Question 8 of 9: Slope-deflection — symmetric two-chord frame

Question 8: Slope-deflection — symmetric two-chord frame (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 rectangular two-chord frame. The top chord runs ①–②–③–④ and the bottom chord ⑤–⑥–⑦–⑧, each in three 10 m bays; two 6 m columns join ②–⑥ and ③–⑦. All four chord ends ①, ④, ⑤ and ⑧ are built in. A uniformly distributed load of 7.8 kN/m acts on the top chord only.

Find. All member end moments, the shear-force and bending-moment diagrams, and the support reactions.

Approach. Structure and load are both symmetric about mid-span, so the response is symmetric: no sidesway, and rotations pair up as \(\theta_{3} = -\theta_{2}\), \(\theta_{7} = -\theta_{6}\). The columns are inextensible, so the two interior joints of each chord drop together by a common amount D. Three unknowns remain: \(\theta_{2}\), \(\theta_{6}\) and D.

1. Set up the kinematics. The chords are inextensible and their ends are built in, so no joint moves horizontally and every column chord rotation is zero. The vertical drop D of joints ② and ③ is shared by ⑥ and ⑦ through the columns, giving $$\psi_{12} = \psi_{56} = -\dfrac{D}{s}, \qquad \psi_{34} = \psi_{78} = +\dfrac{D}{s}, \qquad \psi_{23} = \psi_{67} = \psi_{\text{col}} = 0$$
2. Fixed-end moments. Only the top chord is loaded: $$\text{FEM} = \dfrac{ws^{2}}{12} = \dfrac{7.8 \times 100}{12} = 65.0\ \text{kN}\cdot\text{m}$$ The chord stiffness is \(2EI/s = 0.2EI\) and the column stiffness \(2EI/h = EI/3\).
3. Joint ② and joint ⑥. Writing moment equilibrium at each and using \(\theta_{3} = -\theta_{2}\), \(\theta_{7} = -\theta_{6}\), the fixed-end moments cancel out of the top-joint equation and both reduce to $$1.2667\,\theta_{2} + 0.3333\,\theta_{6} + 0.06D = 0, \qquad 1.2667\,\theta_{6} + 0.3333\,\theta_{2} + 0.06D = 0$$
4. Subtract the pair. The difference gives \(0.9333(\theta_{2}-\theta_{6}) = 0\), so $$\boxed{\theta_{2} = \theta_{6}}$$ The two chords rotate identically at their interior joints. That is a genuinely useful reduction: a three-unknown problem is now a two-unknown one, and either equation then gives \(\theta_{2} = -0.0375D\).
5. The translation equation. Give the four interior joints a virtual downward unit translation and apply virtual work, \(\sum(M_{ij}+M_{ji})\psi^{*}_{ij} + \sum F\!\cdot\!d^{*} = 0\), using the equivalent nodal loads (78 kN at each of ② and ③, not the total 234 kN): $$-0.24\,\theta_{2} - 0.048D + \dfrac{156}{EI} = 0$$ Substituting \(\theta_{2} = -0.0375D\) leaves \(0.039D = 156/EI\).
6. Solve. $$EI\,D = \boxed{4000\ \text{kN}\cdot\text{m}^{3}}, \qquad EI\,\theta_{2} = EI\,\theta_{6} = -150\ \text{kN}\cdot\text{m}^{2}$$
7. End moments — every one an integer. $$M_{12} = \boxed{275.0}, \quad M_{21} = 115.0, \quad M_{23} = 35.0, \quad M_{32} = -35.0$$ $$M_{56} = \boxed{210.0}, \quad M_{65} = 180.0, \quad M_{67} = -30.0, \quad M_{\text{col}} = -150.0 \ \text{(both ends)}$$ all in kN·m. Joint ② checks as \(115 + 35 - 150 = 0\) and joint ⑥ as \(180 - 30 - 150 = 0\).
8. Read the diagrams. Top bay ①–② runs from −275.0 kN·m at the built-in end to +115.0 kN·m at ② with the shear falling from +78.0 kN to exactly zero, so +115.0 is itself the local maximum; contraflexure is at 4.57 m. Bay ②–③ hogs −35.0 kN·m at both ends and sags \(\boxed{+62.5\ \text{kN}\cdot\text{m}}\) at mid-span, its shear running ±39.0 kN. On the unloaded bottom chord the moment is straight everywhere: −210.0 to +180.0 kN·m across ⑤–⑥ at a constant 39.0 kN shear, then a constant +30.0 kN·m with zero shear across the centre bay. Each column carries ±150.0 kN·m in double curvature with a constant shear of 50.0 kN and an axial compression of 39.0 kN.
9. Reactions and the load path. The top built-in ends take 78.0 kN and 275.0 kN·m each, the bottom ends 39.0 kN and 210.0 kN·m each, with horizontal reactions of ±16.67 kN. Two-thirds of the 234 kN goes straight out the loaded chord and one-third is carried down the columns into the unloaded chord — which is why the bottom chord is far from idle despite carrying no load of its own.