Question 5 of 6: Rotating Shaft — Fatigue Diameter and Deflections

Question 5: Rotating Shaft — 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. Self-aligning bearings (simple supports) at $x=0$ and $x=14$ in; a constant $1000$ lb transverse load at $x=18$ in (4 in overhang); torque fluctuating $0\!\to\!2000$ lb·in; $S_{ut}=108$ ksi, $S_y=62$ ksi; design factor $n=2$; $K_f=K_{fs}=1$.

Find. The shaft diameter for $n=2$ in combined fatigue (DE-Goodman), and the corresponding maximum torsional twist and bending deflection.

Approach. Because the shaft rotates under a fixed transverse load, the bending is fully reversed; the torque gives its own mean and alternating parts. Locate the critical section (the right bearing, where $M$ peaks), build the Marin endurance limit, and iterate the DE-Goodman diameter (size factor depends on $d$). Then compute twist $\theta=TL/GJ$ and the overhang deflection.

1. Critical bending moment. Reactions for the overhung load give $R_B=1000(18)/14=1286$ lb up, $R_A=-286$ lb; the bending moment peaks at the right bearing: $$M_{max}=P\times\text{overhang}=1000(4)=4000\ \text{lb}\!\cdot\!\text{in}.$$ Rotation makes this fully reversed: $M_a=4000$, $M_m=0$. The torque splits as $T_m=T_a=1000$ lb·in.
2. Endurance limit (Marin). $S_e'=0.5S_{ut}=54$ ksi. Surface (machined) $k_a=2.70\,S_{ut}^{-0.265}=0.781$; load, temperature and reliability factors $=1$; the size factor $k_b=0.879\,d^{-0.107}$ depends on $d$, so the diameter is found iteratively. Converged $k_b\approx0.851$, giving $$S_e=k_a k_b S_e'=0.781(0.851)(54)=35.9\ \text{ksi}.$$
3. DE-Goodman diameter. With $K_f=K_{fs}=1$ and $M_m=0$, $$d=\left\{\frac{16n}{\pi}\!\left[\frac{\sqrt{4M_a^2+3T_a^2}}{S_e}+\frac{\sqrt{3T_m^2}}{S_{ut}}\right]\right\}^{1/3}.$$ Substituting ($M_a=4000$, $T_a=T_m=1000$ lb·in; $S_e=35{,}900$, $S_{ut}=108{,}000$ psi), $$d=\left\{\frac{16(2)}{\pi}\!\left[\frac{8185}{35{,}900}+\frac{1732}{108{,}000}\right]\right\}^{1/3}=\boxed{1.36\ \text{in}}\ \Rightarrow\ \text{specify }d=1\tfrac38\ \text{in }(1.375).$$
4. Torsional deflection. For $d=1.375$ in, $J=\pi d^4/32=0.351\ \text{in}^4$, $G=11.5\times10^6$ psi. Taking the peak torque over the 18 in drive length, $$\theta=\frac{T L}{GJ}=\frac{2000(18)}{11.5\times10^6(0.351)}=0.0089\ \text{rad}=\boxed{0.51^\circ}.$$
5. Bending deflection. With $I=\pi d^4/64=0.175\ \text{in}^4$, $E=30\times10^6$ psi, the deflection of the overhung end (overhang $a=4$ in, span $L=14$ in) is $$\delta=\frac{P a^2(L+a)}{3EI}=\frac{1000(4^2)(18)}{3(30\times10^6)(0.175)}=\boxed{0.018\ \text{in}}.$$