Question 2 of 7: Plastic design of the rigid frame and the footing at D

Question 2: Plastic design of the rigid frame and the footing at D (12 + 8 = 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. Two-pinned rigid frame A–B–O–C–D of constant $M_p$; $h_{AB}=6$ m, $h_{CD}=8$ m, beam 8 m + 8 m; unfactored loads 80 kN horizontal and 500/400/500 kN vertical at B/O/C; $F_y=350$ MPa; soil bearing capacity 350 kPa; footing concrete $f'_c=30$ MPa, rebar $f_y=400$ MPa.

Find. The plastic moment $M_p$ required at the collapse load, a rolled section that supplies it, and a preliminary reinforced-concrete pad footing at joint D.

[Figure not reproduced: Figure 2 as printed: the two-pinned rigid frame with unequal column heights and the four applied loads. See the official exam paper.]

Approach. Factor the loads, count the independent mechanisms ($N-R$), evaluate the beam, sway and combined mechanisms by virtual work, take the largest $M_p$, verify by a statical check that no section exceeds $M_p$, select a section on $Z\ge M_p/\phi F_y$, then use the collapse reactions at D to size the pad.

Selecting the section for member BOC

1. Factored loads. $$H_f=1.5(80)=120\ \text{kN},\qquad V_{B}=V_{C}=1.5(500)=750\ \text{kN},\qquad V_{O}=1.5(400)=600\ \text{kN}$$
2. Count the independent mechanisms. Two pinned bases give four reaction components against three equations, so the frame is $R=1$ degree redundant. Plastic hinges can form only where the moment can peak: at B, at the load point O and at C, so $N=3$ possible hinge locations. The number of independent mechanisms is $$N-R=3-1=2$$ — one beam mechanism and one sway mechanism, plus their combination.
3. Beam mechanism. Hinges at B, O and C with the mid-point dropping $\delta=8\theta$: $$V_O(8\theta)=M_p\theta+M_p(2\theta)+M_p\theta \;\Rightarrow\; 600(8)=4M_p \;\Rightarrow\; M_p=1200\ \text{kN}\cdot\text{m}$$
4. Sway mechanism. The beam translates by $\Delta$; because the bases are pinned, hinges form only at the column tops, and the columns rotate by $\Delta/6$ and $\Delta/8$: $$H_f\Delta=M_p\left(\frac{\Delta}{6}+\frac{\Delta}{8}\right)\;\Rightarrow\; M_p=\frac{120}{7/24}=411.4\ \text{kN}\cdot\text{m}$$
5. Combined mechanism, cancelling the hinge at B. Superpose the beam mechanism ($\theta$) on the sway mechanism with $\Delta=6\theta$, so the two rotations at B are equal and opposite and that hinge disappears. External work gains the sway term; internal work loses $2M_p\theta$: $$M_p=\frac{V_O(8)+H_f(6)}{4+\left(\tfrac16+\tfrac18\right)(6)-2}=\frac{4800+720}{3.75}=\boxed{1472\ \text{kN}\cdot\text{m}}$$
6. Combined mechanism, cancelling the hinge at C. Taking $\Delta=8\theta$ instead, $$M_p=\frac{4800+960}{4+\left(\tfrac16+\tfrac18\right)(8)-2}=\frac{5760}{4.333}=1329\ \text{kN}\cdot\text{m}$$ which is smaller, so the previous combination governs. By the upper-bound theorem the true collapse load corresponds to the largest $M_p$ among the mechanisms, i.e. $M_p=1472$ kN·m.
7. Statical (lower-bound) check. At collapse the hinges are at O and C. From the free body of column DC, $M_C=8D_x=-M_p$, so $D_x=-184$ kN; horizontal equilibrium then gives $A_x=-120-D_x=+64$ kN. Moments about A give $16D_y+2D_x=17\,520$, hence $D_y=1118$ kN and $A_y=982$ kN. The moment at the cancelled hinge is $$M_B=6A_x=384\ \text{kN}\cdot\text{m}\;<\;M_p$$ and re-computing the moment at O from the left-hand free body returns $-1472$ kN·m exactly. No section exceeds $M_p$, so the mechanism is both kinematically and statically admissible — it is the true collapse mechanism.
8. Select the section. $$Z_{req}=\frac{M_p}{\phi F_y}=\frac{1472\times10^{6}}{0.9(350)}=4673\times10^{3}\ \text{mm}^{3}$$ From the CISC Handbook, W610 × 155 has $Z_x=4750\times10^{3}$ mm$^3$, $A=19\,800$ mm$^2$, $d=611$ mm, $b_f=324$ mm, $t_f=19.0$ mm, $w=12.7$ mm, $I_x=1290\times10^{6}$ mm$^4$, $r_x=255$ mm, $r_y=73.9$ mm. It is a Class 1 section, as plastic design requires, and $$M_r=\phi Z_xF_y=0.9(4750\times10^{3})(350)=\boxed{1496\ \text{kN}\cdot\text{m}}\ \ge\ 1472\ \text{kN}\cdot\text{m}$$ Adopt W610 × 155 throughout, which also satisfies the "constant $M_p$" requirement of the question.

