22-Mec-A4 Design and Manufacture of Machine Elements · May 2016

Question 5 of 6: Continuous Shaft on Three Bearings — Reactions, Bending Moment Diagram, and a Settled Bearing

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

Notes on this paper

Paper format. National Examinations, May 2016 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator. Six questions in two parts: Part A (Q1–Q3, manufacturing-process theory) and Part B (Q4–Q6, machine-element analysis). Candidates answer two from Part A and two from Part B; four questions constitute a complete paper and all questions carry equal value (25 %). All six are solved here.

Reference texts. Kalpakjian & Schmid, Manufacturing Engineering and Technology (Part A); Groover, Fundamentals of Modern Manufacturing (Part A); Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed. (Part B); Hibbeler, Mechanics of Materials, 10th ed. (Q4, Q5); Norton, Machine Design: An Integrated Approach, 6th ed. (Q6).

Check: two corrections carried through Part B. (1) Q5 prints the shaft modulus as “E = 30 ksi”; a 30 ksi modulus is physically impossible for steel and would make the shaft a rubber band, so it is read as the standard E = 30 × 10⁶ psi (30 Mpsi). (2) Q6 supplies the linkage dimensions but not the cross-section of the levers, so the stress in members 2 and 3 is worked for an explicitly stated assumed section and the required section is also reported. Both readings are flagged where they are used, in the spirit of Note 1 on the cover page.

Question 5: Continuous Shaft on Three Bearings — Reactions, Bending Moment Diagram, and a Settled Bearing (25 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
Shaft diameter (continuous, uniform)	$D$	3.00 in
Bearings (immovable, same elevation in part a)	—	A, C, E
Applied loads at B and D	$P$	1000 lbf each
Spans A–B, B–C, C–D, D–E	—	36 in, 24 in, 24 in, 36 in
Overall length A–E	$L$	120 in (60 in + 60 in)
Specific weight of steel	$\gamma$	0.283 lb/in³
Elastic modulus (see callout)	$E$	$30\times10^{6}$ psi
Settlement of bearing C, part (b)	$\Delta$	1/8 in = 0.125 in

Find. (a) The three bearing reactions and the fully dimensioned bending moment diagram, with the shaft's own weight included; (b) the reactions when bearing C sits 0.125 in below A and E.

Check: the printed modulus. The paper states “E = 30 ksi”. A modulus of 30 ksi would make the shaft roughly as stiff as a soft elastomer and the deflections would be a thousand times the shaft diameter, which is nonsense for a steel shaft on bearings. It is read here as the standard value for steel, $E = 30\times10^{6}$ psi. Note that in part (a) the answer is in fact independent of $E$ (both sides of the compatibility equation are proportional to $1/EI$), so only part (b), where a fixed 0.125 in settlement is imposed, actually depends on this reading.

Q5. Top: the three-bearing continuous shaft with its two 1000 lbf loads and its own distributed dead weight. Bottom: the bending moment diagram for part (a), sagging positive, hogging over the centre bearing.

Approach. The shaft is statically indeterminate to the first degree. Release the centre bearing C, compute the downward deflection at C of the resulting simply supported 120 in beam under the two point loads and the distributed dead load, and then restore the geometric condition at C — zero deflection in part (a), a deflection of 0.125 in downward in part (b) — by applying the redundant reaction $R_2$.

