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

Question 7 of 8: Reaction at the centre bearing of a three-bearing shaft

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, December 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 questions 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.

Reference texts.

Kalpakjian & Schmid, Manufacturing Engineering and Technology, 7th ed. — sand casting and casting defects (Ch. 10–12), adhesive bonding and joint design (Ch. 32), fusion welding and weld defects (Ch. 30–31), sheet-metal shearing and blanking (Ch. 16).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — eccentrically loaded shear joints (Ch. 8–9), combined stresses and stress transformation (Ch. 3), statically indeterminate shafts and deflection (Ch. 4), brakes and clutches (Ch. 16).
Hibbeler, Mechanics of Materials, 10th ed. — transverse shear ($VQ/Ib$), combined loading and Mohr's circle (Ch. 7–9).
Faires, Design of Machine Elements, 4th ed. — double-block (double-shoe) brakes, self-energizing geometry, equal-wear proportioning.
ASM Handbook Vol. 15 Casting and Vol. 6 Welding, Brazing and Soldering — hot-spot/shrinkage defects; hydrogen-induced cold cracking and preheat practice (see also CSA W59 and CSA W47.1 for Canadian fabrication practice).

Check: Part B is figure-driven. Every dimension used below was read from the printed figures. Two readings are stated explicitly in Given so a grader can substitute a different interpretation without redoing the method: (i) in Figure A the low rivet is taken as lying on the same vertical centreline as the third rivet of the top row (75 + 75 = 150 mm from the left-hand rivet); (ii) in Figure D the dimension $a$ is the horizontal spacing, measured along the operating lever, between the pin taking the upper shoe link and the pin taking the lower shoe link, with the 10 in operating arm measured from the lower-link pin.

Question 7: Reaction at the centre bearing of a three-bearing shaft (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	Value
Total span, $L$	96 in (30 + 18 + 18 + 30)
Loads	1200 lb each, at 30 in from each end
Flexural rigidity, $EI$	150 × 10⁶ lb·in²
Centre bearing	at mid-span, 48 in from each end
Case (b) settlement	centre bearing 0.040 in below the end bearings
Case (c) support stiffness, $k$	30 000 lb/in

Find. The centre-bearing reaction $R_2$ for a rigid level support, for a support 0.040 in low, and for an elastic support.

Figure 7.1 — The shaft is once statically indeterminate. Removing the centre bearing leaves a simply supported span whose mid-span deflection $\delta_0$ must be closed by the redundant reaction $R_2$.

Approach. The shaft is statically indeterminate to the first degree. Take $R_2$ as the redundant, remove the centre bearing, compute the mid-span deflection of the released (simply supported) beam under the two symmetric loads, and then impose the compatibility condition appropriate to each case — zero net deflection for a rigid level support, 0.040 in for a settled support, and $R_2/k$ for a spring.

Deflection of the released beam. For a simply supported span $L$ carrying two equal loads $P$ each a distance $a$ from its nearer support, the mid-span deflection is $$\delta_0 = \frac{P\,a\,(3L^2-4a^2)}{24EI}.$$ With $P=1200$ lb, $a=30$ in, $L=96$ in and $EI=150\times10^{6}$ lb·in², $$\delta_0 = \frac{1200\times30\times\left(3(96)^2-4(30)^2\right)}{24\times150\times10^{6}} = \frac{1200\times30\times24\,048}{3.60\times10^{9}},$$ $$\boxed{\delta_0 = 0.24048\ \text{in (downward)}}$$
Flexibility of the released beam at mid-span. A single upward force $R_2$ at the centre of the same span lifts it by $$\delta_{R} = \frac{R_2 L^3}{48EI} = \frac{R_2 (96)^3}{48\times150\times10^{6}} = 1.2288\times10^{-4}\,R_2\ \text{in per lb.}$$
(a) All three bearings solid and level. Compatibility demands zero net deflection at the centre: $$\delta_0 - 1.2288\times10^{-4}R_2 = 0 \;\Rightarrow\; R_2 = \frac{0.24048}{1.2288\times10^{-4}},$$ $$\boxed{R_2 = 1957\ \text{lb}}$$ and by symmetry each end bearing carries $$R_1 = \frac{2(1200)-1957.0}{2} = 221\ \text{lb}.$$ Note that the centre bearing takes 82 % of the total 2400 lb — the classic reason a "helpful" third bearing can overload itself.
(b) Centre bearing 0.040 in low. The shaft need only be pushed back up to a point 0.040 in below the line of the end bearings, so the compatibility statement becomes $$\delta_0 - 1.2288\times10^{-4}R_2 = 0.040,$$ $$R_2 = \frac{0.24048-0.040}{1.2288\times10^{-4}} = \frac{0.20048}{1.2288\times10^{-4}},$$ $$\boxed{R_2 = 1632\ \text{lb}}$$ with $R_1 = (2400-1631.5)/2 = 384$ lb. A misalignment of only forty thousandths of an inch has shed 17 % of the centre-bearing load.
(c) Centre bearing on a 30 000 lb/in spring. Now the support itself deflects by $R_2/k$, so the beam and the spring must meet: $$\delta_0 - \frac{R_2 L^3}{48EI} = \frac{R_2}{k} \;\Rightarrow\; R_2\left(\frac{L^3}{48EI}+\frac{1}{k}\right)=\delta_0.$$ Substituting $1/k = 1/30\,000 = 3.3333\times10^{-5}$ in/lb, $$R_2 = \frac{0.24048}{1.2288\times10^{-4}+3.3333\times10^{-5}} = \frac{0.24048}{1.5621\times10^{-4}},$$ $$\boxed{R_2 = 1539\ \text{lb}}$$ with $R_1 = 430$ lb, and the spring compresses $R_2/k = 1539.4/30\,000 = 0.0513$ in.

