) // l e ng t h o f c ho rd i n mm9 D 2 = 10 // d ai me te r o f c u t t e r i n cm

10 o v e rr u n = sqrt ( D 2 * 1 0* d + D 1 * 1 0 * d - d ^ 2) - sqrt ( D 1 * 1 0 * d - d

^2) // mm11 t t = l + ov erru n // t a b l e t r a v e l i n mm

12 N = ( 10 00 * v ) /( % pi * D 2 * 10 ) // r . p .m13 t m = t t /( f *n * N) / / t im e i n min .14 printf ( ” \n The m i l l i n g t i me = %0 . 2 f min . ” , t m )

30



Scilab code Exa 4.20 To find time to grind shaft

1 clc

2 w = 5 0 // w id th o f g r i n d i n g wheep i n mm3 f = w /2 // f e e d i n mm4 t = 0.25 // t o a t a l s t oc k i n mm5 d = 0. 02 5 // d ep th o f c ut i n mm6 n = t / d // number o f c u t s7 v = 1 5 // c u t t i n g s pe ed i n m/ min8 D = 3 8 / / d i a me t e r i n mm9 N = ( 10 00 * v ) /( % pi * D ) // r . p .m.

10 l = 200 // l e ng t h o f p ar t i n mm11 T m = ( l * 10 ) /( f * N ) // min .12 printf ( ” Time r e q u i r e d t o g r i nd t he s h a f t = %0 . 2 f min

. ” , T m )

Scilab code Exa 4.21 To find time to cut threads

1 clc

2 v = 6 // c u t t i n g s pe ed i n m/ min3 n = 5 // number o f c u t s4 D = 4 4 / / d i a me t e r i n mm5 N = ( 10 00 * v ) /( % pi * D ) // r . p .m6 f = 0.5 // f e e d i n cm7 l = 8.9 // l e n g th o f c u t i n cm8 T m = ( l *n ) /( f *N ) // t im e i n min9 printf ( ” \n Time t o c u t t h e t h r e a d s = %0 . 2 f min ” , Tm

)10 / / A nswers v ar y d ue t o ro und o f f e r r o r

31



Scilab code Exa 4.22 find time to produce one piece

1 clc

2 v t = 40 // c u t t in g s pe ed f o r t u rn i ng i n m/ min3 v s = 8 // c u tt i n g s p ee d f o r c u t t i ng and k n u r li n g i n

m/min4 f t = 0.4 / / f e e d f o r t u r ni n g i n mm/ r e v .5 f f = 0.2 // f e e d f o r f or m in g i n mm/ r e v6 d 1 = 25 / / d i a m et e r i n mm7 l 1 = 50 / / mm8 N 1 = 1 00 0* v t / ( %p i * d1 ) / / s p i n d l e s p ee d i n r e v . / min .9 t im e1 = l 1 /( f t * N1 ) // min .

10 t t = 2* t im e1 / / t o t a l t im e i n min .11 d 2 = 15 / / mm12 N 2 = 1 00 0* v t / ( %p i * d2 ) / / r e v / m in .13 l 2 = 30 / / mm14 t im e2 = l 2 /( f t * N2 ) // min .15 e ft = 0 .1 5 / / end f o rm i n g t im e i n min .16 d 3 = 10 / / mm17 N 3 = 1 00 0* v s / ( %p i * d3 ) // r e v . / min .18 l 3 = 15 / / mm19 f = 1.5 / / f e e d i n min .20 t i m e3 = l 3 /( f * N3 )

// min .21 N 4 = 1 00 0* v s / ( %p i * d1 ) // r e v . / min .22 l 4 = 10 / / mm23 t im e4 = l 4 /( f t * N4 ) // min .24 t im e5 = 0 .1 5 // t im e f o r c h am f er i ng i n min .25 D a ve = d 1 /2 / / mm26 N 5 = 1 00 0* v t / ( %p i * D av e ) // r . p .m.27 t im e6 = D av e / ( N5 * f f ) // min ,28 t m t = t t + t i me 2 + t i m e 3 + t i me 4 + t i m e5 + t i m e 6 + e f t // t o t a l

m a ch i ni n g t i me i n min .29 t = 0.05 // min .

30 h t = t im e5 + 6 * t im e6 + 4 * t +3 * t / / h a nd l in g t im e i n min .31 t o t = ht + tm t // t o t a l h a nd l i ng t i me i n min .32 c t = 1 5* t o t / 10 0 / / c o n t i n g en c y i n min .33 t c t = t ot + ct / / t o t a l c y c l e t i m e i n min .34 s t = 60 // s e t up t i m e f o r t u r r e t l a th e

32



35 p = 100 // t o t a l p i e c e s

36 s tp = st / p // s e t up t i me p er p i e c e i n min .37 t pt = t ct + st p // T ot al p r o d uc ti o n t im r p er p i e c e i nmin .

38 printf ( ” \n T ot al p r o du c ti o n t im r p er p i e c e = %0 . 2 f min” , tpt )

39 / / An swers v ar y due t o ro un d o f f e r r o r

33



Chapter 5

Economics of tooling

Scilab code Exa 5.1 To find value of machine tool

1 clc

2 C o = 2 50 00 0 // o r i g i n a l v al ue o f machi ne t o o l i n Rs3 C s = 2 50 00 // s a l v a ge v al ue i n Rs4 n = 2 0 // u s e f u l l i f e i n y e a rs5 d = ( Co - C s) / n // d e p r e c i a t i o n p er y e ar i n Rs6 v 1 = Co - 10* d // v al ue o f machi ne t o o l a t t he end

o f 10 y ea r s i n Rs7 s = C o - C s / / sum a t t he end o f u s e f u l l i f e i n Rs8 i = 8/ 10 0 // a n nu a l i n t e r s t r a t e9 D = ( s* i ) /( (1 + i ) ^n - 1) // a nn ua l d e p o s i t

10 a = D * (( 1+ i )^ 10 - 1) / i // amount a t t h e end o f 1 0y e ar s i n Rs

11 v2 = Co - a // v al ue a t t h e e nd o f 10 y ea r s12 printf ( ” \n V al ue o f machi ne a t t he end o f 10 y e ar s

t hr ou gh 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” , v 1 )

13 printf ( ” \n V al ue o f machi ne a t t he end o f 10 y e ar st h ro u g h s i n k i n g f un d meth od = Rs %d” , v 2 )


34



Scilab code Exa 5.2 To find annual investment

1 clc

2 p = 2 00 00 0 // p r e s e n t wor th i n Rs3 i = 1 0 // a n nu a l i n t e r e s t r a te4 i = 1 0/ 10 0

5 n = 2 0 // number o f y e a r s6 a 1 = ( p * i) / ( (1 + i ) ^n - 1) // a nn ua l i n ve s tm e nt u s i ng

s i n k i n g f un d f a c t o r i n Rs7 a 2 = ( p * i * ( i + 1) ^ n ) / ( ( i + 1) ^ n - 1 ) / / a n nu a l i n v es t m en t

u si ng c a p i t a l r ec ov e r y f a c t o r i n Rs8 printf ( ” \ nAnnua l i n v e s t me n t u s i n g s i n k i n g f un d

f a c t o r = R s % d /− pe r y ea r ” , a 1 )

9 printf ( ” \ nAnnual i n ve st m en t u s i ng c a p i t a l r e c o ve r yf a c t o r = R s % d /− pe r y ea r ” , a 2 )


Scilab code Exa 5.3 find project is economical or not

1 clc

2 // c as h i n f l o w s3 a = 21 24 0 // a nn ua l r ev en u e i n Rs4 i = 1 0 // a n nu a l i n t e r e s t r a te5 i = 1 0/ 10 0

6 n = 5 // p er od i n y e a r s7 f 1 = 80 00 // s a l v a ge v al ue i n Rs8 p 1 = ( a * (( i + 1) ^ n - 1) ) / ( i *( i + 1) ^ 5 ) / / a n nu a l r e ve n ue

i n Rs9 p 2 = f 1 /( i +1 ) ^5 / / p r e s e n t w or th i n Rs

10 t1 = p1 + p2 // t o t a l c a s h i n f l o w s i n Rs11 // c as h o ut f l o w s12 I = 40 00 0 // i n ve s tm e nt i n Rs

35



13 f 2 = 1 20 00 // a n n ua l p aymen t i n Rs

14 p 3 = ( f 2 * ( ( 1+ i ) ^ 5 - 1 ) ) / ( i * (1 + i ) ^ 5 ) // a n n u a l p ay me nt si n Rs15 t 2 = I + p3 // t o t a l c a s h ou t f l o w s i n Rs16 printf ( ” \ n To ta l c as h i n f l o w s = Rs %0 . 2 f \ n T o ta l c a s h

o u t f l o w s = Rs %0 . 2 f ” , t1 , t2 )

17 disp ( ” S i n ce c as h o ut f l o ws a r e more t ha n c as h i nf l ow s t h e r e f o r e p r o j e c t i s no t e co no mi ca l ”)


Scilab code Exa 5.4 selection of economical machine

1 clc

2 / / M a c h i n e A3 f 1 = 20 00 // a nn ua l b e n e f i t from b e t t e r p r o d uc ti o n

q u a l i t y i n Rs4 i = 1 0 // i n t e r e s t r a t e5 i = 1 0/ 10 0

6 f 2 = 1 20 00 // s a l v a ge v al ue i n Rs7 f 3 = 80 00 // o p e r a t i n g and m ai nt en an ce c o s t i n Rs

8 I 1 = 1 00 00 0 / / i n i t i a l c o s t i n Rs9 n = 5 // y e ar s

10 p 1 = ( f 1 * ( ( 1+ i ) ^ n - 1 ) ) / ( i *( i + 1 ) ^ n )

11 p 2 = f 2 /( 1+ i ) ^n

12 c1 = p1 + p2 // c a s h i n f l o w s i n Rs13 p 3 = ( f 3 * ( ( 1+ i ) ^ n - 1 ) ) / ( i *( i + 1 ) ^ n )

14 c2 = I1 + p3 // c a sh ou t f l ow s i n Rs15 Pa = c1 - c 2 / / n e t P .W. i n Rs16 / / M a c h i n e B17 I 2 = 6 00 00 / / i n i t i a l c o s t i n Rs

18 f 4 = 1 60 00 // o p e r a t i ng and m ai nt en an ce c o s t i n Rs19 f 5 = 1 40 00 // r e c o n d i t i o n i n g a t t he end o f t h i r dy ea r i n Rs

20 p 4 = ( 1 6 00 0 * ( (1 + i ) ^ 5 - 1 ) ) / ( i * (1 + i ) ^ 5 )

21 p 5 = f 5 /( 1+ i ) ^5

36



22 Pb = - I2 - p4 - p5 / / n e t P .W. i n Rs

23 printf ( ” \n N et P .W. of M ac hi ne A= R s %0. 2 f \n N et P .W.o f M a ch i ne B = Rs%0 . 2 f ” , Pa , Pb )

24 disp ( ” I t i s c l e a r t ha t Net P .W o f Machine A i s l e s sn a g a t iv e a s compared t o t h at o f Machine B ,t h e r e f o r e Ma ch in e A i s e co no mc al . ” )


Scilab code Exa 5.5 selection of machine

1 clc

2 / / m a c h i n e A3 c 1 = 2 00 00 // manual c o s t i n Rs4 c 2 = 4 00 00 // o p er a ti n g c o st i n Rs5 n 1 = 2 / / m ac hi ne l i f e i n y e a r s6 i = 1 0 // i n t e r e s t r a t e7 i = 1 0/ 10 0

8 c r f 1 = ( ( 1+ i ) ^ n 1 - 1 ) / ( i *( i + 1 ) ^ n 1 ) // c a p i t a l r e c o v e r yf a c t o r

9 p w 1 = c 1 + c2 * c r f1 // p r e s e nt worth i n Rs

10 / / m ac hi ne B11 c 3 = 5 00 00 // manual c o s t i n Rs12 c 4 = 3 00 00 // o p er a ti n g c o st i n Rs13 n 2 = 4 / / m ac hi ne l i f e i n y e a r s14 i = 1 0/ 10 0 // i n t e r e s t r a t e15 c r f 2 = ( ( 1+ i ) ^ n 2 - 1 ) / ( i *( i + 1 ) ^ n 2 ) // c a p i t a l r e c o v e r y

f a c t o r16 p w 2 = c 3 + c4 * c r f2 // p r e s e n t worth i n Rs f o r 4 y e a r s17 p w3 = ( p w2 * c r f1 ) / c rf 2 // p r e s e n t worth i n Rs f o r 2

y e a r s

18 printf ( ” \n P .W. o f e x pe n se 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 ” , p w 1 , p w 3 )

19 disp ( ”As t h e e xp en se s o f machi ne B a re l e s s , s ot h i s i s e co no mi ca l ”)


37



Scilab code Exa 5.6 selection of economical machine

1 clc

2 / / M a c h i n e A3 i = 8 // / / i n t e r e s t r a t e4 i = i / 1 0 0 // i n t e r e s t r at e5 n 1 = 10 / / e co no mi c l i f e i n y e a r s6 C R F 1 = i * ( 1 + 0 .0 8 ) ^ n 1 / ( ( 1+ i ) ^ n 1 - 1 ) // c a p i t a l

r e co v e r y f a c t o r7 p 1 = 4 60 00 // f i r s t c o s t i n Rs8 s 1 = 80 00 // s a l v a ge v al ue i n Rs9 o 1 = 1 00 00 // o p e r a t i n g c h ar g es i n Rs

10 AC1 = ( p1 - s1 )* CR F1 + s1 * i + o1 // a nn ua l c o s t i n Rs11 / / M a c h i n e B12 n 2 = 15 / / e co no mi c l i f e i n y e a r s13 C R F 2 = i * ( 1 + 0 .0 8 ) ^ n 2 / ( ( 1+ i ) ^ n 2 - 1 ) // c a p i t a l

r e co v e r y f a c t o r14 p 2 = 6 00 00 // f i r s t c o s t i n Rs15 s 2 = 1 00 00 // s a l v a ge v al ue i n Rs16 o 2 = 92 00 // o p e ra t i ng c ha r ge s i n Rs17 AC2 = ( p2 - s2 )* CR F2 + s2 * i + o2 // a nn ua l c o s t i n Rs18 printf ( ” \n Annu al c o s t o f m ac hi ne A = Rs %0 . 2 f \n

An nu al c o s t o f M ac hi ne B = R s %0 . 2 f ” , AC1 , A C2 )

19 disp ( ” Machine B w i l l be e c on o mi c al ” )


Scilab code Exa 5.7 find ERR and economicality of project

1 clc

2 a = 1 00 00 0 / / E j ( p / f , e% , j ) i n Rs3 n = 5 // l i f e i n y e ar s

38



4 e = 2 0 / / M . A . R . R .

5 e = e / 1 0 0 / / M . A . R . R .6 i = e

7 A = 32 00 0 // s a v i n gs i n Rs8 s = 20 00 0 // s a l v a ge v al ue i n Rs9 b = ( ( A * ( (( i + 1 ) ^ n ) - 1 ) / i ) +s ) / a / / ( F/ p , I , 5 )

10 i 2 = ( b ) ^( 1/ n ) -1 // i n t e r n a l r a t e o f r e t ur n11 printf ( ” \n ERR = %0.4 f \n I n t e r n a l r a t e o f r et ur n =

%0 . 2 f p e r c e n t ” , b , i 2 * 1 0 0 )

12 disp ( ” S i n ce I n t e r n a l r a t e o f r et ur 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 ” )

Scilab code Exa 5.9 find ERR and economicality of project

1 clc

2 e = 2 0 // M , A . R . R .3 i = e // i n t e r e s t r s t e4 i = i / 1 0 0

5 n = 5 // l i f e i n y e ar s6 s = 32 00 0 // a nn ua l n et s a v i n g s i n R s

7 p = 1 00 00 0 // p r e s e n t wor th i n Rs8 S = 20 00 0 // s a l v a ge v al ue i n Rs9 a = ( p - S) * ( i /( (1 + i ) ^n - 1 ) ) / / ( p−s ) (A/F , e% , n )

10 E = (s - a) /p / / E . R . R . R11 printf ( ” \n ERRR = %0. 2 f pe r c e n t ” , E * 1 00 )

12 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 ti s f e a s i b l e . ” )

Scilab code Exa 5.10 To determine acceptance of machine

1 clc

2 / / m ac hi ne A3 r _e 1 = 9 60 0 / / c as h f l o w i n Rs

39



4 p 1 = 4 60 00 / / i n t i a l c o s t i n Rs

5 s = 0 // s a l v a g e v al u e6 e = 8 / / M . A . R . R7 e = e / 1 0 0

8 i = 8 // i n ve s tm e nt r a t e9 i = i / 1 0 0

10 n = 6 // l i f e i n y e ar s11 x = i / ( (1 + i ) ^n - 1)

12 E RR R1 = ( r _e 1 - ( p1 - s )* x )/ p1

13 / / m a c h i n e B14 r _e 2 = 7 20 0 / / c as h f l o w i n Rs15 p 2 = 3 20 00 / / i n t i a l c o s t i n Rs

16 E RR R2 = ( r _e 2 - ( p2 - s )* x )/ p217 printf ( ” \n ERRR1 = %0. 2 f pe r c e n t \n ERRR2 = %0 . 2 f

p e r c e n t ” , E RR R1 * 1 00 , E RR R2 * 1 0 0)

18 disp ( ” Only m ac hi ne B i s a c c e p t e b l e ” )

Scilab code Exa 5.11 find investment cost and unamortized value

1 clc

2 p m v = 1 50 00 // p r e s e nt m arket v a lu e i n Rs3 s s = 60 00 // sum n eed ed t o make i t s e r v i c e a b l e i n Rs4 ic = ss + pmv // i nv es tm en t c o st i n Rs5 p b v = 3 00 00 // p r e s e nt book v al ue i n Rs6 s v = 1 50 00 // s a l v a ge v al ue i n Rs7 ui = pbv - sv // u n am o rt i ze d i n v es t m en t i n Rs8 printf ( ” \n I n ve s tm e nt c o s t = Rs %d\n U n a m o r ti z e d

i n v e s t m e n t = R s % 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 m ac hi ne

40



3 p mp = 1 00 00 0 // p r e s e n t market p r i c e i n Rs

4 i o = 5 00 00 // i mm ed ia te o v e r h a ul i n g i n Rs5 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 ea r s6 a o c = 5 00 00 // a nn ua l o p e ra t i ng c o s t i n Rs7 s v o = 1 00 00 // s a l v a ge v al ue a f t e r o v e r h a u l in g i n Rs8 pc = io + pmp // p r e s e n t c o st i n Rs9 i = 1 0 // i n t e r e s t r a t e

10 i = 1 0/ 10 0

11 c rf 1 = ( i * (1 + i ) ^ as l ) / (( 1+ i ) ^ a sl - 1 ) // c a p i t a lr e co v e r y f a c t o r

12 AC1 = ( pc - svo ) * cr f1 + svo * i + aoc // a v e ra ge c o s ti n Rs

13 / / p r o p os e d m ac hi ne14 n = 1 0 / / e x p ec t e d e co no mi c l i f e i n y e a r s15 i c = 3 00 00 0 / / i n i t i a l c o s t i n Rs16 s v = 1 00 00 0 // s a l v a ge v al ue i n Rs17 o = 30 00 0 // a nn ua l o p e ra t i ng c o s t i n Rs18 c rf 2 = ( i * (1 + i ) ^1 0) / ( (1 + i ) ^1 0 - 1 )

19 A C 2 = ( i c - s v ) * c r f 2 + s v * i + o // a ve ra ge c o st i nRs

20 printf ( ” E x i s t i n g m ac hi ne = Rs %0 . 3 f \n P r op o se dmac hi ne = Rs %0. 2 f ” , AC1 , AC2 )

21 disp (” S i n ce t he e q u i v a l e n t a nn ua l c o s t o f p ro po se dmachi ne i s l e s s than t ha t o f t h e e x i s t i n g machine

, t h e r e f o r e , t h e r e p l a c e m e n t i s j u s t i f i e d . ” )


Scilab code Exa 5.15 Determine economic repair life

1 clc

2 c = 20 00 0 // f i r s t c o s t o f ma ch in e i n Rs3 s = 1000 // s c r a p v al ue i n machi ne i n Rs4 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 Rs5 n = sqrt ( 2 * ( c - s ) / b ) // y e a rs6 printf ( ” \n Number o f y e a rs o f e co no mi 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 C n = 7 20 00 // c o s t o f new m achine i n s t a l l e d andt o ol e d i n Rs

3 C o = 2 80 00 // c o s t o f new machime i n s t a l l e d andt o ol e d i n Rs

4 p = 1 6 // h ou rl y p i e c e s5 N n = 2 20 0* p // e s t i m a t ed a n nu a l p r o d u c ti o n on new

mac hi ne6 K o = 1 72 00 // p r e s e n t book v al u e o f o l d machi ne i n

Rs7 S o = 64 00 // s c ra p v al u e o f o l d machi ne i n R s8 S n = 80 00 // p r o ba bl e s c r a p v al ue 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 Rs9 o co = 2.5 // o p r e a t o r c o s t p er hour

10 m c o = 4 8 // m ac hi ne c o s t11 r o = 10 // p r od u ct i o n r a t e p er hour

