## SAMPLE PROBLEM 8.1, Page number 520¶

In [1]:
import math
from sympy import symbols

#Variable declaration
Ma=160*0.375                                                       # Torque on rod AC(kN.m)
Va = 160                                                           # Force(kN)
S=511*(pow(10,-6))                                                 # Inertia(m**3)
yb=90.4                                                            # Distance(mm)
c=103                                                              # Distance(mm)
I=52.9*(pow(10,-6))                                                # Moment oof inertia(m**4)
t=0.00787                                                          # Distance(m)

#Calculation
#Normal Stresses on Transverse Plane
#At point a:
Sa=round((Ma/S)/(1000.0),1)                                          # Stress at A(MPa)
#At point b:
Sb=round((Sa*(yb/c)),0)                                            # Stress at B(MPa)

#Shearing Stresses on Transverse Plane
#At point a:
Q=0
ta=0
#At point b:
Q=round((206*12.6*96.7)/(1000.0),0)
Q=(Q)/(pow(10.0,6))
tb=round(((Va*Q)/(I*t))/(pow(10.0,3)),1)                             # Shearing stress(MPa)

#Principal Stress at Point b
Smax=round((1/2.0)*Sb + math.sqrt(pow((1/2.0)*Sb,2) + pow(tb,2)),1)    # Stress maximum(MPa)

# Result
print ('The specification, smax<= 150 MPa, is not satisfied' )

The specification, smax<= 150 MPa, is not satisfied


## SAMPLE PROBLEM 8.2, Page number 521¶

In [2]:
import math
from sympy import symbols

#Variable declaration
Aweb=8.40                                                             # Area of cross section(in**2)
S=127
yb=9.88                                                               # Distance(in)
c=10.50                                                               # Distance(in)
V=12.2                                                                # Force(kips)

#Calculation
#Reactions at A and D
sumMd=0                                                               # Equilibrium equations
sumMa=0                                                               # Equilibrium equations

#Shear and Bending-Moment Diagrams
Mmax=2873                                                             # Maximum moment(kip.ft)
Vmax=43                                                               # Maximum force(kips)

#Section Modulus
Sall=24                                                               # Maximum stress(ksi)
Smin=round((Mmax/Sall),1)                                             # Minimum section modulus(in**3)

#Selection of Wide-Flange Shape

#   W24*68          154
#   W21*62          127
#   W18*76          146
#   W16*77          134
#   W14*82          123
#   W12*96          131

# We now select the lightest shape available, namely W21*62

#Shearing Stress
tm=round(Vmax/Aweb,2)                                                # Shearing stress(ksi)

#Principal Stress at Point b
Sa=round(Mmax/S,1)                                                   # Stress at A(ksi)
Sb=round(Sa*(yb/c),1)                                                # Stress at B(ksi)

tb=round(V/Aweb,2)
smax=round((1/2)*Sb + math.sqrt(pow((1/2)*Sb,2) + pow(tb,2)),1)

# Result
print ('We now select the lightest shape available, namely :-')
print('W21*62')

We now select the lightest shape available, namely :-
W21*62


## SAMPLE PROBLEM 8.3, Page number 522¶

In [3]:
import math
from sympy import symbols

# Variable declaration
P=30                                                                      # Power(kW)
tall=50                                                                   # Shearing stress(MPa)
f=8                                                                       # Frequency(Hz)
Te=597                                                                    # Torque at E(N.m)

# Calculation
# Torques Exerted on Gears
Te=(P/(2.0*(math.pi)*f))                                                    # Torque exerted on gear E(N.m)
Fe=Te/rE                                                                  # Tangential force(kN)
Tc=20/(2.0*(math.pi)*8)                                                     # Torque(N.m)
Fc=6.63                                                                   # Force(kN)
Td=10/(2.0*(math.pi)*8)                                                     # Torque(N.m)
Fd=2.49                                                                   # Force(kN)

# Critical Transverse Section
Tor=math.sqrt(1160.0**2 + 373.0**2 + 597.0**2)                                  # Maximum value of torque(N.m)

# Diameter of Shaft.
c=((((27.14*(pow(10,-6)))*(2/(math.pi)))**(1/3.0))*(1000.0))*2                # Diameter(mm)

# Result
print ('Diamter of shaft = %lf mm' %c)

Diamter of shaft = 51.704299 mm


## Example 8.01, Page number 529¶

In [5]:
import math
from sympy import symbols

#Variable declaration
n=-1
P1=15                                                                            # Force(kN)
P2=18                                                                            # Force(kN)
a=50                                                                             # Distance(mm)
b=60                                                                             # Distance(mm)
c=0.020                                                                          # Distance(m)
F=P1                                                                             # Force(kN)
V=P2                                                                             # Force(kN)
t=0.040                                                                          # Distance(m)
Iz=125.7*(pow(10,-9))                                                            # Moment of inertia(m**4)

