Question 3 of 7: Three-span continuous welded plate girder

Question 3: Three-span continuous welded plate girder (15 + 5 = 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 for a welded I-girder that satisfy flexure at the critical midspan, shear at the supports and the CSA S16 Cl. 14.6 interaction, and the sustained displacement at the midpoint of the central span.

Approach. Solve the two-fold indeterminate beam with the three-moment equation, draw the moment diagram, size the plates for the largest sagging moment, then check shear, interaction and lateral–torsional buckling of the 3 m segment. For part (b) superpose the midspan sagging deflection of span \(BC\) on the upward deflection caused by its two hogging end moments.

1. Support moments. For a central point load, the three-moment load term is \(6A\bar{x}/L = 3PL^2/8\) from either side. Symmetry gives \(M_B = M_C = M\), so for the pair of spans \(AB\)–\(BC\):
   $$2M_B(L_1 + L_2) + M_C L_2 = -\left[\frac{3(300)(12)^2}{8} + \frac{3(200)(12)^2}{8}\right]$$ $$48M + 12M = -27\,000 \quad\Longrightarrow\quad \boxed{M_B = M_C = -450\ \text{kN}\cdot\text{m}\ \text{(hogging)}}$$
2. Span moments and reactions. Adding the free bending moment to the linear end-moment diagram,
   $$M_{\text{mid},AB} = \frac{300(12)}{4} + \frac{0 + (-450)}{2} = 675\ \text{kN}\cdot\text{m} \qquad M_{\text{mid},BC} = \frac{200(12)}{4} - 450 = 150\ \text{kN}\cdot\text{m}$$

   and \(R_A = 150 + 450/12 = 187.5\ \text{kN}\), \(R_B = R_C = 212.5\ \text{kN}\), \(R_D = 187.5\ \text{kN}\). The outer spans carry the critical moment; the middle span is almost unstressed.

Bending moment diagram under the specified loads. The 675 kN·m sagging moment in the outer spans governs the design; the middle span carries only 150 kN·m.
3. Trial section and factored moments. Try flanges 2–220 × 16 with a 700 × 8 web: \(d = 732\ \text{mm}\), \(A = 12\,640\ \text{mm}^2\), 99.2 kg/m (\(w = 0.973\ \text{kN/m}\), factored 1.217 kN/m), \(Z = 3.500\times10^6\ \text{mm}^3\), \(I_x = 1131\times10^6\ \text{mm}^4\), \(r_y = 47.4\ \text{mm}\). A uniform load on the same three-span layout gives \(M_{\text{support}} = 14.4w = 17.5\) and \(M_{\text{mid}} = 10.8w = 13.1\) kN·m, so
   $$M_f = 1.5(675) + 13.1 = 1025.6\ \text{kN}\cdot\text{m} \qquad M_{f,B} = 1.5(450) + 17.5 = 692.5\ \text{kN}\cdot\text{m}$$
4. Classify and check flexure. Flange \(b/t = 6.62 \le 7.75\) (Class 1); web \(h/w = 87.5\), between \(1100/\sqrt{F_y} = 58.8\) and \(1700/\sqrt{F_y} = 90.9\), so the web is Class 2 and the section develops \(Z\). It is also below \(1900/\sqrt{F_y} = 101.6\), so the Cl. 14.3.4 plate-girder flange-stress reduction does not apply.
   $$M_r = \phi ZF_y = 0.90(3.500\times10^6)(350)/10^6 = 1102.6\ \text{kN}\cdot\text{m}$$
5. Lateral–torsional buckling of the critical 3 m segment. The segment runs from the 3 m brace to the load point, where \(M = 843.8\) and 1012.5 kN·m (factored, live only), single curvature, so \(\kappa = -0.833\) and
   $$\omega_2 = 1.75 + 1.05\kappa + 0.3\kappa^2 = 1.083$$ $$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^2 I_yC_w} = 2500\ \text{kN}\cdot\text{m}\ >\ 0.67M_p$$ $$M_r = 1.15\phi M_p\left(1 - \frac{0.28M_p}{M_u}\right) = 1094.0\ \text{kN}\cdot\text{m} \ \ge\ 1025.6\ \text{kN}\cdot\text{m}\quad(\text{utilisation } 0.937)$$
6. Shear. The largest shear is at \(A\): \(V_f = 1.5(187.5) + 8.8 = 290.0\ \text{kN}\); just left of \(B\), \(V_f = 174.6\ \text{kN}\). For an unstiffened web (\(k_v = 5.34\)), \(h/w = 87.5 > 621\sqrt{k_v/F_y} = 76.7\), so elastic buckling governs:
   $$F_s = F_{cre} = \frac{180\,000k_v}{(h/w)^2} = \frac{180\,000(5.34)}{87.5^2} = 125.5\ \text{MPa}$$ $$V_r = \phi A_w F_s = 0.90(700)(8)(125.5)/10^3 = 632.7\ \text{kN}\ \ge\ 290.0\ \text{kN}$$

   No intermediate transverse stiffeners are needed for strength.

7. Moment–shear interaction. The worst simultaneous pair is at \(B\):
   $$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727(0.628) + 0.455(0.276) = 0.582 \le 1.0$$

   Neither trigger of Cl. 14.6 (\(V_f/V_r > 0.60\) and \(M_f/M_r > 0.75\)) is even reached, so interaction is not critical.

8. Bearing stiffeners. The factored reaction at \(B\) is 331.9 kN. A pair of 100 × 12 plates acting with \(25w = 200\ \text{mm}\) of web gives \(A = 4000\ \text{mm}^2\), \(r = 47.4\ \text{mm}\), \(KL/r = 11.1\) and \(C_r = 1254\ \text{kN}\) — ample. Fit them at all four supports and under the three loads.

9. Part (b): displacement at the midpoint of \(BC\). Steel does not creep, so there is no long-term multiplier: the sustained displacement equals the elastic displacement under the sustained loads. Span \(BC\) carries a 200 kN central load between two 450 kN·m hogging end moments, and those end moments bend the span upward:
   $$\delta = \frac{PL^3}{48EI} - \frac{M_eL^2}{8EI} = \frac{200\times10^3(12\,000)^3}{48(2.262\times10^{14})} - \frac{450\times10^6(12\,000)^2}{8(2.262\times10^{14})}$$ $$\delta = 31.83 - 35.81 = -3.98\ \text{mm}$$

   The girder self weight adds \(5wL^4/384EI = 1.16\ \text{mm}\) down and its own end moments take 1.12 mm back up, a net 0.05 mm. Hence

   $$\boxed{\delta_{BC} = -3.93\ \text{mm}\ \text{(3.9 mm UPWARD)}}$$

   The middle span rises because its own load is small (200 kN against 300 kN in the neighbours) while the end moments it inherits are large; the crossover is at \(M_e = PL/6 = 400\) kN·m, and here \(M_e = 450\) kN·m. The magnitude is \(L/3050\), far inside the \(L/360 = 33\ \text{mm}\) serviceability limit, but it must be allowed for when setting cambers and levelling the deck.