Scilab Textbook Companion for Advanced Strength and Applied Elasticity by A. C. Ugural and S. K. Fenster 1 Created by Reshma Sunil Konjari MTech Electrical Engineering VIT University College Teacher None Cross-Checked by None July 31, 2019 1 Funded by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab codes written in it can be downloaded from the ”Textbook Companion Project” section at the website http://scilab.in
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Scilab Textbook Companion forAdvanced Strength and Applied Elasticity
by A. C. Ugural and S. K. Fenster1
Created byReshma Sunil Konjari
MTechElectrical Engineering
VIT UniversityCollege Teacher
NoneCross-Checked by
None
July 31, 2019
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: Advanced Strength and Applied Elasticity
Author: A. C. Ugural and S. K. Fenster
Publisher: Prentice Hall
Edition: 2
Year: 1981
ISBN: 0713134364
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
1 Analysis of Stress 5
2 Strain and stress strain relations 7
3 Two dimensional problems in elasticity 10
4 Mechanical behaviour of materials 12
5 Bending of beams 20
6 Torsion of prismatic bars 24
7 Numerical methods 27
8 Axisymmetrically loaded members 38
9 Beams on elastic foundations 45
11 Elastic stability 47
12 Plastic behavior of solids 49
3
List of Scilab Codes
Exa 1.1 Find stress . . . . . . . . . . . . . . . . . . 5Exa 1.2 Normal and shear stresses . . . . . . . . . . 6Exa 2.1 Principal strains . . . . . . . . . . . . . . . 7Exa 2.2 Principal stresses and their directions . . . . 8Exa 3.5 Size of the contact area and maxi contact
Exa 8.5 Determine maxi stress and radial displacement 42Exa 8.6 Distribution stress and disk profile . . . . . 44Exa 9.1 Calculate maxi deflections and force per unit
24 disp(gammamax ,”max s h e a r s t r a i n i s= ”)25 thetap=atand (577/320) /2
26 disp(thetap ,” o r i e n t a t i o n s o f p r i n c i p a l axe s i s= ”)// or
27 thetap1=atand (577/320) *2
28 disp(thetap1 ,” o r i e n t a t i o n s o f p r i n c i p a l axe s i s= ”)29 sigma1 =(200*10^9/(1 -0.09))*( epsilon1 +0.3* epsilon2)
30 disp(sigma1 ,” p l ane s t r e s s s i s Pa= ”)31 sigma2 =(200*10^3/(1 -0.09))*( epsilon2 +0.3* epsilon1)
32 disp(sigma2 ,” p l ane s t r e s s s i s MPa= ”)33
34 taumax =(200*10^9/(2*(1+0.3)))*gammamax
35 disp(taumax ,” p l ane s t r e s s s i s MPa= ”)
10
Chapter 3
Two dimensional problems inelasticity
Scilab code Exa 3.5 Size of the contact area and maxi contact pressure
