Question 1 of 7: Welded Stiffened-Web Plate Girder

Question 1: Welded Stiffened-Web Plate Girder (12 + 5 + 3 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. A doubly built-in 12 m girder carrying three point loads, to be made from plate with the page-1 steel grade.

Find. A welded plate-girder cross-section, plus a transverse-stiffener layout, that satisfies CSA S16 in flexure (Cl 14.3.4), in shear with tension-field action (Cl 13.4.1.1) and in combined flexure and shear (Cl 14.6).

[Figure not reproduced: Figure 1 as read from the exam drawing sheet, with the factored (1.5 ×) bending-moment diagram of the built-in beam beneath it. Both the peak moment and the peak shear occur at the supports, which is what makes part (c) bite. See the official exam paper.]

Approach. Factor the loads, get the built-in moment and shear envelope by superposing fixed-end moments, choose a deep slender web with modest flanges, then take the three parts in the order the code applies them — reduced flexural resistance for a slender web, tension-field shear resistance, and finally the interaction equation that only bites when both ratios are high.

1. Factor the loads and find the design actions. For a point load $P$ at distance $a$ from the left end of a built-in span $L$ (with $b = L - a$), the fixed-end moments are $M_{A} = -Pab^{2}/L^{2}$ and $M_{D} = +Pa^{2}b/L^{2}$. Superposing the three factored loads $1.5\times300 = 450$ kN, $1.5\times400 = 600$ kN and $1.5\times300 = 450$ kN gives $M_{A} = -(759.4 + 900.0 + 253.1) = -1912.5$ kN.m and, by symmetry, $M_{D} = +1912.5$ kN.m, with end shears $V = (450+600+450)/2 = 750$ kN.
2. Add the girder self weight. The trial section below has $A_{g} = 20\,000$ mm$^{2}$, so $w = 20\,000\times10^{-6}\times7850\times9.81\times10^{-3} = 1.54$ kN/m, factored to $1.25\times1.54 = 1.93$ kN/m. A built-in span under a uniform load adds $wL^{2}/12 = 1.93\times12^{2}/12 = 23.1$ kN.m at each end and $wL/2 = 11.6$ kN to each end shear, so the design actions become $$\boxed{M_{f} = 1935.6\ \text{kN.m}\qquad V_{f} = 761.6\ \text{kN}}$$ Both peak at the same section, the support, which is exactly why part (c) is asked.
3. Choose a trial cross-section. Take a web 1000 mm deep by 8 mm thick with 300 mm by 20 mm flanges, giving an overall depth $d = 1040$ mm. Then $$I_{x} = \frac{8\times1000^{3}}{12} + 2\left[\frac{300\times20^{3}}{12} + 300\times20\times510^{2}\right] = 3.788\times10^{9}\ \text{mm}^{4}$$ and the elastic section modulus is $S_{x} = 2I_{x}/d = 7.285\times10^{6}$ mm$^{3}$. The web area is $A_{w} = 8000$ mm$^{2}$ and each flange area $A_{f} = 6000$ mm$^{2}$.
4. Classify the plate elements. The flange projection ratio is $b/2t = 300/40 = 7.50$, inside the Class 1 limit $145/\sqrt{F_{y}} = 7.75$, so the flange never governs. The web slenderness is $h/w = 1000/8 = 125$, far beyond the $1014/\sqrt{F_{y}} = 54.2$ at which an unstiffened web can still reach shear yield, but well within the Cl 14.3.1 ceiling of $83\,000/F_{y} = 237$ that applies once transverse stiffeners are provided. The girder is therefore a stiffened-web design, as the question directs.
5. Part (a) — flexural resistance with a slender web (Cl 14.3.4). A slender web sheds compressive stress to the flanges, so S16 reduces the elastic resistance. With $\phi S_{x}F_{y} = 0.9\times7.285\times10^{6}\times350 = 2295$ kN.m and $M_{f}/(\phi S_{x}) = 1935.6\times10^{6}/(0.9\times7.285\times10^{6}) = 295.2$ MPa, $$M_{r} = \phi S_{x}F_{y}\left[1 - 0.0005\frac{A_{w}}{A_{f}}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_{f}/(\phi S_{x})}}\right)\right]$$ $$M_{r} = 2295\left[1 - 0.0005\times1.333\times(125 - 110.6)\right] = 2295\times0.9904 = \boxed{2273\ \text{kN.m}}$$ Since $M_{f}/M_{r} = 1935.6/2273 = 0.852 \lt 1.0$, flexure is satisfied. The reduction is only 1 per cent here because the web is not far past the threshold slenderness.
6. Part (b) — shear resistance with tension-field action (Cl 13.4.1.1). Adopt intermediate stiffeners at $a = 1000$ mm, so $a/h = 1.0$ and the shear buckling coefficient is $k_{v} = 5.34 + 4/(a/h)^{2} = 9.34$. Because $h/w = 125$ exceeds $621\sqrt{k_{v}/F_{y}} = 101.4$, the elastic branch applies: $$F_{cri} = \frac{180\,000\,k_{v}}{(h/w)^{2}} = \frac{180\,000\times9.34}{125^{2}} = 107.6\ \text{MPa}$$ The post-buckling tension field adds $$f_{t} = \frac{0.50F_{y} - 0.866F_{cri}}{\sqrt{1 + (a/h)^{2}}} = \frac{175.0 - 93.2}{\sqrt{2}} = 57.9\ \text{MPa}$$ so that $F_{s} = 107.6 + 57.9 = 165.5$ MPa and $$V_{r} = \phi A_{w}F_{s} = 0.9\times8000\times165.5 = \boxed{1191\ \text{kN}}$$ giving $V_{f}/V_{r} = 761.6/1191 = 0.639$.
7. Check the end panel separately. A tension field has to anchor against an adjacent panel, so S16 does not allow it in an end panel; the end panel must carry its shear on buckling alone. Tightening the end spacing to $a = 800$ mm raises $k_{v}$ to $5.34 + 4/0.8^{2} = 11.59$ and $F_{cri}$ to $180\,000\times11.59/125^{2} = 133.5$ MPa, so $V_{r} = 0.9\times8000\times133.5 = 961$ kN, comfortably above the 761.6 kN that acts there. Without that tightening the 1000 mm end panel would return only 775 kN and the margin would nearly vanish.
8. Size the intermediate stiffeners. The projection must satisfy $b_{s}\ge 50 + h/30 = 83$ mm and $b_{s}/t_{s}\le200/\sqrt{F_{y}} = 10.7$; a pair of 90 mm by 10 mm plates, one each side of the web, satisfies both at $b_{s}/t_{s} = 9.0$. Their second moment of area about the web centreline is $2t_{s}b_{s}\left[b_{s}^{2}/12 + (b_{s}/2 + w/2)^{2}\right] = 5.54\times10^{6}$ mm$^{4}$, far above the Cl 14.5.3 requirement of $(h/50)^{4} = 1.6\times10^{5}$ mm$^{4}$. Stiffeners resisting a tension field need not be welded to the tension flange.
9. Part (c) — combined flexure and shear (Cl 14.6). The interaction clause applies whenever a tension field is used and both ratios are high, which is the case here ($M_{f}/M_{r} = 0.852 \gt 0.75$ and $V_{f}/V_{r} = 0.639 \gt 0.60$): $$0.727\frac{M_{f}}{M_{r}} + 0.455\frac{V_{f}}{V_{r}} = 0.727(0.852) + 0.455(0.639) = \boxed{0.910 \le 1.0}$$ The girder therefore passes all three parts, with the interaction check the closest of the three at 91 per cent utilisation.