22-Mec-A4 Design and Manufacture of Machine Elements · May 2013

Question 6 of 8: Minimum shaft diameter by static yield and by fatigue

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examination, 07-Mec-A4 Design and Manufacture of Machine Elements, May 2013 — 3 hours, open book, any non-communicating calculator permitted. Eight questions on six pages, divided into Part A (manufacturing processes, Q1–Q4) and Part B (machine-element design, Q5–Q8). The rubric asks for three from Part A and two from Part B, five questions constituting a complete paper, all of equal value (20 % each). All eight questions are solved here, because this document is a study resource rather than an examination script.

Reference texts.

Kalpakjian & Schmid, Manufacturing Engineering and Technology, 7th ed. — grinding (Ch. 26), sheet-metal forming and the forming-limit diagram (Ch. 16), sand casting and casting defects (Ch. 11–12).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — combined stresses and static failure theories (Ch. 3, 5), fatigue (Ch. 6), shaft design (Ch. 7), bolted joints in shear (Ch. 8), welded joints (Ch. 9).
Hibbeler, Mechanics of Materials, 10th ed. — stress transformation and combined loading.
ASM Handbook Vol. 15, Casting — gas, penetration and mould-wall-movement defects in no-bake sand systems.

Check: Part B is figure-driven. Every dimension used below was read from the printed figures (Fig. S4–S7). Where the drawing dimensions a distance from a face rather than from a bolt centre (Q7), the reading is stated explicitly in Given so a grader can substitute a different interpretation without redoing the method.

Question 6: Minimum shaft diameter by static yield and by fatigue (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.

Quantity	Symbol	Value
Torque delivered to A	$T$	3000 lbf·in
Spur-gear pitch diameter at D	$d_p$	6.00 in
Pressure angle	$\phi$	$20^\circ$
Bearing span B–C	$\ell$	10.0 in
Overhang C–D	$a$	4.00 in
Yield strength	$S_y$	60.0 kpsi
Tensile strength	$S_{ut}$	80.0 kpsi
Design factor	$n_d$	2.50
Sharp shoulder fillet: bending / torsion concentration	$K_t,\ K_{ts}$	2.7, 2.2

Find. The minimum diameter of the 10-in span between the bearings, first from a static distortion-energy yield check at the critical section and then from a completely-reversed-bending fatigue check using the DE-Goodman criterion, both at $n_d = 2.5$.

Fig. 6-1 — Shaft layout of Fig. S5 and the resulting bending-moment diagram. The gear load is overhung, so the bending moment grows linearly from zero at bearing B to its maximum at bearing C, which is also where the sharp shoulder fillet sits.

Approach. Resolve the gear mesh force from the transmitted torque, build the bending-moment diagram to locate the critical section, then size the diameter twice — once from the distortion-energy yield criterion with the stress concentrations applied statically, and once from the DE-Goodman fatigue criterion with the rotating shaft's completely reversed bending and steady torque.