#Calculation
#Internal Forces in Given Section
T=P2*a                                                                           # Torque(N.m)
My=P1*a                                                                          # Moment(N.m)
Mz=P2*b                                                                          # Moment(N.m)
# Case(a) Normal and Shearing Stresses at Point K
# Geometric Properties of the Section
A=(math.pi)*(c**2)                                                               # Area of cross section(m**2)
Iy=(1/4.0)*(math.pi)*(c**4)                                                        # Moment of inertia(m**4)
Jc=(1/2.0)*(math.pi)*(c**4)                                                        # Moment of inertia(m**4)
Q=(A/2.0)*((4*c)/(3.0*(math.pi)))
t=2*c                                                                            # Distance(m)
# Normal Stresses
Sx=(n*(F/A))/(1000.0) + ((My*c)/(Iy))/(1000000.0)                                    # Normal stress(MPa)
# Shearing Stresses
txyV=((V*Q)/(Iz*t))/(1000.0)                                                       # Shearing stress(MPa)
txytwist=((n*(T*c))/(Jc))/(1000000.0)                                              # Shearing stress(MPa)
txy=(txyV + txytwist)                                                            # Shearing stress(MPa)
# Case(b) Principal Planes and Principal Stresses at Point K
OC=CD=(1/2.0)*(107.4)                                                              # Stress(MPa)
DX=52.5                                                                          # Stress(MPa)
phyp=44.4/2.0                                                                      # Angle(degree)
R=math.sqrt(53.7**2 + 52.5**2)                                                        # Stress(MPa)
Smax=OC+R                                                                        # Maximum principal stress(MPa)
Smin=OC-R                                                                        # Minimum principal stress(MPa)
# Case(c) Maximum shearing stress at point k
tmax=75.1                                                                        # Shearing stress(MPa)

# Result
print ('Case(a) Normal stress = %lf MPa' %Sx)
print ('Case(a) Shearing stress = %lf MPa' %txy)
print ('Case(b) Principal axis angle = %lf degree' %phyp)
print ('Case(b) Maximum principal stress at point k = %lf MPa' %Smax)
print ('Case(b) Minimum principal stress at point k = %lf MPa' %Smin)
print ('Case(c) Maximum shearing stress at point k = %lf MPa' %tmax)

Case(a) Normal stress = 107.429587 MPa
Case(a) Shearing stress = -52.526646 MPa
Case(b) Principal axis angle = 22.200000 degree
Case(b) Maximum principal stress at point k = 128.799534 MPa
Case(b) Minimum principal stress at point k = -21.399534 MPa
Case(c) Maximum shearing stress at point k = 75.100000 MPa


## SAMPLE PROBLEM 8.5, Page number 532¶

In [6]:
import math

#Variable declaration
#Free Body. Entire Crankshaft
Vx=-30                                                                        # Force(kN)
P=50                                                                          # Force(kN)
Vz=-75                                                                        # Force(kN)
Mx=(50)*(0.130) - (75)*(0.2)                                                  # Moment(kN.m)
My=0                                                                          # Moment
Mz=30*0.1                                                                     # Moment(kN.m)
A=0.040*0.140                                                                 # Area(m**2)
Ix=(1/12.0)*(0.040)*(pow(0.140,3))                                              # Moment of inertia(m**4)
Iz=(1/12.0)*(pow(0.040,3))*(0.140)                                              # Moment of inertia(m**4)
a=0.020                                                                       # Distance(m)
b=0.025                                                                       # Distance(m)
t=0.040                                                                       # Distance(m)
OC=33.0                                                                       # Stress(MPa)

#Calculation
#Normal Stress at H
Sy=round(((P/A) + ((Mz)*a)/Iz + ((Mx)*b)/Ix)/(1000.0),0)                        # Normal stress at H(MPa)

#Shearing Stress at H
Q=(0.040*0.045*0.0475)
tyz=round((((-(Vz)*(Q))/(Ix*t))/1000.0),2)                                      # Shearing stress at H(MPa)

#Principal Stresses, Principal Planes, and Maximum Shearing Stress at H.
phyp=27.96/2.0
R=math.sqrt(33**2 + 17.52**2)
Smax=OC+R
Smin=OC-R

# Result
print ('Normal stress at H = %lf MPa' %Sy)
print ('Shearing stress at H = %lf MPa' %tyz)
print ('Principal axis angle = %lf degree' %phyp)
print ('Maximum shearing stress at point k = %lf MPa' %R)
print ('Maximum principal stress at point k = %lf MPa' %Smax)
print ('Minimum principal stress at point k = %lf MPa' %Smin)

Normal stress at H = 66.000000 MPa
Shearing stress at H = 17.530000 MPa
Principal axis angle = 13.980000 degree
Maximum shearing stress at point k = 37.362420 MPa
Maximum principal stress at point k = 70.362420 MPa
Minimum principal stress at point k = -4.362420 MPa