Chapter 7 - Keys & Couplings

exa 7.1 Pg 195

In [4]:
from __future__ import division
from math import sqrt,pi,floor
from numpy import roots
# Given Data
P=20## kW
N=240## rpm
tau_s=45## MPa
tau_b=30## MPa
sigma_b=60## MPa
sigma_cs=2*tau_s## MPa
tau_ci=15## MPa
#Tmax=1.25*Tm
mu=0.15## coefficient of friction

#SHAFT DIAMETER
# P= 2*pi*N*Tm/60/1000 
Tm=P/(2*pi*N/60/1000)## N.m
Tmax=1.25*Tm## N.m
# pi*d**3*tau_s/16= Tmax
d=(Tmax/(pi*tau_s/16)*1000)**(1/3)## mm
print ' shaft diameter = %.2f mm. Use d = 50 mm.'%(d)
d=50## mm

# HUB DIAMETER
# Tmax=pi/16*((d1**4-d**4)/d1)*tau_h
tau_h=tau_ci## MPa
#d1*(Tmax/(pi/16)/tau_h)-d1**4=d**4 -- eqn(1)
Tmax=Tmax*1000## N.mm
p=[1, 0 ,0 ,-Tmax/(pi*tau_h/16), -d**4] ## polynomial coefficients from eqn(1)
d1=roots(p)## roots of poly 
d1=d1[0]## mm (taking +ve value)
d1=100## mm (empirically adopted) 
t1=(d1-d)/2## mm (thickness of hub)
print ' \n thickness of hub = %.f mm'%(t1)
d4=d+t1## mm (diameter of recess in flanges)
print ' \n diameter of recess in flanges = %.f mm'%(d4)
d3=4*d## mm (outside diameter of protecting flange)
print ' \n outside diameter of protecting flange = %.f mm'%(d3)

# Hub length
b=d/4## mm (width of key)
l=1.5*d## mm (length of key)
print ' \n width of key = %.1f mm. Use b = 15 mm'%(b)
b=15## mm
print ' \n length of key = %.f mm.'%(l) 
t=b## mm (thickness for square key)
print ' \n thickness for square key = %.f mm'%(t)
print ' \n Hub length = %.f mm'%(l)

#Number of bolts
n=floor(4*d/150+3)## no. of bolts
print ' \n Number of bolts = %.f'%(n)

# Bolt diameter
r2=1.5*d## mm
F=Tmax/r2/n## N
#pi/4*db**2*tau_b=F
db=sqrt(F/(pi/4*tau_b))## mm
print ' \n Bolt diameter = %.2f mm. Use db=12 mm'%(db)
bolt_dia=db##mm

# Bolt diameter based on Tensile load
r3=d3/2## mm
r4=d4/2## mm
rf=2/3*((r3**3-r4**3)/(r3**2-r4**2))## mm
#Tmax=n*mu*Pi*rf## N
Pi=Tmax/(n*mu*rf)## N
# Pi=pi/4*db**2*sigma_t
sigma_t=sigma_b## MPa
db=sqrt(Pi/(pi/4*sigma_t))## mm 
print ' \n Bolt diameter (based on Tensile load) = %.1f mm. Use db=15 mm'%(db)
db=15## mm (adopted)

# Flange thickness
t2=0.5*t1+6## mm (empirically)
print ' \n Flange thickness = %.1f mm. Use t=20 mm'%(t2)
t2=20## mm (adopted)
#F=n*db*t2*sigma_c
sigma_ci=F/n/db/t2## MPa
#2*pi*d1**2*tau*t2/4=Tmax
tau=Tmax/(2*pi*d1**2*t2/4)## MPa
print ' \n permissible bearing stress in flange = %.2f MPa < 30 MPa'%(sigma_ci)
print ' \n shearing of the flange at the junction with hub = %.2f MPa < 15 MPa.'%(tau)
print '  Values are acceptable.'

