Question 2 of 7: Beam-column check of CD; reinforced concrete footing at D

Question 2: Beam-column check of CD; reinforced concrete footing at D (14 + 6 = 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. Member CD from Question 1: welded I, flanges $250 \times 16$, web $370 \times 10$, $d = 402$ mm, $A = 11\,700$ mm², $Z = 1886\times10^3$ mm³, $I_x = 340.4\times10^6$ mm⁴, $r_x = 170.6$ mm, $r_y = 59.7$ mm, height 8 m, braced out of plane at mid-height. At collapse it carries $C_f = 937.5$ kN and $M_{fx} = M_p = 562.5$ kN·m. Soil bearing capacity $q_{\text{allow}} = 400$ kPa; concrete $f'_c = 30$ MPa, rebar $f_y = 400$ MPa.

Find. Whether CD satisfies the three CSA S16 Cl. 13.8.2 interaction cases; if not, a section that does — and then the plan size, depth and reinforcement of a spread footing at D.

Approach. Evaluate cross-sectional strength, overall member strength and lateral–torsional buckling strength in turn; then take the base reactions from an elastic frame analysis (service loads for bearing, factored loads for the structural design of the footing) and size the pad for pressure, one-way shear, punching shear and flexure.

1. Part (a) — effective lengths. The frame is unbraced in its own plane, so $K_x$ comes from the sway alignment chart. With a fixed base $G_A = 0$ and, at the top, $$G_B = \frac{\sum I_c/L_c}{\sum I_b/L_b} = \frac{340.4\times10^6/8000}{783.5\times10^6/15\,000} = 0.814 \;\Rightarrow\; K_x = 1.13$$ giving $KL/r_x = 1.13(8000)/170.6 = 53.1$. Out of plane the member is braced at mid-height, so $KL/r_y = 4000/59.7 = 67.0$, which governs.
2. Axial resistances. With $n = 1.34$, $\lambda = (KL/r)\sqrt{F_y/\pi^2E} = 67.0\sqrt{350/(\pi^2 \times 200\,000)} = 0.892$, $$C_r = \phi A F_y (1 + \lambda^{2n})^{-1/n} = 0.9(11\,700)(350)(1 + 0.892^{2.68})^{-1/1.34} = 2441\ \text{kN}$$ For the cross-sectional case, $\lambda = 0$ and $C_r = \phi A F_y = 3686$ kN. The moment resistance is $M_{rx} = \phi Z F_y = 594$ kN·m.
3. Interaction, Cl. 13.8.2. With $C_e = \pi^2EI_x/L^2 = 10\,498$ kN the amplifier $\omega_1/(1 - C_f/C_e)$ falls below unity, so $U_{1x} = 1.0$ is used. Case (a), cross-sectional strength: $$\frac{C_f}{\phi AF_y} + 0.85U_{1x}\frac{M_{fx}}{M_{rx}} = \frac{937.5}{3686} + 0.85\frac{562.5}{594} = 0.254 + 0.805 = 1.06$$ Case (b), overall member strength: $$\frac{937.5}{2441} + 0.805 = 0.384 + 0.805 = 1.19$$ $$\boxed{\text{Both exceed 1.0 — the Question 1 section is NOT adequate as a beam-column}}$$ This is the expected outcome: Question 1 sized the column for flexure alone with only 6 % spare, and 937.5 kN of axial load consumes far more than that.
4. Upsize the column. Try flanges $300 \times 20$ and web $400 \times 12$, so $d = 440$ mm, $A = 16\,800$ mm², $Z = 3000\times10^3$ mm³, $I_x = 593.6\times10^6$ mm⁴, $r_x = 188.0$ mm, $r_y = 73.2$ mm; $b/t = 7.2$ and $h/w = 33.3$ keep it Class 1. Now $\phi AF_y = 5292$ kN, $M_{rx} = 945$ kN·m, $KL/r_y = 54.6$ governs, $\lambda = 0.727$ and $C_r = 4060$ kN: $$\text{case (a):}\ \ \frac{937.5}{5292} + 0.85\frac{562.5}{945} = 0.177 + 0.506 = 0.68$$ $$\text{case (b):}\ \ \frac{937.5}{4060} + 0.506 = 0.231 + 0.506 = 0.74$$
5. Case (c), lateral–torsional buckling. Over the 4 m braced length, with $I_y = 90.1\times10^6$ mm⁴, $J = 1.83\times10^6$ mm⁴ and $C_w = 3.97\times10^{12}$ mm⁶, $$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^2 I_y C_w} = 2648\ \text{kN}\cdot\text{m} \gg 0.67M_p$$ so Cl. 13.6 returns the full plastic value, $M_r = \phi Z F_y = 945$ kN·m, and case (c) also gives 0.74. $$\boxed{\text{Adopt flanges } 300 \times 20,\ \text{web } 400 \times 12,\ d = 440\ \text{mm} \ (\text{governing ratio } 0.74)}$$ As a bonus this section also satisfies the elastic factored moment at C, 713 kN·m, without relying on redistribution.
6. Part (b) — base reactions. An elastic analysis of the portal with the sections now fixed ($I_b = 783.5\times10^6$, $I_c = 593.6\times10^6$ mm⁴) gives, at D:

