Question 3 of 7: Maximum factored axial load on the 18 m bridge

Question A3: Maximum factored axial load on the 18 m bridge (20 marks: 4 + 4 + 12)

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 force $P_f$ the bridge can carry as a horizontal strut, given that its point of application is eccentric.

Approach. An eccentric axial force is a beam-column: replace it by $C_f = P_f$ acting at the centroid together with $M_{fx} = P_f e$, get the compressive resistance from S16 Cl.13.3 (using an effective area, because A2 showed the section is Class 4), take the moment resistance for the fibre the eccentric moment actually compresses — the bottom plate, because C lies below the centroid — and solve the Cl.13.8 interaction equations for the $P_f$ that just makes them equal to unity.

1. Radii of gyration, and which axis governs. $$r_x = \sqrt{\frac{14.16 \times 10^9}{106\,261}} = 365\ \text{mm}, \qquad r_y = \sqrt{\frac{144.9 \times 10^9}{106\,261}} = 1168\ \text{mm}$$ Both ends hinged gives $K = 1.0$, so $$\frac{KL}{r_x} = \frac{18\,000}{365} = 49.3 \qquad \frac{KL}{r_y} = \frac{18\,000}{1168} = 15.4$$ The x–x axis governs buckling, and $49.3 \leq 200$, satisfying S16 Cl.10.4.2.1.
2. Slenderness parameter. $$\lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^2 E}} = 49.3\sqrt{\frac{350}{\pi^2 (200\,000)}} = 0.657$$ An intermediate column: neither squash-load nor Euler behaviour, so the empirical curve matters.
3. Effective area for a Class 4 section in compression. Under uniform compression every plate is stressed, so all three families reduce (S16 Cl.13.5(c)(iii) widths, applied per Cl.13.3.5):
   • bottom flange, two edges supported: $670(20)/18.71 = 716.3$ mm effective (from 2500) $\rightarrow 14\,325$ mm$^2$;
   • each top flange, one edge supported: 447.6 mm effective (from 500) $\rightarrow 8952$ mm$^2$;
   • each web, two edges supported: 716.3 mm effective (from 906.5) $\rightarrow 14\,326$ mm$^2$.
   $$A_{\text{eff}} = 14\,325 + 2(8952) + 2(14\,326) = 60\,880\ \text{mm}^2 \quad (57\ \text{per cent of } A_g)$$ $\lambda$ is still computed on the gross section, as Cl.13.3.5 requires.
4. Compressive resistance. With $n = 1.34$ for a fabricated section (S16 Cl.13.3.1), $$C_r = \phi A_{\text{eff}} F_y \left(1 + \lambda^{2n}\right)^{-1/n} = 0.90 (60\,880)(350)\left(1 + 0.657^{2.68}\right)^{-1/1.34} \times 10^{-3}$$ $$\boxed{C_r = 15\,552\ \text{kN}}$$ and with $\lambda = 0$, for the cross-sectional-strength check, $C_{r0} = \phi A_{\text{eff}} F_y = 19\,177$ kN.
5. Moment amplification. The eccentricity is the same at both ends and on the same side, so the member bends in single curvature with equal end moments: $\kappa = -1$ and $\omega_1 = 0.6 - 0.4\kappa = 1.0$. The Euler load in the plane of bending is $$C_{ex} = \frac{\pi^2 E I_x}{(KL)^2} = \frac{\pi^2 (200\,000)(14.16 \times 10^9)}{18\,000^2} \times 10^{-3} = 86\,204\ \text{kN}$$ $$U_{1x} = \frac{\omega_1}{1 - C_f/C_{ex}} \;\geq\; 1.0$$
6. Interaction (S16 Cl.13.8.3, since the section is not a Class 1 I-section). With $M_{fx} = C_f (0.080)$ in kN·m. Because C is 80 mm below the centroid, this moment adds compression to the bottom fibre, so the flexural resistance is the one computed in the A2 callout for the 2500 mm plate in compression (effective width 716 mm, top plates in tension): $M_{rx} = 5715$ kN·m. The 7245 kN·m of A2 belongs to the top fibre and would be unconservative here. Both checks take the form $$\frac{C_f}{C_r} + \frac{U_{1x} M_{fx}}{M_{rx}} \leq 1.0$$ using $C_{r0}$ for cross-sectional strength and $C_r$ for overall member strength. Solving each for the load that makes the left-hand side exactly 1.0: $$\text{cross-sectional strength:}\ P_f = 14\,493\ \text{kN} \qquad \text{overall member strength:}\ P_f = 12\,396\ \text{kN}$$
7. Governing answer. The overall member check governs, as it must for a member of any real slenderness: $$\boxed{P_{f,\max} = 12\,400\ \text{kN}\ \ (12.4\ \text{MN})}$$ At that load $U_{1x} = 1/(1 - 12\,396/86\,204) = 1.168$ and $M_{fx} = 12\,396(0.080) = 992$ kN·m, so the interaction reads $0.797 + 0.203 = 1.000$. Note how the split falls: the axial term takes four-fifths of the capacity and the eccentricity one-fifth, even though 80 mm is small against a section 900 mm deep. Ignoring the eccentricity altogether would have returned 15 552 kN — a 25 per cent over-estimate — and using the top-fibre 7245 kN·m from A2 would have returned 12 938 kN, 4 per cent unconservative.
8. Lateral–torsional buckling. Bending is about the strong-in-depth axis of a very wide trough whose $I_y$ is ten times $I_x$; the section cannot buckle laterally under moment about x–x, so the third check of Cl.13.8 does not control.