17. Shaft Design
Introduction
Objectives
„
Compute forces acting on shafts from gears, pulleys, and
sprockets.
•
Find bending moments from gears, pulleys, or sprockets that
are transmitting loads to or from other devices.
•
Determine torque in shafts from gears, pulleys, sprockets,
clutches, and couplings.
•
„
Compare combined stresses to suitable allowable stresses,
including any required stress reduction factors such as stress
concentration factors and factors of safety.
•
Determine suitability of shaft design and/or necessary size of
shafting.
•
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1
Diameter
increments (in.)
Upto 3
1/16
3 to 5
1/8
5 to 8
1/4
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„
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2
Torsion of circular shafts
Standard diameters of shafts
Diameter (in.)
„
Shaft must have adequate torsional strength to
transmit torque and not be over stressed.
Shafts are mounted in bearings and transmit power
through devices such as gears, pulleys, cams and
clutches.
Components such as gears are mounted on shafts
using keys.
Shaft must sustain a combination of bending and
torsional loads.
3
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4
5
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6
Torsion of circular shafts
Angle of twist, θ =
„
„
„
„
„
TL
GJ
θ = the angle of twist (radians)
T = the applied torque (in(in-lb.)
L = shaft length (in.)
J = polar moment on inertia of the shaft cross section
(in4)
G = shear modulus of elasticity of the shaft material
(lb/in2)
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1
Torsional Shear Stresses
Shear Stress in a shaft
Torque
„
„
„
„
„
Torsional shear stress, SS =
Torque
16 T
„ Shear stress, SS =
π D3
Where
„ T = torque
„ D = diameter of the shaft =
Tc
J
4
J = Polar moment of inertia = π × d
32
c = radius of the shaft
T = Torque
d = diameter of shaft
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„
„
„
„
„
„
„
Fn = Ft tan θ
9
„
„
Torque, T = Ft r and r = Dp /2
Combining the above we can write
2T
2 P × 63,000
=
Dp
Dp n
10
Loads from Bevel gears
An additional axial force will be acting on the shaft
because of the bevel angle
For the pinion it is relatively small, and can be
neglected.
For the larger gear it will be significant and will be
larger than the radial separating force.
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Tn
or T = 63,000 P
63,000
n
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Loads from Bevel gears
„
Power, P =
Ft =
Ft
cos θ
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8
Forces on spur gear teeth
Ft = Transmitted force
Fn = Normal force or separating
force
Fr = Resultant force
θ = pressure angle
Fr =
16 T
π SS
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Forces on spur gear teeth
„
3
„
„
„
„
„
„
Force transmitted, Fn = Ft tan θ cos γ
θ = Pressure angle
γ = Cone angle
Axial Force, Fa = Ft tan θ sin γ
2
Resultant Force, Fr =
Ft + F2
F = Fn or Fa depending on whichever is larger
11
2
Loads from Worm gears
Loads from Worm gears
To
Driving force on the worm gear, Ft =
rwg
„ To = Output torque
Ft sin φ
„ Separating force, Fs =
cos φ cos λ - f sin λ
where
„ λ = lead angle
„ ϕ = normal pressure angle
„ f = coefficient of friction
„
Axial
Driving
Separating
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Loads from Worm gears
14
Loads from Worm gears
„
Axial force on the worm gear
⎛ cos φ sin λ + f cos λ ⎞
⎟⎟
Fa(gear) = Ft(gear) ⎜⎜
⎝ cos φ cos λ - f sin λ ⎠
where
„ λ = lead angle
„ ϕ = normal pressure angle
„ f = coefficient of friction
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Loads from Belts and Chains
„
„
„
„
„
Bending of circular shafts
For a belt, Total load, Ft = Ff + Fb
Net driving force, Fd = Ff – Fb
Driving torque, T = Fd r
r = effective radius of pulley or sprocket
For a chain Fb = 0
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„
„
„
17
Shafts transmit power through gears and
pulleys
These produce bending load in addition to
torsion
Use strength of material approach to calculate
the reaction forces and bending moments
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18
3
Bending of circular shafts
Bending of circular shafts
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Shaft Design Problems
„
„
„
„
„
„
20
Example Problem 17-1: Design Stresses in Shafts
• Shaft shown drives a gear set that is transmitting 5
hp at 1750 rpm.
Step 1: Calculate the torque on the shaft from power
Step 2: Find the torsional stress in the shaft
Step 3: Calculate the loads coming from gears, belts
or chains
Step 4: Calculate the bending moment due to the
acting forces. If necessary combine the forces.
Step 5: Calculate the bending stress in the shaft
Step 6: Combine the bending stress and the torsional
stress using the theories discussed in chapter 4
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21
Example Problem 17-1: Design Stresses in Shafts (cont’d.)
• Shaft is supported in self-aligning ball bearings
and gears are both 10 pitch, 40 tooth, 20° spur
gears.
• Find torsional and bending stresses in shaft.
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22
Example Problem 17-1: Design Stresses in Shafts (cont’d.)
•
Find the torsional stress in the shaft.
− First find Z':
•
Find the torsion in the shaft:
(Appendix 3)
(2-6)
Z' =
Tn
hp =
63,000
Z' =
then:
T =
63,000 hp
n
π (.75 in)3
16
(3-6)
Ss =
63,000 (5)
1750
Ss =
T = 180 in-lb
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16
Z' = .083 in3
(17-1)
T =
π D3
T
Z'
180 in-lb
.083 in3
Ss = 2170 lb/in2
23
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4
Example Problem 17-1: Design Stresses in Shafts (cont’d.)
