Plastic design of the rigid frame and its footing (12 + 8 marks): 16-Civ-B2 Advanced Structural Design, May 2013, Q2

Question 2: Plastic design of the rigid frame and its footing (12 + 8 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 pin-based steel rigid frame of Figure 2: a 6 m column AB, a 16 m beam BOC carried on that column and on an 8 m column CD whose base D sits 2 m below A, loaded by 80 kN horizontally at B together with 500 kN at B, 400 kN at mid-span O and 500 kN at C. The frame is to have a single plastic moment capacity throughout, and the soil beneath the bases will carry 350 kPa.

Given data — rigid frame
Quantity	Symbol	Value
Height of column AB	$h_{1}$	6 m
Height of column CD	$h_{2}$	8 m
Beam BOC, two equal halves	$L/2$	8 m each
Horizontal load at B	$H$	80 kN
Vertical loads at B, O, C	$P$	500, 400, 500 kN
Bases A and D	—	pinned
Allowable bearing pressure	$q_{a}$	350 kPa

Find. The plastic moment capacity required at collapse, a rolled section for member BOC, and a preliminary reinforced concrete spread footing for the pinned base.

Figure 2 — pin-based rigid frame with unequal column heights.

Check — which joint carries the footing. The printed question asks for a footing “at joint O”, but O is the mid-span node of the beam and has no support beneath it; the only foundations in Figure 2 are the pins at A and D. The footing is therefore designed at D, which attracts the larger reaction (745 kN against 655 kN at A) and so envelopes both bases. A single footing type can then be used throughout.

Approach. Find the collapse load factor from the upper-bound theorem by scanning every combination of the sway and beam mechanisms, confirm the answer with a statically admissible moment field, size the beam from the factored plastic moment, then proportion the footing on service reactions for bearing and on factored reactions for its own strength.

Part (a) — independent mechanisms. The frame is one degree statically indeterminate, so collapse needs two plastic hinges. Sway alone (hinges at B and C) rotates the columns by $\Delta/h_{1}$ and $\Delta/h_{2}$, so $80\Delta=M_{p}\Delta(1/6+1/8)$ and $M_{p}=274.3\ \text{kN}\cdot\text{m}$. The beam mechanism alone (hinges at B, O and C) gives $400\delta=M_{p}(4\delta/8)$, hence $M_{p}=800\ \text{kN}\cdot\text{m}$.
Combined mechanism. Writing $t=\delta/\Delta$, the internal work per unit sway is $|1/6-t/8|+3t/8+1/8$ while the external work is $80+400t$, so $$M_{p}(t)=\frac{80+400t}{\left|\dfrac{1}{6}-\dfrac{t}{8}\right|+\dfrac{3t}{8}+\dfrac{1}{8}}$$ The modulus sign is the whole story: the hinge at B carries the difference of the two rotations and disappears when they cancel, at $t=8/6=4/3$. Scanning $t$ confirms the maximum sits exactly there, and $$\boxed{M_{p}=\frac{80+533.3}{0.625}=981.3\ \text{kN}\cdot\text{m}}$$ the largest of the three and therefore the governing value.
Confirm by statics. The upper-bound theorem only guarantees an answer that is not unsafe, so check that a moment field exists in equilibrium with the loads and nowhere greater than $M_{p}$. Imposing $M_{O}=M_{p}$ and global equilibrium gives $D_{x}=-122.7\ \text{kN}$, $D_{y}=745.3\ \text{kN}$, $A_{x}=42.7\ \text{kN}$ and $A_{y}=654.7\ \text{kN}$; the column moment at C is $8(122.7)=981.3\ \text{kN}\cdot\text{m}=M_{p}$ and at B it is only $6(42.7)=256\ \text{kN}\cdot\text{m}$. The field is admissible, so 981.3 is the true collapse value, not merely an upper bound.
Required section. Factoring, $M_{p,f}=1.5(981.3)=1472\ \text{kN}\cdot\text{m}$, so the beam needs $$Z_{x}\ \ge\ \frac{M_{p,f}}{\phi F_{y}}=\frac{1472\times10^{6}}{0.9(350)}=4673\times10^{3}\ \text{mm}^{3}$$ A W690x152 supplies $Z_{x}=4987\times10^{3}\ \text{mm}^{3}$, so $\boxed{\phi M_{p}=1571\ \text{kN}\cdot\text{m}\ \ge\ 1472\ \text{kN}\cdot\text{m}}$. Its flange ratio $b/2t=5.93$ and web ratio $h/w=49.3$ are both inside the Class 1 limits, which plastic design requires so that the hinges can rotate without local buckling.
Part (b) — footing size from bearing. Bearing is a serviceability check, so use unfactored reactions: 744.7 kN vertical with 117.2 kN horizontal at the pin. Try 2.0 m square by 0.6 m deep. Adding the footing weight and a little soil over the ledges gives $P=814\ \text{kN}$, while the horizontal thrust acting over the footing depth gives $M=70.3\ \text{kN}\cdot\text{m}$ and $e=86\ \text{mm}$, well inside the middle third $L/6=333\ \text{mm}$, so the whole base stays in contact. Then $$q_{\max}=\frac{P}{A}\left(1+\frac{6e}{L}\right)=\frac{814}{4.0}(1.257)=\boxed{256\ \text{kPa}\ \le\ 350\ \text{kPa}}$$
Footing thickness and flexure. Under factored load the pressure runs from 358 to 200 kPa. The cantilever from a 700 mm pedestal face is 650 mm and carries $M_{f}=72.1\ \text{kN}\cdot\text{m}$ per metre. With $d=515\ \text{mm}$ this needs only $415\ \text{mm}^{2}/\text{m}$, far less than the shrinkage and temperature minimum $0.002A_{g}=1200\ \text{mm}^{2}/\text{m}$, so the minimum governs: 20M at 250 mm each way, top and bottom.
Punching shear. On a perimeter at $d/2$ from the pedestal, $b_{o}=4(700+515)=4860\ \text{mm}$ and $V_{f}=705\ \text{kN}$, against $V_{r}=0.38\lambda\phi_{c}\sqrt{f^{\prime}_{c}}\,b_{o}d=3386\ \text{kN}$. Two-way shear is nowhere near critical, which is the usual outcome once a footing has been made thick enough to develop the dowel bars, and one-way shear is even less demanding because the critical section falls only 135 mm from the edge.