12 ocn = 2 // o p r e a t o r c o s t p er hour13 m c n = 6 2 // m ac hi ne c o s t14 r n = 16 // p r od u ct i o n r a t e p er hour15 P o = ( o co + m co ) / r o // l a b ou r and m ach ine c o s t p er

u n i t on o l d ma ch in e i n Rs16 P n = ( o cn + m cn ) / r n // l a b ou r and m ach ine c o s t p er

u n i t on new m ac hi ne i n Rs17 i = 6 // i n t e r e s t on i nv es tm en t18 i = i / 1 0 0

19 t = 6 // a nn ua l t a x es

20 t = t / 1 0 021 d = 1 0 // a nn ua l a l lo w a n ce f o r d e p r e c i a t i o n22 d = d / 1 0 0

23 m = 3 // a nn ua l a l l o wa n c e f o r m ai nt en an ce24 m = m / 1 0 0

42



25 n = ( ( Cn - S n ) +( Ko - S o ) ) /( ( Nn * ( Po - P n ) ) - C n *( i + t +d + m ))

26 printf ( ” \n The n umber o f y e a r s i n w hi ch t h e newm ac hi ne w i l l p ay 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 = 80 00 0 // c o s t o f new m achi ne i n s t a l l e d andt o ol e d i n Rs

3 nel = 2 // number o f e n g in e l a t h e s4 c = 3 20 00 * ne l // f i r s t c o s t o f e ng i ne l a t h e5 N = 4000 // a n n ua l p r o du ct io n o f t u r r e t l a t h e6 n = 3800 // a nn ua l p r od u ct i o n i n e ng i ne l a t h e7 n hp1 = 4 / / hp m ot or8 L = 2 25 6* n hp 1 // a n n ua l l ab ou r c o st o f t u r r e t l a t h e9 w = 5 // wage i n p er h ou r

10 t im e = 2 30 0 // h o ur s11 l = t im e * ne l *w // l ab ou r c o st o f e ng i n e l a t h e12 n h p2 = 2 .5 / / hp m ot or13 p r = 0. 35 // power r a t e i n kwh

14 p = ( n e l * n hp 2 * 7 4 6 * t i me * p r ) / 1 0 00 // p ower c o s t15 P = ( n h p1 * 7 46 * t i m e * p r ) / 1 00 0 // po we r c o s t16 F = 480 / / s a v i n g17 I = 6/100 // i n t e r e s t r a t e18 T = 4/ 10 0 // t a x r a t e19 D = 1 0/ 10 0 // a l l ow an c e f o r d e p r e c i a t i o n i n e ng in e

l a t h e20 M = 6/ 10 0 // a l l o wa n c e f o r m ai nt en an ce i n e n gi n e

l a t h e21 B = 5 5/ 10 0 // l a bo ur burden i n e ng i ne l a t h e

22 i = 6/100 // i n t e r e s t r a t e23 t = 4/ 10 0 // t a x r a t e24 d = 1 0/ 10 0 // a l l ow an 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 e25 m = 6/ 10 0 // a l lo w a n ce f o r m ai nt en an ce i n t u r r e t

43



l a t h e

26 X = ( L + B *L + P + C*( I+ T+ D+ M) - F )/ N27 x = ( l+ l* B + p + c *( i+ t+ d+ m) )/ n

28 printf ( ” \n U ni t p r o du ct io n c o st on t u r r e t l a t he = Rs%0 . 2 f p er p i e c e \n U ni t p r od u ct i o n c o s t on e ng i nel a t h e = Rs %0 . 2 f p e r p i e c e ” , X , x )

29 disp ( ” T u rr et l a t h e w i l l be m ore e c on o mi c al th an twoe n gi n e l a t h e ” )


Scilab code Exa 5.18 Calculate maximum investment on turret lathe

1 clc

2 X = 9.16 // p ro du ct io n c o st on t u r r e t l a t h e3 N = 4000 // a nn u al r e q u ir e m e nt4 c = X * N // c os t f o r 4 000 p i e c e s on t u r r e t l a t h e5 n = 3800 // p r o d uc t i o n o f e ng i ne l a t h e6 l = 23 00 0 // l a bo u r c o s t7 p = 3002 // p ower c o s t8 i = 6 // i n t e r e s t r a t e

9 i = i / 1 0 010 t = 4 // t a x r a t e11 t = t / 1 0 0

12 d = 1 0 // a ll ow an ce 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 e13 d = d / 1 0 0

14 m = 6 // a l lo w an c e f o r m ai nt en an ce i n t u r r e t l a t h e15 m = m / 1 0 0

16 b = 5 5/ 10 0 / / l a b o u r b u rd e n17 a = i +t +d +m

18 t c = 6 40 00 // f i r s t c o s t o f e ng i ne l a t h e

19 c 1 = ( N *( l * ( 1 + b ) + p ) ) /n + ( t c * a ) // c o s t f o r e ng in el a t h e20 s = c1 - c // s a v i n g s21 amt = s /a // a mount i n v e st e d i n t u r r e t l a t h e o ve r

t h e c os t o f e ng in e l a th e

44



22 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 ve r t he

c o s t o f e ng i ne l a t h e = Rs %d” , amt )23 / / A nswers v ar y d ue t o ro und o f f e r r o r

Scilab code Exa 5.19 To find years for new machine

1 clc

2 C n = 6 00 00 // c o s t o f new ma ch in e3 S n = 50 00 // s c r ap v a lu e o f new m ach ine

4 S o = 10 00 // s c ra p v al u e o f o l d machi ne5 N n = 2 00 00 0 // a n n u al p r o d u c t io n6 I = 1 0 // i n t e r e s t r a t e7 I = I / 1 0 0

8 M = 7 // a l l ow a n ce f o r m ai nt en an e9 M = M / 1 0 0

10 T = 6 // a nn ua l t a x es11 T = T / 1 0 0

12 D = 1/10 // a l l ow an ce f o r d e p r e c i a t i o n13 l co = 300 // l a bo u r c h ar g es f o r o l d machi ne14 m = 1 2 / / m on th s

15 r c o = 1 50 00 // r un ni ng c h a r ge s f o r o l d ma ch in e16 p r o = 5 00 00 // p r o d uc t i o n r a t e f o r o l d machine17 l cn = 500 // l a b ou r c h a rg e s f o r new ma ch in e18 r c n = 1 00 00 // r un ni ng c h a r ge s f o r o l d ma ch in e19 p rn = 2 00 00 0 // p ro du ct i o n r a te f 20 Po = ( lco * m + rco ) /pro // l a b ou r and m ach ine c o s t on

o l d m ac hi ne21 Pn = ( lcn * m + rcn ) /prn // l a b ou r and m ach ine c o s t on

new mac hi ne22 n = (( C n - S n ) - S o ) / (( N n * ( Po - P n ) ) - C n * ( I + T + D +M ) ) / / y e a r s

23 printf ( ” \n Y ea rs i n wh ich new m ach ine w i l l pay f o ri t s e l f = %0. 2 f y e ar s ” , n )

45



Scilab code Exa 5.20 To find cost and pieces

1 clc

2 a = 1.50 / / s a v i ng i n l a b ou r3 b = 5 5 / 1 0 0 // b ur de n a p p l i e d on l a b ou r4 T = 4/ 10 0 // a l lo w a n ce f o r t a xe s5 M = 5/ 10 0 // a l l o wa n c e f o r m ai nt en an ce6 I = 8/ 10 0 // i n t e r e s t r at e7 D = 5 0/ 10 0 // a l l ow an c e f o r d e p r e c i a t i o n8 H = 2 // y e ar s t o a mo r t i z e t he i nv es tm en t9 S = 5 0 // y e a r l y c o s t f o r s e t up

10 C = 3000 / / f i r s t c o s t

11 N 1 = ( C * ( I + T + M +D ) + S ) / ( a * (1 + b ) ) // a n nu al p r o d u c t io nwhen 1 r un i s made

12 r = 5 // number o f r u n s13 N 2 = ( C * ( I + T + M +D ) + S * r ) /( a * ( 1 + b ) ) // a n n ua l

p r o d u c ti o n when 1 r un i s made14 D 1 = 1 00 /1 00 // a l l ow an c e f o r d e p r e c i a t i o n15 N 3 = ( C * ( I + T + M + D1 ) + S ) / ( a * (1 + b ) ) / / p r o d u c ti o n when D

= 1 0 016 n 1 = 15 30 // p i e c e s17 C 1 = ( n 1 * ( a * (1 + b ) ) - S ) /( I + T + M + D 1 ) // e c o n om i c a l

i n v e s t m e n t18 n 2 = 950 // p i e c e s19 a 1 = 2 // l a bo u r c o s t20 r 1 = 6 // number o f r u n s21 S 1 = r1 *S // y e ar l y c o st22 V = n 2 * a 1 * (1 + b ) - C * ( I + T +M + D ) - S 1 // p r o f i t23 printf ( ” \n Number o f p i e c e s when o ne r un i s made and

c o s t i s Rs 3 00 0 = %d p i e c e s ” , N 1 )

24 printf ( ” \n Annu al p r o d u c ti o n when 5 r u ns a r e madep e r y e a r = %d p i e c e s ” , N 2 )

25 printf ( ” \ nAnnua l p r o d u c ti 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 ” , N 3 )26 printf ( ” \ n Ec on om ic al i n v es t m en t when 1 53 0 p i e c e s f o r

s i n g l e run w i t h s a vi n gs Rs 1 . 5 0 p er p i e c e = Rs%d” , C 1 )

27 printf ( ” \ nAnnua l p r o f i t when 9 50 p i e c e s made p e r

46



y ea r i n 6 r un s a nd s av i n g i n l ab ou r c o st Rs 2 p e r

p i e c e = Rs %d p er y ea r ” , V )28 / / ’ Answers v a ry due t o round o f f e r ro r ’

Scilab code Exa 5.21 To find number of components

1 clc

2 a = 0. 12 5 // s a vi n g i n l a bo u r c o s t p er u n it3 b = 0.4 // o ve rh ea d a p p l i e d on d i r e c t l a bo u r s av ed

4 D = 1/2 // a l l ow an c e f o r d e p r e c i a t i o n5 C = 2400 / / f i r s t c o s t6 I = 6/ 10 0 // i n t e r e s t r at e7 T = 4/ 10 0 // a l lo w a n ce f o r t a xe s8 M = 1 0/ 10 0 // a l l ow a n ce f o r m ai nt en an ce9 S = 8 0 // c os t o f s e t up

10 N = ( C *( I + T +D + M )+ S ) /( a * ( 1+ b ) ) // p i e c e s p er y ea r11 t = N *2 // t o t a l number o f p i e c e s12 printf ( ” \n T o ta l number o f p i e c e s p ro du ce d = %d” , t

)


Scilab code Exa 5.22 To find number of components

1 clc

2 a = 0. 12 5 // s a v i n g i n l ab ou r c o st p er u n i t3 b = 0.4 // o ve rh ea d a p p l i e d on d i r e c t l a bo u r s av ed4 D = 1/2 // a l l ow an c e f o r d e p r e c i a t i o n5 C = 2400 / / f i r s t c o s t

6 I = 6/ 10 0 // i n t e r s t r at e7 T = 4/ 10 0 // a l lo w a n ce f o r t a xe s8 M = 1 0/ 10 0 // a l l ow a n ce f o r m ai nt en an ce9 n = 6 // number o f b a c h es

10 S = 8 0 // c os t o f s e t up

47



11 s 1 = S * n // t o t a l s e t up c o s t

12 N = ( C * ( I + T + D+ M ) + s 1 ) /( a * ( 1 + b ) ) // p i e c e s13 t = N *2 // t o t a l number o f p i e c e s14 printf ( ” \n T o ta l number o f p i e c e s p ro du ce d = %d” , t

)


Scilab code Exa 5.23 To find time and profit

1 clc2 C 1 = 20 00 / / f i r s t c o st s ma l l t o o l i n Rs3 N = 5000 // p a r t s p er y ea r4 n = 5 // number o f b a t c h es5 S = 50* n // c os t o f s e t up6 a = 0.15 // s av i n g i n l ab ou r c o st p er u ni t7 b = 5 0/ 10 0 // burde n a p p l i e d on d i r e c t l ab o ur s av ed8 I = 1 0/ 10 0 // i n t e r e s t r a t e9 T = 5/ 10 0 // a l lo w a n ce f o r t a x

10 M = 1 0/ 10 0 // a l l ow a n ce f o r m ai nt en an ce11 H = C 1 / (( N * a * ( 1 + b ) ) - ( C1 * ( I + T + M ) ) - S ) // y e ar s

12 C 2 = 16 00 // c os t o f f i x t u r e13 D = 1/ H // a l l ow an c e f o r d e p r e c i a t i o n14 V = N * a *( 1+ b ) - C2 * ( I +T + D +M ) - S // p r o f i t15 printf ( ” \n Number o f y e a r s t ak en by f i x t u r e o f Rs

2 0 00 = %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 = R s %d” , H ,V )

Scilab code Exa 5.24 To find minimum number of components

1 clc

2 c 1 = 3 // machi ne c o s t p er component u s in g e x i s t i n ge ui p me nt i n Rs

3 c 2 = 1 // machi ne c o st u si ng f i x t u r e i n R s

48



4 s = c 1 - c 2 // s a vi n g i n machi ne c o s t p er p i e c e

5 f = 1 00 0 // c os t o f f i x t u r e i n R s6 N = f /2 / / c o m p on e n ts7 printf ( ” \n M inimum n um be r o f c o m p on e n t s t o b e

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

Scilab code Exa 5.25 To calculate number of pieces

1 clc2 C = 1000 // c os t o f f i x t u r e3 C o = 700 // c os t o f o l d f i x t u r e4 C s = 250 // s c r a p v al ue5 a = 1 0 // s a vi n g p er p i e c e i n p a i sa6 a = a / 1 0 0

7 b = 3 0 // o ve rh ea d a p p li e d on d i r e c t l a bo u r s av ed8 b = b / 1 0 0

9 I = 8 // i n t e r e s t r a t e10 I = I / 1 0 0

11 M = 3 // a l l ow a n ce f o r m ai nt en an ce

12 M = M / 1 0 013 T = 1 2 // a l l ow an ce f o r t a x14 T = T / 1 0 0

15 H = 3/2 // a m o r t i za t i o n16 D = 1/ H // a l l ow an c e f o r d e p r e c i a t i o n17 N = ( C * ( I + T + D+ M ) + ( Co - C s ) * I ) / ( a * ( 1+ b ) ) // p i e c e s p e r

y e a r18 printf ( ” \n Number o f p i e c e s w hi ch must be p ro du ce d

t o b re ak e ve n s o t h at 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 er y ea r ” , N )


Scilab code Exa 5.26 To find cost for new fixture

49



1 clc

2 N = 9000 // number o f p i e c e s3 C o = 700 // c os t o f o l d f i x t u r e4 C s = 250 // s c r a p v al ue5 a = 1 0 // s a vi n g p er p i e c e i n p a i sa6 a = a / 1 0 0



11 M = 3 // a l l ow a n ce f o r m ai nt en an ce12 M = M / 1 0 0

13 T = 1 2 // a l l ow an ce f o r t a x14 T = T / 1 0 0

15 H = 3/2 // a m o r t i za t i o n16 D = 1/ H // a l l ow an c e f o r d e p r e c i a t i o n17 C = ( N * a * ( 1+ b ) - ( Co - C s ) * I ) / ( I + T + D + M ) // c o s t i n Rs18 printf ( ” \n Co st f o r new f i x t u r e = Rs %d” , C )


Scilab code Exa 5.27 find time to amortize fixture

1 clc

2 n = 6500 // y e a r l y p r o du c ti o n3 c = 1350 // c os t o f f i x t u r e4 a = 1 0 // s a vi n g p er p i e c e i n p a i sa5 a = a / 1 0 0


8 I = 8 // i n t e r e s t r a t e

9 I = I / 1 0 010 M = 3 // a l l ow a n ce f o r m ai nt en an ce11 M = M / 1 0 0


50



14 c o = 700 // c os t o f o l d f i x t u r e

15 c s = 250 // s c r a p v al ue16 H = ( c ) / (( n * a * ( 1 + b ) ) - I *( c o - c s ) - c *( I + T + M ) ) //a m ot i za t io n i n y e ar s

17 printf ( ” \n Time t ak en t o a m or t iz e t he f i x t u r e = %0 . 1f y e a rs ” , H )

Scilab code Exa 5.28 To find profit

1 clc2 n = 9000 // p ro du ct io n o f p i e c e s p e r y ea r3 c = 1000 // f i x t u r e c o s t s4 C o = 700 // c os t o f o l d f i x t u r e5 C s = 250 // s c r a p v al ue6 a = 1 0 // s a vi n g p er p i e c e i n p a i sa7 a = a / 1 0 0



12 M = 3 // a l l ow a n ce f o r m ai nt en an ce13 M = M / 1 0 0


16 h = 1.5 // a m o r t i za t i o n17 D = 1/ h // a l l ow an c e f o r d e p r e c i a t i o n18 V = ( n * a * ( 1+ b ) ) - ( c * ( I + T + D+ M ) ) - ( ( Co - C s ) * I ) // p r o f i t19 printf ( ” \n p r o f i t = Rs %d ” , V )


Scilab code Exa 5.29 To find BEP Cost and Components

1 clc

51



2 f c1 = 1 00 00 0 // f i x e d c o s t i n Rs

3 v c1 = 100 // v a r i a b l e c o s t i n Rs p e r u ni t4 s p = 200 / / s e l l i n g p r i c e i n Rs p er u ni t5 q 1 = f c1 / ( sp - v c 1 ) // q ua n ti t y o f p r o d uc ti o n a t b re ak

e ve n p o i n t6 f c2 = 1 25 00 0 // f i x e d c o s t i n Rs7 v c2 = 90 // v a r i a b l e c os t i n Rs p e r u ni t8 q 2 = f c2 / ( sp - v c 2 ) // q ua n ti t y o f p r o d uc ti o n a t b re ak

e ve n p o i n t9 p = 20 00 0 // p r o f i t i n Rs

10 q3 = ( fc1 + p )/( sp - vc1 ) // q u an t i t y o f p r o d uc t i o n a tp r o f i t o f Rs 2 00 00

11 printf ( ” \n B rea k e ven p o i nt = %d p i e c e s \n I f f i x e dc o s t i s 1 2 5 00 0 and v a r i a b l e c o s t i s Rs 9 0 p e ru n i t t he n b re ak e ve n p o i n t = %d p i e c e s \n Numbero f c omponent s t o g e t p r o f i t o f Rs 2 00 00 = %dp i e c e s ” , q1 , q2 , q3 )

Scilab code Exa 5.30 To find break even point

1 clc2 f c 1 = 1 20 00 // f i x e d c o st f o r machi ne A i n R s3 f c 2 = 4 80 00 // f i x e d c o st f o r machi ne B i n Rs4 n 1 = 6 // u n i t p ro du ct i o n c o s t i n Rs p e r p i e c e f o r

m a c h i n e A5 n 2 = 1.2 // u n i t p ro du ct io n c o st i n Rs p e r p i e c e f o r

m a c h i n e B6 q = ( fc2 - f c1 ) / ( n1 - n 2 ) // b re ak e ve n p o i nt7 printf ( ” \n B rea k e ve 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 / / c ap st an l a t h e

3 t c1 = 300 // t o t a l c os t i n Rs4 m c1 = 2.5 // m a t e r i a l c o s t p e r p i e c e i n Rs5 o lc1 = 5 // o p er a t i on l ab ou r c o st p er hour i n Rs6 ct1 = 5 // c y c l e t im e p er p i e c e i n min .7 s lc1 = 20 // s e t t i n g up l ab ou r c o s t i n Rs p er hour8 st1 = 1 // s e t t i n g up t i m e i n ho ur9 m o 1 = 3 00 / 10 0 // m ach in e o ve r h ea ds o f o p e r a t i on

l a bo u r c o s t10 o 1 = m o1 * o lc 1 // o ve rh ea ds o f c ap st an l a t h e i n Rs

p e r h ou r11 fc1 = tc1 + s lc1 * st1 + o1 * st1 // f i x e d c os t o f

c ap st an l a t h e i n Rs12 v c1 = m c1 + ( o lc 1 * ct 1) / 60 + ( o1 * ct 1 ) /6 0 // v a r i a b l e

c o st i n Rs13 // Au to ma ti c ( s i n g l e s p i n d l e )14 t c2 = 300 // t o t a l c os t i n Rs15 c c2 = 1 50 0 // c o st o f cams i n R s16 m c2 = 2.5 // m a t e r i a l c o s t p e r p i e c e i n Rs17 o lc2 = 2 // o p er a t i on l ab ou r c o st p er hour i n Rs18 ct2 = 1 // c y c l e t im e p er p i e c e i n min .19 s lc2 = 20 // s e t t i n g up l ab ou r c o s t i n Rs p er hour20 st2 = 8

// s e t t i n g up t i m e i n ho ur21 m o 2 = 1 0 00 / 10 0 // m achi ne o v er h ea ds o f o p e r a t i o nl a bo u r c o s t

22 o 2 = m o2 * o lc 2 // o ve rh ea ds o f s i n g l e s p i n d l e i n Rsp e r h ou r

23 fc2 = tc2 + cc2 + sl c2 * st2 + o2 * st2 // f i x e d c os t o f s i n g l e s p i n d l e i n Rs

24 v c2 = m c2 + ( o lc 2 * ct 2) / 60 + ( s lc 2 ) /6 0 // v a r i a b l ec o st i n Rs

25 q = ( fc2 - f c1 ) / ( vc1 - v c2 ) // b re ak e ve n q u a n ti t y26 printf ( ” \n B rea k e ve n q u a n t i t y f o r a co mpo nen t w hi ch

can be p ro du ce d on e i t h e r t he c ap st an l a t h e o rs i n g l e s p i n d l e a u to m at ic = %d p i e c e s ” , q )

53



Scilab code Exa 5.32 To do break even analysis

1 clc

2 / / E ng in e l a t h e3 t = 1 2 // t im e / p i e c e i n min .4 l = 7 / / o v er h ea d c o s t / h r5 o = 4 // d i r e c t l a bo u r c o s t / hr6 s = 2 // s e t up t i m e i n hour7 s r = 8 // s e t up r a t e p e r8 / / t u r r e t l a t h e9 T = 5 / / t im e / p i e c e i n min .

