Chapter 4 - Design Against Fluctuating Load

exa 4.1 Pg 102

In [1]:
from __future__ import division
from sympy import symbols,solve
P=6## kN

#dimensions of plate
r=5##mm
d=40##mm
D=50##mm
d0=10##mm
w=40##mm
Sut=200##MPa
n=2.5## factor of safety

#Fillet - 
rBYd=r/d#
DBYd=D/d#
Kt=1.75## factor
print ' for stepped plate under tension, Kt=%.2f for r/d = %.3f & D/d = %.2f '%(Kt,rBYd,DBYd)
t=symbols('t')
sigma_max = Kt*P/t## N per mm sq.

# Hole -
d0BYw=d0/w#
Kt=2.42## factor 
print ' \n for finite width plate under tension with a hole, Kt=%.2f for d0/w = %.2f'%(Kt,d0BYw)
sigma_max_into_t = Kt*P/(w-d0)##N/mm sq.

#Design stress
sigma_d = Sut/n## MPa
#putting sigma_max=sigma_d
t=sigma_max_into_t/sigma_d*1000## mm
print ' \n Thickness of plate = %.2f mm or %.f mm'%(t,t)

 for stepped plate under tension, Kt=1.75 for r/d = 0.125 & D/d = 1.25 
 
 for finite width plate under tension with a hole, Kt=2.42 for d0/w = 0.25
 
 Thickness of plate = 6.05 mm or 6 mm

exa 4.2 Pg 104

In [2]:
from __future__ import division
from math import pi
# Given Data
rBYd=0.1#
DBYd=1.2#
P=3## kN
Syt=300##MPa
n=3## factor of safety
#dimensions of plate
l1=400##mm
l2=300##mm
l3=400##mm


sigma_d=Syt/n## MPa
Kt=1.65## factor for circular fillet radius member
Rp=P/2##kN (bearing reaction due to symmetry)
Mf=Rp*l1## kN.mm (bending moment at fillet)
Mc=P*(l1+l2+l3)/4## kN.mm (bending moment at centre)

#Fillet
#sigma_max=Kt*32*Mf/(pi*d**3)
sigma_max_into_d_cube_1 = Kt*32*Mf*1000/pi


#Centre
#sigma_max=32*Mc/(pi*d**3)
sigma_max_into_d_cube_2 = Kt*32*Mf*1000/pi
sigma_max_into_d_cube=max(sigma_max_into_d_cube_1,sigma_max_into_d_cube_2)## (getting max)

#putting sigma_max=sigma_d
t=(sigma_max_into_d_cube/sigma_d)**(1/3)## mm
print ' \n Diameter of axle = %.1f mm'%(t)

 
 Diameter of axle = 46.5 mm

exa 4.3 Pg 105

In [3]:
from __future__ import division

# Given Data
Sut=440##MPa
d=25##mm
R=95/100## reliability
Kt=1.8## stress concentration factor
q=0.86## sensitivity factor

Se_dash = 0.5*Sut## MPa

# for machined surface
ka=0.82## surface finish factor
kb=0.85## size factor
kc=0.868## reliability factor
kd=1## temperature factor
ke=0.577## load factor

Kf=1+q*(Kt-1)## fatigue strength factor
kf=1/Kf ## fatigue strength reduction factor
Se=ka*kb*kc*kd*ke*kf*Se_dash## (MPa) Endurance limit
print ' \n Endurance limit = %.2f MPa'%(Se)

 
 Endurance limit = 45.50 MPa

exa 4.4 Pg 105

In [4]:
from __future__ import division

# Given Data
Sut=440##MPa
w=60##mm
d=12## mm
P=20## kN
q=0.8## sensitivity factor
R=90/100## reliability
n=2## factor of safety

Kt=2.52## stress concentration factor
Se_dash = 0.5*Sut## MPa
# for hot rollednormalized condition
ka=0.67## surface finish factor
kb=0.85## size factor (assuming t<50 mm)
kc=0.897## reliability factor
kd=1## temperature factor
ke=0.9## load factor
dBYw=d/w# #(for circular hole)

Kf=1+q*(Kt-1)## fatigue strength factor
kf=1/Kf ## fatigue strength reduction factor
Se=ka*kb*kc*kd*ke*kf*Se_dash## (MPa) Endurance limit
sigma_d=Se/n## MPa (design stress)
# sigma_max=P/(w-d)/t
sigma_max_into_t = P*1000/(w-d)#
# putting sigma_max=sigma_d
t=sigma_max_into_t/sigma_d## mm
print ' \n Thickness of plate = %.2f mm or 20 mm'%(t)

