Question 20.5: A pair of worm and worm wheel is designated as, 1/30/10/10 T...

\text { Given } \quad n_{1}=1200 rpm \quad z_{1}=1 \quad z_{2}=30 \text { teeth } .

q = 10 m = 10 mm
Step I Permissible torque on worm wheel

i=\frac{z_{2}}{z_{1}}=\frac{30}{1}=30 .

n_{1}=1200 rpm \quad n_{2}=\frac{1200}{i}=\frac{1200}{30}=40 rpm .

d_{2}=m z_{2}=10(30)=300 mm .

\tan \gamma=\frac{z_{1}}{q}=\frac{1}{10}=0.1 \quad \text { or } \quad \gamma=5.71^{\circ} .

From Eq. (20.20),

F=2 m \sqrt{(q+1)}                 (20.20),

F=2 m \sqrt{(q+1)}=2(10) \sqrt{(10+1)} .

= 66.33 mm.

From Eqs (20.13) and (20.14),

c=0.2 m \cos \gamma                   (20.13).

d_{a 1}=m(q+2)                   (20.14).

c=0.2 m \cos \gamma=0.2(10) \cos (5.71)=1.99 mm .

d_{g 1}=m(q+2)=10(10+2)=120 mm .

From Eq. (20.21),

l_{r}=\left(d_{a 1}+2 c\right) \sin ^{-1}\left[\frac{F}{\left(d_{a 1}+2 c\right)}\right]                 (20.21).

l_{r}=\left(d_{a 1}+2 c\right) \sin ^{-1}\left[\frac{F}{\left(d_{a 1}+2 c\right)}\right] .

=(120+2 \times 1.99) \sin ^{-1}\left[\frac{66.33}{(120+2 \times 1.99)}\right] .

= 69.988 mm
For case-hardened carbon steel 14C6 (Table 20.2),

Table 20.2 Values of bending stress factor S_{b}

S_{b 1}=28.2 .

For centrifugally cast phosphor-bronze,

S_{b 2}=7.0 .

From Fig. 20.14,

X_{b 1}=0.25 \text { for } n_{1}=1200 rpm .

X_{b 2}=0.48 \text { for } n_{2}=40 rpm .

From Eqs (20.35) and (20.36),

\left(M_{t}\right)_{1}=17.65 X_{b 1} S_{b 1} m 1_{r} d_{2} \cos \gamma                   (20.35).

\left(M_{t}\right)_{2}=17.65 X_{b 2} S _{b 2} m 1_{r} d_{2} \cos \gamma                 (20.36).

=17.65(0.25)(28.2)(10)(69.988)(300) \cos (5.71) .

= 25 996 711 N-mm                               (a).

=17.65(0.48)(7.0)(10)(69.998)(300) \cos (5.71) .

= 12 389 922 N-mm                   (b).

The lower value of the torque on the worm wheel is 12 389 922 N-mm.
Step II Power transmitting capacity based on beam strength

kW =\frac{2 \pi n_{2}\left(M_{t}\right)}{60 \times 10^{6}}=\frac{2 \pi(40)(12389922)}{60 \times 10^{6}}=51.9 .