10 L = 5 / / o v er h ea d c o s t / h r11 O = 8 // d i r e c t l a bo u r c o s t / hr12 S = 8 // s e t up t i m e i n hour13 S R = 8 // s e t up r a t e p e r14 q = 6 0 *( S * S R - s * s r ) /( t * ( l + o ) - T *( L + O ) ) // b r ea k e ve n

p o i n t15 q = round ( q )

16 printf ( ” \n B rea k e ve 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 clc

2 f c 1 = 8 00 00 // f i x e d c os t f o r t u r r e t l a t h e i n Rs3 f c 2 = 3 20 00 // f i x e d c o s t f o r e ng in e l a t h e i n Rs4 n 1 = 16 // p r o du c t io n o f p i e c e s p er y ea r i n t u r r e t

l a t h e

5 n 2 = 10 // p r o du c t io n o f p i e c e s p er y ea r i n e ng in el a t h e

6 vc1 = 2 // o p e r a t o r s c o st i n t u r r e t l a t h e7 v c2 = 2.5 // o p e r a t o r s c o st i n e ng i n e l a t h e8 Q = poly (0 , ’Q ’ )

54



9 Q = roots ( ( f c 1 + 1 / n 1 * v c 1 * Q ) - ( f c 2 + 2 . 5 * Q / 1 0 ) )

10 printf ( ” \n B rea k e ve n p o i n t = %d p i e c e s ” , Q )

Scilab code Exa 5.34 To determine the point

1 clc

2 s t1 = 15 // s e t up t i me f o r e ng i n e l a t h e i n min .3 u t1 = 15 // u n i t t i m e f o r e ng i n e l a t h e i n min .4 s t2 = 90 // s e t up t ime f o r a ut om at ic l a t h e i n min .

5 u t2 = 1.5 // u ni t t i m e f o r e ng i n e l a t h e i n min .6 q = ( st2 - s t1 ) / ( ut1 - u t2 ) // q u an t i t y o f p r o d uc t i o n7 printf ( ” \n The p o i nt a t whi ch t he a ut om a ti c l a t h e

w i l l b e j u s t i f i e d = %0 . 2 f ” , q )


Scilab code Exa 5.35 To find quantity of pieces

1 clc2 / / A ut om at ic l a t h e3 p = 3 0 // number o f p i e c e s p ro du ce d p er ho ur4 l = 4 // l ab ou r r a te p er ho ur i n R s5 d = 4.50 // h o ur l y d e p r e c i a t i o n r a t e p er machine i n

hour6 s = 4 // s e t up t i m e i n hour7 / / t u r r e t l a t h e8 P = 1 0 // number o f p i e c e s p ro du ce d p er ho ur9 L = 4 // l ab ou r r a te p er ho ur i n R s

10 D = 1.50 // h o ur l y d e p r e c i a t i o n r a t e p er machine i n

hour11 S = 2 // s e t up t i m e i n hour12 q = ( P * p * ( S * L+ S * D - s *l - s * d ) ) / ( P *( l + d ) - p * ( L + D) ) //

q ua nt i t y o f p i e c e s a t b r e a k e v e n p oi nt

55



13 printf ( ” \n Q ua nt it y o f p i e c e s a t Break e ve n p o in t =

%d p i e c e s ” , q )

Scilab code Exa 5.36 To determine quantity of production

1 clc

2 P a = 8.4 // u n i t t o o l p r o c e s s c o st f o r method A inRs

3 P b = 14 .8 // u ni t t o o l p r oc e ss c o st f o r method B i n

Rs4 T a = 64 80 // t o t a l t o o l c o s t f o r method A i n Rs5 T b = 16 16 // t o t a l t o o l c o st f o r method B i n Rs6 q = ( Ta - T b ) /( Pb - P a ) // b re ak ev en p o i nt7 printf ( ” \n Q ua nt it y o f p r o du c ti o n a t b re ak e ve 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 clc

2 / / m ac hi ne A3 i c 1 = 5 00 00 / / i n i t i a l c o s t4 h oc1 = 10 // h ou rl y o p e ra t i ng c h ar g es5 p p1 = 5 // p i e c e s p ro du ce d p er ho ur6 i = 1 5 // i n t e r e s t r a t e7 i = i / 1 0 0

8 o h = 20 00 // o p e r a t i n g h ou rs9 f c1 = i c1 * i // f i x e d c o s t

10 v c1 = oh * h oc 1 // v a r i a b l e c o s t

11 t c1 = f c1 + vc 1 // t o t a l c ha r g e s12 a o 1 = oh * pp 1 // a n nu a l o u tp ut13 c 1 = t c1 / ao 1 // c o s t p e r u n i t14 / / m ac hi ne B15 i c 2 = 8 00 00 / / i n i t i a l c o s t

56



16 h oc2 = 8 // h o ur l y o p e r a t i n g c h ar g es

17 p p2 = 8 // p i e c e s p ro du ce d p er ho ur18 f c2 = i c2 * i // f i x e d c o s t19 v c2 = oh * h oc 2 // v a r i a b l e c o s t20 t c2 = f c2 + vc 2 // t o t a l c ha r g e s21 a o 2 = oh * pp 2 // a n nu a l o u tp ut22 c 2 = t c2 / ao 2 // c o s t p e r u n i t23 printf ( ” \n ( i ) C os t p er u n i t f o r m achi ne A = Rs %0 . 2

f \n C os t p e r u n i t m ac hi ne B = Rs %0 . 2 f ” , c 1 , c 2 )

24 disp ( ” m achi ne B w i l l be p r e f e r r e d ” )

25 / / m ac hi ne A26 ao3 = 4 00 0 // a n nu a l o u tp ut

27 o c 3 = a o3 * h o c1 / p p 1 // o p e ra t i ng c h ar g es28 t c3 = o c3 + fc 1 // t o t a l a nn ua l c ha rg e29 c 3 = t c3 / ao 3 / / c o s t / p i e c e30 / / m ac hi ne B31 ao4 = 4 00 0 // a n nu a l o u tp ut32 o c 4 = a o4 * h o c2 / p p 2 // o p e ra t i ng c h ar g es33 t c4 = o c4 + fc 2 // t o t a l a nn ua l c ha rg e34 c 4 = t c4 / ao 4 / / c o s t / p i e c e35 printf ( ” \n ( i i ) Co st p er u n i t f o r ma ch in e A = Rs %0

. 2 f \n C os t p e r u n i t m ac hi ne B = Rs %0 . 2 f ” , c 3 , c 4 )

36 disp (” m achi ne A w i l l be p r e f e r r e d ”

)

37 A = h oc 1 / pp 1 // o p e ra t i ng c o s t p er p i e c e on machineA

38 B = h oc 2 / pp 2 // o p e ra t i ng c o s t p er p i e c e on machineB

39 Q = fc2 - fc1 // a nn u al p r o d u c t io n40 printf ( ” \n ( i i i ) Annu al p r o d u c ti o n t o make c o s t p e r

p i e c e e q ua l f o r two m ac hi ne s = %d p i e c e s ” , Q )

Scilab code Exa 5.38 To find BEP and various sales

1 clc

2 a s = 8 00 00 // a nn u al s a l e s i n Rs

57



3 v c = 6 40 00 // v a r i a b l e e xp en se s i n R s

4 c = 16 00 0 // c o n t r i b ut i o n i n Rs5 f c = 2 40 00 // f i x e d e xp en se s i n Rs6 l = 8000 // l o s s e s i n Rs7 p = 9000 // p r o f i t i n Rs8 s1 = fc + vc // s a l e s a t B . E . P i n Rs9 s2 = ( fc + vc + p )/ 0.945 // s a l e s a t n et i ncome o f

Rs9000 and c o r p o r a te t ax r a t e b e in g 5 . 5%10 q = 10 00 0 // q ua n t it y o f u n i t s11 s p = ( fc + vc ) /q / / s e l l i n g p r i c e p e r u ni t i n Rs12 printf ( ” \n S a l e s a t b re ak e v e n p o in t = %d u n i t s ” ,

s1 )

13 printf ( ” \n S a l e s a t n et i ncome o f Rs9000 andc o r p o r at e t a x r a t e b ei ng 5 . 5 = Rs %0 . 2 f \n S a l e sp er u n i t i f B . E . P b ro ug ht down t o 1 00 00 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 clc

2 f c = 5 50 00 // f i x e d c o s t i n Rs3 v c = 45 // v a r i a b l e c o s t p e r p i ec e i n Rs4 s p = 100 / / s e l l i n g p r i c e p e r p i e c e i n Rs5 p = ( vc / sp ) * 10 0 // p e r c e nt ag e o f v a r i a b l e c o st t o6 p m = 100 - p // p r o f i t margi n7 b e p = ( ( 5 5 00 0 / 5 5) * 1 00 ) / 1 0 0 // B re ak e ve n p o i n t8 printf ( ” \n B rea k e ve n p o i n t = %d p i e c e s ” , b ep )

Scilab code Exa 5.40 To calculate economic lot size

1 clc

2 f 1 = 335 // f i x e d c o s t i n Rs f o r c a ps t a n l a th e3 k = 0.25 // s t o ck c a r r y i n g f a c t o r i n p a i s e p e r p i e c e

58



4 k = k / 1 0 0

5 N1 = sqrt ( f 1 / k ) // p i e c e s f o r c ap st an l a t h e6 a 1 = 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 ps t anl a t h e

7 t c1 = a 1 + f1 / N 1 +k * N1 // t o t a l c o s t f o r c a ps t a n l a th e8 f 2 = 21 20 // f i x e d c os t i n Rs f o r t u r r e t l a t he9 N2 = sqrt ( f 2 / k ) // p i e c e s f o r t u r r e t l a t h e

10 a 2 = 2 .8 63 // v a r i a b l e c o s t p e r p i ec e f o r t u r r e tl a t h e

11 t c2 = a 2 + f2 / N 2 +k * N2 // t o t a l c os t f o r t u r r e t l a t h e12 printf ( ” \n T o t a l c o s t p e r p i e c e f o r c a ps t an l a t h e =

Rs %0. 2 f \n T ot 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 / / A nswers v ar y d ue t o ro und o f f e r r o r

Scilab code Exa 5.41 To find EOQ and total cost

1 clc

2 R = 5 0 0 // c o s t o f o r d e r i n g i n Rs p e r o rd er3 A = 1 2 0 0 0 / / a n nu a l c on su mp ti on u n i t s

4 C = 3 . 0 0 // u ni t c o s t o f i t e m5 K =3 // u n i t s t o r a g e c o st6 I 1 = 0 . 2 // i n t e r e s t r a t e7 function y = f ( N )

8 function G = f 2 ( N )

9 G = C * A + I 1 * C * N / 2 + K * N / 2 + A * R / N // t o t a l c o s t p e r y e a r10 e n d f u n c t i o n

11 y = d e r i v a t i v e ( f 2 , N )

12 e n d f u n c t i o n

13 funcprot (0)

14 N = fsolve ( 2 0 0 0 , f )15 O = A / N // number o f o r d e r s16 N 1 = 24 00 // u n i t s17 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 I 2 = ( 2* R * A ) /( C * N1 ^ 2 )

19 printf ( ” \n Eco no mi c o r d e r q u a n t i t y = %d u n i t s \n T ot lc o s t = Rs %d p er y ea r \n I = %0 . 4 f ” , N 1 , t c , I 2 )

20 disp ( ” I t i s c l e a r t h a t i nv en to ry c os t w i l l g e ti n c r e as e d v er y g r e a t l y ” )

Scilab code Exa 5.42 Determine optimum lot size

1 clc

2 A = 40 00 0 // number o f u n i t s p er y ea r3 I = 2 5 // c a r r y i n g c o st i n p e r c e nt4 I = I / 1 0 0

5 C 1 = 8 // c o st f o r 0 < N < 10 00 p er u ni t i n Rs6 C 2 = 7.5 // c o st f o r 1000 < N < 10 00 0 p er u n i t i n Rs7 C 3 = 7. 25 // c os t f o r N >= 10 00 0 p er u n i t i n Rs8 R = 250 // o r de r i n g c o st p er o r d er i n Rs9 N = 10 00 0 // u n i t s

10 N1 = sqrt ( 2 * R * A / ( I * C 3 ) ) // o pt im al q u an t it y f o rl o w e st c u rv e

11 G 1 = C 3 * A +( A * R ) / N + I * C3 * N / 2 // t o t a l c os t i n Rs

12 N2 = sqrt ( 2 * R * A / ( I * C 2 ) ) // o pt im al q u an t it y f o rh i g h e r c ur ve

13 G 2 = C 2 * A +( A * R ) / N 2 + I * C2 * N 2 / 2 // t o t a l c os t i n Rs14 N3 = sqrt ( 2 * R * A / ( I * C 1 ) ) // o pt im al q u an t it y f o r

h i g h e s t c ur ve15 G 3 = C 1 *A + ( A* R ) +1 // t o t a l c os t i n Rs16 printf ( ” \n T o ta l c o st f o r l o w e s t c o st c ur ve = Rs %0

. 2 f \n T ot al c o s t f o r n ex t h i gh e r c ur ve = Rs %0 . 2 f \n T ot al c o s t f o r h i g h e st c ur ve = Rs %0 . 2 f ” , G1

, G 2 , G 3 )

17 disp ( ” Comparing a l l t o t a l c o st l o w e s t i s Rs3 0 0 , 0 62 . 50 f o r an o r de r q ua n ti t y o f 1 0 , 0 00 . ” )

18 disp ( ”N = 1 0 , 00 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 clc

2 c = 50 00 0 / / c o m po n e n ts3 R = 5 0 0 // c o s t o f o r d e r i n g i n Rs p e r o rd er4 A = 1 2 0 0 0 / / a n nu a l c on su mp ti on u n i t s5 C = 3 . 0 0 // u ni t c o s t o f i t e m6 K = 1 . 5 0 // u n i t s t o r a g e c o st7 I = 0 . 2 // i n t e r e s t r a t e8 function y = f ( N )

9 function G = f 2 ( N )

10 G = 0 . 0 2 * N + 1 5 0 0 0 0 0 / N


12 y = d e r i v a t i v e ( f 2 , N )


14 funcprot (0)

15 N = fsolve ( 2 0 0 0 , f )

16 l = c / N // number o f l o t s17 l = ceil ( l )

18 l s = c / l // l o t s i z e19 printf ( ” \n The l o t s i z e = %d c o mp on en ts ” , l s )

61



Chapter 9

Limits Tolerences and Fits

Scilab code Exa 9.1 To find allowance and tolerence

1 clc

2 h 1 = 3 7. 52 // h ig h l i m i t o f h ol e i n mm3 h 2 = 3 7. 50 // low l i m i t o f h o l e i n mm4 s 1 = 3 7. 47 // h i g h l i m i t o f s h a f t i n mm5 s 2 = 3 7. 45 // low l i m i t o f s h a f t i n mm6 h t = h1 - h2 // h o le t o l e r e n c e i n mm

7 s t = s1 - s2 // s h a f t t o l e r e n c e i n mm8 a = h2 - s1 // a l l o w a n c e i n mm9 printf ( ” \n H ol 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 ow anc e = %0. 2 f mm” , ht

, st , a)

Scilab code Exa 9.2 Determine dimensions of shaft and hole

1 clc

2 t = 0. 07 5 // t o l e r e n c e i n mm3 h 2 = 75 // l ow l i m i t o f h o l e i n mm4 a = 0.10 // a l l o w a n c e i n mm

62



5 h 1 = h2 +t // h ig h l i m i t o f h ol e i n mm

6 s 1 = h2 - a // h i g h l i m i t o f s h af t i n mm7 s 2 = s1 - t // l ow l i m i t o f s h a f t i n mm8 printf ( ” \n High l i m i t o f h o l e = %0 . 3 f mm\n H ig h

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 clc2 t = 0. 22 5 // t o l e r e n c e i n mm3 h 2 = 75 // l ow l i m i t o f h o l e i n mm4 a = 0 .0 37 5 // i n t e r f e r e n c e i n mm5 h 1 = h2 +t // h ig h l i m i t o f h ol e i n mm6 s 2 = h1 +a // l ow l i m i t o f s h a f t i n mm7 s 1 = s2 +t // h i g h l i m i t o f s h af t i n mm8 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 . 4f mm” , h1 , s2 , s1 )

Scilab code Exa 9.4 Calculate fundamental deviations and tolerences

1 clc

2 s 1 = 50 // d i am e te r o f s t e p1 i n mm3 s 2 = 80 // d i am e te r o f s t e p2 i n mm4 d = ( s 1 * s2 ) ^ ( 1/ 2) / / mm5 i = ( 0 . 45 * ( d ) ^ ( 1 /3 ) + 0 . 0 0 1* d ) / 1 0 ^ 3 / / mm6 t 1 = 25* i // t o l e r e n c e f o r h o l e i n mm

7 t 2 = 16* i // t o l e r e n c e f o r s h a f t i n mm8 a 1 = 0 // f un da me nt al d e v i a t i o n f o r h o le i n mm9 a 2 = 5 .5 *( d ) ^ 0 .4 1 // f un da me nt al d e v i a t i o n f o r s h a f t

i n m i cr o ns10 a 2 = a2 / 10 ^4 / / mm

63



11 h 1 = 60 // l ow l i m i t o f h o l e i n mm

12 h 2 = h1 + t1 // h ig h l i m i t o f t o l e r e n c e i n mm13 s1 = h1 - t 2 // h i g h l i m i t o f s h a f t i n mm14 s 2 = s1 - t2 // low l i m i t o f s h a f t i n mm15 printf ( ” \n T o l er e n ce f o r h o l e = %0 . 3 f mm\n T o l er e n ce

f o r s h a f t = %0 . 3 f mm” , t1 , t2 )

16 printf ( ” \n Fu nd am en ta l d e v i a t i o n f o r h o l e = %0 . 2 f mm\n F un da men ta l 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 H igh 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 / / A nswers v ar y d ue t o ro und o f f e r r o r

Scilab code Exa 9.5 Find tolerences limits and clearance

1 clc

2 b = 3 0 // b a s i c s i z e i n mm3 s 1 = 0 .0 05 // maximum l i m i t o f s h a f t i n mm

4 s 2 = 0 .0 18 // minimum l i m i t o f s h a f t i n mm5 h 1 = 0 .0 20 // maximum l i m i t o f h o l e i n mm6 h 2 = 0.0 // minimum l i m i t o f h o l e i n mm7 t 1 = s2 - s1 // s h a f t t o l e r e n c e i n mm8 t 2 = h1 - h2 // h o le t o l e r e n c e i n mm9 S h = b -s1 // h i g h l i m i t o f s h af t i n mm

10 S l = b -s2 // l ow l i m i t o f s h a f t i n mm11 H h = b +h1 // h i gh l i m i t o f h o l e i n mm12 H l = b +h2 // l ow l i m i t o f h o l e i n mm13 c 1 = Hh - Sl / / maximum c l e a r a n c e i n mm

14 c 2 = Hl - Sh / / m inimum c l e a r a n c e i n mm15 printf ( ” \n B a s i c s i z e = %d mm\n S ha f t t o l e r e n c e = %0. 3 f mm\n H o le t o l e r e n c e = %0 . 3 f mm” , b , t 1 , t 2 )

16 printf ( ” \n High l i m i t o f s h a f t = %0 . 3 f mm\n Lowl i m i t o f s h a f t = %0 . 3 f mm\n H igh l i m i t o f h ol e =

64



%0 . 3 f mm \n Low l i m i t o f h o l e = %0 . 3 f mm” , S h , S l ,

H h , H l )17 printf ( ” \n Maximum c l e a r a nc e = %0. 3 f mm\n Minimumc l e a r a n c e = %0 . 3 f mm” , c 1 , c 2 )

Scilab code Exa 9.6 Determine limits of shaft and hole

1 clc

2 m in c = 0 .0 1 / / minimum c l e a r a n c e i n mm

3 b s = 25 // b a s i c s i z e i n mm4 m ax c = 0 .0 2 / / maximum c l e a r a n c e i n mm5 x = poly (0 , ’ x ’ )

6 y = 1 . 5 * x

7 x = roots ( y + 0 . 0 1 + x - 0 . 0 2 )

8 y = horner ( y , x )

9 // h o l e b a s i s s ys te m10 l o w_ h1 = bs // low l i m i t o f h ol e i n mm11 h ig h_ h1 = bs + y // h i gh l i m i t o f h o l e i n mm12 u _ s = l ow _h 1 - m i n c // u ppe r l i m i t o f s h a f t i n mm13 l ow _s 1 = u_s - x // l ow er l i m i t o f s h a f t i n mm

