Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2014 — 07-Mec-B1 Advanced Machine Design. Open book, 3 hours, 100 marks. Part I (Problems 1–2) compulsory; answer three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Reference texts: R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design, 10th ed. (shafts & fatigue §6–7, journal bearings §12, clutches & brakes §16); R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design (lubrication, brakes); R. C. Hibbeler, Mechanics of Materials (beam deflection, impact).
Given. A rotating steel shaft carries two gears; the left gear (250 lb) sits 4 in from the left bearing and the right gear (750 lb) is overhung 4 in beyond the right bearing.
Given data
Quantity
Value
Left radial load (gear at $x=4$ in)
250 lb
Right radial load (overhung gear at $x=16$ in)
750 lb
Bearing span (left $x=0$, right $x=12$ in)
12 in
Torque range $T_{\min}\to T_{\max}$
−200 → 400 lb·in
$S_{ut}$, $S_y$
108 ksi, 62 ksi
Fatigue safety factor $n$
2
Find. The minimum shaft diameter giving $n=2$ against fatigue by the DE–Goodman criterion, and the maximum bending deflection of the resulting shaft.
Loading of the two-gear shaft: simple supports at the bearings ($x=0,\,12$ in); overhung 750 lb gear at $x=16$ in.
Approach. Find the bearing reactions and the bending-moment diagram, identify the critical section, split bending and torque into alternating and mean parts (a rotating shaft makes the bending fully reversed), size the diameter with the DE–Goodman equation using Marin-corrected endurance, then double-integrate $M/EI$ for the deflection.
Bearing reactions. Taking moments about the left bearing, $R_B = \dfrac{250(4)+750(16)}{12} = 1083.3\ \text{lb}$, and $R_A = 250+750-1083.3 = -83.3\ \text{lb}$ (the left bearing pulls down, reacting the overhung load).
Bending-moment diagram. The moment magnitude grows from the left bearing to a peak at the right (overhang-side) bearing, where $M = 750\times(16-12) = 3000\ \text{lb}\!\cdot\!\text{in}$; it returns to zero at the overhung gear. The critical section is therefore the right bearing seat.
Alternating / mean components. Because the shaft rotates, the steady radial loads produce fully reversed bending, so $M_a = 3000\ \text{lb}\!\cdot\!\text{in}$, $M_m = 0$. The fluctuating torque gives $T_a = \tfrac{1}{2}(400-(-200)) = 300$ and $T_m = \tfrac{1}{2}(400+(-200)) = 100\ \text{lb}\!\cdot\!\text{in}$.
Endurance limit (Marin). With $S'_e = 0.5S_{ut} = 54$ ksi, machined surface factor $k_a = 2.70\,S_{ut}^{-0.265} = 0.781$, and size factor $k_b = 0.879\,d^{-0.107}$ iterated with the diameter, the corrected endurance limit converges to $S_e \approx 36.4\ \text{ksi}$.
DE–Goodman diameter. Solving Shigley’s distortion-energy–Goodman relation for $d$,
$$d = \left(\frac{16\,n}{\pi}\left\{\frac{\sqrt{4(K_fM_a)^2+3(K_{fs}T_a)^2}}{S_e} + \frac{\sqrt{4(K_fM_m)^2+3(K_{fs}T_m)^2}}{S_{ut}}\right\}\right)^{1/3}$$
with $K_f=K_{fs}=1$ gives $d = 1.19\ \text{in}$, so
$$\boxed{\,d = 1.25\ \text{in}\ \ (1\tfrac{1}{4}\text{ in preferred size}),\quad n=2\,}$$
Maximum bending deflection. For $d = 1.25$ in, $I = \pi d^4/64 = 0.120\ \text{in}^4$. Double-integrating $M(x)/EI$ (with $E = 30\times10^6$ psi, $y=0$ at both bearings) locates the largest deflection at the overhung gear tip: $y_{\max} = 0.0158\ \text{in} = 0.40\ \text{mm}$.
Check / assumptions: No fillet-radius or keyway geometry is given at the bearing seat, so $K_f=K_{fs}=1$ was used; a shoulder fillet or keyway there would raise $K_f\approx1.6$–$2.0$ and require a larger diameter. Reliability is unspecified, so $k_e=1$ (50%). Standard machined-surface constants ($a=2.70$, $b=-0.265$, ksi) are assumed.
