from __future__ import division
from math import pi,tan,atan,sqrt
# Given Data
d=26## mm
p=5## mm
W=10## kN
Do=50## mm
Di=20## mm
mu=0.2## coefficient of thread friction
mu_c=0.15## coefficient of collar friction
N=15## rpm
pb=6## MPa
dm=d-p/2## mm
dc=d-p## mm
t=p/2##mm
l=2*p## mm
alfa=atan(l/(pi*dm))*180/pi## degree
fi=atan(mu)*180/pi## degree
Tf=W*dm/2*tan(pi/180*(alfa+fi))## N.mm
Tc=mu_c*W/4*(Do+Di)## N.mm
T=Tf+Tc## N.mm
print ' \n (a) Stress in the screw'
sigma_c=4*W*10**3/(pi*dc**2)## N/mm.sq.
print ' \n Direct compressive stress = %.2f N/mm.sq'%(sigma_c)
tau=16*T*10**3/(pi*dc**3)##N/mm.sq.
print ' \n Tortional shear stress = %.2f N/mm.sq'%(tau)
tau_max=sqrt(sigma_c**2/4+tau**2)##MPa
print ' \n Maximum shear stress = %.2f N/mm.sq'%(tau_max)
n=W*10**3/(pi*dm*t*pb)#
print ' \n\n (b) number of threads of nut in engagement = %.f'%(n)
from __future__ import division
from math import atan,tan,pi,sqrt,cos
# Given Data
d=50## mm
p=8## mm
W=2## kN
Do=100## mm
Di=50## mm
mu=0.15## coefficient of thread friction
mu_c=0.10## coefficient of collar friction
N=25## rpm
two_beta=29## degree
dm=d-p/2## mm
dc=d-p## mm
t=p/2##mm
l=2*p## mm
alfa=atan(p/(pi*dm))*180/pi## degree
mu_e=mu/cos(pi/180*two_beta/2)## virtual coefficient of friction
fi=atan(mu_e)*180/pi## degree
Tf=W*dm/2*tan(pi/180*(alfa+fi))## N.mm
Tc=mu_c*W/4*(Do+Di)## N.mm
T=Tf+Tc## N.mm
P=2*pi*N*T/(60*10**3)## kW
print ' \n (a) Power required = %.3f kN'%(P)
To=W*dm/2*tan(pi/180*alfa)## N.mm
eta=To/T*100## % (efficiency)
print ' \n (b) Efficiency of screw = %.2f %%'%(eta)
from __future__ import division
from math import sqrt,pi,atan,tan,cos,ceil
# Given Data
d=10## mm
p=3## mm
mu=0.15## coefficient of thread friction
mu_c=0.20## coefficient of collar friction
dc=15## mm
F=60## N
W=4## kN
two_beta=30## degree
h=25## mm
lf=150## mm (screw free length)
dm=d-p/2## mm
alfa=atan(p/(pi*dm))*180/pi## degree
mu_e=mu/cos(pi/2*two_beta/2)## virtual coefficient of friction
fi=atan(mu_e)*180/pi## degree
Tf=W*10**3*dm/2*tan(pi/180*(alfa+fi))## N.mm
Tc=mu_c*W*10**3/2*dc## N.mm
T=Tf+Tc## N.mm
#F*l=T
l=T/F## mm (Length of handle)
print ' \n (a) Length of handle = %.1f mm'%(l)
print ' \n\n (b) Maximum shear stress in screw'
print ' \n Section 1-1 : '
dc=d-p##mm
tau=16*T/(pi*dc**3)## N/mm.sq.
M=F*lf## N.mm
sigma_b=32*M/(pi*dc**3)## N/mm.sq.
