Question 1 of 7: Welded Stiffened-Web Plate Girder

Question 1: Welded Stiffened-Web Plate Girder (12 + 5 + 3 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. A welded plate-girder cross-section plus a transverse-stiffener layout that satisfies CSA S16 in flexure (Cl 14.3.4), in shear with tension-field action (Cl 13.4.1.1) and in combined flexure–shear (Cl 14.6).

[Figure not reproduced: Figure 1 as printed on page 4 of the examination paper. See the official exam paper.]

Approach. Exploit the symmetry of Figure 1 to reduce the analysis to a single fixed-ended span, then size a slender-web girder and check the three limit states the question names, in the order it names them.

1. Analyse the girder. The structure and the loading are both symmetric about B, so the rotation at B vanishes, θ_B = 0, and each 8 m span behaves as a beam built in at both ends carrying a central point load. For that case $$M_{\text{end}} = M_{\text{centre}} = \frac{PL}{8}, \qquad V = \frac{P}{2}$$ With α = 1.5 on the 500 kN loads, $P_f = 750\ \text{kN}$, and adding the girder self weight $w_f = 1.25 \times 2.2 = 2.75\ \text{kN/m}$ (fixed-ended: $wL^2/12$ at the ends, $wL^2/24$ at midspan), $$\boxed{M_f = 764.7\ \text{kN}\cdot\text{m} \ \ (\text{hogging at A, B, C}), \quad M_f = 757.3\ \text{kN}\cdot\text{m}\ (\text{sagging}), \quad V_f = 386.0\ \text{kN}}$$ The prop reaction at B is $2 \times 386.0 = 772.0$ kN.
2. Choose a trial cross-section. For a 16 m girder a web depth near L/20 is economical. Take a web 800 × 6 with flanges 200 × 14, giving $d = 828$ mm. From first principles, $$I_x = \frac{w h^3}{12} + 2\left[\frac{b_f t_f^3}{12} + b_f t_f\left(\frac{h+t_f}{2}\right)^2\right] = 1183.7 \times 10^6\ \text{mm}^4, \qquad S_x = \frac{2I_x}{d} = 2859 \times 10^3\ \text{mm}^3$$ The actual self weight is 0.80 kN/m, comfortably inside the 2.2 kN/m allowed for stiffeners, splices and bracing.
3. Classify the plate elements. The flange projection gives $b/t = 100/14 = 7.14$, against the Class 2 limit $170/\sqrt{F_y} = 9.09$ — so the flange is Class 2, which is all that is needed, because Clause 14.3.4 forces an elastic (not plastic) moment resistance regardless. The web slenderness is $h/w = 800/6 = 133.3$, within the Clause 14.3.1 ceiling $83\,000/F_y = 237$ for a stiffened web. Its Class 3 limit, however, is $$\frac{1900}{\sqrt{M_f/(\phi S_x)}} = \frac{1900}{\sqrt{297.2}} = 110.2 \;<\; 133.3$$ so the web is slender and the girder must be designed as a plate girder rather than a beam.
4. Part (a) — flexural resistance, Clause 14.3.4. A slender web sheds stress to the flanges as it buckles, and S16 accounts for that with a reduction on the elastic resistance: $$M_r = \phi S_x F_y\left[1 - 0.0005\,\frac{A_w}{A_f}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_f/(\phi S_x)}}\right)\right]$$ With $A_w/A_f = 4800/2800 = 1.714$, $$M_r = 0.9(2859\times10^3)(350)\left[1 - 0.0005(1.714)(133.3 - 110.2)\right] = \boxed{882.8\ \text{kN}\cdot\text{m}}$$ so $M_f/M_r = 764.7/882.8 = 0.866$ and flexure is satisfied. Lateral–torsional buckling is not an issue: with bracing every 2 m, $M_u = 3792$ kN·m, more than four times the flexural resistance, so Clause 13.6 never governs.
5. Part (b) — shear resistance with tension-field action, Clause 13.4.1.1. Space the interior transverse stiffeners at $a = 1200$ mm, i.e. $a/h = 1.5$, so that $$k_v = 5.34 + \frac{4}{(a/h)^2} = 7.118$$ Since $h/w = 133.3$ exceeds $621\sqrt{k_v/F_y} = 88.6$, the web buckles elastically and the post-buckling tension field is mobilised: $$F_{cre} = \frac{180\,000\,k_v}{(h/w)^2} = 72.1\ \text{MPa}, \qquad F_t = \frac{0.50F_y - 0.866F_{cre}}{\sqrt{1+(a/h)^2}} = 62.5\ \text{MPa}$$ $$V_r = \phi A_w (F_{cre} + F_t) = 0.9(4800)(134.5) = \boxed{581\ \text{kN}} \;\ge\; V_f = 386\ \text{kN}$$ The end panel has no adjacent panel to anchor the tension field, so it is proportioned on $F_{cre}$ alone. Reducing it to $a = 700$ mm gives $k_v = 10.98$, $F_{cre} = 111.1$ MPa and $V_r = 480$ kN, still above 386 kN.
6. Part (c) — flexure–shear interaction, Clause 14.6. Where a girder relies on tension-field action the flange must also anchor that field, and S16 penalises the combination: $$0.727\,\frac{M_f}{M_r} + 0.455\,\frac{V_f}{V_r} \le 1.0$$ $$0.727\left(\frac{764.7}{882.8}\right) + 0.455\left(\frac{386.0}{581}\right) = 0.630 + 0.302 = \boxed{0.932 \le 1.0}$$ The section passes, but with only 7 % in hand — this check, not flexure, is what fixes the flange size.
7. Detail the stiffeners. At B the girder delivers a 772 kN reaction into the prop, so a bearing stiffener is required. Two plates 110 × 12 acting with $25w$ of web give $A = 3540$ mm² and $r = 57.1$ mm; over $KL = 0.75h = 600$ mm the slenderness is 10.5 and $C_r = 1111$ kN. Allowing a 20 mm cope, the bearing resistance is $B_r = 1.5\phi_{bi}A F_y = 907$ kN. Both exceed 772 kN. Intermediate stiffeners need only stiffness, $I_{st} \ge b\,w^3 j = 86\,400$ mm⁴ with $j = 0.5$ at $a/h = 1.5$; a pair of 65 × 8 plates supplies $1.68 \times 10^6$ mm⁴, and $b/t = 8.1 \le 200/\sqrt{F_y} = 10.7$.

Final Results