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 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