Question 6 of 7: Thermal stresses in a three-bar assemblage

Question 6: Thermal stresses in a three-bar assemblage (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. Three bars in parallel between a fixed wall (node 1) and a rigid end plate (node 2), subjected to a uniform temperature drop.

Find. (a) the axial displacement of node 2; (b) the axial stress in the aluminum bar and in each brass bar.

Approach. Model each bar as a one-dimensional element sharing nodes 1 and 2, so the three elements act in parallel; assemble the single free degree of freedom, apply the thermal load vector, solve for the common displacement, and recover each stress from the mechanical part of the strain.

1. Set up the model. All three bars connect the same two nodes, so they are three elements in parallel with one active degree of freedom, $u_{2}$. Node 1 is fixed against the wall; the rigid end plate forces all three bars to share the same elongation, which is exactly the compatibility that the shared node enforces automatically.
2. Compute the element stiffnesses. For a bar element $k = AE/L$:$$k_{\text{brass}} = \frac{(5\times10^{-4})(100\times10^{9})}{2.5}= 2.0\times10^{7}\ \text{N/m}, \qquad k_{\text{alum}} = \frac{(10\times10^{-4})(70\times10^{9})}{2.5}= 2.8\times10^{7}\ \text{N/m}$$
3. Assemble the reduced stiffness. Parallel elements add:$$K_{22} = 2k_{\text{brass}} + k_{\text{alum}}= 2(2.0\times10^{7}) + 2.8\times10^{7} = 6.8\times10^{7}\ \text{N/m}$$
4. Form the thermal load vector. A temperature change produces the equivalent nodal force $\{f_{T}\} = EA\alpha\,\Delta T\,\{-1,\;+1\}^{T}$ for each bar. At node 2:$$f_{\text{brass}} = (100\times10^{9})(5\times10^{-4})(20\times10^{-6})(-20)= -20\,000\ \text{N (each bar)}$$$$f_{\text{alum}} = (70\times10^{9})(10\times10^{-4})(23\times10^{-6})(-20)= -32\,200\ \text{N}$$ so $F_{2} = 2(-20\,000) + (-32\,200) = -72\,200$ N.
5. Solve for the node-2 displacement. With one equation in one unknown:$$\boxed{\;u_{2} = \frac{F_{2}}{K_{22}} = \frac{-72\,200}{6.8\times10^{7}} = -1.0618\times10^{-3}\ \text{m} = -1.0618\ \text{mm}\;}$$ The assembly shortens by about 1.06 mm, the minus sign indicating movement back towards the wall.
6. Separate mechanical from thermal strain. The total strain is common to all three bars, $\varepsilon = u_{2}/L = -4.2471\times10^{-4}$, but only the mechanical part carries stress:$$\sigma = E\bigl(\varepsilon - \alpha\,\Delta T\bigr)$$
7. Evaluate the two stresses. For brass, $\alpha\Delta T = -4.0000\times10^{-4}$, and for aluminum $\alpha\Delta T = -4.6000\times10^{-4}$, so$$\sigma_{\text{brass}} = 100\times10^{9}\bigl(-4.2471\times10^{-4} + 4.0000\times10^{-4}\bigr) = -2.4706\ \text{MPa}$$$$\sigma_{\text{alum}} = 70\times10^{9}\bigl(-4.2471\times10^{-4} + 4.6000\times10^{-4}\bigr) = +2.4706\ \text{MPa}$$
8. Check the internal equilibrium. The end plate carries no external load, so the three bar forces must sum to zero:$$\boxed{\;2A_{\text{brass}}\sigma_{\text{brass}} + A_{\text{alum}}\sigma_{\text{alum}} = 2(-1\,235.3) + 2\,470.6 = 0\ \text{N}\;}$$

The signs deserve a sentence of interpretation, because they are the part most often reported backwards. Left free, brass would contract by $\alpha\Delta T L = -1.000$ mm and aluminum by $-1.150$ mm. The assembly settles at −1.0618 mm, between the two, so the brass bars are forced to contract more than they want and end in compression, while the aluminum bar is held back from contracting as far as it wants and ends in tension. The larger expansion coefficient always ends in tension on cooling.

Note also that no external restraint is needed to generate these stresses. Node 2 is free to move, so the assembly as a whole is unrestrained and the bar forces are entirely self-equilibrating: 1 235 N of compression in each brass bar balanced by 2 471 N of tension in the aluminum. This is the classic distinction between the fully restrained case, where the stress would be $-E\alpha\Delta T$ and reach 40 MPa in the brass and 32.2 MPa in the aluminum, and the compatibility-driven case here, where the stress is an order of magnitude smaller and depends on the stiffness ratio of the bars.