Question 6 of 9: Slope deflection — symmetric portal with a thermal elongation

Question 6: Slope deflection — symmetric portal with a thermal elongation (22 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A symmetric portal with pinned bases at joints 1 and 4, columns 6 m high and a beam 2–3 spanning 16 m. The beam carries 30 kN/m over its whole length and, in addition, its centre line grows 4 mm because of a temperature rise. All members share $EI = 6.0 \times 10^{6}$ kN·m$^{2}$ and are inextensible for the purposes of the analysis.

Find. The end moments, reactions, and the shear force and bending moment diagrams with their maximum and minimum ordinates.

Approach. Symmetry makes the sway known rather than unknown: because the bases cannot move, the beam’s 4 mm of growth must be shared equally, pushing each column top outwards by 2 mm. That leaves a single rotational unknown, so one joint equilibrium equation solves the whole frame.

1. Reduce the kinematics using symmetry. The frame, the load and the imposed elongation are all symmetric about mid-span, so $\theta_3 = -\theta_2$ and no anti-symmetric sway can develop. The bases are pinned but immovable, and the beam is inextensible apart from its prescribed growth, so $$\Delta = \frac{\delta_T}{2} = \frac{4}{2} = \boxed{2\ \text{mm outwards at each column top}}$$ This is the crucial step: the thermal effect enters as a prescribed chord rotation, not as a load, and the sway ceases to be an unknown. Only $\theta_2$ remains.
2. Write the beam equation. With $\theta_3 = -\theta_2$ and no relative transverse movement of the beam ends, $$M_{23} = \frac{2EI}{L}\left(2\theta_2 + \theta_3\right) + \frac{wL^{2}}{12} = \frac{2EI}{L}\theta_2 + \frac{wL^{2}}{12}$$ Numerically $2EI/L = 2(6.0\times10^{6})/16 = 7.5 \times 10^{5}$ and $wL^{2}/12 = 30(256)/12 = 640$ kN·m.
3. Write the column equation. Base 1 is pinned, so its moment is identically zero and the modified form applies with the chord rotation $\psi = \Delta/h$ produced by the outward movement of joint 2: $$M_{21} = \frac{3EI}{h}\left(\theta_2 - \frac{\Delta}{h}\right)$$ Numerically $3EI/h = 3(6.0\times10^{6})/6 = 3.0 \times 10^{6}$ and $\Delta/h = 0.002/6 = 3.3333 \times 10^{-4}$ rad, so the bracketed sway term alone contributes $3.0\times10^{6} \times 3.3333\times10^{-4} = 1000$ kN·m.
4. Impose equilibrium at joint 2. The joint carries only the column and the beam, so $M_{21} + M_{23} = 0$: $$3.0\times10^{6}\left(\theta_2 - 3.3333\times10^{-4}\right) + 7.5\times10^{5}\,\theta_2 + 640 = 0$$ $$3.75\times10^{6}\,\theta_2 = 1000 - 640 = 360 \qquad\Longrightarrow\qquad \theta_2 = 9.60 \times 10^{-5}\ \text{rad}$$ The rotation is small and positive; had the thermal term been absent it would have come out negative, which is the tell that the temperature rise dominates the joint.
5. Back-substitute for the end moments. $$M_{23} = 7.5\times10^{5}(9.60\times10^{-5}) + 640 = 72 + 640 = \boxed{712\ \text{kN}\cdot\text{m}}$$ and $M_{21} = -712$ kN·m, so by symmetry the beam hogs 712 kN·m at both ends and each column carries a moment rising linearly from zero at its pin to 712 kN·m at the joint.
6. Separate the two causes as a check. Re-running the same equation with $\Delta = 0$ gives $\theta_2 = -1.7067\times10^{-4}$ and $M_{23} = 512$ kN·m, i.e. a thrust of $512/6 = 85.33$ kN; running it with $w = 0$ gives 200 kN·m, i.e. 33.33 kN. The closed forms confirm both: $$H_{\text{load}} = \frac{wL^{3}}{8h^{2}+12hL} = \frac{30(4096)}{288+1152} = 85.33\ \text{kN}, \qquad H_{\text{temp}} = \frac{\delta_T\,EI}{\tfrac{2}{3}h^{3} + h^{2}L} = \frac{0.004(6.0\times10^{6})}{720} = 33.33\ \text{kN}$$ and the two add to $H = 118.67$ kN, exactly $712/6$. The thermal term contributes 28 % of the thrust here — a 4 mm movement is worth a third as much as the whole 480 kN of gravity load.
7. Complete the reactions and the diagrams. Vertical equilibrium and symmetry give $V_1 = V_4 = wL/2 = \boxed{240\ \text{kN}}$, and the horizontal thrust is $H = 118.67$ kN inwards at each pin, which is also the constant shear in each column. The beam shear runs linearly from $+240$ kN at joint 2 to $-240$ kN at joint 3, vanishing at mid-span, where $$M_{\text{mid}} = \frac{wL^{2}}{8} - Hh = \frac{30(256)}{8} - 712 = 960 - 712 = \boxed{248\ \text{kN}\cdot\text{m sagging}}$$ Setting $-712 + 240x - 15x^{2} = 0$ places the points of contraflexure at $x = 3.934$ m and $x = 12.066$ m.