Question 1 of 7: Continuous welded plate girder (12 + 6 + 2 marks)

Question 1: Continuous welded plate girder (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 continuous welded plate girder of two equal 12 m spans, built into rock at A and at C and carried on a simple interior support at B, each span loaded by two 400 kN point loads at its third points. The compression flange is laterally supported at 2 m intervals.

Find. A welded plate-girder cross-section with a stiffened web that satisfies flexure, shear, and their interaction, together with the stiffener spacing and the bearing stiffeners at the supports.

Approach. Exploit the symmetry of the structure and its loading to reduce each span to a fixed-ended beam, factor the resulting moment and shear, then proportion the web for shear (with transverse stiffeners and tension-field action) and the flanges for flexure, closing with the moment–shear interaction check that a stiffened girder demands.

1. Reduce the structure by symmetry. The girder and its loading are symmetric about B, so the rotation at B is zero. Each span therefore behaves exactly as a beam fixed at both ends, and the whole analysis collapses to one 12 m fixed-ended span carrying 400 kN at 4 m and at 8 m.
2. Fixed-end moments. For a single load $P$ at $a$ from the left end and $b$ from the right, $M_{A}=Pab^{2}/L^{2}$ and $M_{B}=Pa^{2}b/L^{2}$. Superposing the two loads, $$M_{A}=\frac{400(4)(8)^{2}}{12^{2}}+\frac{400(8)(4)^{2}}{12^{2}}=711.11+355.56=1066.67\ \text{kN}\cdot\text{m}$$ and by symmetry the same magnitude arises at the far end, so every support carries $\boxed{M=1066.7\ \text{kN}\cdot\text{m}}$ (hogging).
3. Shears and the span moment. Symmetry gives an end reaction of 400 kN in each span, so the shear between the two loads vanishes and the moment is constant over the middle 4 m: $M_{\text{span}}=400(4)-1066.67=533.3\ \text{kN}\cdot\text{m}$. The interior support B therefore delivers 800 kN to its bearing while the girder itself never sees more than 400 kN of shear. Factoring, $M_{f}=1.5(1066.7)=1600\ \text{kN}\cdot\text{m}$ and $V_{f}=1.5(400)=600\ \text{kN}$.
4. Trial plates and slenderness limits. Take a web $1200\times8$ and flanges $250\times16$, giving an overall depth of 1232 mm (about $L/10$, the usual economic range for a stiffened girder). The web slenderness $h/w=150$ is comfortably inside the CSA S16 Clause 14.3.1 ceiling $83\,000/F_{y}=237$ for a girder with transverse stiffeners, but it exceeds $1900/\sqrt{F_{y}}=101.6$, so the web is slender and the moment resistance must be reduced for web bend-buckling. The flange ratio $b/2t=7.81$ satisfies the Class 2 limit $170/\sqrt{F_{y}}=9.09$; a Class 1 flange buys nothing here because a slender web forces the elastic resistance in any case.
5. Flexural resistance. The elastic section modulus of the trial plates is $S=6\,671\times10^{3}\ \text{mm}^{3}$, so $\phi SF_{y}=0.9(6.671\times10^{6})(350)=2101\ \text{kN}\cdot\text{m}$. Clause 14.3.4 then reduces this for the slender web: $$M_{r}=\phi SF_{y}\left[1-0.0005\frac{A_{w}}{A_{f}}\left(\frac{h}{w}-\frac{1900}{\sqrt{M_{f}/\phi S}}\right)\right]=2101\left[1-0.0005(2.4)(150-116.4)\right]$$ giving $\boxed{M_{r}=2017\ \text{kN}\cdot\text{m}\ \ge\ M_{f}=1600\ \text{kN}\cdot\text{m}}$, a 26 percent margin.
6. Lateral–torsional buckling. With the compression flange held every 2 000 mm, $M_{u}=(\omega_{2}\pi/L_{b})\sqrt{EI_{y}GJ+(\pi E/L_{b})^{2}I_{y}C_{w}}=12\,570\ \text{kN}\cdot\text{m}$, an order of magnitude above $M_{r}$. Lateral buckling is not a design condition and the note on the figure is doing real work: without that bracing the answer would be a much heavier flange.
7. Web shear with transverse stiffeners. At $h/w=150$ the web is well past $621\sqrt{k_{v}/F_{y}}$, so elastic buckling governs the pre-buckling strength and post-buckling tension-field action may be counted in interior panels. With stiffeners at $a=h=1200\ \text{mm}$, $k_{v}=5.34+4/(a/h)^{2}=9.34$, $F_{cre}=180\,000k_{v}/(h/w)^{2}=74.7\ \text{MPa}$ and $f_{t}=(0.5F_{y}-0.866F_{cre})/\sqrt{1+(a/h)^{2}}=78.0\ \text{MPa}$, so $F_{s}=152.7\ \text{MPa}$ and $V_{r}=\phi A_{w}F_{s}=1319\ \text{kN}$.
8. End panels. An end panel has no adjacent panel to anchor the tension field, so it must be designed on $F_{cre}$ alone. Closing the first stiffener to $a=900\ \text{mm}$ raises $k_{v}$ to 12.45 and gives $F_{s}=99.6\ \text{MPa}$, hence $\boxed{V_{r}=861\ \text{kN}\ \ge\ V_{f}=600\ \text{kN}}$. Every panel adjacent to A, B and C is therefore 900 mm; the remainder are at 1 200 mm.
9. Moment–shear interaction. Where tension-field action is relied upon, Clause 14.6 requires $0.727M_{f}/M_{r}+0.455V_{f}/V_{r}\le1.0$. The worst combination is the first tension-field panel, 1.2 m from a support, where $M_{f}=880\ \text{kN}\cdot\text{m}$ with $V_{f}=600\ \text{kN}$: $$0.727\left(\frac{880}{2017}\right)+0.455\left(\frac{600}{1319}\right)=0.317+0.207=\boxed{0.524\ \le\ 1.0}$$
10. Bearing stiffeners. The interior support delivers $1.5(800)=1200\ \text{kN}$. A pair of $110\times16$ plates acting with $25w=200\ \text{mm}$ of web forms a strut of area $5120\ \text{mm}^{2}$ and $r=55.6\ \text{mm}$; with an effective length $0.75h$, $\lambda=0.216$ and $C_{r}=1593\ \text{kN}$. Milled to bear on the flange over $2(100)(16)=3200\ \text{mm}^{2}$, $B_{r}=1.5\phi_{be}AF_{y}=1344\ \text{kN}$. Both exceed 1 200 kN, so the same pair serves at A, B and C, and lighter pairs are placed under each 400 kN load.