Slope-deflection — trapezoidal frame with a settling support: 07-Str-A4, December 2018, Q5

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.

Given data
Quantity	Symbol	Value
Leg length (rise 3.6 m, run 4.8 m)	L₁₂, L₃₄	6.0 m
Top member	L₂₃	8.0 m
Imposed settlement of joint ③	δ	12.0 mm downward
Flexural rigidity, all members	EI	3.6 × 10⁵ kN·m²
Applied loads	—	none

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

Question 5 — roller, pin, roller, roller. Five reaction components against three equations makes the frame two degrees indeterminate, which is exactly why an imposed settlement stresses it.

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.

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.
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.
Write the two joint equations. With $3EI/L_{\text{leg}} = EI/2$ and $2EI/L_{23} = EI/4$, joint ② and joint ③ give $$\theta_{2} + 0.25\theta_{3} = -1.125 \times 10^{-3}, \qquad 0.25\theta_{2} + \theta_{3} = +0.125 \times 10^{-3}$$
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.
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.
Shears from the end moments. No member carries a span load, so each shear is constant and equal to $(M_{ij} + M_{ji})/L$: $$V_{12} = \dfrac{222.0}{6} = 37.00\ \text{kN}, \quad V_{23} = \dfrac{222.0 + 372.0}{8} = 74.25\ \text{kN}, \quad V_{34} = \dfrac{372.0}{6} = 62.00\ \text{kN}$$
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.

Question 5 bending moment. Every member is straight (no span loads) and the diagram closes to zero at the two roller ends.

Question 5 — results
Quantity	Value
Outward movement of joint ④	9.00 mm
θ₂	−1.2333 × 10⁻³ rad
θ₃	+0.4333 × 10⁻³ rad
Moment at joint ②	222.0 kN·m (both members)
Moment at joint ③	372.0 kN·m (both members) — maximum
Moment at joints ① and ④	0 (roller ends)
Shear, leg ①–②	37.00 kN
Shear, top member	74.25 kN (maximum)
Shear, leg ③–④	62.00 kN
Reactions ① / ② / ③ / ④	−46.25 / +120.50 / −151.75 / +77.50 kN

Topic. Slope-deflection under a prescribed support movement, with no applied load.

Key relations. $M_{ij} = \frac{2EI}{L}(2\theta_{i} + \theta_{j} - 3\psi_{ij}) + \text{FEM}_{ij}$; modified stiffness $3EI/L$ for a member released at its far end; chord rotation $\psi_{ij} = (\mathbf{d}_{j}-\mathbf{d}_{i})\cdot\mathbf{e}_{2}/L$; inextensible-leg kinematics $u = \delta h/r$; member shear $V = (M_{ij}+M_{ji})/L$.

Why this works. A load-free indeterminate structure has no “primary” force field at all — everything is driven by geometry, so the displacement bookkeeping is the analysis. Inextensibility is what converts a single prescribed drop into a complete displacement field: the top member forbids horizontal movement of the two upper joints, and the right leg then converts the vertical drop into a horizontal spread at the foot. Because the whole answer scales linearly with EI, a stiff frame is punished for settlement in a way a flexible one is not, and 372 kN·m out of a 12 mm movement is not an arithmetic error — it is why differential settlement governs so many serviceability checks.

Common pitfall. Inverting the sign of a chord rotation. The mechanical check is that the arithmetic must give hold-down reactions at two supports and uplift at the others, summing to zero; if the four reactions do not sum to zero, or if a member comes out with a moment at a roller end, the sign convention has been mixed. The other trap is forgetting that the outer joints are free to rotate, and using $4EI/L$ instead of the modified $3EI/L$, which inflates every moment.

Source. Hibbeler, Structural Analysis, 10th ed., §11.3–11.4; Kassimali, Structural Analysis, 6th ed., §13.3–13.4; Ghali, Neville and Brown, Structural Analysis, 7th ed., Ch. 11.

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

Question 5: Slope-deflection — trapezoidal frame with a settling support (18 marks)