tau_max=sqrt((sigma_b/2)**2+tau**2)## MPa
print ' \n Maximum shear stress = %.2f MPa'%(tau_max)
print ' \n Section 2-2 : '
sigma_c=4*W*10**3/(pi*dc**2)## N/mm.sq. (Direct compressive stress)
tau2=16*Tc/(pi*dc**3)#### N/mm.sq. (Tortional shear stress)
tau_max=sqrt((sigma_c/2)**2+tau2**2)## MPa
print ' \n Maximum shear stress = %.2f MPa'%(tau_max)
#h=n*p## height of nut
n=ceil(h/p)## no. of threads
t=p/2## mm (thickness of threads)
pb=W*10**3/(pi*dm*t*n)## MPa
print ' \n\n (b) Bearing pressure on threads = %.1f MPa'%(pb)
from __future__ import division
from math import sqrt,pi,atan,tan,cos,ceil
# Given Data
W=25## kN
two_beta=29## degree
v=0.96## m/min
mu=0.14## coefficient of thread friction
Di=30## mm
Do=66## mm
mu_c=0.15## coefficient of collar friction
d=36## mm
p=6## mm
Sut=630## MPa
Syt=380## MPa
dm=d-p/2## mm
dc=d-p## mm
l=2*p## mm
alfa=atan(l/(pi*dm))*180/pi## degree
mu_e=mu/cos(pi/180*two_beta/2)## virtual coefficient of friction
fi=atan(mu_e)*180/pi## degree
Tf=W*10**3*dm/2*tan(pi/180*(alfa+fi))## N.mm
Tc=mu_c*W*10**3/4*(Do+Di)## N.mm
T=Tf+Tc## N.mm
N=v*10**3/l## rpm
P=2*pi*N*T/(60*10**3)*10**-3## kW
print ' \n Power required to drive the slide = %.2f kN'%(P)
sigma_c=4*W*10**3/(pi*dc**2)## MPa
tau=16*T/(pi*dc**3)## MPa
sigma1=1/2*(sigma_c+sqrt(sigma_c**2+4*tau**2))## MPa
tau_max=sqrt((sigma_c/2)**2+tau**2)## MPa
n_t=Syt/sigma1## factor of safety in tension
print ' \n factor of safety in tension = %.2f '%(n_t)
n_s=Syt/2/tau_max## factor of safety in shear
print ' \n factor of safety in shear = %.2f '%(n_s)
# Note- Answer in the textbook are not accurate.
from __future__ import division
from math import sqrt,pi,atan,tan,cos,ceil
# Given Data
d=12## mm
dc=10## mm
p=2## mm
Do=10##mm
mu=0.15## coefficient of thread friction
mu_c=0.18## coefficient of collar friction
F=100## N
l=150## mm
dm=dc+p/2## mm
alfa=atan(p/(pi*dm))*180/pi## degree
fi=atan(mu)*180/pi## degree
TfByW=dm/2*tan(pi/180*(alfa+fi))## where TfByW = Tf/W
TcByW=mu_c/3*Do## where TcByW = Tc/W
TByW=TfByW+TcByW## N.mm (total torque at B-B)
Tapplied=F*l## N.mm (torque applied by the operator)
#putting T= Tapplied
W= Tapplied/TByW## N
print ' \n (a) Clamping force between the jaws = %.f N'%(W)
eta=W*dm/2*tan(pi/180*alfa)/Tapplied*100## %
print ' \n (b) Efficiency of vice = %.2f %%'%(eta)
Tf=TfByW*W## N.mm
print ' \n (c) Torque at A-A, Tf = %.1f N.mm & Torque at B-B = %.f N.mm'%(Tf,Tapplied)
