Question 1 of 7: Welded stiffened-web plate girder ABC

Question 1: Welded stiffened-web plate girder ABC (12 + 5 + 3 = 20 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 two-span girder ABC read from Figure 1: rigid (fixed) supports at A and C, a simple support at B, spans \(AB = BC = 8\ \text{m}\) each made of two 4 m bays, and a 500 kN point load at the centre of each span. The figure notes lateral support at 2 m centres.

Find. A welded plate-girder cross-section, together with a transverse stiffener arrangement, that satisfies flexure, shear and the CSA S16 flexure–shear interaction rule.

Approach. Exploit the symmetry of Figure 1 to reduce each span to a fixed-ended beam, size the flanges from the elastic section modulus with the CSA S16 Cl. 14.3.4 slender-web reduction, then choose a stiffener spacing that develops enough tension-field shear, and close with the Cl. 14.6 interaction check at the support where \(M_f\) and \(V_f\) peak together.

1. Reduce the structure by symmetry. The girder, its supports and its loading are all symmetric about B, so the rotation at B is zero. Each span therefore behaves as a beam built in at both ends and carrying a central point load: $$M_{\text{end}} = M_{\text{mid}} = \frac{PL}{8}, \qquad V = \frac{P}{2}$$ This also tells us that the hogging moment over the interior support B equals the hogging moment at the built-in ends A and C, so a single section serves the whole girder.
2. Trial section and self weight. Take a welded I-section with a web \(850 \times 6\) and flanges \(220 \times 14\), giving an overall depth \(d = 878\ \text{mm}\) and \(h/w = 850/6 = 141.7\). Its area is \(A = 2(220)(14) + 850(6) = 11\,260\ \text{mm}^2\), which weighs 0.87 kN/m; allowing for stiffeners and welds, take \(w_D = 1.00\ \text{kN/m}\). The gross second moment of area and elastic section modulus are $$I_x = \frac{6(850)^3}{12} + 2\left[\frac{220(14)^3}{12} + 220(14)\left(432\right)^2\right] = 1.4568 \times 10^{9}\ \text{mm}^4, \qquad S_x = \frac{2I_x}{d} = 3.318 \times 10^{6}\ \text{mm}^3$$
3. Factored design actions. With \(P_f = 1.5(500) = 750\ \text{kN}\) and \(w_f = 1.25(1.00) = 1.25\ \text{kN/m}\), $$M_f^{\text{sup}} = \frac{P_fL}{8} + \frac{w_fL^2}{12} = \frac{750(8)}{8} + \frac{1.25(8)^2}{12} = \boxed{756.7\ \text{kN}\cdot\text{m}}$$ and correspondingly \(M_f^{\text{mid}} = 750 + 1.25(8)^2/24 = 753.3\ \text{kN}\cdot\text{m}\), \(V_f = 375 + 5.0 = 380.0\ \text{kN}\). The reaction delivered to the interior support is \(2V_f = 760\ \text{kN}\), which will size the bearing stiffener there.
4. (a) Flexure — slender-web reduction. On the gross section \(\phi S_x F_y = 0.9(3.318\times10^6)(350) = 1045.3\ \text{kN}\cdot\text{m}\). Because $$\frac{h}{w} = 141.7 \;>\; \frac{1900}{\sqrt{M_f/(\phi S)}} = \frac{1900}{\sqrt{253.4}} = 119.4$$ the web is slender and CSA S16 Cl. 14.3.4 reduces the moment resistance by the amount the web sheds to the flanges: $$M_r' = M_r\left[1 - 0.0005\,\frac{A_w}{A_f}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_f/\phi S}}\right)\right] = 1045.3\left[1 - 0.0005(1.656)(22.3)\right] = \boxed{1026.0\ \text{kN}\cdot\text{m}}$$ Since \(M_f = 756.7 \le 1026.0\), flexure is satisfied at 74 % of capacity. Lateral–torsional buckling does not intrude: at \(L_b = 2\ \text{m}\), \(M_u = 5341\ \text{kN}\cdot\text{m}\) and the Cl. 13.6 inelastic expression is capped at \(\phi M_y = 1045.3\ \text{kN}\cdot\text{m}\), so the 2 m bracing noted on the figure is exactly what makes the thin flange work.
5. (b) Shear — tension-field action. Place intermediate transverse stiffeners at \(a = 2000\ \text{mm}\), so \(a/h = 2.353\) and $$k_v = 5.34 + \frac{4}{(a/h)^2} = 6.0625$$ With \(h/w = 141.7 > 621\sqrt{k_v/F_y} = 81.7\) the web buckles elastically, so $$F_{cri} = \frac{180\,000\,k_v}{(h/w)^2} = 54.37\ \text{MPa}, \qquad F_t = \frac{0.50F_y - 0.866F_{cri}}{\sqrt{1 + (a/h)^2}} = 50.03\ \text{MPa}$$ Adding the post-buckling tension field to the buckling stress, \(F_s = 104.41\ \text{MPa}\) and $$V_r = \phi A_w F_s = 0.9(850)(6)(104.41) = \boxed{479.2\ \text{kN}} \;>\; V_f = 380.0\ \text{kN}$$ The end panels at A and C cannot anchor a tension field, so they are designed on \(F_{cri}\) alone; closing them to \(a = 700\ \text{mm}\) raises \(k_v\) to 11.87 and gives \(V_r = 488.8\ \text{kN}\), again above 380 kN.
6. (c) Flexure–shear interaction. At the built-in ends and at B the peak moment and the peak shear act on the same cross-section, so Cl. 14.6 applies: $$0.727\frac{M_f}{M_r'} + 0.455\frac{V_f}{V_r} = 0.727\left(\frac{756.7}{1026.0}\right) + 0.455\left(\frac{380.0}{479.2}\right) = 0.536 + 0.361 = \boxed{0.897 \le 1.0}$$ The section passes with 10 % in hand, and the interaction — not flexure or shear alone — is what governs the design.
7. Stiffener details. Intermediate pairs need \(I_s \ge a w^3 j = 2000(6)^3(0.5) = 2.16 \times 10^{5}\ \text{mm}^4\); a pair of \(80 \times 8\) plates supplies \(3.05 \times 10^{6}\ \text{mm}^4\). At A, B and C a bearing pair \(130 \times 14\) acting with \(25w\) of web gives \(C_r = \phi A F_y = 1430\ \text{kN}\) and a bearing resistance \(B_r = 1.5\phi_{bi}AF_y = 1388\ \text{kN}\), both comfortably above the 760 kN reaction at B.