Resolve the gear tooth force from the transmitted torque. The tangential component does the work, $$W_t = \frac{T}{d_p/2} = \frac{3000}{3.00} = 1000\ \text{lbf}$$ and the pressure angle sets the radial component and the resultant, $$W_r = W_t\tan\phi = 1000\tan 20^\circ = 364\ \text{lbf}, \qquad F = \frac{W_t}{\cos\phi} = 1064\ \text{lbf}$$ Both components are transverse to the shaft, so their resultant of 1064 lbf is the single bending load to carry forward.
Build the bending-moment diagram and locate the critical section. The gear sits 4 in outboard of bearing C, so this is an overhung load. Taking moments about B, $$R_C = F\,\frac{\ell + a}{\ell} = 1064\,\frac{14}{10} = 1490\ \text{lbf}, \qquad R_B = R_C - F = 426\ \text{lbf}\ (\text{downward})$$ The bending moment is zero at B, rises linearly across the span and peaks at C: $$M = F\,a = 1064\,(4.00) = 4257\ \text{lbf}\!\cdot\!\text{in}$$ Since the torque of 3000 lbf·in is transmitted from D through the whole span to A, and the sharp bearing-shoulder fillet is at C, the critical section is unambiguously the shoulder at C.
Adopt the stress-concentration factors for a sharp shoulder fillet. With no fillet radius specified, the standard first-iteration estimates for a sharp shoulder fillet on a shaft are $K_t = 2.7$ in bending and $K_{ts} = 2.2$ in torsion. Because the radius is unknown, notch sensitivity cannot be evaluated, so the conservative choice $K_f = K_t$ and $K_{fs} = K_{ts}$ is used for the fatigue analysis as well.
Part (a): size the diameter from static distortion-energy yield. For a round section carrying bending $M$ and torque $T$ with the concentrations applied, the distortion-energy effective stress is $$\sigma' = \frac{16}{\pi d^{3}}\sqrt{4\left(K_t M\right)^{2} + 3\left(K_{ts} T\right)^{2}}$$ Setting $\sigma' = S_y/n_d$ and solving for the diameter, $$d = \left[\frac{16\,n_d}{\pi S_y}\sqrt{4\left(K_t M\right)^{2}+3\left(K_{ts}T\right)^{2}}\right]^{1/3}$$ Substituting $K_tM = 2.7(4257) = 11\,494$ and $K_{ts}T = 2.2(3000) = 6600$ lbf·in gives a radical of $25\,670$ lbf·in and $$\boxed{\,d_{\text{static}} = \left[\frac{16(2.50)}{\pi(60\,000)}\,(25\,670)\right]^{1/3} = 1.76\ \text{in}\,}$$
Part (b): identify the fatigue stress components. The shaft rotates while the gear load stays fixed in space, so every surface fibre in the span sees the bending stress swing from tension to compression once per revolution. The bending is therefore completely reversed and the torque, delivered steadily, is purely mean: $$M_a = 4257\ \text{lbf}\!\cdot\!\text{in}, \quad M_m = 0, \qquad T_m = 3000\ \text{lbf}\!\cdot\!\text{in}, \quad T_a = 0$$ This is the classic shaft-design load case and is the reason the fatigue check, not the static check, governs almost every rotating shaft.
Estimate the corrected endurance limit. Starting from the rotating-beam estimate for steel with $S_{ut} < 200$ kpsi, $$S_e' = 0.5\,S_{ut} = 40.0\ \text{kpsi}$$ The machined-surface factor is $k_a = a\,S_{ut}^{\,b} = 2.00\,(80)^{-0.217} = 0.773$, and the size factor for a round rotating section is $k_b = 0.879\,d^{-0.107}$. Load, temperature and reliability factors are taken as unity (bending, room temperature, 50 % reliability as the question implies no other requirement). Because $k_b$ depends on the answer, the diameter must be iterated.
Solve the DE-Goodman diameter equation iteratively. With only alternating bending and only mean torque, the DE-Goodman shaft equation reduces to $$d = \left\{\frac{16\,n_d}{\pi}\left[\frac{2\,K_f M_a}{S_e} + \frac{\sqrt{3}\,K_{fs}T_m}{S_{ut}}\right]\right\}^{1/3}$$ Starting from a trial $d = 2.0$ in and iterating $k_b \rightarrow S_e \rightarrow d$ to convergence gives $k_b = 0.802$, hence $$S_e = 0.773\,(0.802)\,(40.0) = 24.8\ \text{kpsi}$$ $$\boxed{\,d_{\text{fatigue}} = 2.39\ \text{in}\,}$$
Select and interpret. Fatigue governs by a wide margin, requiring 2.39 in against the 1.76 in from static yield — a 36 % larger diameter and roughly twice the cross-sectional area. The bending term contributes about 86 % of the bracket in the DE-Goodman expression, confirming that reversed bending, amplified by the sharp fillet, is the design driver. Specifying the next convenient size, $d = 2.5$ in, satisfies both criteria with margin. The single most valuable design change would be to specify a generous shoulder fillet radius instead of a sharp one: dropping $K_f$ from 2.7 to about 1.7 would reduce the required fatigue diameter to roughly 2.1 in.

Quantity	Symbol	Result
Tangential gear force	$W_t$	1000 lbf
Radial gear force	$W_r$	364 lbf
Resultant gear force	$F$	1064 lbf
Maximum bending moment (at C)	$M$	4257 lbf·in
Transmitted torque	$T$	3000 lbf·in
Corrected endurance limit	$S_e$	24.8 kpsi
(a) Minimum diameter, static DE yield	$d$	1.76 in
(b) Minimum diameter, DE-Goodman fatigue	$d$	2.39 in
Recommended specified size	$d$	2.50 in

Topic: Rotating-shaft design for combined steady torque and completely reversed bending.

Key relations: $W_t = 2T/d_p$, $W_r = W_t\tan\phi$; static DE sizing $d = \left[\frac{16n}{\pi S_y}\sqrt{4(K_tM)^2+3(K_{ts}T)^2}\right]^{1/3}$; Marin factors $S_e = k_ak_bS_e'$ with $k_a = aS_{ut}^b$ and $k_b = 0.879d^{-0.107}$; DE-Goodman $d = \left\{\frac{16n}{\pi}\left[\frac{2K_fM_a}{S_e}+\frac{\sqrt3 K_{fs}T_m}{S_{ut}}\right]\right\}^{1/3}$.

Why this works: The essential insight is kinematic rather than algebraic. The gear load is fixed in direction while the shaft turns beneath it, so a material point travels from the tension side to the compression side and back on every revolution: bending is fully alternating with zero mean. The torque, in contrast, is delivered continuously in one sense and is therefore fully mean with zero alternating part. Feeding that split into a mean–alternating failure criterion is exactly what the DE-Goodman shaft equation does, and it is why the result is so much larger than the static answer even though the loads are identical.

Common pitfall: Placing the maximum bending moment at mid-span. The load here is overhung, so the moment is zero at the far bearing and maximum at the near one; a mid-span assumption both misplaces the critical section and under-predicts the moment. Two further traps: forgetting that $k_b$ depends on the unknown diameter and so must be iterated, and applying the concentration factors to only one of the two load components.

Source: Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §7-4 (shaft design for stress, DE-Goodman), Table 7-1 (first-estimate stress-concentration factors for shaft shoulders), and §6-9 (Marin surface and size factors).

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