# Chapter3-Linear Stress-Strain Temperature Relations¶

## Ex1-pg88¶

In [1]:
##calculate the maximum circumferential stress
## initialization of variables
import math
## part (a)
E=72. ## G Pa
v=0.33 ## Poisoon's ratio
h=2. ## mm
R=600. ## mm
##calculations
sig_cir=E*h/(2.*(1.-v**2)*R)
## results
print('\n part (a) \n')
print"%s %.2f %s"%(' The maximum circumferential stress is ',sig_cir*10**3,' M Pa')

 part (a)

The maximum circumferential stress is  134.67  M Pa


## Ex7-pg99¶

In [2]:
##calculate the stress in aluminum and steel cylinder
## initialization of variables
import math
tR=0.02 ## t/R ration
E_A=69. ##G Pa
v_A=0.33  ## Poisson's ratio
alpha_A=21.6*10**-6 ## /degree Celcius (Coefficient of expansion)
E_S=207. ## G Pa
v_S=0.280
alpha_S=10.8*10**-6 ## /degree Celcius (Coefficient of expansion)
## calculations
## Sig_LA=a*p+b*delT+c*sig_thS
## Sig_LS=v_S*Sig_thS+d*delT
E_S=E_S*10**9
E_A=E_A*10**9
a=1./tR*E_A/E_S
b=-2./3.*alpha_S*E_S
c=-E_A/E_S
d=-alpha_S*E_S
## SigthS=e*p+f*delT
## SigthA=g*p+h*delT
e=37.16
f=0.8639*10**6
g=1./tR-e
h=-f
## results
p=689.4 ## kPa
delT=100. ## degree Celcius
p=p*10**3 ## Pa
SigthA=g*p+h*delT
SigthS=e*p+f*delT
Sig_LA=a*p+b*delT+c*SigthS
Sig_LS=v_S*SigthS+d*delT
print"%s %.2f %s %.2f %s "%('Thus, for p = ',p/10**3,' k Pa' and 'delT = ',delT,' degree celcius \n')
print"%s %.2f %s %.2f %s "%(' SigthA = ',SigthA/10**6,' M Pa'and   'Sig_LA = ',Sig_LA/10**6,' M Pa \n')
print"%s %.2f %s %.2f %s "%(' SigthS = ',SigthS/10**6,' M Pa'and    'Sig_LS = ',Sig_LS/10**6,' M Pa')

Thus, for p =  689.40 delT =  100.00  degree celcius

SigthA =  -77.54 Sig_LA =  -174.89  M Pa

SigthS =  112.01 Sig_LS =  -192.20  M Pa


## Ex8-pg100¶

In [3]:
##calculate the  orientation of the principal stress and deterimine stress components and orientation of the principal axes of strain
## initialization of variables
import math
import numpy
## Material constants
Ex=14700. ## M Pa
Ey=1000. ## M Pa
Ez=735. ## M Pa
Gxy=941. ## M Pa
Gxz=1147. ## M Pa
Gyz=103. ## M pa
Vxy=0.292
Vxz=0.449
Vyz=0.39
## Stresses at a point
Sxx=7. ## M pa
Syy=2.1 ## M Pa
Szz=-2.8 ##M Pa
Sxy=1.4 ## M Pa
Sxz=0. ##M Pa
Syz=0. ## M Pa
## part (a)
th=1/2.*math.atan(2.*Sxy/(Sxx-Syy))*180./math.pi
I1=Sxx+Syy+Szz
I2=Sxx*Syy-Sxy**2+Szz*Sxx-Sxz**2+Szz*Syy-Syz**2
M=numpy.matrix([[Sxx, Sxy, Sxz],
[Sxy, Syy, Syz],
[Sxz, Syz, Szz]])
I3=numpy.linalg.det(M)
p=([1, -I1, I2 ,-I3])
S=numpy.roots(p)
## results
print('Part (a) \n')
print"%s %.2f %s"%('The maximum principal stress is S1 = ',S[0],' M Pa')
print"%s %.2f %s"%('\n and occurs in direction th = ',th,' degrees')
print"%s %.2f %s %.2f %s "%('\n and the intermediate principal stress S2 = ',S[2],' M Pa'and 'occurs in the direction th = ',th+90,' degrees \n')
print"%s %.2f %s"%(' The minimum principal stress is S3 = Szz = ',S[1],' M Pa')
Ex=Ex*10**6
Ey=Ey*10**6
Ez=Ez*10**6
Gxy=Gxy*10**6
Gxz=Gxz*10**6
Gyz=Gyz*10**6
## part (b) is to find strains
Exx=Sxx/Ex-Vxy*Syy/Ey-Vxz*Szz/Ez
Eyy=-Vxy*Sxx/Ex+Syy/Ey-Vyz*Szz/Ez
Ezz=-Vxz*Sxx/Ex-Vyz*Syy/Ey+Szz/Ez
Exy=Sxy/Gxy
Exz=Sxz/Gxz
Eyz=Syz/Gyz
print('\n Part (b)')
print('\n The srains are')
print"%s %.2e %s %.2e %s %.2e %s"%('\n Exx = ',Exx,'' and'Eyy =',Eyy,''and 'Ezz = ',Ezz,'')
print"%s %.2e %s %.2d %s %.2d %s"%('\n Exy = ',Exy,''and '  Exz = ',Exz,''and  'Eyz = ',Eyz,'')
## Wrong Exx value in the textbook
th=1/2.*math.atan(Exy/(Exx-Eyy))
th=th*180./math.pi
th2=th+90.
print('\n part (c)')
print"%s %.2f %s %.2f %s "%('\n theta = ',th,'' and 'or theta=',th2,'degrees')
## Wrong theta too since Example  given in textbook is wrong

Part (a)

The maximum principal stress is S1 =  7.37  M Pa

and occurs in direction th =  14.87  degrees

and the intermediate principal stress S2 =  1.73 occurs in the direction th =  104.87  degrees

The minimum principal stress is S3 = Szz =  -2.80  M Pa

Part (b)

The srains are

Exx =  1.57e-09  3.45e-09  -4.84e-09

Exy =  1.49e-09  00  00

part (c)

theta =  -19.23  70.77 degrees