Force Analysis Of Spur Gears 2a2v5y

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