Ch:13 Cotter and knuckle joints

exa 13-1 - Page 371

In [1]:
from math import sqrt, pi
F=25*10**3#
sigat=50#
Ta=40#
pa=80#
d=sqrt((4*F)/(pi*sigat))#
d=26#
t=d/4#
t=7#
d1=1.2*d#
d1=32#
pc=F/(d1*t)#
t=10#
c=0.75*d#
c=20#
d2=44#
tw=(d2-d1)/2#
b=F/(2*t*Ta)#
b=34#
a=0.5*d#
d3=(F/(pa*t))+d1#
d3=64#
e=F/(Ta*(d3-d1))#
d4=sqrt((F*4/(pi*pa))+d1**2)#
d4=40#
f=0.5*d#
sigbc=3*F*d3/(t*b**2*4)#
print " d is %0.0f mm     "%(d)#
print "\n d1 is %0.0f mm     "%(d1)#
print "\n d2 is %0.0f mm     "%(d2)#
print "\n d3 is %0.0f mm     "%(d3)#
print "\n d4 is %0.0f mm     "%(d4)#
print "\n sigbc is %0.1f MPa     "%(sigbc)#

 d is 26 mm     

 d1 is 32 mm     

 d2 is 44 mm     

 d3 is 64 mm     

 d4 is 40 mm     

 sigbc is 103.0 MPa

exa 13-2 - Page 372

In [2]:
from math import sqrt, pi
P=40*10**3#
sigut=490#
FOS=4#
sigts=sigut/FOS#
sigcs=1.4*sigts#
sigs=0.8*sigts#
d=sqrt((4*P)/(pi*sigts))#
d=21#
sigcc=1.4*330/4#
Tc=0.8*330/4#
t=d/3#
b=P/(2*t*Tc)#
b=31#
t=10#
d1=28#
d2=40#
c=d/2#
c=15#
a=P/(2*(d2-d1)*98)#
a=20#
L=(2*a)+(2*b)+(2*c)+(2*3)#
print " d is %0.0f mm     "%(d)#
print "\n d1 is %0.0f mm     "%(d1)#
print "\n t is %0.0f mm     "%(t)#
print "\n b is %0.0f mm     "%(b)#
print "\n d2 is %0.0f mm     "%(d2)#
print "\n L is %0.0f mm     "%(L)#

 d is 21 mm     

 d1 is 28 mm     

 t is 10 mm     

 b is 31 mm     

 d2 is 40 mm     

 L is 138 mm

exa 13-3 - Page 372

In [3]:
from math import sqrt, pi
P=40*10**3#
sigt=60#
sigc=125#
T=45#
a=sqrt(P*3/(2*sigt))#
a=33#
t=a/3#
b=P/(4.5*t*T)#
b=20#
b1=1.25*b#
t1=P*3/(4*a*sigt)#
t1=16#
l2=P/(2*2*T*t1)#
l2=14#
l1=P/(2*a*T)#
l1=14#
l3=(0.6*a)#
l3=20#
l4=11#
sigcr=P/(t*a)#
sigcr1=P/(2*t1*t)#
print " a is %0.0f mm     "%(a)#
print "\n t is %0.0f mm     "%(t)#
print "\n t1 is %0.0f mm     "%(t1)#
print "\n b is %0.0f mm     "%(b)#
print "\n b1 is %0.0f mm     "%(b1)#
print "\n l1 is %0.0f mm     "%(l1)#
print "\n l2 is %0.0f mm     "%(l2)#
print "\n l3 is %0.0f mm     "%(l3)#
print "\n l4 is %0.0f mm     "%(l4)#
print "\n sigcr is %0.1f MPa     "%(sigcr)#
print "\n sigcr1 is %0.1f MPa     "%(sigcr1)#

 a is 33 mm     

 t is 11 mm     

 t1 is 16 mm     

 b is 20 mm     

 b1 is 25 mm     

 l1 is 14 mm     

 l2 is 14 mm     

 l3 is 20 mm     

 l4 is 11 mm     

 sigcr is 110.0 MPa     

 sigcr1 is 113.0 MPa

exa 13-4 - Page 373

In [4]:
P=50*10**3#
sigp=380#
FOS=4#
sigca=80#
Ta=50#
sigta=sigp/FOS#
At=P/sigta#
d=30#
d1=1.5*d#
t=P/(sigca*d1)#
t=14#
A=(pi*(d1**2)/4)-(d1*t)#
#let tearing stress be sigt
sigt=P/A#
b=P/(2*t*Ta)#
b=36#
print " d is %0.0f mm     "%(d)#
print "\n sigt is %0.1f MPa     "%(sigt)#
print "\n b is %0.0f mm     "%(b)#
  
#The answer to tearing stress in bolt 'sigt' is calculated incorrectly in the book.

 d is 30 mm     

 sigt is 52.1 MPa     

 b is 36 mm