# Check for crushing of bolt
#n*db*t2*sigma_cb*d2/2=Tmax
d2=d1+d## mm
db=bolt_dia##mm
sigma_cb=Tmax/(n*db*t2*d2/2)## MPa
print ' \n permissible crushing strength of bolts = %.1f MPa < 60 MPa.'%(sigma_cb)
print '  Hence design is safe.'

# Thickness of protecting flange
t3=0.5*t2## mm
print ' \n Thickness of protecting flange = %.f mm'%( t3)
# Hub overlap 
ho=3## mm (min)
print ' \n Hub overlap = %.f mm (min)'%(ho)
#Note - Answer for **Bolt diameter based on Tensile load** is calculated wrong in the textbook(error in Pi calculation).

 shaft diameter = 48.29 mm. Use d = 50 mm.
 
 thickness of hub = 25 mm
 
 diameter of recess in flanges = 75 mm
 
 outside diameter of protecting flange = 200 mm
 
 width of key = 12.5 mm. Use b = 15 mm
 
 length of key = 75 mm.
 
 thickness for square key = 15 mm
 
 Hub length = 75 mm
 
 Number of bolts = 4
 
 Bolt diameter = 11.86 mm. Use db=12 mm
 
 Bolt diameter (based on Tensile load) = 21.9 mm. Use db=15 mm
 
 Flange thickness = 18.5 mm. Use t=20 mm
 
 permissible bearing stress in flange = 2.76 MPa < 30 MPa
 
 shearing of the flange at the junction with hub = 3.17 MPa < 15 MPa.
  Values are acceptable.
 
 permissible crushing strength of bolts = 14.0 MPa < 60 MPa.
  Hence design is safe.
 
 Thickness of protecting flange = 10 mm
 
 Hub overlap = 3 mm (min)

exa 7.2 Pg 200

In [2]:
from __future__ import division
from numpy import roots,pi,sqrt
# Given Data
P=30## kW
N=750## rpm
#Tmax=1.2*Tm## MPa
tau_s=35## MPa
tau_b=35## MPa
tau_k=35## MPa
sigma_cs=70## MPa
sigma_ck=70## MPa
sigma_cb=70## MPa
tau_ci=15## MPa
pb=0.8## MPa

#sigma_cs=2*tau_s## MPa

#Tmax=1.5*Tm
mu=0.15## coefficient of friction

#SHAFT DIAMETER
# P= 2*pi*N*Tm/60/1000 
Tm=P/(2*pi*N/60/1000)## N.m
Tmax=1.2*Tm## N.m
# pi*d**3*tau_s/16= Tmax
d=(Tmax/(pi*tau_s/16)*1000)**(1/3)## mm
print ' shaft diameter = %.2f mm. Use d = 42 mm.'%(d)
d=42## mm

# HUB DIAMETER
# Tmax=pi/16*((d1**4-d**4)/d1)*tau_h
tau_h=tau_ci## MPa
#d1*(Tmax/(pi/16)/tau_h)-d1**4=d**4 -- eqn(1)
Tmax=Tmax*1000## N.mm
p=[1, 0 ,0 ,-Tmax/(pi*tau_h/16) ,-d**4] ## polynomial coefficients from eqn(1)
d1=roots(p)## roots of poly 
d1=d1[0]## mm (taking +ve value)
d1=2*d## mm (empirically adopted) 
t1=(d1-d)/2## mm (thickness of hub)
print ' \n thickness of hub = %.f mm'%(t1)
#d4=d+t1## mm (diameter of recess in flanges)
#print ' \n diameter of recess in flanges = %.f mm'%(d4)
d3=4*d## mm (outside diameter of protecting flange)
print ' \n outside diameter of protecting flange = %.f mm. Use 170 mm'%(d3)
d3=170## mm (adopted)

