Question 7 of 9: Slope-deflection — gable frame with load and support settlement

Question 7: Slope-deflection — gable frame with load and support settlement (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 two-member gable: joint ① is built in at ground level, joint ② is the apex 5 m higher carrying a roller support, and joint ③ is a roller back at ground level. Each rafter spans 12 m horizontally, so each is 13 m long. The uniformly distributed load of 8.45 kN/m is drawn as a horizontal band across the full 24 m and is therefore per metre of horizontal projection.

Find. The end moments, reactions, and the shear-force and bending-moment diagrams of both members with their maxima and minima.

Approach. Resolve the projected load into the rafter transverse intensity, get the fixed-end moments from the projection, obtain the translations from inextensibility (there is no free sway unknown), then solve two joint equations for the two unknown rotations.

1. Reduce the projected load. A vertical load of w per horizontal metre becomes \(w\cos^{2}\alpha\) per metre normal to a rafter, with \(\cos\alpha = 12/13\): $$w_{\perp} = 8.45\left(\dfrac{12}{13}\right)^{2} = 7.20\ \text{kN/m}$$ and the fixed-end moment collapses to a projection-only formula: $$\text{FEM} = \dfrac{w_{\perp}L^{2}}{12} = \dfrac{w a^{2}}{12} = \dfrac{8.45 \times 144}{12} = \boxed{101.40\ \text{kN}\cdot\text{m}}$$ Note which length goes where: the projection sets the fixed-end moment, the true 13 m length sets the member stiffness. Interchanging them is the standard error on sloping members.
2. Fix the translations by inextensibility. Joint ① is built in, so member ①–② can only stay 13 m long if $$u_{2} = -\dfrac{h}{a}v_{2} = \dfrac{5}{12}(0.012) = \boxed{5.00\ \text{mm}}$$ and applying the same condition to the second rafter, with \(v_{3} = 0\) at its roller, gives \(u_{3} = 10.00\) mm. There is no independent sway unknown: the two rollers plus inextensibility determine every translation from the single imposed settlement.
3. Chord rotations. $$\psi_{12} = \dfrac{(0.005)(-5) + (-0.012)(12)}{13^{2}} = -1.000 \times 10^{-3}, \qquad \psi_{23} = +1.000 \times 10^{-3}$$ Equal and opposite — the frame flattens symmetrically about the apex.
4. Write the slope-deflection equations. With \(k = 2EI/L = 8200\) kN·m and \(\theta_{1} = 0\) at the built-in end, $$M_{12} = k(\theta_{2} - 3\psi_{12}) + 101.40, \qquad M_{21} = k(2\theta_{2} - 3\psi_{12}) - 101.40$$ $$M_{23} = k(2\theta_{2} + \theta_{3} - 3\psi_{23}) + 101.40, \qquad M_{32} = k(2\theta_{3} + \theta_{2} - 3\psi_{23}) - 101.40$$
5. Two equilibrium conditions. Joint ② must balance and joint ③ is a roller end, so $$M_{21} + M_{23} = 0 \;\Rightarrow\; 4\theta_{2} + \theta_{3} = 0 \;\Rightarrow\; \theta_{3} = -4\theta_{2}$$ $$M_{32} = 0 \;\Rightarrow\; 2\theta_{3} + \theta_{2} = 3\psi_{23} + \dfrac{101.40}{k}$$ The first is remarkably clean: the fixed-end moments and the chord-rotation terms cancel identically between the two members because \(\psi_{12} = -\psi_{23}\).
6. Solve. $$-7\theta_{2} = 3(0.001) + 0.0123659 \;\Rightarrow\; \theta_{2} = -2.1951 \times 10^{-3}\ \text{rad}, \qquad \theta_{3} = +8.7805 \times 10^{-3}\ \text{rad}$$
7. End moments and reactions. $$M_{12} = \boxed{108.0\ \text{kN}\cdot\text{m}}, \quad M_{21} = -112.8, \quad M_{23} = +112.8, \quad M_{32} = 0$$ $$V_{1} = 50.30\ \text{kN}, \quad V_{2} = 111.20\ \text{kN}, \quad V_{3} = 41.30\ \text{kN}$$ which sum to \(8.45 \times 24 = 202.8\) kN. Both rollers are vertical-only and the load is vertical, so the horizontal reaction at the built-in end is exactly zero — a one-line check that the whole solution passes.
8. Diagrams. Member ①–② starts at −108.0 kN·m, reaches a sagging peak of +41.71 kN·m at 6.45 m along the rafter, and ends at −112.8 kN·m at the apex, its transverse shear running from +46.43 kN to −47.17 kN. Member ②–③ starts at −112.8 kN·m, peaks at \(\boxed{+100.93\ \text{kN}\cdot\text{m}}\) at 7.71 m and closes at zero on the roller, with shear from +55.48 kN to −38.12 kN.
9. Separate the two causes. Running the load alone and the settlement alone shows how modest the settlement contribution is here: the base moment splits 86.91 kN·m from the load and 21.09 kN·m from the 12 mm drop. Quoting the split costs one extra analysis and tells the designer whether tightening the foundation tolerance would be worth anything.