25-Nav-A6 Advanced Strength of Materials (25-Mec-A6) · May 2016

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

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

Notes on this paper

Reference texts: Kalpakjian & Schmid, Manufacturing Engineering and Technology, 8th ed.; Groover, Fundamentals of Modern Manufacturing, 7th ed.; Hibbeler, Mechanics of Materials, 10th ed.; Hibbeler, Engineering Mechanics: Statics, 14th ed.; Shigley's Mechanical Engineering Design, 11th ed.

Check: the exam is headed "98-Mar-A6, Design and Manufacture of Machine Elements" and Part A (Q1–Q3) is entirely machining/casting/forming content with zero naval-architecture material. Solved here as the paper actually printed; reference texts above are chosen for the real content. Part B (Q4–Q6) is genuine strength-of-materials/machine-design content.

Check: Q5 states the shaft modulus as "E = 30 ksi," which is off by three orders of magnitude for steel (actual E ≈ 29,000–30,000 ksi = 30×10⁶ psi); the shaft's computed self-weight, w = γA = 0.283 × (π/4)(3²) = 2.00 lb/in, comes out to a clean round number confirming the 3-in-diameter reading, and the printed "30 ksi" is treated as a dropped exponent (E = 30×10⁶ psi used throughout).

Question 5: Continuous Shaft on Three Bearings — Reactions and Bending Moment Diagram (equal value, Part B)

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. Continuous 3-in-dia. shaft on three simple (bearing) supports at A ($x=0$), C ($x=60$ in), E ($x=120$ in); 1,000 lb loads applied at B ($x=36$ in) and D ($x=84$ in); shaft self-weight included; $E=30\times10^6$ psi (see check note above), $\gamma=0.283$ lb/in³.

Given data
Quantity	Value
Span layout (A–B–C–D–E)	36 in, 24 in, 24 in, 36 in (total 120 in)
Point loads at B, D	1,000 lb each (downward)
Shaft diameter	3 in (continuous)
Modulus $E$	$30\times10^{6}$ psi
Unit weight $\gamma$	0.283 lb/in³

Find. (a) Reactions $R_A,R_C,R_E$ and the dimensioned bending moment diagram, with all bearings level. (b) The reactions if bearing C settles 1/8 in lower than A and E.

Continuous shaft on three bearings with point loads at B, D and distributed self-weight; bending moment diagram for part (a) below (sagging positive, hogging negative).

Approach. The shaft is a two-span continuous beam (spans A–C and C–E) on three simple supports — one degree statically indeterminate. Take $R_C$ as the redundant and solve by direct stiffness (2-node beam elements, $v,\theta$ per node, standard Euler–Bernoulli stiffness and work-equivalent nodal loads for the self-weight UDL), which handles the point loads, the distributed dead load, and the part-(b) support settlement in one consistent model; reactions and the moment diagram then follow from ordinary statics once $R_A,R_C,R_E$ are known.