#Key size & Hub length
b=d/4## mm (width of key)
l=1.5*d## mm (length of key)
print ' \n width of key = %.1f mm. Use b = 12 mm'%(b)
b=12## mm
print ' \n length of key = %.f mm.'%(l) 
t=b## mm (thickness for square key)
print ' \n thickness for square key = %.f mm'%(t)
print ' \n Hub length = %.f mm'%(l)

#Number of bolts
n=(0.04*d+3)## no. of bolts
print ' \n Number of bolts = %.2f. Use n=6'%(n)
n=6## adopted

# Bolt diameter
db=0.5*d/sqrt(n)## mm
print ' \n Bolt diameter = %.2f mm. Use db=20 mm for design purpose'%(db)
db=20##mm (adopted)
bolt_dia=db##mm
dsb=24## mm(shank diameter of bolt for design)

# Outer diameter of rubber bush
trb=2## mm (thickness of rubber bush)
tbb=6## mm (thickness of brass bush)
d3=dsb+2*trb+2*tbb## mm 
d2=d1+d3+2*tbb## mm (pitch circle diameter of bolts)
print ' \n pitch circle diameter of bolts = %.f mm '%(d2)

# Check of shear in bolt
F=2*Tmax/n/d2## N
#pi/4*db*2*tau=F
tau=F/(pi/4*db**2)##MPa
print ' \n Permissible shear stress in bolts = %.2f MPa < 35 MPa. Hence design is safe.'%( tau)

# Length of brush
pb=0.8## MPa(bearing pressure of brush)
#F=l2*d3*pb#
l2=F/d3/pb## mm
print ' \n length of bush = %.f mm'%(l2)

# Check for pin in bending
c=5## mm (clearance between two flanges)
l3=(l2-c)/2+c##mm
#M=pi/32*db**3*sigma_b & M=F*l3
sigma_b = F*l3/(pi/32*db**3)## MPa
print ' \n Bending stress in pin = %.1f MPa'%(sigma_b)

# Maximum shear stress in pin
tau_max=sqrt((sigma_b/2)**2+tau**2)##MPa
print ' \n Maximum shear stress in pin = %.2f MPa < 35 MPa. Hence design is safe.'%(tau_max)

# Flange thickness
t2=0.5*t1+6## mm (empirically)
print ' \n Flange thickness = %.1f mm. Use t=18 mm'%(t2)
t2=18## mm (adopted)
tau=Tmax/(2*pi*d1**2*t2/4)## MPa
print ' \n shearing of the flange at the junction with hub = %.2f MPa < 15 MPa.'%(tau)
print '  Values are acceptable.'

#Note - Answer in llast part is calculated wrong in the textbook(error in calculation).

 shaft diameter = 40.55 mm. Use d = 42 mm.
 
 thickness of hub = 21 mm
 
 outside diameter of protecting flange = 168 mm. Use 170 mm
 
 width of key = 10.5 mm. Use b = 12 mm
 
 length of key = 63 mm.
 
 thickness for square key = 12 mm
 
 Hub length = 63 mm
 
 Number of bolts = 4.68. Use n=6
 
 Bolt diameter = 8.57 mm. Use db=20 mm for design purpose
 
 pitch circle diameter of bolts = 136 mm 
 
 Permissible shear stress in bolts = 3.58 MPa < 35 MPa. Hence design is safe.
 
 length of bush = 35 mm
 
 Bending stress in pin = 28.7 MPa
 
 Maximum shear stress in pin = 14.78 MPa < 35 MPa. Hence design is safe.
 
