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{60000 H}{\pi d n}[/latex] (13–36)

[latex]W_{t}=\frac{60000 H}{\pi d_{2} n}=\frac{60000(2.5)}{\pi(50)(1750)}=0.546 kN[/latex]

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

[latex]F_{23}^{r}=F_{23}^{t} an 20^{\circ}=(0.546) an 20^{\circ}=0.199 kN[/latex]

and so

[latex]F_{23}=\frac{F_{23}^{t}}{\cos 20^{\circ}}=\frac{0.546}{\cos 20^{\circ}}=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_{43}^{t}=0.546 kN \quad F_{43}^{r}=0.199 kN \quad 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_{b 3}^{x}=-\left(F_{23}^{t}+F_{43}^{r} ight)=-(-0.546+0.199)=0.347 kN[/latex]

[latex]F_{b 3}^{y}=-\left(F_{23}^{r}+F_{43}^{t} ight)=-(0.199-0.546)=0.347 kN[/latex]

The resultant shaft reaction is

[latex]F_{b 3}=\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...

Related Answered Questions