--- title: "Steel rod inside copper tube" subject: "Strength of Materials" chapter: 2 key: "semester_4/strength_of_materials_mechanical:steel-rod-copper-tube-composite" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/steel-rod-copper-tube-composite site: BCT Engineering --- # Steel rod inside copper tube Strength of Materials · Chapter 2: Stress and strain – axial load - Asked in: year not given, 5+3 marks - Video: [COMPOSITE/COMPOUND BARS](https://www.youtube.com/watch?v=eSLLf4fi6fk) — EASY SCIENCE PREP - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/steel-rod-copper-tube-composite - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-2 ## Question A steel rod 30 mm in diameter is placed concentrically inside a copper tube of 30 mm inside and 50 mm outside diameter. Both are 600 mm long and are loaded through rigid end plates by an axial compressive load of 150 kN. Es = 200 GPa, Ec = 100 GPa. (a) Find the stress in each material, the load carried by each, and the shortening. (b) The temperature then rises by 40 degrees C. Find the final stress in each material. Take alpha for steel = 12 x 10^-6 and for copper = 17 x 10^-6 per degree C. ## Short answer From strain compatibility, load stresses: σs = 112.3 MPa, σc = 56.2 MPa (both compressive); loads Ps = 79.4 kN, Pc = 70.6 kN; shortening 0.337 mm. For ΔT = 40°C, internal thermal force 13.31 kN, final stresses: σs = –93.5 MPa, σc = –66.8 MPa. ## Answer #### Section areas $A_s = \frac{\pi}{4}(30)^2 = 706.9\ \text{mm}^2$, $A_c = \frac{\pi}{4}(50^2 - 30^2) = 1256.6\ \text{mm}^2$. #### (a) Load sharing Equilibrium: $P_s + P_c = 150\times10^3$ N. Compatibility (same shortening, same length): $\varepsilon_s = \varepsilon_c$, so $$ \frac{\sigma_s}{E_s} = \frac{\sigma_c}{E_c} \;\Rightarrow\; \sigma_c = \sigma_s\frac{E_c}{E_s} = 0.5\,\sigma_s $$ $$ \sigma_s A_s + \sigma_c A_c = 150000 \;\Rightarrow\; \sigma_s\,(706.9 + 0.5\times1256.6) = 150000 $$ $$ \sigma_s = 112.3\ \text{MPa}, \qquad \sigma_c = 56.2\ \text{MPa} $$ $$ P_s = 112.3\times706.9 = 79.4\ \text{kN}, \qquad P_c = 56.2\times1256.6 = 70.6\ \text{kN} $$ (check: $79.4 + 70.6 = 150$ kN) $$ \delta = \frac{\sigma_s L}{E_s} = \frac{112.3\times600}{200000} = 0.337\ \text{mm (shortening)} $$ #### (b) Temperature rise of 40 degrees C Copper tends to expand more than steel, but they are held together. Copper is held back (extra compression) and steel is pulled along (extra tension). The internal forces are equal and opposite, $P_t$: $$ \alpha_c\Delta T L - \frac{P_t L}{A_cE_c} = \alpha_s\Delta T L + \frac{P_t L}{A_sE_s} $$ $$ P_t = \frac{(\alpha_c - \alpha_s)\Delta T}{\dfrac{1}{A_cE_c} + \dfrac{1}{A_sE_s}} = \frac{5\times10^{-6}\times 40}{\dfrac{1}{1256.6\times10^5} + \dfrac{1}{706.9\times2\times10^5}} = 13.31\ \text{kN} $$ Thermal stresses: steel $= +13310/706.9 = +18.8$ MPa (tension), copper $= -13310/1256.6 = -10.6$ MPa (compression). Final stresses (add to the load stresses, compression negative): | Material | Load | Thermal | Final | |---|---|---|---| | Steel | $-112.3$ | $+18.8$ | $-93.5$ MPa | | Copper | $-56.2$ | $-10.6$ | $-66.8$ MPa | Check: $-93.5\times706.9 - 66.8\times1256.6 = -150$ kN. **Answer: (a) $\sigma_s = 112.3$ MPa, $\sigma_c = 56.2$ MPa (both compressive), $P_s = 79.4$ kN, $P_c = 70.6$ kN, shortening $0.337$ mm; (b) final $\sigma_s = -93.5$ MPa, $\sigma_c = -66.8$ MPa.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What compatibility condition is used to share the 150 kN between steel and copper? A. Equal stress in both materials B. Equal load in both materials C. Equal strain in both materials D. Equal area of both materials Answer: C. Both are held by rigid end plates and shorten equally, so sigma_s/E_s = sigma_c/E_c. (Key point: Compatibility condition) 2. With Es = 200 GPa and Ec = 100 GPa, what is the relation between the stresses? A. sigma_c = 2 sigma_s B. sigma_c = 0.5 sigma_s C. sigma_c = sigma_s D. sigma_c = 0.25 sigma_s Answer: B. sigma_c = sigma_s x Ec/Es = 0.5 sigma_s, so the stiffer steel carries the higher stress. (Key point: Stress ratio) 3. Why do thermal stresses arise when the temperature rises by 40 degrees C? A. Steel expands more than copper B. Copper tends to expand more than steel but they are held together, so copper is compressed and steel stretched C. Both materials expand equally so nothing happens D. The end plates absorb the expansion Answer: B. alpha_c > alpha_s: copper is held back (extra compression) and steel is pulled along (extra tension), with equal and opposite forces. (Key point: Effect of temperature rise) 4. How are the final stresses obtained in part (b)? A. Use only the thermal stresses B. Replace the load stresses by the thermal stresses C. Add the thermal stresses algebraically to the load stresses D. Multiply the load stresses by alpha dT Answer: C. Superposition: steel -112.3 + 18.8 = -93.5 MPa and copper -56.2 - 10.6 = -66.8 MPa. (Key point: Final stresses) ## Related in Strength of Materials - [Tubular column load and eccentric loading](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Atube-column-and-eccentric-short-column) — Strength of Materials, chapter 10 - [Elongation of a stepped steel bar](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Astepped-bar-axial-elongation) — Strength of Materials, chapter 2 ## Related in other subjects - [Two copper, one steel rod: 3000 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Atwo-copper-one-steel-rods-3000kg) — Strength of Materials, chapter 3, asked 2× - [Two copper, one steel rod: 5000 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Atwo-copper-one-steel-rods-5000kg) — Strength of Materials, chapter 3, asked 2× - [Hollow steel tube strut: critical, safe load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-steel-strut-both-hinged-safe) — Strength of Materials, chapter 8 - [Steel tube with asbestos: heat loss, interface temp](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Asteel-tube-asbestos-heat-loss) — Fundamental of Thermodynamics & Heat Transfer, chapter 7