# Note- Answer in the textbook are not accurate.
from __future__ import division
from math import sqrt,pi,atan,tan,cos,ceil
# Given Data
W=100## kN
lift=400## mm
sigma_ts=100## MPa
sigma_cs=100## MPa
tau_s=60## MPa
tau_tn=50## MPa
sigma_cn=45## MPa
tau_n=40## MPa
pb=15## MPa
mu=0.2## coefficient of thread friction
mu_c=0.15## coefficient of collar friction
#sigma_cs=4*W/(pi*dc**2)
dc=sqrt(4*W*10**3/(pi*sigma_cs))## mm
print ' \n Screw Diameter-\n Core diameter of screw, dc=%.2f mm. Use dc=40 mm'%(dc)
dc=40## mm
p=7## mm (for normal series square threads)
d=dc+p##mm
print ' \n outside diameter = %.f mm'%(d)
dm=dc+p/2## mm
print ' \n mean diameter = %.1f mm'%(dm)
t=p/2## mm
print ' \n thread thickness = %.1f mm'%(t)
print ' \n Maximum tensile & shear stress in screw -'
sigma_c=4*W*1000/pi/dc**2## MPa
alfa=atan(p/(pi*dm))*180/pi## degree
fi=atan(mu)*180/pi## degree
Tf=dm*W*10**3/2*tan(pi/180*(alfa+fi))## where TfByW = Tf/W
tau=16*Tf/(pi*dc**3)## MPa
sigma12=(1/2)*(sigma_c+sqrt(sigma_c**2+4*tau**2))## MPa
print ' \n Maximum tensile stress = %.f MPa < %.f MPA. Hence design is safe.'%(sigma12,sigma_ts)
tau_max=sqrt((sigma_c/2)**2+tau**2)## MPa
print ' \n Maximum shear stress = %.2f MPa < %.f MPA. Hence design is safe.'%(tau_max,tau_s)
print ' \n Height of nut-'
n=W*10**3/(pi/4)/pb/(d**2-dc**2)## no. of threads
n= ceil(n)## no. of threads (rounding)
h=n*p## mm
print ' \n h=%.f mm'%(h)
print ' \n Check for stress in screw and nut'
tau_screw=W*10**3/(pi*n*dc*t)## MPa
print ' \n shear stress in screw = %.2f MPa < %.f MPa'%(tau_screw,tau_s)
tau_nut=W*10**3/(pi*n*d*t)## MPa
print ' \n shear stress in nut = %.2f MPa < %.f MPa'%(tau_nut,tau_n)
print ' \n These are within permissible limits. Hence design is safe.'
print ' \n Nut collar size-'
# pi/4*(D1**2-d**2)*sigma_tn=W
D1=sqrt(W*10**3/(pi/4)/tau_tn+d**2)## mm
print ' \n Inside diameter of collar = %.2f mm. Use D1=70 mm'%(D1)
D1=70##mm (adopted for design)
# pi/4*(D2**2-D1**2)*sigma_cn=W
D2=sqrt(W*10**3/(pi/4)/sigma_cn+D1**2)## mm
print ' \n Outside diameter of collar = %.2f mm. Use D2=90 mm'%(D2)
D2=90##mm (adopted for design)
# pi*D1*tc*tau_n=W
tc=W*10**3/(pi*D1*tau_n)## mm
print ' \n thickness of nut = %.2f mm. Use tc=12 mm.'%(tc)
tc=12## mm (adopted for design)
print ' \n Head Dimensions-'
D3=1.75*d## mm
print ' \n Diameter of head on top of screw = %.2f mm. use D3=84 mm.'%(D3)
D3=84## mm (adopted for design)
D4=D3/4## mm
print ' \n pin diameter in the cup = %.f mm'%(D4)
print ' \n Torque required between cup and head-'
Tc=mu_c*W*10**3/3*((D3**3-D4**3)/(D3**2-D4**2))## N.mm
print ' \n Tc=%.f N.mm (acc. to uniform pressure theory)'%(Tc)
T=Tf+Tc## N.mm
print ' \n Total Torque, T=%.f N.mm'%(T)
F=300## N (as a normal person can apply 100-300 N)
l=T/F##mm
print ' \n length of lever = %.f mm. Use 3300 mm'%(l)
M=F*l## N.mm
dl=(32*M/pi/sigma12)**(1/3)## mm
print ' \n Diameter of lever, dl=%.1f mm. Use dl=48 mm.'%(dl)
dl=48## mm (adopted for design)
H=2*dl## mm
print ' \n Height of head, H=%.f mm'%(H)
print ' \n Check for screw in buckling-'
L=lift+0.5*h## mm
K=dc/4## mm
C=0.25## spring index
sigma_y=200## MPa
Ac=pi/4*dc**2##mm.sq.
