Question 7 of 8: Thermal stress and joint movement in a doubly built-in two-rod assembly

Question 7: Thermal stress and joint movement in a doubly built-in two-rod assembly (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 two-segment bar built in at both ends and heated uniformly; the assembly is once statically indeterminate in the axial direction.

Find. (a) the axial stress in each rod after the temperature rise; (b) the direction and magnitude of the movement of joint B.

Approach. The supports prevent any change in overall length, so the free thermal expansion of the two rods must be exactly cancelled by the elastic contraction produced by a single internal force; that compatibility condition gives the force, and the same force applied over rod (1) alone gives the movement of B.

1. Part (a) — recognise the redundancy and choose the release. Two immovable supports and one axial degree of freedom make the bar once indeterminate. Releasing the support at C lets the bar expand freely; the redundant is the axial force $F$ that must then be applied to push the end back to C. Because no load acts between A and C, that same force runs through both rods.
2. Free thermal expansion of the released bar. Each rod lengthens by $\alpha\,\Delta T\,L$: $$\delta_T = \Delta T\bigl(\alpha_1 L_1 + \alpha_2 L_2\bigr) = 65\bigl[(10\times10^{-6})(1500) + (19\times10^{-6})(1000)\bigr]$$$$\delta_T = 65\,(0.015 + 0.019) = 65(0.034) = 2.210\ \text{mm}$$ Rod (2) contributes slightly more than rod (1) despite being two-thirds its length, because its expansion coefficient is nearly twice as large.
3. Axial flexibility of the two rods in series. $$\frac{L_1}{E_1A_1} + \frac{L_2}{E_2A_2} = \frac{1500}{160\,000(1200)} + \frac{1000}{110\,000(2500)} = 7.8125\times10^{-6} + 3.6364\times10^{-6}$$$$= 11.4489\times10^{-6}\ \text{mm/N}$$ Rod (1) supplies more than two-thirds of the flexibility, being longer and of smaller area.
4. Compatibility fixes the redundant force. The total change of length between the two rigid supports is zero: $$\delta_T + F\left(\frac{L_1}{E_1A_1}+\frac{L_2}{E_2A_2}\right) = 0 \;\Rightarrow\; F = -\frac{2.21}{0.000011449}$$$$F = \boxed{-193\,032\ \text{N}} \;=\; 193.0\ \text{kN compressive}$$ The negative sign simply records that the rods are pushing outward against supports that will not move, so the internal force is compressive, as physical sense demands for a heated restrained bar.
5. Axial stresses follow from the common force. The force is the same in both rods but the areas differ, so the stresses do not: $$\begin{aligned}\sigma_1 &= \frac{F}{A_1} = \frac{-193\,032}{1200} = \boxed{-160.86\ \text{MPa}}, \\ \sigma_2 &= \frac{F}{A_2} = \frac{-193\,032}{2500} = \boxed{-77.21\ \text{MPa}}\end{aligned}$$ Both are compressive; rod (1) is stressed slightly more than twice as heavily as rod (2), in exact inverse proportion to the areas.
6. Part (b) — movement of the joint at B. Take rod (1) alone, measuring displacement positive to the right (from A toward C). Its end at B moves by its own free expansion plus its elastic shortening under $F$: $$\delta_B = \alpha_1\Delta T L_1 + \frac{F L_1}{E_1A_1} = (10\times10^{-6})(65)(1500) + \bigl(-193\,032\bigr)\bigl(7.8125\times10^{-6}\bigr)$$$$\delta_B = 0.9750 - 1.5081 = \boxed{-0.5331\ \text{mm}}$$ so joint B moves 0.5331 mm to the left, that is back toward the support at A.
7. Check the same displacement from the other side. Working from C through rod (2) must give the identical answer with the opposite sign convention: $$\delta_B = -\left[\alpha_2\Delta T L_2 + \frac{F L_2}{E_2A_2}\right] = -\bigl[1.2350 - 0.7019\bigr] = -0.5331\ \text{mm}\ \checkmark$$ The two routes agree, which confirms both the redundant force and the flexibility arithmetic. Physically, B moves toward A because rod (2) wants to expand more than rod (1) does, and the compressive force is not enough to reverse that difference.