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Frame deflection due to temperature variation

Short answer

With outer face +20°C, inner +10°C, α=12×10⁻⁶/°C and depth 0.2 m, the free end deflects 10.8 mm to the left and 24.96 mm downward due to axial expansion and curvature.

Calculate horizontal and vertical deflection at the free end of the given frame due to temperature variation as stated in figure. The thickness of member is 20 cm and coefficient of linear expansion is 12×10−6/∘12\times10^{-6}/^\circC. [Figure: frame with fixed support A, left column AB, top member BC of 5 m, right column CD hanging 3 m below C (free end D; total frame height 3 m + 3 m); temperature marked +20°C on the outer face of the left column and +10°C inside the frame.]

Given data and assumptions

Frame fixed at A: left column AB 6 m, top member BC 5 m, right column CD hanging 3 m (D free). Take A(0,0), B(0,6), C(5,6), D(5,3). Temperature: +20∘+20^\circC on the outer face and +10∘+10^\circC on the inner face of all members (assumed uniform along each member). Depth d=0.20d = 0.20 m, α=12×10−6\alpha = 12\times10^{-6}/∘^\circC.

Principle

In each member the temperature varies linearly across the depth:

  • Mean temperature rise Tm=20+102=15∘T_m = \dfrac{20 + 10}{2} = 15^\circC. This causes an axial strain ε=αTm=1.8×10−4\varepsilon = \alpha T_m = 1.8\times10^{-4} (each element lengthens by αTm ds\alpha T_m\,ds).
  • Temperature difference ΔT=20−10=10∘\Delta T = 20 - 10 = 10^\circC. This causes a curvature κ=α ΔTd=12×10−6×100.2=6×10−4\kappa = \dfrac{\alpha\,\Delta T}{d} = \dfrac{12\times10^{-6}\times10}{0.2} = 6\times10^{-4} per m, with the hotter (outer) fibres longer.

By the virtual work (unit load) method, with virtual axial force nn and moment mm from a unit load at D:

Δ=∫n αTm ds+∫m α ΔTd ds\Delta = \int n\,\alpha T_m\,ds + \int m\,\frac{\alpha\,\Delta T}{d}\,ds

(The equivalent geometric approach: each element's curvature rotates the rest of the frame about that element, and each element's axial elongation moves D along the member.)

Evaluation (outer face hotter in every member)

Axial (mean temperature) effects on D: AB (6 m) lifts D by 1.8×10−4×6000=1.081.8\times10^{-4}\times6000 = 1.08 mm; BC (5 m) moves D to the right by 0.900.90 mm; CD (3 m) lowers D by 0.540.54 mm. Net: Δh=+0.90\Delta_h = +0.90 mm (right), Δv=+0.54\Delta_v = +0.54 mm (up).

Curvature effects: with the outer face hotter, every element rotates clockwise by κ ds\kappa\,ds. The movement of D is (horizontal = κ∫(yD−y) ds\kappa\int(y_D - y)\,ds, vertical = −κ∫(xD−x) ds-\kappa\int(x_D - x)\,ds):

MemberHorizontalVertical
ABκ∫06(3−y) dy=0\kappa\int_0^6(3-y)\,dy = 0−κ∫065 dy=−30κ-\kappa\int_0^6 5\,dy = -30\kappa
BCκ∫05(3−6) dx=−15κ\kappa\int_0^5(3-6)\,dx = -15\kappa−κ∫05(5−x) dx=−12.5κ-\kappa\int_0^5(5-x)\,dx = -12.5\kappa
CDκ∫36(3−y) dy=−4.5κ\kappa\int_3^6(3-y)\,dy = -4.5\kappa0
Sum−19.5κ=−11.70-19.5\kappa = -11.70 mm−42.5κ=−25.50-42.5\kappa = -25.50 mm

Total:

Δh=0.90−11.70=−10.80 mm,Δv=0.54−25.50=−24.96 mm\Delta_h = 0.90 - 11.70 = -10.80\ \text{mm},\qquad \Delta_v = 0.54 - 25.50 = -24.96\ \text{mm}

Answer: horizontal deflection of D =10.8= 10.8 mm towards the left; vertical deflection =24.96= 24.96 mm downward. (The results assume the 20∘^\circC face is on the outside of the frame for every member.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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