Question 5 of 9: Moment distribution — symmetric gable with overhangs

Question 5: Moment distribution — symmetric gable with overhangs (18 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 symmetric two-rafter gable. The apex, joint 3, is a pin support 5 m above the springing level; joints 2 and 4 are roller supports 12 m each side of the apex measured horizontally; the rafters continue 2 m beyond each roller to the free tips 1 and 5. The rafter slope is therefore 5 in 12, so every 12 m of plan is 13 m of rafter. A load of 10 kN/m of plan covers the whole left rafter from tip 1 to the apex, a 52 kN point load acts 6 m (in plan) to the right of the apex, and a 9 kN…6 kN vertical load acts at the right-hand tip 5.

Find. The end moments by moment distribution, the reactions, and the shear force and bending moment diagrams with their maximum and minimum ordinates.

Approach. The two overhangs are determinate cantilevers, so replace each by the known moment it applies at its support; the remainder is a two-span continuous member over the apex with moment-free ends, which moment distribution balances in a single cycle. All fixed-end moments and all statical calculations are done on the horizontal projection, because a vertical load per metre of plan produces exactly the same moment field as it would on the equivalent horizontal beam.

1. Establish the geometry and the projection rule. With a rise of 5 m over a 12 m half-span the rafter length is $\sqrt{12^{2}+5^{2}} = 13$ m, so $\cos\theta = 12/13$. A vertical intensity $w$ per metre of plan becomes $w\cos^{2}\theta$ per metre of rafter measured normal to the member, and therefore $$\text{FEM} = \frac{w\cos^{2}\theta\,L_s^{2}}{12} = \frac{w}{12}\left(\frac{L_p}{L_s}\right)^{2}L_s^{2} = \frac{w L_p^{2}}{12}$$ where $L_p$ is the plan length. The same cancellation makes every point-load fixed-end moment a function of plan distances only. Member stiffness, however, still uses the true length 13 m — the two lengths are not interchangeable.
2. Reduce the overhangs to applied moments. The left overhang carries $10 \times 2 = 20$ kN whose resultant is 1 m in plan from joint 2, and the right overhang carries the 6 kN tip load 2 m in plan from joint 4: $$M_{2}^{\text{ov}} = 20(1) = \boxed{20\ \text{kN}\cdot\text{m}}, \qquad M_{4}^{\text{ov}} = 6(2) = \boxed{12\ \text{kN}\cdot\text{m}}$$ both hogging. Bending moments in a sloping member are still (force) × (horizontal lever arm) for a vertical force, which is why plan distances appear here too.
3. Compute the fixed-end moments of the two main spans. Span 2–3 carries the UDL over its full 12 m of plan and span 3–4 carries the 52 kN load at mid-plan: $$\text{FEM}_{23} = +\frac{wL_p^{2}}{12} = \frac{10(144)}{12} = +120, \qquad \text{FEM}_{32} = -120$$ $$\text{FEM}_{34} = +\frac{P a b^{2}}{L_p^{2}} = \frac{52(6)(36)}{144} = +78, \qquad \text{FEM}_{43} = -78$$ in kN·m, counter-clockwise positive.
4. Release joints 2 and 4 first. At joint 2 the only stiff member is span 2–3, and the overhang fixes the required value $M_{23} = +20$. Correcting from $+120$ to $+20$ is a change of $-100$, half of which carries over to the apex: $$M_{32} = -120 + \tfrac{1}{2}(-100) = -170$$ At joint 4 the required value is $M_{43} = -12$; correcting from $-78$ is a change of $+66$, and carrying half over, $$M_{34} = +78 + \tfrac{1}{2}(+66) = +111$$
5. Balance the apex. Both spans now have released far ends, so each offers the modified stiffness $3EI/13$; the two are identical, so the distribution factors are 0.5 and 0.5. The unbalanced moment at joint 3 is $$M_{32} + M_{34} = -170 + 111 = -59\ \text{kN}\cdot\text{m}$$ Distributing $+59$ equally and carrying nothing back (both far ends are released) closes the distribution in one cycle: $$M_{32} = -170 + 29.5 = \boxed{-140.5\ \text{kN}\cdot\text{m}}, \qquad M_{34} = 111 + 29.5 = +140.5\ \text{kN}\cdot\text{m}$$ so the apex carries a hogging moment of 140.5 kN·m.
6. Recover the shears and reactions span by span. Working on the plan projection with $M(x) = M_0 + Vx - wx^{2}/2$ for span 2–3, $M(0) = -20$ and $M(12) = -140.5$ give $V = 49.958$ kN just right of joint 2, so with the 20 kN from the overhang $$R_2 = 49.958 + 20 = \boxed{69.958\ \text{kN}}$$ For span 3–4, $M(0) = -140.5$ and $M(12) = -12$ with the 52 kN at mid-plan give $V = 36.708$ kN just right of the apex, and the shear immediately left of the apex is $49.958 - 120 = -70.042$ kN, so $$R_3 = 36.708 + 70.042 = \boxed{106.75\ \text{kN}}, \qquad R_4 = (52 - 36.708) + 6 = \boxed{21.292\ \text{kN}}$$
7. Check global equilibrium and the horizontal reaction. The applied load totals $10(14) + 52 + 6 = 198$ kN and the reactions sum to $69.958 + 106.75 + 21.292 = 198.0$ kN. Because every applied load is vertical and the apex pin is the only support that can take horizontal force, $H_3 = 0$; a non-zero value would signal an arithmetic error rather than arch action.
8. Locate the diagram extremes. In span 2–3 the shear vanishes at $x = 49.958/10 = 4.996$ m from joint 2, where $$M_{\max} = -20 + \frac{49.958^{2}}{2(10)} = \boxed{+104.79\ \text{kN}\cdot\text{m}}$$ and the moment passes through zero at $x = 0.418$ m and $x = 9.574$ m. Span 3–4 has no distributed load, so its moment runs straight from $-140.5$ at the apex to $$M_P = -140.5 + 36.708(6) = \boxed{+79.75\ \text{kN}\cdot\text{m}}$$ under the point load and on to $-12$ at joint 4, crossing zero 3.827 m past the apex and again 5.215 m past the load.