Question 8 of 8: Moment Distribution of a Symmetric Four-Column Frame

Question 8: Moment Distribution of a Symmetric Four-Column 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. A continuous roof beam 26 m long carrying 12 kN/m over its whole length, supported by four columns. Measuring from the left end ②: the beam meets the column head ③ at 8 m, an internal hinge at 10 m, a second internal hinge at 16 m, the column head ⑤ at 18 m and the right end ⑥ at 26 m. The outer columns ①–② and ⑦–⑥ are 4 m high and pinned at their bases; the inner columns ④–③ and ⑧–⑤ are 4.8 m high and built in at their bases. The layout 8 | 2 | 6 | 2 | 8 is symmetric about the centre line at 13 m, as is the loading.

Find. The member-end moments by moment distribution, and the shear force and bending moment diagrams with maximum and minimum ordinates labelled.

Approach. Use the symmetry the question points at: it rules out sidesway, so half the frame can be distributed on its own with only two rotating joints. Detach the drop-in span between the hinges first, carry its reaction onto the 2 m cantilever as a known moment, then distribute.

1. Symmetry removes sidesway. A lateral translation of the beam level would be antisymmetric, and an antisymmetric response cannot follow from a symmetric structure under symmetric load. The inextensible drop-in span additionally forces the two halves to translate together. Both arguments give the same conclusion: $\Delta = 0$, so the chord rotations vanish and no sway correction is needed. Only the joint rotations at ② and ③ are unknown in the left half, and the right half mirrors them.
2. Detach the drop-in span. The 6 m reach between the two hinges is a simply supported beam, so $$R = \frac{wL}{2} = \frac{12(6)}{2} = 36\ \text{kN at each hinge}, \qquad M_{\text{mid}} = \frac{wL^2}{8} = \frac{12(6)^2}{8} = \boxed{+54\ \text{kN.m}}$$
3. Reduce the 2 m cantilever to a known joint moment. Between joint ③ and the hinge the beam carries its own 12 kN/m plus the 36 kN handed over by the drop-in span, so the hogging moment it delivers to the joint is $$M_{\text{cant}} = 12(2)(1) + 36(2) = 24 + 72 = 96\ \text{kN.m}$$ The cantilever has no rotational stiffness, so it takes no share of any balancing moment; it appears in the distribution only as this fixed 96 kN.m demand on joint ③.
4. Stiffnesses and distribution factors. Taking $EI = 1$ and using the modified stiffness $3EI/L$ for the pin-based outer column, $$K_{21} = \tfrac{3}{4} = 0.750, \quad K_{23} = \tfrac{4}{8} = 0.500, \quad K_{32} = 0.500, \quad K_{34} = \tfrac{4}{4.8} = 0.833$$ which give distribution factors of $0.600$ to the column and $0.400$ to the beam at joint ②, and $0.375$ to the beam and $0.625$ to the column at joint ③. Carry-over is one half to the fixed base and zero to the pin.
5. Fixed-end moments and the distribution. The 8 m span gives $\mathrm{FEM}_{23} = -12(8)^2/12 = -64$ kN.m and $\mathrm{FEM}_{32} = +64$ kN.m; nothing else is loaded. Releasing joints ② and ③ alternately — balance, carry over, repeat — the cycle converges to $$\boxed{M_{21} = +36.16,\quad M_{23} = -36.16,\quad M_{32} = +83.53,\quad M_{34} = +12.47,\quad M_{43} = +6.23\ \text{kN.m}}$$ Both joints check: $36.16 - 36.16 = 0$ at ②, and $83.53 + 12.47 = 96.00$ at ③, the cantilever demand exactly.
6. Shears in the 8 m span. The sagging moment there runs from $-36.16$ kN.m at ② to $-83.53$ kN.m at ③, so writing $M(x) = -36.16 + V_2x - 6x^2$ and forcing $M(8) = -83.53$ gives $V_2 = +42.08$ kN, falling to $-53.92$ kN at ③. Zero shear occurs at $42.08/12 = 3.51$ m, where $$M_{\max} = -36.16 + \frac{42.08^2}{2(12)} = \boxed{+37.62\ \text{kN.m}}$$
7. Shears in the cantilever and reactions. Just right of ③ the shear is $36 + 12(2) = 60$ kN, falling to 36 kN at the hinge. The vertical reaction at the column head ③ is therefore $60 - (-53.92) = 113.92$ kN, and at ② it is 42.08 kN; by symmetry the right half repeats them. Total: $2(42.08 + 113.92) = 312$ kN, which is $12(26)$, the whole load. The column shears are $36.16/4 = 9.04$ kN in the pinned column and $(6.23 + 12.47)/4.8 = 3.90$ kN in the fixed one, so each half sheds 12.94 kN horizontally, taken as axial thrust in the beam and balanced by the mirror-image half.
8. Assemble the diagrams. The beam moment starts at $-36.16$ kN.m over the pinned column, swings to $+37.62$ kN.m at 3.51 m, reaches $-83.53$ kN.m just left of ③ and jumps to $-96.00$ kN.m just right of it — the 12.47 kN.m step being the moment the fixed column delivers — then rises to zero at the hinge, arches to $+54$ kN.m at the centre of the drop-in span, and mirrors itself over the right half. The shear diagram is antisymmetric about the centre line, running $+42.08 \to -53.92$ kN over the 8 m span, jumping to $+60$ kN at ③ and falling to $\pm 36$ kN across the drop-in span. Column moments are $+36.16$ kN.m at the head of the outer column falling to zero at its pin, and $+12.47$ kN.m at the head of the inner column falling to $+6.23$ kN.m at its fixed base.