from __future__ import division
from sympy import symbols,solve
# Given Data
P=30## kN
Sut=350## MPa
n=2.5## factor of safety
sigma_w=Sut/n## MPa (Working stress for the link)
t=symbols('t')## thickness of link
A=2.5*t**2## mm.sq.
I=t*(2.5*t)**3/12## mm**4 (Moment of Inertia about N-A)
sigma_d=P/A## N/mm.sq.
e=10+1.25*t##mm
M=P*10**3*e## N.mm
sigma_t=M*1.25*t/I## N/mm.sq.
#maximum tensile stress at the top fibres = sigma_d+sigma_t=sigma_w ...eqn(1)
expr=sigma_d+sigma_t-sigma_w ## expression of polynomial from above eqn.
t=solve(expr)## solving the equation (as denominator will me be multiplied by zero on R.H.S)
t=t[0]## mm # discarding -ve roots
print ' dimension of cross section of link, t=%.f mm. Adopt t=21 mm. '%(t)
from __future__ import division
from sympy import symbols,solve
from math import sin,cos,pi
# Given Data
P=6## kN
alfa=30## degree
Sut=250## MPa
n=2.5## factor of safety
sigma_w=Sut/n## MPa (Working stress for the link)
PH=P*10**3*cos(pi/180*alfa)## kN
PV=P*10**3*sin(pi/180*alfa)## kN
t=symbols('t')## thickness of link
A=2*t*t## mm.sq.
sigma_d=PH/A## N/mm.sq.
M=PH*100+PV*250## N.mm
I=t*(2*t)**3/12## mm**4 (Moment of Inertia)
sigma_t=M*t/I## N/mm.sq.
#maximum tensile stress at the top fibres = sigma_d+sigma_t=sigma_w ...eqn(1)
expr=sigma_d+sigma_t-sigma_w ## expression of polynomial from above eqn.
t=solve(expr,'t')## solving the equation (as denominator will me be multiplied by zero on R.H.S)
t=t[0]## mm # discarding -ve roots
print ' dimension of cross section of link, t=%.f mm.'%(t)
from __future__ import division
from sympy import symbols,solve
# Given Data
P=20## kN
Sut=300## MPa
n=3## factor of safety
sigma_w=Sut/n## MPa (Working stress for the link)
t=symbols('t')## thickness of link
A=4*t*t## mm.sq.
sigma_d=P*10**3/A## N/mm.sq.
e=6*t##mm
M=P*10**3*e## N.mm
z=t*(4*t)**2/6## mm**3 (section modulus at x1-x2)
sigma_b=M/z## N/mm.sq.
#maximum tensile stress at x1 = sigma_d+sigma_b=sigma_w ...eqn(1)
expr=sigma_d+sigma_b-sigma_w ## expression of polynomial from above eqn.
t=solve(expr,'t')## solving the equation (as denominator will me be multiplied by zero on R.H.S)
t=t[1]## mm # discarding -ve roots
print ' dimension of cross section of link, t=%.2f mm. Use 23 mm.'%(t)
from __future__ import division
from sympy import symbols,solve
from math import ceil
# Given Data
P=15## kN
sigma_t=20## MPa
sigma_c=60## MPa
n=3## factor of safety
a=symbols('a')## from the diagram.
# Area of cross section
A1=2*a*a## mm.sq.
A2=2*a*a/2## mm.sq.
A=A1+A2## mm.sq.
# Location of neutral axis
#3*a**2*y_bar=2*a**2*a/2+a**2*(a+a/2)
y_bar=(2*a**2*a/2+a**2*(a+a/2))/(3*a**2)## mm
# Moment of Inertia about neutral axis N-A
I=2*a*a**3/12+2*a**2*(y_bar-0.5*a)**2+2*((a/2)*(a**3/12)+(a**2/2)*(1.5*a-y_bar)**2)## mm**4
yt=y_bar##mm
yc=2*a-y_bar## mm
e=y_bar-0.5*a##mm
M=P*10**3*e## N.mm
sigma_d=P*10**3/A## N/mm.sq.
sigma_t1=M*yt/I##N/mm.sq.
sigma_c1=M*yc/I##N/mm.sq.
sigma_r_t=sigma_d+sigma_t1## N/mm.sq. (sigma_r_t=resultant tensile stress at AB=sigma_d+sigma_t)
sigma_r_c=sigma_c1-sigma_d## N/mm.sq. (sigma_r_t=resultant tensile stress at AB=sigma_d+sigma_t)
#equating resulting tensile stress with given value sigma_t-sigma_r_t=0...eqn(1)
expr1=sigma_t-sigma_r_t## expression of polynomial from above eqn.
