--- title: "Steel-aluminium bar with temperature rise: stresses" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:steel-aluminium-bar-temperature-supports" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-aluminium-bar-temperature-supports site: BCT Engineering --- # Steel-aluminium bar with temperature rise: stresses Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2076 Asoj, 10 marks - Video: [Thermal STRESSES and STRAINS Induced in a Composite Bar.](https://www.youtube.com/watch?v=TDMduOLMOaY) — George Kamiti - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-aluminium-bar-temperature-supports - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question A composite bar made up of steel and aluminum is rigidly fixed between two supports as shown in figure. The two bars are free of stress at initial temperature of $25^\circ\text{C}$. Find the stresses in the two bars when the temperature increases to $50^\circ\text{C}$ if, i) The support are unyielding ii) The supports move away from each other by 0.1 mm. [Given: $E_s = 200\ \text{GPa}$, $E_A = 70\ \text{GPa}$, $\alpha_s = 13\times10^{-6}/^\circ\text{C}$, $\alpha_A = 23.1\times10^{-6}/^\circ\text{C}$] [Figure: steel bar of $4\ \text{cm}$ diameter and length 50 cm, joined to an aluminium bar of $6\ \text{cm}$ diameter and length 70 cm, both fixed between rigid supports at the ends.] ## Short answer The compressive force is found by equating prevented thermal expansion to elastic shortening. For unyielding supports, stresses are 81.61 MPa in steel and 36.27 MPa in aluminium. When supports move 0.1 mm apart, stresses reduce to 67.21 MPa and 29.87 MPa, all compressive. ## Answer The bars are joined in series between rigid supports. On heating, both expand; the supports resist this, and the same compressive force $P$ acts in both bars. #### Data $$ A_s = \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2,\qquad A_a = \frac{\pi}{4}(60)^2 = 2827.43\ \text{mm}^2,\qquad \Delta T = 50-25 = 25^\circ\text{C} $$ #### Free thermal expansion $$ \begin{aligned} \delta_{T,s} &= 13\times10^{-6}\times25\times500 = 0.1625\ \text{mm}\\ \delta_{T,a} &= 23.1\times10^{-6}\times25\times700 = 0.4042\ \text{mm}\\ \delta_T &= 0.5667\ \text{mm} \end{aligned} $$ Flexibility of the two bars in series: $$ \frac{L_s}{A_sE_s}+\frac{L_a}{A_aE_a} = \frac{500}{1256.64\times200\times10^3}+\frac{700}{2827.43\times70\times10^3} = 5.5262\times10^{-6}\ \text{mm/N} $$ #### (i) Unyielding supports $$ P\times5.5262\times10^{-6} = 0.5667 \Rightarrow P = 102556.7\ \text{N} = 102.56\ \text{kN} $$ $$ \sigma_s = \frac{102556.7}{1256.64} = 81.61\ \text{N/mm}^2,\qquad \sigma_a = \frac{102556.7}{2827.43} = 36.27\ \text{N/mm}^2\ \text{(both compressive)} $$ #### (ii) Supports move apart by 0.1 mm Only the expansion not taken up by the movement of the supports is prevented: $\delta = 0.5667 - 0.1 = 0.4667$ mm. $$ P = \frac{0.4667}{5.5262\times10^{-6}} = 84461.1\ \text{N} = 84.46\ \text{kN} $$ $$ \sigma_s = 67.21\ \text{N/mm}^2,\qquad \sigma_a = 29.87\ \text{N/mm}^2\ \text{(compressive)} $$ **Answer: (i) steel 81.61 N/mm², aluminium 36.27 N/mm²; (ii) steel 67.21 N/mm², aluminium 29.87 N/mm² (all compressive).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens when the bars are heated between rigid supports? A. Expansion is prevented, so the same compressive force P acts in both bars B. Tensile force develops in both C. Steel is in tension and aluminium in compression D. No stress develops Answer: A. Series bars carry the same force; blocked expansion makes it compressive. (Key point: Thermal force) 2. What is the total free thermal expansion? A. alpha dT only for aluminium B. alpha_s L_s dT + alpha_a L_a dT = 0.5667 mm C. (alpha_s + alpha_a) L dT / 2 D. 0.1625 mm Answer: B. 0.1625 (steel) + 0.4042 (aluminium) = 0.5667 mm. (Key point: Free expansion) 3. What equation gives P for unyielding supports? A. P = E alpha dT A for steel only B. P = sum of E alpha dT A C. P (L_s/A_sE_s + L_a/A_aE_a) = total free expansion D. P x flexibility = 0 Answer: C. The forced compression must exactly cancel the free expansion. (Key point: Compatibility, unyielding) 4. How does support movement of 0.1 mm change the analysis? A. Only 0.5667 - 0.1 = 0.4667 mm of expansion is resisted B. 0.5667 + 0.1 mm is resisted C. The stresses become zero D. The stresses double Answer: A. Part of the expansion is accommodated by the support movement, so stresses fall to 67.21 and 29.87 N/mm^2. 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