Wcr=Ac*sigma_y*(1-(sigma_y/4/C/pi**2/(200*10**3))*(L/K)**2)/1000## kN
print ' \n Buckling or critical load for screw, Wcr = %.f kN > 100kN'%(Wcr)
To=W*10**3*dm/2*tan(pi/180*alfa)## N.mm
eta=To/T*100## %
print ' \n Efficiency of screw = %.1f %%'%(eta)
print ' \n Body dimensions-'
D5=1.5*D2## mm
t2=2*tc## mm
t3=0.25*d##mm
D6=2.25*D2## mm
print ' \n Diameter of body at top, D5 = %.f mm'%( D5)
print ' \n Thickness of base, t2 = %.f mm'%( t2)
print ' \n Thickness of body, t3 = %.f mm'%( t3)
print ' \n Inside diameter of bottom, D6 = %.1f mm. Use D6=205 mm.'%( D6)
D6=205## mm (adopted for design)
D7=1.75*D6## mm
hb=lift+h+100## mm
print ' \n Outside diameter at the bottom, D7 = %.2f mm. Use 360 mm.'%( D7)
print ' \n Height of body = %.f mm. Use 600mm'%(hb)
from __future__ import division
from math import sqrt,pi,atan,tan,cos,ceil
# Given Data
two_beta=30## degree
W=400*10**3## N
d=100## mm
p=12## mm
mu=0.15## coefficient of thread friction
dm=d-p/2## mm
dc=d-p## mm
l=2*p## mm
alfa=atan(l/pi/dm)*180/pi## degree
mu_e=mu/cos(pi/180*two_beta/2)## virtual coefficient of friction
fi=atan(mu)*180/pi## degree
Tf=W*dm/2*tan(pi/180*(alfa+fi))## N.mm (Frictional torque for raising load)
T=W*dm/4*tan(pi/180*fi)## N.mm
To=W*dm/2*tan(pi/180*alfa)## N.mm (Torque without friction)
eta1=To/Tf*100## %
print ' \n Efficiency during raising the load = %.2f %%'%(eta1)
eta2=T/To*100## %
print ' \n Efficiency during lowering the load = %.2f %%'%(eta2)
# Note - answer & solution is wrong in the textbook.
from __future__ import division
from math import sqrt,pi,atan,tan,cos,ceil
# Given Data
d=70## mm
mu=0.13## coefficient of thread friction
mu_c=0.15## coefficient of collar friction
Do=90## mm
Di=26## mm
L=450## mm
# C-25 steel screw
sigma_t1=275## MPa
sigma_c1=275## MPa
tau1=137.5## MPa
# Phosphor bronze nut
sigma_t2=100## MPa
sigma_c2=90## MPa
tau2=80## MPa
pb=15##MPa
n=2## factor of safety
#screw
sigma_ts=137.5## MPa
sigma_cs=137.5## MPa
tau_s=68.75## MPa
#Nut
sigma_tn=50## MPa
sigma_cn=45## MPa
tau_n=40## MPa
p=10## mm (for normal series square threads)
dc=d-p##mm
dm=d-p/2##mm
t=p/2##mm
alfa=atan(p/pi/dm)*180/pi## degree
fi=atan(mu)*180/pi## degree
K=dc/4## mm
C=0.25## spring index
sigma_y=275## MPa
Ac=pi/4*dc**2##mm.sq.
Wcr=Ac*sigma_y*(1-(sigma_y/4/C/pi**2/(200*10**3))*(L/K)**2)## N
print ' \n (a) Safe Capacity of press or critical load for the screw = %.f N'%(Wcr)
n=Wcr/(pi*dm*t*pb)## no. of threads
n=ceil(n)## rounding
h=n*p## mm
print ' \n (b) Height of nut, h=%.f mm'%(h)
W=Wcr## N
Tf=W*dm/2*tan(pi/180*(alfa+fi))/1000## N.mm (Frictional torque)
Tc=mu_c*W/4*(Do+Di)/1000## N.mm (Collar torque)
T=Tf+Tc## N.mm
print ' \n (c) Necessary torsional moment or total torque = %.2f N.mm'%(T)
# Note - answer in the textbook is wrong.
from __future__ import division
from math import pi,tan,atan,sqrt
# Given Data
d=26## mm
L=0.25##m
F=300## N
mu=0.14## coefficient of thread friction
p=5## mm (for normal series)
dc=d-p## mm
dm=d-p/2## mm
l=2*p## mm
alfa=atan(l/pi/dm)*180/pi## degree
fi=atan(mu)*180/pi## degree
To=F*L## N.m (Torque applied by the operator)
#Tf=W*dm/2*tand(alfa+fi)## N.mm
# And Tf=To
W=To*1000/(dm/2*tan(pi/180*(alfa+fi)))## N
print ' The force required for the job is : %.f N'%(W)
