Thermal stresses in a restrained two-segment rod: 22-Mec-B9 Advanced Engineering Structures, May 2018, Q1

Question 1: Thermal stresses in a restrained two-segment rod (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. Two uniform linearly elastic rods are welded at B and built in to rigid supports at A and C, so the assembly cannot change its overall length. The temperature of both rods is raised by $\Delta T = 60\,{}^{\circ}\text{C}$.

Quantity	Rod (1)	Rod (2)
Modulus E	150 GPa	95 GPa (see the note above)
Area A	2000 mm²	2600 mm²
Length L	1800 mm	1000 mm
Expansion coefficient α	8 × 10⁻⁶ /°C	9 × 10⁻⁶ /°C

Find. (a) the axial stress carried by each rod after the 60 °C rise, and (b) the direction and magnitude of the movement of the welded joint B.

Question 1 — the two-segment rod between rigid supports at A and C. Rod (1) is the longer, softer, slimmer segment; the weld at B is the joint whose movement part (b) asks for.

Approach. The structure is statically indeterminate to the first degree: release the support at C, let the bar grow freely by $\alpha \Delta T L$, then apply the single unknown axial force F that pulls the free end back to its original position, and read the movement of B off whichever segment is convenient.

Free thermal growth of each segment. With the support at C removed, each rod lengthens by $\delta_T = \alpha\,\Delta T\,L$: $$\delta_{T1} = (8\times10^{-6})(60)(1800) = 0.864\ \text{mm}, \qquad \delta_{T2} = (9\times10^{-6})(60)(1000) = 0.540\ \text{mm}$$ so the released bar would grow by 1.404 mm in total.
Flexibility of the two rods in series. Because the same axial force runs through both segments, their flexibilities add: $$f = \frac{L_1}{A_1E_1} + \frac{L_2}{A_2E_2} = \frac{1800}{(2000)(150\,000)} + \frac{1000}{(2600)(95\,000)} = 6.000\times10^{-6} + 4.049\times10^{-6} = 1.0049\times10^{-5}\ \text{mm/N}$$ Working consistently in N and mm keeps every modulus in MPa.
Compatibility gives the redundant force. The rigid supports demand zero net change of length, so the mechanical shortening must cancel the thermal growth: $$\delta_{T1} + \delta_{T2} + F f = 0 \quad\Longrightarrow\quad F = -\frac{1.404}{1.0049\times10^{-5}} = \boxed{-139.7\ \text{kN}}$$ The minus sign says the bar is in compression, which is what a restrained expansion must produce.
Axial stresses (part a). One force, two areas: $$\sigma_1 = \frac{F}{A_1} = \frac{-139\,721}{2000} = -69.9\ \text{MPa}, \qquad \sigma_2 = \frac{F}{A_2} = \frac{-139\,721}{2600} = -53.7\ \text{MPa}$$ Both rods are in compression; the slimmer rod (1) is the more highly stressed, and neither stress is anywhere near yield for a structural alloy.
Movement of joint B (part b). Track segment (1) from the fixed support at A: it grows thermally and is squashed elastically, and the movement of B is what survives: $$u_B = \delta_{T1} + \frac{F L_1}{A_1E_1} = 0.864 + (-139\,721)(6.000\times10^{-6}) = 0.864 - 0.838 = \boxed{+0.026\ \text{mm}}$$ The result is positive, so B moves 0.026 mm to the right, that is toward C.
Independent check from the other end. Walking back from C gives $u_B = -(\delta_{T2} + FL_2/A_2E_2) = -(0.540 - 0.566) = +0.026\ \text{mm}$, the same number from different data — the arithmetic and the sign convention are both sound.

The movement is tiny because the two segments are so nearly balanced: rod (1) wants to grow 0.864 mm and is pushed back 0.838 mm, so only 26 microns of net motion survive. That near-cancellation is exactly why the joint position is a sensitive question and the modulus ambiguity matters — taking the printed 95,000 GPa literally makes rod (2) effectively rigid, the force rises to 234 kN, and B is dragged 0.54 mm the other way.

Quantity	Worked value (E₂ = 95 GPa)	Printed value taken literally (E₂ = 95,000 GPa)
Axial force in both rods	139.7 kN compression	233.8 kN compression
Stress in rod (1), σ₁	−69.9 MPa	−116.9 MPa
Stress in rod (2), σ₂	−53.7 MPa	−89.9 MPa
Movement of joint B	0.026 mm to the right	0.539 mm to the left

Topic. Statically indeterminate axial members under a uniform temperature change.

Key relations. Free thermal growth $\delta_T=\alpha\,\Delta T\,L$; axial flexibility $f=L/AE$; compatibility $\sum\delta_T + Ff = 0$ for a fully restrained bar; $\sigma = F/A$.

Why this works. A temperature rise produces no stress at all in an unrestrained bar — stress appears only when a support refuses the free expansion. Removing one support turns the problem into a determinate one whose single unknown, the redundant force, is fixed by the geometric statement that the overall length cannot change. Because that force is common to both segments, the two flexibilities simply add, and the internal joint then sits wherever the thermal growth and the elastic squash of the first segment leave it. The answer for B is a small difference between two much larger numbers, which is the usual signature of a restrained-expansion problem.

Common pitfall. Adding the two stresses as though each rod carried its own force, or assuming that the stiffer rod must carry more load. Equilibrium at the weld admits only one force; the areas differ, so the stresses differ but the force does not. A second trap is forgetting that B can move even when A and C cannot.

Source. Hibbeler, Mechanics of Materials, 10th ed., §4.4–4.6 (statically indeterminate axially loaded members and thermal stress); Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed., ch. 3.

Question 1 of 8: Thermal stresses in a restrained two-segment rod

Question 1: Thermal stresses in a restrained two-segment rod (20 marks)