A3 — Maximum factored load on the box section used as a column: 16-Civ-A2 Elementary Structural Design, December 2016, Q3

Question 3: A3 — Maximum factored load on the box section used as a column (8 + 12 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 Question 2 box section (A = 38 400 mm², I_x = 1477 × 10⁶, I_y = 1082 × 10⁶ mm⁴, M_rx = 2179, M_ry = 1889 kN·m, F_y = 350 MPa) as a 6 m column, fixed at the base and free to translate at the top. A vertical factored load P_f acts at point A, 0.5 m from the centroid O along a line inclined 30° to the x-x axis.

Find. The maximum factored axial load P_f that satisfies the S16 beam-column interaction equations.

Figure A3 — the 6 m column and the plan position of the load point A, 0.5 m from O on a line 30° above the x-x axis.

Approach. Resolve the eccentricity into components about the two axes so that both moments become fixed multiples of P_f; then every term of the S16 13.8.2 interaction equation is proportional to P_f, and the equation inverts directly for the maximum load.

Resolve the eccentricity. The load point sits at $$e_x=0.5\cos30^\circ=0.433\ \text{m},\qquad e_y=0.5\sin30^\circ=0.250\ \text{m}$$ An offset in the x direction bends the column about y-y, and vice versa, so $$M_{fy}=0.433P_f,\qquad M_{fx}=0.250P_f$$
Effective length. Fixed at the base, free to translate (and rotate) at the top is the classic sway cantilever, $K=2.0$, so $KL=12\,000$ mm. The radii of gyration are $r_x=196.1$ mm and $r_y=167.8$ mm, so the weak direction governs buckling: $$\frac{KL}{r_y}=\frac{12\,000}{167.8}=71.5$$
Compressive resistance (S16 Clause 13.3.1, $n=1.34$): $$\lambda=\frac{KL}{r_y}\sqrt{\frac{F_y}{\pi^2E}}=71.5\sqrt{\frac{350}{\pi^2(200\,000)}}=0.952$$ $$C_r=\phi AF_y\bigl(1+\lambda^{2n}\bigr)^{-1/n}=0.90(38\,400)(350)(1+0.952^{2.68})^{-1/1.34}=\boxed{7561\ \text{kN}}$$ For reference, buckling about x-x would give 8608 kN and the squash load is $C_{r0}=\phi AF_y=12\,096$ kN.

Two interaction checks are needed. The cross-sectional strength check uses the unreduced squash load with moment magnifiers, while the overall member strength check uses the buckling resistance with $U_1=1.0$, because S16 Clause 13.8.4 sets $U_1=1.0$ for a member in a sway frame (the $P$–$\Delta$ effect is already carried by the notional-load or amplified-sway analysis). Since this is a plate assembly and not an I-shape, the coefficients $\beta$ in all terms are 1.0 rather than 0.85 and 0.6.

Overall member strength, Clause 13.8.2(b). $$\frac{P_f}{C_r}+\frac{M_{fx}}{M_{rx}}+\frac{M_{fy}}{M_{ry}}\le1.0$$ $$P_f\left[\frac{1}{7561}+\frac{0.250}{2179}+\frac{0.433}{1889}\right]=P_f\left(4.763\times10^{-4}\right)\le1.0$$ $$P_f\le\boxed{2100\ \text{kN}}$$
Cross-sectional strength, Clause 13.8.2(a). The Euler loads for magnification (K = 1 for this check) are $$C_{ex}=\frac{\pi^2EI_x}{L^2}=80\,992\ \text{kN},\qquad C_{ey}=\frac{\pi^2EI_y}{L^2}=59\,301\ \text{kN}$$ With $\omega_1=1.0$ (uniform moment over the height) the magnifiers converge to $U_{1x}=1.029$ and $U_{1y}=1.040$, and $$P_f\left[\frac{1}{12\,096}+\frac{1.029(0.250)}{2179}+\frac{1.040(0.433)}{1889}\right]\le1.0\ \Rightarrow\ P_f\le2277\ \text{kN}$$
Lateral-torsional buckling, Clause 13.8.2(c). The section is a closed box, so its torsional stiffness is enormous and $M_{rx}$ is not reduced; check (c) therefore reduces to the same expression as check (b) and gives the same 2100 kN.

The overall member strength check governs by about 8 %, which is the expected signature of a slender sway column: the axial term is evaluated against the buckling resistance rather than the squash load, and that single change is worth 4500 kN of capacity. At the governing load the axial term contributes 0.278 of the unity total and the two moment terms 0.241 and 0.481 — so this column is really a beam-column dominated by the y-y moment produced by the 433 mm offset.

Result	Value
Eccentricity components	e_x = 433 mm, e_y = 250 mm
Effective length	K = 2.0, KL = 12.0 m, KL/r_y = 71.5
Slenderness parameter	λ = 0.952
Compressive resistance, C_r	7561 kN
Check 13.8.2(a), cross-section	P_f = 2277 kN
Check 13.8.2(b), overall member	P_f = 2100 kN (governs)
Maximum factored load	P_f = 2100 kN

Topic. Biaxial beam-column design of a sway column by the S16 interaction equations.

Key relations. $e_x=e\cos\theta$, $e_y=e\sin\theta$; $C_r=\phi AF_y(1+\lambda^{2n})^{-1/n}$; $U_1=\omega_1/(1-C_f/C_e)$ for check (a) but $U_1=1.0$ for sway members in checks (b) and (c); $\beta=1.0$ for sections other than Class 1 and 2 I-shapes.

Why this works. Every action is a fixed multiple of the single unknown P_f, because the eccentricity is geometric. The interaction expression is then linear in P_f and inverts in one step — no iteration is needed except for the magnifiers in check (a), which converge in two or three passes.

Common pitfall. Using $U_1=\omega_1/(1-C_f/C_e)$ in the overall-member check for a sway column. That double-counts the second-order effect and can cut the answer by 10–15 %; S16 Clause 13.8.4 is explicit that $U_1=1.0$ there.

Source. CSA S16:19 Clauses 13.3.1, 13.8.2 and 13.8.4.

Question 3 of 7: A3 — Maximum factored load on the box section used as a column

Question 3: A3 — Maximum factored load on the box section used as a column (8 + 12 marks)