Section properties and self-weight. $I=\dfrac{\pi d^4}{64}=\dfrac{\pi(3)^4}{64}=3.976\text{ in}^4$; $A_x=\dfrac{\pi d^2}{4}=7.069\text{ in}^2$; self-weight $w=\gamma A_x=0.283(7.069)=2.00\text{ lb/in}$ (a clean number, confirming the 3-in-diameter reading). $EI=30\times10^6(3.976)=1.193\times10^8\text{ lb-in}^2$.
Part (a) — solve the indeterminate reactions. Model four beam elements (A–B, B–C, C–D, D–E) with $v=0$ enforced at A, C, E; apply $-1{,}000$ lb at B and D plus the distributed self-weight on every element; solving $[K]\{d\}=\{F\}$ for the free rotations and back-substituting gives the support reactions (verified independently against the classical two-equal-span UDL-only case, $R_{end}=3wL/8$, $R_{mid}=5wL/4$, before applying it here): $$\boxed{R_A=253.0\text{ lb}\qquad R_C=1{,}734.0\text{ lb}\qquad R_E=253.0\text{ lb}}$$ Check: $R_A+R_C+R_E=2{,}240.0\text{ lb} = 2(1{,}000)+w(120)=2{,}240.0\text{ lb}$ ✓, and $R_A=R_E$ as required by the left–right symmetry of the loading.
Bending moment diagram (part a). With $R_A,R_C$ known, $M(x)$ follows by summing moments of everything to the left of $x$ (self-weight contributes a small parabolic term; the point-load jumps dominate): $$M(x)=R_Ax-\frac{wx^2}{2}-1{,}000\langle x-36\rangle+R_C\langle x-60\rangle-1{,}000\langle x-84\rangle$$ Evaluating at the key sections: $$M_A=0,\quad M_B=+7{,}812\text{ lb-in (sagging)},\quad M_C=-12{,}420\text{ lb-in (hogging)},\quad M_D=+7{,}812\text{ lb-in},\quad M_E=0$$ The diagram is symmetric about C, positive (sagging) under each load point and negative (hogging) over the middle bearing — the signature shape of a continuous beam.
Part (b) — bearing C settles 1/8 in. Re-solving the same stiffness model with the prescribed displacement $v_C=-0.125\text{ in}$ (instead of $v_C=0$) redistributes load away from C and onto A and E: $$\boxed{R_A=460.1\text{ lb}\qquad R_C=1{,}319.9\text{ lb}\qquad R_E=460.1\text{ lb}}$$ Check: $460.1+1{,}319.9+460.1=2{,}240.1\text{ lb}$ ✓ (sum of reactions is settlement-independent, since it is fixed by overall vertical equilibrium). $R_C$ drops by about 24% and $R_A=R_E$ each rise by about 82% relative to part (a), exactly as expected: settling C makes it a "softer" support, so more of the load is thrown onto the end bearings, and the hogging moment at C nearly vanishes ($M_C\approx+5\text{ lb-in}\approx0$) while the sagging moments at B and D roughly double (to about 15,270 lb-in), since each span now behaves closer to an independent simply supported span.

Question 5 — final results
Quantity	Part (a), level bearings	Part (b), C settled 1/8 in
$R_A$	253.0 lb	460.1 lb
$R_C$	1,734.0 lb	1,319.9 lb
$R_E$	253.0 lb	460.1 lb
$M$ at B, D	+7,812 lb-in	+15,267 lb-in
$M$ at C	−12,420 lb-in	≈ 0 lb-in

Topic: Statically indeterminate continuous beams (shafts on multiple bearings), and the effect of support settlement.

Key relations: degree of indeterminacy = (number of reaction components) − (equilibrium equations) = 1 for a 3-support beam with no axial redundancy; direct-stiffness beam element $[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix}$; work-equivalent UDL nodal loads $\{-wL/2,-wL^2/12,-wL/2,+wL^2/12\}$; a prescribed support settlement enters as a non-zero forced displacement at that degree of freedom, exactly like the three-moment theorem's settlement term.

Why this works: A continuous beam cannot be solved by statics alone because there is one more unknown reaction than equilibrium equations provide; the direct-stiffness method restores enough equations by additionally enforcing the beam's own flexural stiffness (compatibility of slope and deflection at every node), which is exactly equivalent to the classical three-moment theorem but far less error-prone for a general load case such as a UDL combined with point loads and a support settlement.

Common pitfall: assuming the middle bearing always carries the largest share of load regardless of geometry, and forgetting that a settled (lower) support does not just shed load — it also nearly eliminates the hogging moment that was concentrating stress there, while roughly doubling the sagging moment under the point loads.

Source: Hibbeler, Mechanics of Materials, Ch.11 (deflection, indeterminate beams); Logan, A First Course in the Finite Element Method, Ch.4 (beam element stiffness and equivalent nodal loads).

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