Question 2 — plastic design and footing
Quantity	Result
Governing mechanism	combined sway and beam, hinges at O and C
Plastic moment, working loads	$M_{p}=981.3\ \text{kN}\cdot\text{m}$
Plastic moment, factored	$M_{p,f}=1472\ \text{kN}\cdot\text{m}$
Required plastic modulus	$Z_{x}\ge4673\times10^{3}\ \text{mm}^{3}$
Section adopted for BOC	W690x152, $Z_{x}=4987\times10^{3}\ \text{mm}^{3}$, $\phi M_{p}=1571\ \text{kN}\cdot\text{m}$
Service base reaction at D	$745\ \text{kN}$ vertical, $117\ \text{kN}$ horizontal
Footing	2.0 m square, 0.6 m thick, $q_{\max}=256\ \text{kPa}$
Footing reinforcement	20M at 250 mm each way (minimum governs)
Two-way shear	$V_{f}=705\ \text{kN}$ against $V_{r}=3386\ \text{kN}$

Topic. Plastic (limit) design of rigid frames, and spread footings on cohesionless soil.

Key relations. Upper bound: $\sum W\delta=\sum M_{p}\theta$ over each mechanism, and the correct $M_{p}$ is the largest such value; lower bound: any moment field in equilibrium with $|M|\le M_{p}$ everywhere proves the answer safe. Footing: $q=\dfrac{P}{A}\left(1\pm\dfrac{6e}{L}\right)$ valid while $e\le L/6$.

Why this works. A frame collapses when enough hinges have formed to turn it into a mechanism, and each candidate mechanism gives an equation of virtual work. Because the sway and beam mechanisms share the hinge at B, combining them cancels that hinge and removes its internal work while keeping both external loads — which is precisely why the combination is more critical than either parent.

Common pitfall. Stopping at the two independent mechanisms. Here the beam mechanism alone gives $800\ \text{kN}\cdot\text{m}$ and the combination gives 981, so a frame designed on the parent mechanisms would be 23 percent under strength.

Source. CSA S16 Clause 8.6 and Annex D; Beedle, Plastic Design of Steel Frames, Chapters 4 and 5; CSA A23.3 Clauses 10.5, 13.3 and 15; Das, Principles of Foundation Engineering, Chapter 3.

Question 2 of 7: Plastic design of the rigid frame and its footing (12 + 8 marks)

Question 2: Plastic design of the rigid frame and its footing (12 + 8 marks)