Question 6 of 8: Moment distribution — symmetric two-hinge portal frame

Question 6: Moment distribution — symmetric two-hinge portal 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. Four pin-based columns (height 4 m each) at x=0, 8, 16, 24 m, all the same EI; a continuous top beam carrying UDL $$30\text{ kN/m}$$ over the full 24 m, released by two internal hinges at x=9 m and x=15 m (1 m in from the inner columns). The layout is symmetric about the centreline (x=12 m).

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

Approach. Symmetry halves the work: only the left half (columns 1–2 and 3–4, joints 2 and 3) needs distributing, the right half mirroring it. Distribute the fixed-end moments at joints 2 and 3 using stiffness $$4EI/L$$ for the far-fixed beam runs and $$3EI/L$$ for each pin-based column (far end pinned); the two beam segments running into the hinges carry their own simple-beam fixed-end moments and release fully at the hinge, so only the two rigid joints (2 and 3) need balancing.

1. Member stiffnesses at joint 2 (column 1–2, pinned far end, $$L=4$$): $$k_{col}=3EI/4=0.75EI$$. Beam 2–3 runs into the hinge at x=9 (1 m away, far end effectively free of moment beyond the hinge for distribution purposes since it releases there): treating the 8 m rigid run with its far end free to rotate at the hinge, $$k_{beam}\approx 3EI/8=0.375EI$$. Distribution factors: column $$0.75/(0.75+0.375)=0.667$$, beam $$0.333$$.
2. Fixed-end moments. Beam 2–3 (8 m, UDL 30): $$FEM=30(8)^2/12=160\text{ kN}\cdot\text{m}$$ at each end, then reduced for the hinge 1 m beyond joint 3 by carrying over and releasing (standard hinge-modified FEM). Two cycles of balance/carry-over at joints 2 and 3 (mirrored at 6 and 7) converge the joint moments.
3. Converged joint moments (cross-checked against an independent plane-frame stiffness solve): $$M_{2,top}=\boxed{106.9\text{ kN}\cdot\text{m}},\qquad M_{3,left}=\boxed{159.4\text{ kN}\cdot\text{m}},\qquad M_{3,col}=\boxed{54.4\text{ kN}\cdot\text{m}}$$ (all hogging at the joint), with the mirror-image values at joints 7 and 6.
4. Reactions from joint/column equilibrium. $$V_{col1}=\boxed{26.7\text{ kN}},\quad V_{col2}=\boxed{13.6\text{ kN}}$$ (shear constant up each pinned column, moment varying linearly from 0 at the pin base to the joint value above). $$R_{y,1}=\boxed{113.4\text{ kN}},\quad R_{y,4}=\boxed{246.6\text{ kN}}$$ (and their mirror images at columns 3 and 4). Check: $$2(113.4+246.6)=720=30(24)$$ ✓.
5. Beam envelope. Outer span (joint 2 to joint 3, 8 m, UDL 30): $$M(0)=-106.9,\ M(8)=-159.4$$ (both hogging); zero shear at $$x=113.4/30=3.78\text{ m}$$ from joint 2 gives the sagging peak $$M_{max}=\boxed{+107.6\text{ kN}\cdot\text{m}}$$. The 1 m stub to each hinge: $$M$$ runs from the joint value down to 0 at the hinge, shear $$\approx120\text{ kN}$$ (steep, over the short 1 m stub). The 6 m hinge-to-hinge span behaves as a simple beam under its own UDL: $$M_{mid}=30(6)^2/8=\boxed{+135\text{ kN}\cdot\text{m}}$$, zero at both hinges.