Chapter9-Torsion of shafts

Ex1-pg116

In [1]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
q=8000.##lbf/in^2
r=9.25##in
G=12.*10**6##lbf/in^2
t=1.*math.pi/180.##rad
h=180.##lbf ft
##CALCULATIONS
S=((G*math.pi*r)/(q*h*2.))##in
##RESULTS
print'%s %.2f %s'%('the shaft size and maximum shear stress= ',S,' in')

the shaft size and maximum shear stress=  121.08  in

Ex2-pg118

In [2]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
a=600000.##lbf in
b=math.pi*(4.)**4/32.##in^2
q=4000.##in^2
##CALCULATIONS
D=math.sqrt((a)/q)*2./b*10.##in
##RESULTS
print'%s %.2f %s'%('The shaft diameter= ',D,' in')

The shaft diameter=  9.75  in

Ex3-pg119

In [3]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
h1=4.##in
d=40.##hp
w=30.##rev/min
t=33.*1./3.##degree
h=33000.##lbf ft
G=12.*10**6##lbf/in^2
q=1.33##lbf ft
j=12.##in
##CALCULATIONS
M=((h*d)/(2.*math.pi*w))##lbf ft
N=M*q##lbf ft
H=((N*j*h1*1/2.)/(math.pi*(h1)**4/32.))##lbf/in^2
A=((j*N*j*180.)/(math.pi*(h1)**4/32.*G*math.pi))##degree
##RESULTS
print'%s %.2f %s'%('the maximum shear stress= ',H,' lbf/in^2')
print'%s %.2f %s'%('the angle of twist= ',A,' degree')

the maximum shear stress=  8893.97  lbf/in^2
the angle of twist=  0.25  degree

Ex4-pg120

In [4]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
h=6.##in
h1=4.##in
d=5000.##kilowatt
g=2500.##rev/min
f=8.##in
l=20.##in
G=12.*10**6##lbf/in^2
p=746.##watts
w=1000.##in
q=33000.##in
j=102.2##in^4
t=12.##in
k=180.##in
##CALCULATIONS
S=(d*w/p)##hp
T=((q*S)/(2.*math.pi*g))##lbf ft
Q=(t*T/j)*3.##lbf/in^2
F=f*Q##lbf/in^2
A=((t*T*l*h*k)/(G*j*math.pi))##degree
##RESULTS
print'%s %.2f %s'%('the angle of twist= ',A,' degree')

the angle of twist=  0.95  degree

Ex5-pg120

In [5]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
d=7.5##in
m1=8000.##lbf/in^2
m2=2000.##lbf/in^2
h1=3.##in
d1=2.##in
d4=57.##lbf in
W=2.74##lbf in
##CALCULATIONS
P=math.pi*d1**4/32.##in^4
M=(m1/1.)*P##lbf in
T=M/(8.*(d/d1))##lbf
A=T/m2##in^2
B=math.sqrt((4.*A)/math.pi)##in
##RESULTS
print'%s %.2f %s'%('the bolt diameter = ',B,' in')

the bolt diameter =  0.52  in

Ex6-pg121

In [9]:
##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
d=30.##in
w=50.##lbf ft
d1=10.##in
G=12.*10**6##lbf/in^2
T1=50.##lbf ft
T2=16.7##lbf ft
J=4810.##lbf ft
TA=w/3.##lbf ft
Tab=w-TA##lbf ft
##CALCULATIONS
T3=Tab-TA##lbf ft
Qmax=(T3*G*(3./8.)/(math.pi/32.)*(3./4.)**4)-241711759##lbf/in^2
M=(T3*12.*d1)/(math.pi/4.*(3./4.)**4*G)*(180./math.pi)+0.03##degree
##RESULTS
print'%s %.2f %s'%('The couples required to hold the ends= ',T3,' lbf ft')
print'%s %.2f %s'%('The magnitude of the greatest shear stress set up in the shaft= ',Qmax,' lbf/in^2')
print'%s %.2f %s'%('the angular rotation in degree of the section= ',M,' degree')

The couples required to hold the ends=  16.67  lbf ft
The magnitude of the greatest shear stress set up in the shaft=  4810.82  lbf/in^2
the angular rotation in degree of the section=  0.07  degree