Question 2 of 7: Moments of resistance of the built-up bridge section

Question A2: Moments of resistance of the built-up bridge section (20 marks: 6 + 7 + 7)

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 of the trough section about its own centroidal axes.

Approach. Build the section from four component rectangles, locate the centroid, accumulate $I_x$ and $I_y$ using the rotated-rectangle formulae for the inclined webs, classify every plate element against S16 Table 2, and then take $M_r = \phi F_y S$ on whatever section — gross or effective — the classification allows.

1. Web geometry. Clear depth between flanges $h_w = 900 - 2(20) = 860$ mm; horizontal run over that depth $300 (860/900) = 286.7$ mm; sloping length $$L_w = \sqrt{860^2 + 286.7^2} = 906.5\ \text{mm},\qquad \cos\theta = \frac{860}{906.5} = 0.9487,\quad \sin\theta = 0.3162$$ so the webs stand at $\theta = 18.4^{\circ}$ from the vertical.
2. Areas. Bottom flange $A_1 = 2500(20) = 50\,000$ mm$^2$; each top flange $A_2 = 500(20) = 10\,000$ mm$^2$; each web $A_3 = 906.5(20) = 18\,130$ mm$^2$. Hence $$A_g = 50\,000 + 2(10\,000) + 2(18\,130) = 106\,261\ \text{mm}^2$$
3. Centroid above the soffit. Taking first moments about the soffit, with the component centroids at 10, 890 and 450 mm, $$\bar y = \frac{50\,000(10) + 2(10\,000)(890) + 2(18\,130)(450)}{106\,261} = \frac{34.617 \times 10^6}{106\,261}$$ $$\boxed{\bar y = 325.8\ \text{mm above the soffit},\qquad y_{\text{top}} = 900 - 325.8 = 574.2\ \text{mm}}$$ The centroid sits low because the 2500 mm bottom plate is half the area of the section — the “varies” dimension on the exam figure is this quantity.
4. Second moment about x–x. For a thin rectangle of length $L$ and thickness $t$ inclined at $\theta$ to the vertical, the own centroidal value is $I = \tfrac{1}{12}L^3 t\cos^2\theta + \tfrac{1}{12}L t^3 \sin^2\theta$, giving $1.118 \times 10^9$ mm$^4$ per web. Adding the parallel-axis terms, $$I_x = \underbrace{4.985 \times 10^9}_{\text{bottom flange}} + \underbrace{6.367 \times 10^9}_{\text{two top flanges}} + \underbrace{2.795 \times 10^9}_{\text{two webs}}$$ $$\boxed{I_x = 14.16 \times 10^9\ \text{mm}^4}$$
5. Second moment about y–y. The bottom plate now bends about its own strong direction, and the top flanges sit 1550 mm off the axis (1250 + 300), the web centroids 1393 mm off it: $$I_y = \frac{20(2500)^3}{12} + 2\left[\frac{20(500)^3}{12} + 10\,000(1550)^2\right] + 2\left[0.125 \times 10^9 + 18\,130(1393)^2\right]$$ $$\boxed{I_y = 144.9 \times 10^9\ \text{mm}^4}$$ Ten times $I_x$ — the section is a wide, shallow trough, which is exactly what a footbridge deck wants.
6. Elastic section moduli. $$S_{x,\text{top}} = \frac{14.16 \times 10^9}{574.2} = 24.66 \times 10^6\ \text{mm}^3, \qquad S_{x,\text{bot}} = \frac{14.16 \times 10^9}{325.8} = 43.46 \times 10^6\ \text{mm}^3$$ About y–y the extreme fibre is the outer tip of a top plate at $1250 + 300 + 250 = 1800$ mm, so $$S_y = \frac{144.9 \times 10^9}{1800} = 80.51 \times 10^6\ \text{mm}^3$$
7. Classify every plate element (S16 Table 2, $\sqrt{F_y} = 18.71$). This is the step the marks turn on, because a 20 mm plate is thin relative to these widths.
   • Top-flange outstand, supported along one edge: $b/t = (500 - 20)/2 \div 20 = 12.0$ against a Class 3 limit of $200/\sqrt{F_y} = 10.7$ — Class 4 (marginally).
   • Webs in flexure, supported along two edges: $h/w = 860/20 = 43.0$ against a Class 1 limit of $1100/\sqrt{F_y} = 58.8$ — Class 1.
   • Bottom flange, supported along two edges: $b/t = 125$ against a Class 3 limit of $670/\sqrt{F_y} = 35.8$ — grossly Class 4, but it lies in the tension zone for sagging bending about x–x, where classification does not apply.
   The section is therefore Class 4 and S16 Cl.13.5(c)(iii) applies: reduce the Class 4 compression elements to an effective width of $200t/\sqrt{F_y}$ where supported along one edge and $670t/\sqrt{F_y}$ where supported along two, then use the elastic modulus of the reduced section.
8. Effective section for bending about x–x (top flanges in compression). The effective outstand is $200(20)/18.71 = 213.8$ mm each side, so each top plate reduces from 500 mm to $2(213.8) + 20 = 447.6$ mm. Repeating steps 3 to 6 on the reduced section gives $\bar y_e = 314.4$ mm, $I_{xe} = 13.47 \times 10^9$ mm$^4$ and $S_{xe,\text{top}} = 23.00 \times 10^6$ mm$^3$, whence $$M_{rx} = \phi F_y S_{xe} = 0.90 \times 350 \times 23.00 \times 10^6 \times 10^{-6}$$ $$\boxed{M_{rx} = 7245\ \text{kN}\cdot\text{m}}$$ Had the flanges satisfied Class 3, the gross value would be $0.90(350)(24.66 \times 10^6) = 7768$ kN·m — the Class 4 penalty is only 7 per cent, because the flanges are barely over the limit and the section is web-and-bottom-plate dominated.
9. Bending about y–y. Now the whole of one web is in near-uniform compression, a plate supported along two edges with $L_w/t = 45.3$ against the Class 3 limit of 35.8, so it too reduces — to $670(20)/18.71 = 716.3$ mm effective, taken adjacent to the two supported edges. With the top flanges reduced as before, $I_{ye} = 125.1 \times 10^9$ mm$^4$ and $S_{ye} = 69.50 \times 10^6$ mm$^3$: $$M_{ry} = 0.90 \times 350 \times 69.50 \times 10^6 \times 10^{-6}$$ $$\boxed{M_{ry} = 21\,893\ \text{kN}\cdot\text{m}}$$