14 / / s h a f t b a s i s s y st e m15 h i gh _s = bs // h i g h l i m i t o f s h af t i n mm16 l ow _s 2 = bs - x // low l i m i t o f s h a f t i n mm17 l ow _ h2 = b s + mi nc // low l i m i t o f h ol e i n mm18 h i gh _ h2 = l ow _ h2 + y // h ig h l i m i t o f h ol e i n mm19 printf ( ” H ol e b a s i s s ys te m \n Lower l i m i t o f h o l e =

%d mm\n H ig he r l i m i t o f h o l e = %0 . 3 f mm\n H i gh erl 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” , l o w_ h 1 , h i g h _ h1 , u _ s , l o w _ s 1 )

20 printf ( ” \n S ha f t b a s i s sy st e m \n h ig h l i m i t o f s h a f t

= %0. 3 f mm\n l ow er l i m i t o f s h a f t = %0 . 3 f mm\nl o we r l i m i t o f h o l e = %0 . 3 f mm\n u ppe r l i m i t o f ho l e = %0. 3 f mm” , h ig h_ s , l o w_ s2 , l o w_ h2 , h i g h _ h2 )

65



Scilab code Exa 9.7 Determine dimensions of shaft and hole

1 clc

2 b s = 100 // b a s i c s i z e i n mm3 s 1 = 120 // d i am e te r o f s t e p1 i n mm4 s 2 = 80 // d i am e te r o f s t e p2 i n mm5 d = ( s 1 * s2 ) ^ ( 1/ 2) / / mm6 d = ceil ( d )

7 i = ( 0 . 45 * ( d ) ^ ( 1 /3 ) + 0 . 0 0 1* d ) / 1 0 ^ 3 / / mm8 t 1 = 16* i // t o l e r e n c e f o r h o l e i n mm9 t 2 = 25* i // t o l e r e n c e f o r s h a f t i n mm

10 G = ( 2. 5* ( d ) ^ 0. 34 ) / 1 0^ 3 // f un da me nt al d e v i a t i o n f o rh o l e i n mm

11 e = ( 11 *( d ) ^ 0 .1 1) / 1 0 ^3 // f un da me nt al d e v i a t i o n f o rs h a f t i n m ic ro ns

12 / / H ol e13 L Lh = bs + G // l ow er l i m i t o f h o l e i n mm14 H L h = L Lh + t1 // h i g h e r l i m i t o f h o l e i n mm15 / / s h a f t

16 U Ls = bs - e // u ppe r l i m i t o f s h a f t i n mm17 L L s = ULs - t 2 // l ow er l i m i t o f s h a f t i n mm18 printf ( ” \n l ow er l i m i t o f h o le = %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 pp er l i m i t o f s h af t =%0 . 3 f mm\n l o we r l i m i t o f s h a f t = %0 . 3 f mm” ,

L L h , H L h , U L s , L L s )


Scilab code Exa 9.8 Determine size of bearing and journal

1 clc

2 t b = 0 .0 05 // t o l e r e n c e on b e ar i ng i n mm3 t j = 0 .0 04 // t o l e r e n c e on j o u r n a l i n mm

66



4 a = 0. 00 2 / / a l l o w a n c e i n mm

5 / / h o l e −b a s i s s ys te m6 b = 100 // b a s i c s i z e i n mm7 B l = b // l ow er l i m i t o f b e a r i ng i n mm8 B h = Bl + tb // h i g h e r l i m i t o f b e a r i ng i n mm9 J h = 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

10 Jl1 = Jh - tj // l ow e r l i m i t o f j o u r n a l i n11 / / s h a ft −b a s i s s ys te m12 J u = b // upp e r l i m i t o f j o u r n a l i n mm13 J l2 = Ju - t j // l ow er l i m i t o f j o u r n a l i n mm14 B l = Ju +a // l ow er l i m i t o f b e a r i n g i n mm15 B u = Bl + tb // u ppe r l i m i t o f b e ar i ng i n mm

16 printf ( ” \n H ol e b a s i s s ys te m \n L ower l i m i t o f j o u r n a l = %d mm\n H ig he r l i m i t o f b e ar i ng = %0 . 3 f

mm\n H ig he r l i m i t o f j o u r n a l = %0 . 3 f mm \n L ow erl i m i t o f j o u r n a l = %0 . 3 f mm” , B l , Bh , J h , J l 1 )

17 printf ( ” \n s h a f t b a s i s sy s t e m \n u ppe r l i m i t o f j o u r n a l = %0 . 3 f mm\n l ow er l i m i t o f j o u rn a l = %0. 3 f mm\n l o we r l i m i t o f b e a ri n g = %0 . 3 f mm\nu pp er l i m i t o f b e a r i n g = %0 . 3 f mm” , Ju , J l2 , B l , B u

)

Scilab code Exa 9.9 Determine size of two mating parts

1 clc

2 / / H ol e−b a s i s s ys te m3 b = 100 // b a s i c s i z e i n mm4 i 1 = 0. 12 / / maximum i n t e r f e r e n c e i n mm5 i 2 = 0. 05 / / minimum i n t e r f e r n c e i n mm6 t = ( i1 - i 2) /2 // t o l e r e n c e i n mm

7 S h = b +i1 // up pe r l i m i t o f s h a f t i n mm8 H l = b // l ow er l i m i t o f h o l e i n mm9 H h = b + t // h i g h er l i m i t o f h ol e i n mm

10 S l1 = Sh - t // l ow er l i m i t o f s h a f t i n mm11 / / s h a ft −b a s i s s ys te m

67



12 S u = b // upp e r l i m i t o f s h a f t i n mm

13 Sl2 = b -t // l ow er l i m i t o f s h a f t i n mm14 Hl1 = b - i1 // l ow er l i m i t o f h o l e i n mm15 H u = Hl 1+ t // h i g h e r l i m i t o f h o l e i n mm16 printf ( ” \n H ol e b a s i s s ys te m \n u pp er l i m i t o f s h af t

= %0. 3 f mm\n l ow er l i m i t o f h o le = %0 . 3 f mm\nh 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 sh af t = %0. 3 f mm” , S h , Hl , H h , S l 1 )

17 printf ( ” \n S ha f t b a s i s sy st e m \n u ppe r l i m i t o f s h a f t = %0 . 3 f mm\n l ow er l i m i t o f s h a f t = %0 . 3 f

mm\n l ow er l i m i t o f h o l e = %0 . 3 f mm\n up pe r l i m i to f h o l e = %0 . 3 f mm” , S u , S l2 , H l1 , H u )

Scilab code Exa 9.10 Determine size of hole and shaft

1 clc

2 a a = 0. 04 / / a v er a ge a l l o wa n c e i n mm3 a = 0. 01 2 / / a l l o w a n c e i n mm4 M ax = aa + a / / maximum a l l o w a n c e i n mm5 M in = aa - a / / minimum a l l o w a n c e i n mm

6 t = ( Max - M in ) /3 // t o l e r e n c e i n mm7 t s = t // t o l e r e n c e i n s ha t i n mm8 t h = 2* t // t o l e r e n c e i n h o l e i n mm9 b = 100 // b a s i c s i z e i n mm

10 H l = b // l ow er l i m i t o f h o l e i n mm11 H u = b +th // up pe r l i m i t o f h ol e i n mm12 S u = b - 0 .0 28 // u ppe r l i m i t o f s h a f t i n mm13 S l = Su - ts // l ow er l i m i t o f s h a f t i n mm14 printf ( ” \n l ow er l i m i t o f h o le = %d mm\n u ppe r l i m i t

o f h o l e = %0 . 3 f mm\n uppe r l i m i t o f s h a f t = %0 . 3

f mm\n l o we r l i m i t o f s h a f t = %0 . 3 f mm” , H l , H u , S u, S l )

68



Chapter 11

Surface finish

Scilab code Exa 11.1 Calculate CLA value

1 clc

2 v = 15 00 0 // v e r t i c a l m a g ni f i c a t i o n3 h = 100 // h o r i z o n t a l m a g ni f i c a t i o n4 l = 0.8 // s a mp l in g l e n g t h i n mm5 a 1 = 160 // a r e a a bo ve datum l i n e i n mmˆ 26 a 2 = 90 // a r e a a bo ve datum l i n e i n mmˆ 2

7 a 3 = 180 // a r e a a bo ve datum l i n e i n mmˆ 28 a 4 = 50 // a r e a a bo ve datum l i n e i n mmˆ 29 a 5 = 95 / / a r e a b el ow datum l i n e i n mmˆ 2

10 a 6 = 65 / / a r e a b el ow datum l i n e i n mmˆ 211 a 7 = 170 // a r e a b el o w d atum l i n e i n mmˆ 212 a 8 = 150 // a r e a b el o w d atum l i n e i n mmˆ 213 a = ( a 1 + a 2 + a3 + a 4 + a 5 + a 6 + a7 + a 8 ) / ( v * h )

14 C LA = a / l

15 printf ( ” \n C . L . A v a l u e = %0 . 2 f ∗10 ˆ −6 m ” , C LA * 1 0 0 0 )

Scilab code Exa 11.2 Calculate average and rms value

69



1 clc

2 / / from f i g u r e 1 1. 2 33 y 1 = 0. 15 // mu m4 y 2 = 0. 25 // mu m5 y 3 = 0. 35 // mu m6 y 4 = 0. 25 // mu m7 y 5 = 0. 30 // mu m8 y 6 = 0. 15 // mu m9 y 7 = 0. 10 // mu m

10 y 8 = 0. 30 // mu m11 y 9 = 0. 35 // mu m12 y 10 = 0 .1 0 // mu m

13 y 1s qr = y1 ^ 2 // mu m14 y 2s qr = y2 ^ 2 // mu m15 y 3s qr = y3 ^ 2 // mu m16 y 4s qr = y4 ^ 2 // mu m17 y 5s qr = y5 ^ 2 // mu m18 y 6s qr = y6 ^ 2 // mu m19 y 7s qr = y7 ^ 2 // mu m20 y 8s qr = y8 ^ 2 // mu m21 y 9s qr = y9 ^ 2 // mu m22 y 1 0 sq r = y 10 ^ 2 // mu m23 n = 1 0

24 y n = ( y 1 + y 2 + y3 + y 4 + y 5 + y6 + y 7 + y 8 + y 9 + y 10 ) / n //a r i t h m e t i c a v er a ge i n mu m

25 r ms = sqrt ( ( y 1 s q r + y 2 s q r + y 3 s q r + y 4 s q r + y 5 s q r + y 6 s q r +

y 7 s q r + y 8 s q r + y 9 s q r + y 1 0 s q r ) / n ) / / r .m. s v a l u e i nmu m

26 printf ( ” \n The a r i t h m e t i c a v e ra g e = %0 . 2 f ∗10ˆ −6 m \nThe r . m. s . v a l u e = %0 . 3 f ∗10 ˆ −6 m” , y n , r m s )

70



Chapter 13

Analysis of metal forming

processes

Scilab code Exa 13.1 To find drawing load

1 clc

2 s ig ma _0 = 2 40 / / N/mmˆ23 d 1 = 5 / / i n i t i a l w i r e d i a m e t e r i n mm4 d 0 = 5.5 // f i n a l w ir e d ia me te r i n mm5 x = d1 /d0 / / mm6 alpha = 8 // a n g l e o f c on t a c t7 a lp ha = a lp ha * % pi / 1 80

8 m u = 0.1 // c o e f f i c i e n t o f f r i c t i o n9 B = mu *cotg ( a l p h a )

10 s i gm a _d = ( s i g ma _ 0 * (1 + B ) *( 1 -( x ) ^ (2 * B )) ) / B / / N/mmˆ211 l = 3 // d i e l an d i n mm12 m u = 0.1 // c o e f f i c i e n t o f f r i c t i o n13 r 1 = d1 /2 / / mm14 s ig ma _t = s ig ma _0 - ( s ig ma _0 - s ig ma _d ) / exp ( ( 2 * m u * l )

/ r 1 ) / / N/mmˆ215 d l = s i gm a _t * % p i *( r 1 ) ^2 // d ra wi n g l oa d i n N16 printf ( ” \n T o ta l d ra wi ng l o a d = %0 . 1 f N” , d l )


71



Scilab code Exa 13.2 Calculate drawing force

1 clc

2 a lp ha = 15 // a ng l e o f c on t a c t3 a lp ha = a lp ha * % pi / 1 80

4 b ita = 0 // d e gr e e5 m u = 0.1 // c o e f f i c i e n t o f f r i c t i o n6 m u1 = mu

7 m u2 = mu

8 h 1 = 1. 75 / / mm9 h 0 = 2.5 / / mm

10 B = ( mu 1 + mu 2 )/( tan ( a l p h a ) -tan ( b i t a ) )

11 y 1 = ( 1+ B ) * (1 - ( h 1 / h0 ) ^ B )/ B // s i gm a d / s i g ma 0 f o rp l u g m a n d r e l s i n N/mmˆ 2

12 z = 1 /( ( h 0 / h1 ) - 1)

13 y2 = log10 ( z ) // s ig ma d / s i g ma 0 f o r m ovab le m an dr el si n N/mmˆ2

14 printf ( ” \n The p i p e d ra wi ng f o r c e f o r c e on p lu gm a n d r e l s = %0 . 3 f \n The p i p e d ra wi ng f o rc w on

m a n d r e l s = %0 . 3 f ” , y 1 , y 2 )15 disp ( ” Use o f mo vabl e m an dr el s u b s t a n t i a l l y r e d uc e s

d ra wi ng f o r c e ” )

Scilab code Exa 13.3 find neutral section slips and pressure

1 clc

2 h 0 = 25 // t h i c k n e s s o f p l a t e i n mm

3 h 1 = 20 / / mm4 d e l ta _ h = h0 - h 1 // mm5 s i gm a = 1 00 // maximum pr e s s ur e i n N/mmˆ26 D = 500 // r o l l e d d ia me te r i n mm7 r = D /2 // r o l l e d r a di u s i n mm

72



8 a l ph a = acos ( 1 - ( d e l t a _ h / D ) ) // a ng l e o f c o n t a c t i n

r a d i a n s9 mu = tan ( a l p h a ) // c o e f f i c i e n t o f f r i c t i o n10 H o = 2* sqrt ( r / h 1 ) * atan ( sqrt ( r / h 1 ) * m u )

11 H n = ( Ho - ( log ( h 0 / h 1 ) ) / m u ) / 2

12 t h et a = sqrt ( h 1 / r ) * tan ( sqrt ( h 1 / r ) * ( H n / 2 ) ) // r a d ia n13 h n = h1 + r * th eta ^ 2 // n e u t ra l s e c t i o n i n mm14 x = hn /h0

15 b s = (1 - x ) *1 00 // b ac kw ar d s l i p16 y = hn /h1

17 f s = ( y -1 ) *1 00 // f or wa rd s l i p18 s i g ma 0 = 2 * s ig ma / sqrt (3)

19 p n = s ig ma 0 * hn * exp ( m u * H n ) / h 1 //N/mmˆ220 printf ( ” \n N e u t r a l s e c t i o n = %0 . 1 f mm” , h n )

21 printf ( ” \n Backward s l i p = %0 . 1 f p e r c e n t \n F o rw a rds l i p = %0 . 1 f p e r c e n t ” , bs , f s)

22 printf ( ” \n Maximum p r e s s u r e = %0. 1 f N/mmˆ2” , p n )

23 // ’ Answers v ar y due t o ro un d o f f e r r or ’

Scilab code Exa 13.4 To determine maximum force

1 clc

2 D o = 250 / / d i a me t er i n mm3 h o = 250 // h i e g h t i n mm4 d el ta _h = 1 00 / / mm5 h = 150 / / mm6 s i gm a0 = 55 / / N/mmˆ27 d = Do *sqrt ( h o / ( h o - d e l t a _ h ) ) / / d i a me t er i n mm8 m u = 0. 42 // c o e f f i c i e n t o f f r i c t i o n9 R = 16 2. 5 / / mm

10 p a = s i g ma 0 / 2 * ( h / ( mu * R ) ) ^ 2 *( % e ^ ( 2 * m u * R / h ) - 2* m u * R / h-1) / / N/mmˆ211 p = p a * %p i *( R ) ^2 // f o r c e i n kN12 printf ( ” \n F o r c e = %d kN ”, p / 1 0 0 0 )

73



Scilab code Exa 13.5 Determine sticking radius and total load

1 clc

2 d = 150 / / d i a m et e r i n mm3 h = 1 0 // t h i c k n e s s i n mm4 R = d /2 // r a d i u s i n mm5 m u = 0.2 // c o e f f i c i e n t o f f r i c t i o n6 s ig ma _0 = 2 00 / / N/mmˆ27 R s = R - ( h /(2 * mu )) *log (1/( sqrt ( 3 ) * m u ) ) // s t i c k i n g

r a d i u s i n mm8 P s = s ig ma _0 * exp ( 2 * m u * ( R - R s ) / h ) // p r e s s u re a t

s t i c k i n g r a d i u s i n N/mmˆ 29 function y = f ( r )

10 y = 2 * % p i * r * s i g m a _ 0 * exp ( 2 * m u / h * ( R - r ) )


12 L _ sl d = intg ( 4 8 . 5 , 7 5 , f )

13 L _s ld = L _s ld / 1 0 00 // l oa d o n s l i d i n g p o r t i o n i n kN14 P c = P s + ( 2* s i gm a_ 0 * Rs ) /( h *sqrt ( 3 ) ) // p r e s s u re a t

c e nt r e i n N/mmˆ2

15 L _ s p = ( P c + P s ) * %p i * ( R s ) ^ 2 / ( 2* 1 0 0 0) // l oa d ons t i c k i n g p o rt i on i n kN

16 F_l = L _s ld + L _sp // t o t a l f o r g i n g l oa d i n kN17 printf ( ” \n S t i c k i n g r a d i u s = %0 . 1 f mm \n T ot al

f o r g i n g l o a d = %0 . 3 f MN” , R s , F _l / 1 0 0 0 )

18 // ’ Ans wers v ar y due t o ro un d o f f e r r or ’

Scilab code Exa 13.7 To find drawing load and power

1 clc

2 R A = 0. 30

3 d = 1 2 / / d i a me t e r i n mm4 alpha = 6 // a ng l e o f c on t a c t i n d e g re e

74



5 a lp ha = 6 * %p i / 18 0 // a n g l e o f c o n t a c t i n r ad ia n

6 m u = 0. 10 // c o e f f i c i e n t o f f r i c t i o n7 s ig ma _0 = 2 40 // N/mmˆ28 B = mu *cotg ( a l p h a )

9 x = 1 - RA

10 s ig ma _ d = ( s i g ma _ 0 * (1 + B ) *( 1 -( x ) ^ B) ) / B / / N/mmˆ211 r1 = sqrt ( x ) * ( d / 2 ) / / mm12 l = s ig m a_ d * % pi * ( r 1 )^ 2 // l oa d i n kN13 i ta = 98 // e f f i c i e n c y14 i ta = i ta / 10 0

15 s = 2.3 // d ra wi ng s p ee d i n m/ s16 P = ( l* s )/ it a / / kW

17 printf ( ” \n D ra wi ng l o a d = %0 . 2 f kN\n Power o f m ot or= %0 . 2 f kW” , l / 10 00 , P /1 00 0 )


Scilab code Exa 13.8 calculate drawing load and power rating

1 clc

2 m u1 = 0 .1 5 // c o e f f i c i e n t o f f r i c t i o n

3 m u2 = 0 .1 8 / / c o e f f i c i e n t o f f r i c t o n4 a lp ha = 14 // a ng l e o f c on t a c t i n d e g re e5 a lp ha = a lp ha * % pi / 1 80

6 b ita = 10 / / s em i−c o ne a n gl e i n d e g r e e7 b it a = b it a * %p i / 18 0

8 s i g ma _ 0 = 1 .4 0 / / kN/mmˆ29 h 0 = 1.5 //mm

10 h 1 = 1 / / mm11 B = ( mu 1 + mu 2 )/( tan ( a l p h a ) + tan ( b i t a ) )

12 s i g m a d = ( s i g m a_ 0 * ( 1 + B ) * (1 - ( h 1 / h 0 ) ^ B ) ) / B / / d r aw i ng

s t r e s s i n kN /mmˆ 213 d 1 = 11 // o u t s i d e d i am et er i n mm14 t = 1 // t h i c k n e s s i n mm15 d 2 = d1 - 2* t / / mm16 a = ( % pi * ( ( d1 ) ^ 2 -( d 2 ) ^2 ) ) /4 / / a r e a i n mmˆ 2

75



17 l = s ig ma d *a // l oa d i n kN

18 s = 0.65 // d ra wi ng s p ee d i n m/ s19 w = l * s // work i n kJ / s20 p = w / / p ow er i n kW21 printf ( ” \n D ra wi ng l o a d = %0 . 3 f kN\n P ower r a t i n g o f

motor = %0. 2 f kW” , l , p )


Scilab code Exa 13.9 To calculate forging loads

1 clc

2 s ig ma _0 = 50 // p r e s s u r e a t s t a r t i n MPa3 B = 0.9 // w i d th i n m4 h 1 = 0.2 // t h i c k n e ss i n m5 b = 0.3 // t o o l b i t e i n m6 / / At commencement o f f o r g i n g7 F L = s i g ma _ 0 * B * b * ( 1+ ( b / ( 4 * h 1 ) ) ) // f o r g i n g l o ad i n

