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

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