7. Plan size from bearing and contact. Try $4.2 \times 2.6 \times 0.80$ m, adding the pad weight $10.92(0.8)(24) = 210$ kN and the 10 kN steel column. The moment migrates to the underside, $M = 362 + 105(0.8) = 446$ kN·m against $P = 862$ kN, so $e = 0.517$ m. The middle-third limit is $B/6 = 0.70$ m, so contact is full and $$q = \frac{P}{A} \pm \frac{M}{S} = \frac{862}{10.92} \pm \frac{446}{7.644} = 79.0 \pm 58.3\ \text{kPa}$$ $$\boxed{q_{\max} = 137\ \text{kPa} \le 400\ \text{kPa}, \qquad q_{\min} = +21\ \text{kPa} \ (\text{no uplift})}$$ Note that the plan size is set by the eccentricity, not by the 400 kPa: at only 34 % of the allowable pressure the footing is still as small as the middle-third rule permits.
8. Factored pressure for the structural design. Excluding the pad weight, which passes straight into the soil, $M = 543 + 157(0.8) = 669$ kN·m and $e = 0.694$ m, just inside $B/6$: $$q_{f} = \frac{963}{10.92} \pm \frac{669}{7.644} = 88.2 \pm 87.5 \;\Rightarrow\; q_{f,\max} = 176\ \text{kPa},\quad q_{f,\min} = 0.8\ \text{kPa}$$
9. One-way shear. With 75 mm cover, $d = 700$ mm and $d_v = \max(0.9d,\,0.72h) = 630$ mm. The critical section lies $d$ from the 600 mm base plate, i.e. 1.10 m from the toe, where the pressure block delivers $V_f = 437$ kN. For a slab without stirrups the size-effect factor is $\beta = 230/(1000 + d_v) = 0.141$: $$V_r = \phi_c\beta\sqrt{f'_c}\,b d_v = 0.65(0.141)(5.477)(2600)(630) = 823\ \text{kN} \;\geq\; 437\ \text{kN}$$
10. Two-way (punching) shear. The critical perimeter at $d/2$ is $b_o = 2(1300 + 1200) = 5000$ mm; with $\beta_c = 1.2$ and the depth correction $1300/(1000 + d)$, $$v_c = 0.38\lambda\phi_c\sqrt{f'_c}\left(\frac{1300}{1700}\right) = 1.035\ \text{MPa} \;\Rightarrow\; V_r = 1.035(5000)(700) = 3622\ \text{kN}$$ against a demand of about 826 kN — ample, as it usually is once one-way shear has fixed the depth.
11. Flexure. Integrating the trapezoidal pressure about the face of the base plate, 1.8 m from the toe, $M_f = 635$ kN·m. The required steel, $2711$ mm², is less than the shrinkage-and-temperature minimum for a footing, $0.002 A_g = 0.002(2600)(800) = 4160$ mm², so the minimum governs: $$\boxed{9\text{--}25\text{M bottom bars (}4500\ \text{mm}^2\text{) in the 4.2 m direction},\ M_r = 1042\ \text{kN}\cdot\text{m}}$$ In the 2.6 m direction the 1.05 m cantilever needs only 204 kN·m, so minimum steel governs there too: 14–25M at 300 mm. Bars are hooked at the ends to develop $f_y$ within the 1.8 m projection.