Beam mid-span deflection with temperature gradient

Given data

Simply supported steel beam, span $L = 5$ m, depth $d = 0.3$ m, UDL $w = 20$ kN/m. Top surface $40^\circ$C, bottom $30^\circ$C, linear variation. $\alpha = 11.7\times10^{-6}$/$^\circ$C, $E = 210$ GN/m$^2 = 210\times10^6$ kN/m$^2$, $I = 15000$ cm$^4 = 1.5\times10^{-4}$ m$^4$.

$$EI = 210\times10^6\times1.5\times10^{-4} = 31\,500\ \text{kNm}^2$$

Virtual system

Unit load at mid-span: $m = \dfrac{x}{2}$ for $0\le x\le L/2$ (sagging), maximum $m = L/4 = 1.25$.

(i) Deflection due to loads

Real moment: $M = \dfrac{wLx}{2} - \dfrac{wx^2}{2}$.

$$\Delta_1 = \int_0^L\frac{Mm}{EI}dx = \frac{5wL^4}{384EI} = \frac{5\times20\times5^4}{384\times31500} = 5.167\times10^{-3}\ \text{m} = 5.167\ \text{mm}\ (\downarrow)$$

(ii) Deflection due to temperature gradient

Mean temperature rise causes only axial expansion (no deflection). Temperature difference $\Delta T = 40 - 30 = 10^\circ$C gives a curvature

$$\kappa = \frac{\alpha\,\Delta T}{d} = \frac{11.7\times10^{-6}\times10}{0.3} = 3.9\times10^{-4}\ \text{per m}$$

The top is hotter, so the top fibres lengthen more and the beam bends upward (hogging curvature). The virtual work equation gives (negative for an upward movement):

$$\Delta_2 = -\int_0^L m\,\kappa\,dx = -\kappa\times(\text{area of the } m \text{ diagram})$$

The $m$ diagram is a triangle of base 5 m and height 1.25 m, so its area is $\tfrac12\times5\times1.25 = 3.125$ m$^2$:

$$\Delta_2 = -3.9\times10^{-4}\times3.125 = -1.219\times10^{-3}\ \text{m} = -1.219\ \text{mm}\ (\uparrow)$$

Total

$$\Delta = 5.167 - 1.219 = 3.948\ \text{mm}$$

Answer: mid-span deflection $= 3.95$ mm downward (load $5.17$ mm down, temperature $1.22$ mm up).