Two-Segment Rod — Uniform Heating Between Rigid Supports: 22-Mec-A7 Advanced Strength of Materials, December 2013, Q1

Question 1: Two-Segment Rod — Uniform Heating Between Rigid Supports (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 two-segment rod A–B–C built in between rigid walls, both segments heated uniformly by ΔT = 60 °C.

Given data
Rod (1), A–B	E₁ = 135 GPa, A₁ = 520 mm², L₁ = 120 mm, α₁ = 7.5×10⁻⁶/°C
Rod (2), B–C	E₂ = 85 GPa, A₂ = 950 mm², L₂ = 90 mm, α₂ = 12.5×10⁻⁶/°C
Temperature rise	ΔT = +60 °C (both rods)
Constraint	Rigid supports at A and C ⇒ total elongation = 0

Find. (a) the axial stresses σ₁, σ₂; (b) the direction and magnitude of the displacement of the welded joint B.

Series rod fixed at A and C; heating induces a common internal force N (statically indeterminate).

Approach. The two rods carry the same internal force N (series equilibrium). Because both ends are held, the free thermal growth plus the mechanical (force) deformation of the pair must sum to zero — one compatibility equation gives N, hence the stresses; the motion of B is the net stretch of rod 1.

Compatibility of the constrained pair. Free thermal expansion is cancelled by the flexibility of the two segments under the common force N: $$\alpha_1L_1\,\Delta T+\alpha_2L_2\,\Delta T+N\!\left(\frac{L_1}{A_1E_1}+\frac{L_2}{A_2E_2}\right)=0.$$
Thermal and flexibility terms. Numerically, $$\alpha_1L_1\Delta T+\alpha_2L_2\Delta T=0.0540+0.0675=0.1215\ \text{mm},$$ $$\frac{L_1}{A_1E_1}+\frac{L_2}{A_2E_2}=1.709\times10^{-6}+1.115\times10^{-6}=2.824\times10^{-6}\ \tfrac{\text{mm}}{\text{N}}.$$
Internal force. Solving, $$\boxed{N=-\frac{0.1215}{2.824\times10^{-6}}=-43.0\ \text{kN}}$$ negative ⇒ the constrained heating puts the rod pair into compression.
Axial stresses (part a). The same force divides by each area: $$\sigma_1=\frac{N}{A_1}=\frac{-43\,025}{520}=-82.7\ \text{MPa},\qquad \sigma_2=\frac{N}{A_2}=\frac{-43\,025}{950}=-45.3\ \text{MPa}.$$ Both are compressive, the smaller-area rod (1) being the more highly stressed.
Motion of joint B (part b). Measuring displacement positive toward C (right), B moves by the net elongation of rod 1: $$u_B=\alpha_1L_1\Delta T+N\frac{L_1}{A_1E_1}=0.0540-0.0736=\boxed{-0.0195\ \text{mm}}$$ The negative sign means B moves to the left by about 0.020 mm: rod 1 — the stiffer, lower-expansion segment — is compressed more than it thermally grows, so the weld is pulled back toward A. (Checking from the C side, rod 2's net change is +0.0195 mm, confirming the closure u_A=u_C=0.)

Results — Question 1
Quantity	Value
Common internal force N	−43.0 kN (compression)
Stress in rod (1), σ₁	−82.7 MPa
Stress in rod (2), σ₂	−45.3 MPa
Displacement of joint B	0.0195 mm to the left

Topic: Statically indeterminate axial members with a thermal load (a series-of-two-materials bar between rigid walls).

Key relations: equilibrium gives one internal force N common to both segments; compatibility Σ(αLΔT + NL/AE) = 0 supplies the second equation. Stress is N/A in each segment; a joint displacement is the accumulated (thermal + mechanical) elongation up to that joint.

Why this works: the structure has one redundant reaction, so a single compatibility statement — that the built-in ends cannot move — closes the problem. Superposing free thermal growth (force-free) with the deformation produced by the redundant force is the standard force method for indeterminate bars.

Common pitfall: forgetting that N is the same in both segments (they are in series, not parallel), or dropping the sign of the thermal term so that heating is mistaken for tension. A constrained temperature rise always produces compression.

Source: Hibbeler, Mechanics of Materials 10th ed., §4.4–4.6 (force method and thermal stress in axial members).

Question 1 of 7: Two-Segment Rod — Uniform Heating Between Rigid Supports

Question 1: Two-Segment Rod — Uniform Heating Between Rigid Supports (20 marks)