Question 20.6: Assume the data of Example 20.5 for a pair of worm gears. De...

\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
For the given pair of worm gears,

d_{2}=300 mm .

\text { For }(q=10) \text { and }\left(z_{1}=1\right), \text { the zone factor } Y_{z} \text { from } .

Table 20.4 is given by

Table 20.4 Values of the zone factor Y_{z}

Y_{z}=1.143 .

For case-hardened carbon steel 14C6 (Table 20.3),

Table 20.3 Values of the Surface Stress Factor S_{c}

S_{c 1}=4.93 .

For centrifugally cast phosphor-bronze,

S_{c 2}=1.55 .

From Eq. (20.33),

V_{s}=\frac{\pi d_{1} n_{1}}{60000 \cos \gamma}             (20.33).

V_{s}=\frac{\pi d_{1} n_{1}}{60000 \cos \gamma}=\frac{\pi(10 \times 10)(1200)}{60000 \cos (5.71)}=6.315 m / s .

(Fig. 20.15),

X_{c 1}=0.112 .

\text { For } V_{s}=6.315 m / s \quad \text { and } \quad n_{1}=40 rpm .

X_{c 2}=0.26 .

From Eqs (20.38) and (20.39),

\left(M_{t}\right)_{3}=18.64 X_{c 1} S_{c 1} Y _{ z }\left(d_{2}\right)^{1.8} m                 (20.38).

\left(M_{t}\right)_{4}=18.64 X_{c 2} S_{c 2} Y _{ z }\left(d_{2}\right)^{1.8} m                   (20.39).

=18.64(0.112)(4.93)(1.143)(300)^{1.8}(10) .

= 3 383 570.4 N-mm             (a)

= 2 469 535.8 N-mm                (b)
The lower value of torque on worm wheel is 2 469 535.8 N-mm.
Step II Power transmitting capacity based on wear strength

kW =\frac{2 \pi n_{2}\left(M_{t}\right)}{60 \times 10^{6}}=\frac{2 \pi(40)(2469535.8)}{60 \times 10^{6}}=10.34 .