Question 1 of 7: Plastic design of the steel rigid frame, and its footing

Question 1: Plastic design of the steel rigid frame, and its footing (12 + 5 + 3 = 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.

Find. The reference plastic moment \(M_p\) that just produces collapse under the factored loads, the welded plate sections that supply \(0.8M_p\) in the beam and \(1.5M_p\) in the columns, and plan dimensions for the spread footing at \(A\).

[Figure not reproduced: Figure 1 — loaded steel rigid frame, as printed on page 4 of the paper. Both bases are fixed; the 60 kN horizontal load acts at mid-height of the right column. See the official exam paper.]

Approach. Enumerate the independent collapse mechanisms, take the largest required \(M_p\) from virtual work (upper bound), confirm it with a statically admissible moment field that nowhere exceeds the member capacities (lower bound), then choose Class 1 welded I-sections and check the bracing that the assumed hinges require.

1. Count the independent mechanisms. Possible hinge positions are the two bases \(A\) and \(F\), the two beam ends \(B\) and \(D\) (the beam is the weaker member at those joints, so the hinge forms in the beam), the beam midspan \(C\), and \(E\) where the horizontal load is applied — six positions. The frame is three times statically indeterminate, so there are \(6 - 3 = 3\) independent mechanisms; a fourth (the combined mode) is their superposition. Trial sections are needed first only because the beam self weight enters the work equation, so start with the beam plate girder assumed below and confirm it at step 5.
2. Beam mechanism. Hinges at \(B\), \(C\) and \(D\), all in the \(0.8M_p\) beam, with rotation \(\theta\) at the ends and \(2\theta\) at midspan. The 400 kN loads sit over the columns and do no work; the factored self weight \(1.25w_{sw} = 0.822\ \text{kN/m}\) moves through an average of \(2.5\theta\):
   $$W_{\text{ext}} = 1.5(200)(5\theta) + 0.822(10)(2.5\theta) = 1520.6\,\theta \qquad W_{\text{int}} = 4(0.8M_p)\theta$$

   Equating them,

   $$\boxed{M_p = \frac{1520.6}{3.2} = 475.2\ \text{kN}\cdot\text{m}}$$
3. The other three mechanisms are far less critical. The sway mechanism (hinges at \(A\), \(F\) at \(1.5M_p\) and at \(B\), \(D\) at \(0.8M_p\)) gives \(M_p = 1.5(60)(4)/(2 \times 1.5 + 2 \times 0.8) = 78.3\ \text{kN}\cdot\text{m}\); the column mechanism (hinges at \(F\), \(E\), \(D\)) gives 67.9 kN·m; and the combined mechanism, formed by adding beam and sway and cancelling the hinge at \(B\), gives
   $$M_p = \frac{1520.6 + 1.5(60)(4)}{3.2 + 4.6 - 1.6} = \frac{1880.6}{6.2} = 303.3\ \text{kN}\cdot\text{m}$$

   The largest requirement governs, so the beam mechanism controls and \(M_p = 475.2\ \text{kN}\cdot\text{m}\). The lateral load is simply too small, relative to the 200 kN at midspan, to drive the frame sideways.

Independent mechanisms. The beam mechanism requires the largest M_p and therefore governs; combining it with sway lowers the requirement to 303 kN·m.
4. Required member capacities. With the ratios printed on the figure,
   $$M_{p,\text{beam}} = 0.8(475.2) = 380.1\ \text{kN}\cdot\text{m} \qquad M_{p,\text{col}} = 1.5(475.2) = 712.8\ \text{kN}\cdot\text{m}$$
5. Select the beam. A Class 1 section is mandatory for plastic design, and welded plate construction suits the welded corner detailed in Question 2. Try flanges 2–180 × 14 with a 350 × 10 web, giving \(d = 378\ \text{mm}\), \(A = 8540\ \text{mm}^2\), 67.0 kg/m (\(w_{sw} = 0.657\ \text{kN/m}\), the value used at step 2), and
   $$Z = b_f t_f (d - t_f) + \frac{t_w h_w^2}{4} = 180(14)(364) + \frac{10(350)^2}{4} = 1.224 \times 10^6\ \text{mm}^3$$ $$M_r = \phi Z F_y = 0.90(1.224\times10^6)(350) = 385.4\ \text{kN}\cdot\text{m} \ \ge\ 380.1\ \text{kN}\cdot\text{m}\quad(\text{utilisation } 0.986)$$

   Local buckling: flange \(b/t = (180-10)/2/14 = 6.07 \le 145/\sqrt{350} = 7.75\) and web \(h/w = 35.0 \le 1100/\sqrt{350} = 58.8\), so the section is Class 1 in both elements.