MN8 / / At c om pl et i o n o f f o r g i n g9 h 2 = 0.1 // t h i c k n e ss i n m

10 s i g ma _ 0c = 1 50 // p r e s s u r e a t c o mp l et i on i n MPa11 F L 2 = s i g ma _ 0 c * B * b * ( 1+ ( b / ( 4 * h 2 ) ) ) // f o r g i n g l oa d i n

MN12 printf ( ” \n F or gi ng l oa d a t s t a r t o f f o r g i n g = %0 . 4 f

MN\n F or gi ng l oa d a t c om pl et i o n o f f o r g i n g = %0 . 3f MN” , FL , FL2 )

Scilab code Exa 13.10 Determine extrusion load

1 clc

2 s ig ma _0 = 2 50 / / N/mmˆ23 d 1 = 5 / / i n i t i a l w i r e d i a m e t e r i n mm4 d 0 = 15 // f i n a l w ir e d ia me te r i n mm

76



5 r 0 = d0 /2

6 r 1 = d1 /27 x = ( r0 / r1 ) ^2 / / mm8 a lp ha = 45 // a ng l e o f c on t a c t9 a lp ha = a lp ha * % pi / 1 80

10 m u = 0.1 // c o e f f i c i e n t o f f r i c t i o n11 B = mu *cotg ( a l p h a )

12 s ig m a_ x 0 = ( s i g ma _0 * ( 1 + B) * (1 - ( x ) ^B ) ) /B / / N/mmˆ213 s i gm a_ x 0 = - s i gm a _x 0

14 l = 37.5 / / l e n g t h 0 f b i l l e t i n mm15 t a u1 = s i gm a_ 0 / 2 // Mpa16 P e = s ig ma _x 0 + ( 4* t au 1 *l )/ d0 // e x t r us i o n p r e s s ur e

in Mpa17 e l = P e * %p i *( r 0 ) ^2 // e x t r us i o n l oa d i n MN18 printf ( ” \n E x t r u s i o n l o a d = %d MN” , e l / 1 0 00 0 )

Scilab code Exa 13.11 To find roll pressures

1 clc

2 h 0 = 4. 05 // t h i c k n e s s o f p l a te i n mm

3 h 1 = 3. 55 / / mm4 D = 500 // r o l l e d d ia me te r i n mm5 r = D /2 // r o l l e d r a di u s i n mm6 m u = 0. 04 // c o e f f i c i e n t o f f r i c t i o n7 s i gm a = 2 10 / / N/mmˆ28 d e l ta _ h = h0 - h 1 // mm9 p = 2* s ig ma / sqrt (3) / / N/mmˆ2

10 a l ph a = acos ( 1 - ( d e l t a _ h / D ) ) // a n g l e o f c o n t a c t11 H o = 2* sqrt ( r / h 1 ) * atan ( sqrt ( r / h 1 ) * a l p h a )

12 Hn1 = ( Ho - ( log ( h 0 / h 1 ) ) / m u ) / 2

13 t h et a = sqrt ( h 1 / r ) * tan ( sqrt ( h 1 / r ) * ( H n 1 / 2 ) ) //r a d i a n s14 hn = h1 + 2* r*(1 - cos ( t h e t a ) ) / / mm15 pn1 = p * hn * exp ( m u * H n 1 ) / h 1 / / r o l l p r e s s u r e i n N/mmˆ 216 / / 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



0 . 4

17 m u2 = 0.4 // c o e f f i c i e n t o f f r i c t i o n18 Hn2 = ( Ho - ( log ( h 0 / h 1 ) ) / m u 2 ) / 2

19 t h et a = sqrt ( h 1 / r ) * tan ( sqrt ( h 1 / r ) * ( H n 2 / 2 ) ) //r a d i a n s

20 hn2 = h1 + r * th et a ^2 // mm21 p n 2 = ( p * hn 2 *exp ( m u 2 * H n 2 ) ) / h 1 // r o l l p r e s s u r e i n N/

mmˆ222 // c ) i f t e n s i o n i s a p p l i e d o f 35 N/mmˆ223 s ig ma _f = 35 // f r o n t t e n s i o n i n N/mmˆ 224 p n3 = ( p - s ig m a_ f ) * hn * exp ( m u * H n 1 ) / h 1 // r o l l r e s s u r e

i n N/mmˆ2

25 printf ( ” \n ( a ) R ol 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 R o l l p r e s s ur e a t n e u t r a l p la n e =%0 . 2 f N/mmˆ2 ” , p , p n 1 )

26 printf ( ” \n ( b ) R ol l p r e s s u r e a t n e u t r al p oi nt whenco− e f f i c i e n t of f r i c t i o n i s 0. 40 = %0. 2 f N/mmˆ2”, pn2 )

27 printf ( ” \n ( c ) R o l l p r e s s u r e when 3 5 N/mmˆ 2 t e n s i o ni 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/mmˆ 2 ” , pn3

)


Scilab code Exa 13.12 Determine neutral plane

1 clc

2 h 1 = 6. 35 // t h i c k n e s s i n mm3 m u = 0.2 // c o e f f i c i e n t o f f r i c t i o n4 r = 5 0 // r o l l e d r a d i u s i n cm5 r = r *10 / / mm

6 R = 3 0 // r e du c t io n i n p e rc e n t7 h 0 = h 1 * 10 0/ (1 00 - R ) / / mm8 d e l ta _ h = h0 - h 1 // mm9 a l ph a = acos ( 1 - ( d e l t a _ h / ( 2 * r ) ) ) // a ng l e o f c on t a c t

10 H o = 2* sqrt ( r / h 1 ) * atan ( sqrt ( r / h 1 ) * a l p h a )

78



11 H n = ( Ho - ( log ( h 0 / h 1 ) ) / m u ) / 2

12 t h et a = sqrt ( h 1 / r ) * tan ( sqrt ( h 1 / r ) * ( H n / 2 ) ) // n e u t ra lp l a ne i n r a di a n s13 t he ta = t he ta * 1 8 0/ % p i // n e ut r al p la ne i n d e g r e e s14 printf ( ” \n N e ut r al p l an e = %0 . 2 f d e g re e ” , t he ta )


79



Chapter 14

Theory of metal cutting

Scilab code Exa 14.1 calculate the tool life

1 clc

2 v 1 = 18 // c u t t i n g s pe ed i n m/ min3 t 1 = 3 / / t o o l l i f e i n h ou rs4 n = 0. 12 5 / / e x po n en t5 c = v1 * ( t1 * 60 ) ^n // c o n st a n t6 v 2 = 24 // c u t t i n g s pe ed i n m/ min

7 t = ( c / v2 ) ^ ( 1/ 0 .1 2 5) // t o o l l i f e i n min .8 printf ( ” T ool l i f e = %d min . ” , t )

Scilab code Exa 14.2 Calculate the optimum cutting speed

1 clc

2 c_t = 8 // t o o l c ha ng e t im e i n min .3 r_t = 5 // t o o l r e −g r i n d t im e i n min .4 m r_c = 5 // ma ch in e r un ni ng c o s t p er h ou r5 d = 3 0 // t o t a l d e pr e c i a t i o n p e r r e −g ri nd i n p a i s a6 n = 0.25 / / e x po n en t7 c = 150 // c o n st a n t

80



8 c _ c = m r_ c * c _t / 6 0 // t o t a l c h an g e c o st i n Rs

9 r _ c = m r_ c * r _t / 6 0 // r e g r i n d c o s t i n Rs10 c t = c _c + r _c + d / 10 0 // t o o l i n g c o s t i n Rs11 c m = m r_ c /6 0 // m ac hi ni ng c o s t i n Rs12 v = c * (( c m * n) / ( ct * (1 - n ) ) ) ^n // c u t t i n g s pe ed 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 )

Scilab code Exa 14.3 To find different orthogonal cutting picture

1 clc

2 m u1 = 0 .1 5 // c o e f f i c i e n t o f f r i c t i o n3 m u2 = 0 .1 8 / / c o e f f i c i e n t o f f r i c t o n4 a lp ha = 14 // a ng l e o f c on t a c t i n d e g re e5 a lp ha = a lp ha * % pi / 1 80

6 b ita = 10 / / s em i−c o ne a n gl e i n d e g r e e7 b it a = b it a * %p i / 18 0

8 s i g ma _ 0 = 1 .4 0 / / kN/mmˆ29 h 0 = 1.5 //mm

10 h 1 = 1 / / mm

11 B = ( mu 1 + mu 2 )/( tan ( a l p h a ) + tan ( b i t a ) )12 s i g m a d = ( s i g m a_ 0 * ( 1 + B ) * (1 - ( h 1 / h 0 ) ^ B ) ) / B / / d r aw i ng

s t r e s s i n kN /mmˆ 213 d 1 = 11 // o u t s i d e d i am et er i n mm14 t = 1 // t h i c k n e s s i n mm15 d 2 = d1 - t / / mm16 a = ( % pi * ( ( d1 ) ^ 2 -( d 2 ) ^2 ) ) /4 / / a r e a i n mmˆ 217 l = s ig ma d *a // l oa d i n kN18 s = 0.65 // d ra wi ng s p ee d i n m/ s19 w = l * s // work i n kJ / s

20 p = w / / p ow er i n kW21 printf ( ” \n D ra wi ng l o a d = %0 . 3 f kN\n P ower r a t i n g o f motor = %0. 2 f kW” , l , p )

22 clc

23 t = 0. 12 7 // u ncu t c h ip t h i c k n e s s i n mm

81



24 b = 6.35 // wi dt h o f c ut i n mm

25 v = 2 // c u t t i n g s pe ed i n m/ s26 a lp ha = 10 // r a ke a ng l e i n d e g r ee s27 f c = 567 // c u t t i n g f o r c e i n N28 f t = 227 // t h r u s t f o r c e i n N29 t c = 0 .2 28 // c hi p t h i c k n e s s i n mm30 r = t / tc // c hi p t h i c k n e ss r a t i o31 a lp ha = a lp ha * % pi / 1 80 // r ak e a ng l e i n r a di a ns32 p hi = atan ( r * cos ( a l p h a ) / ( 1 - ( r * sin ( a l p h a ) ) ) ) // s h e ar

a n g l e33 p hi 1 = p hi * 1 8 0/ % p i // s h ea r a n gl e34 printf ( ” \n S he ar a n g le = %0 . 2 f d e g re e ” , p hi 1 )

35 m u = (( f c * sin ( a l p h a ) + f t * cos ( a l p h a ) ) / ( f c * cos ( a l p h a ) - f t* sin ( a l p h a ) ) ) // c o e f f i c i e n t o f f r i c t i o n

36 b i ta = atan ( m u ) / / f r i c t i o n a n g l e37 b it a = b it a * 1 80 /( % p i )

38 printf ( ” \n F r i c t i o n a n gl e = %0 . 2 f d e g r e e ” , b i t a )

39 f s = fc *cos ( p h i ) - f t *sin ( p h i ) // s h ea r f o r c e i n N40 ta us = ( fs * sin ( p h i ) ) / ( b * t ) // s he ar s t r e s s41 printf ( ” \n S h e a r s t r e s s = %0 . 1 f N/mmˆ 2 ” , t au s )

42 c p = fc * v /1 00 0 // c u t t i n g power i n kw43 printf ( ” \n C u t ti n g p ow er = %0 . 3 f kw ” , c p )

44 v c = v * r // c hi p v e l o c i t y i n m/ s45 printf ( ” \n Chip v e l o c i t y = %0 . 3 f m/ s ” , v c )

46 ss = cotg ( ph i ) + tan ( p h i - a l p h a ) // s he ar s t r a i n47 printf ( ” \n s h e ar s t r a i n = %0 . 3 f ” , s s )

48 spl = t /sin ( p h i ) // s h ea r p la ne l e ng t h49 v s = v * cos ( a l p h a ) / cos ( p h i - a l p h a ) // s he ar v e l o c i t y50 S = vs * 10 / sp l // s he ar s t r a i n r a t e51 S = S *1 0^ 3 // s he a r s t r a i n r a t e52 printf ( ” \n S h e a r s t r a i n r a t e = %. 3 f s ˆ−1” , S )


Scilab code Exa 14.4 To find tool life

82



1 clc

2 v = 3 0 // c u t t i n g s pe ed i n m/ min3 f e ed = 0 .3 / / f e e d r a t e i n mm/ r e v .4 d = 2.5 // d ep th o f c ut i n mm5 t = 6 0 // t o o l l i f e i n min .6 c = v * t ^ 0 . 13 * f e e d ^ 0 . 77 * d ^ 0 . 3 7 // c o n st a n t7 printf ( ” \n c o n st a n t = %0 . 2 f ” , c )

8 v 2 = v *1. 2 // c u t t i n g s pe ed i n m/ min9 t 2 = ( c / ( v 2 * f ee d ^ 0 . 7 7* d ^ 0 . 3 7 ) ) / / t o o l l i f e when

c u t t i n g s pe ed i n c r e a s e d by 20% i n min .10 t 2 = t 2 ^ (1 / 0. 1 3)

11 f 2 = f ee d *1 .2 / / f e e d r a t e i n mm/ r e v .

12 t 3 = ( c /( v * d ^ 0. 37 * f 2 ^ 0. 7 7) ) // t o o l l i f e when f e e dr a t e i n c r e a s e d by 20% i n min .

13 t 3 = t 3 ^ (1 / 0. 1 3)

14 d 2 = d *1. 2 // d ep th o f c ut i n mm15 t 4 = ( c / ( v * f ee d ^ 0 . 7 7* d 2 ^ 0 . 3 7 ) ) / / t o o l l i f e when

d ep th o f c ut i n c r e a s e d by 20% i n min .16 t 4 = t 4 ^ (1 / 0. 1 3)

17 t 5 = ( c / ( v 2 * d2 ^ 0 . 3 7* f 2 ^ 0 . 7 7 ) ) / / t o o l l f e i n min .18 t 5 = t 5 ^ (1 / 0. 1 3)

19 printf ( ” \n To ol l i f e when c u t t i n g s pe ed i n c r e a s e d by

20 = %0. 2 f min . ” , t 2 )

20 printf ( ” \n To ol 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 . ” , t 3 )

21 printf ( ” \n To ol l i f e when d ep th o f c ut i n c r e a s e d by20 = %0. 2 f min . ” , t 4 )

22 printf ( ” \n Too l l i f e when a l l t ak en t o g e t h e r a f t e ri n c r e a s i n g by 2 0 = %0 . 2 f min . ” , t 5 )


Scilab code Exa 14.5 find force and coefficient of friction

1 clc

2 t = 0.25 // u nc ut c h ip t h i c k n e s s i n mm

83



3 b = 2.5 // w id th o f c ut i n mm

4 v = 2.5 // c u t t in g s pe ed i n m/ s5 a lp ha = 10 // r a ke a ng l e i n d e g r ee s6 f c = 11 30 // c u t t i n g f o r c e i n N7 f t = 295 // t h r u s t f o r c e i n N8 t c = 0. 45 // c hi p t h i c k n e s s i n mm9 r = t / tc // c hi p t h i c k n e ss r a t i o

10 a lp ha = a lp ha * % pi / 1 80 // r ak e a ng l e i n r a di a ns11 p hi = atan ( ( r * cos ( a l p h a ) ) / ( 1 - r * sin ( a l p h a ) ) ) // s h e ar

a n g l e12 p hi 2 = p hi * 1 8 0/ % p i // s h ea r a n gl e13 f s = fc *cos ( phi ) - ft * sin ( p h i ) / / s he ar f o r c e i n N

14 printf ( ” \n F or ce o f s h ea r a t s h ea r p l a ne = %0 . 2 f N”, f s )

15 mu = atan ( ( f c * sin ( a l p h a ) + f t * cos ( a l p h a ) ) / ( f c * cos (

a l p h a ) - f t * sin ( a l p h a ) ) ) // f r i c t i o n a n g l e l e16 printf ( ” \n F r i c t i o n a n gl e = %0 . 3 f d e g r e e ” , m u )


Scilab code Exa 14.6 To find terms of orthogonal cutting

1 clc

2 t = 0.2 // u nc ut c h ip t h i c k n e s s i n mm3 a lp ha = 15 // r a ke a ng l e i n d e g r ee s4 t c = 0. 62 // c hi p t h i c k n e s s i n mm5 r = t / tc // c hi p t h i c k n e ss r a t i o6 c rc = 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 t7 printf ( ” \n C ut ti ng r a t i o = %0 . 3 f \n Chip r e d u c t io n co

− e f f i c i e n t = %0 . 1 f ” , r , crc )

8 a lp ha = a lp ha * % pi / 1 80 // r ak e a ng l e i n r a di a ns

9 p hi = atan ( r * cos ( a l p h a ) / ( 1 - r * sin ( a l p h a ) ) ) // s h e ara n g l e10 p h i = p hi * 1 80 / % pi // s h ea r a n gl e11 printf ( ” \n S he ar a n g le = %0 . 2 f d e g re e ” , phi )

12 ss = cotg ( p h i * % pi / 1 8 0 ) + tan ( ( p h i * % p i ) / 1 8 0 - ( a l p h a *

84



% p i ) / 1 8 0 ) // s he ar s t r a i n

13 printf ( ” \n s h e ar s t r a i n = %0 . 3 f ” , s s )14 // ’ Ans wers v ar y due t o ro un d o f f e r r or ’

Scilab code Exa 14.7 To solve tool life equation

1 clc

2 v 1 = 25 // c u t t i n g s pe ed i n m/ min3 t 1 = 90 // t o o l l i f e i n min .

4 v 2 = 35 // c u t t i n g s pe ed i n m/ min5 t 2 = 20 // t o o l l i f e i n min6 n = log ( v 2 / v 1 ) / log ( t 1 / t 2 ) / / e x po n en t7 C = v1 * ( t1 ) ^n // c o n st a n t8 t = 6 0 // t o o l l i f e i n min .9 v = C /( t) ^n / / c u t t i n g s p ee d i n m/ min .

10 printf ( ” \n n = %0 . 3 f \n C = %0 . 1 f \n C ut ti ng s pe ed =%0 . 2 f m/min . ” , n , C ,v )


Scilab code Exa 14.8 Determine normal and tangential force

1 clc

2 t = 0.5 // u nc ut c h ip t h i c k n e s s i n mm3 b = 3 // w id th o f c ut i n mm4 a lp ha = 15 // r a ke a ng l e i n d e g r ee s5 a lp ha = a lp ha * % pi / 1 80 // r ak e a ng l e i n r a di a ns6 r = 0. 38 3 // c hi p t h i c k n e s s r a t i o7 m u = 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 o n t o o l

f a c e8 b i ta = atan ( m u ) / / f r i c t i o n a n g l e9 t au = 280 // y i e l d s t r e s s i n N/mmˆ2

10 p hi = atan ( ( r * cos ( a l p h a ) ) / ( 1 - r * sin ( a l p h a ) ) ) // s h e ara n g l e

85



11 f c = ( t a u * b * t ) /( s e c ( b it a - a l ph a ) * cos ( p h i + b i t a - a l p h a ) *

sin ( p h i ) ) // c u t t i n g f o r c e i n N12 f t = ( fc * (mu - tan ( a l p h a ) ) ) / ( 1 + m u * tan ( a l p h a ) ) //t h r u s t f o r c e i n N

13 F = fc *sin ( a l p h a ) + f t * cos ( a l p h a ) // t a n g e n t i a l f o r c eon t o o l f a c e i n N

14 F = ceil ( F )

15 N = fc *cos ( a l p h a ) - f t * sin ( a l p h a ) // no rmal f o r c e ont o o l f a c e i n N

16 printf ( ” T a n ge nt i al f o r c e on t o o l f a c e = %d N\nn or ma l f o r c e on t o o l f a c e = %0 . 1 f N” , F , N )


Scilab code Exa 14.9 To find cutting and thrust force

1 clc

2 t = 0.25 // u nc ut c h ip t h i c k n e s s i n mm3 b = 0.5 // w i d th o f c ut i n cm4 v = 8.2 / / c u t t i n g s p ee d i n m/ min .5 a lp ha = 20 // r a ke a ng l e i n d e g r ee s

6 a l p h a2 = a l ph a * % p i / 1 80 // r ak e a n g l e i n r a d i an s7 r = 0. 35 1 // c u t t i n g r a t i o8 p hi = atan ( r * cos ( a l p h a 2 ) / ( 1 - r * sin ( a l p h a 2 ) ) ) // s h e ar

a ng le i n r a d ia ns9 p hi 2 = p hi * 1 8 0/ % p i // s he ar a n g l e i n d e g r e e s

10 a l p h a2 = a l ph a * % p i / 1 80 // r ak e a n g l e i n r a d i an s11 b it a = 3 5+ a l ph a - p h i2 // d e g r e es12 s = cotg ( p hi ) + tan ( p h i - a l p h a 2 ) // s he ar s t r a i n13 e = s / sqrt (3) // n a tu r a l s t r a i n14 s ig ma = 7 84 *( e ) ^ 0 .1 5 // t e n s i l e p r o p e r ty i n N/mmˆ 2

15 t a u = s ig ma / sqrt (3) // y i e l d s h ea r s t r e s s i n N/mmˆ216 A s = ( b *1 0* t ) /sin ( p h i ) // s h e a r p l a n e a r e a i n mmˆ 217 Fs = ta u* As // s he ar g o r ce i n N18 R = Fs /cos ( p h i + ( b i t a * % p i / 1 8 0 ) - a l p h a 2 )

