Question 8 of 9: Slope deflection — symmetric two-storey, two-bay frame

Question 8: Slope deflection — symmetric two-storey, two-bay 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 symmetric frame two storeys high and two bays wide. Each bay spans 10 m and each storey is 5 m. The three column bases (joints 1, 8 and 6) are pinned. The outer columns run the full two storeys with relative stiffness $EI$; the centre column rises only to the mid level (joint 7), also $EI$. The mid-level beam 2–7–5 has relative stiffness $1.5EI$ per span and carries 55.2 kN/m; the top beam 3–4 spans the full 20 m with relative stiffness $2EI$ and carries 27.6 kN/m.

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

Approach. Exploit the mirror symmetry to cut the unknowns to two joint rotations, use the modified stiffness at the pinned bases and at the symmetric mid-span condition, then solve the two joint equilibrium equations simultaneously.

1. Exploit symmetry to reduce the unknowns. The frame and the loading are mirror-symmetric about the centre column, so the rotation of the centre joint vanishes, $\theta_7 = 0$, and the right-hand joints mirror the left, $\theta_5 = -\theta_2$ and $\theta_4 = -\theta_3$. Symmetric loading also excludes sidesway. The three pinned bases carry no moment, so their rotations are absorbed into modified stiffnesses. Only $\theta_2$ and $\theta_3$ remain, so a $22$-mark frame collapses to two equations.
2. Compute the fixed-end moments. Both beams carry full-length uniform loads: $$\text{FEM}_{27} = \frac{w_{\text{mid}}L^{2}}{12} = \frac{55.2(100)}{12} = 460\ \text{kN}\cdot\text{m}$$ $$\text{FEM}_{34} = \frac{w_{\text{top}}L^{2}}{12} = \frac{27.6(400)}{12} = 920\ \text{kN}\cdot\text{m}$$ The centre column and both outer column segments carry no span load, so their fixed-end moments are zero.
3. Assemble the slope-deflection equations at joint 2. Column 1–2 has a pinned far end, so it takes the modified stiffness $3EI/5 = 0.6EI$; column 2–3 is a normal member with $2EI/5 = 0.4EI$; and beam 2–7 has a far end whose rotation is zero by symmetry, so it behaves as a normal member with $2(1.5EI)/10 = 0.3EI$: $$M_{21} = 0.6EI\,\theta_2, \qquad M_{23} = 0.8EI\,\theta_2 + 0.4EI\,\theta_3, \qquad M_{27} = 0.6EI\,\theta_2 + 460$$
4. Assemble the equations at joint 3. The top beam is a symmetric member with $\theta_4 = -\theta_3$, so its two rotation terms partly cancel: $$M_{32} = 0.8EI\,\theta_3 + 0.4EI\,\theta_2, \qquad M_{34} = \frac{2(2EI)}{20}\left(2\theta_3 + \theta_4\right) + 920 = 0.2EI\,\theta_3 + 920$$
5. Write and solve the two joint equations. Summing the moments at each joint to zero, $$\text{joint 2:}\quad 2.0EI\,\theta_2 + 0.4EI\,\theta_3 = -460$$ $$\text{joint 3:}\quad 0.4EI\,\theta_2 + 1.0EI\,\theta_3 = -920$$ Eliminating $\theta_2$ gives $0.92EI\,\theta_3 = -828$, hence $$\boxed{EI\,\theta_3 = -900}, \qquad EI\,\theta_2 = -230 - 0.2(-900) = \boxed{-50}$$ The upper joint rotates eighteen times as much as the lower one, because the 20 m top beam is far more flexible than the two 10 m mid-level spans working together.
6. Back-substitute for every end moment. $$M_{21} = 0.6(-50) = -30, \qquad M_{23} = 0.8(-50) + 0.4(-900) = -400$$ $$M_{32} = 0.8(-900) + 0.4(-50) = -740, \qquad M_{34} = 0.2(-900) + 920 = +740$$ $$M_{27} = 0.6(-50) + 460 = +430, \qquad M_{72} = 0.3(-50) - 460 = -475$$ Joint 2 balances exactly, $-30 - 400 + 430 = 0$, and joint 3 balances, $-740 + 740 = 0$.
7. Note the centre column result. Its base is pinned, its top rotation is zero by symmetry and it carries no span load, so $$M_{87} = M_{78} = \boxed{0}$$ The centre column takes no bending at all — it is a pure axial strut. This is a useful and slightly counter-intuitive consequence of symmetry, and a good check that the symmetry conditions were applied correctly.
8. Recover the shears. For the mid-level span, $M(x) = -430 + Vx - 27.6x^{2}$ with $M(10) = -475$ gives $V = 271.5$ kN at joint 2 and $552 - 271.5 = 280.5$ kN at joint 7; the shear vanishes at $x = 4.918$ m where $$M_{\max} = -430 + \frac{271.5^{2}}{2(55.2)} = \boxed{+237.7\ \text{kN}\cdot\text{m}}$$ The top beam is symmetric, so its shear is $\pm 276$ kN and its mid-span moment is $$M_{\text{mid}} = \frac{27.6(400)}{8} - 740 = 1380 - 740 = \boxed{+640\ \text{kN}\cdot\text{m}}$$ Column shears follow from $(M_{\text{top}} + M_{\text{bot}})/h$: $(400+740)/5 = 228$ kN in the upper outer column and $30/5 = 6$ kN in the lower one.
9. Assemble the reactions and check. The outer base carries the mid-level beam shear plus the axial force delivered down the upper column, which is the top-beam shear: $$V_1 = V_6 = 271.5 + 276 = \boxed{547.5\ \text{kN}}, \qquad V_8 = 280.5 + 280.5 = \boxed{561\ \text{kN}}$$ and $2(547.5) + 561 = 1656$ kN, which equals the total applied load $27.6(20) + 55.2(20) = 552 + 1104 = 1656$ kN. Horizontally each outer base carries the lower column shear of 6 kN inwards while the centre base carries none, so horizontal equilibrium is satisfied. The 228 kN of shear in the upper columns is balanced internally by 222 kN of axial force in the mid-level beam plus the 6 kN base thrust.