 Flange thickness = 16.5 mm. Use t=18 mm
 
 shearing of the flange at the junction with hub = 2.30 MPa < 15 MPa.
  Values are acceptable.

exa 7.3 Pg 204

In [1]:
from __future__ import division
from math import sqrt,pi
# Given Data
n=8## no. of spline
d=52## mm
D=60## mm
pm=6## MPa
mu=0.06## coefficient of friction
N=320## rpm
P=20## kW

T=60*10**3*P/2/pi/N## N.m
l=8*T*10**3/pm/n/(D**2-d**2)## mm
print ' length of hub = %.f mm'%(l)
Rm=(D+d)/4## mm
F=T*10**3/Rm## N
Ff=mu*F##N (Force of friction)
print ' \n Force required to shift the connection = %.1f N'%(Ff)

 length of hub = 111 mm
 
 Force required to shift the connection = 1278.9 N

exa 7.4 Pg 204

In [4]:
from __future__ import division
from math import pi
# Given Data
d=75## mm
tau=50## MPa
sigma_c=75## MPa
print ' for key to be equally strong in shear & crushing - \n'
b=d/4## mm
print '  b= %.2f mm. Use b=20 mm.'%(b)
b=20##mm
#2*b/t=sigma_c/tau for key to be equally strong in shear & crushing
t=2*b/(sigma_c/tau)## mm
print ' \n t=%.2f mm. Use t=27 mm'%(t)
l= pi*d**2/8/b## mm (for key to be equally strong in shear as shaft)
print ' for key to be equally strong in shear as shaft - \n'
print '  l=%.1f mm. Use l=115 mm'%(l)

 for key to be equally strong in shear & crushing - 

  b= 18.75 mm. Use b=20 mm.
 
 t=26.67 mm. Use t=27 mm
 for key to be equally strong in shear as shaft - 

  l=110.4 mm. Use l=115 mm

exa 7.6 Pg 205

In [7]:
from __future__ import division
from math import pi,ceil,sqrt
from numpy import roots
# Given Data
P=135## kW
N=120## rpm
tau_s=55## MPa
tau_b=45## MPa
tau_ci=175## MPa
sigma_ci=75## MPa

#sigma_cs=2*tau_s## MPa

#Tmax=1.5*Tm
mu=0.15## coefficient of friction

#SHAFT DIAMETER
# P= 2*pi*N*Tm/60/1000 
Tm=P/(2*pi*N/60/1000)## N.m
# pi*d**3*tau_s/16= Tm
d=(Tm/(pi*tau_s/16)*1000)**(1/3)## mm
d=ceil(d)
print ' shaft diameter = %.2f mm.'%(d)
Tmax=Tm## N.m

# HUB DIAMETER
# Tmax=pi/16*((d1**4-d**4)/d1)*tau_h
tau_h=tau_ci## MPa
#d1*(Tmax/(pi/16)/tau_h)-d1**4=d**4 -- eqn(1)
Tmax=Tmax*1000## N.mm
p=[1, 0, 0, -Tmax/(pi*tau_h/16), -d**4] ## polynomial coefficients from eqn(1)
d1=roots(p)## roots of poly 
d1=d1[0]## mm (taking +ve value)
d1=2*d## mm (empirically adopted) 
t1=(d1-d)/2## mm (thickness of hub)
print ' \n thickness of hub = %.f mm'%(t1)
d4=d+t1## mm (diameter of recess in flanges)
print ' \n diameter of recess in flanges = %.f mm'%(d4)
d3=4*d## mm (outside diameter of protecting flange)
print ' \n outside diameter of protecting flange = %.f mm.'%(d3)

#Key size & Hub length
b=d/4## mm (width of key)
l=1.5*d## mm (length of key)
print ' \n width of key = %.1f mm.'%(b)
print ' \n length of key = %.f mm.'%(l) 
t=b## mm (thickness for square key)
print ' \n thickness for square key = %.f mm'%(t)
print ' \n Hub length = %.f mm'%(l)

#Number of bolts
n=ceil(4*d/150+3)## no. of bolts
print ' \n Number of bolts = %.2f.'%(n)

# Bolt diameter
r2=1.5*d## mm
F=Tm*1000/r2/n##N
#(pi/4)*db**2*tau_b=F
db=sqrt(F/((pi/4)*tau_b))## mm
print ' \n Bolt diameter = %.2f mm. Use db=20 mm for design purpose'%(db)
db=20## mm (adopted for design)
bolt_dia=db##mm

