Question 2 of 7: A2. Moments of resistance of a built-up bridge section (6 + 7 + 7 marks)

Question 2 — A2. Moments of resistance of a built-up bridge section (6 + 7 + 7 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. $M_{rx}$ and $M_{ry}$, the factored moments of resistance about the two centroidal axes.

Approach. Compute the gross section properties by the parallel-axis theorem, classify every plate element against S16 Table 2, and — because several elements fall outside Class 3 — recompute the section properties on the code's effective widths, so that $M_r = \phi S_e F_y$.

1. Set out the plate elements. Each inclined web runs 860 mm vertically and 300 mm horizontally, so its developed length is $$L_w = \sqrt{860^2 + 300^2} = 910.8\ \text{mm}, \qquad A_w = 910.8(20) = 18\,216\ \text{mm}^2$$ The bottom flange contributes $2500(20) = 50\,000$ mm$^2$ and the two top flanges $2 \times 500(20) = 20\,000$ mm$^2$, for a gross area $$A_g = 50\,000 + 2(18\,216) + 20\,000 = 106\,433\ \text{mm}^2$$
2. Locate the centroid and compute the gross second moments. Taking first moments about the soffit, with the bottom flange at 10 mm, the webs at 450 mm and the top flanges at 890 mm, $$\bar{y} = \frac{50\,000(10) + 36\,433(450) + 20\,000(890)}{106\,433} = 326.0\ \text{mm}$$ so the extreme fibres lie 326.0 mm below and 574.0 mm above the x-x axis. Applying the parallel-axis theorem (the inclined webs contribute $A_w h_v^2/12$ about x-x and $A_w b_h^2/12$ about y-y, where $h_v$ and $b_h$ are their vertical and horizontal projections), $$\boxed{I_x = 14.16\times10^{9}\ \text{mm}^4, \qquad I_y = 146.2\times10^{9}\ \text{mm}^4}$$ giving $r_x = 364.8$ mm and $r_y = 1172$ mm. The gross elastic moduli are $S_{x,top} = 24.67\times10^{6}$ mm$^3$ and $S_y = 81.22\times10^{6}$ mm$^3$.
3. Classify the plate elements. With $F_y = 350$ MPa the Class 3 limits of S16 Table 2 are $200/\sqrt{F_y} = 10.69$ for a projecting element, $1900/\sqrt{F_y} = 101.6$ for an element supported along both edges in flexure, and $670/\sqrt{F_y} = 35.8$ for one supported along both edges in uniform compression. The actual ratios are $$\frac{b}{t}\Big|_{\text{top flange outstand}} = \frac{240}{20} = 12.0, \qquad \frac{b}{t}\Big|_{\text{web}} = \frac{910.8}{20} = 45.5, \qquad \frac{b}{t}\Big|_{\text{bottom flange}} = \frac{2500}{20} = 125$$ The webs are Class 1 in flexure (45.5 < 58.8), but the top-flange outstands exceed 10.69 and the very wide bottom flange exceeds every limit. The section is therefore Class 4, and S16 Clause 13.5(c) requires $M_r = \phi S_e F_y$ with $S_e$ computed on effective widths.
4. Bending about x-x (sagging, top flange in compression). Only the top-flange outstands need reducing; the wide bottom flange is in tension and is fully effective. The effective outstand is $200t/\sqrt{F_y} = 213.8$ mm, so each top flange becomes $2(213.8) + 20 = 447.6$ mm wide. Recomputing, $$I_{x,e} = 13.48\times10^{9}\ \text{mm}^4, \qquad \bar{y}_e = 314.7\ \text{mm}, \qquad S_{e,top} = \frac{13.48\times10^{9}}{585.3} = 23.03\times10^{6}\ \text{mm}^3$$ $$\boxed{M_{rx} = \phi S_e F_y = 0.90(23.03\times10^{6})(350) = 7\,256\ \text{kN}\cdot\text{m}}$$ Had every element been compact this would have risen only to 7 772 kN·m, so the slender outstands cost about 7 %.
5. Bending about y-y. Now the bottom flange spans across the direction of the stress gradient and behaves as a plate supported along both edges in flexure; its effective width is $1900t/\sqrt{F_y} = 2031$ mm. The webs, nearly uniformly stressed at this distance from the y-y axis, reduce to $670t/\sqrt{F_y} = 716$ mm, and the top-flange outstands reduce as before. The extreme fibre is the outer tip of a top flange, 1 800 mm from the axis: $$I_{y,e} = 113.7\times10^{9}\ \text{mm}^4, \qquad S_{e,y} = \frac{113.7\times10^{9}}{1800} = 63.14\times10^{6}\ \text{mm}^3$$ $$\boxed{M_{ry} = 0.90(63.14\times10^{6})(350) = 19\,889\ \text{kN}\cdot\text{m}}$$

The section is roughly 2.7 times stronger about the vertical axis than about the horizontal one, which is exactly what a wide, shallow tub is for: it carries the deck loads about x-x while its great lateral stiffness about y-y suppresses lateral-torsional buckling entirely. That second point matters in Question A3, where the same section is used as an 18 m compression member.