1 clc;
2
3 E=210 //GPa4 v=0.3
5 r1=0.4 //m r a d i u s6 r2=0.3 //m c r o s s r a d i u s7 P=90 //kN compre s s i on l oad8 1/r1 ’==1/r2 ’==0
9
10 m=4/((1/ r1)+(1/r2))
11 disp(m)
12 A=(1/2) *((1/r1)+(1/r2))
13 disp(A)
14 B=(1/2) *((1/r1) -(1/r2))
15 disp(B)
16 coss =(((1/ r1) -(1/r2))/((1/ r1)+(1/r2)))
17 disp(coss ,” co s ap lha i s= ”)18 n=(4*E*10^9) /(3*(1 -v^2))
19 disp(n,”n i s ”)
11
20 s=acosd(coss)
21 disp(s,” s i s a lpha v a l u e = ”) // ans i s 8 1 . 7 9 d e g r e ebut he r e s i n c e co sa i s i n n e g a t i v e we g e t ans as9 8 . 2 1
22 ca =1.1040 // from the i n t e r p o l a t i n g t a b l e23 cb =0.9112 // from the i n t e r p o l a t i n g t a b l e24 a=ca *(90000*m/n)^(0.33)
25 disp(a,” s emiaxe s o f the e l l i p t i c a l c o n t a c t a r ea i nmeter i s= ”)
26 b=cb *(90000*m/n)^(0.33)
27 disp(b,” s emiaxe s o f the e l l i p t i c a l c o n t a c t a r ea i nmeter i s= ”)
28 sigmac =1.5*(90000/( %pi*a*b))
29 disp(sigmac ,”max c o n t a c t p r e s s u r e i n Pa i s= ”) //t e x t book ans i s wrong
12
Chapter 4
Mechanical behaviour ofmaterials
Scilab code Exa 4.1 Calculate diameter
1 clc;
2
3 sigmayp =350 //MPa4 sigma3 =0
5 M=8 //kN6 Mt=24 //kNm7 N=2
8 v=0.3
9
10 // sigma= My/ I ==32M/%pid ˆ311 // tau= Mt∗ r /J ==16Mt/%pid ˆ312 // s igma1 =(16∗(M+s q r t (Mˆ2+Mtˆ2) ) ) /( %pi∗d ˆ3)13 // s igma2 =(16∗(M−s q r t (Mˆ2+Mtˆ2) ) ) /( %pi∗d ˆ3)14
15 // s o l u t i o n a : max p r i n c i p a l s t r e s s t h e o r y16 // ( 1 6∗ (M+s q r t (Mˆ2+Mtˆ2) ) ) /( %pi∗d ˆ3)=sigmayp /N17
18 a=(16*(M+sqrt(M^2+Mt^2)))/%pi
19 disp(a)
13
20 b=sigmayp *10^6/N
21 disp(b)
22 d=(a/b)^(1/3)
23 disp(d,” d iamete r o f the bar i n meter i s= ”)24
25 // s o l u t i o n b : max s h e a r i n g s t r e s s th e o r y26
27 c=(32* sqrt(M^2+Mt^2))/%pi
28 disp(c)
29 d=(c/b)^(1/3)
30 disp(d,” d iamete r o f the bar i n meter i s= ”)31
32 // s o l u t i o n c : max p r i n c i p a l s t r a i n t he o r y33 // e p s i l o n 1 =(sigma1−v ( s igma2+sigma3 ) ) /E=e p s i l o n y p /N=
sigmayp /EN Or34 // sigma1−v ( s igma2+sigma3 )=b g i v e n35 // s igma1=b+v ( s igma2+sigma3 ) s u b s t i t u t i n g the
v a l u e s o f the s igma 1 ,2 and 3 we g e t36 // ( 1 6∗ (M+s q r t (Mˆ2+Mtˆ2)−vM−v∗ s q r t (Mˆ2+Mtˆ2) ) ) /( %pi∗d
13 // l o c a t i o n o f the c e n t r o i d14 z=(A1*z1+A2*z2)/(A1+A2)
15 disp(z,” i n meter i s= ”)16
17 Iz =(0.02*(0.13) ^3) /12+ (0.13*0.02*(0.04) ^2)
