Plastic design of the steel rigid frame and its footing: 16-Civ-B2 Advanced Structural Design, May 2015, Q1

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. The single-bay portal of Figure 1, fixed at both bases A and F, with the geometry, loads and relative plastic capacities tabulated below.

Quantity	Value
Beam BD span, with C at mid-span	10 m (5 m + 5 m)
Column height AB and DF	8 m, with E at mid-height of DF
Vertical loads (unfactored)	400 kN at B, 200 kN at C, 400 kN at D
Horizontal load (unfactored)	60 kN at E, acting outwards
Plastic capacity of beam BD	0.8 M_p
Plastic capacity of columns AB, DF	1.5 M_p
Steel	F_y = 350 MPa, φ = 0.90
Allowable bearing pressure	400 kPa

Find. The reference plastic moment M_p at which the factored loads just produce a collapse mechanism, the rolled sections that deliver it, and a spread footing at A that keeps the service bearing pressure below 400 kPa.

[Figure not reproduced: Figure 1 — loaded steel rigid frame (loads as printed, unfactored). The beam is the weaker member at 0.8 M p ; both columns are 1.5 M p . See the official exam paper.]

Approach. Factor the loads, enumerate the independent collapse mechanisms (beam, sway, column) and their significant combination by virtual work, take the largest required M_p, confirm it with a statically admissible moment field (lower-bound theorem), then choose Class 1 sections and size the footing from the service base reactions.

Part (a) — factor the applied loads. Every load on Figure 1 is multiplied by 1.5: $$P_B = P_D = 1.5(400) = 600\ \text{kN},\qquad P_C = 1.5(200) = 300\ \text{kN},\qquad H_E = 1.5(60) = 90\ \text{kN}$$ The frame is fixed at A and F, so it is three times statically indeterminate and a complete mechanism needs four hinges. The candidate hinge sections are A, B, C, D, E and F, giving 6 − 3 = 3 independent mechanisms.
Beam mechanism. Hinges form at B, C and D. With a rotation $\theta$ at each end and $2\theta$ at mid-span, the mid-span load moves $5\theta$ (metres). Because the beam (0.8 M_p) is weaker than the columns, every hinge at a joint forms in the beam: $$300(5\theta) = 0.8M_p\theta + 0.8M_p(2\theta) + 0.8M_p\theta = 3.2 M_p \theta$$ $$\boxed{M_p = \frac{1500}{3.2} = 468.75\ \text{kN}\cdot\text{m}}$$
Sway (panel) mechanism. Hinges at A, B, D and F, with the beam translating $\Delta = 8\theta$ and the load point E, at mid-height, moving $4\theta$: $$90(4\theta) = 1.5M_p\theta + 0.8M_p\theta + 0.8M_p\theta + 1.5M_p\theta = 4.6M_p\theta \;\Rightarrow\; M_p = \frac{360}{4.6} = 78.3\ \text{kN}\cdot\text{m}$$ The horizontal load is small relative to the gravity loads, so sway alone is far from governing.
Column mechanism on DF. With the beam held, the 90 kN load at mid-height can form hinges at D, E and F: $$90(4\theta) = 0.8M_p\theta + 1.5M_p(2\theta) + 1.5M_p\theta = 5.3M_p\theta \;\Rightarrow\; M_p = 67.9\ \text{kN}\cdot\text{m}$$
Combined beam-plus-sway mechanism. Adding the beam and sway mechanisms cancels the two hinges at B (one rotation is the reverse of the other), so $$1500\theta + 360\theta = (3.2 + 4.6 - 1.6)M_p\theta = 6.2M_p\theta \;\Rightarrow\; M_p = \frac{1860}{6.2} = 300\ \text{kN}\cdot\text{m}$$ The combination is less demanding than the beam mechanism alone, because the extra internal work of the four sway hinges outweighs the modest external work of the 90 kN load.
Governing requirement. The upper-bound theorem requires the largest value obtained from any mechanism, so the beam mechanism governs: $$\boxed{M_p = 468.75\ \text{kN}\cdot\text{m}}\qquad M_{p,\text{beam}} = 0.8M_p = 375\ \text{kN}\cdot\text{m},\qquad M_{p,\text{col}} = 1.5M_p = 703.1\ \text{kN}\cdot\text{m}$$ This is a partial mechanism: only the beam collapses, and the columns remain elastic. An incremental elastic–plastic analysis confirms the sequence C (λ = 0.855), D (λ = 0.961), B (λ = 1.000), with collapse reached exactly at the full factored load.
Lower-bound (static) check. The collapse moment field is $$M_A = 74.3,\quad M_B = M_C = M_D = 375,\quad M_E = 150.4,\quad M_F = 434.3\ \text{kN}\cdot\text{m}$$ Every value satisfies $|M| \le$ the section capacity (703.1 in the columns, 375 in the beam) and the field is in equilibrium with the factored loads, so the answer is exact rather than merely an upper bound. The base reactions that go with it are $V_A = V_F = 750$ kN, $H_A = 56.2$ kN and $H_F = -146.2$ kN.

Bending moments at collapse. The three beam hinges form the mechanism; the columns stay below their 703 kN·m capacity everywhere.

