Question 9.2: Shaft Design for Combined Bending and Torsion The gear A is ......

a. Conditions of equilibrium are applied to Figure 9.3(b) to find tangential force F_D acting on gear D . Then support reactions are determined using equilibrium conditions and marked on the figure. Referring to Figure 9.3(a), we thus have T_E=F_D(2)=480(2)=960  lb \cdot in .

b. Observe from Figure 9.3(c) through Figure 9.3(e) that, since M_C>M_D , the critical section where largest value of the stress is expected to occur is at C. Through the use of Equation (9.7), we have

\frac{S_y}{n}-\frac{32}{\pi D^3}\left[M^2+T^2\right]^{1 / 2}        (9.7)

D=\left[\frac{32 n}{\pi S_y} \sqrt{M_C^2+T_C^2}\right]^{1 / 3}        (a)

Substituting the numerical values results in

D=\left[\frac{32(1.6)}{\pi(43.5)} \sqrt{(1.8)^2+(1.44)^2}\right]^{1 / 3}=0.952 \text { in. }

Comment: It is interesting to note that, similar to the distortion energy criterion, Equation (9.8) gives D=0.936 in. Thus, a standard diameter of 1.0 in. shaft can be safely used.

\frac{S_y}{n}=\frac{32}{\pi D^3}\left[M^2+\frac{3}{4} T^2\right]^{1 / 2}       (9.8)