Question 6 of 8: Indeterminate frame by moment distribution (22 marks)

Question 6 — Indeterminate frame by moment distribution (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 frame whose horizontal member runs from a free tip at joint 1 through joint 2 to joint 3 and on to joint 4. A 12 m column drops from joint 3 to a fixed base at joint 5.

Find. Shear-force and bending-moment diagrams for every member, with the maximum and minimum ordinates labelled.

Approach. Check the degree of indeterminacy and whether sidesway can occur, compute fixed-end moments, then distribute and carry over at the two rotating joints until the unbalanced moments vanish; finally recover member shears from free-body equilibrium.

With $m=4$, $j=5$, $r = 2+1+3 = 6$ and no releases, $\mathrm{DSI} = 3(4)+6-3(5) = 3$. The pin at joint 2 and the fixed base at joint 5 both block horizontal translation of the beam, so there is no sidesway and the distribution runs on joint rotations alone. The 3 m overhang is statically determinate and simply applies a known moment at joint 2.

1. Reduce the overhang. The cantilever 1–2 carries the UDL directly to joint 2: $$M_{21} = -\frac{wL^{2}}{2} = -\frac{24(3)^{2}}{2} = \boxed{-108\ \text{kN}\cdot\text{m}}$$ together with a shear of $24(3) = 72\ \text{kN}$. This moment is fixed and takes no part in the distribution; it simply has to be balanced by span 2–3.
2. Fixed-end moments for the loaded spans. For the central point load on span 2–3 and the UDL on span 3–4, $$\mathrm{FEM}_{23} = \pm\frac{PL}{8} = \pm\frac{112(9)}{8} = \pm126\ \text{kN}\cdot\text{m}$$ $$\mathrm{FEM}_{34} = \pm\frac{wL^{2}}{12} = \pm\frac{24(9)^{2}}{12} = \pm162\ \text{kN}\cdot\text{m}$$ The column carries no transverse load, so its fixed-end moments are zero.
3. Distribution factors. With equal $EI$ throughout, relative stiffnesses are $I/L$: span 2–3 gives $1/9$, span 3–4 gives $1/9$ (modified to $\tfrac{3}{4}(1/9)$ because joint 4 is a roller, a pinned far end) and the column gives $1/12$. At joint 3 these yield distribution factors of roughly 0.42 to span 2–3, 0.32 to span 3–4 and 0.26 to the column; joint 2 distributes wholly into span 2–3 because the overhang moment is fixed.
4. Distribute and carry over. Balancing joints 2 and 3 in turn, carrying over one half to each fixed far end and iterating until the residuals fall below 0.1 kN·m, the final member end moments settle at $$M_{23} = +108, \quad M_{32} = -171, \quad M_{34} = +207, \quad M_{43} = 0, \quad M_{35} = -36, \quad M_{53} = +18$$ all in kN·m. Joint 3 balances exactly: $-171 + 207 - 36 = 0$.
5. Member shears from free bodies. Span 2–3 carries end shears of $+49$ kN and $-63$ kN either side of the 112 kN load; span 3–4 runs from $+131$ kN at joint 3 to $-85$ kN at the roller; the column carries a constant $4.5$ kN. Summing at the supports, $$R_2 = 72 + 49 = \boxed{121\ \text{kN}}, \quad R_4 = \boxed{85\ \text{kN}}, \quad R_5 = 63 + 131 = \boxed{194\ \text{kN}}$$ with a horizontal thrust of 4.5 kN at the base opposed by the pin at joint 2, and a base moment of 18 kN·m. Total vertical reaction is $121+85+194 = 400\ \text{kN}$, matching the applied $72+112+216 = 400\ \text{kN}$.

1. Locate the sagging peak in span 3–4. Measuring $s$ from joint 3, the shear $131 - 24s$ vanishes at $s = 5.458\ \text{m}$, where $$M = -207 + 131(5.458) - 12(5.458)^{2} = \boxed{+150.5\ \text{kN}\cdot\text{m}}$$ and the moment returns to zero at the roller, as it must.
2. Read the step at joint 3. The beam moment is $-171$ kN·m just left of joint 3 and $-207$ kN·m just right of it. The difference of 36 kN·m is exactly the moment carried away down the column — the visible signature of joint equilibrium on a bending-moment diagram. Along the column the moment runs linearly from 36 kN·m at the top to 18 kN·m of opposite sense at the base, passing through zero 4 m above the foundation.