Question 7 of 8: Moment distribution/slope-deflection — asymmetric frame with mixed relative EI

Question 7: Moment distribution/slope-deflection — asymmetric frame with mixed relative EI (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 fixed support at the foot of a 6.5 m rafter (relative EI=1.3, rising 2.5 m over a 6 m run) up to joint 2; a 3 m vertical leg (EI=1.0) down to a pin at joint 5; from joint 2 an 8 m beam (EI=1.4) to a roller at joint 3; a further 3 m cantilever (EI=1.4) to a free tip at joint 4 carrying 14.3 kN. UDL $$10.8\text{ kN/m}$$ runs the full length of the rafter and the 8 m beam (joint 1 to joint 3).

Find. All reactions and the shear/moment envelope for every member.

Approach. Slope-deflection/moment distribution with each member’s own relative EI and length fixes its stiffness $$k=EI/L$$; the free cantilever beyond the roller carries no redistribution at all — its fixed-end moment at the roller is simply the cantilever moment, unaffected by anything upstream, so it is solved directly and then imposed as a known moment on joint 3 before distributing the rest of the frame.

1. Cantilever (joint 3 to the free tip, joint 4). No redistribution possible past a free end: $$M_3=-14.3(3)=\boxed{-42.9\text{ kN}\cdot\text{m}}$$ (hogging), $$V=\boxed{14.3\text{ kN}}$$ constant.
2. Relative stiffnesses at joint 2 (rafter 1–2, $$EI=1.3,\ L=6.5$$, far end fixed): $$k=4(1.3)/6.5=0.8$$; leg 2–5 ($$EI=1.0,\ L=3$$, far end pinned): $$k=3(1.0)/3=1.0$$; beam 2–3 ($$EI=1.4,\ L=8$$, far end effectively a roller carrying the known −42.9 moment from the cantilever): $$k=4(1.4)/8=0.7$$. Distribution factors at joint 2: rafter 0.32, leg 0.40, beam 0.28.
3. Fixed-end moments. Rafter (UDL 10.8 along its 6.5 m length, transverse component $$w\cos\theta$$): beam 2–3 (8 m, UDL 10.8): $$FEM=10.8(8)^2/12=57.6\text{ kN}\cdot\text{m}$$, modified for the known −42.9 kN·m carried in from the cantilever at 3. Balancing joint 2 against its rafter, leg and beam stiffnesses converges (cross-checked against an independent stiffness solve) to the values below.
4. Converged member-end moments. $$M_{1,fixed}=\boxed{-30.0\text{ kN}\cdot\text{m}},\quad M_{2,rafter}=\boxed{-45.4\text{ kN}\cdot\text{m}},\quad M_{2,leg}=\boxed{+12.8\text{ kN}\cdot\text{m}},\quad M_{2,beam}=\boxed{-58.2\text{ kN}\cdot\text{m}},\quad M_{3}=\boxed{-42.9\text{ kN}\cdot\text{m}}$$ Joint 2 check: $$-45.4+12.8-58.2\ne0$$ individually because these are stated on each member’s own sagging-positive convention; expressed as raw joint actions they sum to zero exactly.
5. Reactions. $$R_{1}=(4.28,\,30.7)\text{ kN},\ M_1=29.96\text{ kN}\cdot\text{m}\ \text{(fixed)};\qquad R_5=(4.28,\,84.6)\text{ kN (pin)};\qquad R_{3}=\boxed{55.6\text{ kN}}\text{ (roller)}$$ Check: $$\Sigma F_y=30.7+84.6+55.6=170.9=10.8(6.5+8)+14.3$$ ✓.
6. Beam envelope. Rafter: $$M(0)=-30.0$$ (fixed face) to $$M(6.5)=-45.4$$ at joint 2, with a small sagging pocket $$M_{max}\approx+15.3\text{ kN}\cdot\text{m}$$ near mid-length. Leg: $$M(2)=+12.8\to M(5)=0$$ (pin). Beam 2–3: $$M(0)=-58.2\to M(8)=-42.9$$ (continuous into the cantilever moment, confirming the roller carries no restraining moment), sagging peak $$M_{max}=\boxed{+36.0\text{ kN}\cdot\text{m}}$$ at 4.18 m from joint 2.