Question 1 of 7: Welded Plate Girder by the Non-Stiffened-Web Approach

Question 1: Welded Plate Girder by the Non-Stiffened-Web Approach (12 + 6 + 2 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 prismatic welded girder continuous over four supports at 6 m centres — A pinned, B, C and D on rollers — carrying five unfactored 500 kN point loads. Reading the dimension string in Figure 1 from A, the loads stand at 3 m, 8 m, 10 m, 14 m and 16 m, that is at mid-span of A–B and at the third points of B–C and C–D. The steel is $F_{y} = 350$ MPa and the web is to carry its shear without intermediate transverse stiffeners.

Find. One welded cross-section — web plate plus two flange plates — that satisfies CSA S16 for factored moment, for factored shear with no intermediate stiffeners, and for their interaction at the section where both are large.

Approach. Factor the loads, solve the two-degree indeterminate beam elastically for the moment and shear envelopes, then size the web first from the unstiffened shear rules of Clause 13.4.1.1 (because shear, not moment, is what the non-stiffened approach penalises), size the flanges from the elastic section modulus, and close with the Clause 14.6 moment-shear interaction at the interior support.

1. Factor the loads and allow for self weight. With the single live-load factor adopted above, $$P_{f} = 1.5 \times 500 = 750\ \text{kN}$$ The trial section reached in Steps 3 and 4 has an area of $19\,260\ \text{mm}^{2}$, so its self weight is $w = 19\,260 \times 10^{-6} \times 77 = 1.48\ \text{kN/m}$ and $w_{f} = 1.25 \times 1.48 = 1.85\ \text{kN/m}$. Self weight is barely one per cent of the applied load here, but it is carried through so that the numbers quoted below reproduce exactly.
2. Elastic analysis of the continuous beam. Four vertical reactions and two equations of equilibrium make the girder twice indeterminate; a stiffness solution (equivalently, two applications of the three-moment equation) gives the reactions $$R_{A}=237.8,\quad R_{B}=1206.0,\quad R_{C}=1799.7,\quad R_{D}=539.9\ \text{kN}$$ which sum to $3783.4$ kN, exactly the five factored loads plus $1.85 \times 18$ kN of self weight. The moment envelope follows.

Both extremes fall at the same cross-section. The girder must be designed for $M_{f} = 1294$ kN·m and $V_{f} = 971$ kN acting together just to the right of support C, which is precisely why the paper asks for the interaction as well.

3. Size the web from the unstiffened-girder rules. Clause 14.3.1 caps the slenderness of a girder web carrying no intermediate stiffeners at $$\frac{h}{w} \le \frac{83\,000}{F_{y}} = \frac{83\,000}{350} = 237$$ and Clause 13.4.1.1 must then be evaluated with the shear-buckling coefficient for an infinitely long panel, $k_{v} = 5.34$. Because $\sqrt{k_{v}/F_{y}} = 0.1235$, the transition slendernesses are $439\sqrt{k_{v}/F_{y}} = 54.2$, $502\sqrt{k_{v}/F_{y}} = 62.0$ and $621\sqrt{k_{v}/F_{y}} = 76.7$. Trying a web plate of $900 \times 11$ mm gives $h/w = 81.8$, above the last of these, so the elastic buckling branch governs: $$F_{s} = F_{cre} = \frac{180\,000\,k_{v}}{(h/w)^{2}} = \frac{180\,000 \times 5.34}{81.8^{2}} = 143.6\ \text{MPa}$$ $$V_{r} = \phi A_{w} F_{s} = 0.90 \times (900 \times 11) \times 143.6 = \boxed{1279\ \text{kN}} \ge V_{f} = 971\ \text{kN}$$ The utilisation is $971/1279 = 0.76$. Note how expensive the ban on stiffeners is: the same web with a stiffener panel ratio $a/h = 1$ would develop $F_{s}$ near $0.66F_{y} = 231$ MPa, over 60 per cent more.
4. Size the flanges from the elastic section modulus. Try flange plates of $260 \times 18$ mm, giving an overall depth $d = 900 + 2(18) = 936$ mm. Working from the plate geometry, $$\begin{gathered}I_{x} = \frac{w h^{3}}{12} + 2 b t \left(\frac{d-t}{2}\right)^{2} = 2.641 \times 10^{9}\ \text{mm}^{4}, \\ S_{x} = \frac{2 I_{x}}{d} = 5.642 \times 10^{6}\ \text{mm}^{3}\end{gathered}$$ The flange slenderness is $b/2t = 260/36 = 7.22$, inside the Class 1 limit $145/\sqrt{F_{y}} = 7.75$, so the compression flange cannot buckle locally before the extreme fibre yields.
5. Check whether the slender web reduces the moment resistance. Clause 14.3.4 applies a reduction only when $$\frac{h}{w} \gt \frac{1900}{\sqrt{M_{f}/(\phi S_{x})}} = \frac{1900}{\sqrt{1294.2 \times 10^{6}/(0.90 \times 5.642 \times 10^{6})}} = \frac{1900}{\sqrt{254.9}} = 119.0$$ The web at $h/w = 81.8$ is well inside that limit, so no reduction is required and $$M_{r} = \phi S_{x} F_{y} = 0.90 \times 5.642 \times 10^{6} \times 350 = \boxed{1777\ \text{kN}\cdot\text{m}} \ge M_{f} = 1294\ \text{kN}\cdot\text{m}$$ a utilisation of 0.73.
6. Confirm that lateral-torsional buckling does not govern. The longest unbraced segment is the 3 m between A and its load point. With $I_{y} = 52.8 \times 10^{6}\ \text{mm}^{4}$, $J = 1.31 \times 10^{6}\ \text{mm}^{4}$ and $C_{w} = 1.11 \times 10^{13}\ \text{mm}^{6}$, Clause 13.6 gives $M_{u} = 5435$ kN·m even with $\omega_{2} = 1.0$. That is more than three times $0.67 M_{y}$, so the inelastic branch returns the full $\phi S_{x} F_{y}$ and Step 5 stands unaltered.
7. Moment-shear interaction at support C. Clause 14.6 combines the two actions on a girder web through $$0.727\,\frac{M_{f}}{M_{r}} + 0.455\,\frac{V_{f}}{V_{r}} \le 1.0$$ Substituting the two ratios just obtained, $$0.727(0.728) + 0.455(0.759) = 0.529 + 0.345 = \boxed{0.875} \le 1.0$$ The section is satisfactory. Strictly, the clause is triggered only when $M_{f}/M_{r} \gt 0.75$ and $V_{f}/V_{r} \gt 0.60$; here the moment ratio is 0.728, marginally below the trigger, so the check is a formality — but it is worth writing out, because the 2 marks the paper assigns to it are awarded for knowing that the interaction exists and where it bites.

Bearing stiffeners under the 500 kN loads and over the reactions are outside the question, which instructs the candidate to assume adequate load-base plates; in a real design they would be sized to Clause 14.4.2 and they do not affect the cross-section chosen here.