In Fig. 13–38 a 1-hp electric motor runs at 1800 rev/min in the clockwise direction, as viewed from the positive x axis. Keyed to the motor shaft is an 18-tooth helical pinion having a normal pressure angle of 20◦, a helix angle of 30◦, and a normal diametral pitch of 12 teeth/in. The hand of the

Question

In Fig. 13–38 a 1-hp electric motor runs at 1800 rev/min in the clockwise direction, as viewed from the positive x axis. Keyed to the motor shaft is an 18-tooth helical pinion having a normal pressure angle of 20°, a helix angle of 30°, and a normal diametral pitch of 12 teeth/in. The hand of the helix is shown in the figure. Make a three-dimensional sketch of the motor shaft and pinion, and show the forces acting on the pinion and the bearing reactions at A and B. The thrust should be taken out at A.

Accepted Answer

From Eq. (13–19) we find

[latex]\cos ψ =\frac{ an \phi _{n}}{ an \phi _{t}}[/latex] (13–19)

[latex]\phi _{t} = an^{−1} \frac{ an \phi_{ n}}{\cos ψ} = an^{−1} \frac{ an 20°}{\cos 30°} =22.8°[/latex]

Also, [latex]P_{t} = P_{n} \cos ψ = 12 \cos 30° = 10.39[/latex] teeth/in. Therefore the pitch diameter of the pinion is [latex]d_{p}[/latex] = 18/10.39 = 1.732 in. The pitch-line velocity is

[latex]V =\frac{πdn}{12} =\frac{π(1.732)(1800)}{12} = 816 ft/min[/latex]

The transmitted load is

[latex]W_{t} =\frac{33 000H}{V} =\frac{(33 000)(1)}{816} = 40.4 lbf[/latex]

From Eq. (13–40) we find

[latex]W_{r} = W_{t} tan \phi_{t}[/latex]
[latex]W_{a} = W_{t} an ψ[/latex]
[latex]W =\frac{W_{t}}{\cos \phi_{n} \cos ψ}[/latex] (13–40)

[latex]W_{r} = W_{t} an \phi_{t} = (40.4) an 22.8° = 17.0 lbf[/latex]
[latex]W_{a} = W_{t} an ψ = (40.4) an 30° = 23.3 lbf[/latex]
[latex]W =\frac{W_{t}}{\cos \phi_{n} \cos ψ }=\frac{40.4}{\cos 20° \cos 30°} = 49.6 lbf[/latex]

These three forces, [latex]W_{r}[/latex] in the −y direction, [latex]W_{a}[/latex] in the −x direction, and [latex]W_{t}[/latex] in the +z direction, are shown acting at point C in Fig. 13–39. We assume bearing reactions at A and B as shown. Then [latex]F^{x}_{A} = W_{a}[/latex] = 23.3 lbf. Taking moments about the z axis,

[latex]−(17.0)(13) + (23.3) \left( \frac{1.732}{2} ight) + 10F^{y}_{B} = 0[/latex]

or [latex]F^{y}_{B}[/latex] = 20.1 lbf. Summing forces in the y direction then gives [latex]F^{y}_{A}[/latex] = 3.1 lbf. Taking moments about the y axis, next

[latex]10F^{z}_{B} − (40.4)(13) = 0[/latex]

or [latex]F^{z}_{B}[/latex] = 52.5 lbf. Summing forces in the z direction and solving gives [latex]F^{z}_{A}[/latex] = 12.1 lbf.
Also, the torque is [latex]T = W_{t}d_{p}/2[/latex] = (40.4)(1.732/2) = 35 lbf · in.
For comparison, solve the problem again using vectors. The force at C is

[latex]W = −23.3i − 17.0j + 40.4k lbf[/latex]

Position vectors to B and C from origin A are

[latex]R_{B} = 10 i R_{C} = 13i + 0.866j[/latex]

Taking moments about A, we have

[latex]R_{B} × F_{B} + T + R_{C} ×W = 0[/latex]

Using the directions assumed in Fig. 13–39 and substituting values gives

[latex]10 i × (F^{y}_{B}j − F^{z}_{B}k)− T i + (13i + 0.866j) × (−23.3i − 17.0j + 40.4k) = 0[/latex]

When the cross products are formed, we get

[latex](10F^{y}_{B}k + 10F^{z}_{B}j)− T i + (35i − 525j − 201k) = 0[/latex]

whence T = 35 lbf · in, [latex]F^{y}_{B}[/latex] = 20.1 lbf, and [latex]F^{z}_{B}[/latex] = 52.5 lbf.

Next,

[latex]F_{A} = −F_{B} −W, and so F_{A} = 23.3i − 3.1j + 12.1k lbf.[/latex]

Question 13.9: In Fig. 13–38 a 1-hp electric motor runs at 1800 rev/min in ...

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