Question 6 of 8: Two-storey frame by slope-deflection

Question 6: Two-storey frame by slope-deflection (20 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 frame consisting of a continuous vertical column carrying two 12 m beams that cantilever to the left.

Find. The end moments in every member, and shear force and bending moment diagrams labelled with the maximum and minimum ordinates.

Approach. Show first that the frame cannot sway, so the only kinematic unknowns are the rotations at the two beam-to-column joints; write the slope-deflection equations with the modified stiffness $3EI/L$ for the two column lengths that end at a pinned support; solve the two joint-equilibrium equations; then back-substitute for the member end moments and diagrams.

1. Rule out sidesway. The column is pinned at both ends and is inextensible, so nodes 3 and 5 cannot move vertically. Both beams are horizontal and inextensible and are built in at their far ends, so nodes 3 and 5 cannot move horizontally either. Every joint translation is therefore prevented and the only unknowns are $\theta_{3}$ and $\theta_{5}$. This is what the phrase "members are inextensible" in the question is there to license.
2. Fixed-end moments. Only the two beams are loaded, each by a uniform 6 kN/m over 12 m: $$\text{FEM} = \frac{wL^{2}}{12} = \frac{6(12)^{2}}{12} = 72\ \text{kN}\cdot\text{m}$$ so $M^{F}_{23} = M^{F}_{45} = -72$ and $M^{F}_{32} = M^{F}_{54} = +72\ \text{kN}\cdot\text{m}$, taking clockwise moments on the member ends as positive.
3. Slope-deflection equations. With relative $EI$ values and no joint translation, and using the modified stiffness $3EI/L$ for members 1–3 and 5–6 because each has a pinned far end carrying no other member, $$M_{32} = \tfrac{2(3EI)}{12}(2\theta_{3}) + 72 = EI\theta_{3} + 72, \qquad M_{31} = \tfrac{3EI}{3}\theta_{3} = EI\theta_{3}$$ $$M_{35} = \tfrac{2EI}{3}(2\theta_{3} + \theta_{5}), \qquad M_{53} = \tfrac{2EI}{3}(2\theta_{5} + \theta_{3})$$ with the mirror-image pair at node 5.
4. Joint equilibrium. Setting $M_{31} + M_{32} + M_{35} = 0$ and $M_{53} + M_{54} + M_{56} = 0$ gives $$\tfrac{10}{3}EI\theta_{3} + \tfrac{2}{3}EI\theta_{5} = -72, \qquad \tfrac{2}{3}EI\theta_{3} + \tfrac{10}{3}EI\theta_{5} = -72$$ The two equations are symmetric, so $$\boxed{EI\theta_{3} = EI\theta_{5} = -18\ \text{kN}\cdot\text{m}}$$ Both joints rotate anticlockwise, which is what a downward load on a beam cantilevering to the left must do.
5. Member end moments. Substituting back, $$M_{23} = -81, \quad M_{32} = +54, \quad M_{31} = -18, \quad M_{35} = -36\ \text{kN}\cdot\text{m}$$ $$M_{53} = -36, \quad M_{54} = +54, \quad M_{45} = -81, \quad M_{56} = -18\ \text{kN}\cdot\text{m}$$ with $M_{13} = M_{65} = 0$ at the two pins. Both joint sums check to zero: $54 - 18 - 36 = 0$. An independent direct-stiffness solution of the same frame reproduces all eight moments exactly.
6. Beam shear and moment diagrams. For each 12 m beam, taking moments about one end, $$R_{\text{column end}} = \frac{M_{23} + M_{32} + wL^{2}/2}{L} = \frac{-81 + 54 + 432}{12} = 33.75\ \text{kN}$$ so the shear runs from $+38.25$ kN at the built-in end to $-33.75$ kN at the column, crossing zero at $x = 38.25/6 = 6.375$ m. The moment there is $$M_{\max} = -81 + 38.25(6.375) - 3(6.375)^{2} = \boxed{+40.92\ \text{kN}\cdot\text{m}}$$ between hogging values of $\boxed{-81\ \text{kN}\cdot\text{m}}$ at the built-in end and $-54\ \text{kN}\cdot\text{m}$ at the column.
7. Column shear and moment diagrams. Each column length carries no transverse load, so its shear is constant and equal to the sum of its end moments divided by its length: $$V_{13} = \frac{0 + 18}{3} = 6\ \text{kN}, \qquad V_{35} = \frac{36 + 36}{3} = \boxed{24\ \text{kN}}, \qquad V_{56} = \frac{18 + 0}{3} = 6\ \text{kN}$$ The moment varies linearly from $0$ to $-18\ \text{kN}\cdot\text{m}$ over 1–3, from $-36$ to $+36\ \text{kN}\cdot\text{m}$ over 3–5 with a point of contraflexure exactly at mid-height, and from $-18\ \text{kN}\cdot\text{m}$ back to $0$ over 5–6.