# Flange thickness
t2=0.5*t1+6## mm (empirically)
print ' \n Flange thickness = %.1f mm. Use t=20 mm'%(t2)
#F=n*db*t2*sigma_c
sigma_ci=F/n/db/t2## MPa
#2*pi*d1**2*tau*t2/4=Tmax
tau=Tmax/(2*pi*d1**2*t2/4)## MPa
print ' \n permissible bearing stress in flange = %.2f MPa < 75 MPa'%(sigma_ci)
print ' \n shearing of the flange at the junction with hub = %.2f MPa < 175 MPa.'%(tau)
print '  Values are acceptable.'

# Check for crushing of bolt
#n*db*t2*sigma_cb*d2/2=Tmax
d2=d1+d## mm
db=bolt_dia##mm
sigma_cb=Tmax/(n*db*t2*d2/2)## MPa
print ' \n permissible crushing strength of bolts = %.2f MPa < 60 MPa.'%(sigma_cb)
print '  Hence design is safe.'
# Thickness of protecting flange
t3=0.5*t2## mm
print ' \n Thickness of protecting flange = %.f mm'%( t3)
# Hub overlap 
ho=3## mm (min)
print ' \n Hub overlap = %.f mm (min)'%(ho)

 shaft diameter = 100.00 mm.
 
 thickness of hub = 50 mm
 
 diameter of recess in flanges = 150 mm
 
 outside diameter of protecting flange = 400 mm.
 
 width of key = 25.0 mm.
 
 length of key = 150 mm.
 
 thickness for square key = 25 mm
 
 Hub length = 150 mm
 
 Number of bolts = 6.00.
 
 Bolt diameter = 18.38 mm. Use db=20 mm for design purpose
 
 Flange thickness = 31.0 mm. Use t=20 mm
 
 permissible bearing stress in flange = 3.21 MPa < 75 MPa
 
 shearing of the flange at the junction with hub = 5.52 MPa < 175 MPa.
  Values are acceptable.
 
 permissible crushing strength of bolts = 19.25 MPa < 60 MPa.
  Hence design is safe.
 
 Thickness of protecting flange = 16 mm
 
 Hub overlap = 3 mm (min)

exa 7.7 Pg 208

In [9]:
from __future__ import division
from math import sqrt,pi
# Given Data
d=50## mm
tau=42## MPa
sigma_c=72## MPa
print ' for key to be equally strong in shear & crushing - \n'
b=d/4## mm
print '  b= %.2f mm. Use b=15 mm.'%(b)
b=15##mm
#2*b/t=sigma_c/tau for key to be equally strong in shear & crushing
t=2*b/(sigma_c/tau)## mm
print ' \n t=%.2f mm. Use t=20 mm'%(t)
l= pi*d**2/8/b## mm (for key to be equally strong in shear as shaft)
print ' \n for key to be equally strong in shear as shaft - \n'
print '  l=%.2f mm. Use l=70 mm'%(l)

 for key to be equally strong in shear & crushing - 

  b= 12.50 mm. Use b=15 mm.
 
 t=17.50 mm. Use t=20 mm
 
 for key to be equally strong in shear as shaft - 

  l=65.45 mm. Use l=70 mm

exa 7.8 Pg 208

In [11]:
from __future__ import division
from math import sqrt,pi,ceil

# Given Data
d=25## mm
N=550## rpm
P=12## kW
sigma_yt=400## N/mm.sq.
sigma_yc=400## N/mm.sq.
n=2.5## factor of safety

# P= 2*pi*N*T/(60*10**3)
T=P/(2*pi*N/(60*10**3))## N.m
tau=0.5*sigma_yt## MPa
tau_d=tau/n## N/mm.sq.
print ' design shear stress = %.f N/mm.sq.'%(tau_d)
sigma_d=sigma_yc/n## N/mm.sq.
print ' \n design crushing strength = %.f N/mm.sq.'%(sigma_d)
b=d/4##mm
print ' \n width of key = %.f mm. Use 7mm'%(b)
b=ceil(d/4)## mm
t=b## mm
print ' \n thickness of key = %.f mm.'%(t)
l_s=2*T*10**3/(d*b*tau_d)## mm (length of key based on shear failure)
print ' \n length of key based on shear failure = %.2f mm or %.f mm'%(l_s, l_s)
l_c=4*T*10**3/(d*t*sigma_d)## mm (length of key based on crushing failure)
print ' \n length of key based on crushing failure = %.2f mm or %.f mm'%(l_c, l_c)