19 F c = R * cos ( ( b i t a * % p i / 1 8 0 ) - a l p h a 2 ) // c u t t i n g f o r c e

86



i n N

20 F t = R * sin ( ( b i t a * % p i / 1 8 0 ) - a l p h a 2 ) // t h r u s t f o r c e i nN21 printf ( ” \n C u tt i ng f o r c e = %0 . 1 f N\n Th ru st f o r c e =

%0. 1 f N” , Fc , Ft )


Scilab code Exa 14.10 find terms of orthogonal rake system

1 clc2 f = 0.2 / / f e e d i n mm/ r e v .3 t = 0.2 // u nc ut c h ip t h i c k n e s s i n mm4 a lp ha = 10 // r a ke a ng l e i n d e g r ee s5 f c = 16 00 // c u t t i n g f o r c e i n N6 f t = 850 // t h r u s t f o r c e i n N7 t c = 0. 39 // c hi p t h i c k n e s s i n mm8 r = t / tc // c hi p t h i c k n e ss r a t i o9 d = 2 // d ep th o f c ut i n mm

10 b = 2 // mm11 a l p h a2 = a l ph a * % p i / 1 80 // r ak e a n g l e i n r a d i an s

12 p hi = atan ( r * cos ( a l p h a 2 ) / ( 1 - r * sin ( a l p h a 2 ) ) ) // s h e ara ng l e i n r a d i an s

13 p hi 2 = p hi * 1 8 0/ % p i // s he ar a n g l e i n d eg re e14 f s = fc *cos ( p h i ) - f t *sin ( p h i ) // s h ea r f o r c e i n N15 f n = fc *sin ( p h i ) + f t *cos ( p h i ) // nor mal f o r c e i n N16 f = fc *sin ( a l p h a 2 ) + f t * cos ( a l p h a 2 ) / / f r i c t i o n f o r c e

i n N17 m u = (( f c * tan ( a l p h a 2 ) + f t ) / ( f c - f t * tan ( a l p h a 2 ) ) ) //

k i n e t i c c o e f f i c i e n t of f r i c t i o n18 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/mmˆ 2

19 printf ( ” \n S he ar f o r c e = %d N\n No rma l 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 cc o e f f i c i e n t o f f r i c t i o n = %0. 3 f ” , fs , f n ,f ,

mu )

20 printf ( ” \n S p e c i f i c c u t t i n g e n er g y = %d N/mmˆ2 ” , s )

87




Scilab code Exa 14.11 Calculate CLA

1 clc

2 c s = 20 // s i d e c u t t i n g e d g e a ng l e i n d eg r e e3 c e = 30 // end c u t t i n g e d ge a n gl e i n d e g r e e4 f = 0.1 / / f e e d i n mm/ r e v .5 r = 3 // n os e r a d i u s i n mm

6 c s 2 = c s * %p i / 18 0 // s i d e c u tt i n g e d g e a n g l e i nr a d i a n s7 c e 2 = c e * %p i / 18 0 // end c u t t i n g e dg e a n gl e i n

r a d i a n s8 h = (1 -cos ( c e2 ) )* r + f * sin ( c e 2 ) * cos ( c e2 ) - sqrt ( ( 2 * f

* r * ( sin ( c e 2 ) ) ^ 3 ) - ( ( f ^ 2 ) * ( sin ( c e 2 ) ) ^ 4 ) )

9 Ra = h /4 // C en tr e l i n e a v er a ge r o ug h ne s s i n mm10 printf ( ” \n C en tr e l i n e a v er a ge r o ug h ne s s = %0 . 2 f

∗10 ˆ −6m” , R a * 10 ^ 3)


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 ng le i n d eg re e3 a lp ha = 10 // o r th o go n al r ak e a n gl e i n d e g r e e4 l em da = 75 // p r i n c i p a l c u t t i n g e d g e a ng l e i n d eg r e e5 a lp ha = a lp ha * % pi / 1 80 // o r th o go n al r ak e a n gl e i n

r a d i a n6 l em da = l em da * % pi / 1 80 // p r i n c i p a l c u t t i ng e d g e

a n gl e i n r a di a n7 a l p ha _ b = atan ( cos ( l e m d a ) * tan ( a l p h a ) + sin ( l e m d a ) * tan (

i ) ) // b ack r ak e a n gl e i n r a d ia n s

88



8 a l p h a _b = a l p ha _ b * 1 8 0/ % p i // back r ak e a n gl e i n

d e g r e e9 a l p ha _ s = atan ( sin ( l e m d a ) * tan ( a l p h a ) -cos ( l e m d a ) * tan (

i ) ) // s i d e r a ke a ng l e i n r a di a ns10 a l p h a _s = a l p ha _ s * 1 8 0/ % p i // s i d e r a ke a ng le i n

d e g r e e11 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 ak e a n g le = %0 . 2 f d e g re e ” , a l p ha _ b , a l p h a _ s )

Scilab code Exa 14.13 Calculate inclination and rake angle

1 clc

2 a lp ha b = 8 // back r ak e i n d e g r e e3 a lp ha s = 4 // s i d e r a ke i n d eg re e4 c s = 15 // s i d e c u t t i n g e d g e a ng l e i n d eg r e e5 lemda = 90 - cs // a pp ro ac h a n g l e i n d e gr e e6 a l p h ab = a l ph a b * % p i / 18 0 // back r ak e i n r ad i an7 a l p h as = a l ph a s * % p i / 18 0 // s i d e r ak e i n r ad i a n8 c s = c s * %p i / 18 0 // s i d e c u t t i n g e dg e a ng l e i n r ad i a n9 l em da = l em da * % pi / 1 80 // a pp ro ac h a n g le i n r a d ia n

10 a l ph a = atan ( tan ( a l p h a s ) *sin ( l e m d a ) + tan ( a l p h a b ) * cos (l e m d a ) ) // o r th o go n al r ak e a n gl e i n r a di a n

11 a lp ha = a lp ha * 1 8 0/ % p i // o r th o go n al r ak e a n gl e i nd e g r e e

12 i = atan ( sin ( l e m d a ) * tan ( a l p h a b ) -cos ( l e m d a ) * tan (

a l p h a s ) ) // i n c l n a t i o n a ng l e i n r ad i a n13 i = i * 18 0/ % pi // i n c l n a t i o n a n g l e i n d eg re e14 printf ( ” \n O th og on al 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 gl e = %0 . 1 f d e g r ee ” , alpha , i )

Scilab code Exa 14.14 find different powers and resistance

1 clc

89



2 c s = 15 // s i d e c u t t i n g e d g e a ng l e i n d eg r e e

3 v = 0.2 // c u t t in g s pe ed i n m/ s4 f = 0.5 / / f e e d r a t e i n mm/ r e v .5 d = 3.2 // d ep th o f c ut i n mm6 f c = 1 5 9 3* ( f ) ^ 0 . 8 5* ( d ) ^ 0 . 98 // c u t t i n g f o r c e i n N7 p c = fc * v /1 00 0 // c u t t i n g power i n kw8 i t a _m t = 0 .8 5 // e f f i c i e n c y o f l a t he9 p m = pc / i ta _m t / / m ot or p ow er i n kw

10 a = f * d // a r ea o f u nc ut c h i o i n mmˆ211 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/mmˆ 212 p = pc / (a * v) / / u n i t p o we r i n W/ (mmˆ 3 ) ∗ s13 printf ( ” \n C u t ti n g p ow er = %0 . 3 f kw\n M ot or p ow er =

%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 ni t pow er = %0. 3 f W/(mmˆ3) ∗ s ” , pc , p m , r ,p

)


Scilab code Exa 14.15 Calculate percentage increase in tool life

1 clc

2 C = 4003 n = 0 . 5

4 a =2 / / ( T1 /T2 ) ˆ n5 b = 2 ^ ( 1 / n ) / / T26 i = ( b -1 ) *1 00 // p e rc e n t ag e i n c r e a s e7 printf ( ” \n P e rc en t ag e i n c r e a s e = %d p e r ce n t ” , i )

Scilab code Exa 14.16 To find percentage of total energy

1 clc

2 t = 0. 12 7 // u ncu t c h ip t h i c k n e s s i n mm3 b = 6.35 // wi dt h o f c ut i n mm4 v = 1.20 / / c u t t i n g s p ee d i n m/ min .

90



5 a lp ha = 10 // r a ke a ng l e i n d e g r ee s

6 f c = 5 56 .2 5 // c u t t i n g f o r c e i n N7 f t = 2 22 .5 0 // t h r u s t f o r c e i n N8 t c = 0 .2 29 // c hi p t h i c k n e s s i n mm9 r = t / tc // c hi p t h i c k n e ss r a t i o

10 R = sqrt ( ( f c ^ 2 ) + ( f t ^ 2 ) )

11 b ita = ( acos ( f c /R ) ) + a lp ha * % p i / 18 0 //12 f = R * sin ( b i t a ) //13 f e = f * r / / f r i c t i o n e n e r g y14 p = ( f* r *1 00 ) / fc // p e r c e nt ag e o f f r i c t o n e n r gy and

t o t a l e ne rg y15 printf ( ” \n The p e rc e n t ag e o f t o t a l e ne rg y t ha t g oe s

i n t o o ve rc om in g f r i c t i o n a t t o o l c h ip i n t e r f a c e =%0 . 2 f p e r c e n t ” , p )


Scilab code Exa 14.17 To find power and different energies

1 clc

2 D = 300 / / d i a m et e r i n mm

3 r = 4 5 / / r e v / m in .4 d = 2 // d ep th o f c ut i n mm5 f = 0.3 / / f e e d i n mm/ r e v6 f c = 18 50 // c u t t i n g f o r c e i n N7 f f = 450 // f e ed f o r c e i n N8 V = 2 .5 *1 0^ 6 / / m e t a l r em ov ed i n mm9 v = ( % pi * D * r) / ( 6 0* 1 00 0 ) // c u t t i n g v e l o c i t y i n m/ s

10 p c = f c* v /1 00 0 / / c u t t i n g power i n kW11 f v = f * r / 60 * 10 00 // f e e d v e l o c i t y i n m/ s12 f p = fv * ff // f e ed power i n W

13 m rr = d * f *v * 6 0* 1 00 0 // mmˆ3 / min .14 c e = p c * 10 0 0* 6 0/ m r r / / s p e c i f i c c u t t i n g e n er g y i n W∗s /mmˆ2

15 E = c e *V / ( 3 60 0 *1 0 00 ) / / e n e r g y c on su me d16 printf ( ” \n Po wer c o n s um p t i on = %0 . 2 f W\n S p e c i f i c

91



c u t t i n g e n er g y = %0 . 2 f W∗ s /mmˆ3 \ n E n e rg y c on s um ed

= %0 . 3 f kWh” , p c , c e , E )17 // ’ Ans wers v ar y due t o ro un d o f f e r r or ’

Scilab code Exa 14.18 Determine components of force and power

1 clc

2 D = 100 / / d i a m et e r i n mm3 c s = 30 // s i d e c u t t i n g e d g e a ng l e i n d eg r e e

4 lemda = 90 - cs // a pp ro ac h a n g l e i n d e gr e e5 d = 2.5 // d ep th o f c ut i n mm6 f = 0. 12 5 / / f e e d i n mm/ r e v .7 N = 300 / / t u r ni n g s pe ed o f j o b i n r ev . / min .8 m u = 0.6 // c o e f f i c i e n t o f f r i c t i o n9 t au = 400 // u l t i m a te s h ea r s t r e s s i n Mpa

10 b i ta = a ta nd ( m u ) // f r i c t i o n a n g l e i n r a d i a n11 a l ph as = 10 // s i d e r ak e a ng l e12 a lp ha b = 6 // ba ck r a ke a n g le13 a l ph a = a t an d ( t a n d ( a l ph a s ) * s i nd ( l e m d a ) + t a nd ( a l p h a b ) *

c o s d ( l e m d a ) ) // o r th o go n al r ak e a n gl e i n d e g r ee

14 phi = 45 - ( bita - alpha ) // s h ea r a n gl e15 F c = t a u * d *f / ( s e cd ( b i ta - a l p ha ) * c o s d ( p h i + bi ta - a l p ha ) *

s i n d ( p h i ) ) // c u t t i n g f o r c e i n N16 F t = F c * ta nd ( b it a - a l ph a ) // t h r u s t component i n N17 F f = F t * si nd ( l e md a ) // f ee d f o r c e a lo n g a x i s o f j o b

i n N18 R f = F t * co sd ( l e md a ) // r a d i a l f o r c e normal t o a x i s

o f j o b i n N19 v = % pi * D * N / (1 0 00 * 60 ) // v e l o c i t y i n m/ s20 p = Fc *v // p ower i n w a tt s

21 printf ( ” \n C ut ti ng f o r c e = %d N\n T hr us t f o r c e = %0. 3 f N\n Feed f o r c e = %0 . 1 f N\n R ad ia l f o r c e = %0. 3 f N\n C u t ti n g p ow er = %d w a t ts ” , F c , Ft , F f , R f , p

)


92



93



Chapter 15

Design and manufacture of

cutting tools

Scilab code Exa 15.1 calculate horsepower at cutter and motor

1 clc

2 w = 1 0 // w i dt h o f c ut i n cm3 h = 0.32 // de pt h o f c ut i n cm4 n = 8 // number o f t e e t h i n c u t t e r5 f t = 0 .0 33 // f e e d r a t e p er t oo th6 N = 200 // c u t t e r s pe ed i n rpm7 i ta = 6 0/ 10 0 // e f f i c i e n c y8 f = n * f t * N / / f e e d r a t e i n cm/ min .9 mrr = w *h *f // m e ta l r e mo v a l r a t e i n cm ˆ3 / min .

10 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 . 311 h pc = m rr / k // h or se po we r a t c u t t e r12 h pm = h pc / it a // h o rs e po w er a t mo to r13 printf ( ” \n H or se po we r a t c u t t e r = %0 . 2 f \n H o r se p o we r

a t m o to r = %0 . 2 f ” , hpc , hpm )


94



Scilab code Exa 15.2 Determine broaching power and Design broach

1 clc

2 l = 3 5 // l e ng t h o f b or e i n mm3 v = 0.15 // c u t t i n g s pe ed i n m/ s4 t 1 = 0. 01 // u pp er l i m i t i n mm5 t 2 = 0. 05 // u pp er l i m i t i n mm6 D = 32 .2 5 // f i n i s h e d b ro ac h i n mm7 D 1 = 3 2. 25 + t2 / / mm8 d = 32 .7 5 // f i n i s h d ia me te r i n mm9 d1 = 32.75 + t1 // f i n i s h d ia me te r o f h o l e i n mm

10 s = 0.05 / / mm

11 B = 1.30 // b lu nt b ro ac h f a c t o r12 c = 4 5 / / s p e c i f i c c u t t i n g f o r c e i n N/mmˆ 213 n = 3 // number o f t e e t h c u t t i n g a t a t i me14 F = n * % pi * d 1 *s * c *B // f o r c e n e e d e d f o r b ro ac hi ng i n

N15 b p = F * v /1 00 0 / / B r oa c hi n g p ower i n kw16 / / b ro ac h d e s i g n17 p = 1. 75 * sqrt ( l ) // p i t c h i n mm18 t he ta = 10 // f a c e a ng l e i n d eg r e e19 a l ha 1 = 1 .5 // r e l i e f a n g le f o r r o ug h in g i n d e gr e e20 a l ha 2 = 1 .0

/ / r e l i e f a n gl e f o r f i n i s h i n g i n d e g r e e21 w = 0.3* p // w id th o f l an d i n mm22 h = 0.4* p // de pt h o f c u t t i n g t e et h i n mm23 r = 0.3* p / / t o o t h f i l l e t r a d i u s i n mm24 T = ( d1 - D 1) /2 / / mm25 n = T / s // number o f c u t t i n g t e e t h26 n = round ( n )

27 l = ( n +7) * p // l e ng t h o f t oo th ed p o r ti o n o f b ro ac h i nmm

28 printf ( ” \n ( i ) Br oa c hi n g pow er = %0. 4 f kW” , b p )

29 disp ( ” ( i i ) B r oa c h D e s i g n ” )

30 printf ( ” ( a ) P i t ch d i a me t er = %0 . 2 fmm\n ( b ) wi dt h o f l and = %0. 2 f mm \n de pt h of 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 )

31 printf ( ” \n ( c ) Length o f t oo th ed p o r ti o n o f b ro ac h =

95



%d mm” , l )


Scilab code Exa 15.3 Estimate moment thrust force and power

1 clc

2 H b = 200 // b r i n e l l h ar dn es s3 d = 12.7 / / d i a me t er i n mm4 f = 0. 25 4 / / f e e d i n mm/ r e v .

5 N = 100 // rpm6 M = ( H b *( d ) ^ 2* f ) /8 / / moment i n k g f −mm7 k = 1.1 / / m a t e r i a l f a c t o r8 p = ( 1 . 25 * ( d ) ^ 2 * k * N * ( 0 .0 5 6 + 1. 5 * f ) ) / ( 1 0) ^ 5 / / p ow er

in kW9 T 1 a = ( 1. 7* M ) / d // t h r us t f o r c e k gf

10 T 1 b = ( 3. 5* M ) / d // k gf 11 T 1 = ( T 1a + T 1b ) / 2 // a v e r ag e12 w = 0. 14 * d // t h i c k n e s s i n mm13 T 2 a = ( 0 .1 * % p i * ( w ) ^ 2* H b ) / 4 // t h r u s t f o r c e k g f 14 T 2 b = ( 0 .2 * % p i * ( w ) ^ 2* H b ) / 4 // t h r u s t f o r c e k g f

15 T 2 = ( T 2a + T 2b ) / 2 // a v e r ag e16 a vg = T1 + T2 // k gf 17 t h r u s t = 1 . 16 * k * d * ( 1 00 * f ) ^ 0 . 8 5 // k gf 18 printf ( ” \n Moment = %0. 1 f k gf −mm\n P ow er = %0 . 3 f hp\

n A ve ra ge f o r c e = %d k g f \n T hr us t f o r c e = %0 . 1 f k g f ” , M, p , avg , t hr us t)


Scilab code Exa 15.4 Design shell inserted blade reamer

1 clc

2 d = 5 5 / / d i a me t e r i n mm3 u l = 0 .0 35 // u ppe r l i m i t i n mm

96



4 l l = 0 .0 00 // l ow er l i m i t i n mm

5 D m ax = d + ul / / maximum d i a m e t e r o f h o l e i n mm6 D m in = d + ll // minimum d i a m e t e r o f h o l e i n mm7 I T = 0 .0 35 // h o le t o l e r e n c e i n mm8 d m ax = D ma x - 0 .1 5 * I T // m aximum d i a m e t e r o f r e a me r i n

mm9 d m in = d ma x - 0 .3 5 * I T // minimum d i a m e t e r o f r e am e r i n

mm10 l = ( ( d /4 ) +( d / 3) ) / 2 // l en g t h o f g ui di ng s e c t i o n i n

mm11 Z = 1.5* sqrt ( d ) + 2 // number o f t e e t h12 Z = ceil ( Z )

13 printf ( ” \n 1 D ia me te r o f r ea me r \n Maximum di am et ero f r e a me r = %0 . 3 f mm \n M inimum d i a m e t e r o f r e ame r = %0. 3 f mm \n 2 Back t a p e r = 0 . 0 5 mm \n 3V al ue s o f v a r i o u s a n g l es \n Rake a n g l e = 5 d e g re e

\n Pl an a pp ro ac h a n g l e = 45 d e g re e \n C i r c u l a rl an d = 0 . 2 5 t o 0 . 5 0 mm \n S ec on da ry c l e a r a n c ea n g le = 1 0 d e g re e \n 4 L en gth o f r ea me r \n L e ng t h

o f f l u t e d p o r ti o n = 8 2 . 5 mm \n L en gt h o f r ea mi nga l l o w a n c e = 0 . 1 8 mm \n L engt h o f c u t t i n g s e c t i o n

= 2 . 2 5 mm \n Le ng th o f g u i d in g s e c t i o n = %d mm \

n 5 Number o f t e e t h = %d” , d ma x , d m in , l , Z )

14 / / A nswer v ar y due t o ro un d o f f e r r o r

Scilab code Exa 15.5 To design single point cutting tool

1 clc

2 P m = 10 // power o f motor i n kw3 v = 4 0 / / c u t t i n g s p ee d i n m/ min .

4 i ta = 70 // e f f i c i e n c y5 i ta = i ta / 10 0

6 Pc = Pm * ita

7 F c = ( P c * 10 0 0* 6 0) / v // c u t t i ng f o r c e8 s ig ma b = 2 50 // s t r e s s i n Mpa

97



9 B = sqrt ( ( F c * 1 . 2 5 * 6 ) / ( s i g m a b * 1 . 6 ) ) // w id th o f s ha nk

i n mm10 h = 1.6* B // h i e g ht o f s ha nk i n mm11 l = 1. 25 * h // s ha nk o v e ra n g i n mm12 printf ( ” \n The w i dt h o f s h an k = %0 . 1 f mm\n H ei gh t o f

shank = %0. 2 f mm\n S ha nk o v e r h a n g = %0 . 1 f mm” ,

B , h , l )


Scilab code Exa 15.8 find various terms for stainless steel

1 clc

2 l = 150 / / l e n g th i n mm3 D = 12 .7 0 / / d i a m et e r i n mm4 d i a = 1 2. 19 // d ia me te r on c e n tr e l a t h e i n mm5 N = 400 / / s p i n d l e s pe ed i n r ev . / min6 s = 2 03 .2 0 / / a x i a l s p e e d i n mm/ min.∗####7 v = ( % pi * D * N) / ( 1 00 0 *6 0 ) // c u t t i n g v e l o c i t y i n m/ s8 d = ( D- d ia ) /2 // d ep th o f c ut i n mm9 f = s / N / / f e e d i n mm/ r e v .

