A pair of worm and worm wheel is designated as
3/60/10/6
The worm is transmitting 5 kW power at 1440 rpm to the worm wheel. The coefficient of friction is 0.1 and the normal pressure angle is 20°. Determine the components of the gear tooth force acting on the worm and the worm wheel.

Question

Accepted Answer

[latex] ext { Given } kW =5 \quad n=1440 rom \quad \mu=0.1 [/latex].

[latex] \alpha=20^{\circ} \quad z_{1}=3 \quad z_{2}=60 ext { teeth } \quad q=10 [/latex].

m = 6 mm
Step I Components of tooth force acting on worm

[latex] d_{1}=q m=10(6)=60 mm [/latex].

[latex] an \gamma=\frac{z_{1}}{q}=\frac{3}{10}=0.3 \quad ext { or } \quad \gamma=16.7^{\circ} [/latex].

[latex] M_{t}=\frac{60 imes 10^{6}( kW )}{2 \pi n_{1}}=\frac{60 imes 10^{6}(5)}{2 \pi(1440)} [/latex].

[latex] =33157.28 N - mm [/latex].

From Eq. (20.29),

[latex] \left(P_{1} ight)_{t}=\frac{2 M_{t}}{d_{1}} [/latex] (20.29).

[latex] \left(P_{1} ight)_{t}=\frac{2 M_{t}}{d_{1}}=\frac{2(33157.28)}{60}=1105.24 N [/latex] (a).

From Eqs (20.30),

[latex] \left(P_{1} ight)_{a}=\left(P_{1} ight)_{t} imes \frac{(\cos \alpha \cos \gamma-\mu \sin \gamma)}{(\cos \alpha \sin \gamma+\mu \cos \gamma)} [/latex] (20.30).

[latex] \left(P_{1} ight)_{a}=\left(P_{1} ight)_{t} imes \frac{(\cos \alpha \cos \gamma-\mu \sin \gamma)}{(\cos \alpha \sin \gamma+\mu \cos \gamma)} [/latex].

[latex] =1105.24 imes \frac{[\cos (20) \sin (16.7)-0.1 \sin (16.7)]}{[\cos (20) \sin (16.7)+0.1 \cos (16.7)]} [/latex].

= 2632.55 N (b)
From Eq. (20.31),

[latex] \left(P_{1} ight)_{r}=\left(P_{1} ight)_{t} imes \frac{\sin \alpha}{(\cos \alpha \sin \gamma+\mu \cos \gamma)} [/latex] (20.31),

[latex] \left(P_{1} ight)_{r}=\left(P_{1} ight)_{t} imes \frac{\sin \alpha}{(\cos \alpha \sin \gamma+\mu \cos \gamma)} [/latex].

[latex] =1105.24 imes \frac{\sin (20)}{[\cos (20) \sin (16.7)+0.1 \cos (16.7)]}[/latex].

= 1033.35 N (c)
Step II Components of tooth force acting on worm wheel
The force components acting on the worm wheel are as follows (Eqs. 20.22 to 20.24):

[latex] \left(P_{2} ight)_{t}=\left(P_{1} ight)_{a} [/latex]. (20.22)

[latex] \left(P_{2} ight)_{a}=\left(P_{1} ight)_{t} [/latex]. (20.23)

[latex] \left(P_{2} ight)_{r}=\left(P_{1} ight)_{r} [/latex]. (20.24)

[latex] \left(P_{2} ight)_{t}=\left(P_{1} ight)_{a}=2632.55 N [/latex].

[latex] \left(P_{2} ight)_{a}=\left(P_{1} ight)_{t}=1105.24 N [/latex].

[latex] \left(P_{2} ight)_{r}=\left(P_{1} ight)_{r}=1033.35 N [/latex].

Question 20.2: A pair of worm and worm wheel is designated as 3/60/10/6 The...

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