Question 1 of 7: Thermal restraint of three bars in series

Question 1: Thermal restraint of three bars in series (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 solid circular bars are welded end to end and the resulting assembly is built into unyielding walls at both extremities, so neither end can move. The whole assembly is then heated through the same temperature rise.

Find. The axial force carried by each of the three bars once the assembly has reached the higher temperature.

Approach. Superpose the unrestrained thermal growth of the chain on the elastic response to an unknown restraint force, then impose the single compatibility statement that the two walls have not moved apart.

1. Reduce each bar to an area. The bars are solid and circular, so $$A_i=\frac{\pi D_i^{2}}{4}$$ which gives $A_1 = 7853.98\ \text{mm}^2$, $A_2 = 4417.87\ \text{mm}^2$ and $A_3 = 1963.50\ \text{mm}^2$. The heaviest bar has four times the area of the lightest, and that ratio is what will separate the stresses at the end.
2. Add up the free thermal expansion. If the right-hand wall were removed, each bar would grow by $\alpha_i L_i \Delta T$ and the growths would simply accumulate: $$\delta_T=\sum_i \alpha_i L_i \Delta T = 24\left(23.6{\times}10^{-6}(600)+18.7{\times}10^{-6}(500)+17.3{\times}10^{-6}(500)\right)$$ The three contributions are 0.3398 mm, 0.2244 mm and 0.2076 mm, so $\delta_T = 0.7718\ \text{mm}$. This is the elongation the walls must undo.
3. Write the axial flexibility of each bar. A bar of length $L$, area $A$ and modulus $E$ stretches by $FL/AE$ under a force $F$, so its flexibility is $$f_i=\frac{L_i}{A_iE_i}$$ giving $f_1 = 1.0913{\times}10^{-6}$, $f_2 = 0.9431{\times}10^{-6}$ and $f_3 = 1.3403{\times}10^{-6}\ \text{mm/N}$. Their sum is $\sum f_i = 3.3747{\times}10^{-6}\ \text{mm/N}$. Note that the stiffest material, bar 3 at 190 GPa, is the most flexible member, because its small diameter more than cancels the high modulus.
4. Impose compatibility and solve for the restraint force. The walls do not move, so the free growth must be exactly cancelled by the elastic shortening produced by a compressive force $F$ common to every bar: $$\delta_T-F\sum_i f_i=0 \qquad\Longrightarrow\qquad F=\frac{\delta_T}{\sum_i f_i}=\frac{0.7718}{3.3747{\times}10^{-6}}$$ $$\boxed{F = 228\,711\ \text{N} = 228.7\ \text{kN (compression)}}$$
5. State the force in each member. The bars are in series and no load is introduced at the joints, so equilibrium of any joint gives the same force on both sides of it. Every member therefore carries the identical 228.7 kN of compression; there is nothing further to distribute. The question asks for "force in each member", and the correct answer is that all three forces are equal — the material properties decide the magnitude, not the split.
6. Convert to stresses, which are not equal. Dividing the common force by each area, $$\sigma_i=\frac{F}{A_i}\quad\Longrightarrow\quad \sigma_1=29.12,\quad \sigma_2=51.77,\quad \sigma_3=116.48\ \text{MPa}$$ all compressive. The slender bar 3 is stressed four times harder than bar 1 and is the member that would be checked against yielding or buckling in a real design.
7. Locate the joints, as a check. Walking from the left-hand wall in the order drawn, each interface moves by the free growth of the bars behind it minus their elastic shortening: $$u_k=\sum_{i\le k}\left(\alpha_i L_i \Delta T - F f_i\right)$$ The bar 3/bar 2 joint sits at $-0.0989$ mm and the bar 2/bar 1 joint at $-0.0902$ mm, both moving towards the left wall, and the accumulated total closes on zero at the right-hand wall. That closure is the arithmetic check on the whole solve.