22-Mec-A4 Design and Manufacture of Machine Elements · December 2014

Question 5 of 6: Three-bearing shaft — rigid, settled and elastic centre support

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

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator permitted. Six questions divided into Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine elements); candidates answer two from Part A and two from Part B, and all questions carry equal value (25 % each). All six questions are worked here.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology — Ch. 10–12 (casting processes and design), Ch. 1 and 3 (structure, cold work, annealing), Ch. 21–22 (machining and tool life).
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design — Ch. 3 (stress, Mohr's circle, beam deflection), Ch. 4 (statically indeterminate shafts), Ch. 16 (clutches and brakes).
R. C. Hibbeler, Mechanics of Materials — Ch. 6–9 (bending, transverse shear, combined loading, stress transformation).
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering — Ch. 7 (dislocations, strengthening, recovery and recrystallization).

it reports a single 5,200 N load and a dimension chain of 750 + 450 + 450 = 1,650 mm that "contradicts" the 2,400 mm overall. Q5 is solved against the drawing, not the caption. Likewise in Q4 the 3,000 lb load is read from the drawing as a horizontal force applied through the bolted plate on the neutral axis, and the section as a 8 in deep I-beam (¹⁄₂ + 3¹⁄₂ + 3¹⁄₂ + ¹⁄₂).

Question 5: Three-bearing shaft — rigid, settled and elastic centre support (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
Span between the end bearings	$L$	2,400 mm
Each load, from its nearer end bearing	$a$	750 mm
Load at each of the two stations	$P$	5,200 N, downward
Shaft diameter	$d$	80 mm
Young's modulus (steel)	$E$	207,000 MPa
Centre-bearing settlement, part (b)	$\Delta$	1 mm
Elastic support stiffness, part (c)	$k$	5,200 N/mm

Find. The centre-bearing reaction $R_2$ for a rigid level support, for a support set 1 mm low, and for an elastic support.

Three-bearing shaft: two 5,200 N loads, span 750 + 450 + 450 + 750 = 2,400 mm.

Approach. The shaft is statically indeterminate to the first degree. Release the centre bearing, compute the mid-span sag $\delta_0$ under the two loads, then restore compatibility by applying $R_2$ upward; the three parts differ only in the compatibility condition imposed at the centre.

