Question 5 of 9: Slope-deflection — trapezoidal frame with a settling support

Question 5: Slope-deflection — trapezoidal frame with a settling support (18 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 trapezoidal frame. Joint ① is a roller at ground level; joint ② is a pin support at the top-left corner; joint ③ is a roller support at the top-right corner; joint ④ is a roller at ground level. The inclined legs rise 3.6 m over a 4.8 m run (6.0 m long, a 3-4-5 triangle) and the top member spans 8.0 m.

Find. The end moments, reactions, shear-force and bending-moment diagrams produced by the imposed settlement alone.

Approach. With no external load the entire answer comes from kinematics: fix the imposed and inextensibility-driven displacements first, convert them to chord rotations, then write two joint-moment equations in the two unknown rotations. Both outer joints are single-member ends free to rotate, so use the modified stiffness 3EI/L there and drop their rotations from the unknowns.

1. Fix the displacement field. The pin at ② holds that joint in both directions. The top member is inextensible and horizontal, so \(u_{3} = u_{2} = 0\); the left leg is inextensible with both its ends already fixed vertically, so \(u_{1} = 0\). Only joint ④ can move, and the inextensible right leg forces $$u_{4} = \dfrac{\delta\,h}{r} = \dfrac{0.012 \times 3.6}{4.8} = \boxed{9.00\ \text{mm outward}}$$ This spreading of the base is the physical heart of the question: a 12 mm drop at the top pushes the foot out by 9 mm.
2. Convert to chord rotations. Using \(\psi_{ij} = (\mathbf{d}_{j} - \mathbf{d}_{i})\cdot\mathbf{e}_{2}/L\) with \(\mathbf{e}_{2}\) the member axis turned 90° counter-clockwise, $$\psi_{12} = 0, \qquad \psi_{23} = -\dfrac{\delta}{L_{23}} = -1.500 \times 10^{-3}, \qquad \psi_{34} = \dfrac{\delta}{r} = +2.500 \times 10^{-3}$$ The left leg is unaffected because neither of its ends moves; the right leg feels the largest rotation of the three.
3. Write the two joint equations. With \(3EI/L_{\text{leg}} = EI/2\) and \(2EI/L_{23} = EI/4\), joint ② and joint ③ give $$\theta_{2} + 0.25\theta_{3} = -1.125 \times 10^{-3}, \qquad 0.25\theta_{2} + \theta_{3} = +0.125 \times 10^{-3}$$
4. Solve. $$\theta_{2} = -1.2333 \times 10^{-3}\ \text{rad}, \qquad \theta_{3} = +0.4333 \times 10^{-3}\ \text{rad}$$ The two rotations have opposite signs, which is the frame folding about the settling corner.
5. Recover the end moments. $$M_{21} = \dfrac{3EI}{L}\theta_{2} = \boxed{-222.0\ \text{kN}\cdot\text{m}}, \qquad M_{23} = +222.0\ \text{kN}\cdot\text{m}$$ $$M_{32} = \boxed{+372.0\ \text{kN}\cdot\text{m}}, \qquad M_{34} = -372.0\ \text{kN}\cdot\text{m}$$ with \(M_{12} = M_{43} = 0\) at the two roller ends. Each joint balances exactly, which is the only free check available on a load-free problem.
6. Shears from the end moments. No member carries a span load, so each shear is constant and equal to \((M_{ij} + M_{ji})/L\): $$V_{12} = \dfrac{222.0}{6} = 37.00\ \text{kN}, \quad V_{23} = \dfrac{222.0 + 372.0}{8} = 74.25\ \text{kN}, \quad V_{34} = \dfrac{372.0}{6} = 62.00\ \text{kN}$$
7. Reactions, and why they must sum to zero. $$R_{1} = -46.25\ \text{kN}, \quad R_{2} = +120.50\ \text{kN}, \quad R_{3} = -151.75\ \text{kN}, \quad R_{4} = +77.50\ \text{kN}$$ There is no applied load, so this set is self-equilibrating: the four values sum to zero and their moments about any point sum to zero. Two of the four are hold-downs — a settlement of a redundant support does not push down on that support, it drags the whole frame with it, and the anchorage has to be designed for the reversal.