Question 6 of 8: Sway Frame with an Overhang by the Slope-Deflection Method

Question 6: Sway Frame with an Overhang by the Slope-Deflection Method (21 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.

Find. All member end moments, the sway of the frame, the support reactions, and the shear force and bending moment diagrams with their extreme ordinates.

Approach. Reduce the determinate overhang to a known joint moment and shear, treat the roller end with modified stiffness, and take the two unknowns as \(\theta_2\) and the sway \(\Delta\). The second equation is the horizontal shear condition on the single column, which because no horizontal load is applied reads simply “column shear = 0”.

1. Dispose of the overhang. The 2 m cantilever beyond joint 2 is determinate. It hands the joint$$M_{23} = -\frac{w a^{2}}{2} = -\frac{2(2)^{2}}{2} = -4.00\ \text{kN}\cdot\text{m},\qquad V_{2^+} = +wa = +4.00\ \text{kN}$$and adds no degree of freedom to the problem.
2. Count the unknowns. Joint 1 is a roller at a beam end, so \(M_{12} = 0\) and \(\theta_1\) can be eliminated with the modified stiffness \(3EI/L\). Joint 4 is built in. Vertical inextensibility of the column fixes \(v_2 = 0\), and there is no vertical restraint to fight, so the unknowns are$$\theta_2 \quad\text{and}\quad \Delta \ (\text{the horizontal sway of the beam line}),\qquad \psi_{24} = \Delta/4$$
3. Slope-deflection equations. With the fixed-end moment of a propped span, \(wL^{2}/8\), for the modified member 2–1:$$M_{21} = \tfrac{3EI}{12}\theta_2 + \frac{wL^{2}}{8} = 0.25EI\,\theta_2 + 36.0$$$$M_{24} = \tfrac{2EI}{4}\left(2\theta_2 - 3\psi\right) = EI\theta_2 - 1.5EI\psi,\qquad M_{42} = \tfrac{2EI}{4}\left(\theta_2 - 3\psi\right) = 0.5EI\theta_2 - 1.5EI\psi$$
4. The sway (shear) equation. No horizontal force is applied anywhere and the roller cannot supply one, so the storey shear carried by the only column must vanish:$$V_{\text{col}} = -\frac{M_{24}+M_{42}}{4} = 0 \;\Longrightarrow\; 1.5EI\theta_2 - 3EI\psi = 0 \;\Longrightarrow\; \psi = 0.5\,\theta_2$$This is the key simplification: the frame does sway, but it sways exactly enough to leave the column shear-free.
5. Moment equilibrium at joint 2 and the solution.$$M_{21} + M_{23} + M_{24} = 0$$$$0.25EI\theta_2 + 36.0 - 4.0 + EI\theta_2 - 1.5EI(0.5\theta_2) = 0 \;\Longrightarrow\; 0.5EI\theta_2 = -32.0$$$$EI\,\theta_2 = \boxed{-64.0\ \text{kN}\cdot\text{m}^{2}},\qquad EI\,\Delta = 4EI\psi = \boxed{-128\ \text{kN}\cdot\text{m}^{3}}$$The negative sway means the beam line translates to the left. No numerical \(EI\) is quoted in the question, so the displacement is properly reported as \(128/EI\) metres to the left.
6. End moments. Substituting back,$$M_{21} = 0.25(-64) + 36 = \boxed{+20.0\ \text{kN}\cdot\text{m}},\qquad M_{23} = -4.0,\qquad M_{24} = -64 + 48 = -16.0$$$$M_{42} = -32 + 48 = +16.0\ \text{kN}\cdot\text{m}$$Joint 2 balances: \(20.0 - 4.0 - 16.0 = 0\). Because \(M_{24}+M_{42} = 0\), the column carries a constant moment of 16.0 kN·m over its whole height and no shear at all — the shape a careless sketch always gets wrong.
7. Reactions and shears. Working along the beam with \(M_{\text{sag}}(2) = -20.0\),$$R_1 = \frac{wL^{2}/2 - 20.0}{L} = \frac{144-20}{12} = \boxed{10.33\ \text{kN (up)}}$$$$V(12^-) = 10.33 - 24 = -13.67\ \text{kN},\qquad V(12^+) = +4.00\ \text{kN}$$so the jump of 17.67 kN at joint 2 is the axial force taken by the column, which also equals \(2(14) - 10.33\). At the built-in base$$N_4 = 17.67\ \text{kN},\qquad H_4 = 0,\qquad M_4 = 16.0\ \text{kN}\cdot\text{m}$$
8. Extreme ordinates of the diagrams. In the main span the shear passes through zero at$$x = \frac{R_1}{w} = \frac{10.33}{2} = 5.167\ \text{m from joint 1}$$where the sagging moment peaks at$$M_{\max} = R_1 x - \frac{w x^{2}}{2} = \boxed{+26.69\ \text{kN}\cdot\text{m}}$$The moment then falls to \(-20.0\) at joint 2, and on the overhang runs parabolically from \(-4.0\) at joint 2 to zero at the free tip. The column is a constant \(-16.0\) throughout.
Question 6 — bending moment diagram for the beam 1–2–3, sagging positive; the column carries a constant −16 kN·m over its full height.
Question 6 — shear force diagram for the beam; the 17.67 kN jump at joint 2 is the axial force taken by the column.