Question 9.2: Shaft Design for Combined Bending and Torsion The gear A is ...
Shaft Design for Combined Bending and Torsion
The gear A is attached to the AISI 1010 CD steel shaft AB of yield strength S_{y} that carries a vertical load of 360 lb (Figure 9.3a). The shaft is fitted with gear D that forms a set with gear E.
Find: (a) The value of the torque T_{E} applied on the gear E to support the loading and reactions at the bearings and (b) the required shaft diameter D, applying the maximum shear stress failure criterion.
Given: S_{y} = 300/6.895 = 43.5 ksi (from Table B.3).
Assumptions: The bearings at B and C are taken as simple supports. A safety factor of n = 1.6 is to be used with respect to yielding.
a. Conditions of equilibrium are applied to Figure 9.3b 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 9.3a, we thus have T_{E}= F_{D}(2) = 480(2) = 960 lb·in.
b. Observe from Figure 9.3c through e that, since M_{C} > M_{D} , 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
\begin{aligned} D &=\left[\frac{32(1.6)}{\pi(43.5)} \sqrt{(1.8)^2+(1.44)^2}\right]^{1 / 3} \\ &=0.952 in . \end{aligned}
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)