Shaft Calculation
In determining shaft and bearing loads for the bevel gear application, the usual
Practice is to use tangential or transmitted load that would occur if all the forces
were concentrated at the midpoint of the tooth.
Determination of Torque from the equation below:
T.S.R (r) = R/V
R=1.5 m
= 5.75 rad/sec
V= 3 m/s
Where,
T.S.R (r) = Tip speed ratio
R= Radius of the rotor
= Angular velocity of rotor
V= Velocity of the fluid
r = (1.5 x 5.75)/3
= 2.875
Flow Angle () = 2/3 tan-1(1/r)
= 2/3 tan-1(1/2.875)
= 12.78
For 12 Angle of attack:
= 12.78+
angle)
= 12.78 + 12
= 24.78
(In analysis point of view is added to the real flow
For the values of the coefficients we have:
Cl = coefficient of lift
Cd = coefficient of drag
Cn = coefficient of normal force
Ct = coefficient of tangential force
From the equation below we know,
Cl = 0.304 , Cd = 0.0468
Cn = Cl cos + Cd sin
= 0.304 cos (24.78) + 0.0468 sin (24.78)
= 0.295
Cn = 0.295
Ct = Cl sin Cd cos
= 0.304sin (24.78) 0.0468 cos (24.78)
= 0.107
Ct = 0.107
Again,
a = Axial induction factor
a=
1
4 F sin 2
+1
Cn
BC
Substituting = ( 2 r
where, B = Number of blades
C = Chord Length
= solidity
We get,
1
4 F sin 24.78
+1
9 0.266
0.298
2 r
2
a=
a'
= Angular Induction Factor
1
4 F sin cos
1
BC
C t
2 r
a'
1
4 F sin 24.78 cos 24.78
1
9 0.266
0.107
2 r
f=
B Rr
2 r sin
where, R = Total radius of the rotor
r = local radius
2r
(
)
= 10.736
r
2
1 1
F = cos e
10.736(
2
cos1 e
2r
)
r
Now integrating with respect to T, we get
dT = 4 1000 r 3 v 0 (1a ) a' Fdr
Substituting the value and integrating we get,
Torque (T) = 4423 Nm
So, the transmitted load is given by
w t=
T
ra
Where,
wt = Transmitted load
T = Torque transmitted
ra = Pitch radius at the mid-point of the gear tooth.
To start with the calculation we have taken some assumptions:
For simplicity considering mitre gear in place of bevel gear
Number of teeth in pinion (Np) = Number of teeth in gear (Ng)
Now,
For
ra = 0.065
Pressure angle () = 20
Below are the calculations:
ra = 0.065 m
(Np) = (Ng)
So (Np) / (Ng) =1
Tan = Np / Ng = 1
= 45
= 20
wt = T/ ra
= 4423/0.065 = 68046.1538 N
wt = 68046.1538 N
Tan = 0.3639
Cos = 0.7071
wr = wt Tan Cos ( where wr = Radial load )
= 68046.15 x 0.3639 x 0.7071
= 17509.37 N
wr = 17509.37 N
wa = wr ( As mitre gear has been taken so Np = Ng i.e. Axial load = Radial load)
wa = 17509.37 N
According to the data above and iterations assumption has been taken as
follows:
L = 1.6 m
L1/0.5 = L
L2/0.5 = L
L = 1.6 m
L1 = 0.8 m
L2 = 0.8 m
E = 193 x 109 Mpa (Youngs Modulus)
Mb = wa ra
= 17509.37 N x 0.065
= 569.049 N.m
Mb = 569.049 N.m
By Von Mises stress Theory Diameter of the Shaft can be
determined as:
The maximum bending:
Mmax =
l 2 l 1 w rl 1 M b
l 1 +l 2
+ Mb
0.8 x 0.8 x 17509.2+0.8 x 569.049
1.6
= 7288.19 Nm
al =
sy
nd
where, al = Allowable stress
sy = Tensile Strength
nd = Design Factor
al =
sy
nd
6
205 10
1.5
= 136.66 x 106 pa
= (
M max k t wb 2
T 2
+
)
+3
(
)
d3
d2
d3
32
4
16
0.5
= Von mises stress
T = Torque
Mmax = maximum Bending moment
kt = Stress concentration factor
M max k t wb 2
T 2
(
+
)
+3
(
)
d3
d2
d3
32
4
10
0.5
= (
7288.19 3.4 8754.6
+
)
d3
d2
32
4
2 0.5
4423
2+3(
)
(
d 3 )
10
For stainless Steel Grade 316
= 205 x 106 pa
Now substituting the value in the above equation and solving for the diameter:
D = 0.12341 m
By Shear Stress Theory Diameter of the Shaft can also be
determined as:
Mmax =
l 2 l 1 w rl 1 M b
l 1 +l 2
+ Mb
0.8 x 0.8 x 17509.2+0.8 x 569.049
1.6
= 7288.19 Nm
sy
2 nd
al =
where, al = Allowable stress
sy = Tensile Strength
nd = Design Factor
al=
sy
2 nd
6
205 10
2 1.5
= 68.333 x 106 pa
max=
M max k t wb 2
T 2
(
+
)
+
4(
)
d3
d2
d 3
32
4
16
7288.19 3.4 8754.6 2
4423 2
+
)
+
3(
)
3
2
3
d
d
d
32
4
10
For stainless Steel Grade 316
max = 205 x 106 pa
Now substituting the value in the above equation and solving for the diameter:
D = 0.1235 m
Allowable deflections at bearing C
From the equation of the slope of bearings we get
64 w r l 2 ( l 22l 2 ) + M b ( 6l 1 l2 l 23 l 12 )
0.003 El
0.25
=0
=>
64 [ 17509.37 0.8 ( 0.8 21.62 ) +569.049 ( 6 0.8 1.621.6 23 0.8 2) ]
D
0.003 193 106 1.6
Substituting the above values in the equation we get:
D = 0.1236 m
Allowable deflections at bearing D
From the equation of the slope of bearings we get
0.25
=0
64 w r l 1 ( l 2l12 ) + M b ( l 23 l 12 )
=>
0.003 El
0.25
=0
64 [ 17509.37 0.8 ( 1.62 0.82 ) +569.049 ( 1.62 3 0.82 ) ]
D
0.003 193 106 1.6
0.25
=0
Substituting the above values in the equation we get:
D = 0.1559 m
Deterministic design Results:
Design Criteria
Shaft Diameter
Distortion-energy theory
0.1234 m
Maximum shear theory
0.1235 m
Allowable deflections at bearing C
0.1236 m
Allowable deflections at bearing D
0.1559 m