# Note - answer in the textbook is wrong.
from __future__ import division
from math import pi,tan,atan,sqrt
# Given Data
W=50## kN
lift=200## mm
gc=300## mm (ground clearance)
pb=16## MPa
mu=0.14## coefficient of collar friction
#Screw C-35
Sut=288## MPa
n=3## factor of safety for screw
# Nut : phosphor-bronze
sigma_t=100## MPa
sigma_c=90## MPa
tau=80## MPa
n2=3## factor of safety for nut
sigma_ts=Sut/n## MPa
sigma_cs=Sut/n## MPa
tau_s=sigma_ts/2## MPa
# sigma_cs=4*W/(pi*dc**2)
dc= sqrt(4*W*10**3/(pi*sigma_cs))## mm
print ' \n Screw diameter - \n Core diameter, dc = %.2f mm. Use 30 mm'%(dc)
dc=30## mm (adopted for design)
p=6## mm (for normal series square threads)
d=dc+p##mm
print ' \n outside diameter = %.f mm'%(d)
dm=dc+p/2## mm
print ' \n mean diameter = %.1f mm'%(dm)
t=p/2## mm
print ' \n thread thickness = %.1f mm'%(t)
print ' \n Maximum tensile & shear tress in screw -'
sigma_c=4*W*1000/pi/dc**2## MPa
alfa=atan(p/(pi*dm))*180/pi;# degree
fi=atan(mu)*180/pi; # degree
Tf=dm*W*10**3/2*tan(pi/180*(alfa+fi))## where TfByW = Tf/W
tau=16*Tf/(pi*dc**3)## MPa
sigma12=(1/2)*(sigma_c+sqrt(sigma_c**2+4*tau**2))## MPa
print ' \n Maximum tensile stress = %.1f MPa < %.f MPA. Hence design is safe.'%(sigma12,sigma_ts)
tau_max=sqrt((sigma_c/2)**2+tau**2)## MPa
print ' \n Maximum shear stress = %.2f MPa < %.f MPA. Hence design is safe.'%(tau_max,tau_s)
print ' \n Height of nut-'
n=W*10**3/(pi/4)/pb/(d**2-dc**2)## no. of threads
n= round(n)## no. of threads (rounding)
h=n*p## mm
print ' \n h=%.f mm'%(h)
print ' \n Check for stress in screw and nut'
tau_screw=W*10**3/(pi*n*dc*t)## MPa
print ' \n shear stress in screw = %.2f MPa\n'%(tau_screw)
tau_nut=W*10**3/(pi*n*d*t)## MPa
print ' \n shear stress in nut = %.2f MPa'%(tau_nut)
print ' \n These are within permissible limits. Hence design is safe.'
print ' \n Nut collar size-'
# pi/4*(D1**2-d**2)*sigma_tn=W
D1=sqrt(W*10**3/(pi/4)/(50)+d**2)## mm
print ' \n Inside diameter of collar = %.2f mm. Use D1=52 mm'%(D1)
D1=52##mm (adopted for design)
# pi/4*(D2**2-D1**2)*sigma_cn=W
D2=sqrt(W*10**3/(pi/4)/45+D1**2)## mm
print ' \n Outside diameter of collar = %.1f mm. Use D2=65 mm'%(D2)
D2=65##mm (adopted for design)
# pi*D1*tc*tau_cn=W
tau_cn=40## MPa
tc=W*10**3/(pi*D1*tau_cn)## mm
print ' \n thickness of nut = %.2f mm. Use tc=8 mm.'%(tc)
tc=8## mm (adopted for design)
print ' \n Head Dimensions-'
D3=1.75*d## mm
print ' \n Diameter of head on top of screw = %.2f mm. use D3=64 mm.'%(D3)
D3=64## mm (adopted for design)
D4=D3/4## mm
print ' \n pin diameter in the cup = %.f mm'%(D4)
print ' \n Torque required between cup and head-'
Tc=mu*W*10**3/3*((D3**3-D4**3)/(D3**2-D4**2))## N.mm
print ' \n Tc=%.f N.mm (acc. to uniform pressure theory)'%(Tc)
T=Tf+Tc## N.mm
print ' \n Total Torque, T=%.f N.mm'%(T)
F=300## N (as a normal person can apply 100-300 N)
l=T/F##mm
print ' \n length of lever = %.f mm. Use 1075 mm'%(l)
M=F*l## N.mm
dl=(32*M/pi/sigma12)**(1/3)## mm
print ' \n Diameter of lever, dl=%.1f mm.'%(dl)
H=2*dl## mm
print ' \n Height of head, H=%.f mm'%(H)
print ' \n Check for screw in buckling-'
L=lift+0.5*h## mm
K=dc/4## mm
C=0.25## spring index
sigma_y=288## MPa
Ac=pi/4*dc**2##mm.sq.