 design shear stress = 80 N/mm.sq.
 
 design crushing strength = 160 N/mm.sq.
 
 width of key = 6 mm. Use 7mm
 
 thickness of key = 7 mm.
 
 length of key based on shear failure = 29.76 mm or 30 mm
 
 length of key based on crushing failure = 29.76 mm or 30 mm

exa 7.9 Pg 209

In [13]:
from __future__ import division
from math import sqrt,pi,ceil
# Given Data
d=36## mm
P=15## kW
N=720## rpm
#Tmax=1.25*Tm
sigma_yt=245## MPa (for C20 steel)
n=3## factor of safety
sigma=82## MPa (Design tensile stress)

tau=0.577*sigma## MPa (shear stress)
sigma_u=200## MPa (for FG 200 cast Iron)
n2=5## factor of safety (for FG 200 cast Iron)
tau2=20## MPa shear stress (for FG 200 cast Iron)

# Max. torque transmitted
#P=2*pi*N*Tm/(60*10**3)
Tm=P/(2*pi*N/(60*10**3))*1000## N.mm
Tmax=1.25*Tm## N.mm
print ' \n Maximum transmitted torque = %.f N.mm'%(Tmax)

# Hub diameter
tau_h=20## MPa (permissible shear stress in hub)
#Tmax=(pi/16)*(d1**4-d**4)/d1*tau_h   ...eqn(1)
d1=2*d##mm (empirically)
tau_h=Tmax*1000/((pi/16)*(d1**4-d**4)/d1)## MPa
t1=(d1-d)/2## mm (thickness of hub)
print ' \n Hub diameter = %.f mm'%(d1)
print ' \n Thickness of hub = %.f mm'%(t1)
d4=d+t1## mm
print ' \n Diameter of recess in flanges = %.f mm'%(d4)
d3=4*d##mm
print ' \n Outside diameter of protecting flange = %.f mm'%(d3)

#Hub length
b=d/4## mm (width of key)
l=1.5*d## mm (length of key)
print ' \n width of key = %.1f mm.'%(b)
print ' \n length of key = %.f mm.'%(l) 
t=b## mm (thickness for square key)
print ' \n thickness for square key = %.f mm'%(t)
print ' \n Hub length = %.f mm'%(l)

#Number of bolts
n=ceil(4*d/150+3)## no. of bolts
print ' \n Number of bolts = %.2f.'%(n)

# Bolt diameter
r2=1.5*d## mm
F=Tmax/r2/n##N
#(pi/4)*db**2*tau_b=F
db=sqrt(F/((pi/4)*tau))## mm
print ' \n Bolt diameter = %.2f mm. Use db=6 mm for design purpose'%(db)
db=6## mm (adopted for design)
bolt_dia=db##mm

# Flange thickness
t2=0.5*t1+6## mm (empirically)
print ' \n Flange thickness = %.1f mm. Use t=20 mm'%(t2)
#F=n*db*t2*sigma_c
sigma_ci=F/n/db/t2## MPa
#2*pi*d1**2*tau*t2/4=Tmax
tau=Tmax/(2*pi*d1**2*t2/4)## MPa
print ' \n permissible bearing stress in flange = %.2f MPa < 40 MPa'%(sigma_ci)
print ' \n shearing of the flange at the junction with hub = %.2f MPa < 20 MPa.'%(tau)
print '  Values are acceptable.'

