Question 4 of 7: Plastic Design of the Steel Rigid Frame

Question 4: Plastic Design of the Steel Rigid Frame (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.

Find. (a) the required M_p and rolled sections for BC and AB; (b) the plan size and depth of a spread footing at A.

[Figure not reproduced: Figure 4 as printed on page 4 of the examination paper. See the official exam paper.]

Approach. Enumerate the independent collapse mechanisms, confirm the governing one with a static lower-bound check, then convert M_p into sections and design the footing on the collapse-state reactions.

1. Part (a) — establish what mechanisms are possible. First note a feature of this frame that decides everything: joint B cannot translate horizontally, because the beam BC is axially rigid and C is built into a wall. There is therefore no sway mechanism. The 200 kN horizontal load is resisted by the column bending between a fixed base and a laterally held top, so the only column mechanism is a three-hinge beam-type mechanism within AB itself. That leaves three candidates: the column mechanism, and a beam mechanism with its interior hinge under either the 3 m or the 6 m load.
2. Column mechanism. Hinges at A, at the 5 m load point and at B, all of capacity M_p. With a virtual sway Δ at the load point the chord rotations are $\Delta/5$ each way, so the hinge rotations are $\Delta/5$, $2\Delta/5$ and $\Delta/5$: $$300\Delta = M_p\left(\frac{\Delta}{5} + \frac{2\Delta}{5} + \frac{\Delta}{5}\right) \quad\Longrightarrow\quad M_p = 375\ \text{kN}\cdot\text{m}$$
3. Beam mechanisms. At the corner the hinge forms in the weaker member, so the joint capacity is M_p, not 2M_p. With the interior hinge under the 3 m load and a virtual deflection δ there, the hinge rotations are $\delta/3$ at B, $\delta/2$ at the hinge and $\delta/6$ at C, while the 6 m load moves $\delta/2$: $$675\delta + 675\left(\frac{\delta}{2}\right) = M_p\frac{\delta}{3} + 2M_p\frac{\delta}{2} + 2M_p\frac{\delta}{6} \quad\Longrightarrow\quad M_p = 607.5\ \text{kN}\cdot\text{m}$$ Repeating with the hinge under the 6 m load gives 552.3 kN·m. Combining the beam and column mechanisms does not help here: the hinge at B rotates in the same sense in both, so the rotations add rather than cancel and the combination returns only 504 kN·m. The largest requirement therefore governs: $$\boxed{M_p = 607.5\ \text{kN}\cdot\text{m}}$$
4. Confirm with the lower bound. An upper-bound answer is only an answer once a safe equilibrium state is exhibited. Writing every critical moment in terms of the three redundants at A and minimising M_p subject to $|M| \le$ capacity at A, the 5 m point, B, the two beam load points and C returns exactly 607.5 kN·m, with $V_A = 1507.5$ kN. Upper and lower bounds coincide, so 607.5 kN·m is exact, not merely an estimate.
5. Read off the collapse state. The governing mechanism has only three hinges (B, the 3 m load point and C) in a frame that is three times redundant, so it is a partial mechanism — the column stays elastic and its moments are not determined by statics alone. Stepping the factored load up hinge by hinge with an elastic–plastic analysis (hinges appear at C, then B, then the 3 m point, collapse arriving at a load factor of 1.0000, which is itself a check on the mechanism work) gives $$M_A = 260,\quad M_{5\text{m}} = 316,\quad M_B = 607.5,\quad M_{6\text{m}} = 1012.5,\quad M_C = 1215\ \text{kN}\cdot\text{m}$$ together with $V_A = 1507.5$ kN and a column base shear of 115 kN. These are the actions carried into Questions 5 and 7.
6. Select the beam. The beam must supply $2M_p = 1215$ kN·m, so $$Z_x \ge \frac{1215\times10^6}{0.9(350)} = 3857\times10^3\ \text{mm}^3 \quad\Longrightarrow\quad \boxed{\text{BC: W610} \times 140}$$ which offers $Z_x = 4111 \times 10^3$ mm³ and $M_r = 1295$ kN·m. It is Class 1 ($b/t = 5.18 \le 7.75$, $h/w = 43.7 \le 58.8$), as plastic design requires, and its axial force of 211 kN is only 3 % of $C_y$, so no reduction applies.
7. Select the column. The column carries 1507.5 kN of axial load, which is 23 % of $C_y$ — above the 15 % threshold — so Clause 13.5 reduces the plastic moment: $$M_{pc} = 1.18\,M_p\left(1 - \frac{C_f}{C_y}\right) \le M_p$$ For W310 × 143, $C_y = 6312$ kN and $\phi M_{pc} = 677$ kN·m ≥ 607.5 kN·m, so on strength alone $$\boxed{\text{AB: W310} \times 143 \ \ \text{(provisional — see Question 5)}}$$ Question 5 asks explicitly whether this is adequate as a beam-column; it is not, and the section is revised there.
8. Part (b) — size the footing at A. The base is built in, so the footing carries moment as well as axial load. At service (dividing the collapse actions by 1.5) it takes $N = 1005$ kN, $M = 173$ kN·m and $H = 77$ kN. Try a 2.4 m square pad, 0.8 m thick (self weight 110.6 kN). The moment at the underside is $173 + 77(0.8) = 235$ kN·m, so the eccentricity is 210 mm, inside the middle third ($B/6 = 400$ mm), and $$q = \frac{P}{B^2} \pm \frac{6M}{B^3} = 193.7 \pm 101.9 \quad\Longrightarrow\quad \boxed{q_{\max} = 296\ \text{kPa} \le 350\ \text{kPa}, \quad q_{\min} = 92\ \text{kPa} > 0}$$ The whole base stays in contact with the soil. A 2.2 m pad would reach 359 kPa, so 2.4 m is the smallest practical size.
9. Check the footing internally. With 75 mm cover the effective depth is $d = 710$ mm. Punching around a 600 mm pedestal gives a critical perimeter $b_o = 5240$ mm and $V_f = 1058$ kN against $V_r = 0.38\lambda\phi_c\sqrt{f'_c}\,b_o d = 5033$ kN — not remotely critical, and one-way shear less so (68 vs 409 kN/m). Flexure on the 0.9 m cantilever gives $M_f = 152$ kN·m/m, requiring only 638 mm²/m, so the minimum-steel rule governs: $A_s = 0.002A_g = 1600$ mm²/m, supplied by 20M at 180 mm each way, bottom.

Final Results