Question 4 of 7: Two-span welded plate girder — flexure, stiffened web, interaction

Question 4: Two-span welded plate girder — flexure, stiffened web, interaction (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. Two equal spans A–B and B–C of 13 m, each carrying a specified 400 kN load at its centre (6.5 m from each support); $F_y = 350$ MPa. Factored load per span $P = 1.5(400) = 600$ kN. The compression flange is braced at the supports and at the load points, i.e. every 6.5 m.

Find. A welded cross-section adequate in flexure, a transverse-stiffener layout for the web, and the moment–shear interaction check at the interior support.

Approach. Analyse the continuous beam for the hogging moment and shear at B, which occur together and therefore govern; choose plate sizes; classify the section, reduce the moment resistance for the slender web (Cl. 14.3.4) and for lateral–torsional buckling (Cl. 13.6); size the stiffener spacing from Cl. 13.4.1.1; then combine the two in Cl. 14.6.

1. Part (a) — continuous-beam analysis. For two equal spans with a central point load in each, superposition of the standard single-span result $3PL/32$ gives $$M_B = -\frac{3PL}{16} = -\frac{3(600)(13)}{16} = -1462.5\ \text{kN}\cdot\text{m}$$ and the end reaction is $P/2 - |M_B|/L = 300 - 112.5 = 187.5$ kN, so the span moment is $187.5(6.5) = 1218.75$ kN·m. The shear just inside B is $600 - 187.5 = 412.5$ kN.
2. Add the girder self-weight. The section selected below weighs 1.31 kN/m; allowing 10 % for stiffeners and factoring at 1.25 gives $w_f = 1.80$ kN/m, which adds $w_fL^2/8 = 37.9$ kN·m at B and $5w_fL/8 = 14.6$ kN at B. The design actions are therefore $$\boxed{M_f = 1500\ \text{kN}\cdot\text{m} \ \text{and}\ V_f = 427\ \text{kN, both at the interior support B}}$$ with a span moment of 1238 kN·m, and a support reaction $R_B = 854$ kN.
3. Trial plates and classification. Take flanges $280 \times 16$ and web $1000 \times 8$, so $d = 1032$ mm and $A = 16\,960$ mm². Then $I_x = 2979\times10^6$ mm⁴ and $$S_x = \frac{I_x}{d/2} = \frac{2979\times10^6}{516} = 5773\times10^3\ \text{mm}^3, \qquad \phi S_x F_y = 1819\ \text{kN}\cdot\text{m}$$ The flange, $b/t = 8.5$, is Class 2 ($\le 170/\sqrt{F_y} = 9.09$); the web, $h/w = 125$, exceeds the Class 3 limit $1900/\sqrt{F_y} = 101.6$, so this is a genuine slender-web plate girder and the stiffened design the question calls for is the right vehicle.
4. Lateral–torsional buckling over the 6.5 m braced length. With $I_y = 58.6\times10^6$ mm⁴, $J = 0.935\times10^6$ mm⁴ and $C_w = 15.12\times10^{12}$ mm⁶, and with the segment from B to the load point bent in double curvature ($\kappa = -0.825$, giving $\omega_2 = 2.5$ at the cap), $$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^2I_yC_w} = 3649\ \text{kN}\cdot\text{m} > 0.67M_y$$ $$M_r = 1.15\phi M_y\left(1 - \frac{0.28M_y}{M_u}\right) = 1767\ \text{kN}\cdot\text{m} \le \phi M_y = 1819\ \text{kN}\cdot\text{m}$$
5. Slender-web reduction, Cl. 14.3.4. A slender web sheds some of its share of the compression block to the flanges, and the code accounts for it with $$M'_r = M_r\left[1 - 0.0005\frac{A_w}{A_f}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_f/\phi S}}\right)\right]$$ Here $A_w/A_f = 8000/4480 = 1.79$ and $M_f/\phi S = 289$ MPa, so the bracket term is $125 - 111.8 = 13.2$ and the reduction is only 1.2 %: $$\boxed{M'_r = 1767(0.988) = 1746\ \text{kN}\cdot\text{m} \ \ge\ M_f = 1500\ \text{kN}\cdot\text{m}}$$
6. Part (b) — why stiffeners are needed. With no intermediate stiffeners, $k_v = 5.34$ and $h/w = 125$ exceeds $621\sqrt{k_v/F_y} = 76.7$, so the web is in the elastic-buckling range: $$F_{cre} = \frac{180\,000k_v}{(h/w)^2} = \frac{180\,000(5.34)}{15\,625} = 61.5\ \text{MPa}, \qquad V_r = \phi A_wF_s = 0.9(8000)(61.5) = 443\ \text{kN}$$ That is only 3.7 % above the 427 kN demand — no margin at all for a girder this slender, and it takes no advantage of tension-field action.
7. Stiffener spacing. Adopt $a = 1500$ mm, i.e. $a/h = 1.5$, so $k_v = 5.34 + 4/(1.5)^2 = 7.12$ and $k_a = 1/\sqrt{1 + (a/h)^2} = 0.555$. With $F_{cre} = 82.0$ MPa the post-buckling (tension-field) term applies: $$F_s = F_{cre} + k_a(0.50F_y - 0.866F_{cre}) = 82.0 + 0.555(175 - 71.0) = 139.7\ \text{MPa}$$ $$\boxed{V_r = 0.9(1000 \times 8)(139.7) = 1006\ \text{kN} \ \ge\ 427\ \text{kN}}$$ Intermediate stiffeners need $I_{st} \ge (h/50)^4 = 1.6\times10^5$ mm⁴; a single-sided $90 \times 10$ plate gives $wb^3/3 = 2.43\times10^6$ mm⁴, so use $90 \times 10$ plates on one side at 1500 mm centres.
8. Bearing stiffeners. At B the reaction is 854 kN. A pair of $125 \times 16$ plates acting with $25w$ of web gives $A = 5600$ mm² and, over $0.75h$, $KL/r$ is so small that $C_r = 1755$ kN; the bearing check with a 15 mm clip gives $B_r = 1.5\phi A F_y = 1663$ kN. Both exceed 854 kN, so a single pair suffices at each support and at each load point.
9. Part (c) — moment–shear interaction, Cl. 14.6. Because the maximum moment and the maximum shear occur at the same section, the web must be checked for both at once: $$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727\frac{1500}{1746} + 0.455\frac{427}{1006} = 0.625 + 0.193$$ $$\boxed{= 0.82 \le 1.0 \quad \text{— the section is adequate}}$$ The flexure term dominates, which tells the designer that if this girder had to be lightened, the flanges — not the web — are where the material is doing the work.