Confirm the geometry closes. The dimension chain along the shaft is $750 + 450 + 450 + 750 = 2{,}400$ mm, which matches the stated overall length exactly. The two 5,200 N loads are therefore symmetric about mid-span, and so is the whole problem — the two end reactions must be equal.
Section property. For the solid 80 mm shaft, $$I = \frac{\pi d^{4}}{64} = \frac{\pi (80)^{4}}{64} = 2.0106 \times 10^{6}\ \text{mm}^4, \qquad EI = 4.162 \times 10^{11}\ \text{N}\cdot\text{mm}^2 .$$
Mid-span sag with the centre bearing released. For a simply supported beam of span $L$ carrying a single load $P$ at distance $a$ from one end (with $a \le L/2$), the deflection at mid-span is $$\delta_c = \frac{P a\,(3L^{2} - 4a^{2})}{48EI}.$$ Substituting $P = 5{,}200$ N, $a = 750$ mm, $L = 2{,}400$ mm, $$\delta_c = \frac{5{,}200(750)\left[3(2{,}400)^2 - 4(750)^2\right]}{48(4.162\times10^{11})} = 2.934\ \text{mm}.$$ The second load is the mirror image of the first about mid-span and so contributes exactly the same amount there, giving $$\delta_0 = 2 \times 2.934 = \boxed{5.868\ \text{mm}} .$$
Flexibility of the shaft at mid-span. A unit upward force applied at mid-span of the released beam lifts that point by $$f = \frac{L^{3}}{48EI} = \frac{(2{,}400)^{3}}{48(4.162\times10^{11})} = 6.920\times10^{-4}\ \text{mm/N}.$$ This single coefficient carries all three parts of the question.
General compatibility statement. The centre of the shaft must finish level with wherever the centre bearing actually sits. Writing $\Delta$ for a settlement of the bearing seat and $R_2/k$ for the compression of an elastic support, $$\delta_0 - R_2 f = \Delta + \frac{R_2}{k} \qquad \Longrightarrow \qquad \boxed{R_2 = \frac{\delta_0 - \Delta}{f + 1/k}} .$$ Each part is now a matter of setting $\Delta$ and $k$ appropriately. Note that the shaft flexibility $f$ and the support flexibility $1/k$ add, because the two act in series.
(a) All bearings rigid and level. Here $\Delta = 0$ and the support is rigid, so $1/k = 0$: $$R_2 = \frac{5.868}{6.920\times10^{-4}} = \boxed{8{,}480\ \text{N}} .$$ Vertical equilibrium then gives the end reactions, $R_1 = (2 \times 5{,}200 - 8{,}480)/2 = 960$ N each. The centre bearing carries about 82 % of the total 10,400 N load, which is characteristic of a stiff central support on a symmetric two-span shaft.
(b) Centre bearing 1 mm low. The seat is displaced downward by $\Delta = 1$ mm, so the shaft need only be pushed back the remaining 4.868 mm: $$R_2 = \frac{5.868 - 1.000}{6.920\times10^{-4}} = \boxed{7{,}036\ \text{N}} ,$$ with $R_1 = 1{,}682$ N at each end. A misalignment of one millimetre — easily produced by a machining error or by foundation settlement — sheds about 17 % of the centre-bearing load onto the end bearings.
(c) Centre bearing on an elastic support. Now $\Delta = 0$ but $1/k = 1/5{,}200 = 1.923\times10^{-4}$ mm/N, so $$R_2 = \frac{5.868}{6.920\times10^{-4} + 1.923\times10^{-4}} = \frac{5.868}{8.843\times10^{-4}} = \boxed{6{,}636\ \text{N}} ,$$ with $R_1 = 1{,}882$ N at each end. The spring compresses $R_2/k = 6{,}636/5{,}200 = 1.276$ mm, and the shaft centre therefore ends up 1.276 mm below the end-bearing line — consistent with $\delta_0 - R_2 f = 5.868 - 4.592 = 1.276$ mm, which closes the compatibility check.

The comparison is instructive. The elastic support sheds more load than the 1 mm settlement does, and the reason is visible in the last calculation: the spring deflects 1.276 mm, which is more than the 1 mm by which the seat was lowered in part (b). The general lesson is that the load taken by a redundant support is governed by the ratio of the support's flexibility to the structure's, and a support that is soft relative to the shaft simply does not attract load however carefully it is aligned.

Case	Condition at the centre bearing	$R_2$ (N)	$R_1$ each (N)
(a)	Rigid, level with the end bearings	8,480	960
(b)	Rigid, seat 1 mm low	7,036	1,682
(c)	Elastic, $k = 5{,}200$ N/mm	6,636	1,882

Topic. Statically indeterminate shafts on three bearings; compatibility, support settlement and elastic supports.

Key relations. Mid-span deflection under an off-centre load, $\delta_c = Pa(3L^2 - 4a^2)/48EI$; mid-span flexibility under a central load, $f = L^3/48EI$; compatibility with settlement and an elastic support, $R_2 = (\delta_0 - \Delta)/(f + 1/k)$; $I = \pi d^4/64$ for a solid round shaft.

Why this works. Superposition converts a redundant structure into a determinate one plus a correction. Removing the centre bearing leaves a simply supported beam whose sag is easy to write down; the redundant reaction is then whatever force is needed to restore the geometric condition at that point. Writing the condition in the general form $\delta_0 - R_2 f = \Delta + R_2/k$ handles all three parts with a single equation, because a rigid support is just the limit $k \to \infty$ and a level support is just $\Delta = 0$. The observation that flexibilities in series add is what makes the elastic-support case no harder than the rigid one.

Common pitfall. Using $PL^3/48EI$ for the sag under the off-centre loads — that formula applies only to a load at mid-span, and here the loads sit at 750 mm, so the $Pa(3L^2-4a^2)/48EI$ form is required. A second error is to add the support flexibility to the wrong side, or to treat the spring as though it were in parallel with the shaft. A third is to forget that the two symmetric loads each contribute the full $\delta_c$ at mid-span, so $\delta_0$ is twice the single-load value.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, Ch. 4.4–4.10 (deflection, superposition, statically indeterminate problems) and Table A-9 (beam deflection formulas).

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