 
 Thickness of plate = 18.26 mm or 20 mm

exa 4.5 Pg 107

In [5]:
from __future__ import division
from math import log10
# Given Data
Sut=650##MPa
N=10**5## cycles
Se_dash = 0.5*Sut## MPa
of=5## unit
ob=6##unit
bf=ob-of## unit
be=3##unit

# calculating endurance section wise
OE=log10(Se_dash)#
OA=log10(0.9*Sut)#
AE=OA-OE#
#log10_Sf=OD=OE+ED=OE+FC
log10_Sf=OE+(bf/be)*AE#
Sf=10**log10_Sf# # (MPa) Endurance
print ' \n Endurance of specimen = %.2f MPa'%(Sf)

 
 Endurance of specimen = 395.34 MPa

exa 4.6 Pg 108

In [6]:
from __future__ import division
from math import log10,pi
# Given Data
Sut=540##MPa
N=10**4## cycles
q=0.85## sensitivity factor
R=90/100## reliability
P=1500## N
l=160## mm

Se_dash = 0.5*Sut## MPa
# for cold drawn steel
ka=0.79## surface finish factor
kb=0.85## size factor (assuming t<50 mm)
kc=0.897## reliability factor
kd=1## temperature factor
ke=1## load factor

Kt=1.33## under bending

Kf=1+q*(Kt-1)## fatigue strength factor
kf=1/Kf ## fatigue strength reduction factor
Se=ka*kb*kc*kd*ke*kf*Se_dash## MPa( Endurance limit)

of=4## unit
ob=6##unit
bf=ob-of## unit
be=3##unit

# calculating endurance section wise
OE=log10(Se)#
OA=log10(0.9*Sut)#
AE=OA-OE#
#log10_Sf=OD=OE+ED=OE+FC
log10_Sf=OE+(bf/be)*AE#
Sf=10**log10_Sf# # (MPa) Endurance

MB=P*l## N.mm
# 32*MB/pi/d**3 = Sf
d=(32*MB/pi/Sf)**(1/3)
print ' \n diameter of beam %.f mm'%(d)

 
 diameter of beam 20 mm

exa 4.7 Pg 110

In [7]:
from __future__ import division
from math import pi,atan
# Given Data
Sut=600##MPa
Syt=380##MPa
q=0.9## sensitivity factor
R=90/100## reliability
n=2## factor of safety
Pmin=-100## N
Pmax=200## N
l=150## mm

Se_dash = 0.5*Sut## MPa
# for cold drawn steel
ka=0.76## surface finish factor
kb=0.85## size factor (assuming t<50 mm)
kc=0.897## reliability factor
kd=1## temperature factor
ke=1## load factor

Kt=1.4## under bending

Kf=1+q*(Kt-1)## fatigue strength factor
kf=1/Kf ## fatigue strength reduction factor
Se=ka*kb*kc*kd*ke*kf*Se_dash## MPa( Endurance limit)
Mmax=Pmax*l## N.mm
Mmin=Pmin*l## N.mm
Mm=(Mmax+Mmin)/2## N.mm
Ma=(Mmax-Mmin)/2## N.mm
theta=atan(Ma/Mm)*180/pi## degree

#equation of Goodman - sigma_m/Sut+sigma_a/Se=1
#here sigma_a/sigma_m=3
sigma_m=1/(1/Sut+3/Se)##MPa
sigma_a=3*sigma_m## MPa

sigma_da=sigma_a/n## MPa
#sigma_da=32*Ma/pi/d**3
d=(32*Ma/pi/sigma_da)**(1/3)## mm 
print ' \n diameter d at fillet cross section = %.f mm'%(d)

 
 diameter d at fillet cross section = 16 mm

exa 4.8 Pg 112

In [8]:
from __future__ import division
from math import pi,tan,atan
# Given Data
Sut=500##MPa
Syt=300##MPa
R=90/100## reliability
n=2## factor of safety
Tmin=-200## N.m
Tmax=500## N.m

Se_dash = 0.5*Sut## MPa
# for cold drawn steel
ka=0.80## surface finish factor
kb=0.85## size factor (assuming t<50 mm)
kc=0.897## reliability factor
kd=1## temperature factor
ke=0.577## load factor

