Question 1 of 7: Two-span continuous welded plate girder

Question 1: Two-span continuous welded plate girder (12 + 6 + 2 = 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.

Find. Plate sizes (web depth and thickness, flange width and thickness) for a welded I-girder that satisfies flexure at the interior support, shear at the interior support, and the CSA S16 Cl. 14.6 moment–shear interaction.

Approach. Analyse the symmetric two-span beam elastically by the three-moment equation to obtain the governing hogging moment over \(B\) and the coincident shear, factor the load effects, then choose plate proportions that keep the flanges Class 1 and the web Class 2 so the full plastic moment \(\phi Z F_y\) is available without a Cl. 14 web-slenderness reduction.

1. Elastic analysis of the two-span beam. For two equal spans \(L\) each carrying a central point load \(P\), Clapeyron's three-moment equation with \(M_A = M_C = 0\) gives
   $$2M_B(L + L) = -\left(\tfrac{3}{8}PL^2 + \tfrac{3}{8}PL^2\right) \;\Rightarrow\; M_B = -\frac{3PL}{16}$$
   Substituting \(P = 500\ \text{kN}\) and \(L = 16\ \text{m}\),
   $$M_B = -\frac{3(500)(16)}{16} = -1500\ \text{kN}\cdot\text{m}\quad\text{(hogging)}$$
2. Reactions and span moment. Taking span \(AB\) as a simple beam with the end moment applied,
   $$R_A = \frac{P}{2} + \frac{M_B}{L} = 250 - \frac{1500}{16} = 156.25\ \text{kN}$$
   so the sagging moment under the load is \(M_{mid} = R_A(L/2) = 156.25 \times 8 = 1250\ \text{kN}\cdot\text{m}\), and the shear immediately left of \(B\) is \(V = 500 - 156.25 = 343.75\ \text{kN}\). The interior reaction is \(R_B = 687.5\ \text{kN}\). The support moment governs the design.
3. Trial plate proportions. A span-to-depth ratio near \(L/17\) is economical for a two-span girder, so try a web \(900 \times 10\ \text{mm}\) with flanges \(300 \times 20\ \text{mm}\), overall depth \(d = 940\ \text{mm}\). The plate properties follow directly:
   $$A = 2(300)(20) + (900)(10) = 21\,000\ \text{mm}^2, \qquad m = 164.9\ \text{kg/m}$$
   $$Z = b_f t_f (d - t_f) + \frac{t_w h_w^2}{4} = 300(20)(920) + \frac{10(900)^2}{4} = 7.545 \times 10^{6}\ \text{mm}^3$$
4. Section classification (CSA S16 Table 2). The flange projection and the web are checked against the welded-section limits:
   $$\frac{b}{t} = \frac{(300 - 10)/2}{20} = 7.25 \;\le\; \frac{145}{\sqrt{350}} = 7.75 \quad\text{(Class 1 flange)}$$
   $$\frac{h}{w} = \frac{900}{10} = 90.0 \;\le\; \frac{1700}{\sqrt{350}} = 90.9 \quad\text{(Class 2 web)}$$
   The section is therefore Class 2 in flexure and the plastic moment may be used. Keeping \(h/w\) below \(1900/\sqrt{F_y} = 101.6\) also means this is not a slender-web girder, so the Cl. 14.3.4 moment reduction \(M_r' = M_r\left[1 - 0.0005\,\frac{A_w}{A_f}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_f/\phi S}}\right)\right]\) does not apply.
5. Factored load effects. The girder self-weight is \(w = 164.9 \times 9.81/1000 = 1.617\ \text{kN/m}\) (dead). For two equal spans under a UDL, \(M_B = wL^2/8\) and the maximum span moment is \(9wL^2/128\). Combining with the factored point loads,
   $$M_{f,B} = 1.5(1500) + \frac{1.25(1.617)(16)^2}{8} = 2250 + 64.7 = \boxed{2314.7\ \text{kN}\cdot\text{m}}$$
   $$M_{f,mid} = 1.5(1250) + \frac{9(1.25)(1.617)(16)^2}{128} = 1911.4\ \text{kN}\cdot\text{m}$$
   $$V_f = 1.5(343.75) + \frac{5(1.25)(1.617)(16)}{8} = 515.6 + 20.2 = 535.8\ \text{kN}$$
6. Flexural resistance. With the compression flange braced every 2 m the unbraced length is far below the limit at which lateral-torsional buckling reduces the resistance: from Cl. 13.6 with \(\omega_2 = 1.0\), \(M_u = 20\,606\ \text{kN}\cdot\text{m} \gg 0.67M_p = 1769\ \text{kN}\cdot\text{m}\), so
   $$M_r = \phi Z F_y = 0.90(7.545\times10^{6})(350) = \boxed{2376.7\ \text{kN}\cdot\text{m}} \;\ge\; M_{f,B} = 2314.7\ \text{kN}\cdot\text{m}$$
   The girder is 97.4 % utilised in flexure — an economical section.
7. Shear resistance (Cl. 13.4.1.1). With no transverse stiffeners the shear-buckling coefficient is \(k_v = 5.34\). The slenderness thresholds are \(439\sqrt{k_v/F_y} = 54.2\), \(502\sqrt{k_v/F_y} = 62.0\) and \(621\sqrt{k_v/F_y} = 76.7\); since \(h/w = 90 > 76.7\) the web is in the elastic-buckling range and, with \(k_a = 0\) for an unstiffened web,
   $$F_s = F_{cre} = \frac{180\,000\,k_v}{(h/w)^2} = \frac{180\,000(5.34)}{90^2} = 118.7\ \text{MPa}$$
   $$V_r = \phi A_w F_s = 0.90(900)(10)(118.7) = \boxed{961.1\ \text{kN}} \;\ge\; V_f = 535.8\ \text{kN}$$
   No intermediate transverse stiffeners are required for shear; bearing stiffeners are still needed at \(A\), \(B\) and \(C\) and under each 500 kN load.
8. Moment–shear interaction (Cl. 14.6). The clause is triggered when both \(V_f > 0.60V_r\) and \(M_f > 0.75M_r\). Here \(V_f/V_r = 0.558\) is marginally below the trigger, but the check is performed for completeness because moment and shear peak at the same section:
   $$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727(0.974) + 0.455(0.558) = \boxed{0.962} \;\le\; 1.0$$
   The interaction is satisfied with 4 % reserve.