Question 7 of 7: Two-span continuous welded plate girder

Question 7: Two-span continuous welded plate girder

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. Two 12 m spans, pinned at A and on rollers at B and C. Unfactored loads of 300 kN at the mid-point of span AB and 200 kN at the mid-point of span BC. Lateral support at 2.0 m intervals; $F_y = 350$ MPa.

Find. A welded plate-girder section adequate in bending, in shear and in their interaction, using the stiffened-web approach.

Approach. Analyse the continuous beam, choose a deliberately slender web so transverse stiffeners can be used, then check flexure with the Cl 14.3.4 reduction, shear with tension-field action, and the two together.

1. Analyse the two-span beam. For a point load at the mid-point of one span, the hogging moment at the interior support is $3PL/32$; the two loads together give $-(3 \times 300 \times 12 + 3 \times 200 \times 12)/32 = -562.5$ kN·m unfactored. Factoring by 1.5 and adding the girder's own 0.91 kN/m gives $$\begin{gathered}M_B = -864.3 \text{ kN}\cdot\text{m}, \\ M_{span} = +938.4 \text{ kN}\cdot\text{m}, \\ V_f = 303.9 \text{ kN}\end{gathered}$$ with a reaction at B of 533 kN. The sagging moment in the heavier span slightly exceeds the hogging moment at the support, so both must be designed for.
2. Choose a web that suits the stiffened-web approach. The whole point of the question is a web too slender to carry shear by beam action alone. Take a web 800 × 6 mm, so $h/w = 133.3$. That is far above the Class 3 limit of $1900/\sqrt{F_y} = 101.6$, making the web Class 4 and the member a plate girder governed by Cl 14, but well inside the Cl 14.3.1 ceiling of $83\,000/F_y = 237$ that applies when transverse stiffeners are provided.
3. Size the flanges. Try 220 × 16 mm flanges, giving an overall depth of 832 mm, $A = 11840$ mm$^2$, $I = 1.4281 \times 10^{9}$ mm$^4$ and $S = 3.4328 \times 10^{6}$ mm$^3$. The flange slenderness $b/2t = 6.88$ is inside the Class 1 limit of 7.75, so the flanges themselves are compact and the section's weakness is entirely in the web.
4. Check flexure with the slender-web reduction. A Class 3 section would give $M_r = \phi S F_y = 1081$ kN·m, but Cl 14.3.4 reduces this whenever $h/w$ exceeds $1900/\sqrt{M_f/(\phi S)} = 109.0$, because the slender web sheds compression into the flanges: $$M_r' = M_r\left[1 - 0.0005 \frac{A_w}{A_f}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_f/\phi S}}\right)\right] = 1063 \text{ kN}\cdot\text{m}$$ a reduction of only 1.7 %, so $$\boxed{M_r' = 1063 \text{ kN}\cdot\text{m} \; \gt \; M_f = 938.4 \text{ kN}\cdot\text{m}}$$ a utilisation of 0.882. Lateral torsional buckling does not intervene: with bracing at 2.0 m, $L/r_y = 40.8$ and Cl 13.6 returns the full Class 3 capacity.
5. Design the web for shear using tension-field action. Space the intermediate transverse stiffeners at $a = 1200$ mm, i.e. $a/h = 1.5$, so $$k_v = 5.34 + \frac{4}{(a/h)^2} = 7.118$$ Since $h/w = 133.3$ exceeds $621\sqrt{k_v/F_y} = 88.6$, the web buckles elastically and $$F_{cri} = \frac{180\,000 k_v}{(h/w)^2} = 72.1 \text{ MPa}$$ The post-buckling tension field adds $k_a(0.50 F_y - 0.866 F_{cri})$ with $k_a = 1/\sqrt{1 + (a/h)^2} = 0.555$, giving $F_s = 134.5$ MPa and $$\boxed{V_r = \phi A_w F_s = 581 \text{ kN} \; \gt \; V_f = 303.9 \text{ kN}}$$ a utilisation of 0.523. Note that $F_{cri}$ alone would give only 311 kN: the tension field is contributing half the resistance, which is exactly why the stiffeners are needed.
6. Check the moment-shear interaction. Where tension-field action is relied upon, Cl 14.6 requires $$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} \le 1.0$$ Taking the largest moment and the largest shear together, which is conservative because they do not occur at the same section, gives $$0.727 \times 0.882 + 0.455 \times 0.523 = 0.879 \le 1.0$$
7. Design the stiffeners. Provide intermediate transverse stiffeners on one side of the web at 1200 mm centres throughout, and bearing stiffeners in pairs at A, B, C and under each load. The interior reaction is 533 kN; a pair of 100 × 14 mm plates has $b/t = 7.1$ against the Cl 14.4.2 limit of $200/\sqrt{F_y} = 10.7$, and acting with the adjacent $25w$ of web gives an effective column of 3700 mm$^2$ at $KL/r = 11.4$, so $C_r = 1160$ kN — a ratio of 0.459. The fitted bearing ends give $B_r = 1.5\phi A F_y = 1323$ kN, a ratio of 0.403.
8. Detail the welds. Continuous 6 mm fillet welds each side connect the web to the flanges; the horizontal shear flow $VQ/I$ at the flange-web junction is well within their capacity at the maximum shear. Stop the intermediate stiffeners short of the tension flange by four to six times the web thickness so they do not create a fatigue-sensitive detail at the point of maximum tensile stress.