Question 4 of 7: Plastic design of the steel rigid frame, and its footing

Question 4: Plastic design of the steel rigid frame, and its footing (16 + 4 = 20 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. From Figure 4: an L-shaped rigid frame with a 10 m column AB fixed at A and a 9 m beam BC fixed at C. A 200 kN horizontal load acts at mid-height of the column, a 600 kN load acts vertically at the joint B, and two 450 kN loads act on the beam at 3 m and 6 m from B. The column has capacity \(M_p\) and the beam \(2M_p\).

Find. (a) steel sections for BC and AB from a plastic collapse analysis; (b) the plan size and thickness of the footing at A.

Approach. Because C is fully fixed and the beam is axially rigid, joint B cannot translate, so there is no sway mechanism; enumerate the remaining independent mechanisms, take the largest required \(M_p\), confirm it with a statically admissible moment field, select Class 1 sections, and size the footing from the service base moment and axial load.

1. Factored loads and possible hinge locations. \(H_f = 1.5(200) = 300\) kN, \(P_{B,f} = 900\) kN and the two beam loads become 675 kN each. Hinges can form at A, at the column load point D (5 m), at B, at E (3 m along the beam), at F (6 m) and at C — six locations against three redundancies, so three independent mechanisms. At the corner B only two members meet, so the joint moment cannot exceed the weaker member's capacity, \(M_p\).
2. Why there is no sway mechanism. A sway would require B to move horizontally, but the beam is axially rigid and C is fixed, so the 200 kN is carried into C as beam axial force. The column therefore behaves as a member restrained in position at both ends, and its mechanism is a "beam" mechanism with hinges at A, D and B.
3. Mechanism 1 — column. Rotate AD by \(\theta\) so D moves \(5\theta\) sideways, DB rotating back by \(\theta\): $$M_p\theta + M_p(2\theta) + M_p\theta = 300(5\theta) \;\Rightarrow\; M_p = \frac{1500}{4} = 375\ \text{kN}\cdot\text{m}$$
4. Mechanisms 2 and 3 — beam, hinge at E then at F. With a hinge at E the segment BE rotates \(\theta\) and EC rotates \(0.5\theta\), so the hinge rotations are \(\theta\) at B, \(1.5\theta\) at E and \(0.5\theta\) at C, and the load at F drops \(1.5\theta\): $$M_p\theta + 2M_p(1.5\theta) + 2M_p(0.5\theta) = 675(3\theta) + 675(1.5\theta) \;\Rightarrow\; M_p = \frac{3037.5}{5} = \boxed{607.5\ \text{kN}\cdot\text{m}}$$ Repeating with the hinge at F gives \(11M_p\theta = 6075\theta\), i.e. \(M_p = 552.3\ \text{kN}\cdot\text{m}\). Combining the column and beam mechanisms cannot help: both rotate the corner hinge in the same sense, so nothing cancels there.
5. Confirm by the lower-bound theorem. The kinematic method gives an upper bound on the collapse load, so the required \(M_p\) is the largest of the three values, 607.5 kN·m. Check that a statically admissible field exists: put \(M_C = -2M_p = -1215\), \(M_E = +1215\) and \(M_B = -607.5\); equilibrium of the beam then requires \(V_C = 742.5\ \text{kN}\), which leaves \(M_F = -1215 + 3(742.5) = 1012.5\ \text{kN}\cdot\text{m} \le 2M_p\). Nowhere is a capacity exceeded, so 607.5 kN·m is exact.
6. (a) Beam BC. The beam needs \(2M_p = 1215\ \text{kN}\cdot\text{m}\), so $$Z_{\text{req}} = \frac{1215 \times 10^{6}}{0.9(350)} = 3.86 \times 10^{6}\ \text{mm}^3$$ Adopt a welded I with flanges \(250 \times 22\) and web \(560 \times 10\) (\(Z = 3.985 \times 10^{6}\ \text{mm}^3\), \(d = 604\ \text{mm}\)), giving \(M_r = \phi ZF_y = \boxed{1255\ \text{kN}\cdot\text{m}} \ge 1215\). It is Class 1 (\(b/t = 5.45 < 7.75\); \(h/w = 56.0 < 58.8\)), which plastic design requires, and its shear resistance \(V_r = 1128\ \text{kN}\) covers the 742.5 kN at C.
7. (a) Column AB. Plastic design asks only for \(M_p = 607.5\), so \(Z_{\text{req}} = 1.93 \times 10^{6}\ \text{mm}^3\); flanges \(200 \times 20\) with a \(400 \times 10\) web give \(Z = 2.08 \times 10^{6}\) and \(M_r = \boxed{655\ \text{kN}\cdot\text{m}}\). Question 5 shows that this section, chosen on flexure alone, is not adequate once its axial load is included — the two questions are deliberately sequential and the point is developed there.
8. Base actions for the footing. At collapse the column is a propped cantilever carrying 300 kN at mid-height with the hinge moment \(M_p\) at B, so \(M_A = -3H_fL/16 + M_p/2 = -258.8\ \text{kN}\cdot\text{m}\) and the base shear is 115.1 kN; both \(|M_A|\) and the mid-height moment (316.9 kN·m) stay below \(M_p\), confirming that no further hinge forms. Bearing pressure, however, is a serviceability check, so use the elastic service analysis of the frame (slope deflection, no sway): \(M_A = 167.8\ \text{kN}\cdot\text{m}\), beam shear at B = 369.0 kN, hence a column axial of \(600 + 369.0 + 9.2 = 978\ \text{kN}\) including the column's own weight.
9. (b) Footing size. Try a square pad \(2.2 \times 2.2 \times 0.65\ \text{m}\), weighing 75.5 kN, so \(P = 1054\ \text{kN}\) and $$e = \frac{M_A}{P} = \frac{167.8}{1054} = 0.159\ \text{m} \;<\; \frac{B}{6} = 0.367\ \text{m}$$ so the whole base stays in contact and $$q_{\max} = \frac{P}{B^2}\left(1 + \frac{6e}{B}\right) = \boxed{312\ \text{kPa}} \le 350\ \text{kPa}, \qquad q_{\min} = 123\ \text{kPa} > 0$$ For thickness, check two-way shear at ULS: with \(d = 555\ \text{mm}\) the critical perimeter is \(b_o = 3500\ \text{mm}\), \(V_f = 1283\ \text{kN}\) and \(V_c = 0.38\phi_c\lambda\sqrt{f'_c}\,b_od = 2628\ \text{kN}\), so 650 mm is ample.