a1=solve(expr1,'a')## solving the equation (as denominator will me be multiplied by zero on R.H.S)
a1=a1[1]## mm # discasrding -ve roots
print ' Equating resultant tensile stress gives, a = %.2f mm'%(a1)
#equating resulting compressive stress with given value sigma_c-sigma_c_t=0...eqn(1)
expr2=sigma_c-sigma_r_c## expression of polynomial from above eqn.
a2=solve(expr2,'a')## solving the equation (as denominator will me be multiplied by zero on R.H.S)
a2=a2[1]## mm # discarding -ve roots
print ' \n Equating resultant compressive stress gives, a = %.2f mm'%(a2)
a=ceil(a1)##mm
print ' \n dimension of cross section of link, a=%.2f mm. adopt a=%.f mm.'%(a1,a)
from __future__ import division
from math import pi,sqrt,ceil
# Given Data
Syt=760## MPa
M=15## kN.m
T=25##kN.m
n=2.5## factor of safety
E=200## GPa
v=0.25## Poisson's ratio
sigma_d=Syt/n## MPa
# let d is diameter of the shaft
sigma_b_into_d_cube=32*M*10**6/pi## N/mm.sq. (where sigma_b_into_d_cube = sigma_d*d**3)
tau_into_d_cube=16*T*10**6/pi#d**3## N/mm.sq. (where tau_into_d_cube = tau*d**3)
sigma1_into_d_cube=sigma_b_into_d_cube/2+1/2*sqrt(sigma_b_into_d_cube**2+4*tau_into_d_cube**2) # # (where sigma1_into_d_cube=sigma1*d**3)
sigma2_into_d_cube=sigma_b_into_d_cube/2-1/2*sqrt(sigma_b_into_d_cube**2+4*tau_into_d_cube**2)# # (where sigma2_into_d_cube=sigma2*d**3)
print ' \n (i) Maximum shear stress theory'
tau_max_into_d_cube=(sigma1_into_d_cube-sigma2_into_d_cube)/2# #(where tau_max_into_d_cube = tau_max*d**3)
d=(tau_max_into_d_cube/(sigma_d/2))**(1/3)##mm
print ' diameter of shaft, d=%.1f mm or %.f mm'%(d,ceil(d))
print ' \n (ii) Maximum strain energy theory'
#sigma1**2+sigma2**2-2*v*sigma1*sigma2=sigma_d**2
d=((sigma1_into_d_cube**2+sigma2_into_d_cube**2-2*v*sigma1_into_d_cube*sigma2_into_d_cube)/sigma_d**2)**(1/6)
print ' diameter of shaft, d=%.1f mm'%(d)
print ' \n Adopt d=100mm'
from __future__ import division
from math import sqrt,pi
# Given Data
N=200## rpm
P=200## kW
tau_d=42## Mpa
W=900## N
L=3## m
sigma_t=56## MPa
sigma_c=56## MPa
T=P*60*10**3/(2*pi*N)## N.m
M=W*L/4## N.m
Te=sqrt(M**2+T**2)## N.m
#Te=(pi/16)*d**3*tau_d
d=(Te/((pi/16)*tau_d)*1000)**(1/3)## mm
print ' \n Using equivalent torque equation,\n shaft diameter d = %.f mm'%(d)
Me=(1/2)*(M+sqrt(M**2+T**2))## N.m
#Me=(pi/32)*d**3*sigma_d
d=(Me/((pi/32)*sigma_c)*10**3)**(1/3)##mm
print ' \n Using equivalent bending moment equation,\n shaft diameter d = %.2f mm or %.f mm'%(d, ceil(d))
print ' \n Adopt d=105 mm.'
from __future__ import division
from math import sqrt,pi
# Given Data
M=15## N.m
P=5## kW
N=500## rpm
tau_d=40## Mpa
sigma_d=58## MPa
T=P*60*10**3/(2*pi*N)## N.m
Te=sqrt(M**2+T**2)## N.m
#Te=(pi/16)*d**3*tau_d
d=(Te/((pi/16)*tau_d)*1000)**(1/3)## mm
print ' \n Using equivalent torque equation,\n shaft diameter d = %.f mm'%(d)
Me=(1/2)*(M+sqrt(M**2+T**2))## N.m
#Me=(pi/32)*d**3*sigma_d
d=(Me/((pi/32)*sigma_d)*10**3)**(1/3)##mm
print ' \n Using equivalent bending moment equation,\n shaft diameter d = %.2f mm or %.f mm'%(d, ceil(d))
print ' \n Adopt d=23 mm.'
from __future__ import division
from math import sqrt,pi
from sympy import symbols,solve
# Given Data
d=4## cm
M=15000## N.cm
Syt=20000## N/cm.sq.
print ' \n (i) Maximum Principal Stress Theory-'
z=pi*d**3/32## cm.cube.
sigma_b=M/z## N/cm.sq.
T=symbols('T')
tau=16*T/(pi*d**3)## N/cm.sq.
