Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, 16-Mec-B1 Advanced Machine Design, May 2018. Open book, 3 hours, 100 marks. Part I (Problems 1 & 2) is compulsory; candidates answer only three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Given. A steel shaft on two supports carries a constant transverse load with a fluctuating torque; the design target is an infinite-life fatigue safety factor of 2. From the figure the supports are at 0 and 8 in and the 400 lb load is at 10 in (it overhangs the right bearing).
Given data
Transverse load $F$
400 lb (constant magnitude, rotating)
Bearing span / load station
supports at 0 and 8 in; load at 10 in (2 in overhang)
Torque range
$T = -500 \to +500\ \text{lb}\cdot\text{in}$
Material
$S_{ut}=108\ \text{kpsi}$, $S_y=62\ \text{kpsi}$
Design factor / life / reliability
$n=2$, infinite life, 50% ($k_e=1$), no stress raiser ($K_f=K_{fs}=1$)
Find. The minimum shaft diameter $d$ giving $n=2$ against fatigue for infinite life by the ASME-elliptic criterion.
Simply-supported shaft with a 2 in overhang: the 400 lb load sits outboard of bearing B, so the peak bending moment occurs at B.
Approach. Locate the peak bending moment (at the outboard bearing), resolve it and the torque into alternating and mean components, build the Marin endurance limit $S_e$, and solve the ASME-elliptic equation iteratively for $d$ (the size factor $k_b$ depends on $d$).
Reactions and peak bending moment. With supports at $A(0)$ and $B(8)$ and the load at 10 in, taking moments about $A$: $R_B(8)=400(10)$, so $R_B=500$ lb and $R_A=400-500=-100$ lb. The load overhangs $B$ by 2 in, so the maximum moment is at $B$:
$$M_{\max}=F\,(10-8)=400\times 2=800\ \text{lb}\cdot\text{in}.$$
Alternating / mean components. Because the shaft rotates under a fixed-direction load, the bending stress at any surface fibre fully reverses each revolution: $M_a=800\ \text{lb}\cdot\text{in}$, $M_m=0$. The torque swings symmetrically about zero, so $T_a=\tfrac{500-(-500)}{2}=500\ \text{lb}\cdot\text{in}$ and $T_m=0$.
Surface and endurance limit. For $S_{ut}=108\ \text{kpsi}\lt 200$, $S_e'=0.5S_{ut}=54\ \text{kpsi}$. Machined surface: $k_a=2.70\,S_{ut}^{-0.265}=2.70(108)^{-0.265}=0.781$. Load $k_c=1$ (bending basis), reliability $k_e=1$ (50%). Size factor $k_b=0.879\,d^{-0.107}$ is carried through the iteration:
$$S_e=k_a\,k_b\,S_e'.$$
ASME-elliptic diameter equation. With $K_f=K_{fs}=1$ and $M_m=T_m=0$,
$$\frac{1}{n}=\frac{16}{\pi d^{3}}\sqrt{4\!\left(\frac{M_a}{S_e}\right)^{2}+3\!\left(\frac{T_a}{S_e}\right)^{2}}\;\Rightarrow\; d=\left[\frac{16n}{\pi S_e}\sqrt{4M_a^{2}+3T_a^{2}}\right]^{1/3}.$$
The load term is $\sqrt{4(800)^2+3(500)^2}=\sqrt{3.31\times10^{6}}=1819\ \text{lb}\cdot\text{in}$.
Iterate for $d$. Starting at $d=1$ in and updating $k_b$ each pass, the diameter converges in a few steps to $k_b=0.902$, $S_e=0.781(0.902)(54\,000)=38\,000\ \text{psi}$, and
$$\boxed{d=0.787\ \text{in}}$$
so select the next standard size, $d=\tfrac{7}{8}\ \text{in}=0.875\ \text{in}$.
Verify the selected size. At $d=0.875$ in, $k_b=0.879(0.875)^{-0.107}=0.891$, $S_e=37\,600\ \text{psi}$, and the realised factor of safety is
$$n=\left[\frac{16}{\pi d^{3}}\sqrt{4(M_a/S_e)^2+3(T_a/S_e)^2}\right]^{-1}=2.72\gt 2\ \checkmark.$$
Check: the figure gives no fillet, keyway or shoulder, so per the statement $K_f=K_{fs}=1$ are used. A real bearing seat with a press fit or keyway would add $K_f\approx1.6$–$2$, roughly doubling the required section modulus — the design should be revisited once the actual geometry is fixed.
