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Steel-aluminium bar with temperature rise: stresses

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.

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∘C25^\circ\text{C}. Find the stresses in the two bars when the temperature increases to 50∘C50^\circ\text{C} if, i) The support are unyielding ii) The supports move away from each other by 0.1 mm. [Given: Es=200 GPaE_s = 200\ \text{GPa}, EA=70 GPaE_A = 70\ \text{GPa}, αs=13×10−6/∘C\alpha_s = 13\times10^{-6}/^\circ\text{C}, αA=23.1×10−6/∘C\alpha_A = 23.1\times10^{-6}/^\circ\text{C}] [Figure: steel bar of 4 cm4\ \text{cm} diameter and length 50 cm, joined to an aluminium bar of 6 cm6\ \text{cm} diameter and length 70 cm, both fixed between rigid supports at the ends.]

The bars are joined in series between rigid supports. On heating, both expand; the supports resist this, and the same compressive force PP acts in both bars.

Data

As=π4(40)2=1256.64 mm2,Aa=π4(60)2=2827.43 mm2,ΔT=50−25=25∘CA_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

δT,s=13×10−6×25×500=0.1625 mmδT,a=23.1×10−6×25×700=0.4042 mmδT=0.5667 mm\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:

LsAsEs+LaAaEa=5001256.64×200×103+7002827.43×70×103=5.5262×10−6 mm/N\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×5.5262×10−6=0.5667⇒P=102556.7 N=102.56 kNP\times5.5262\times10^{-6} = 0.5667 \Rightarrow P = 102556.7\ \text{N} = 102.56\ \text{kN} σs=102556.71256.64=81.61 N/mm2,σa=102556.72827.43=36.27 N/mm2 (both compressive)\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: δ=0.5667−0.1=0.4667\delta = 0.5667 - 0.1 = 0.4667 mm.

P=0.46675.5262×10−6=84461.1 N=84.46 kNP = \frac{0.4667}{5.5262\times10^{-6}} = 84461.1\ \text{N} = 84.46\ \text{kN} σs=67.21 N/mm2,σa=29.87 N/mm2 (compressive)\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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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