Question 3 of 7: A3. Maximum factored axial load on the 18 m bridge (4 + 4 + 12 marks)

Question 3 — A3. Maximum factored axial load on the 18 m bridge (4 + 4 + 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.

Find. The largest factored axial compression $P_f$ the member can sustain, given that the eccentricity makes it a beam-column.

Approach. The eccentric load is statically equivalent to an axial force at the centroid plus a constant moment $P_f e$ about x-x. Compute $C_r$ on the effective area (the section is Class 4 in uniform compression), compute $M_{rx}$ for the fibre the eccentric moment compresses, then solve the S16 Clause 13.8.3 interaction equation for the load that makes it exactly unity.

1. Identify the actions. Shifting $P_f$ from C up to the centroid introduces a moment about the x-x axis of magnitude $M_{fx} = P_f e = 0.080 P_f$ (kN·m for $P_f$ in kN). Because the load sits below the centroid, this moment puts the wide bottom flange into additional compression — that fibre, not the top, governs the flexural term.
2. Effective area for compression. Under uniform compression every element is checked against $670/\sqrt{F_y} = 35.8$ (both edges) or $200/\sqrt{F_y} = 10.69$ (outstand). The bottom flange ($b/t = 125$), the webs ($45.5$) and the top-flange outstands ($12.0$) all exceed their limits, so each reduces to its effective width — 716.3 mm for the both-edge elements and 447.6 mm overall for each top flange: $$A_{eff} = 716.3(20) + 2(716.3)(20) + 2(447.6)(20) = 60\,880\ \text{mm}^2$$ a little over half the gross area. This is the price of a very wide, thin bottom plate in compression.
3. Compression resistance. Buckling is about the weaker x-x axis ($r_x = 364.8$ mm << $r_y$): $$\frac{KL}{r_x} = \frac{18\,000}{364.8} = 49.3, \qquad \lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^2 E}} = 49.3\sqrt{\frac{350}{\pi^2 (200\,000)}} = 0.657$$ $$C_r = \phi A_{eff} F_y (1 + \lambda^{2n})^{-1/n}, \quad n = 1.34$$ $$\boxed{C_r = 0.90(60\,880)(350)(1.301)^{-0.746} = 15\,549\ \text{kN}}$$
4. Moment resistance for the governing fibre. With the bottom flange in compression it reduces to its 716.3 mm effective width while the top flanges, now in tension, stay fully effective. The effective section gives $I_x = 8.804\times10^{9}$ mm$^4$ with the centroid 485.3 mm above the soffit, so $$S_{e,bot} = \frac{8.804\times10^{9}}{485.3} = 18.14\times10^{6}\ \text{mm}^3, \qquad M_{rx} = 0.90(18.14\times10^{6})(350) = 5\,715\ \text{kN}\cdot\text{m}$$ Note how much smaller this is than the 7 256 kN·m of Question A2 — compressing the wide bottom plate is far less efficient than compressing the narrow top flanges.
5. Apply the beam-column interaction. For a Class 4 section S16 Clause 13.8.3 requires $$\frac{C_f}{C_r} + \frac{U_{1x} M_{fx}}{M_{rx}} \le 1.0$$ with $U_{1x} = \omega_1/(1 - C_f/C_{ex})$ but not less than 1.0 for the overall member check. The Euler load is $C_{ex} = \pi^2 E I_x / (KL)^2 = 86\,284$ kN, so the amplification is negligible and $U_{1x} = 1.0$ governs. Lateral-torsional buckling need not be checked: with $r_y = 1172$ mm the section is enormously stiff laterally. Substituting $M_{fx} = 0.080P_f$ and solving, $$\frac{P_f}{15\,549} + \frac{0.080 P_f}{5\,715} = 1.0 \;\Longrightarrow\; P_f(6.43\times10^{-5} + 1.40\times10^{-5}) = 1$$ $$\boxed{P_{f,max} = 12\,400\ \text{kN}}$$
6. Confirm the split. At that load the axial term contributes 0.797 and the moment term 0.203 of the unity check, with $M_{fx} = 12\,398(0.080) = 992$ kN·m. An 80 mm eccentricity — under 9 % of the section depth — therefore consumes a fifth of the capacity, which is the standard warning that eccentricity is expensive in compression members.