11-7 Force Analysis of Spur Gears Gear Free Body Diagrams ...

11-7 Force Analysis of Spur Gears
Gear Free Body Diagrams
pressure angle �
� 2
� 3
3 pitch circle
3
b
line of action
a a
2 2
• Resolve applied force into radial and tangential directions
r
Fa2 Transmitted Load Wt = F t 32
• Constant speed situation
�
F a2
t
F32 a
F 32
t
Fa2 T = d
2 Wt
T = Ta2
T a2
• If we let the pitch line velocity be V = ω d
2 ,
1
2
r
F32 �
F 23
F a2
T b3
b
T a2
F b3
F 32

• In SI units:
where
Power = Force × Velocity
Wt = (60)(1000)H
πωd
- Wt = tangential force in N or kN
- H =powerinWorkW
- ω = rotational speed in rpm
- d = diameter in mm
• Note that the formula above takes into account the unit conversions
• In MathCAD, you can use
Wt = 2H
ωd
in whatever units you select.
• MathCAD will take care of units, as long as you specify units for all
variables.
• In US units:
where
H = WtV
33000
= Tω
63000
- Wt = tangential force in lbf
- H = horsepower in HP
- V = tangential speed in ft/min = πdω
12
- d = diameter in in
2

- ω = rotational speed in rpm
• We also have
T = 63000H
ω
• The radial force acting on the gear is given by
Wr = Wt tan φ
• The total force acting on the gear is
W = Wt
cos φ
Example
Shaft a has a power input of 75 kW at a speed of 1000 rpm counterclockwise.
Gears have a module of 5 mm and a 20◦ pressure angle. Gear
3 is an idler (to change direction of the output).
• Find the force F3b that gear 3 exerts on shaft b
• Find the torque T4c that gear 4 exerts on shaft c
4
3
2
3
c
b
a T 17T
34T
51T

Pitch diameters
• gear 2: d2 =
• gear 3: d3 =
• gear 4: d4 =
Wt =
F t 32 =
F r 32 =
F23 =
• Gear 3 is an idler so it transmits no power to its shaft
t
F43 t
F23 F 43
4
r
F32 F 23
y
Fb3 b
r
F23 r
F43 t
F32 F 32
F b3
x
Fb3

Hence
for equilibrium
in the x direction only
F t 43 =
F r 43 =
F x b3 − F t 23 − F t 43 =0 ⇒
F y
b3 + F r 23 − F r 43 =0 ⇒ F y
b3 =0kN
T c4
� 4
F3b =
y
Fc4 c
r
F34 F c4
x
Fc4 F 34
t
F34 Tc4 − F t 34d4
2 =0
Tc4 =
5