A 2-tooth right-hand worm transmits 1 hp at 1200 rev/min to a 30-tooth worm gear. The gear has a transverse diametral pitch of 6 teeth/in and a face width of 1 in. The worm has a pitch diameter of 2 in and a face width of 21/2 in. The normal pressure angle is 141/2 ◦ . The materials and quality of

Question

A 2-tooth right-hand worm transmits 1 hp at 1200 rev/min to a 30-tooth worm gear. The gear has a transverse diametral pitch of 6 teeth/in and a face width of 1 in. The worm has a pitch diameter of 2 in and a face width of [latex]2 \frac{1}{2}[/latex] in. The normal pressure angle is [latex]14 \frac{1}{2}^{\circ}[/latex] . The materials and quality of work needed are such that curve B of Fig. 13–42 should be used to obtain the coefficient of friction.

(a) Find the axial pitch, the center distance, the lead, and the lead angle.

(b) Figure 13–43 is a drawing of the worm gear oriented with respect to the coordinate system described earlier in this section; the gear is supported by bearings A and B. Find the forces exerted by the bearings against the worm-gear shaft, and the output torque.

Accepted Answer

(a) The axial pitch is the same as the transverse circular pitch of the gear, which is

[latex]p_{x}=p_{t}=\frac{\pi}{P}=\frac{\pi}{6}=0.5236 ext { in }[/latex]

The pitch diameter of the gear is [latex]d_{G}=N_{G} / P=30 / 6=5[/latex] in. Therefore, the center distance is

[latex]C=\frac{d_{W}+d_{G}}{2}=\frac{2+5}{2}=3.5 in[/latex]

From Eq. (13–27), the lead is

[latex]L=p_{x} N_{W}[/latex] (13–27)

[latex]L=p_{x} N_{W}=(0.5236)(2)=1.0472 ext { in }[/latex]

Also using Eq. (13–28), find

[latex] an \lambda=\frac{L}{\pi d_{W}}[/latex] (13–28)

[latex]\lambda= an ^{-1} \frac{L}{\pi d_{W}}= an ^{-1} \frac{1.0472}{\pi(2)}=9.46^{\circ}[/latex]

(b) Using the right-hand rule for the rotation of the worm, you will see that your thumb points in the positive z direction. Now use the bolt-and-nut analogy (the worm is righthanded, as is the screw thread of a bolt), and turn the bolt clockwise with the right hand while preventing nut rotation with the left. The nut will move axially along the bolt toward your right hand. Therefore the surface of the gear (Fig. 13–43) in contact with the worm will move in the negative z direction. Thus, the gear rotates clockwise about x, with your right thumb pointing in the negative x direction.

The pitch-line velocity of the worm is

[latex]V_{W}=\frac{\pi d_{W} n_{W}}{12}=\frac{\pi(2)(1200)}{12}=628 ft / min[/latex]

The speed of the gear is [latex]n_{G}=\left(\frac{2}{30} ight)(1200)=80 rev / min[/latex]. Therefore the pitch-line velocity of the gear is

[latex]V_{G}=\frac{\pi d_{G} n_{G}}{12}=\frac{\pi(5)(80)}{12}=105 ft / min[/latex]

Then, from Eq. (13–47), the sliding velocity [latex]V_{S}[/latex] is found to be

[latex]V_{S}=\frac{V_{W}}{\cos \lambda}[/latex] (13–47)

[latex]V_{S}=\frac{V_{W}}{\cos \lambda}=\frac{628}{\cos 9.46^{\circ}}=637 ft / min[/latex]

Getting to the forces now, we begin with the horsepower formula

[latex]W_{W t}=\frac{33000 H}{V_{W}}=\frac{(33000)(1)}{628}=52.5 lbf[/latex]

This force acts in the negative x direction, the same as in Fig. 13–40. Using Fig. 13–42, we find f = 0.03. Then, the first equation of group (13–42) and (13–43) gives

[latex]\begin{aligned}W &=\frac{W^{x}}{\cos \phi_{n} \sin \lambda+f \cos \lambda} \&=\frac{52.5}{\cos 14.5^{\circ} \sin 9.46^{\circ}+0.03 \cos 9.46^{\circ}}=278 lbf\end{aligned}[/latex]

Also, from Eq. (13–43),

[latex]\begin{array}{l}W^{x}=W\left(\cos \phi_{n} \sin \lambda+f \cos \lambda ight) \W^{y}=W \sin \phi_{n} \W^{z}=W\left(\cos \phi_{n} \cos \lambda-f \sin \lambda ight)\end{array}[/latex] (13 43)

[latex]\begin{aligned}W^{y} &=W \sin \phi_{n}=278 \sin 14.5^{\circ}=69.6 lbf \W^{z} &=W\left(\cos \phi_{n} \cos \lambda-f \sin \lambda ight) \&=278\left(\cos 14.5^{\circ} \cos 9.46^{\circ}-0.03 \sin 9.46^{\circ} ight)=264 lbf\end{aligned}[/latex]

At this point a three-dimensional line drawing should be made in order to simplify the work to follow. An isometric sketch, such as the one of Fig. 13–44, is easy to make and will help you to avoid errors.

We shall make B a thrust bearing in order to place the gearshaft in compression.

Thus, summing forces in the x direction gives

[latex]F_{B}^{x}=-52.5 lbf[/latex]

Taking moments about the z axis, we have

[latex]-(52.5)(2.5)-(69.6)(1.5)+4 F_{B}^{y}=0 \quad F_{B}^{y}=58.9 lbf[/latex]

Taking moments about the y axis,

[latex](264)(1.5)-4 F_{B}^{z}=0 \quad F_{B}^{z}=99 ext { lbf }[/latex]

These three components are now inserted on the sketch as shown at B in Fig. 13–44. Summing forces in the y direction,

[latex]-69.6+58.9+F_{A}^{y}=0 \quad F_{A}^{y}=10.7 lbf[/latex]

Similarly, summing forces in the z direction,

[latex]-264+99+F_{A}^{z}=0 \quad F_{A}^{z}=165 lbf[/latex]

These two components can now be placed at A on the sketch. We still have one more equation to write. Summing moments about x,

[latex]-(264)(2.5)+T=0 \quad T=660 lbf \cdot ext { in }[/latex]

It is because of the frictional loss that this output torque is less than the product of the gear ratio and the input torque.

Question 13.10: A 2-tooth right-hand worm transmits 1 hp at 1200 rev/min to ...

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