10 D a ve = ( D + di a ) /2 // a v e r ag e d i a m et e r i n mm11 V = % pi * D av e *N

12 a = d * f // a r ea o f c ut i n mmˆ213 mrr = a *V / / m e t a l r e m o va l r a t e i n mmˆ 3 / min .14 T = l /( f* N) / / m ac hi ne t i m i ng i n min .15 c = 5 6 // c o ns t an t fr om t a b l e16 p = d *f * v *6 0* c // power i n w at ts17 o me ga = ( 2* % p i *N ) / 60 // rpm18 t = p / om eg a // t o r qu e i n Nm19 F c = ( 2* t * 1 00 0) / D a ve // c u t t i n g f o r c e i n N

20 printf ( ” \n C u t ti n g s p e ed = %0 . 2 f m/ s \n MRR = %d mmˆ3/ min . \ n Time t o c u t = %0 . 2 f min . \ n P ow er = %0 . 1f w at ts \ n C u tt i n g f o r c e = %d N” , v , mrr , T ,p ,

Fc )

21 / / A ns wer s a r e g i v e n wrong i n b oo k

98



Scilab code Exa 15.9 To find MRR power and torque

1 clc

2 f = 0.2 / / f e e d i n mm/ r e v .3 N = 800 / / s p i n d l e s p ee d i n r e v . / min .4 d = 1 0 // d oa me te r o f h o l e i n mm5 m rr = % pi * ( d ^2 ) * f* N /4 // m e ta l r e mo v a l r a t e i n mmˆ 3/

min .

6 m r r = m rr / 60 // mmˆ3 / s7 p = 0 .5 * mrr // c u t t i n g power from t a b l e 1 4 . 2 i n

w a t t s8 o me ga = 2 * %p i * N /6 0 // rpm9 T = p / om eg a // t or qu e i n N .m

10 printf ( ” \n MRR = %0 . 2 f mmˆ3 / s \n C u t t in g p ow er = %0 . 3f w at ts \n T o rq u e = %0 . 2 f N . m” , mr r , p , T)

Scilab code Exa 15.10 find MRR power torque and time

1 clc

2 l = 300 / / l e n g th i n mm3 w = 100 / / w id t h i n mm4 f = 0.25 / / f e e d i n mm/ t o o t h5 d = 3.2 // d ep th o f c ut i n mm6 D = 5 0 // c u t t e r d i am et er i n mm7 n = 2 0 // number o f c u t t e r t e e t h8 N = 100 / / c u t t e r s p ee d i n r e v . / min .9 t f = f *n *N

/ / t a b l e f e e d i n mm/ min .10 m r r = w * d* tf / / m e ta l r e mo v al r a t e i n mmˆ 3/ min .11 m r r = m rr / 60 // mmˆ3 / s12 p = 6* mrr // c u t t i n g power from t a b l e 1 4 .2 i n w at ts13 o me ga = 2 * %p i * N /6 0 // rpm

99



14 T = p / om eg a // t or qu e i n N .m

15 a tt = sqrt ( ( D * d ) - ( d ^ 2 ) ) // added t a b l e t r a v e l i n mm16 t = ( l+ a tt ) / tf / / c u t t i n g t im e i n min .17 t = t *60 // s18 printf ( ” \n MRR = %0 . 2 f mmˆ3 / s \n C u t t in g p ow er = %d

w a t t s \n T or qu e = %0 . 2 f N . m\n C u tt i ng t im e = %0 . 1f s ” , m rr , p , T , t )

100



Chapter 16

Gear manufacture

Scilab code Exa 16.1 Calculate settings of gear tooth

1 clc

2 n = 3 4 // number o f t e e t h s3 m = 5 / / m od ul e i n mm4 w = m * n *sin ( % p i / ( n * 2 ) ) / / t o o th t h i c k n e s s i n mm5 h = m *( 1+ (n *(1 - cos ( % p i / ( n * 2 ) ) ) / 2 ) ) // c h o r da l

addendum in mm

6 printf ( ” \n To ot h t h i c k n e s s = %0 . 3 f mm\n C h or d aladdendum = %0 . 3 f mm” , w , h)


101



Chapter 17

Thread manufacturing

Scilab code Exa 17.1 Calculate best wire size

1 clc

2 d = 8 0 // o u t s i d e d i am e te r i n mm3 p = 6 // p i t c h d i am e te r i n mm4 d = 0 .5 77 4* p // b es t w ir e s i z e i n mm5 printf ( ” \n B es 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 clc

2 D = 2 0 / / d i a me t e r i n mm3 p = 2.5 // p i t c h d i am et er i n mm4 d = 0 .5 77 4* p / / mm5 W = D + 3* d - 1 .5 1 56 * p // b es t w ir e s i z e i n mm6 printf ( ” \n B es t w i re s i z e = %0 . 3 f mm\n D i s ta n ce o v er

w i r e s = %0. 3 f mm” , d, W )


102



Chapter 21

Statical quality control

Scilab code Exa 21.1 Construct R and X chart

1 clc

2 c l f ( )

3 n = 1 0 // number o f s a m p l es4 A 2 = 0 .5 77

5 D 3 = 0

6 D 4 = 2 .1 15

7 // n umber o f d e f e c t i v e s8 x 1 = 1 1. 27 4

9 x 2 = 1 1. 24 6

10 x 3 = 1 1. 20 4

11 x 4 = 1 1. 29 4

12 x 5 = 1 1. 25 2

13 x 6 = 1 1. 23 8

14 x 7 = 1 1. 23 0

15 x 8 = 1 1. 27 6

16 x 9 = 1 1. 20 8

17 x 10 = 1 1. 26 618 r 1 = 0. 15

19 r 2 = 0. 20

20 r 3 = 0. 33

21 r 4 = 0. 46

103



22 r 5 = 0. 10

23 r 6 = 0. 1524 r 7 = 0. 20

25 r 8 = 0. 23

26 r 9 = 0. 50

27 r10 = 0 .3 0

28 x = x 1 + x 2 + x3 + x 4 + x 5 + x6 + x 7 + x 8 + x 9 + x 10

29 r = r 1 + r 2 + r3 + r 4 + r 5 + r6 + r 7 + r 8 + r 9 + r 10

30 Xa vg = x /n

31 Ra vg = r /n

32 / / f o r X c h ar t33 u cl 1 = X av g + A2 * R av g

34 l cl 1 = X av g - A2 * R av g35 / / f o r R c h ar t36 u c l2 = D 4 * Ra vg

37 l c l2 = D 3 * Ra vg

38 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 F o r R c h a r t s \n UCl= %0. 3 f \n LCL = %0. 3 f ” , u cl 1 , l c l1 , u cl 2 , l c l 2 )

39 // X c h a rt40 x = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 1 0 ] ;

41 y

= [ 1 1 . 2 7 4 , 1 1 . 2 4 6 , 1 1 . 2 0 4 , 1 1 . 2 9 4 , 1 1 . 2 5 2 , 1 1 . 2 3 8 , 1 1 . 2 3 0 , 1 1 . 2 7 6 , 1 1

42 plot ( x , y )

43 xtitle ( ”X c h a r t ” , ”Sampl e No . ” , ”X” )

44 // R c h a rt45 xset ( ”window” ,1 )

46 z =

[ 0 . 1 5 , 0 . 2 0 , 0 . 3 3 , 0 . 4 6 , 0 . 1 0 , 0 . 1 5 , 0 . 2 0 , 0 . 2 3 , 0 . 5 0 , 0 . 3 0 ]

47 plot ( x , z )

48 xtitle ( ” R c h a r t ” , ”Sampl e no . ” , ”R” )

Scilab code Exa 21.2 Construct the control charts

104



1 clc

2 c l f ( )3 n = 100 // t o t a l number o f sub g ro up s4 s = 1 0 // number o f s a m p l es5 // n umber o f d e f e c t i v e s6 d 1 = 3

7 d 2 = 2

8 d 3 = 3

9 d 4 = 5

10 d 5 = 3

11 d 6 = 3

12 d 7 = 2

13 d 8 = 414 d 9 = 3

15 d10 = 2

16 d = d 1 + d 2 + d3 + d 4 + d 5 + d6 + d 7 + d 8 + d 9 + d 10 // t o t a l numbero f d e f e c t i v e s

17 p 1 = d /( n *s ) // a ve ra ge f r a c t i o n o f d e f e c t i v e s18 s i g ma p 1 = sqrt ( p 1 * ( 1 - p 1 ) / n )

19 uc l1 = p1 + 3* s ig ma p1

20 lc l1 = p1 - 3* s ig ma p1

21 // c o n t r o l c ha rt f o r f r a c t i o n d e f e c t i v e s22 x = linspace ( 0 , 1 0 , 1 0 )

23 y = linspace ( 0 , 0 . 0 8 1 , 1 0 )

24 plot ( x , y )

25 xtitle ( ” C on tr ol c ha 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 ct i o n d e f e c t i v e s ” )

26 // p e r ce n t d e f e c t i v e ( mean )27 p1 = p1 * 100

28 s i g ma p 2 = sqrt ( p 1 * ( 1 0 0 - p 1 ) / n )

29 uc l2 = p1 + 3* s ig ma p2

30 lc l2 = p1 - 3* s ig ma p2

31 printf ( ” \n C o nt ro l l i m i t s \n F ra 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 er ce nt d e f e c t i v e s \nUCL = %0.1 f \n LCL = %0. 1 f ” , u cl 1 , l c l1 , u cl 2 , l c l2

)

32 // c o n t r o l c ha r t f o r p e r ce nt d e f e c t33 xset ( ”window” ,1)

105



34 z = linspace ( 0 , 8 . 1 , 1 0 )

35 plot ( x , z )36 xtitle ( ” C o nt ro l c h ar t f o r p e rc e n t d e f e c t s ” , ”Sampl eno . ” , ” p e r c e n t d e f e c t s ” )

37 / / ’ Answers v a ry due t o round o f f e r ro r ’

Scilab code Exa 21.4 Calculate poisson probabilities

1 clc

2 n = 1000 // number o f u n i t s3 s = 4 / / r an do m s a m p l e4 d = 5 0 // d e f e c t i v e s5 z = d *s /n

6 p p0 = exp ( - 0 . 2 ) * 1 // p oi ss on p r o b a b i l i t i e s f o r 0d e f e c t i v e s

7 p p1 = exp ( - 0 . 2 ) * ( z ) // p oi ss on p r o b a b i l i t i e s f o r 1d e f e c t i v e s

8 p p2 = exp ( - 0 . 2 ) * ( z ^ 2 / f a c t o r i a l ( 2 ) ) // p o i ss o np 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

9 p p3 = exp ( - 0 . 2 ) * ( z ^ 3 / f a c t o r i a l ( 3 ) ) // p o i ss o n

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 s10 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 ” , p p0 , p p1 , p p2 ,

p p 3 )

Scilab code Exa 21.5 Calculate probabilities of defective items

1 clc

2 d = 5 0 // d e f e c t i v e s3 l = 1000 // l o t o f p i e c e s4 p = d / l // p r o a b i l i t y o f an e ve nt h ap pe n in g5 q = 1 - p // p r o a b i l i t y o f an e ve nt no t h ap pe ni ng6 n = 4 // sa mp le s i z e

106



7 p 0 = 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

8 p 1 = 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 s9 p 2 = 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 s10 p 3 = 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 s11 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 , p 1 , p2 , p 3 )

Scilab code Exa 21.6 Determine producers and consumers risk

1 clc2 / / p ro d uc er ’ s r i s k3 n = 7 1 // s am pl e s i z e4 A Q L = 0 .0 05

5 L TP D = 0 .0 5

6 l _s = 500 // l o t s i z e7 z 1 = n * AQL / / mean number o f d e f e c t s8 p p1 = exp ( - z 1 ) + z 1 * exp ( - z 1 ) // p o i ss o n p r o a b i l i t y f o r

1 o r l e s s d e f e c t i v e9 a lp ha = ( 1 - pp 1 ) *1 00 / / p r od u ce r ’ s r i s k

10 a l ph a = ceil ( a l p h a )

11 / / c on su me r ’ s r i s k12 z2 = n * LT PD / / mean number o f d e f e c t s13 p p2 = exp ( - z 2 ) + z 2 * exp ( - z 2 ) // p o i ss o n p r o a b i l i t y f o r

1 o r l e s s d e f e c t i v e14 b i ta = p p2 * 1 00 / / c on su me r ’ s r i s k15 printf ( ” \n P ro d uc er s r i s k = %d p e r ce n t \n C onsume rs

r i s k = %0 . 2 f p e r c e n t ” , a lp ha , b it a )

Scilab code Exa 21.7 Evaluate preliminary and revised control limits

1 clc

2 t d 1 = 2 0 // t o t a l number o f d ays3 n 1 = 200 // sa mpl e s i z e

107



4 // n umber o f d e f e c t i v e s

5 d 1 = 106 d 2 = 15

7 d 3 = 10

8 d 4 = 12

9 d 5 = 11

10 d 6 = 9

11 d 7 = 22

12 d 8 = 4

13 d 9 = 12

14 d 10 = 24

15 d 11 = 21

16 d 12 = 1517 d13 = 8

18 d 14 = 14

19 d15 = 4

20 d 16 = 10

21 d 17 = 11

22 d 18 = 11

23 d 19 = 26

24 d 20 = 13

25 d = d 1 + d 2 + d3 + d 4 + d 5 + d6 + d 7 + d 8 + d 9 + d 10 + d 1 1 + d 1 2 + d 13 + d 1 4 +

d 1 5 + d 1 6 + d 1 7 + d 1 8 + d 1 9 + d 2 0 // t o t a l number o f d e f e c t i v e s

26 p 1 = d /( n 1* t d1 ) // a ve r a ge f r a c t i o n o f d e f e c t i v e s27 s i g ma p 1 = sqrt ( p 1 * ( 1 - p 1 ) / n 1 )

28 uc l1 = p1 + 3* s ig ma p1

29 lc l1 = p1 - 3* s ig ma p1

30 // r e v i s e d c o n t r o l l i m i t s31 t d2 = 18 // t o t a l number o f d ay s32 D = d - ( d10 + d19 ) // number o f d e f e c t s33 p 2 = D /( n 1* t d2 )

34 s i g ma p 2 = sqrt ( p 2 * ( 1 - p 2 ) / n 1 )

35 uc l2 = p2 + 3* s ig ma p236 lc l2 = p2 - 3* s ig ma p2

37 printf ( ” \n P r e l i mi 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 is ed c o n t r o l l i m i t s \n UCL

= %0. 3 f \n LCL = %0. 3 f ” , u cl 1 , l c l1 , u cl 2 , l c l 2 )

108



Scilab code Exa 21.8 Find control limits for c chart

1 clc

2 n 1 = 15 // t o t a l number o f sub g ro up s3 // n umber o f d e f e c t i v e s4 d 1 = 77

5 d 2 = 64

6 d 3 = 75

7 d 4 = 938 d 5 = 45

9 d 6 = 61

10 d 7 = 49

11 d 8 = 65

12 d 9 = 45

13 d 10 = 77

14 d 11 = 59

15 d 12 = 54

16 d 13 = 84

17 d 14 = 40

18 d 15 = 92

19 d = d 1 + d 2 + d3 + d 4 + d 5 + d6 + d 7 + d 8 + d 9 + d 10 + d 1 1 + d 1 2 + d 13 + d 1 4 +

d15 // t o t a l number o f d e f e c t i v e s20 c 1 = d /n1

21 ucl1 = c1 + 3* sqrt ( c 1 )

22 lcl1 = c1 - 3* sqrt ( c 1 )

23 // r e v i s e d c o n t r o l l i m i t s24 n 2 = 12 // t o t a l number o f sub g ro up s25 D = d - ( d 4 + d 1 4 + d 1 5 ) // number o f d e f e c t s26 c 2 = D /n2

27 ucl2 = c2 + 3* sqrt ( c 2 )28 lcl2 = c2 - 3* sqrt ( c 2 )

29 printf ( ” \n P r e l i mi 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 is ed c o n t r o l l i m i t s \n UCL

= %0. 3 f \n LCL = %0. 3 f ” , u cl 1 , l c l1 , u cl 2 , l c l 2 )

109



Scilab code Exa 21.9 find control limits for charts

1 clc

2 n = 2 0 // number o f s a m p l es3 A = 1. 34 2

4 A 1 = 1 .5 96

5 A 2 = 0 .5 77

6 d 2 = 2 .3 26

7 d 3 = 0 .8 648 D 1 = 0

9 D 2 = 4 .9 18

10 D 3 = 0

11 D 4 = 2 .1 15

12 // n umber o f d e f e c t i v e s13 x 1 = 32 90

14 x 2 = 31 80

15 x 3 = 33 50

16 x 4 = 34 70

17 x 5 = 30 80

18 x 6 = 32 40

19 x 7 = 32 60

20 x 8 = 33 10

21 x 9 = 36 40

22 x 10 = 4 11 0

23 x 11 = 3 22 0

24 x 12 = 3 59 0

25 x 13 = 4 27 0

26 x 14 = 4 04 0

27 x 15 = 3 58 0

28 x 16 = 3 50 029 x 17 = 3 57 0

30 x 18 = 3 56 0

31 x 19 = 2 74 0

32 x 20 = 3 20 0

110



33 r 1 = 560

34 r 2 = 41035 r 3 = 200

36 r 4 = 300

37 r 5 = 90

38 r 6 = 650

39 r 7 = 890

40 r 8 = 410

41 r 9 = 11 20

42 r 10 = 520

43 r 11 = 580

44 r 12 = 670

45 r 13 = 48046 r 14 = 250

47 r 15 = 170

48 r 16 = 670

49 r 17 = 440

50 r 18 = 660

51 r 19 = 560

52 r 20 = 590

53 x = x 1 + x 2 + x3 + x 4 + x 5 + x6 + x 7 + x 8 + x 9 + x 10 + x 1 1 + x 1 2 + x 13 + x 1 4 +

x 1 5 + x 1 6 + x 1 7 + x 1 8 + x 1 9 + x 2 0

54 r = r 1 + r 2 + r3 + r 4 + r 5 + r6 + r 7 + r 8 + r 9 + r 10 + r 1 1 + r 1 2 + r 13 + r 1 4 +

r 1 5 + r 1 6 + r 1 7 + r 1 8 + r 1 9 + r 2 0

55 Xa vg = x /n

56 Ra vg = r /n

57 / / f o r X c h ar t58 u cl 1 = X av g + A2 * R av g

59 l cl 1 = X av g - A2 * R av g

60 / / f o r R c h ar t61 u c l2 = D 4 * Ra vg

62 l c l2 = D 3 * Ra vg

63 // R e vi se d c o n t r o l l i m i t s

64 n 1 = 1565 n 2 = 19

66 X = ( x - ( x 5 + x1 0 + x1 3 + x1 4 + x1 9 ) )/ n 1

67 R = ( r - ( r9 )) /n2

68 / / f o r X c h ar t

111



69 ucl3 = X + A2 *R

70 lcl3 = X - A2 *R71 / / f o r R c h ar t72 uc l4 = D4 *R

73 lc l4 = D3 *R

74 printf ( ” \n P r e l i mi n a r y c o n t r o l l i m i t s \n For Xc h a r t s \n UCL = %0.2 f \n LCL = %0.2 f \n F o r Rc h a r t s \n UCl = %0. 3 f \n LCL = %0.3 f \n R e vi s edc o nt r ol l i m i t s \n F o r X c h a r t \n UCL = %0.3 f \nLCL = %0. 3 f \n F o r R c h a r t s \n UCl = %0. 3 f \n LCL= %0. 3 f ” , u cl 1 , l c l1 , u cl 2 , l c l2 , u cl 3 , u c l3 , u cl 4 ,

l c l 4 )

75 / / ’ Answers v a ry due t o round o f f e r ro r ’

Scilab code Exa 21.10 Determine producers and consumers risk

1 clc

2 c l f ( )

3 n = 5 0 // s am pl e s i z e4 r n = 2 // r e j e c t i o n number

5 A QL = 0 .0 26 L TP D = 0 .0 8

7 / / P ro du ce r ’ s r i s k8 z 1 = n * AQL // mean number o f d e f e c t i v e s9 p p1 = exp ( - z 1 ) + z 1 * exp ( - z 1 ) // p o i ss o n p r o a b i l i t y f o r

1 o r l e s s d e f e c t i v e10 a lp ha = ( 1 - pp 1 ) *1 00 / / p r od u ce r ’ s r i s k11 / / c on su me r ’ s r i s k12 z2 = n * LT PD // mean n umber o f d e f e c t i v e s13 b ita = ( exp ( - z 2 ) + z 2 *exp ( - z 2 ) ) * 1 0 0 / / c on su me r ’ s r i s k