+(0.15*(0.02) ^3) /12+ (0.15*0.02*(0.035) ^2)
18 disp(Iz,” I z i n meter ˆ4 i s= ”)19 Iy =(0.02*(0.13) ^3) /12+ (0.13*0.02*(0.04) ^2)
+(0.15*(0.02) ^3) /12+ (0.15*0.02*(0.035) ^2)
21
20 disp(Iy,” Iy i n meter ˆ4 i s= ”)21 Iyz =0+A1 *0.04*( -0.035) +0+A2*( -0.035) *0.03
22 disp(Iyz ,” I y z i n meter ˆ4 i s= ”)23 // the tap =( atand ((−2∗ I y z ) /( Iz−Iy ) ) ) /224 // d i s p ( the tap )25 I1=(Iz+sqrt (0+(6.79*10^ -6) ^2))
26 disp(I1,” I1 i n meter ˆ4 i s= ”)27 I2=(Iz-sqrt (0+(6.79*10^ -6) ^2))
28 disp(I2,” I2 i n meter ˆ4 i s= ”)29 My1 =11000* sind (45)
30 disp(My1 ,”My1 i n Nm i s ”)31 Mz1 =11000* sind (45)
32 disp(Mz1 ,”Mz1 i n Nm i s ”)33
34 sigmaxA =((My1*(zA))/I1) -((Mz1*yA)/I2)
35 disp(sigmaxA ,” sigmaxA i n MPa i s ”)36 sigmaxB =0-((My1*yB)/I2)
37 disp(sigmaxB ,” sigmaxB i n MPa i s ”)38
39 My=0
40 y=((Mz*Iyz)*z)/(Mz*Iy) // . . . . . . . e q u a l to z =−1.71y41 disp(y)
Scilab code Exa 5.4 Determine flange
1 clc
2
3 t=1.25 //mm4 y=15.87 //mm5 z=5.28 //mm6 Iy =4765.62 //mmˆ47 Iz =21054.69 //mmˆ48 Iyz =3984.37 //mmˆ49 thetap =13.05 // d e g r e e
10 Iy1 =3828.12 //mmˆ4
22
11 Iz1 =21953.12 //mmˆ412 s=12.5
13
14 // tau =(Vy/ I z 1 ∗ t ) ∗ s ∗ t (19 .55+ s ∗ a s i n d ( 1 3 . 0 5 ) /2) . . . .e q u a t i o n 1
15 //F1=i n t e g r a t e ( ( tau ∗ t ) ds )16 x=integrate( ’ ( 0 ) ’ , ’ s ’ ,0,1)17 //F1 =0.0912∗Vy1 s u b s t i t u t i n g the
v a l u e o f tau we g e t F118 //Vy1∗ ez1 =37.5∗F1 s u b s t i t u t i n g the
v a l u e o f F1 we ge t ez119 ez1 =37.5*0.0912
20 disp(ez1 ,” the d i s t a n c e i n mm i s= ”)21
22 // tau =(Vz1/ Iy1 ∗ t ) ∗ s ∗ t (12.05− s ∗ a s i n d ( 1 3 . 0 5 ) /2) . . . .e q u a t i o n 2
23 //F1=i n t e g r a t e ( ( tau ∗ t ) ds )24 x=integrate( ’ ( 0 ) ’ , ’ s ’ ,0,1)25 //F1 =0.204∗Vz1 s u b s t i t u t i n g the
v a l u e o f tau we g e t F126 //Vz1∗ ey1 =37.5∗F1 s u b s t i t u t i n g the
v a l u e o f F1 we ge t ez127 ey1 =37.5*0.204
28 disp(ey1 ,” the d i s t a n c e i n mm i s= ”)
Scilab code Exa 5.5 Determine area and tangential stress
1 clc
2
3 P=70 //kN4 c=0.05 //m5 c1=c
6 c2=c
7 R=0.1+0.05
8 A=0.005
23
9
10 //m=(−1/(2∗ c ) ) ∗ i n t e g r a t e ( ( y/R+y ) dy )11 x=integrate( ’ (−c ) ’ , ’ c ’ ,0,1)12 m=-1+(R/2*c)*log((R+c)/(R-c))
13 disp(m)
14 //m=(−1/(2∗ c ) ) ∗ i n t e g r a t e ( ( y/R)−(y ˆ2/Rˆ2) +(y ˆ3/Rˆ3)−(y ˆ4/Rˆ4) + . . . . . ) dy )
15 m= -1+(3/2)*log(2)
16 disp(m)
17
18 M=P*R
19 disp(M)
20 sigmatheta1 =(-P*c2)/(m*A*(R-c1))
21 disp(sigmatheta1 ,” s t r e s s i n Pa i s= ”)22 sigmatheta2 =(P*c2)/(m*A*(R+c2))