Wcr=Ac*sigma_y*(1-(sigma_y/4/C/pi**2/(200*10**3))*(L/K)**2)/1000## kN
print ' \n Buckling or critical load for screw, Wcr = %.f kN > 50kN'%(Wcr)
print ' \n Hence design is safe.'
from __future__ import division
from math import pi,tan,atan,sqrt,ceil
# Given Data
d=32## mm
p=5## mm
W=12## kN
D3=50## mm
D4=20## mm
mu=0.15## coefficient of thread friction
mu_c=0.20## coefficient of collar friction
N=24## rpm
pb=6## N/mm.sq.
tau_s=30## MPa
tau_n=30## MPa
dm=d-p/2## mm
dc=d-p## mm
t=p/2## mm
l=2*p##mm
alfa=atan(l/pi/dm)*180/pi## degree
fi=atan(mu)*180/pi## degree
Tf=W*10**3*dm/2*tan(pi/180*(alfa+fi))## N.mm
Tc=mu_c*W*10**3/4*(D3+D4)## N.mm
T=Tf+Tc## N.mm
print ' \n (i) Torque required to rotate the screw = %.f N.mm'%(T)
print ' \n (ii) Stresses induced in screw - '
sigma_c=4*W*10**3/(pi*dc**2)## N/mm.sq.
print ' \n Direct compressive stress = %.2f N/mm.sq'%(sigma_c)
tau=16*T/(pi*dc**3)## N/mm.sq.
print ' \n Tortional shear stress = %.2f N/mm.sq'%(tau)
tau_max=sqrt((sigma_c/2)**2+tau**2)## MPa
print ' \n Maximum shear stress = %.2f MPa < %.f MPa'%(tau_max,tau_s)
print ' \n Hence design is safe.'
n=W*10**3/(pi*dm*t*pb)## no. of threads
n=ceil(n)## rounding
h=n*p##mm
print ' \n (iii) Height of nut = %.f mm'%(h)
from __future__ import division
from math import pi,tan,atan,sqrt,ceil
# Given Data
W=100## kN
lift=260## mm
pb=15## N/mm.sq.
mu=0.15## coefficient of thread friction
mu_c=0.20## coefficient of collar friction
#Screw
Suts=800## N/mm.sq.
sigma_ss=340## N/mm.sq.
ns=4## factor of safety
#Nut
Sutn=552## N/mm.sq.
sigma_sn=260## N/mm.sq.
nn=5## factor of safety
sigma_ts=Suts/ns## N/mm.sq.
sigma_cs=Suts/ns## N/mm.sq.
tau_s=sigma_ss/ns## N/mm.sq.
sigma_tn=Sutn/nn## N/mm.sq.
sigma_cn=Sutn/nn## N/mm.sq.
tau_n=sigma_sn/nn## N/mm.sq.
#sigma_cs=4*W/(pi*dc**2)
dc=sqrt(4*W*10**3/(pi*sigma_cs))## mm
print ' \n Screw Diameter-\n Core diameter of screw, dc=%.2f mm. Use dc=33 mm'%(dc)
dc=33## mm
p=7## mm (for normal series square threads)
d=dc+p##mm
print ' \n outside diameter = %.f mm'%(d)
dm=dc+p/2## mm
print ' \n mean diameter = %.1f mm'%(dm)
t=p/2## mm
print ' \n thread thickness = %.1f mm'%(t)
print ' \n Maximum stresses in screw -'
sigma_c=4*W*1000/pi/dc**2## MPa
alfa=atan(p/(pi*dm))*180/pi## degree
fi=atan(mu)*180/pi## degree
Tf=dm*W*10**3/2*tan(pi/180*(alfa+fi))## where TfByW = Tf/W
tau=16*Tf/(pi*dc**3)## MPa
sigma12=(1/2)*(sigma_c+sqrt(sigma_c**2+4*tau**2))## MPa
print ' \n Maximum tensile stress = %.1f N/mm.sq. < %.f N/mm.sq.. Hence design is safe.'%(sigma12,sigma_ts)
tau_max=sqrt((sigma_c/2)**2+tau**2)## MPa
print ' \n Maximum shear stress = %.2f N/mm.sq. < %.f N/mm.sq.. Hence design is safe.'%(tau_max,tau_s)
print ' \n Height of nut-'
n=W*10**3/(pi/4)/pb/(d**2-dc**2)## no. of threads
n= ceil(n)## no. of threads (rounding)
h=n*p## mm
print ' \n h=%.f mm. Use 120 mm.'%(h)
h=120## mm
print ' \n Check for stress in screw and nut'
tau_screw=W*10**3/(pi*n*dc*t)## MPa
print ' \n shear stress in screw = %.2f MPa < %.f MPa'%(tau_screw,tau_s)
tau_nut=W*10**3/(pi*n*d*t)## MPa
print ' \n shear stress in nut = %.2f MPa < %.f MPa'%(tau_nut,tau_n)
print ' \n These are within permissible limits. Hence design is safe.'
