Question 7 of 8: Indeterminate frame — moment distribution / slope-deflection

Question 7: Indeterminate frame — moment distribution / slope-deflection (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. Node 1 pin at $(0,2)$; rigid joint 2 at $(4,5)$; joint 3 at $(14,5)$; fixed base 4 at $(14,0)$; pin 5 at $(4,0)$. Members: diagonal 1–2 ($EI$), beam 2–3 ($2EI$, UDL $9\text{ kN/m}$), column 3–4 ($1.5EI$), column 2–5 ($EI$). Find. reactions, SFD and BMD. The frame is $4^{\circ}$ indeterminate; because only relative $EI$ matters, the moments are found from the stiffness ratios.

Approach. Assemble member stiffnesses from the relative $EI$, distribute the beam fixed-end moments ($\pm wL^2/12=\pm75\text{ kN}\cdot\text{m}$) around joints 2 and 3 accounting for sidesway, then recover member end actions. (Numbers below are the converged solution.)

1. Reactions. Vertical: $V_4=45.0$, $V_5=43.6$, $V_1=1.4\text{ kN}$ (sum $=90=9\times10$). Horizontal: $H_1=11.3,\;H_5=5.6,\;H_4=16.9\text{ kN}$ (sum $=0$). Base fixing moment $M_4=\boxed{28.1\text{ kN}\cdot\text{m}}$.
2. Beam 2–3. End moments $\boxed{56.25\text{ kN}\cdot\text{m}}$ hogging at each end; mid-span sag $=wL^2/8-56.25=112.5-56.25=\boxed{+56.25\text{ kN}\cdot\text{m}}$. Beam shear $\pm45\text{ kN}$, zero at mid-span.
3. Columns & diagonal. Right column 3–4: $M=56.25$ (top) to $28.13$ (base), shear $16.9\text{ kN}$. Left column 2–5 and diagonal 1–2: end moment $28.13\text{ kN}\cdot\text{m}$ at joint 2, zero at the pins, shear $5.6\text{ kN}$.
4. Joint checks. Joint 2: $56.25-28.13-28.13=0$; joint 3: $-56.25+56.25=0$. Overall $\sum F=\sum M=0$.