Question 3 of 7: Plastic Design of the Rigid Frame and its Welded Knee

Question 3: Plastic Design of the Rigid Frame and its Welded Knee (16 + 4 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. An L-shaped rigid frame, built in at both ends, whose member strengths are prescribed in the ratio 1 : 1.5.

Find. (a) the plastic moment the frame actually needs and rolled sections that supply it, and (b) a welded knee connection at B that can deliver the beam's plastic moment into the column.

Approach. Count the redundants, enumerate the independent mechanisms and every combination of them, take the largest required M_p, then confirm the answer with a lower-bound moment field obtained by stepping the load up hinge by hinge. Select sections on strength and on the Class 1 requirement that plastic design imposes, and size the connection on the beam's own capacity rather than on the demand.

1. Part (a) — establish the degree of redundancy. Two built-in supports give six reaction components against three equations of equilibrium, so the frame is three times redundant and a complete mechanism needs four hinges. A partial mechanism, in which only part of the frame moves, may need fewer — and that is what happens here.
2. Beam mechanism. Joint B cannot move vertically (the column is axially rigid) and cannot move horizontally (the beam is axially rigid and A is built in), so hinges at A, at mid-span and at B alone form a valid mechanism. Note that the hinge at B forms in the beam, the weaker of the two members meeting there. With a virtual rotation $\theta$ at each end and $2\theta$ at mid-span, and mid-span deflection $4\theta$: $$525(4\theta) = M_{p}\theta + M_{p}(2\theta) + M_{p}\theta \Rightarrow M_{p} = \frac{2100}{4} = 525\ \text{kN.m}$$ The 750 kN load at B does no work because B does not move.
3. Column mechanism. Hinges at C, at the 150 kN load point and at B, with the load point displacing $5\theta$, give $$150(5\theta) = 1.5M_{p}\theta + 1.5M_{p}(2\theta) + M_{p}\theta \Rightarrow M_{p} = \frac{750}{5.5} = 136.4\ \text{kN.m}$$ far less demanding than the beam mechanism, because the column is both stronger and more lightly loaded.
4. Test every combination, and watch the sign at B. Let the column-mechanism rotations be $r$ times the beam-mechanism rotations. At the shared hinge B the beam mechanism puts tension on the outside of the knee while the column mechanism puts it on the inside, so the two rotations subtract and the hinge partially cancels: $$M_{p}(r) = \frac{2100 + 750r}{3 + |1-r| + 4.5r}$$ This is a maximum at $r = 0$. Combining the mechanisms therefore lowers the required capacity (380 kN.m at $r = 1$), the opposite of the usual portal-frame result, so $$\boxed{M_{p} = 525\ \text{kN.m}\quad\text{and}\quad 1.5M_{p} = 787.5\ \text{kN.m}}$$
5. Confirm with a lower bound. Because the collapse mechanism is partial, the column moments are not fixed by statics and must be traced. Stepping the factored loads up on an elastic-plastic model, hinges appear at A at load factor 0.849, at mid-span at 0.898 and at B at 1.0000 — collapse landing exactly on the factored load, which confirms the mechanism analysis. The moment field at collapse is $-525$ at A, $+525$ at mid-span, $-525$ at B, $+103$ at mid-height of the column and $-19$ at C, every value within capacity, so the answer is exact rather than merely an upper bound.
6. Extract the member forces the sections must carry. From the beam free body with $M = -525$ at A and $+525$ at mid-span, $V_{A} = (525 + 525)/4 = 262.5$ kN, so the beam delivers $525 - 262.5 = 262.5$ kN to the joint and the column axial force is $262.5 + 750 = 1012.5$ kN. The column shear is $(103.1 + 525)/5 = 125.6$ kN in the upper half and $-24.4$ kN in the lower half, the difference being the 150 kN applied load.
7. Select the beam. Strength requires $Z_{x}\ge M_{p}/(\phi F_{y}) = 525\times10^{6}/(0.9\times350) = 1.667\times10^{6}$ mm$^{3}$. W460 by 82 gives $Z_{x} = 1.810\times10^{6}$ mm$^{3}$ and $\phi M_{p} = 570$ kN.m, is Class 1 on both flange ($b/2t = 5.97$) and web ($h/w = 43.2$), and satisfies the Cl 13.7 limit on unbraced length next to a plastic hinge: with double curvature ($\kappa = +1$) $L_{cr} = r_{y}(25\,000 + 15\,000\kappa)/F_{y} = 42.4\times40\,000/350 = 4847$ mm against the 4000 mm actually provided.
8. Select the column, where the section class governs. Strength alone needs only $Z_{x}\ge2.50\times10^{6}$ mm$^{3}$, which W360 by 147 satisfies at $\phi M_{p} = 885$ kN.m. But plastic design requires Class 1 sections throughout, and the Class 1 flange limit is $b/2t\le145/\sqrt{350} = 7.75$: W360 by 147 is 9.34 (Class 3), W360 by 162 is 8.51 and even W360 by 179 is 7.80, all Class 2. The lightest Class 1 shape in the series is W360 by 196 at $b/2t = 7.14$, giving $\phi M_{p} = 1199$ kN.m. It is 52 per cent stronger than required, and the surplus is the price of the class requirement, not of the analysis.
9. Part (b) — set the design force for the knee at B. A connection at a plastic hinge is sized on the capacity actually installed, not on the demand, so that the hinge forms in the member and not in the joint. With $\phi M_{p} = 570$ kN.m for the beam, each flange delivers $$T_{f} = \frac{570\times10^{6}}{d - t} = \frac{570\times10^{6}}{460 - 16} = 1284\ \text{kN}$$
10. Weld the beam to the column. Complete-joint-penetration groove welds from each beam flange to the column flange develop the flange directly and are the standard detail for a moment knee. The web carries only the 262.5 kN beam shear: two 8 mm fillets along the 428 mm web depth give $2\times428\times1.244 = 1065$ kN using $V_{r} = 0.67\phi_{w}A_{w}X_{u}$ with $\phi_{w} = 0.67$, so 8 mm is ample and is set by the minimum size for a 16 mm flange rather than by strength.
11. Provide continuity stiffeners. Opposite each beam flange the column needs a stiffener to carry the 1284 kN across the column web. Taking the stiffeners alone, $A_{st}\ge1284\times10^{3}/(0.9\times350) = 4077$ mm$^{2}$; a pair of 150 mm by 16 mm plates gives 4800 mm$^{2}$ with $b/t = 9.4$ inside the $200/\sqrt{F_{y}} = 10.7$ limit.
12. Check the panel zone and add the diagonal. The shear carried across the joint panel is the flange force less the column shear, $1284 - 126 = 1159$ kN, against a web resistance of $$V_{r} = 0.55\phi F_{y}d_{c}w = 0.55\times0.9\times350\times372\times16.4 = 1057\ \text{kN}$$ The web is 10 per cent short, so a diagonal stiffener across the panel takes the balance. The diagonal is $\sqrt{460^{2} + 372^{2}} = 592$ mm long with $\cos\theta = 372/592 = 0.629$, so it must carry $(1159-1057)/0.629 = 162$ kN, needing $162\times10^{3}/(0.9\times350) = 513$ mm$^{2}$. A pair of 100 mm by 10 mm plates (2000 mm$^{2}$) is the smallest practical size and is adopted.