Chapter5-Uniaxial Deformations¶

Ex1-pg139¶

In [1]:
import math
##initialization of variables
#find the depth of clay  bed
l=20. ##cm
dL=1. ##m
dl=0.004 ##cm
##calculations
L=l*dL/dl ##m
##results
print'%s %.2f %s'%('The depth of the clay bed is ',L,' m')

The depth of the clay bed is  5000.00  m


Ex2-pg140¶

In [2]:
import math
##initialization of variables
#find the total extension of the rod and draw the force and extension diagrams
A=1. ##unit area
E=2.*10**6 ##kg/cm^2
## calculations
db=3000.*90./(A*E)
dc=db+5000.*60./(A*E)
dd=dc+4000.*30./(A*E)
##results
print'%s %.2e %s %.2e %s %.2e %s '%('The extension of the rod in part AB is ',db,' cm'and'in part BC is ',dc,' cm'and' \n and in part CD is ',dd,' cm')

The extension of the rod in part AB is  1.35e-01 in part BC is  2.85e-01
and in part CD is  3.45e-01  cm


Ex3-pg141¶

In [1]:
import math
##initialization of variables
#findthe extension under its own weight
A=3. ##cm^2
L=18. ##m
E= 2*10**6 ##kg/cm^2
r=7833. ##kg/m^3
##calculations
e=r*(L*100)**2./(2*E*10**6)
## results
print'%s %.4f %s'%('The elongation is ',e,' cm')

The elongation is  0.0063  cm


Ex4-pg142¶

In [4]:
import math
##initialization of variables
#find the extension under a concertrated load of 3 tonne at the bottom
P=3 ##tonne
E=2*10**6 ##kg/cm^2
d_0= 1. ##cm
d_l=2.8 ##cm
## calculations
e=4*P*1000.*d_l*10**3/(d_l**2*math.pi*E*(1-((d_l-d_0)/d_l)))
##results
print'%s %.2f %s'%('The total elongation is ',e,' cm')

The total elongation is  1.91  cm


Ex6-145¶

In [3]:
import math
##initialization of variables
import numpy
from numpy import linalg
import scipy as Sci
from scipy import linalg
P=10 ##tonne
import numpy as np
E=2*10**6 ##kg/cm^2
## calculations
## We have to solve linear system Ax=B
A=numpy.matrix([[1, 1, 1, 0], [3, 1, -3, 0],[-2, 2, 0, -E],[0, -1, 2, -E]])
B=numpy.matrix([[P*10**3],[0],[0],[0]])
x=numpy.dot(np.linalg.inv(A),B)
W1=x[0,0]/1000.
W2=x[1,0]/1000.
W3=x[2,0]/1000.
th=x[3,0]
##results
print'%s %.2f %s %.2f %s %.2f %s '%('The load taken by each rod is',W1,' tonne'and'',W2,' tonne'and'',W3,'tonne')
print'%s %.3e %s'%('\n and the slope is theta = ',th,' radians')

The load taken by each rod is 2.33  4.00  3.67 tonne

and the slope is theta =  1.667e-03  radians


Ex8-pg147¶

In [15]:
import math
## initialization of variables
#calculate the safe load taken by the column
b=30. ## cm
h=30. ##cm
n=6.
A=36. ##cm^2
ss_s=1500. ##kg/cm^2
ss_c=60. ##kg/cm^2
Er=15. ## Elasticity ratio
## calculations
L=A*Er*ss_c+(b*h-A)*ss_c
## results
print'%s %.2f %s'%('The safe load is ',L,'.kg')

The safe load is  84240.00 .kg


Ex9-pg148¶

In [5]:
#find the stress in steel and concerete after redisturbution of stress in steel and cncerete
## initiaization of variables
import math
gs_b=10. ##cm
gs_h=10. ##cm
d_b=2. ##cm
d_h=2. ##cm
As= 1. ##cm^2
s=10000. ##kg/cm^2
## part (a)
Es=2*10**6 ##kg/cm^2
Ec=2*10**5 ##kg/cm^2
## calculations
e=s/Es
Ac=gs_b*gs_h-(d_b*d_h)
e_c=e*Es*As/(Ec*Ac+Es*As)
s_c=Ec*e_c
e_s=e-e_c
s_s=Es*e_s
## results
print'%s %.2f %s %.2f %s '%('part (a) \n The stress in steel and concrete are respectively ',s_s,''and '',s_c,' kg/cm^2')
## part(b)
P=8000. ##kg
## calculations
e_c=(e*Es*As-P)/(Ec*Ac+Es*As)
e_s=e-e_c
s_c=Ec*e_c
s_s=Es*e_s
## results
print'%s %.2f %s %.2f %s'%('\n part (b) \n The stress in steel and concrete are respectively ',s_s,''and  '',s_c,'kg/cm^2')

part (a)
The stress in steel and concrete are respectively  9056.60  94.34  kg/cm^2

part (b)
The stress in steel and concrete are respectively  9811.32  18.87 kg/cm^2


