Cantilever tip deflection, temperature gradient

Given data

Cantilever AB, fixed at A, length $L = 3$ m, rectangular section 5 cm $\times$ 30 cm (depth $d = 0.30$ m). Bottom temperature $200^\circ$C, top $50^\circ$C, $\alpha = 12\times10^{-6}$/$^\circ$C. The temperature varies linearly over the depth.

Temperature effect

Mean temperature $T_m = \dfrac{200 + 50}{2} = 125^\circ$C: this causes only axial elongation, no vertical displacement.

The temperature difference across the depth, $\Delta T = 200 - 50 = 150^\circ$C, gives a uniform curvature:

$$\kappa = \frac{\alpha\,\Delta T}{d} = \frac{12\times10^{-6}\times150}{0.30} = 6\times10^{-3}\ \text{per m}$$

The bottom fibres are hotter, so they lengthen more and the beam curves upward (concave up, i.e. a sagging-type curvature).

Virtual work

Unit vertical load at B: $m = x$ ($x$ from B). The unit downward load causes hogging, while the thermal curvature is sagging, so B moves opposite to the unit load (upward). In magnitude:

$$\Delta_B = \int_0^L m\,\kappa\,dx = \kappa\int_0^3 x\,dx = \kappa\,\frac{L^2}{2}$$ $$\Delta_B = 6\times10^{-3}\times\frac{3^2}{2} = 0.027\ \text{m}$$

Answer: vertical displacement of the free end $= 27$ mm upward. (The mean temperature only lengthens the beam horizontally, by $\alpha T_m L = 12\times10^{-6}\times125\times3 = 4.5$ mm if measured from 0$^\circ$C; it does not move B vertically.)