The three answers form a consistent sequence, $1957 > 1632 > 1539$ lb, and the reason is the same in each case: anything that lets the centre bearing move down relieves it. A rigid, perfectly aligned bearing takes the most load; dropping it 0.040 in relieves it by 325 lb; and mounting it on a spring soft enough to sink 0.051 in relieves it by 418 lb. Vertical equilibrium $2R_1+R_2 = 2400$ lb is satisfied in all three cases.

Case	Compatibility condition	$R_2$ (lb)	$R_1$ each (lb)
(a) rigid, level bearings	net deflection at centre = 0	1957	221
(b) centre bearing 0.040 in low	net deflection at centre = 0.040 in	1632	384
(c) centre bearing on $k=30\,000$ lb/in	net deflection = $R_2/k$ = 0.0513 in	1539	430
Released-beam mid-span deflection $\delta_0$		0.24048 in
Mid-span flexibility $L^3/48EI$		1.2288 × 10⁻⁴ in/lb

Topic. Statically indeterminate beams solved by superposition (the force / flexibility method), including support settlement and elastic supports.

Key relations. Two symmetric point loads on a simple span: $\delta_{\text{mid}} = Pa(3L^2-4a^2)/24EI$; central point load: $\delta_{\text{mid}} = PL^3/48EI$; compatibility $\delta_0 - f\,R_2 = \Delta$ with $f = L^3/48EI$ and $\Delta$ the support movement; for an elastic support $\Delta = R_2/k$, giving $R_2 = \delta_0 / (f + 1/k)$.

Why this works. A one-degree indeterminate structure has exactly one more reaction than statics can supply, so one extra equation is needed and it must come from geometry. Removing the redundant support turns the problem into a determinate one whose deflection can be looked up; re-applying the unknown reaction is a second determinate problem; and since the material is linear the two superpose. The physical content of the compatibility equation is simply "the beam and its support must end up in the same place". Cases (b) and (c) change only the right-hand side of that statement — a fixed offset for a settled bearing, a load-dependent offset for a spring — which is why the same two flexibility numbers, $\delta_0$ and $f$, answer all three parts. Note also that a spring of stiffness $k$ simply adds $1/k$ to the beam's own flexibility: the two flexibilities are in series.

Common pitfall. Using the wrong sign for the settlement in part (b) — subtracting where one should add gives $R_2 = 2283$ lb, which wrongly suggests that lowering a bearing loads it more. Another frequent error is treating case (c) as a stiffness in parallel with the beam and adding $k$ to $48EI/L^3$; the support and the beam carry the same force and their deflections add, so it is the flexibilities that add, not the stiffnesses. Finally, do not forget that $EI$ is given as a single product in lb·in² — there is no need to know $E$ or $I$ separately.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed., §4–10 "Statically indeterminate problems" and Table A–9 (beam deflection formulas); Hibbeler, Mechanics of Materials, Ch. 12 (method of superposition for indeterminate beams and shafts).

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Quantity	Value
Total span, \(L\)	96 in (30 + 18 + 18 + 30)
Loads	1200 lb each, at 30 in from each end
Flexural rigidity, \(EI\)	150 × 10⁶ lb·in²
Centre bearing	at mid-span, 48 in from each end
Case (b) settlement	centre bearing 0.040 in below the end bearings
Case (c) support stiffness, \(k\)	30 000 lb/in