Establish the section properties and the dead load. For a solid 3 in circular shaft, $$I = \frac{\pi D^{4}}{64} = \frac{\pi (3)^{4}}{64} = 3.976\ \text{in}^{4}, \qquad A = \frac{\pi D^{2}}{4} = 7.069\ \text{in}^{2}$$ and the dead load per unit length is $$w = \gamma A = (0.283)(7.069) = 2.000\ \text{lbf/in}$$ a convenient round figure. Over the 120 in length the shaft therefore weighs 240.0 lbf, so the total downward load on the system is $$W = 2P + wL = 2000 + 240.0 = 2240.0\ \text{lbf}$$
Choose the redundant and compute the deflection at C in the released structure. Removing the bearing at C leaves a simply supported beam of span $L = 120$ in with supports at A and E. The deflection at midspan from the uniform dead load is $$\delta_w = \frac{5wL^{4}}{384EI} = \frac{5(2.000)(120)^{4}}{384\,(30\times10^{6})(3.976)} = 0.04528\ \text{in}$$ and the deflection at midspan from a point load $P$ placed a distance $a$ from the near support (with $a \leq L/2$) is $Pa(3L^{2} - 4a^{2})/(48EI)$. With $a = 36$ in, and doubling for the symmetric pair of loads, $$\delta_P = 2\,\frac{Pa\left(3L^{2} - 4a^{2}\right)}{48EI} = 2\,\frac{(1000)(36)\left(3(120)^{2} - 4(36)^{2}\right)}{48(30\times10^{6})(3.976)} = 0.47806\ \text{in}$$ Adding them, the released beam sags at C by $$\delta_C = 0.04528 + 0.47806 = 0.52334\ \text{in}$$
Compute the flexibility at C. A single force applied at the midspan of a simply supported beam produces a midspan deflection of $R_2L^{3}/(48EI)$, so the deflection per unit force is $$f_C = \frac{L^{3}}{48EI} = \frac{(120)^{3}}{48(30\times10^{6})(3.976)} = 3.0177\times10^{-4}\ \text{in/lbf}$$
Part (a): impose zero deflection at C. With all three bearings at the same elevation the shaft must be straight at C, so the redundant must exactly cancel $\delta_C$: $$R_2 f_C = \delta_C \quad \Longrightarrow \quad R_2 = \frac{0.52334}{3.0177\times10^{-4}}$$ $$\boxed{R_2 = 1734\ \text{lbf (up, at bearing C)}}$$ Vertical equilibrium and symmetry then give the two end reactions: $$R_1 = \frac{W - R_2}{2} = \frac{2240.0 - 1734.0}{2} = \boxed{R_1 = 253\ \text{lbf (up, at A and at E)}}$$ The centre bearing carries 77 % of the total load — the characteristic result for a two-span continuous beam, and a reminder that a middle bearing must be sized for far more than an equal share.
Part (a): build the bending moment diagram. Taking $x$ from A and sagging moments positive, on $0 \le x \le 36$ the only actions are $R_1$ and the dead load, so $M(x) = R_1 x - wx^{2}/2$ and at B $$M_B = (253.0)(36) - \frac{(2.000)(36)^{2}}{2} = 9108 - 1296 = 7812\ \text{lbf}\cdot\text{in}$$ Continuing to C, the 1000 lbf load at B now contributes: $$M_C = (253.0)(60) - (1000)(24) - \frac{(2.000)(60)^{2}}{2} = 15\,180 - 24\,000 - 3600 = -12\,420\ \text{lbf}\cdot\text{in}$$ $$\boxed{M_B = M_D = +7812\ \text{lbf}\cdot\text{in}, \qquad M_C = -12\,420\ \text{lbf}\cdot\text{in}}$$ By symmetry $M_D = M_B$, and the moment is zero at both ends. The diagram is a gentle parabola from A up to the peak at B, a steep near-straight fall through zero (at $x \approx 45.4$ in) to the hogging peak over the centre bearing, then the mirror image. The governing moment is therefore the hogging moment at C, not the sagging moment under the loads.
Part (a): the resulting bending stress. With $c = D/2 = 1.5$ in, $$\sigma_{\max} = \frac{M_C\,c}{I} = \frac{12\,420 \times 1.5}{3.976} = 4686\ \text{psi} = 4.69\ \text{ksi}$$ which is very low, confirming that this shaft is sized by stiffness or by shaft-dynamics considerations rather than by bending strength.
Part (b): impose a 0.125 in settlement at C. If the centre bearing sits 0.125 in below A and E, the shaft only has to be pulled up to that lower elevation, so the compatibility condition becomes $$R_2 f_C = \delta_C - \Delta = 0.52334 - 0.125 = 0.39834\ \text{in}$$ $$R_2 = \frac{0.39834}{3.0177\times10^{-4}} = \boxed{R_2 = 1320\ \text{lbf}}$$ and by equilibrium $$R_1 = \frac{2240.0 - 1319.9}{2} = \boxed{R_1 = 460\ \text{lbf (up, at A and at E)}}$$
Interpret the settlement result. The change is directly proportional to the settlement: the centre reaction drops by exactly $$\Delta R_2 = \frac{\Delta}{f_C} = \frac{0.125}{3.0177\times10^{-4}} = 414\ \text{lbf}$$ and each end reaction picks up half of that, 207 lbf. Recomputing the moments with $R_1 = 460.1$ lbf gives $M_B = M_D = +15\,267\ \text{lbf}\cdot\text{in}$ and $M_C = +5\ \text{lbf}\cdot\text{in}$, i.e. the hogging moment over the centre bearing has been almost exactly eliminated and the shaft now behaves like two nearly independent simple spans. The maximum bending stress rises to $15\,267(1.5)/3.976 = 5760\ \text{psi}$, a 23 % increase. Continuing the trend, a settlement of $\delta_C = 0.523$ in would unload the centre bearing completely and the shaft would lift clear of it. This sensitivity — 414 lbf of reaction change for one-eighth of an inch of misalignment — is precisely why multi-bearing shafts are so demanding of installation alignment.

