Pinion 2 in Fig. 13–34a runs at 1750 rev/min and transmits 2.5 kW to idler gear 3. The teeth are cut on the 20° full-depth system and have a module of m = 2.5 mm. Draw a free-body diagram of gear 3 and show all the forces that act upon it.

Question

Accepted Answer

The pitch diameters of gears 2 and 3 are

[latex]d_2 = N_{2}m = 20(2.5) = 50 mm[/latex]
[latex]d_3 = N_{3}m = 50(2.5) = 125 mm[/latex]

From Eq. (13–36) we find the transmitted load to be

[latex]W_{t} =\frac{60 000H}{πdn}[/latex] (13–36)

[latex]W_{t} =\frac{60 000H}{πd_{2}n}=\frac{60 000(2.5)}{π(50)(1750)} = 0.546 kN[/latex]

Thus, the tangential force of gear 2 on gear 3 is [latex]F^{t}_{23}[/latex] = 0.546 kN, as shown in Fig. 13–34b. Therefore

[latex]F^{r}_{23} = F^{t}_{23} tan 20° = (0.546) tan 20° = 0.199 kN[/latex]

and so

[latex]F_{23} =\frac{F^{t}_{23}}{cos 20°} =\frac{0.546}{cos 20°} = 0.581 kN[/latex]

Since gear 3 is an idler, it transmits no power (torque) to its shaft, and so the tangential reaction of gear 4 on gear 3 is also equal to [latex]W_{t}[/latex]. Therefore

[latex]F^{t}_{43} = 0.546 kN F^{r}_{43} = 0.199 kN F_{43} =0.581 kN[/latex]

and the directions are shown in Fig. 13–34b.
The shaft reactions in the x and y directions are

[latex]F^{x}_{b3} = −(F^{t}_{23} + F^{r}_{43})= −(−0.546 + 0.199) = 0.347 kN[/latex]
[latex]F^{y}_{b3} = −(F^{r}_{23} + F^{t}_{43})= −(0.199 − 0.546) = 0.347 kN[/latex]

The resultant shaft reaction is

[latex]F_{b3} =\sqrt{(0.347)^{2} + (0.347)^{2}} = 0.491 kN[/latex]

These are shown on the figure.

Question 13.7: Pinion 2 in Fig. 13–34a runs at 1750 rev/min and transmits 2...

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