Question 3 of 6: Rotating shaft with an overhung load — fatigue diameter and deflections (20 marks)

Question 3: Rotating shaft with an overhung load — fatigue diameter and deflections (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A rotating steel shaft on two self-aligning bearings, loaded transversely beyond the right-hand bearing and carrying a pulsating torque; the figure gives the three stations along the shaft.

Find. (1) the shaft diameter that gives a fatigue safety factor of 2, and (2) the corresponding maximum bending deflection and maximum angle of twist.

Approach. Recognise that the load overhangs bearing B, so the bending moment peaks at the bearing; because the shaft rotates under a constant-direction load, that bending is fully reversed while the torque pulses about a non-zero mean. Size the shaft with the DE-Goodman shaft equation, iterating on the size factor, then compute the overhang deflection and the twist at the selected standard diameter.

1. Support reactions. Bearings at $x = 0$ and $x = 14$ in are simple supports; the 1000 lb load sits 4 in outboard of the right bearing. $$R_B = \frac{P\,x_P}{x_B} = \frac{1000(18)}{14} = 1286\text{ lb}\ (\uparrow), \qquad R_A = P - R_B = -286\text{ lb}\ (\downarrow)$$ The left bearing carries a downward (hold-down) reaction — typical of an overhung layout and the reason the left bearing must be retained axially and radially in both directions.
2. Critical bending moment. Beyond bearing B the moment builds linearly from zero at the load; between the bearings it decays linearly to zero at A. The maximum is therefore at bearing B, where the overhang is $a = 18-14 = 4$ in. $$M_{\max} = P\,a = 1000 \times 4 = 4000\text{ lb}\!\cdot\!\text{in}$$
3. Resolve the loading into mean and alternating components. The transverse load keeps a constant magnitude and direction while the shaft turns, so every surface fibre sees tension and compression once per revolution: the bending is fully reversed. The torque, by contrast, pulses between 0 and 2000 lb·in about a positive mean. $$M_a = 4000\text{ lb}\!\cdot\!\text{in}, \qquad M_m = 0$$ $$T_a = \frac{T_{\max}-T_{\min}}{2} = 1000\text{ lb}\!\cdot\!\text{in}, \qquad T_m = \frac{T_{\max}+T_{\min}}{2} = 1000\text{ lb}\!\cdot\!\text{in}$$
4. Endurance limit with Marin factors. For steel with $S_{ut} = 108\text{ ksi} \lt 200$ ksi the rotating-beam limit is $S'_e = 0.5S_{ut} = 54$ ksi. Take a machined surface, room temperature, and 50 % reliability as the baseline of the shaft equation: $$k_a = a\,S_{ut}^{\,b} = 2.70\,(108)^{-0.265} = 0.781, \qquad k_b = 0.879\,d^{-0.107}\ \ (0.3 \le d \le 2\text{ in})$$ $$S_e = k_a k_b S'_e$$ Because $k_b$ depends on the diameter being sought, the sizing is iterative.
5. Assemble the von Mises moment groups. The DE (distortion-energy) shaft equation combines bending and torsion into equivalent alternating and midrange terms: $$\sqrt{4(K_f M_a)^2 + 3(K_{fs}T_a)^2} = \sqrt{4(4000)^2+3(1000)^2} = 8185\text{ lb}\!\cdot\!\text{in}$$ $$\sqrt{4(K_f M_m)^2 + 3(K_{fs}T_m)^2} = \sqrt{0+3(1000)^2} = 1732\text{ lb}\!\cdot\!\text{in}$$
6. DE-Goodman diameter, iterated on the size factor. Shigley's first-cycle-free design form is $$d = \left\{\frac{16n}{\pi}\left[\frac{\sqrt{4(K_fM_a)^2+3(K_{fs}T_a)^2}}{S_e} + \frac{\sqrt{4(K_fM_m)^2+3(K_{fs}T_m)^2}}{S_{ut}}\right]\right\}^{1/3}$$ Starting from a trial $d = 1.00$ in and re-evaluating $k_b$ each pass, the iteration converges in a few cycles to $$\boxed{\;d_{req} = 1.356\text{ in}\;}$$
7. Select a standard size and confirm the factor. Round up to the next standard fractional shaft diameter, $d = 1\tfrac{3}{8}$ in $= 1.375$ in. At that size $k_b = 0.879(1.375)^{-0.107} = 0.847$, $S_e = 35.7$ ksi, and back-substitution gives $$\boxed{\;d = 1\tfrac{3}{8}\text{ in, } n_f = 2.09 \;\gt\; 2 \;\checkmark\;}$$
8. First-cycle yield check. Fatigue sizing must be backed by a static check at the peak of the load cycle. At $d = 1.375$ in, $$\sigma_b = \frac{32M_{\max}}{\pi d^3} = 15\,674\text{ psi}, \qquad \tau = \frac{16T_{\max}}{\pi d^3} = 3919\text{ psi}$$ $$\sigma' = \sqrt{\sigma_b^2 + 3\tau^2} = 17\,067\text{ psi} \;\Longrightarrow\; n_y = \frac{62\,000}{17\,067} = 3.63$$ Yielding is not the governing mode, as expected for a fatigue-driven design.
9. Bending deflection at the overhung load. For a simply supported beam of span $L$ with a load $P$ on an overhang $a$, the tip deflection is $\delta = Pa^2(L+a)/(3EI)$. With $E = 30\times10^6$ psi and $I = \pi d^4/64 = 0.1755\text{ in}^4$, $$\delta = \frac{1000(4)^2(14+4)}{3(30\times10^6)(0.1755)}$$ $$\boxed{\;\delta_{bend} = 0.0182\text{ in} = 0.463\text{ mm (downward, at the load)}\;}$$
10. Angle of twist at peak torque. With $G = 11.5\times10^6$ psi and $J = 2I = 0.3509\text{ in}^4$, taking the torque as carried over the 18 in from the drive end to the load station, $$\theta = \frac{T_{\max}L_T}{GJ} = \frac{2000(18)}{(11.5\times10^6)(0.3509)} = 0.00892\text{ rad}$$ $$\boxed{\;\theta_{max} = 0.511^\circ\;}$$

Both deflections are comfortably inside normal machinery limits. The bending deflection of 0.018 in at the overhang is well under the 0.001 in-per-inch-of-span guideline often applied to gear and pulley shafts, and the 0.51° twist is far below the 1° per foot of length rule of thumb for power-transmission shafting — here 0.51° over 18 in is 0.34° per foot.