--- title: "Beam mid-span deflection with temperature gradient" subject: "Theory of Structures I" chapter: 3 key: "semester_4/theory_of_structures_i:beam-midspan-defl-temperature" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-midspan-defl-temperature site: BCT Engineering --- # Beam mid-span deflection with temperature gradient Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method - Asked in: 2071 Magh, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-midspan-defl-temperature - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3 ## Question Use virtual work method to determine the mid-span deflection for a simply supported steel beam of depth 300 mm carrying a superimposed udl of 20 kN/m over a span of 5 m, if the temperature of the top surface is 40°C and at bottom surface is 30°C. Assume the temperature to vary linearly over the depth of the beam. Take coefficient of thermal expansion $= 11.7\times10^{-6}/^\circ$C, $E = 210\ \text{GN/m}^2$ and moment of inertia $= 15000\ \text{cm}^4$. ## Short answer Unit load at midspan. UDL deflection = 5.167 mm down. Temperature difference 10°C over 300 mm depth yields curvature 3.9e-4 /m, causing 1.219 mm upward. Total midspan deflection = 3.95 mm downward. ## Answer #### 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The top is at 40°C and the bottom at 30°C. What does the mean temperature cause? A. Mid-span deflection B. Only axial expansion, no deflection C. Curvature D. Rotation of the supports Answer: B. A uniform rise only changes the length; deflection comes from the temperature difference across the depth. (Key point: Mean temperature) 2. What is the curvature from the temperature gradient? A. alpha x 10 / 0.3 = 3.9 x 10^-4 per m B. alpha x 10 x 0.3 C. alpha x 35 / 0.3 D. alpha x 40 / 0.3 Answer: A. kappa = alpha DeltaT / d = 11.7e-6 x 10 / 0.3 = 3.9e-4 per m. (Key point: Curvature) 3. The top is hotter than the bottom. How does the beam deflect due to temperature? A. Sideways B. No deflection for a simply supported beam C. Downward, sagging D. Upward, because the top fibres lengthen more Answer: D. The hotter top lengthens more, giving hogging curvature, so the mid-span moves upward and the term is subtracted. (Key point: Direction of thermal deflection) 4. With load deflection 5.167 mm down and thermal 1.219 mm up, what is the net mid-span deflection? A. 6.386 mm down B. 1.22 mm up C. 3.95 mm down D. 3.95 mm up Answer: C. 5.167 - 1.219 = 3.948 mm downward. The thermal part is kappa x 3.125 (area of the triangular m diagram). 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