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_2m[/latex] = 20(2.5) = 50 mm

[latex]d_3 = N_3m[/latex] = 50(2.5) = 125 mm

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

[latex]W_t=\frac{60 000H}{\pi d_2n}=\frac{60 000(2.5)}{\pi (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}[/latex] tan 20° = (0.546) tan 20° = 0.199 kN

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}[/latex] = 0.546 kN [latex]F^r_{43}[/latex] = 0.199 kN [latex]F_{43}[/latex] = 0.581 kN

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...

Related Answered Questions