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## Q. 12.9

Figure 12.19 shows a power-transmitting shaft. Each pulley has 750 mm in diameter and ratio of tight-side tension to slack-side tension is 3:1. The shaft is transmitting 37.5 kW power at 300 rpm. Calculate the shaft diameter required, if $\tau_{\text {working }}=42$ MPa. Power is input to the pulley C horizontally and is taken off from the pulley D vertically. ## Verified Solution

Power transmitted is

$\text { Power }=\frac{2 \pi T N}{60}=37.5 \times 10^3$

Therefore,

$\text { Torque }=T=\frac{(60)(37.5)\left(10^3\right)}{(2 \pi)(300)} N m =1193.66 Nm$

$\text { But } T=\left(P_1-P_2\right) r \text {. Thus, }$

$P_1-P_2=\frac{T}{r}=\frac{1193.66}{0.375}$

or              $P_1-P_2=3183.1 N$                (1)

Also,          $\frac{P_1}{P_2}=3$                (2)

Solving Eqs. (1) and (2), we get $P_1=4774.65 N \text { and } P_2=1591.55 N$. Now, shear stress induced on the shaft due to T is

$\frac{16 T}{\pi d^3}=\frac{(16)(1193.66)\left(10^3\right)}{\pi d^3}=\frac{19.099\left(10^6\right)}{\pi d^3} MPa$

where d is in mm.
Let us now take care of the bending loads $P_1+P_2=6366.2$ N applied horizontally at C and vertically at D causing bending in both planes. Let us now draw the bending moment diagrams sequentially to address these bendings as shown in Figure 12.20.

From the above set of figures, it is obvious that resultant bending moment is maximum at point C, where

$M_{ C }=\sqrt{1.194^2+0.4775^2}\left(10^6\right) N mm =1.286\left(10^6\right) N mm$

Normal stress due to $M_{ C }$ is

$\frac{32 M_{ C }}{\pi d^3}=\frac{41.152\left(10^6\right)}{\pi d^3} MPa$

and maximum shear stress due to the combined loading of bending and torsion is

$\tau_{\max }=\sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup +\tau_{x y}^2}=\sqrt{20.576^2+19.099^2} \frac{10^6}{\pi d^3}$

Now,            $\tau_{\max }=\frac{28.07\left(10^6\right)}{\pi d^3} MPa =42.0$

So,        d = 59.699 mm ≅ 60 mm

The required shaft diameter is 60 mm. 