# Check for crushing of bolt
#n*db*t2*sigma_cb*d2/2=Tmax
d2=d1+d## mm
db=bolt_dia##mm
sigma_cb=Tmax/(n*db*t2*d2/2)## MPa
print ' \n permissible crushing strength of bolts = %.2f MPa < 82 MPa.'%(sigma_cb)
print '  Hence design is safe.'
# Thickness of protecting flange
t3=0.5*t2## mm
print ' \n Thickness of protecting flange = %.f mm'%( t3)
# Hub overlap 
ho=3## mm (min)
print ' \n Hub overlap = %.f mm (min)'%(ho)

 
 Maximum transmitted torque = 248680 N.mm
 
 Hub diameter = 72 mm
 
 Thickness of hub = 18 mm
 
 Diameter of recess in flanges = 54 mm
 
 Outside diameter of protecting flange = 144 mm
 
 width of key = 9.0 mm.
 
 length of key = 54 mm.
 
 thickness for square key = 9 mm
 
 Hub length = 54 mm
 
 Number of bolts = 4.00.
 
 Bolt diameter = 5.57 mm. Use db=6 mm for design purpose
 
 Flange thickness = 15.0 mm. Use t=20 mm
 
 permissible bearing stress in flange = 3.20 MPa < 40 MPa
 
 shearing of the flange at the junction with hub = 2.04 MPa < 20 MPa.
  Values are acceptable.
 
 permissible crushing strength of bolts = 12.79 MPa < 82 MPa.
  Hence design is safe.
 
 Thickness of protecting flange = 8 mm
 
 Hub overlap = 3 mm (min)

exa 7.10 Pg 212

In [14]:
from __future__ import division
from math import sqrt,pi,ceil
# Given Data
d=35## mm
d2=125## mm
n=6## factor of safety
T=800## N.m
N=350## rpm
tau_s=63## MPa
tau_b=56## MPa
tau_CI=10## MPa
tau_k=46## MPa

# Diameter of bolts:
F=2*T*10**3/d2/n## N
#pi/4*db**2*tau_b=F
db=sqrt(F/(pi/4*tau_b))## mm
print ' \n (i) Diameter of bolts = %.2f mm. Use 8 mm.'%(db)

# Flange thickness
d1=2*d## mm
#T=pi/2*d1**2*t2*tau_CI
t2=T*1000/(pi/2*d1**2*tau_CI)## mm
print ' \n (ii) Flange thickness = %.1f mm. Use t2 = 12 mm'%(t2)
t2=12## mm

#Key dimensions
b=10## mm (width of key)
t=7## mm (from tables)
#T=l*b*tau_k*d/2
l=T*10**3/(b*tau_k*d/2)## mm
l=ceil(l)## mm
print ' \n (iii) Length of key = %.f mm\n\t\td=%.f mm\n\t\tb=%.f mm'%(l,d,b)

# Hub length
lh=l## mm (length of hub)
print ' \n (iv) Hub length = %.f mm'%(l)
tau_c=T*10**3/(pi/16*(d1**4-d**4)/d1)## N/mm.sq.
print ' \n shear stress in hub = %.2f N/mm.sq.'%(tau_c)
print ' It is nearly equal to %.f N/mm.sq.'%(tau_CI)
print ' \n hence design parameters are fine.'

# Power transmitted
P=2*pi*N*T/60/10**3## kW
print ' \n (v) Power transmitted = %.2f kW'%(P)

 
 (i) Diameter of bolts = 6.96 mm. Use 8 mm.
 
 (ii) Flange thickness = 10.4 mm. Use t2 = 12 mm
 
 (iii) Length of key = 100 mm
		d=35 mm
		b=10 mm
 
 (iv) Hub length = 100 mm
 
 shear stress in hub = 12.67 N/mm.sq.
 It is nearly equal to 10 N/mm.sq.
 
 hence design parameters are fine.
 
 (v) Power transmitted = 29.32 kW