23 disp(sigmatheta2 ,” s t r e s s i n Pa i s= ”)
24
Chapter 6
Torsion of prismatic bars
Scilab code Exa 6.2 Shearing stress and angle of twist
1 clc
2
3 G=28 //GPa4 t1 =0.012
5 t2 =0.006
6 t3=0.01
7 t4 =0.006
8 A=0.125
9 h=226000 //N/m10 Mt=2*A*h
11 disp(Mt,” a p p l i e d t o r qu e i n Nm i s=”)12
13 tau1=(h/t1)
14 disp(tau1 ,” s h e a r i n g s t r e s s i n Pa i s= ”)15 tau2=(h/t2)
16 disp(tau2 ,” s h e a r i n g s t r e s s i n Pa i s= ”)17 tau3=(h/t3)
18 disp(tau3 ,” s h e a r i n g s t r e s s i n Pa i s= ”)19 tau4=(h/t4)
20 disp(tau4 ,” s h e a r i n g s t r e s s i n Pa i s= ”)21
25
22 // t h e t a =(h/2∗G∗A) i n t c ( ( 1 / t ) ds )23 theta=(h/(2*G*10^9*A))*((0.25/ t1)+2*(0.5/ t2)+(0.25/
t3))
24 disp(theta ,” a n g l e o f t w i s t per u n i t l e n g t h i n rad /mi s= ”)
Scilab code Exa 6.4 Maxi longitudinal stress and angle of twist
10 // s o l u t i o n a : e x a c t s o l u t i o n11 //p=Mh/ I12 // sigmax=−(y/h ) ∗p13 sigmay =0
14 tauxy=0
15 // d e r i v a t i v e ( u , x )=−(yp/Eh)16 // d e r i v a t i v e ( v , y ) =(v∗y∗p ) /(Eh)17 // d e r i v a t i v e ( u , y )+d e r i v a t i v e ( v , x )=018 //u=−(p/E∗h ) ∗x∗y // f o r u
30 // s o l u t i o n d : maxi p r i n c i p a l s t r a i n t h eo r y31 // ( s igmatheta−v∗ s i gmar ) /E=sigmayp /E32 pi=sigmayp /(sigmatheta -v*sigmar)
33 disp(pi,” i n MPa i s= ”)34
35 // s o l u t i o n e : o c t a h e d r a l s h e a r i n g s t r e s s t h e o r y :36 pi=(sqrt (2)*sigmayp)/sqrt(( sigmatheta -sigmar)^2+
sigmar ^2+(- sigmatheta)^2)
37 disp(pi,” i n MPa i s= ”)
Scilab code Exa 8.3 Distribution of tangential stress
11 disp(p,” the c o n t a c t p r e s s u r e i n Pa i s= ”) //t ex tbook ans i s wrong
12
13 p=12.3*10^6
14 sigmatheta=p*((b^2+c^2)/(c^2-b^2)) // where r =0.2
41
15 disp(sigmatheta ,” t a n g e n t i a l s t r e s s e s i n the o u t e rc y l i n d e r i n Pa i s= ”)
16 sigmatheta1 =(2*p*b^2)/(c^2-b^2) // where r =0.2517 disp(sigmatheta1 ,” t a n g e n t i a l s t r e s s e s i n the o u t e r
c y l i n d e r i n Pa i s= ”)18 sigmatheta3 =-(2*p*b^2)/(b^2-a^2) // where r =0.1519 disp(sigmatheta3 ,” t a n g e n t i a l s t r e s s e s i n the i n n e r
c y l i n d e r i n Pa i s= ”)20 sigmatheta4=-p*((b^2+a^2)/(b^2-a^2)) // where r =0.221 disp(sigmatheta4 ,” t a n g e n t i a l s t r e s s e s i n the i n n e r
c y l i n d e r i n Pa i s= ”)
Scilab code Exa 8.4 Determine shrinking allowance and maxi stress