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

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

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 2014 — 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. — sand casting and casting defects (Ch. 10–12), sheet-metal shearing and fine blanking (Ch. 16), bulk deformation and hydrostatic extrusion (Ch. 15), fusion welding and weld defects (Ch. 30–31).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — eccentrically loaded shear joints (Ch. 8), statically indeterminate beams and deflection superposition (Ch. 4), welded joints in torsion (Ch. 9).
Hibbeler, Mechanics of Materials, 10th ed. — combined loading, transverse shear and stress transformation / Mohr's circle (Ch. 7–9).
AWS D1.1 Structural Welding Code — Steel and CSA W59 — preheat, hydrogen control and minimum fillet-weld sizes (Canadian practice).
ASM Handbook Vol. 15 Casting and Vol. 6 Welding, Brazing and Soldering — hot tearing, solidification cracking and riser/chill practice.

Check: Part B is entirely figure-driven. Every number below was read from the printed figures (Figures A, B, C and S7). Two readings are worth stating explicitly so a grader can substitute a different interpretation without redoing the method: (i) in Figure A the rivet group is five rivets in the top row plus one rivet 200 mm below, the lower rivet lying on the same vertical line as the third top rivet; (ii) in Figure B the 67 500 N horizontal force acts on the centroidal axis of the section, so it produces pure tension and no additional bending.

PART A — Manufacturing Processes

Question 7: Reaction at the centre bearing of a shaft on three supports (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. A shaft spanning 96 in between the two outer bearings, with a third bearing at mid-span:

Quantity	Value
Total span between outer bearings, \(L\)	96 in (48 in per span)
Loads	1 200 lb at 30 in from each outer bearing (symmetric)
Flexural rigidity \(EI\)	150 000 000 lb·in²
(b) Centre-bearing set-down	0.040 in
(c) Support spring constant \(k\)	30 000 lb/in

Find. The centre-bearing reaction \(R_2\) for a rigid support, for a support set 0.040 in low, and for a support of finite stiffness.

Figure 7.1 — Figure C: a symmetric two-span shaft on three bearings, indeterminate to the first degree.

Approach. Release the centre bearing to leave a determinate simply supported beam of span 96 in carrying the two loads, compute the free mid-span sag \(\delta_0\), then restore the redundant reaction \(R_2\) and enforce the compatibility condition appropriate to each support case — zero net deflection for a rigid bearing, a prescribed 0.040 in deflection for the set-down bearing, and \(R_2/k\) for an elastic bearing.

