Frame deflection at C due to temperature

Given data and assumptions

Frame fixed at A: column AB 6 m, member BC 4 m. Outer face $-20^\circ$C, inner face $+20^\circ$C, depth $d = 0.30$ m. Coefficient of expansion not given; take $\alpha = 12\times10^{-6}$/$^\circ$C (usual value for steel/concrete).

Temperature effects

• Mean temperature $= \dfrac{-20 + 20}{2} = 0$, so there is no axial deformation.
• Temperature difference across depth $= 40^\circ$C, so curvature

$$\kappa = \frac{\alpha\,\Delta T}{d} = \frac{12\times10^{-6}\times40}{0.30} = 1.6\times10^{-3}\ \text{per m}$$

The inner face is hotter (longer), so each member curves with its inner face convex.

Virtual work

$$\Delta = \int m\,\kappa\,ds$$

Horizontal deflection of C (unit horizontal load at C, $m$ is the vertical distance from C to the section: zero in BC, and $m = 6 - y$ at height $y$ above A in AB):

$$\Delta_h = \kappa\int_0^6 (6-y)\,dy = 1.6\times10^{-3}\times18 = 28.8\ \text{mm}$$

The column AB bends with its inner (right) face longer, so C moves to the left (towards the outer side).

Vertical deflection of C (unit vertical load at C, $m$ = horizontal distance from C: 4 m in all of AB, and $x$ in BC):

$$\Delta_v = \kappa\left[\int_0^6 4\,dy + \int_0^4 x\,dx\right] = 1.6\times10^{-3}\times(24 + 8) = 51.2\ \text{mm}$$

Answer: horizontal deflection of C $= 28.8$ mm to the left; vertical deflection of C $= 51.2$ mm upward.