25-Nav-A5 Ship Design · May 2016

Question 2 of 8: Restrained Two-Segment Rod — Thermal Stress

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

Notes on this paper

National Exams, May 2016 — 98-Mar-A5 Advanced Strength of Materials, 3 hours, open book, non-communicating calculator permitted (any five of the eight problems constitute a complete paper, all equal value; all eight answered below for full study coverage).

Reference texts: Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed.; Hibbeler, Mechanics of Materials, 10th ed.; Timoshenko & Goodier, Theory of Elasticity, 3rd ed.

It is solved as the strength-of-materials exam it actually is.

Question 2: Restrained Two-Segment Rod — Thermal Stress (equal value)

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. Two rods in series (1 from A to B, 2 from B to C) between rigid walls, both heated by $\Delta T=40\,{}^{\circ}\text{C}$.

Given data
Quantity	Rod (1)	Rod (2)
Modulus $E$	175 GPa	100 GPa
Area $A$	530 mm²	800 mm²
Length $L$	175 mm	110 mm
Expansion coeff. $\alpha$	$8.5\times10^{-6}/^{\circ}\text{C}$	$14\times10^{-6}/^{\circ}\text{C}$

Find. (a) $\sigma_1,\sigma_2$; (b) direction and magnitude of the displacement of joint B.

Two rods in series between rigid walls, welded at B; both heated by 40 °C (drawn to approximate length scale).

Approach. The system is internally statically indeterminate: because both ends are rigid, the free thermal elongation of the whole assembly must be cancelled exactly by an axial (compressive) force $F$, common to both rods since they are in series. Solve $F$ from a single compatibility equation, then get stresses and the displacement of B by superposing free expansion and mechanical contraction on rod (1).

Compatibility. Total free thermal growth of the two rods must equal the total mechanical contraction produced by the internal force $F$ that the rigid walls induce: $$\alpha_1\Delta T L_1+\alpha_2\Delta T L_2=F\left(\frac{L_1}{E_1A_1}+\frac{L_2}{E_2A_2}\right)$$
Solve for F. Free expansion $=8.5\times10^{-6}(40)(0.175)+14\times10^{-6}(40)(0.110)=1.210\times10^{-4}\text{ m}$; flexibility $=\dfrac{0.175}{175\times10^9\times530\times10^{-6}}+\dfrac{0.110}{100\times10^9\times800\times10^{-6}}=3.262\times10^{-9}\text{ m/N}$, so $$\boxed{F=\frac{1.210\times10^{-4}}{3.262\times10^{-9}}=37.13\text{ kN (compressive)}}$$
Part (a) — axial stresses. Uniform compressive force through the series system: $$\sigma_1=\frac{F}{A_1}=\frac{37{,}127}{530\times10^{-6}}=70.05\text{ MPa (C)}\qquad \sigma_2=\frac{F}{A_2}=\frac{37{,}127}{800\times10^{-6}}=46.41\text{ MPa (C)}$$
Part (b) — displacement of B. Superpose rod (1)'s free thermal growth and its force-induced shortening, measured from the fixed end A: $$u_B=\alpha_1\Delta T L_1-\frac{FL_1}{E_1A_1}=5.950\times10^{-5}-7.005\times10^{-5}=-1.055\times10^{-5}\text{ m}$$ $$\boxed{u_B\approx-0.0106\text{ mm (joint B moves LEFT, toward A)}}$$ Cross-checking from rod (2) (which must give $-u_B$ measured from C, since C is fixed): $\alpha_2\Delta TL_2-FL_2/(E_2A_2)=+1.055\times10^{-5}\text{ m}$, consistent.

Question 2 — final results
Quantity	Value
Internal axial force	$F=37.13\text{ kN}$ (compressive)
Stress in rod (1)	$\sigma_1=70.05\text{ MPa}$ (compression)
Stress in rod (2)	$\sigma_2=46.41\text{ MPa}$ (compression)
Displacement of joint B	0.0106 mm to the LEFT (toward A)

Topic: Thermally restrained axial members (statically indeterminate by compatibility).

Key relations: free thermal strain $\varepsilon_T=\alpha\Delta T$; combined thermo-mechanical compatibility $\sum(\alpha\Delta T L - FL/EA)=0$ for a series system between rigid walls; $\sigma=F/A$.

Why this works: Rigid supports at both ends forbid any net length change of the assembly, so whatever the rods "want" to expand thermally must be exactly cancelled by elastic shortening under a single unknown internal force common to the whole series chain; one compatibility equation is then enough to solve the one redundant unknown, after which every local quantity (stress, joint displacement) follows from ordinary mechanics of materials.

Common pitfall: assuming the joint necessarily moves toward the rod with the larger free expansion by inspection; here rod (1) is stiffer but has the smaller $\alpha$, and rod (2)'s larger free-expansion "push" toward A wins even though the mechanical contraction terms are also unequal — the sign only resolves after both terms are computed and compared.

Source: Hibbeler, Mechanics of Materials, Ch.4 (axial load, thermal stress, statically indeterminate axial members).

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