Example Problem 17-1: Design Stresses in Shafts (cont’d.)
•
•
Find the resultant force on the shaft:
Find the load at the gear pitch circle:
(12-2)
Fr =
(11-4)
NT
Dp =
Pd
Fr =
40
10
Dp =
Ft
cos θ
90 lb
cos 20°
Fr = 96 lb
Dp = 4 inches
•
Find the maximum moment:
(12-3)
Ft =
Ft =
(Appendix 2)
2T
DP
Mm =
2 (180 in-lb)
4 in
Mm =
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Find the stress:
Z =
Z =
„
M
Z
π D3
26
Combined Stresses in Shafts
Example Problem 17-1: Design Stresses in Shafts (cont’d.)
S =
96 lb (15 in)
4
Mm = 360 in-lb
Ft = 90 lb
•
FL
4
As seen in Chap 4
(Appendix 3)
32
π (.75 in)3
32
Z = .041in3
S =
S =
M
Z
360 in-lb
.041 in3
S = 8780 lb/in2
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Combined maximum shear stress
„
„
„
„
Example Problem 17-2: Combined Stresses in
Shafts
τ = Maximum combined shear stress
1/2
S = normal stress
⎡ 2 ⎛ S ⎞2 ⎤
τ
=
S
+
⎜
⎟
⎢
⎥
S
SS = shear stress
⎝ 2 ⎠ ⎦⎥
⎣⎢
This can be rewritten as
• From previous example problem, find the combined stress using
the maximum shear stress theorem:
(4-5)
⎛
⎝
„
− Substituting stresses from previous example problem:
5.1 2
(T + M 2 ) 1/2
D3
T = Torque in the shaft
M = Maximum moment
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⎛ S ⎞2 ⎞ ½
⎝ ⎠ ⎠
τ = ⎜Ss2 + ⎜ 2 ⎟ ⎟
τ=
„
28
⎛
2
⎛ 8780
τ = ⎜(2170 lb/in2) + ⎜ 2
⎝
⎝
⎞2 ⎞ ½
lb/in2⎟ ⎟
⎠ ⎠
τ = 4900 lb/in
2
− This should be compared to shear stress allowables.
29
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30
5
Example Problem 17-3: Combined Stresses in
Shafts
Maximum Normal Stress Theory
„
„
„
σ=
„
• From Example Problem 17-1, find the combined stress using the
maximum normal stress theory:
σ = equivalent combined normal stress
S = normal stress from bending or axial loads
SS = shear or torsional stress
2
S ⎡ 2 ⎛S⎞ ⎤
± ⎢SS + ⎜ ⎟ ⎥
2 ⎣⎢
⎝ 2 ⎠ ⎦⎥
σ =
1/2
[
σ =
]
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1
2
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Solid Circular shaft
3
2
⎛
⎛ 8780 in 2 ⎞ ⎞⎟
8780 lb / in 2
⎜
⎟
+ ⎜ ( 2170 lb / in 2 ) 2 + ⎜⎜
⎟ ⎟⎟
2
2
⎜
⎝
⎠ ⎠
⎝
σ = 9300 lb / in 2
31
D=
2
– This should be compared to the normal stress allowable.
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D =3
1
– Substituting stresses from Example Problem 17-1:
This can be written as
5.1
σ = 3 M + (T 2 + M 2 )1/2
D
2
⎛
S
⎛S⎞ ⎞
± ⎜ Ss2 + ⎜ ⎟ ⎟
⎜
2
⎝ 2 ⎠ ⎟⎠
⎝
32
Critical speeds of shafts
5.1 2
(T + M 2 ) 1/2
τ
[
5.1
M + (T 2 + M 2 )1/2
σ
]
33
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34
Critical speeds of shafts
„
„
Operating speed should be 20% away from the critical speed.
Vibration frequency, f is given by
1
kg
2π W
f = frequency in cycles per second, Hz
k = force constant, force per inch of deflection
g = acceleration due to gravity, 386.4 in./s2
W = weight in pounds, lb
f=
„
„
„
„
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36
6
„
„
„
Shaft with n concentrated loads
Change the frequency to rpm
Critical speed, Nc = 60 × f
Also k is weight divided by deflection
k=
Nc =
„
W
δ
60
2π
N c = 187.7
Wg
Wδ
N c = 187.7
Rayleigh’
Rayleigh’s equation is used.
W1 δ1 + W2 δ 2 + W3 δ3 + ... + Wn δ n
2
2
37
Example Problem 17-5: Critical Speed
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Example Problem 17-5: Critical Speed (cont’d.)
• Find the estimated critical speed for the shaft in Example Problem 17-1
(assume the entire shaft diameter is ¾ inch).
Nc =
− First, find deflection:
3
(Appendix 2)
δ = – 48 EI
I =
I =
2
1
δ
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FL
2
W1 δ1 + W2 δ2 + W3 δ3 + ... + Wn δ n
π D4
Nc =
(Appendix 3)
64
Nc
π (.75 in)4
188
∂
188
(17-14)
.21
= 410 rpm
64
I = .016 in4
96 lb (15 in)4
δ = – 48 (30 x 106 lb/in2) (.016 in4)
• This is approximate, and additional multiples would exist at 820,
1230, and 1640 rpm.
δ = .21 inch
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7