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' )
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')
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)
rE=0.16 # Radius(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)
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)
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)