Question 4 of 7: Beam-Column Check of BC and the Footing at C

Question 4: Beam-Column Check of BC and the Footing at C (14 + 6 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 W360 by 196 column selected in Question 3, carrying the axial force and moments the collapse analysis returned, in a laterally braced frame.

Find. (a) whether W360 by 196 satisfies the three Cl 13.8.2 beam-column checks, and (b) plan dimensions, thickness and reinforcement for a spread footing under C.

Approach. Run the three interaction checks in the order S16 writes them — cross-sectional strength, overall member strength and lateral-torsional buckling — each with its own compressive resistance and its own moment resistance. Then take the service actions at C for the bearing check, and the factored ones for the structural design of the pad.

1. Part (a) — compute the compressive resistances. The frame is braced, so take $k = 1.0$. Out of plane the column is held at B, at mid-height and at C, giving $kL/r_{y} = 5000/95.9 = 52.1$; in plane it is unbraced over the full height, $kL/r_{x} = 10\,000/159.5 = 62.7$. With $\lambda = (kL/r)\sqrt{F_{y}/\pi^{2}E}$ and $C_{r} = \phi AF_{y}(1+\lambda^{2n})^{-1/n}$, $n = 1.34$: $$C_{r0} = 7824\ \text{kN}\ (\lambda = 0),\quad C_{ry} = 6167\ \text{kN},\quad C_{rx} = 5467\ \text{kN}$$
2. Establish the moment resistance. The critical segment runs from B to mid-height, 5 m long, with end moments $-525$ and $+103$ kN.m, so $\kappa = +0.196$ in double curvature and $\omega_{2} = 1.75 + 1.05\kappa + 0.3\kappa^{2} = 1.968$. The elastic critical moment is $$M_{u} = \frac{\omega_{2}\pi}{L}\sqrt{EI_{y}GJ + \left(\frac{\pi E}{L}\right)^{2}I_{y}C_{w}} = 8022\ \text{kN.m}$$ far above $0.67M_{p} = 893$ kN.m, so Cl 13.6 returns the full plastic value and $M_{rx} = \phi Z_{x}F_{y} = 1199$ kN.m in all three checks. A stocky W360 laterally held every 5 m simply does not buckle laterally.
3. Find the amplification factor. The Euler load in the plane of bending is $C_{e} = \pi^{2}EI_{x}/(kL)^{2} = 12\,467$ kN. Because a concentrated transverse load acts between the supports, Cl 13.8.5 gives $\omega_{1} = 0.85$, so $U_{1x} = 0.85/(1 - 1036/12\,467) = 0.92$; for a braced frame this is taken as not less than 1.0, so $U_{1x} = 1.00$.
4. Run the three Cl 13.8.2 checks. Each has the form $C_{f}/C_{r} + 0.85U_{1x}M_{fx}/M_{rx}\le1.0$: $$\text{(a) cross-section: } \frac{1036}{7824} + \frac{0.85(525)}{1199} = 0.133 + 0.372 = \boxed{0.505}$$ $$\text{(b) overall member: } \frac{1036}{5467} + 0.372 = 0.190 + 0.372 = \boxed{0.562}$$ $$\text{(c) lateral-torsional: } \frac{1036}{6167} + 0.372 = 0.168 + 0.372 = \boxed{0.540}$$ and separately $M_{fx}/M_{rx} = 0.438\le1.0$.
5. Confirm the class under combined loading. The Class 1 web limit tightens with axial load to $\left(1100/\sqrt{F_{y}}\right)\left(1 - 0.39C_{f}/\phi C_{y}\right) = 55.8$, and the actual $h/w = 19.5$, so the section remains Class 1. W360 by 196 is therefore adequate as a beam-column, governed by check (b) at 56 per cent. The margin is large because the Class 1 requirement, not strength, chose the section in Question 3.
6. Part (b) — get the service actions at the base. Bearing pressure is a serviceability question, so the frame is analysed elastically under unfactored loads — and it is still elastic there, since the first hinge did not form until a load factor of 0.849 on the factored loads. That analysis gives $N = 652$ kN, $M_{C} = 75$ kN.m and $H_{C} = 35$ kN at the base.
7. Size the pad on bearing. Assume $q_{a} = 200$ kPa, a routine allowable value for a firm sandy or stiff clayey soil, and state it as an assumption. Try a 2.4 m square pad 650 mm thick: its own weight is $2.4^{2}\times0.65\times24 = 90$ kN, so $N_{\text{tot}} = 742$ kN and the eccentricity, including the horizontal thrust acting over the pad depth, is $e = (75 + 35\times0.65)/742 = 129$ mm, well inside the middle third ($B/6 = 400$ mm). Hence $$q_{\max,\min} = \frac{N}{B^{2}}\left(1\pm\frac{6e}{B}\right) = 129\left(1\pm0.323\right) = 174\ \text{and}\ 90\ \text{kPa}$$ Both are compressive and the peak is inside the 200 kPa allowed. A 2.2 m pad returns 209 kPa and fails, so 2.4 m is the smallest square that works.
8. Check punching shear on the factored pressure. Factored, $N_{f} = 1001$ kN and $M_{f} = 113$ kN.m give $q_{f}$ between 125 and 223 kPa. With 75 mm cover and 20M bars the effective depth is $d = 550$ mm, so the critical perimeter at $d/2$ from a 400 mm square base plate is $b_{o} = 4(400+550) = 3800$ mm and the punching force is $1082$ kN. Then $v_{f} = 1082\times10^{3}/(3800\times550) = 0.52$ MPa against $v_{r} = 0.38\phi_{c}\lambda\sqrt{f'_{c}} = 1.35$ MPa, a ratio of 0.38. One-way shear is even lighter at 0.27.
9. Design the flexural steel. The critical section is at the face of the column, a 1.00 m cantilever, carrying $M_{f} = 243$ kN.m over the 2.4 m width. That needs only $A_{s} = 1367$ mm$^{2}$, but the minimum for a footing, $0.002A_{g} = 0.002\times2400\times650 = 3120$ mm$^{2}$, is more than twice as much, so minimum steel governs: provide 12-20M each way in the bottom, 3600 mm$^{2}$, at roughly 200 mm centres. This is the usual outcome for a pad under a nearly axially loaded column, and it means the thickness is set by shear and cover, not by bending.