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Chapter 11

Q. 11.4

A load P = 30.0 kN is applied to the crank pin of the crank shaft in Figure 11.22 to rotate the shaft at constant speed. The shaft is made of a ductile steel \left(\sigma_{y p}=276  MPa \right) . Using maximum shear stress criterion against yielding, calculate the required shaft diameter. Assume factor of safety used in the design = 2.0.

11.22

Step-by-Step

Verified Solution

The shaft is clearly subjected to bending moment, M = 250P N mm and torque, T = 200P N mm.
Thus,

\sigma_{x x}=\frac{32 M}{\pi d^3} \quad \text { and } \quad \tau_{x y}=\frac{16 T}{\pi d^3}

Maximum shear stress, \tau_{\max } is

\tau_{\max }=\sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup^2+\tau_{x y}^2}

=\left\lgroup \frac{16}{\pi d^3} \right\rgroup \sqrt{M^2+T^2}

For initiation of yielding

\left\lgroup \frac{16}{\pi d^3}\right\rgroup \sqrt{M^2+T^2}=\frac{\tau_{ yp }}{\text { factor of safety }}=\frac{\sigma_{ yp }}{2(\text { factor of safety })}

\Rightarrow \left\lgroup\frac{16}{\pi d^3} \right\rgroup (30) \times 10^{-3} \sqrt{250^2+200^2}=\frac{276}{(2)(2)}

\Rightarrow d=89.166  mm \simeq 89.2  mm

Hence, the required shaft diameter is 89.2 mm.