6. Select the columns. The columns must supply 712.8 kN·m while carrying the factored axial load \(C_f = 1.5(400) + 1.5(200)/2 + 0.822(5) = 754.1\ \text{kN}\). Try flanges 2–240 × 16 with a 460 × 10 web (\(d = 492\ \text{mm}\), \(A = 12\,280\ \text{mm}^2\), 96.4 kg/m):
   $$Z = 240(16)(476) + \frac{10(460)^2}{4} = 2.357 \times 10^6\ \text{mm}^3 \qquad M_r = 742.4\ \text{kN}\cdot\text{m} \ \ge\ 712.8\ \text{kN}\cdot\text{m}$$

   Flange \(b/t = 7.19 \le 7.75\). The Class 1 web limit is reduced by the axial load to \(58.8\left(1 - 0.39 C_f/\phi C_y\right) = 58.8(1 - 0.39 \times 0.195) = 54.3\), and \(h/w = 46.0\), so the web is still Class 1.

7. Confirm the collapse load with a lower-bound check. At collapse the beam carries \(-380.1\) kN·m at \(B\) and \(D\) and \(+380.1\) kN·m at \(C\). Three hinges leave the frame still once redundant, so a family of admissible fields exists; the simplest member of it takes the axial force in the beam as zero, which sends the whole 90 kN factored lateral load down the right column. The resulting moments are 380.1 kN·m throughout column \(AB\), and 380.1 kN·m at \(D\) and \(E\) falling to 20.1 kN·m at \(F\). Every ordinate is below \(1.5M_p = 712.8\) kN·m, so by the uniqueness theorem \(M_p = 475.2\) kN·m is the true collapse value.
A statically admissible moment field at collapse. Because no ordinate reaches the column capacity of 713 kN·m, the beam mechanism is confirmed as the true collapse mode.
8. Bracing the plastic hinges (the note in the paper is not enough). CSA S16 Cl. 13.7 limits the unbraced length next to a plastic hinge to \(L_{cr} = (25\,000 + 15\,000\kappa)r_y/F_y\). With \(r_y = 40.0\ \text{mm}\) and \(\kappa = +1\) (double curvature between \(B\) and \(C\)), \(L_{cr} = 4567\ \text{mm}\) against the 5 m from \(B\) to \(C\). Lateral support at the beam quarter points, 2.5 m apart, is therefore required, and the columns must be braced at \(E\) — see Question 2(b), where the unbraced column fails outright.
9. Beam shear. \(V_f = 1.5(200)/2 + 0.822(5) = 154.1\ \text{kN}\) against \(V_r = \phi A_w (0.66F_y) = 0.90(3500)(231)/10^3 = 727.6\ \text{kN}\) — not close to governing.
10. Part (b): the footing at \(A\). Footings are proportioned on specified loads. The column delivers \(P = 400 + 100 + 0.657(5) + 0.946(8) = 510.9\ \text{kN}\) and, dividing the admissible base moment by the load factor, \(M = 380.1/1.5 = 253.4\ \text{kN}\cdot\text{m}\). Bearing alone would need only \(\sqrt{510.9/400} = 1.13\ \text{m}\) square; the moment, not the bearing pressure, sets the size. Try a 2.6 m square pad 0.6 m thick (\(W_f = 2.6^2(0.6)(24) = 97.3\ \text{kN}\)):
    $$e = \frac{253.4}{510.9 + 97.3} = 0.417\ \text{m} \ \le\ \frac{B}{6} = 0.433\ \text{m} \quad\text{(resultant inside the kern, no uplift)}$$ $$q_{\max} = \frac{P}{B^2}\left(1 + \frac{6e}{B}\right) = \frac{608.2}{6.76}(1.949) = 176.5\ \text{kPa} \qquad q_{\min} = 3.5\ \text{kPa}$$ $$\boxed{2.6\ \text{m} \times 2.6\ \text{m} \times 0.6\ \text{m deep};\ q_{\max} = 176\ \text{kPa} < 400\ \text{kPa}}$$

    The pad is governed by the requirement that the resultant stay within the middle third, not by bearing capacity; the 0.6 m thickness is set by two-way (punching) shear around the base plate and by development of the dowels.