Ses=ka*kb*kc*kd*ke*Se_dash## MPa( Endurance limit)
Sys=ke*Syt## MPa
Tm=(Tmax+Tmin)/2## N.m
Ta=(Tmax-Tmin)/2## N.m
theta=atan(Ta/Tm)*180/pi## degree
Sms=Ses/tan(pi/180*theta)##MPa
Sas=Ses##MPa
tau_da=Sas/n##MPa
#tua_da=16*Ta/pi/d**3
d=(16*Ta*1000/pi/tau_da)**(1/3)##mm
print ' \n diameter of shaft = %.f mm'%(d)

 
 diameter of shaft = 34 mm

exa 4.9 Pg 113

In [9]:
from __future__ import division
from math import pi,log10
# Given Data
Sut=860##MPa
Syt=690##MPa
Pmin=60## N
Pmax=120## N
R=50/100## reliability
l=500##mm
d=10##mm
Se_dash = 0.5*Sut## MPa
# for machines surface
ka=0.70## surface finish factor
kb=0.85## size factor (assuming t<50 mm)
kc=1## reliability factor
kd=1## temperature factor
ke=1## load factor

Se=ka*kb*kc*kd*ke*Se_dash## MPa( Endurance limit)
Mmax=Pmax*l## N.mm
Mmin=Pmin*l## N.mm
Mm=(Mmax+Mmin)/2## N.mm
Ma=(Mmax-Mmin)/2## N.mm
Sm=32*Mm/pi/d**3##MPa
sigma_m=Sm##MPa
Sa=32*Ma/pi/d**3##MPa
sigma_a=Sa##MPa
Sf=Sa*Sut/(Sut-Sm)##MPa

#calculating section
OB=6##unit ref. o at 3
BE=OB-3##unit
OC=Sf## MPa
AE=log10(0.9*Sut)-log10(Se)##MPa
AC=log10(0.9*Sut)-log10(Sf)##MPa
CD=BE*AC/AE##
#log10(N)=3+CD
N=10**(3+CD)## cycle
print ' \n life of the spring, N = %.f cycles'%(N)
#Note : answer in the textbook is wrong.

 
 life of the spring, N = 215630 cycles

exa 4.10 Pg 116

In [10]:
from __future__ import division
from math import sqrt,tan,pi,atan
# Given Data
Sut=600##MPa
Se=280##MPa
sigma_x_min=50## MPa
sigma_x_max=100## MPa
sigma_y_min=20## MPa
sigma_y_max=70## MPa

sigma_xm=(sigma_x_max+sigma_x_min)/2## MPa
sigma_xa=(sigma_x_max-sigma_x_min)/2## MPa
sigma_ym=(sigma_y_max+sigma_y_min)/2## MPa
sigma_ya=(sigma_y_max-sigma_y_min)/2## MPa

# distortion energy theory - 
sigma_m=sqrt(sigma_xm**2+sigma_ym**2-sigma_xm*sigma_ym)## MPa
sigma_a=sqrt(sigma_xa**2+sigma_ya**2-sigma_xa*sigma_ya)## MPa
theta=atan(sigma_a/sigma_m)*180/pi## degree
# Sm/Sut+Sa/Se=1 where Sa=Sm*tan(theta)
Sm=1/(1/Sut+tan(pi/180*theta)/Se)## MPa
Sa=tan(pi/180*theta)*Sm## MPa
n=Sa/sigma_a## factor of safety

print ' \n factor of safety, n = %.2f'%(n)

 
 factor of safety, n = 5.04

exa 4.11 Pg 117

In [11]:
from __future__ import division
from math import sqrt,pi
# Given Data
Sut=600##MPa
Syt=400##MPa
Se=200##MPa
Mmin=200## N.m
Mmax=500## N.m
Tmin=60## N.m
Tmax=180## N.m
n=2## factor of safety

