Question 2 of 7: Two-span welded plate girder — flexure, shear and interaction (12 + 6 + 2 marks)

Question 2: Two-span welded plate girder — flexure, shear and interaction (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 welded plate girder continuous over two 12 m spans, each span carrying two 400 kN loads at its third points, with lateral support to the compression flange at 2 m intervals and no intermediate transverse web stiffeners.

Find. Web and flange plate sizes that satisfy flexure, shear and the shear-moment interaction rule at the interior support, where both actions peak together.

Approach. Analyse the continuous beam elastically for the support moment, then choose a web whose unstiffened shear buckling resistance carries the support shear, keep the web slenderness inside the Class 3 limit so the flexural resistance is not reduced, and close with the Clause 14.6 interaction check.

1. Analyse the continuous beam. With equal spans and identical loading, the three-moment equation applied to the simple-span moment diagram (area $19\,200$ kN·m², centroid at mid-span) gives $$2M_B(2L)=-6\left(\frac{A\bar{x}}{L}+\frac{A\bar{x}}{L}\right) \ \Rightarrow\ M_B=\boxed{-2400\ \text{kN}\cdot\text{m}}$$ The end reaction is then $R_A=600+M_B/L=400$ kN, so the span moments are $M(4\,\text{m})=1600$ kN·m and $M(8\,\text{m})=800$ kN·m, and the shear just inside the interior support is $$V_B=400-2(600)=\boxed{-800\ \text{kN}}$$ Hogging at B therefore governs flexure and coincides with the peak shear.
2. Choose a web on the unstiffened shear resistance. With no intermediate stiffeners the aspect ratio is unbounded, so $k_v=5.34$ and $k_a=0$. Trying a 1200 mm by 12 mm web, $h/w=100$, which exceeds $621\sqrt{k_v/F_y}=76.7$, so shear buckling is elastic: $$F_s=F_{cre}=\frac{180\,000\,k_v}{(h/w)^2}=\frac{180\,000(5.34)}{100^2} =96.1\ \text{MPa}$$ $$V_r=\phi A_wF_s=0.9(1200\times12)(96.1)=\boxed{1246\ \text{kN}}\ \ge\ 800\ \text{kN}$$
3. Proportion the flanges and classify the section. Trying 300 mm by 20 mm flanges, the overall depth is 1240 mm and $$I=\frac{12(1200)^3}{12}+2\left[\frac{300(20)^3}{12}+6000(610)^2\right] =6194\times10^6\ \text{mm}^4,\qquad S=\frac{I}{620}=9.99\times10^6\ \text{mm}^3$$ The flange projection gives $b/2t=7.50\le145/\sqrt{350}=7.75$, so the flange is Class 1; the web at $h/w=100$ is just inside the Class 3 limit $1900/\sqrt{F_y}=101.6$. The section is therefore Class 3 overall and the elastic resistance governs: $$M_r=\phi SF_y=0.9(9.99\times10^6)(350)=\boxed{3147\ \text{kN}\cdot\text{m}}$$
4. Add the girder self weight. The plate area is $26\,400$ mm², i.e. 2.03 kN/m, factored to 2.54 kN/m. On a two-span continuous beam a uniform load contributes $wL^2/8$ at the interior support and $0.625wL$ to the shear there, so $$M_f=2400+45.7=\boxed{2446\ \text{kN}\cdot\text{m}},\qquad V_f=800+19.1=\boxed{819\ \text{kN}}$$ Both remain inside the resistances found above, at 78 and 66 per cent respectively.
5. Check the plate-girder flexural reduction. Clause 14.3.4 reduces $M_r$ only when the web is more slender than $$\frac{1900}{\sqrt{M_f/(\phi S)}}=\frac{1900}{\sqrt{2446\times10^6/(0.9\times9.99\times10^6)}} =115.2$$ Since $h/w=100<115.2$, no reduction applies — the reason for holding the web at 1200 mm by 12 mm rather than going thinner and deeper.
6. Apply the shear-moment interaction rule. At the interior support both ratios are high enough to trigger Clause 14.6: $$0.727\frac{M_f}{M_r}+0.455\frac{V_f}{V_r} =0.727(0.777)+0.455(0.658)=\boxed{0.864}\ \le\ 1.0$$ The girder therefore has 14 per cent reserve in the combined action, which is the controlling check for a non-stiffened web.
7. Check the sagging region and the bearings. The largest sagging moment, $1625$ kN·m at 4 m from an end support, is only 52 per cent of $M_r$, so the uniform section is adequate throughout. The factored interior reaction is $R_B=1638$ kN, against a web bearing resistance of only $$B_r=\phi_{bi}w(N+10t)F_y=0.80(12)(200+200)(350)=1344\ \text{kN}$$ so bearing stiffeners are required at the interior support; the end reactions of 411 kN are carried by the web alone. Interior web crippling under the 600 kN point loads gives $B_r=1398$ kN, so no stiffeners are needed there.