Question 11.6: The shaft shown in Figure 11.23 is supported in flexible bea...
The shaft shown in Figure 11.23 is supported in flexible bearings A and B. Two spur gears are attached at point B and C of the shaft as shown in the figure. Assume that the gears subjected to tangential forces as shown in the side view the shaft. Also, assume that the shaft is made of ductile steel \left(\sigma_{y p}=290 MPa \right) and factor of safety used in the design against the yielding using maximum shearing stress criterion is 1.85. Calculate the shaft diameter.
Clearly, the shaft is subjected to bending due to the tangential loads carried by the gears ignoring the other bending loads. Let us draw the shear force and bending moment diagrams of the shaft as shown in Figure 11.24:
Therefore, M_{\max } is the maximum bending moment on the shaft = 0.54 kN mm = 540 × 10³ N mm at location C. Thus, bending stress on the shaft is
\sigma_{x x}=\frac{32 M}{\pi d^3}
where M = 540 × 10³ N mm. Evidently, torque diagram of the shaft is shown in Figure 11.25.
Clearly, T = 0.45 kN m = 450 × 10³ N mm.
Shear stress on the shaft material is
\tau_{x y}=\frac{16 T}{\pi d^3}
Now, \tau_{\max } is maximum shear stress developed on the shaft and is given by
\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}
=\left\lgroup \frac{16}{\pi d^3} \right\rgroup \sqrt{540^2+450^2}\left(10^3\right) MPa
=\frac{3.58 \times 10^6}{d^3} MPa
Using maximum shear stress criterion and the concept of factor of safety, we get
\tau_{\max }=\frac{\tau_{ yp }}{\text { factor of safety }}
=\frac{\sigma_{ yp }}{(2)(\text { factor of safety })}=\frac{290}{(2)(1.85)}
\frac{3.58 \times 10^6}{d^3}=\frac{290}{(2)(1.85)}
⇒ d = 35.75 mm
Thus, the required shaft diameter is 35.75 mm.
