Homework SoursePlease solve as much as you can 675 cm 65 a plane of Fi and
Solution
solution:
design of shaft for gear
1)as tangential velocity should be 1626 in/s or 41.3 m/s
torque on gear and pully are
P=T*2*pi*N/60
v=pi*d*N/60=41.3
N2=5175.71 rpm
hence torque on gear is
T=2752.7 Nmm
tangential force =Ft=T/dg/2
Ft=36.12 N
radial fotce Fr=Fttan20=13.14 N
resulatnt force=F=38.43 N
where reaction at support are
Ra=15.37 N
Rb=23.058 N
hence moment at gear base is maximum hence
M=F*l
l=1.75 in=44.45 mm
M=780.79 Nmm
for Kb=1.5,Kt=1
by ASME code
equaivalent torque is
Te=((KbM)^2+(KtT)^2)^.5=2991.49 Nmm
here material is 1050cold drawn steel
Sut=790 MPa,Syt=670MPa
tall=.18*Sut=142.5 N/mm2
but for keyway 25% reduction in strength
tall=106.65 N/mm2
hence diametr of shaft is
taa=16Te/pi*d1^3
d1=5.2275 mm approx 6 mm
here for bending strength step shaft is used as per moment distribution
do=1.25*d1=6.53 approx 7 mm
2)design for shaft of pully
as tangential velocity should be 1626 in/s or 41.3 m/s
torque on gear and pully are
P=T*2*pi*N/60
v=pi*d*N/60=41.3
N2=5175.71 rpm
hence torque on gear is
T=2752.7 Nmm
T=(F1-F2)dg/2
F1=5F2
4F2=36.12
F2=9.03 N
F1=45.15 N
F=F1+F2=54.18 N
hence tyorque is
T=2752.7 Nmm
moment is
M=F*l=2408.62 Nmm
l=44.45 mm
for Kb=1.5,Kt=1
by ASME code
equaivalent torque is
Te=((KbM)^2+(KtT)^2)^.5=4542.1 Nmm
here material is 1050cold drawn steel
Sut=790 MPa,Syt=670MPa
tall=.18*Sut=142.5 N/mm2
but for keyway 25% reduction in strength
tall=106.65 N/mm2
hence diametr of shaft is
taa=16Te/pi*d3^3
d3=6 mm approx 6 mm
here for bending strength step shaft is used as per moment distribution
d2=1.25*d3=7.53 approx 8 mm
2)design for square key
w=h=d/4
as both shaft have same diametr 6 mm,same key for both by shear and crushing criteria
material choosing-plain carbon steel
Syt=300 MPa
Nf=3
t=.5*Syt/Nf=50 MPa
Sc=Syt/Nf=100 MPa
where formula are
for calculating length of key is
t=2T/d2wl
and Sc=4T/d2hl
w=h=d2/4=1.5 mm
hence l=12.23 mm
4)design of spur gear
module=5 mm
center distance a<18\'\'<457.2 m
a=mT2(1+G/2)
from speed ratio T1=3T2
G=3
we get
T2=45
T1=15
design for bending strength
Yp=.484-2.87/t1
Yp=.2926
b=1 in=25.4 mm
Sb=200 MPA
SUT=600MPA
Nf=3
Fb=m*b*Sb*Yp=3716.02 N
where effective load
Kv=.1268
Ka=application factor=1.5
Feff=(Ka/kv)Ft=427.11 N
hence factor safety
Nf=Fb/Feff=8.7
where designagainst pitting failure
Fw=bdpQk
Q=2*T2/(T1+T2)=1.5
dp=75 mm
Fw=Fb
K=.16(BHN/100)^2
all putting in equation we get that hardeness of ear should be
BHN=285.09
in this way material of steel with Sut=600 MPa
7)in thisway all parameter are selected for design and bearing selected from diameter of shaft and load condition