print ' \n Nut collar size-'
# pi/4*(D1**2-d**2)*sigma_tn=W
D1=sqrt(W*10**3/(pi/4)/sigma_tn+d**2)## mm
print ' \n Inside diameter of collar = %.2f mm. Use D1=55 mm'%(D1)
D1=55##mm (adopted for design)
# pi/4*(D2**2-D1**2)*sigma_cn=W
D2=sqrt(W*10**3/(pi/4)/sigma_cn+D1**2)## mm
print ' \n Outside diameter of collar = %.2f mm. Use D2=70 mm'%(D2)
D2=70##mm (adopted for design)
# pi*D1*tc*tau_n=W
tc=W*10**3/(pi*D1*tau_n)## mm
print ' \n thickness of nut = %.f mm. Use tc=15 mm.'%(tc)
tc=15## mm (adopted for design)
print ' \n Head Dimensions-'
D3=1.75*d## mm
print ' \n Diameter of head on top of screw = %.2f mm.'%(D3)
D4=D3/4## mm
print ' \n pin diameter in the cup = %.1f mm. Use 20 mm.'%(D4)
D4=20## mm (adopted for design)
print ' \n Torque required between cup and head-'
Tc=mu_c*W*10**3/3*((D3**3-D4**3)/(D3**2-D4**2))## N.mm
print ' \n Tc=%.f N.mm (acc. to uniform pressure theory)'%(Tc)
T=Tf+Tc## N.mm
print ' \n Total Torque, T=%.f N.mm'%(T)
F=300## N (as a normal person can apply 100-300 N)
l=T/F##mm
print ' \n length of lever = %.f mm or %.2f m'%(l,l/1000)
M=F*l## N.mm
sigma_b=100## N/mm.sq. (assumed)
dl=(32*M/pi/sigma_b)**(1/3)## mm
print ' \n Diameter of lever, dl=%.1f mm. Use dl=45 mm.'%(dl)
dl=45## mm (adopted for design)
H=2*dl## mm
print ' \n Height of head, H=%.f mm'%(H)
print ' \n Check for screw in buckling-'
L=lift+0.5*h## mm
K=dc/4## mm
C=0.25## spring index
sigma_y=200## MPa
Ac=pi/4*dc**2##mm.sq.
Wcr=Ac*sigma_y*(1-(sigma_y/4/C/pi**2/(200*10**3))*(L/K)**2)/1000## kN
print ' \n Buckling or critical load for screw, Wcr = %.f kN > 100kN'%(Wcr)
To=W*10**3*dm/2*tan(pi/180*alfa)## N.mm
eta=To/T*100## %
print ' \n Efficiency of screw = %.2f %%'%(eta)
print ' \n Body dimensions-'
D5=1.5*D2## mm
t2=2*tc## mm
t3=0.25*d##mm
D6=2.25*D2## mm
print ' \n Diameter of body at top, D5 = %.f mm'%( D5)
print ' \n Thickness of base, t2 = %.f mm'%( t2)
print ' \n Thickness of body, t3 = %.f mm'%( t3)
print ' \n Inside diameter of bottom, D6 = %.1f mm. Use D6=160 mm.'%( D6)
D6=160## mm (adopted for design)
D7=1.75*D6## mm
hb=lift+h+100## mm
print ' \n Outside diameter at the bottom, D7 = %.2f mm.'%( D7)
print ' \n Height of body = %.f mm.'%(hb)