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 \psi=\frac{ an \phi_{n}}{ an \phi_{t}}[/latex] (13–19)

[latex]\phi_{t}= an ^{-1} \frac{ an \phi_{n}}{\cos \psi}= an ^{-1}\frac{ an 20^{\circ}}{\cos 30^{\circ}}=22.8^{\circ}[/latex]

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

[latex]V=\frac{\pi d n}{12}=\frac{\pi(1.732)(1800)}{12}=816 ft / min[/latex]

The transmitted load is

[latex]W_{t}=\frac{33000 H}{V}=\frac{(33000)(1)}{816}=40.4 lbf[/latex]

From Eq. (13–40) we find

[latex]W=\frac{W_{t}}{\cos \phi_{n} \cos \psi}[/latex] (13–40)

[latex]W_{r}=W_{t} an \phi_{t}=(40.4) an 22.8^{\circ}=17.0 lbf[/latex]

[latex]W_{a}=W_{t} an \psi=(40.4) an 30^{\circ}=23.3 lbf[/latex]

[latex]W=\frac{W_{t}}{\cos \phi_{n} \cos \psi}=\frac{40.4}{\cos 20^{\circ} \cos 30^{\circ}}=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_{A}^{x}=W_{a}=23.3[/latex] lbf. Taking moments about the z axis,

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

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

[latex]10 F_{B}^{z}-(40.4)(13)=0[/latex]

or [latex]F_{B}^{z}=52[/latex] lbf. Summing forces in the z direction and solving gives [latex]F_{A}^{Z}=12.1[/latex] lbf. Also, the torque is [latex]T=W_{t} d_{p} / 2=(40.4)(1.732 / 2)=35[/latex] lbf · in.

For comparison, solve the problem again using vectors. The force at C is

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

Position vectors to B and C from origin A are

[latex]R _{B}=10 i \quad R _{C}=13 i +0.866 j[/latex]

Taking moments about A, we have

[latex]R _{B} imes F _{B}+ T + R _{C} imes W = 0[/latex]

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

[latex]10 i imes\left(F_{B}^{y} j -F_{B}^{z} k ight)-T i +(13 i +0.866 j ) imes(-23.3 i -17.0 j +40.4 k )= 0[/latex]

When the cross products are formed, we get

[latex]\left(10 F_{B}^{y} k +10 F_{B}^{z} j ight)-T i +(35 i -525 j -201 k )= 0[/latex]

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

Next,

[latex]F _{A}=- F _{B}- W , ext { and so } F _{A}=23.3 i -3.1 j +12.1 k lbf[/latex]

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

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