Quantity	Symbol	Part (a): bearings level	Part (b): C low by 1/8 in
Dead load of shaft	$w$	2.000 lbf/in (240.0 lbf total)
End reactions at A and E	$R_1$	253 lbf	460 lbf
Centre reaction at C	$R_2$	1734 lbf	1320 lbf
Moment at B and at D	$M_B, M_D$	+7812 lbf·in	+15 267 lbf·in
Moment at C	$M_C$	−12 420 lbf·in	≈ 0
Maximum bending stress	$\sigma_{\max}$	4.69 ksi	5.76 ksi

Topic. Statically indeterminate continuous beams solved by the force (flexibility) method, including support settlement.

Key relations. Midspan deflections of a simply supported beam: $5wL^{4}/(384EI)$ for a uniform load, $Pa(3L^{2}-4a^{2})/(48EI)$ for a point load at $a \le L/2$, and $PL^{3}/(48EI)$ for a midspan load; compatibility $R\,f = \delta_0 - \Delta$; bending stress $\sigma = Mc/I$ with $I = \pi D^{4}/64$.

Why this works. A three-support beam has one more unknown reaction than statics can supply, so one geometric statement must replace one equilibrium equation. Choosing the centre reaction as the redundant is the natural choice here because the released structure — a simply supported beam — has standard, tabulated deflections, and because the geometry is symmetric so only one compatibility equation is needed. The settlement case then costs nothing extra: settlement simply changes the right-hand side of that one equation from zero to $-\Delta$, which is why the reaction change is exactly linear in the settlement and independent of the applied loads. Note the structural consequence: in part (a) the centre bearing carries more than three times either end bearing, and the largest moment is the hogging moment over it, not the sagging moment under the applied loads.

Common pitfall. Using $PL^{3}/(48EI)$ for the deflection produced by the two off-centre 1000 lbf loads — that formula applies only to a load at midspan and would overstate $\delta_P$ by 25 %. A second pitfall is omitting the dead load, which looks negligible at 2 lbf/in but contributes 240 lbf to the total and shifts the moment at C by about 3600 lbf·in. A third is getting the sign of the settlement term backwards: a support that drops reduces the reaction it must supply.

Source. Hibbeler, Mechanics of Materials, 10th ed., §12.5–12.9 (statically indeterminate beams, method of superposition) and Appendix C (deflection tables); Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §4.10 and Table A-9.

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