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 [latex] \left(\sigma_{y p}=276 MPa \right) [/latex]. Using maximum shear stress criterion against yielding, calculate the required shaft diameter. Assume factor of safety used in the design = 2.0.

The shaft is clearly subjected to bending moment, M = 250P N mm and torque, T = 200P N mm. Thus, [latex] \sigma_{x x}=\frac{32 M}{\pi d^3} \quad \text { and } \quad \tau_{x y}=\frac{16 T}{\pi d^3} [/latex] Maximum shear stress, [latex] \tau_{\max } [/latex] is [latex] \tau_{\max }=\sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup^2+\tau_{x y}^2} [/latex] [latex] =\left\lgroup \frac{16}{\pi d^3} \right\rgroup \sqrt{M^2+T^2} [/latex] For initiation of yielding [latex] \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 })} [/latex] [latex] \Rightarrow \left\lgroup\frac{16}{\pi d^3} \right\rgroup (30) \times 10^{-3} \sqrt{250^2+200^2}=\frac{276}{(2)(2)} [/latex] [latex] \Rightarrow d=89.166 mm \simeq 89.2 mm [/latex] Hence, the required shaft diameter is 89.2 mm.

Question 11.4: A load P = 30.0 kN is applied to the crank pin of the crank ...

Strength of Materials

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.

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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.