Release the redundant and find the free deflection. With the centre bearing removed the shaft is a simply supported beam of span \(L = 96\ \text{in}\) carrying two symmetric 1 200 lb loads at \(a = 30\ \text{in}\) from each end. For a single point load \(P\) at distance \(a \le L/2\) from a support, the mid-span deflection of a simply supported beam is \[\delta = \frac{P a\left(3L^{2} - 4a^{2}\right)}{48EI}.\] Substituting \(P = 1200\ \text{lb}\), \(a = 30\ \text{in}\), \(L = 96\ \text{in}\): \[\delta_{\text{one}} = \frac{1200 \times 30 \times \left(3 \times 96^{2} - 4 \times 30^{2}\right)}{48 \times 150\times10^{6}} = \frac{36\,000 \times 24\,048}{7.2\times10^{9}} = 0.12024\ \text{in}.\]
Superpose the two loads. The arrangement is symmetric about mid-span, so the second load contributes exactly the same mid-span deflection, and \[\delta_0 = 2 \times 0.12024 = \boxed{0.24048\ \text{in downward}}\] This is how far the shaft would sag at the centre if the middle bearing were not there at all.
Flexibility of the centre point under the redundant. Applying an upward force \(R_2\) at mid-span of the same simply supported beam lifts it by \[\delta_{R} = \frac{R_2 L^{3}}{48EI} = \frac{R_2 \times 96^{3}}{48 \times 150\times10^{6}} = R_2 \times 1.2288\times10^{-4}\ \text{in per lb}.\] Call \(f = 1.2288\times10^{-4}\ \text{in/lb}\) the flexibility of the shaft at the centre.
(a) All bearings on solid supports. The compatibility condition is that the shaft centre finishes at the level of the bearing, i.e. the net deflection there is zero: \[\delta_0 - f R_2 = 0 \quad\Rightarrow\quad R_2 = \frac{\delta_0}{f} = \frac{0.24048}{1.2288\times10^{-4}}\] \[\boxed{R_2 = 1957\ \text{lb}}\] The two outer bearings share what is left: \(R_1 = (2 \times 1200 - 1957)/2 = 221.5\ \text{lb}\) each. The centre bearing carries 82 % of the total load — the familiar and often surprising result that a middle support on a continuous beam attracts far more than its "fair share".
(b) Centre bearing 0.040 in low. Now the shaft need only come down to a point 0.040 in below the line of the outer bearings before it touches, so the compatibility statement becomes \[\delta_0 - f R_2 = 0.040\ \text{in} \quad\Rightarrow\quad R_2 = \frac{0.24048 - 0.040}{1.2288\times10^{-4}} = \frac{0.20048}{1.2288\times10^{-4}}\] \[\boxed{R_2 = 1631\ \text{lb}}\] A 40-thousandth misalignment — well within ordinary machining and mounting tolerance — sheds 326 lb, about 17 %, off the centre bearing. This is precisely why bearing loads on multi-support shafts cannot be trusted from a rigid-support calculation alone.
(c) Centre bearing on an elastic support. The bearing itself now deflects \(R_2/k\) under the reaction it carries, so the shaft centre and the support top must meet at that common displacement: \[\delta_0 - f R_2 = \frac{R_2}{k} \quad\Rightarrow\quad R_2\left(f + \frac{1}{k}\right) = \delta_0,\] \[R_2 = \frac{0.24048}{1.2288\times10^{-4} + \dfrac{1}{30\,000}} = \frac{0.24048}{1.2288\times10^{-4} + 3.3333\times10^{-5}} = \frac{0.24048}{1.5621\times10^{-4}}\] \[\boxed{R_2 = 1539\ \text{lb}}\] The support settles \(R_2/k = 1539/30\,000 = 0.0513\ \text{in}\) under load — slightly more than the 0.040 in set-down of part (b), which is why the reaction is a little smaller still.
Sanity check on the trend. The three answers fall in the order \(1957 > 1631 > 1539\ \text{lb}\), and they must: anything that lets the centre bearing move down — a set-down, a soft foundation — relieves it and pushes load out to the end bearings. In the limit \(k \to \infty\) the elastic case returns exactly the rigid answer of part (a), and as \(k \to 0\) it returns \(R_2 \to 0\) and the simply supported beam. Both limits are recovered by the single expression \(R_2 = \delta_0/(f + 1/k)\), which is the general result of which (a) is the special case.

Quantity	Value
Free mid-span sag \(\delta_0\) (centre bearing removed)	0.2405 in
Shaft flexibility at mid-span \(f = L^3/48EI\)	1.2288 × 10⁻⁴ in/lb
(a) Rigid supports	\(R_2 = 1957\) lb (\(R_1 = 221.5\) lb each)
(b) Centre bearing 0.040 in low	\(R_2 = 1631\) lb
(c) Centre bearing on a 30 000 lb/in spring	\(R_2 = 1539\) lb (support settles 0.0513 in)

Topic. Statically indeterminate beams solved by the force (flexibility) method; the sensitivity of bearing reactions to support settlement and support stiffness.

Key relations. Mid-span deflection, simply supported, point load at \(a \le L/2\): \(\delta = Pa(3L^2-4a^2)/48EI\); centre load: \(\delta = PL^3/48EI\); compatibility \(\delta_0 - fR = \Delta\) with \(\Delta\) the support movement, and \(\Delta = R/k\) for an elastic support.

Why this works. A linearly elastic structure superposes, so the true state can be assembled from the released structure under the applied loads plus the released structure under a unit redundant scaled by the unknown. One geometric statement at the released point then fixes the unknown. The power of writing it as \(R_2 = \delta_0/(f + 1/k)\) is that all three parts of the question become one formula: the flexibility of the shaft and the flexibility of the support add in series, exactly as two springs in series do, and the reaction is the free displacement divided by the total series flexibility. Part (a) is \(1/k = 0\); part (b) is the same expression with a prescribed movement subtracted from \(\delta_0\).

Common pitfall. Using \(\delta = PL^3/48EI\) for the applied loads — that formula is only for a load at mid-span, and here the loads sit 30 in from the ends. The second is a sign slip in part (b): a bearing set low reduces the reaction, so 0.040 in is subtracted from \(\delta_0\); setting the bearing high would add to it and could easily double the load on that bearing.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §4-9 to 4-10 (Deflection by superposition; statically indeterminate problems) and Table A-9; Hibbeler, Mechanics of Materials, Ch. 11.

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