Question 2 of 7: Plastic Design of the Same Girder

Question 2: Plastic Design of the Same Girder (12 + 8 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. The same continuous beam and the same factored loads as Question 1 — three 6 m spans, $P_{f} = 750$ kN at 3, 8, 10, 14 and 16 m from A — but the member is now to be a rolled shape selected by plastic analysis, with lateral support at every support and every load point.

Find. The plastic moment $M_{p}$ demanded by the governing collapse mechanism, and a rolled section that supplies it while meeting the Class 1 and bracing requirements that plastic design imposes.

Approach. Enumerate the independent beam mechanisms (one per span — a continuous beam on simple supports has no sway mode), take the largest $M_{p}$ by the upper-bound theorem, confirm it with a lower-bound statically admissible moment field, then select a Class 1 section on $Z_{x}$ and check shear and unbraced length.

1. Count the redundancies and the hinges needed. The beam has four vertical reactions and two equations of equilibrium, so it is twice redundant and a complete mechanism needs three hinges. Each span can form its own beam mechanism, and because every hinge at an interior support rotates in the same sense in the two mechanisms that share it, no combination of span mechanisms can cancel a hinge rotation. For a continuous beam under gravity load, therefore, the independent span mechanisms are the complete list — unlike a portal frame, where the sway combination usually governs.
2. Span A–B: hinge under the load and at B. With a virtual deflection $\delta$ under the 3 m load, the end rotations are $\theta_{A} = \theta_{B} = \delta/3$ and the hinge rotations are $2\delta/3$ under the load and $\delta/3$ at B. Equating work, $$P_{f}\,\delta = M_{p}\left(\frac{2\delta}{3} + \frac{\delta}{3}\right) \;\Rightarrow\; M_{p} = P_{f} = 750\ \text{kN}\cdot\text{m}$$ The pin at A supplies the third release, so only two hinges are needed in this span.
3. Span B–C: hinges at B, under a load, and at C. Placing the sagging hinge under the load 2 m from B and giving it a deflection $\delta$, the load 4 m from B descends $\delta/2$, and $$P_{f}\,\delta + P_{f}\,\frac{\delta}{2} = M_{p}\left(\frac{\delta}{2} + \frac{3\delta}{4} + \frac{\delta}{4}\right) \;\Rightarrow\; M_{p} = \frac{1125}{1.5} = 750\ \text{kN}\cdot\text{m}$$ The mirror-image hinge 4 m from B returns the same value, as symmetry of the loading requires.
4. Span C–D: the end span, and the one that governs. D is an end support, so no hinge can form there and the mechanism needs only two — one at C and one under a load. Putting the sagging hinge under the load 4 m from C, the rotations are $\delta/4$ at C and $3\delta/4$ at the sagging hinge, while the load 2 m from C descends $\delta/2$: $$P_{f}\,\delta + P_{f}\,\frac{\delta}{2} = M_{p}\left(\frac{\delta}{4} + \frac{3\delta}{4}\right) \;\Rightarrow\; M_{p} = \boxed{1125\ \text{kN}\cdot\text{m}}$$ Moving that hinge to the load 2 m from C gives only 900 kN·m, so the outer load controls. The end span demands half again as much as either interior span, because it has one restrained end instead of two.

5. Close the upper bound with a lower-bound check. An upper-bound mechanism is only an answer once a statically admissible moment field exists with $|M| \le M_{p}$ everywhere. Impose $M_{C} = -M_{p}$ and the sagging hinge at 16 m, and span C–D alone gives $R_{D} = M_{p}/2 = 562.5$ kN, hence $M$ at 14 m $= 562.5(4) - 750(2) = 750$ kN·m and $M_{C} = 562.5(6) - 750(4) - 750(2) = -1125$ kN·m, reproducing the assumed value exactly. Re-analysing the remaining two spans elastically with $M_{C} = -1125$ kN·m returns a peak of exactly 1125 kN·m, so the field is admissible and the two bounds coincide: $M_{p} = 1125$ kN·m is the answer, not merely a bound.
6. Select the section. Plastic design requires $\phi Z_{x} F_{y} \ge M_{p}$, so $$Z_{x} \ge \frac{1125 \times 10^{6}}{0.90 \times 350} = 3.571 \times 10^{6}\ \text{mm}^{3}$$ Clause 13.5 also demands a Class 1 section throughout, whose limits at $F_{y} = 350$ MPa are $b/2t \le 145/\sqrt{350} = 7.75$ for the flange and $h/w \le 1100/\sqrt{350} = 58.8$ for a web in flexure. Print the candidates before choosing.

W610×113 falls 10 per cent short. W690×125 has the larger modulus at the same mass but its flange ratio of 7.76 misses Class 1 by a hair, and a section that is only Class 2 may not be used where a plastic hinge must rotate. W610×125 is the lightest Class 1 shape that works.

7. Shear at the collapse state. The largest shear occurs at C in span C–D: $V = 2(750) - 562.5 = 937.5$ kN. For the rolled section $A_{w} = d w = 612 \times 11.9 = 7283\ \text{mm}^{2}$ and $h/w = 48.1 \lt 439\sqrt{k_{v}/F_{y}} = 54.2$, so the web yields in shear rather than buckling: $$V_{r} = \phi A_{w} (0.66 F_{y}) = 0.90 \times 7283 \times 231 = 1514\ \text{kN} \ge 937.5\ \text{kN}$$ a ratio of 0.62.
8. Bracing adjacent to the hinges. Plastic design is only valid if each hinge can rotate without the member buckling laterally. Taking the conservative $\kappa = 0$ form of Clause 13.7, $$L_{cr} = \frac{25\,000\,r_{y}}{F_{y}} = \frac{25\,000 \times 49.9}{350} = 3560\ \text{mm}$$ The braced segments are 3 m in span A–B and 2 m elsewhere, all inside that limit, so the assumed lateral support at joints and load points is sufficient.

Economy is the point of the comparison the paper sets up. The elastic plate girder of Question 1 weighs 151 kg/m; the plastically designed rolled beam weighs 125 kg/m, a saving of 17 per cent, because plastic analysis redistributes the peak support moment of 1294 kN·m into the spans and because a rolled shape earns its shape factor where the plate girder must use $S_{x}$. If a splice is acceptable, spans A–B and B–C could drop to W530×101 on their 750 kN·m demand, but the splice would have to sit well away from C, and for an 18 m girder the single section is almost always cheaper.