Question 3 of 7: Plastic design of the rigid frame and its footing (16 + 4 marks)

Question 3: Plastic design of the rigid frame and its footing (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. A pin-based portal frame 12 m high spanning 10 m, carrying a 150 kN horizontal load at B together with 600 kN at each column head and 400 kN at each of the two interior beam points; the beam is proportioned for 1.5M_p and each column for 2M_p.

Find. (a) The reference plastic moment M_p that just produces collapse under the factored loads, and rolled sections delivering 1.5M_p in the beam and 2M_p in the columns; (b) the plan size, thickness and reinforcement of the spread footing at A.

Approach. Enumerate the independent mechanisms, form the combination that cancels a hinge, then confirm the upper bound with a statically admissible moment field — with pinned bases the frame is only once redundant, so the two-hinge combined mechanism is complete and the two bounds coincide.

1. Part (a) — factor the loads and note where hinges can form. Applying 1.5 to every load on the figure gives $H=225$ kN, 900 kN at each column head and 600 kN at each interior beam point. At a joint the hinge forms in the weaker member, and here the beam ($1.5M_p$) is weaker than the columns ($2M_p$), so every joint hinge is a beam hinge of capacity $1.5M_p$.
2. Evaluate the beam mechanism. With a hinge at the 3 m load, end rotations are $\theta$ at B and $3\theta/7$ at C, and the sagging displacements are $3\theta$ and $9\theta/7$: $$600(3\theta)+600\!\left(\tfrac{9}{7}\theta\right) =1.5M_p\!\left(\theta+\theta+\tfrac{3}{7}\theta+\tfrac{3}{7}\theta\right) \ \Rightarrow\ M_p=\frac{2571.4}{4.286}=\boxed{600\ \text{kN}\cdot\text{m}}$$ The mirror-image hinge at the 7 m load returns the same 600 kN·m, as it must for a symmetrically loaded beam.
3. Evaluate the sway mechanism. The pinned bases need no hinge, so only the two beam ends yield while the frame sways $\Delta=12\theta$: $$225(12\theta)=1.5M_p(\theta+\theta)\ \Rightarrow\ M_p=\frac{2700}{3}=\boxed{900\ \text{kN}\cdot\text{m}}$$
4. Combine to cancel the hinge at B. Adding the beam and sway mechanisms with equal rotations at B removes that hinge from both, saving $2(1.5M_p)\theta$ of internal work while the external work simply adds: $$M_p=\frac{2571.4+2700}{4.286+3-3}=\frac{5271.4}{4.286} =\boxed{1230\ \text{kN}\cdot\text{m}}$$ Combining the sway with the other beam hinge gives only 870 kN·m, so the mechanism above governs. Stopping at the independent mechanisms would have under-designed the frame by 37 per cent.
5. Confirm with the lower bound. Global statics gives $V_D=1770$ kN and $V_A=1230$ kN. Imposing $M=+1.5M_p=1845$ kN·m at the 3 m hinge fixes the redundant, $H_A=-71.25$ kN, hence $H_D=-153.75$ kN, and then $$M_C=12H_D=-1845\ \text{kN}\cdot\text{m}=-1.5M_p$$ exactly, so both hinge conditions are met by one statically admissible field. The remaining moments, $M_B=855$ and $M(7\ \text{m})=765$ kN·m, and the leeward column top at 1845 kN·m against a capacity of 2460 kN·m, all lie inside their capacities. Upper and lower bounds coincide, so $M_p=1230$ kN·m is the exact collapse value.
6. Select the beam. Plastic design requires Class 1 sections (Clause 13.5), and $$Z_{\text{req}}=\frac{1.5M_p}{\phi F_y}=\frac{1845\times10^6}{0.9(350)} =5.86\times10^6\ \text{mm}^3$$ Modelling W610×195 from its nominal plate dimensions (622 × 327, 24.4 mm flange, 15.4 mm web) gives $Z=6.03\times10^6$ mm³, hence $\phi M_p=1900$ kN·m. Its flange $b/2t=6.70$ and web $h/w=37.2$ are both Class 1, and the beam axial force (154 kN, under 2 per cent of $C_y$) needs no moment reduction.
7. Check the beam bracing for plastic design. Clause 13.7 limits the unbraced length adjacent to a hinge to $L_{cr}=(25\,000+15\,000\kappa)r_y/F_y$. With $r_y=75.8$ mm the three segments give $L_{cr}=3908$, 4067 and 6760 mm against actual lengths of 3000, 4000 and 3000 mm. Every segment passes, the 3 m to 7 m segment only just, so the bracing stated in the question is exactly what the plastic design needs.
8. Select the columns. The required capacity is $2M_p=2460$ kN·m, i.e. $Z\ge7.81\times10^6$ mm³. W840×251 (859 × 292, 24.4 mm flange, 15.6 mm web) gives $Z=8.51\times10^6$ mm³ and $\phi M_p=2680$ kN·m, with $b/2t=5.98$ (Class 1) and a web at $h/w=51.9$ against the beam-column Class 1 limit of 54.0 at $C_f/\phi C_y=0.209$. Question 6 examines this member in detail.
9. Part (b) — size the footing on bearing. The base at A is pinned, so it carries axial load and shear only: $V_A=1230/1.5=820$ kN and $H_A=71.25/1.5=47.5$ kN in service. Trying a 1.5 m square, 400 mm thick pad (self weight 21.6 kN), the shear applied at the underside gives $e=47.5(0.4)/841.6=22.6$ mm, well inside the middle third ($B/6=250$ mm), and $$q_{\max}=\frac{841.6}{1.5^2}\!\left(1+\frac{6(0.0226)}{1.5}\right) =\boxed{408\ \text{kPa}}\ \le\ 500\ \text{kPa}$$
10. Check the footing in shear and flexure. The net factored pressure is $q_{nf}=1230/2.25=546.7$ kPa and, with 75 mm cover, $d=315$ mm. Punching on a perimeter $b_0=3960$ mm at $d/2$ from a 900 × 450 mm base plate gives $V_f=722$ kN against $V_r=0.38\phi_c\lambda\sqrt{f'_c}\,b_0d=1688$ kN. One-way shear at $d$ from the plate face gives $V_f=172$ kN against $V_r=294$ kN. The cantilever moment is $$M_f=\frac{546.7(1.5)(0.525)^2}{2}=113\ \text{kN}\cdot\text{m} \ \Rightarrow\ A_s=1082\ \text{mm}^2$$ which is below $A_{s,\min}=0.002A_g=1200$ mm², so minimum steel governs: provide 6–15M each way in the bottom. Sliding is not critical ($0.5\times842=421$ kN against 47.5 kN).
11. State what the wording of part (b) can and cannot mean. The question names support A, which carries the smaller vertical reaction; the base at D takes 1770 kN factored and would need a 1.8 m square on the same bearing pressure. In practice both bases are detailed identically, so the 1.8 m pad would be used throughout; the 1.5 m pad answers the question exactly as asked.