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Chapter 12

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.

12.19

Step-by-Step

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.

12.20