Mm=(Mmax+Mmin)/2## N.mm
Ma=(Mmax-Mmin)/2## N.mm
Tm=(Tmax+Tmin)/2## N.mm
Ta=(Tmax-Tmin)/2## N.mm
# sigma_xm=32*Mm/pi/d**3
sigma_xm_into_d_cube=(32*Mm*1000)/pi#
# sigma_xa=32*Ma/pi/d**3
sigma_xa_into_d_cube=(32*Ma*1000)/pi#
#Txym=16*Tm/pi/d**3
Txym_into_d_cube=16*Tm*1000/pi#
#Txya=16*Ta/pi/d**3
Txya_into_d_cube=16*Ta*1000/pi#
# sigma_m=sqrt(sigma_xm**2+3*Txym**2)
sigma_m_dash=sqrt(sigma_xm_into_d_cube**2+3*Txym_into_d_cube**2)## taken sigma_m_dash = sigma_m*d**(-3) for calculation
# sigma_a=sqrt(sigma_xa**2+3*Txya**2)
sigma_a_dash=sqrt(sigma_xa_into_d_cube**2+3*Txya_into_d_cube**2)## taken sigma_a_dash = sigma_a*d**(-3) for calculation
#tan(theta) = sigma_a/sigma_m
theta = atan(sigma_a_dash/sigma_m_dash)## radian
#Sm/Sut+Sa/Se= 1 where Sa/Sm=0.4348 
Sm= 1/(1/Sut+0.4348/Se)## MPa
Sa=0.4348 * Sm## MPa
#sigma_a=Sa/n
d=(Sa/n/sigma_a_dash)**(1/3)*1000## mm
print ' \n diameter of shaft, d = %.2f mm'%(d)
# Note - Ans in the textbook is wrong.

 
 diameter of shaft, d = 32.71 mm

exa 4.12 Pg 119

In [12]:
from __future__ import division
from math import pi,atan,ceil
# Given Data
Sut=620##MPa
Syt=380##MPa
R=90/100## Reliability
n=2.5## factor of safety
Tmin=-200## N.m
Tmax=400## N.m

Se_dash=0.5*Sut##MPa
# for ground shaft
ka=0.92## surface finish factor
kb=0.85## size factor (assuming t<50 mm)
kc=0.897## reliability factor
kd=1## temperature factor
ke=0.577## load factor
Ses=ka*kb*kc*kd*ke*Se_dash## MPa( Endurance limit)
Sys=ke*Syt## MPa
Tm=(Tmax+Tmin)/2## N.mm
Ta=(Tmax-Tmin)/2## N.mm
theta=atan(Ta/Tm)##radian
Sas=Ses## MPa
Sms=Sas/3## MPa
#Tda=Sas/n=16*Ta/pi/d**3
d=(16*Ta*1000/pi/(Sas/n))**(1/3)## mm
print ' \n diameter of shaft, d = %.2f mm or %d mm'%(d, ceil(d))

 
 diameter of shaft, d = 31.22 mm or 32 mm

exa 4.14 Pg 121

In [13]:
from __future__ import division

# Given Data
sigma_max=300## MPa
sigma_min=-150## MPa
n=1.5## factor of safety


sigma_m=(sigma_max+sigma_min)/2## MPa
sigma_a=(sigma_max-sigma_min)/2## MPa
# Goodman failure line - sigma_m/Sut+sigma_a/Se=1/n
Sut=(sigma_m+sigma_a/(0.5))*n ## putted Se=0.5*Sut
print ' \n Minimum required ultimate strength, Sut = %.1f MPa'%(Sut)

 
 Minimum required ultimate strength, Sut = 787.5 MPa

exa 4.16 Pg 122

In [14]:
from __future__ import division
from math import pi,sqrt
# Given Data
Pmin=-15## kN
Pmax=25## kN
Se_dash=360## MPa
Sy=400## MPa
Ki=1.25## impact factor
n=2.25## factor of safety
ka=0.88## surface finish factor
Kt=2.25## stress concentration factor
Pm=(Pmax+Pmin)/2## kN
Pa=(Pmax-Pmin)/2## kN
q=0.8## sensitivity factor

# sigma_m=4*Pm/pi/d**2
sigma_m_into_d_sq = 4*Pm*1000/pi#
sigma_a_into_d_sq = 4*Pa*1000/pi#
Kf=1+q*(Kt-1)## fatigue strength factor
kf=1/Kf ## fatigue strength reduction factor
kb=0.85## size factor
ke=0.9##load factor
ki=1/Ki## reverse impact factor
Se=ka*kb*ke*kf*ki*Se_dash## MPa
#soderburg failure equation - sigma_m/Sy+sigma_a/Se=1/n
d=sqrt((sigma_m_into_d_sq/Sy+sigma_a_into_d_sq/Se)*n)
print ' \n Size of piston rod, d = %.f mm'%(d)

 
 Size of piston rod, d = 25 mm

exa 4.18 Pg 123

In [15]:
from __future__ import division
from math import pi,sqrt
# Given Data
Pmin=-300## kN
Pmax=700## kN
Se_dash=280## MPa
Sy=350## MPa
Kf=1.8##fatigue strength factor
n=2## factor of safety