14 d 1 = 1 // i nc om in g d e f e c t i v e i n p e rc e n t15 z 3 = n * d1 / 10 0 // a v er a ge number o f d e f e c t i v e16 p p a1 = exp ( - z 3 ) + z 3 * exp ( - z 3 ) // p r o a b i l i t y o f

a c c e p t a n c e17 p p a1 = p pa 1 * 10 0

112



18 p p r1 = 1 00 - p p a1 // p r o a b i l i t y o f r e j e c t i o n

19 A OQ 1 = p pr 1 *0 + p pa 1 * d1 / 10 020 d 2 = 2 // i nc om in g d e f e c t i v e i n p e rc e n t21 z 4 = n * d2 / 10 0 // a v er a ge number o f d e f e c t i v e22 p p a2 = exp ( - z 4 ) + z 4 * exp ( - z 4 ) // p r o a b i l i t y o f


24 p p r2 = 1 00 - p p a2 // p r o a b i l i t y o f r e j e c t i o n25 A OQ 2 = p pr 2 *0 + p pa 2 * d2 / 10 0










44 printf ( ” \n P ro d uc er s r i s k = %0 . 2 f p e r ce n t \nC on su me rs r i s k = %0 . 3 f p e r c e n t ” , a lp ha , b it a )

45 x = [ 1 ,2 , 4 ,6 , 8]

46 y = [ 0 .9 1 , 1 . 4 7 16 , 1 . 62 4 , 1 . 1 9 4 , 0 . 73 3 ]47 plot ( x , y )

48 xtitle ( ”AOQ cu rv e ” , ” P e rc en t d e c t i v e s ” , ”AOQ of l o t ”)

113



Chapter 22

Kinematics of machine tools

Scilab code Exa 22.1 Find range of cutting velocity

1 clc

2 d 1 = 10 / / min . d i a o f c u t t e r i n mm3 d 2 = 60 // max . d i a o f c u t t e r i n mm4 v = 30 e3 // o p e r a t i n g s p ee d i n m/ min5 n1 = v / ( %pi * d2 ) / / n mi n i n rpm6 n2 = v / ( %pi * d1 ) / / n max i n rpm

7 phi = ( n2 / n1 ) ^( 1/ 5)8 s p i n d l e _s p e e ds = zeros ()

9 for i = 0 : 5

10 s p in d le _ sp e ed s ( i + 1) = p hi ^ i * n 1

11 end

12 c ut te r_ di a = v . / ( % pi * s pi nd le _s pe ed s )

13 c l f ( )

14 y = [0; v ]

15 plot ( [0 ; c u tt e r_ d ia ( 1 ) ] , y , [ 0; c u tt e r_ d ia ( 2 ) ] , y ,

[ 0; c u tt e r_ d ia ( 3 ) ] , y , [ 0; c u tt e r_ d ia ( 4 ) ] , y , [ 0;

c u tt e r_ d ia ( 5 ) ] , y , [ 0; c u tt e r_ d ia ( 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 / / f ro m g r ap h18 v ma x1 = 36 // m/min

114



19 v mi n1 = 2 4. 5 // m/min

20 r 1 = v ma x1 - v mi n1 // Range o f c u t t i n g s pe ed f o r 12mm d i a m e t e r i n m/ m in21 v ma x2 = 3 6. 5 // m/min .22 v mi n2 = 26 // m/min .23 r 2 = v ma x2 - v mi n2 // Range o f c u t t i n g s pe ed f o r 36

mm d i a m e t e r i n m/ m in24 printf ( ” \n Range o f c u t t i n g s pe ed f o r 12 mm d ia m et er

= %0. 1 f m/min . \ n R ange o f c u t t i n g s pe e d f o r 36mm di a me t e r = %0. 1 f m/min . ” , r1 , r2 )

Scilab code Exa 22.2 Determine speed ratios and teeth

1 clc

2 m = 2.5 / / m od ul e i n mm3 p hi = 1.2 / / common r a t i o4 n = 150 // s pe ed i n r ev / min . o f d r i v i n g s h a f t5 n 1 = 70 // s pe ed i n r ev / min . o f d r iv e n s h a f t6 n 2 = ( p hi ) ^1 * n1 // s pe ed i n r ev / min . o f d r iv e n s h a f t7 n 3 = ( p hi ) ^2 * n1 // s pe ed i n r ev / min . o f d r iv e n s h a f t

8 n 4 = ( p hi ) ^3 * n1 // s pe ed i n r ev / min . o f d r iv e n s h a f t9 T1 = poly (0 , ’ T1 ’ )

10 t 1 = n 1 / n * T 1

11 T1 = roots ( t 1 + T 1 - 8 0 )

12 t1 = horner ( t 1 , T 1 )

13 T2 = poly (0 , ’ T2 ’ )

14 t 2 = n 2 / n * T 2

15 T2 = roots ( t 2 + T 2 - 8 0 )

16 t2 = horner ( t 2 , T 2 )

17 T3 = poly (0 , ’ T3 ’ )

18 t 3 = n 3 / n * T 319 T3 = roots ( t 3 + T 3 - 8 0 )

20 t3 = horner ( t 3 , T 3 )

21 T4 = poly (0 , ’ T4 ’ )

22 t 4 = n 4 / n * T 4

115



23 T4 = roots ( t 4 + T 4 - 8 0 )

24 t4 = horner ( t 4 , T 4 )25 t1 = floor ( t 1 ) // number o f t e et h on d r i v i n g s h a f t26 T1 = ceil ( T 1 ) // number o f t e et h on d r iv e n s h a f t27 t2 = ceil ( t 2 ) // number o f t e et h on d r i v i n g s h a f t28 T2 = floor ( T 2 ) // number o f t e et h on d r iv e n s h a f t29 t3 = floor ( t 3 ) // number o f t e et h on d r i v i n g s h a f t30 T3 = ceil ( T 3 ) // number o f t e et h on d r iv e n s h a f t31 t4 = ceil ( t 4 ) // number o f t e et h on d r i v i n g s h a f t32 T4 = floor ( T 4 ) // number o f t e et h on d r iv e n s h a f t33 // r u n ni ng s p ee d s34 n 1 = n * t1 / T1

35 n 2 = n * t2 / T236 n 3 = n * t3 / T3

37 n 4 = n * t4 / T4

38 printf ( ” \n Number o f t e e t h on d r i v e r and d r iv 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 ” , t 1 , T 1 , t 2 , T 2 , t 3

, T 3 , t 4 , T 4 )

39 printf ( ” \n The a c t ua l r un ni ng s pe ed o f d r iv e n s h a f tw i l l be : \n n1 = %0 . 2 f r e v / min \n n2 = %0 . 2 f r e v /mi n \n n3 = %0 . 2 f r e v / min \n n4 = %0 . 2 f r e v / min ” ,

n 1 , n 2 , n 3 , n 4 )

40 / / A nswer o f n3 i s g i ve n wrong i n book41 / / A nswer v ar y due t o ro un d o f f e r r o r

Scilab code Exa 22.3 Calculate speed and number of teeths

1 clc

2 z = 6 // number o f s t e p s

3 n 1 = 180 / / r e v / min4 n 2 = 100 / / r e v / min5 R n = n1 / n2

6 p hi = ( R n ) ^( 1/ ( z - 1) ) / / common r a t i o7 n3 = ph i* n2 / / r e v / min

116



8 n 4 = ( p hi ) ^2 * n2 / / r e v / min

9 n 5 = ( p hi ) ^3 * n2 / / r e v / min10 n 6 = ( p hi ) ^4 * n2 / / r e v / min11 n 7 = 225 // s pe e d o f i np ut s h a f t i n r ev / min12 Ta = poly (0 , ’ Ta ’ )

13 t b = n 7 / n 5 * T a

14 Ta = roots ( t b + T a - 5 2 )

15 tb = horner ( t b , T a )

16 tb = ceil ( t b )

17 Tc = poly (0 , ’ Tc ’ )

18 t d = n 7 / n 6 * T c

19 Tc = roots ( t d + T c - 5 2 )

20 td = horner ( t d , T c )21 Tc = ceil ( T c )

22 Te = poly (0 , ’ Te ’ )

23 t f = n 7 / n 1 * T e

24 Te = roots ( t f + T e - 5 2 )

25 tf = horner ( t f , T e )

26 tf = ceil ( t f )

27 Th = poly (0 , ’ Th ’ )

28 t j = n 2 / n 5 * T h

29 Th = roots ( t j + T h - 4 6 )

30 Th = ceil ( T h )

31 tj = horner ( t j , T h )

32 tj = floor ( t j )

33 Ti = poly (0 , ’ T i ’ )

34 t g = n 5 / n 5 * T i

35 Ti = roots ( t g + T i - 4 6 )

36 tg = horner ( t g , T i )

37 printf ( ” \n Ta = %d Tb = %d \n Tc = %d Td = %d \n Te= %d t f = %d \n T h = % d T j = % d \n T i = %d Tg =%d” , T a , tb , T c , t d , T e , tf , t j , T h , T i , t g )


Scilab code Exa 22.4 Calculate common ratio

117



1 clc

2 v = 2 1 // c u t t i n g s pe ed i n r ev / min .3 z = 6

4 d min = 5 / / d a i me t er i n mm5 d max = 20 / / d a i m et e r i n mm6 n m ax = 1 0 00 * v / ( % p i * d m in ) // s p i n d l e s pe ed i n r ev / min

.7 n m in = 1 0 00 * v / ( % p i * d m ax ) // s p i n d l e s pe ed i n r ev / min

.8 p h i = ( n m a x / n m in ) ^ ( 1 / ( z - 1 ) ) / / common r a t i o9 n 1 = nm in / / r e v / min .

10 n2 = ph i* n1 / / r e v / m in .

11 n 3 = ( p hi ) ^2 * n1 / / r e v / m in .12 n 4 = ( p hi ) ^3 * n1 / / r e v / m in .13 n 5 = ( p hi ) ^4 * n1 / / r e v / m in .14 n 6 = ( p hi ) ^5 * n1 / / r e v / m in .15 printf ( ” \n Common r a t i o = %0 . 2 f \n S p i n d le s p ee ds

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 / m in . ” , p h i , n 1 , n 2 , n 3 , n 4 , n 5 , n 6 )


Scilab code Exa 22.5 Calculate gear ratio teeth and speed

1 // from f i g . 2 2 .1 8A2 clc

3 // Th ree g e ar r a t i o s b et we en i n pu t and i n t e r m e d i a tes h a f t

4 n ma x = 1 40 0 / / maximum s p e e d i n r e v / m in .5 i 1 = 1/1

6 i 2 = 1 /1 .2 6

7 i 3 = 1 /( 1 .2 6 ) ^28 / / The t wo r a t i o s b et we en i n t e r m e d i at e and o ut pu ts h a f t

9 i 4 = 1/1

10 i 5 = 1 /( 1 .2 6 ) ^3

118



11 / / number o f t e et h f o r i np ut and i n t e r m e di a t e s h a f t

12 t 1 = 2 7/ 2713 t 2 = 2 4/ 30

14 t 3 = 2 1/ 33

15 // number o f t e e t h f o r o ut pu t and i n t e r m e di a t es h a f t

16 t 4 = 3 4/ 34

17 t 5 = 2 0/ 48

18 / / o u tp ut s p e e d s i n r e v . / min19 n 1 = t 3 * t5 * n ma x

20 n 2 = t 2 * t5 * n ma x

21 n 3 = t 1 * t5 * n ma x

22 n 4 = t 3 * t4 * n ma x23 n 5 = t 2 * t4 * n ma x

24 n 6 = t 1 * t4 * n ma x

25 printf ( ” \n T hree g e ar r a t i o s b et we en i n pu t andi n t e r m e d i a te 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 be t we e n i n t er m e di a t e ando ut pu t s h a f t i 4 = %d i 5 = %0 . 3 f \n nu mb er o f t e et h f o r e ac h p a i r bet w ee n i np ut andi n t e r m e di a t e s h a f t t 1 = 2 7/ 27 , t 2 = 24/ 30 , t 3 =2 1 / 3 3 \n number o f t e e t h f o r e ac h p a i r b et we en

o ut pu t and i n t e r m e d i a te s h a f t = t 4 = 3 4/ 34 , t 5 =2 0 / 4 8 \n Ou tp ut s p e e d s \n n 1 = %d r ev / min , n2 =%d r e v / min , n3 = %d r e v / min \n n 4 = %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 )


119



Chapter 23

Production planning and

control

Scilab code Exa 23.1 Calculate forecast

1 clc

2 d 1 = 90 / / demand f o r f i r s t q u a r t e r3 d 2 = 100 // demand f o r s e co n d q u a r t e r4 d 3 = 80 // demand f o r t h i r d q u a r te r5 s a = ( d 1 + d2 + d 3 ) /3 // s i m pl e a v er a ge6 printf ( ” \n F o r e c a s t = %d” , s a )

Scilab code Exa 23.2 Calculate forecat by SMA method

1 clc

2 d 1 = 300 // demand f o r j u l y3 d 2 = 350 // demand f o r a u g u s t4 d 3 = 400 / / demand f o r s e p te m b er5 d 4 = 500 // demand f o r o c t o b e r6 d 5 = 600 / / demand f o r n ov em be r7 d 6 = 700 / / d emand f o r d e ce m be r

120



8 // a ssu mi ng n = 3 , wher e n i s number o f t i me p e ri o d

9 f o re ca s t = ( d 6 + d5 + d 4 ) /3 // f o r e c a s t10 printf ( ” \n F o r e c a s t = %d” , f o re c as t )

Scilab code Exa 23.3 Calculate forecat by WMA method

1 clc

2 d 1 = 500 // demand f o r o c t o b e r3 d 2 = 600 / / demand f o r n ov em be r

4 d 3 = 700 / / d emand f o r d e ce m be r5 w 1 = 0. 25 // r e l a t i v e w ei gh t w it h d ec em ber6 w 2 = 0. 25 // r e l a t i v e w ei gh t w it h november7 w 3 = 0.5 // r e l a t i v e w e ig ht w i t h o c t ob e r8 f = w1 *d1 + w2 *d2 + w3 *d3 // f o r e c a s t9 printf ( ” \n F o r e c a s t by w e ig h te d moving a v e ra g e = %d”

, f )

Scilab code Exa 23.4 Calculate forecast for january

1 clc

2 a l ph a = 0 .7 // sm oo th in g c o e f f i c i e n t3 d 1 = 250 / / demand f o r n ov em be r4 d 2 = 300 / / d emand f o r d e ce m be r5 f 1 = 200 // f o r e c a s t f o r november6 f 2 = a lp ha * d1 + (1 - a lp ha ) * f1 // f o r e c a s t f o r

de c e mbe r7 f 3 = a lp ha * d2 + (1 - a lp ha ) * f2 // f o r e c a s t f o r j an ua ry8 f3 = ceil ( f 3 )

9 printf ( ” \n F o r ec a s t f o r j a nu a ry = %d u n i t s ” , f 3 )

121



Scilab code Exa 23.5 Calculate total cost

1 clc

2 s = 600 // s e t up c os t p e r l o t i n Rs3 c = 6 // u ni t c o s t o f i t e m i n Rs4 a = 1 00 00 0 / / a n nu a l demand f o r i te m5 i = 2 5 // a nn u al c a r r y i n g c ha r g e s o f a ve ra ge

i n v e n t o r y6 i = 2 5/ 10 0

7 k = c * i // c a r r y i n g c o st f a c t o r i n u n i t / y ea r8 n = sqrt ( 2 * s * a / k ) // most e c on om ic l o t s i z e9 t c = a * c + s * a / n + k * n / 2 // t o t a l c os t i n Rs

10 printf ( ” \n T o ta l c o s t = Rs %0 . 2 f ” , t c )11 / / ’ Answers v a ry due t o round o f f e r ro r ’

Scilab code Exa 23.6 Calculate economical order quantity

1 clc

2 a = 8000 // a nn ua l r e qu i re m en t o f p a r t s3 c = 6 0 // u ni t c o s t o f p a rt i n Rs

4 r = 150 // o r d e r i n g c o s t p e r l o t i n Rs5 i = 3 0 // a nn ua l c a r r yi n g c h ar g es o f a v e ra ge

i n v e n t o r y6 i = 3 0/ 10 0

7 k = i * c // c a r r y i n g c o st p er u ni t p er y ea r8 n = sqrt ( 2 * r * a / k ) // most e c on o mi c al o r d er q u a n ti t y9 printf ( ” \n Most e co n o mi c al o r d e r i n g q u a nt i t y = %d

u n i t s ” , n )

Scilab code Exa 23.7 Calculate economic lot size

1 clc

2 a = 12 00 0 // a n nu a l r e q ui r e m en t

122



3 c = 5 // u ni t c o s t o f p ar t

4 s = 6 0 // s e t up c o s t p e r l o t5 p = 18 75 0 // p r o d uc ti o n r a t e p er y ea r6 i = 2 0 // i n v e n to r y c a r r y i n g c o s t7 i = 2 0/ 10 0

8 k = i * c // c a r r y i n g c o st p er u ni t p er y ea r9 n = sqrt ( 2 * s / ( 1 / a - 1 / p ) * k ) // Most e co no mi c l o t s i z e

10 printf ( ” \n Most e co no mi 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 clc

2 a = 15 62 5 // a nn ua l r e qu i re m en t o f p a r t s3 c = 1 2 // u ni t c o s t o f p a rt i n Rs4 r = 6 0 // o r d e r i n g c o s t p e r l o t i n Rs5 k = 1.2 // i n v e n t o r y c a r r y i n g c o st p er u n i t6 n = sqrt ( 2 * r * a / k ) // e c on o mi c al o r d er q u a n ti t y7 o c = r *a /n // o r de r i n g c o st i n Rs8 c c = k *n /2 // c a r r y i n g c o st i n Rs9 tc = oc + cc // t o t a l i nv en t o ry c o s t i n Rs

10 printf ( ” \n E co no mi ca l o r d er q u a n ti t y = %d u n i t s \no r d er c o s t = Rs %d\n c a r r y i n g c o s t = Rs %d\nT o ta 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 clc

2 / / c as e a

3 a = 5 0 // a nn ua l r e q ui r e me nt o f p a r t s i n t on ne s4 c = 500 // u ni t c o s t o f p ar t i n Rs5 r = 100 // o r de r i n g c o st p er o r d er i n Rs6 i = 2 0 // i n v e n to r y c a r r y i n g c o s t7 i = i / 1 0 0

123



8 d = 2 // d i s c o u nt o f p ur ch as e c o st i n p e r ce nt

9 k = i * c // i n v e n t o r y c a r r y i n g c o st p er u n i t10 n1 = sqrt ( 2 * r * a / k ) // e c on o mi c al o r d er q u a nt i t y11 o c 1 = r * a/ n1 // o r de r i n g c o st i n Rs12 c c 1 = k * n1 / 2 // c a r r y i n g c o st i n Rs13 t c1 = oc1 + cc1 // t o t a l i nv en t o r y c o s t i n Rs14 / / c as e b15 n 2 = 25 // o rd er p e r l o t16 o c 2 = r * a/ n2 // o r de r i n g c o st i n Rs17 c c 2 = k * n2 / 2 // c a r r y i n g c o st i n Rs18 t c2 = oc2 + cc2 // t o t a l i nv en t o r y c o s t i n Rs19 i = tc2 - t c1 // i n c r e a s e i n c o s t i n Rs

20 d _ o = d * c *a / 10 0 // d i s c o u nt o f f e r e d21 printf ( ” \n I n c r e a s e i n i n ve n to r y c o s t = Rs %d\n

D i s c o u nt 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 ep t i n g ” )

Scilab code Exa 23.10 Calculate EOQ and reorder point

1 clc

2 a = 1 00 00 00 // a nn ua l r e qu i re m en t o f p a r t s3 r = 3 2 // o r d e r i n g c o s t p e r l o t i n Rs4 k = 4 // i n v e n t o r y c a r r y i n g c o st p e r u n i t5 d 1 = 250 // number o f w o rk i ng d a ys6 d 2 = 2 // da y s f o r s a f e t y s t o c k7 d 3 = 4 // l e ad t im e i n da ys8 e oq = sqrt ( 2 * r * a / k ) // e co n om i ca l o r d er q u a nt i t y9 o c = r * a/ e oq // o r de r i n g c o st i n Rs

10 c c = k * eo q /2 // c a r r y i n g c o st i n Rs11 tc = oc + cc // t o t a l i nv en t o ry c o s t i n Rs

12 s s = a * d2 / d1 // s a f e t y s t o c k13 r o _p = s s + eo q * d3 // r e o r d e r p o in t14 printf ( ” \n E co no mi c o r d e r q u n a n t i t y = %d c om po ne nt s \

n Re−o r d e r p o i n t = %d c om po ne nt s ” , e oq , r o_ p )

124



Chapter 26

Plant layout

Scilab code Exa 26.1 Calculate number of machine required

1 clc

2 N = 1 00 00 0 // a nn ua l o ut pu t o f p a r ts3 s = 2 // e x pe c te d s c r ap4 t = 105 // e st im at ed t i m e p er p ar t i n s5 i ta = 80 // p r o d uc t i o n e f f i c i e n c y o f machi ne6 a = 2300 // number o f w o rk i ng h o u r s

7 o u t p u t = ( 3 60 0 * i t a ) /( t * 1 0 0 ) // p a rt s r e q u i r e d p erhour

A Textbook of Production Engineering_P. C. Sharma

Documents

Transcript of A Textbook of Production Engineering_P. C. Sharma