Ex10-pg151¶

In [6]:
##calculate temperature which sleeve must be heated if the room temperature is 10c
#and pressure and find axial force necessary to separate the two room temperature  and the temperature at which sleeve will easily come off
## initialization
import math
d=10 ##cm
D=9.99 ##cm
t=3 ##mm
E=1.0*10**6 ##kg/cm^2
a=2.02*10**-5 ## degree/celcius
## part(a)
Tr=10. ##degree C
T=(d-D)/D*1/a
print'%s %.2f %s'%('part(a) \n The sleeve must be heated to ',T+Tr,' degree C or more for this purpose')

##part(b)
s_th=a*T*E
p=s_th*t*2./(d*10.)
print'%s %.2f %s'%('\n part(b) \n The pressure developed between the rod and sleeve is',p,' kg/cm^2')

## part(c)
f=0.2
o=10. ## overlap: cm
A=math.pi*d*o
F=f*p*A
print'%s %.2f %s'%('\n part (c) \n The axial force required is ',F,' kg')

##part (d)
T2=20. ##degree C
a2=1.17*10**-5 ##  /degree C
Ts=(a-a2)*(T2-Tr)*E
Ts=s_th-Ts
p2=p*Ts/s_th
F2=F*Ts/s_th
print'%s %.2f %s'%('\n part(d)\n The pressure developed between the rod and sleeve is',p2,' kg/cm^2')
print'%s %.2f %s'%('\n The axial force required is ',F2,' kg')
##part(e)
T3=Tr+(s_th/((a-a2)*10**6))
print'%s %.2f %s'%('\n part(e) \n The temperature at which the sleeve comes off easily is ',T3,' C')

print('calculations in the text: rounding off errors')

part(a)
The sleeve must be heated to  59.55  degree C or more for this purpose

part(b)
The pressure developed between the rod and sleeve is 60.06  kg/cm^2

part (c)
The axial force required is  3773.68  kg

part(d)
The pressure developed between the rod and sleeve is 54.96  kg/cm^2

The axial force required is  3453.24  kg

part(e)
The temperature at which the sleeve comes off easily is  127.76  C
calculations in the text: rounding off errors


Ex11-pg154¶

In [4]:
import math
##initialization of variables
#calculate the radius of curvature of this strip at temperautre of 93.3
T1=37.8 ## degre C
t=0.355 ##mm
T2=93.3 ## degree C
L=2 ##cm
m=1
n=1.53
a=1.86*10**-5
##calculations
R=2*t*(3*(1+m)**2.+(1+m*n)*(m**2+(m*n)**-1))
R=R/(6.*a*(T2-T1)*(1+m**2))  ## mm
R=R/10.
D=L**2./(8.*R)
## results
print'%s %.2f %s'%('The radius of curvature is ',R,' cm')
print'%s %.4f %s'%('\n The deflection is',D ,' cm')

The radius of curvature is  92.76  cm

The deflection is 0.0054  cm


Ex12-pg155¶

In [7]:
import math
## initialization of variables
#find the energy stored in bolt
L=5. ##cm
D=1.8 ##cm
l=2.5 ##cm
d=1.5 ##cm
F=1 ##tonne
E=2.1*10**6 ##kg/cm^2
## calculations
s1=F*1000.*4./(D**2*math.pi)
s2=F*1000.*4./(d**2*math.pi)
U1=1/2.*s1**2./E
U1=U1*L*D**2*math.pi/4.
U2=1/2.*s2**2./E
U2=U2*l*d**2*math.pi/4.
U=U1+U2
## results
print'%s %.1f %s'%('The energy stored in the bolt is ',U,' kg-cm')

The energy stored in the bolt is  0.8  kg-cm


Ex13-pg159¶

In [13]:
import math
## initialization of variables
#calculate which of the two ways will give stronger joint
t=16. ##mm
Pt=1500. ##kg/cm^2
Ps=1025. ##kg/cm^2
Pb=2360. ##kg/cm^2

##part (a)
p=6. ##cm
r=24. ##mm
d=r/10.+0.15
Ft=t*(p-d)*Pt/10.
Fs=math.pi*d**2*Ps/4.
Fb=d*t*Pb
x=min(Ft,Fs,Fb)
effA=x*100./(p*t/10.*Pt)

##part (b)
p=9. ##cm
r=30. ##mm
d=r/10.+0.2
Ft=t*(p-d)*Pt/10.
Fs=math.pi*d**2*Ps/4.
Fb=d*t*Pb
x=min(Ft,Fs,Fb)
effB=x*100./(p*t/10.*Pt)

## results
print'%s %.2f %s %.2f %s '%('The efficiencies corresponding to cases a and b are ',effA,'' and '',effB,'')
print('\n Hence part b is better than part a')

The efficiencies corresponding to cases a and b are  36.35  38.16

Hence part b is better than part a