Question 5 of 6: Overhung Rotating Shaft — Fatigue Diameter and Deflection

Question 5: Overhung Rotating Shaft — Fatigue Diameter and Deflection (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. Bearing A at $x=0$, bearing B at $a=7$ in; the transverse load $P=500$ lb acts at $b=9$ in — i.e. it overhangs bearing B by $c=b-a=2$ in; the shaft extends to $l=12$ in. Because the load direction is fixed while the shaft turns, the bending is fully reversed.

Find. The shaft diameter for a fatigue safety factor $n=2$ (DE-Goodman), and the maximum deflection at that diameter.

Approach. Statics gives the peak (fully-reversed) bending moment at bearing B; the torque splits into mean and alternating parts. Build the Marin endurance limit for the machined, 99 %-reliable shaft, then solve the DE-Goodman equation for the diameter at $n=2$, round up to a standard size, and finally compute the overhang deflection.

1. Bending moment (fully reversed). The overhung load bends the shaft most at bearing B: $$M_a=P\,c=500\times2=1000\ \text{lb}\cdot\text{in},\qquad M_m=0.$$
2. Torque components. $T_m=\dfrac{T_{\max}+T_{\min}}{2}=\dfrac{600+(-100)}{2}=250$, $\ T_a=\dfrac{T_{\max}-T_{\min}}{2}=\dfrac{600-(-100)}{2}=350\ \text{lb}\cdot\text{in}.$
3. Endurance limit (Marin). $S_e'=0.5S_{ut}=54$ ksi; machined surface $k_a=2.70\,S_{ut}^{-0.265}=0.781$; size $k_b=(d/0.3)^{-0.107}$; load $k_c=1$ (bending); temperature $k_d=1$; reliability 99 % $k_e=0.814$. Thus $S_e=k_a k_b k_e S_e'$ (a function of $d$).
4. DE-Goodman for the diameter. With $K_f=K_{fs}=1$ (no fillet/keyway data given at the bearing seat), $$\frac{1}{n}=\frac{16}{\pi d^{3}}\!\left[\frac{\sqrt{4(K_fM_a)^2+3(K_{fs}T_a)^2}}{S_e}+\frac{\sqrt{4(K_fM_m)^2+3(K_{fs}T_m)^2}}{S_{ut}}\right].$$ Setting $n=2$ and iterating on the size factor gives $$\boxed{\ d=0.90\ \text{in}\ \rightarrow\ \text{use }d=\tfrac{15}{16}=0.9375\ \text{in (next standard)}\ }$$ at which $S_e\approx30.4$ ksi and the realized factor exceeds 2.
5. Maximum deflection. The largest deflection is at the overhung load. For a load $c$ beyond the support, over span $a$, $$\delta=\frac{P\,c^{2}(a+c)}{3EI}=\frac{500(2)^2(7+2)}{3(2.9\times10^{7})\,I},\quad I=\frac{\pi d^4}{64}=0.0379\ \text{in}^4,$$ $$\delta=5.5\times10^{-3}\ \text{in}\ (\approx0.14\ \text{mm}).$$