#sigma1=(1/2)*(sigma_b+sqrt(sigma_b**2+4*tau**2)) # Maximum principal stress
#sigma1=(sigma_b/2+sqrt(sigma_b**2/4+tau**2)) # on solving
#tau=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)
sigma1=Syt## N/cm.sq.
T=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)*(pi*d**3)/16## N.cm.
print ' \n Maximum value of torque, T = %.f N.cm.'%(T)
print ' \n (ii) Maximum Shear Stress Theory'
tau_d=0.5*Syt## N.cm.
#Te=sqrt(M**2+T**2)=(pi/16)*d**3*tau_d
T=sqrt(((pi/16)*d**3*tau_d)**2-M**2)## N.cm.
print ' \n Maximum value of torque, T = %.f N.cm.'%(T)
# Answer in the textbook is not accurate.
from __future__ import division
from math import sqrt,pi
# Given Data
N=200## rpm
P=25## kW
tau_d=42## MPa
W=900## N
L=3## m
Syt=56## MPa
Syc=56## MPa
sigma_d=56## MPa
T=P*60*10**3/(2*pi*N)## N.m
M=W*L/4## N.m
Te=sqrt(M**2+T**2)## N.m
# Te=(pi/16)*d**3*tau_d
d=(Te*10**3/((pi/16)*tau_d))**(1/3)## mm
print ' \n shaft diameter(using equivalent torque)-\n d=%.f mm.'%(d)
Me=(1/2)*(M+sqrt(M**2+T**2))##N.m
# Me=(pi/32)*d**3*sigma_d
d=(Me*10**3/((pi/32)*sigma_d))**(1/3)## mm
print ' \n shaft diameter(using equivalent bending moment)-\n d=%.f mm.'%(d)
print ' \n adopt d=57 mm.'
from __future__ import division
from math import sqrt,pi,cos,sin
from sympy import symbols,solve
# Given Data
sigma_w=60## MPa
F=10## kN
alfa=30## degree
FH=F*sin(pi/180*alfa)## kN
FV=F*cos(pi/180*alfa)## kN
t=symbols('t')## mm
A=t*t## mm.sq.
sigma_d=FV*10**3/A
M=FV*10**3*120+FH*10**3*150## N.mm
I=t*(2*t)**3/12## mm**4
sigma_t=M*t/I## N/mm.sq.
# Tensile stress at A=sigma_d+sigma_t=sigma_w ...eqn(1)
expr = sigma_d+sigma_t-sigma_w## polynomial from above eqn.
t=solve(expr,'t')## roots of the polynomial
t=t[0]## mm # discarding -ve roots
print ' \n value of t = %.1f mm'%(t)
A=2*t**2## mm.sq.
print ' \n Area of cross-section of Hanger, A = %.f mm.sq.'%(A)
# Note-Answer in the textbook is slighly wrong and cross section not calculated.
from __future__ import division
from math import tan,pi,sqrt
# Given Data
P=15## kW
n1=200## rpm
l=600## mm
z2=18## no. of teeth
m2=5## mm
alfa2=14.5## degree
l2=120## mm
z1=72## no. of teeth
m1=5## mm
alfa1=14.5## degree
l1=150## mm
sigma_d=80## MPa
d1=m1*z1## mm
v1=pi*d1*n1/(60*10**3)## m/s
Ft1=10**3*P/v1## N (outwards)
Fr1=Ft1*tan(pi/180*alfa1)## N (Downwards)
d2=m2*z2## mm
v2=pi*d2*n1/(60*10**3)## m/s
Ft2=10**3*P/v2## N (outwards)
Fr2=Ft2*tan(pi/180*alfa2)## N (Upwards)
# RAV*600=Fr1*450+Fr2*120 (Taking moments about bearing B)
RAV=(Fr1*450+Fr2*120)/600## N (Downwards)
RBV=(Fr1-Fr2-RAV)## N (upwards)
MCV=RAV*l1## N.mm
MBV=Fr2*l2## N.mm
# RAH*600=-Ft1*450+Ft2*120 (Taking moments about bearing B)
RAH=(-Ft1*450+Ft2*120)/600## N (Outwards)
RBH=Ft1+Ft2+RAH## N (inwards)
MCH=RAH*l1## N.mm
MBH=Ft2*l2## N.mm
# Resultant Bending Moments
MC=sqrt(MCV**2+MCH**2)## N.mm
MB=sqrt(MBV**2+MBH**2)## N.mm
Mmax=max(MC,MB)## N.mm
T=10**3*P/(2*pi*n1)## N.m
Me=(1/2)*(Mmax+sqrt(Mmax**2+T**2))## N.mm
# Me=(pi/32)*d**3*sigma_d
d=(Me/((pi/32)*sigma_d))**(1/3)
print ' \n shaft diameter is : %.f mm'%(d)