Question 5 of 8: Indeterminate frame by moment distribution / slope-deflection

Question 5: Indeterminate frame by moment distribution / 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 continuous member 1–2–3–4 fixed at 1; a 2 m column 2–5 (pinned base) props joint 2; a roller supports joint 3; and a 4 m cantilever 3–4 ends at the free tip 4. Loads: UDL $9$ kN/m over span 2–3 (16 m, $2EI$) and an 11 kN point load at tip 4. Spans 1–2 ($EI$, 8 m) and 3–4 ($EI$, 4 m).

Find. Member end moments, then the SFD and BMD with extreme ordinates.

Approach. The frame is inextensible and cannot sidesway (joint 2 is held by the axially-rigid beam to the fixed end 1 and by the column to pin 5; joint 3 by the roller). Only two rotations ($\theta_2,\theta_3$) are unknown, so slope-deflection with $\psi=0$ gives two equations. The cantilever applies a fixed end moment $11\times4=44$ kN·m at joint 3.

1. Fixed-end moments. Span 2–3 (UDL): $\text{FEM}_{23}=-\dfrac{wL^2}{12}=-\dfrac{9(16)^2}{12}=-192$, $\text{FEM}_{32}=+192$ kN·m. Cantilever: end moment at 3 $=11(4)=44$ kN·m (hogging).
2. Stiffnesses (relative $EI$). $k_{21}=\tfrac{4EI}{8}=0.5EI$; $k_{25}=\tfrac{3EI}{2}=1.5EI$ (pinned far end); span 2–3 uses $\tfrac{2E(2I)}{16}=0.25EI$ per unit rotation term.
3. Slope-deflection / joint equations. With $a=EI\theta_2,\ b=EI\theta_3$: joint 2: $2.5a+0.25b=192$; joint 3: $0.25a+0.5b=-148$. Solving, $EI\theta_2=112,\ EI\theta_3=-352$.
4. Member end moments. $M_{12}=28,\ M_{21}=56,\ M_{25}=168,\ M_{23}=-224,\ M_{32}=+44,\ M_{34}=-44$ kN·m (checks: joint 2 $56+168-224=0$; joint 3 $44-44=0$). Verified independently by a direct-stiffness 2-D frame analysis.
5. Span 2–3 diagram. $M(x)=-4.5x^2+83.25x-224$ (x from joint 2): hogging $-224$ at 2, a sagging peak ==**$+161$ kN·m at $x=9.25$ m**==, hogging $-44$ at 3. End shears $+83.25$ and $-60.75$ kN.
6. Other members. Column 2–5: moment $168$ (top) $\to0$ (pin), shear $=168/2=84$ kN. Span 1–2: linear $28\to56$ (hog), shear $10.5$ kN. Cantilever: $-44\to0$, shear 11 kN.