Thermal stresses in a two-segment welded rod: 22-Mec-A7 Advanced Strength of Materials, May 2016, Q2

Question 2: Thermal stresses in a two-segment welded rod (equal value)

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.

Rod data
	E (GPa)	A (mm²)	L (mm)	α (10⁻⁶/°C)
Rod 1 (A–B)	175	530	175	8.5
Rod 2 (B–C)	100	800	110	14

Both ends A and C are rigidly fixed; temperature rise ΔT = +40 °C throughout.

Find. (a) axial stress in each rod; (b) direction and magnitude of the displacement of the weld B.

Two rods welded at B between rigid walls; both segments are heated by 40 °C.

Approach. With both ends fixed, the total length is constrained: the free thermal expansion is exactly annulled by an internal axial force N (common to both rods in series). Impose δ₁ + δ₂ = 0, solve for N, then get stresses and the position of B.

Compatibility of the fixed–fixed bar. Each segment’s elongation is thermal plus mechanical; their sum must vanish: $$\delta_1+\delta_2=0,\qquad \delta_i=\alpha_i\,\Delta T\,L_i+\frac{N L_i}{A_i E_i}$$
Solve for the internal force N. $$N=-\frac{\Delta T(\alpha_1 L_1+\alpha_2 L_2)}{\dfrac{L_1}{A_1E_1}+\dfrac{L_2}{A_2E_2}}=-\frac{40(1.4875+1.540)\times10^{-3}}{(1.887+1.375)\times10^{-6}}$$ $$\boxed{N=-37.13\ \text{kN}\quad(\text{compression})}$$ The negative sign confirms the constrained heating puts the bar in compression.
Axial stresses. Same force, different areas: $$\sigma_1=\frac{N}{A_1}=\frac{-37127}{530}=-70.1\ \text{MPa},\qquad \sigma_2=\frac{N}{A_2}=\frac{-37127}{800}=-46.4\ \text{MPa}$$ Both rods are in compression, the smaller rod (1) more highly stressed.
Displacement of the weld B. Measure B from the fixed support A (positive to the right): $$u_B=\delta_1=\alpha_1\Delta T L_1+\frac{N L_1}{A_1E_1}=0.0595-0.0700=-0.0106\ \text{mm}$$ $$\boxed{u_B=0.0106\ \text{mm to the LEFT}}$$ Checking from the C side, δ₂ = +0.01055 mm (net stretch of rod 2), which likewise places B 0.0106 mm left of its original position — the two routes agree.

Question 2 — results
Quantity	Value
Internal force N	37.13 kN (compression)
σ₁ (rod 1)	−70.1 MPa
σ₂ (rod 2)	−46.4 MPa
Movement of B	0.0106 mm to the left

Topic. Statically indeterminate thermal stress in bars connected in series between rigid supports.

Key relations. Segment elongation δ = αΔT·L + NL/(AE); global compatibility Σδ_i = 0; stress σ = N/A.

Why this works. The structure has one redundant (the reaction pair), so equilibrium alone cannot find N. The extra equation is geometric: rigid walls forbid any net change of the assembly length. Because the two rods carry the same axial force (series load path) but expand freely by different amounts, the mechanical strains must redistribute to erase the total thermal growth. Solving the single compatibility equation gives N, from which the individual stresses follow directly from N/A even though the areas and moduli differ. The weld position is then read off either segment’s actual (thermal + elastic) elongation.

Common pitfall. Forgetting that both rods share one N (they are in series, not parallel), or dividing the same N by the wrong area. Another is sign confusion: constrained heating always produces compression, so N and both stresses must come out negative.

Source. Gere & Goodno, Mechanics of Materials §2.5 (thermal effects); Hibbeler §4.6 (statically indeterminate axially loaded members).

Question 2 of 8: Thermal stresses in a two-segment welded rod

Question 2: Thermal stresses in a two-segment welded rod (equal value)