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

\text { Given } kW =5 \quad n=1440 rom \quad \mu=0.1 .

\alpha=20^{\circ} \quad z_{1}=3 \quad z_{2}=60 \text { teeth } \quad q=10 .

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

d_{1}=q m=10(6)=60 mm .

\tan \gamma=\frac{z_{1}}{q}=\frac{3}{10}=0.3 \quad \text { or } \quad \gamma=16.7^{\circ} .

M_{t}=\frac{60 \times 10^{6}( kW )}{2 \pi n_{1}}=\frac{60 \times 10^{6}(5)}{2 \pi(1440)} .

=33157.28 N – mm .

From Eq. (20.29),

\left(P_{1}\right)_{t}=\frac{2 M_{t}}{d_{1}}                   (20.29).

\left(P_{1}\right)_{t}=\frac{2 M_{t}}{d_{1}}=\frac{2(33157.28)}{60}=1105.24 N           (a).

From Eqs (20.30),

\left(P_{1}\right)_{a}=\left(P_{1}\right)_{t} \times \frac{(\cos \alpha \cos \gamma-\mu \sin \gamma)}{(\cos \alpha \sin \gamma+\mu \cos \gamma)}             (20.30).

\left(P_{1}\right)_{a}=\left(P_{1}\right)_{t} \times \frac{(\cos \alpha \cos \gamma-\mu \sin \gamma)}{(\cos \alpha \sin \gamma+\mu \cos \gamma)} .

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

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

\left(P_{1}\right)_{r}=\left(P_{1}\right)_{t} \times \frac{\sin \alpha}{(\cos \alpha \sin \gamma+\mu \cos \gamma)}               (20.31),

\left(P_{1}\right)_{r}=\left(P_{1}\right)_{t} \times \frac{\sin \alpha}{(\cos \alpha \sin \gamma+\mu \cos \gamma)} .

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

= 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):

\left(P_{2}\right)_{t}=\left(P_{1}\right)_{a} .              (20.22)

\left(P_{2}\right)_{a}=\left(P_{1}\right)_{t} .              (20.23)

\left(P_{2}\right)_{r}=\left(P_{1}\right)_{r} .              (20.24)

\left(P_{2}\right)_{t}=\left(P_{1}\right)_{a}=2632.55 N .

\left(P_{2}\right)_{a}=\left(P_{1}\right)_{t}=1105.24 N .

\left(P_{2}\right)_{r}=\left(P_{1}\right)_{r}=1033.35 N .