Two-Segment Rod — Thermal Stress in a Statically Indeterminate Bar: 22-Mec-A7 Advanced Strength of Materials, May 2013, Q2

Question 2: Two-Segment Rod — Thermal Stress in a Statically Indeterminate Bar (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. A composite rod A–B–C built in between two rigid walls, heated uniformly by ΔT = 50 °C.

Given data
Rod (1) A–B	E₁ = 180 GPa, A₁ = 550 mm², L₁ = 150 mm, α₁ = 8×10⁻⁶/°C
Rod (2) B–C	E₂ = 90 GPa, A₂ = 900 mm², L₂ = 100 mm, α₂ = 17×10⁻⁶/°C
Temperature rise	ΔT = +50 °C

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

Two welded rods restrained between rigid walls; the same internal force N passes through both segments.

Approach. The rigid walls prevent any net length change, so a single internal force N (constant through the series) develops such that the combined mechanical + thermal elongation is zero. Solve the compatibility equation for N, divide by area for stress, then integrate the elongation of segment (1) to locate B.

Compatibility. Zero total change in length between the fixed walls: $$\delta_{tot}=\Big(\tfrac{NL_1}{A_1E_1}+\alpha_1L_1\Delta T\Big)+\Big(\tfrac{NL_2}{A_2E_2}+\alpha_2L_2\Delta T\Big)=0.$$
Solve for the internal force. Rearranging, $$N=-\frac{\Delta T\,(\alpha_1L_1+\alpha_2L_2)}{\dfrac{L_1}{A_1E_1}+\dfrac{L_2}{A_2E_2}}=-\frac{50(1.20+1.70)\times10^{-3}}{(1.515+1.235)\times10^{-6}}.$$ $$\boxed{N=-52.73\ \text{kN}\quad(\text{compression})}$$
Axial stresses. Same force, different areas: $$\sigma_1=\frac{N}{A_1}=\frac{-52\,733}{550}=-95.9\ \text{MPa},\qquad \sigma_2=\frac{N}{A_2}=\frac{-52\,733}{900}=-58.6\ \text{MPa}.$$ Both rods are in compression, as expected for restrained heating.
Displacement of joint B. B moves by the actual elongation of segment (1) measured from wall A: $$u_B=\frac{NL_1}{A_1E_1}+\alpha_1L_1\Delta T=(-0.0799)+(0.0600)=-0.0199\ \text{mm}.$$ $$\boxed{u_B \approx 0.020\ \text{mm to the left (toward A)}}$$ The check from the C–side, $-(NL_2/A_2E_2+\alpha_2L_2\Delta T)=-0.0199$ mm, agrees.

Question 2 — results
Quantity	Value
Internal force N	52.73 kN (compression)
Stress σ₁	−95.9 MPa
Stress σ₂	−58.6 MPa
Movement of B	0.020 mm to the left

Topic: Thermal stress in a statically indeterminate axially loaded member.

Key relations: segment change of length $\delta=NL/(AE)+\alpha L\Delta T$; series force continuity $N_1=N_2=N$; fixed-end compatibility $\sum\delta=0$.

Why this works: With no transverse load at B, force equilibrium of the free joint forces the same axial force through both segments. The problem is one-degree indeterminate: statics alone cannot find N, so the geometric constraint (the walls do not move) supplies the extra equation. Because the softer, higher-expansion rod (2) wants to grow more, the wall reaction is compressive; the stiffer, higher-EA rod (1) attracts a larger stress magnitude for the same force. B drifts slightly toward A because rod (2)’s larger free thermal growth pushes the weld back.

Common pitfall: using different forces in the two segments, or forgetting that the free (unrestrained) thermal term $\alpha L\Delta T$ must be added to the mechanical term inside the compatibility equation.

Source: Hibbeler, Mechanics of Materials, §4.6–4.7; Ugural & Fenster, Advanced Strength and Applied Elasticity, Ch. 2.

Question 2 of 8: Two-Segment Rod — Thermal Stress in a Statically Indeterminate Bar

Question 2: Two-Segment Rod — Thermal Stress in a Statically Indeterminate Bar (20 marks)