Question 1 of 7: Stiffened welded plate girder — flexure, shear and interaction

Question 1: Stiffened welded plate girder — flexure, shear and 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. Continuous girder, ends A and C fixed, intermediate roller at B; spans $L=12$ m each; four unfactored point loads $P=400$ kN at the third points of each span; lateral bracing every 2 m; $F_y=350$ MPa, $E=200\,000$ MPa; welded plate girder with transverse web stiffeners.

Find. A welded plate-girder cross-section (web plate, flange plates and transverse stiffener spacing) whose factored moment resistance, factored shear resistance and combined moment–shear resistance all exceed the factored effects at the critical section.

[Figure not reproduced: Figure 1 as printed: the two-span girder, encastré at A and C, roller at B, four 400 kN loads at the span third points. See the official exam paper.]

Approach. Exploit the symmetry of structure and loading about B — the rotation at B is zero, so each span is a fixed-ended beam — obtain $M_f$ and $V_f$ in closed form, size a web plate and flange plates against CSA S16 Cl. 14.3 (flexure, with the Class 4-web reduction), then check shear with tension-field action (Cl. 13.4.1.1) and finally the moment–shear interaction of Cl. 14.6.

(a) Flexure

1. Factor the loads. Treating the printed loads as specified live load and adding an estimated girder self-weight, $$P_f=\alpha_L P=1.5\times 400=600\ \text{kN},\qquad w_f=1.25\times 4.0=5.0\ \text{kN/m}$$
2. Use symmetry to make each span fixed-ended. The structure and the loading are both symmetric about the roller at B, so the tangent to the elastic curve at B is horizontal: $\theta_B=0$. A span whose far end cannot rotate and cannot deflect is, for analysis, encastré. Span AB is therefore a fixed-fixed beam of 12 m carrying two 600 kN loads at its third points.
3. Fixed-end moments. For a single load $P_f$ at distance $a$ from A ($b=L-a$), $$M_A=-\frac{P_f a b^{2}}{L^{2}},\qquad M_B=+\frac{P_f a^{2} b}{L^{2}}$$ For $a=4$ m: $M_A=-600(4)(8)^2/144=-1066.7$ kN·m and $M_B=+600(4)^2(8)/144=+533.3$ kN·m. For $a=8$ m the two swap. Superposing, $$|M_{A}|=|M_{B}|=1066.7+533.3=\boxed{1600\ \text{kN}\cdot\text{m}}$$
4. Add the self-weight moment. A fixed-fixed span under $w_f$ carries $w_fL^2/12$ at the ends, so $$M_f=1600+\frac{5.0(12)^{2}}{12}=1600+60=1660\ \text{kN}\cdot\text{m}$$ The sagging moment between the two loads is, correspondingly, $M_f^{+}=800+30=830$ kN·m — half the support value, so the supports govern.
5. Shear at the critical section. Symmetry gives each end of the span one point load, so $$V_f=P_f+\frac{w_fL}{2}=600+30=630\ \text{kN}$$ and the reaction delivered to the roller at B is $2\times 630=1260$ kN.
6. Choose trial plates. A span/depth ratio near $L/13$ suits a welded girder, so try a web $900\times 8$ mm with flanges $300\times 20$ mm, giving $d=940$ mm. Then $$I_x=\frac{8(900)^{3}}{12}+2\left[\frac{300(20)^{3}}{12}+300(20)\left(\frac{920}{2}\right)^{2}\right]=3.026\times10^{9}\ \text{mm}^{4}$$ $$S_x=\frac{I_x}{d/2}=\frac{3.026\times10^{9}}{470}=6.437\times10^{6}\ \text{mm}^{3}$$
7. Classify the plates. The flange outstand ratio is $b/t=(300-8)/2/20=7.30$, against the Class 1 limit $145/\sqrt{F_y}=7.75$ — the flanges are Class 1. The web slenderness is $h/w=900/8=112.5$, above the Class 3 limit $1900/\sqrt{F_y}=101.6$, so the web is Class 4 and the girder must be designed to Cl. 14 with transverse stiffeners. It is comfortably inside the stiffened-web ceiling $83\,000/F_y=237$.
8. Moment resistance with the Cl. 14.3.4 web reduction. $$M_r=\phi S_xF_y\left[1-0.0005\frac{A_w}{A_f}\left(\frac{h}{w}-\frac{1900}{\sqrt{M_f/(\phi S_x)}}\right)\right]$$ with $A_w=7200$ mm$^2$, $A_f=6000$ mm$^2$ and $M_f/(\phi S_x)=1660\times10^{6}/(0.9\times6.437\times10^{6})=286.5$ MPa, so $1900/\sqrt{286.5}=112.3$ and the bracket is $1-0.0005(1.20)(112.5-112.3)=0.9998$. Hence $$M_r=0.9(6.437\times10^{6})(350)(0.9998)=\boxed{2027\ \text{kN}\cdot\text{m}}\ \ge\ M_f=1660\ \text{kN}\cdot\text{m}$$ The web is right at the threshold where the reduction begins to bite, which is the economical place for it to be.
9. Lateral-torsional buckling is not an issue. With $I_y=2(20)(300)^3/12=90.0\times10^{6}$ mm$^4$ and $A=19\,200$ mm$^2$, $r_y=68.5$ mm, so at 2 m bracing $L_b/r_y=29.2$. The girder is far inside the fully-braced range and $M_r$ above stands.

(b) Shear, and their interaction

1. Set a stiffener spacing. Take transverse stiffeners at $a=1200$ mm, so $a/h=1200/900=1.333\le 3$ and $$k_v=5.34+\frac{4}{(a/h)^{2}}=5.34+2.25=7.59$$
2. Elastic critical shear stress. $$F_{cri}=\frac{180\,000\,k_v}{(h/w)^{2}}=\frac{180\,000(7.59)}{(112.5)^{2}}=107.9\ \text{MPa}$$ Since $h/w=112.5$ exceeds $621\sqrt{k_v/F_y}=91.4$, the web is in the fully post-buckled range and tension-field action may be used.
3. Ultimate shear stress including the tension field. With $k_a=1/\sqrt{1+(a/h)^{2}}=0.600$, $$F_s=F_{cri}+k_a\left(0.50F_y-0.866F_{cri}\right)=107.9+0.600\left(175-93.4\right)=156.9\ \text{MPa}$$ $$V_r=\phi A_wF_s=0.9(7200)(156.9)=\boxed{1016\ \text{kN}}\ \ge\ V_f=630\ \text{kN}$$
4. Moment–shear interaction (Cl. 14.6). Because tension-field action has been mobilised, the girder must satisfy $$0.727\frac{M_f}{M_r}+0.455\frac{V_f}{V_r}\le 1.0$$ $$0.727\left(\frac{1660}{2027}\right)+0.455\left(\frac{630}{1016}\right)=0.595+0.282=\boxed{0.877}\le 1.0$$ so flexure and shear can coexist at the encastré ends without further change.
5. Complete the detailing. Provide intermediate stiffener pairs $90\times10$ mm at 1200 mm centres, fitted to the compression flange and stopped 4$w$ clear of the tension flange; bearing stiffener pairs on the full web depth at A, B, C and under each 600 kN load, welded to both flanges and proportioned as columns of length $0.75h$ with $12w$ of web acting with them; and continuous fillet welds (6 mm) joining flanges to web, sized for the horizontal shear flow $V_fQ/I=630\times10^{3}(6000\times460)/3.026\times10^{9}=575$ N/mm, which is well inside the $2\times 0.67(0.67)(0.707\times6)(490)=1860$ N/mm supplied.