Pm=(Pmax+Pmin)/2## kN
Pa=(Pmax-Pmin)/2## kN
# sigma_m=4*Pm/pi/d**2
sigma_m_into_d_sq = 4*Pm*1000/pi#
sigma_a_into_d_sq = 4*Pa*1000/pi#
kf=1/Kf ## fatigue strength reduction factor
kb=0.85## size factor
ke=0.9##load factor
ka=0.93## surface finish factor
Se=ka*kb*ke*kf*Se_dash## MPa
#Goodman failure equation - sigma_m/Sy+sigma_a/Se=1/n
d=sqrt((sigma_m_into_d_sq/Sy+sigma_a_into_d_sq/Se)*2.25)
print ' \n Suitable diameter of rod, d = %.f mm'%(d)
# Note - Ans in the textbook is wrong.

 
 Suitable diameter of rod, d = 121 mm

exa 4.19 Pg 124

In [16]:
from __future__ import division

# Given Data
w=110## mm
Pmin=98.1## kN
Pmax=250## kN
Se=225## N/mm.sq
Sy=300## N/mm.sq
n=1.5## factor of safety

Pm=(Pmax+Pmin)/2## kN
Pa=(Pmax-Pmin)/2## kN
# sigma_m=Pm/w/t
sigma_m_into_t = Pm/w#
sigma_a_into_t = Pa/w#
#Soderburg failure equation - sigma_m/Sy+sigma_a/Se=1/n
d=(sigma_m_into_t/Sy+sigma_a_into_t/Se)*n*1000## mm
print ' \n thickness of plate, t = %.1f mm'%(d)

 
 thickness of plate, t = 12.5 mm

exa 4.20 Pg 124

In [17]:
from __future__ import division
from math import sqrt,pi
# Given Data
Mmin=200## kN.mm
Mmax=600## kN.mm
Tmin=60## kN
Tmax=180## kN
Su=550## MPa
Sy=400## MPa
Se=0.5*Su## MPa
n=1.5## factor of safety
Ktb=1.5## stress concentration factor in blending
Kts=1.2## stress concentration factor in torsion

Mm=(Mmax+Mmin)/2## kN.mm
Ma=(Mmax-Mmin)/2## kN.mm

#sigma_xm=32*Mm/pi/d**3
sigma_xm_into_d_cube=32*Mm/pi#
#sigma_xa=32*Ma/pi/d**3
sigma_xa_into_d_cube=32*Ma/pi#
Tm=(Tmax+Tmin)/2## kN.mm
Ta=(Tmax-Tmin)/2## kN.mm
Txym_into_d_cube=16*Tm/pi#
Txya_into_d_cube=16*Ta/pi#
# using distortion energy theory
# sigma_m=sqrt(sigma_xm**2+3*Txym**2)
sigma_m_into_d_cube=sqrt(sigma_xm_into_d_cube**2+3*Txym_into_d_cube**2)#
# sigma_a=sqrt((Ktb*sigma_xa)**2+3*(Kts*Txym)**2)
sigma_a_into_d_cube=sqrt((Ktb*sigma_xa_into_d_cube)**2+3*(Kts*Txya_into_d_cube)**2)#
# Sodeburg equation - sigma_m + (Su/Se)*sigma_a=Sy/n
d=((sigma_m_into_d_cube + (Su/Se)*sigma_a_into_d_cube)*1000/(Sy/n))**(1/3)
print ' \n shaft size, d = %.f mm'%(d)

 
 shaft size, d = 34 mm

exa 4.21 Pg 126

In [18]:
from __future__ import division

# Given Data
# Hole -
d=25##mm
w=150##mm
Kt=2.56## stress concentration factor
P=50## kN
sigma_max=100## N/mm.sq
t=Kt*P*1000/(w-d)/sigma_max## mm
print ' Calculating for hole - \n thickness is : %.2f mm'%(t)

# Notch -
d=30##mm
w=120##mm
w=150##mm
Kt=2.3## stress concentration factor
P=50## kN
sigma_max=100## N/mm.sq
t=Kt*P*1000/(w-d)/sigma_max## mm
print ' \n Calculating for notch - \n thickness is : %.2f mm'%(t)
print ' Suggestion, Adopt t = 11 mm'

 Calculating for hole - 
 thickness is : 10.24 mm
 
 Calculating for notch - 
 thickness is : 9.58 mm
 Suggestion, Adopt t = 11 mm