Thermal stress in a two-segment rod between rigid supports: 22-Mec-B9 Advanced Engineering Structures, December 2018, Q1

Question 1: Thermal stress in a two-segment rod between rigid 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.

Quantity	Rod (1)	Rod (2)
Modulus $E$ (GPa)	150	95
Area $A$ (mm$^2$)	2100	2800
Length $L$ (mm)	1500	1100
Expansion coefficient $\alpha$ ($10^{-6}/{}^{\circ}\text{C}$)	5	9

The temperature rise is $\Delta T = +50\,{}^{\circ}\text{C}$, uniform over both rods. Supports A and C are rigid, so the overall length of the assembly cannot change.

Find. (a) the axial stress carried by each rod, and (b) the direction and magnitude of the displacement of the welded joint B.

Question 1 — the two-segment rod restrained between rigid supports at A and C. The heavier outline on rod (2) indicates its larger cross-sectional area. The exaggerated red arrow shows the computed movement of joint B.

Approach. The assembly is a single statically indeterminate axial member: equilibrium gives one common internal force, and the rigid supports supply the one compatibility equation needed to find it.

Recognise that both rods carry the same axial force. Joint B is loaded only by the two rods, so a cut anywhere in the assembly exposes the same internal force $F$:$$F_1 = F_2 = F$$There is no external load between the supports, so $F$ is whatever value the restraint demands. Sign convention: $F$ positive in tension.
Compute the free thermal expansion. If the right-hand support were removed, the bar would simply lengthen by$$\delta_T = \Delta T\,(\alpha_1 L_1 + \alpha_2 L_2) = 50\big[(5\times10^{-6})(1500) + (9\times10^{-6})(1100)\big]\ \text{mm}$$which evaluates to $\delta_T = 50(0.0075 + 0.0099) = 0.870\ \text{mm}$. Rod (2) contributes more than rod (1) despite being shorter, because its expansion coefficient is 1.8 times larger.
Compute the axial flexibility of each rod. The mechanical part of the length change is $FL/(AE)$, so it is convenient to work with the flexibilities$$f_1 = \frac{L_1}{A_1E_1} = \frac{1500}{(2100)(150\,000)} = 4.7619\times10^{-6}\ \text{mm/N},\qquad f_2 = \frac{L_2}{A_2E_2} = \frac{1100}{(2800)(95\,000)} = 4.1353\times10^{-6}\ \text{mm/N}$$Their sum is $f_1+f_2 = 8.8972\times10^{-6}\ \text{mm/N}$. The two rods are of comparable stiffness — the ratio $k_1/k_2 = f_2/f_1 = 0.868$ — which matters in part (b).
Impose compatibility and solve for the force. The distance AC is fixed, so the two length changes must cancel:$$\delta_1 + \delta_2 = 0 \quad\Longrightarrow\quad\delta_T + F\,(f_1+f_2) = 0$$Solving for $F$,$$\boxed{F = -\frac{\delta_T}{f_1+f_2} = -\frac{0.870}{8.8972\times10^{-6}} = -97\,783\ \text{N}}$$The negative sign confirms compression: the supports resist the thermal growth by pushing back with 97.8 kN.
Convert to stresses (part a). Each rod carries the same force over its own area, so the stresses differ:$$\sigma_1 = \frac{F}{A_1} = \frac{-97\,783}{2100},\qquad\sigma_2 = \frac{F}{A_2} = \frac{-97\,783}{2800}$$which give$$\boxed{\sigma_1 = -46.56\ \text{MPa},\qquad\sigma_2 = -34.92\ \text{MPa}}$$Both are compressive, and the narrower rod (1) is the more highly stressed of the two by exactly the inverse area ratio 2800/2100 = 1.333.
Locate joint B (part b). Measuring from the fixed support A, the displacement of B is simply the total length change of rod (1):$$u_B = \alpha_1\,\Delta T\,L_1 + F f_1 = 0.375 - (97\,783)(4.7619\times10^{-6})\ \text{mm}$$The thermal growth of 0.375 mm is slightly smaller than the elastic shortening of 0.4656 mm, so$$\boxed{u_B = -0.0906\ \text{mm}\quad\text{i.e. }0.0906\ \text{mm to the LEFT}}$$
Check the answer from the other support. Working instead from C, rod (2) changes length by$$\delta_2 = \alpha_2\,\Delta T\,L_2 + F f_2 = 0.495 - 0.4044 = +0.0906\ \text{mm}$$Rod (2) therefore lengthens by 0.0906 mm, and because C cannot move, B must be pushed back towards A by that same amount. The two routes agree to every digit, and $\delta_1+\delta_2 = 0$ as required.

The result is worth a sentence of interpretation, because the sign is not obvious from inspection. Rod (2) is the hotter-growing member (larger $\alpha L$) but it is also the more flexible one, and in a series assembly the member that grows more is the one that ends up pushing the joint away from itself. Here rod (2) wins that contest, so B is driven towards A. Had the flexibilities been interchanged B would have moved to the right instead, with the same stresses.

Quantity	Value
Common internal axial force $F$	−97.78 kN (compressive)
(a) Axial stress in rod (1)	−46.56 MPa
(a) Axial stress in rod (2)	−34.92 MPa
(b) Movement of joint B	0.0906 mm, to the left

Topic. Thermal stress in a statically indeterminate axial assembly.

Key relations. Total length change of a bar, $\delta = \alpha\,\Delta T\,L + FL/(AE)$; series equilibrium, $F_1 = F_2$; compatibility for rigid end supports, $\sum \delta_i = 0$; axial stress $\sigma = F/A$.

Why this works. A temperature change is not a load: it imposes a strain, and stress appears only to the extent that the strain is prevented. Superposition makes this precise. Release one support, let the bar expand freely by $\delta_T$, then apply the force $F$ needed to pull the released end back to its original position. Because the material is linear elastic, the two effects simply add, and the single compatibility statement $\delta_1+\delta_2 = 0$ closes the problem. Note that the answer to part (a) is completely independent of which support you release, whereas part (b) is not — the displacement of B must be measured from a stated datum, and quoting it without saying “from A” is meaningless.

Common pitfall. Writing the compatibility condition as $\delta_1 = \delta_2$ rather than $\delta_1 + \delta_2 = 0$. The former is the condition for two parallel members sharing a load; here the rods are in series between two rigid walls, and it is their sum that must vanish. A second trap is to compute $u_B$ from the thermal term alone and report 0.375 mm to the right — the elastic recovery is slightly larger and reverses the sign.

Source. Hibbeler, Mechanics of Materials, 10th ed., §4.6 (statically indeterminate axially loaded members) and §4.7 (thermal stress); Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed., ch. 3.

Question 1 of 8: Thermal stress in a two-segment rod between rigid supports

Question 1: Thermal stress in a two-segment rod between rigid supports (20 marks)