Select the beam. For a plastically designed member CSA S16 Cl 13.5 gives $M_r = \phi Z F_y$, so $$Z_{req} = \frac{375\times10^6}{0.90(350)} = 1.19\times10^6\ \text{mm}^3$$ Take W410×67: modelled from its nominal plate dimensions (410 × 179, flange 14.4, web 8.8) it gives $Z_x = 1.339\times10^6\ \text{mm}^3$ and $M_r = 421.9\ \text{kN}\cdot\text{m} \ge 375$. Flange $b/2t = 6.22 \le 145/\sqrt{350} = 7.75$ and web $h/w = 43.3$, so the section is Class 1 as plastic design requires.
Select the columns. Likewise $$Z_{req} = \frac{703.1\times10^6}{0.90(350)} = 2.23\times10^6\ \text{mm}^3$$ Take W530×92 (533 × 209, flange 15.6, web 10.2): $Z_x = 2.329\times10^6\ \text{mm}^3$, $M_r = 733.6\ \text{kN}\cdot\text{m} \ge 703.1$. Flange $b/2t = 6.70$ and web $h/w = 49.2$ against a Class 1 web limit of $\frac{1100}{\sqrt{F_y}}\left(1-0.39\frac{C_f}{\phi C_y}\right) = 54.2$ at $C_f = 750$ kN, so the columns are Class 1 as well. Their adequacy as beam-columns is confirmed in Question 2(b).
Part (b) — service loads on footing A. Dividing the collapse reactions by the load factor gives the service actions delivered to the base: $$P = \frac{750}{1.5} = 500\ \text{kN},\qquad M = \frac{74.3}{1.5} = 49.5\ \text{kN}\cdot\text{m}, \qquad H = \frac{56.2}{1.5} = 37.5\ \text{kN}$$
Size the pad for bearing. Try a 1.6 m square pad, 450 mm thick, founded 1.2 m below grade. Its own weight is $1.6^2(0.45)(24) = 27.6$ kN and the soil over it $1.6^2(0.75)(18) = 34.6$ kN, so $$P_{tot} = 562.2\ \text{kN},\qquad M_{base} = 49.5 + 37.5(0.45) = 66.4\ \text{kN}\cdot\text{m}, \qquad e = \frac{66.4}{562.2} = 0.118\ \text{m}$$ Since $e = 0.118 < B/6 = 0.267$ m the whole base stays in contact, and $$q_{max,min} = \frac{P_{tot}}{B^2}\left(1 \pm \frac{6e}{B}\right) = 219.6(1 \pm 0.443) \;\Rightarrow\; \boxed{q_{max} = 317\ \text{kPa} < 400\ \text{kPa}}$$ with $q_{min} = 122$ kPa, comfortably in compression.
Check the pad structurally. The factored net pressure varies from 439 kPa to 147 kPa. With $d = 355$ mm the two-way (punching) resistance on a perimeter $b_o = 4(600+355) = 3820$ mm is $$V_r = 0.38\lambda\phi_c\sqrt{f'_c}\,b_o d = 0.38(0.65)\sqrt{30}(3820)(355) = 1835\ \text{kN} \gg V_f = 350\ \text{kN}$$ One-way shear at d from the column face gives $V_f = 102$ kN against $V_r = 425$ kN. The cantilever moment at the face is 79.0 kN·m over the full width, needing only 454 mm²/m, so the minimum $A_s = 0.002(1000)(450) = 900$ mm²/m governs.

Footing at base A — 1600 mm square × 450 mm thick, 15M @ 200 mm each way in the bottom face.

Result	Value
Governing mechanism	beam mechanism, hinges at B, C, D
Required reference capacity M_p	468.75 kN·m
Beam BD (0.8 M_p = 375 kN·m)	W410×67, M_r = 421.9 kN·m
Columns AB, DF (1.5 M_p = 703.1 kN·m)	W530×92, M_r = 733.6 kN·m
Footing at A	1600 × 1600 × 450 mm, 15M @ 200 each way
Maximum service bearing pressure	317 kPa < 400 kPa

Topic. Plastic (limit) design of rigid frames — the kinematic and static theorems.

Key relations. Virtual work $\sum P\delta = \sum M_p\theta$; independent mechanisms $= N_{\text{critical sections}} - N_{\text{redundants}}$; a moment field with $|M| \le M_p$ everywhere is safe (lower bound); $M_r = \phi Z F_y$ for a Class 1 section.

Why this works. The kinematic theorem says any assumed mechanism gives a collapse load at or above the true one, so the mechanism demanding the greatest capacity is the correct answer. The static theorem says any equilibrium moment field that nowhere exceeds capacity is safe. When the two bounds meet, as they do here at 468.75 kN·m, the analysis is exact. Plastic design needs Class 1 sections because the hinges must rotate through large plastic strains without local buckling, and it needs lateral bracing at and near the hinges for the same reason.

Common pitfall. Stopping at the independent mechanisms without testing their combinations. Here the combination happens to be less severe, but on many frames the combined sway-plus-beam mechanism governs and ignoring it under-designs the frame. The other frequent slip is to take the hinge capacity at a joint as the stronger member’s: the hinge always forms in the weaker one.

Source. CSA S16:19 Cl 8.6 and Cl 13.5; Hibbeler, Structural Analysis, 10th ed., Ch. 18; CISC Handbook of Steel Construction, 12th ed., Part 4.

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)