Preliminary design of the reinforced concrete footing at joint D

1. Actions delivered to the footing. From the collapse analysis the factored column reactions at D are $C_f=1118$ kN vertical and $H_f=184$ kN horizontal; the base is pinned, so no moment is transferred at the pin itself. Dividing by 1.5 recovers the specified (service) actions, $P=745$ kN and $H=123$ kN, which are what the soil must be checked against.
2. Size the pad on bearing. Try a 2.0 m square pad, 600 mm thick. Its own weight is $2.0(2.0)(0.6)(24)=57.6$ kN, so $P_{tot}=745+58=803$ kN. The base shear acting over the pad thickness gives $M=123(0.6)=73.6$ kN·m, hence $e=M/P_{tot}=0.092$ m, well inside the middle third ($L/6=0.333$ m). Then $$q_{max}=\frac{P_{tot}}{A}\left(1+\frac{6e}{L}\right)=\frac{803}{4.0}(1.275)=\boxed{256\ \text{kPa}}\ <\ 350\ \text{kPa}$$ with $q_{min}=146$ kPa — the whole base stays in contact.
3. Factored soil pressure for the concrete design. Using the factored column actions on the same footprint, $q_f=1118/4.0=279.5$ kPa with $e_f=184(0.6)/1118=0.099$ m, so $q_{f,max}=362$ kPa.
4. Flexure. With a 500 mm square pedestal the cantilever is $c=(2000-500)/2=750$ mm and, with 75 mm cover and 20M bars, $d=515$ mm. Per metre of width, $$M_f=\frac{q_{f,max}c^{2}}{2}=\frac{362(0.75)^{2}}{2}=102\ \text{kN}\cdot\text{m/m} \;\Rightarrow\; A_s=\frac{102\times10^{6}}{0.85(400)(0.9\times515)}=647\ \text{mm}^{2}/\text{m}$$ This is below the A23.3 minimum for a footing, $A_{s,min}=0.002bh=1200$ mm$^2$/m, which therefore governs: provide 20M at 250 mm each way, bottom (1200 mm$^2$/m).
5. Two-way (punching) shear. The critical perimeter at $d/2$ from the pedestal face is $b_o=4(500+515)=4060$ mm, and $$v_c=0.38\phi_c\lambda\sqrt{f'_c}=0.38(0.65)(1)\sqrt{30}=1.353\ \text{MPa} \;\Rightarrow\; V_r=1.353(4060)(515)=2829\ \text{kN}$$ against $V_f=1118-279.5(1.015)^{2}=830$ kN — ample.
6. One-way shear. At $d$ from the pedestal face, $V_f=362(0.75-0.515)(2.0)=170$ kN. With $d_v=\max(0.9d,0.72h)=464$ mm and $\beta=230/(1000+d_v)=0.157$, $$V_c=\phi_c\beta\lambda\sqrt{f'_c}\,b_wd_v=0.65(0.157)\sqrt{30}(2000)(464)=518\ \text{kN}\ \gg\ 170\ \text{kN}$$ No shear reinforcement is required. A 2.0 m × 2.0 m × 0.6 m pad is adequate; the 184 kN base shear is transferred by friction on the base ($\mu\approx0.45$ gives $0.45\times803=361$ kN available) assisted by passive pressure on the buried face.