11-7 Force Analysis of Spur Gears
Gear Free Body Diagrams
3
b
w3
pressure angle f
3
pitch circle
Tb3
line of action
b
F32
f
w2
F23
a
Fa2
2
• Resolve applied force into radial and tangential directions
t
F32
r
F32
f
r
Fa2
Fb3
Fa2
a
Ta2
t
Fa2
t
Transmitted Load Wt = F32
• Constant speed situation
d
Wt
2
T = Ta2
T =
• If we let the pitch line velocity be V = ω d2 ,
1
F32
2
a
Ta2
2
Power = Force × Velocity
• In SI units:
Wt =
(60)(1000)H
πωd
where
- Wt = tangential force in N or kN
- H = power in W or kW
- 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:
H=
Tω
Wt V
=
33000 63000
where
- Wt = tangential force in lbf
- H = horsepower in HP
- V = tangential speed in ft/min =
- d = diameter in in
2
πdω
12
- ω = rotational speed in rpm
• We also have
63000H
ω
• The radial force acting on the gear is given by
T =
Wr = Wt tan φ
• The total force acting on the gear is
Wt
W =
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
c
51T
3
34T
b
a
2
3
T 17T
Pitch diameters
• gear 2: d2 =
• gear 3: d3 =
• gear 4: d4 =
Wt
t
F32
r
F32
F23
=
=
=
=
t
F32
r
F32
F32
• Gear 3 is an idler so it transmits no power to its shaft
t
F43
r
F43
F43
y
Fb3
b
Fb3
Fb3
F23
t
F23
4
x
r
F23
t
F43
=
Hence
r
F43
=
for equilibrium
t
t
Fb3x − F23
− F43
=0
r
r
Fb3y + F23
− F43
=0
⇒
⇒
Fb3y = 0 kN
F3b =
in the x direction only
Tc4
y
Fc4
c
Fc4
Fc4
x
r
F34
w4
F34
t
F34
Tc4
Tc4
t
F34
d4
=0
−
2
=
5