Question 3 of 7: Plate girder for flexure, shear and their interaction

Question 3: Plate girder for flexure, shear and their 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 12 m girder built in at A and D with a roller prop at mid-span, carrying 500 kN at B (4 m from A) and 500 kN at C (4 m from D). The note on Figure 2 reads "Lateral Support Provided @ 2 m interval", so $L_b = 2$ m throughout. $F_y = 350$ MPa.

Find. Web and flange plate sizes, the stiffener arrangement, and a demonstration that flexure, shear and the moment–shear interaction of Cl 14.6 are all satisfied.

Approach. Exploit symmetry to reduce the four-times redundant girder to a fixed-fixed half span, take the peak hogging moment and the peak shear (which coincide at the prop), then size a welded section, apply the Cl 14.3.4 slender-web reduction, obtain the shear resistance with tension field action and close with the Cl 14.6 interaction.

1. Reduce the structure by symmetry. Structure and loading are symmetric about mid-span, so the slope there is zero while the roller holds the deflection at zero. Each half is therefore a fixed-fixed beam of 6 m carrying one 750 kN load 4 m from the built-in end.
2. Fixed-end moments and reactions. With $a = 4$ m, $b = 2$ m, $L = 6$ m, $$\begin{gathered}M_A = -\frac{P a b^2}{L^2} = -\frac{750(4)(4)}{36} = -333.3\ \text{kN}\cdot \text{m}, \\ M_{prop} = -\frac{P a^2 b}{L^2} = -\frac{750(16)(2)}{36} = -666.7\ \text{kN} \cdot\text{m},\end{gathered}$$ $$R_A = \frac{P b^2(3a + b)}{L^3} = \frac{750(4)(14)}{216} = 194.4\ \text{kN}.$$
3. Design actions. The sagging peak under the load is $M = R_A a + M_A = 194.4(4) - 333.3 = 444.4$ kN·m, and the shear just inside the prop is $V = 750 - 194.4 = 555.6$ kN. Both maxima sit at the prop, so $$\boxed{M_f = 666.7\ \text{kN}\cdot\text{m},\qquad V_f = 555.6\ \text{kN}}$$ and the prop reaction is $2(555.6) = 1111$ kN. Check: $2(194.4) + 1111 = 1500 = 2P_f$.
4. Trial cross-section. A plate girder wants a deep, thin web. Try a web $800 \times 6$ with flanges $220 \times 16$, giving $d = 832$ mm, $$\begin{gathered}I = \frac{6(800)^3}{12} + 2\!\left[\frac{220(16)^3}{12} + 220(16)\!\left(\frac{816}{2}\right)^{\!2}\right] = 1.428\times10^{9}\ \text{mm}^4, \\ S = 3.433\times10^{6}\ \text{mm}^3 .\end{gathered}$$ The web slenderness $h/w = 133.3$ is well inside the Cl 14.3.1 ceiling $83\,000/F_y = 237$, and the flange $b/2t = 6.88$ is Class 1.
5. Flexural resistance with the slender-web penalty. A web this thin sheds compression to the flanges, so Cl 14.3.4 applies when $$\frac{h}{w} \;\gt\; \frac{1900}{\sqrt{M_f/(\phi S)}} = \frac{1900}{\sqrt{215.8}} = 129.3,$$ which it does. With $A_w/A_f = 4800/3520 = 1.364$, $$M_r' = \phi S F_y\!\left[1 - 0.0005\frac{A_w}{A_f} \!\left(\frac{h}{w} - 129.3\right)\right] = 1081.3(0.9973) = 1078\ \text{kN}\cdot\text{m} \;\gt\; 666.7\quad\checkmark$$ The penalty is only 0.3 per cent here because the web is barely over the limit. Lateral-torsional buckling does not arise: with $L_b = 2$ m and $r_y = 49$ mm the segment is effectively fully braced.
6. Shear — the unstiffened web is not enough. With no intermediate stiffeners $k_v = 5.34$ and $$F_{cri} = \frac{180\,000 k_v}{(h/w)^2} = \frac{180\,000(5.34)}{17\,778} = 54.1\ \text{MPa} \;\Rightarrow\; V_r = 0.90(4800)(54.1)/10^3 = 234\ \text{kN},$$ far below $V_f = 555.6$ kN. Transverse stiffeners are mandatory.
7. Shear with stiffeners and tension field action. Set $a/h = 1.0$, i.e. stiffeners at 800 mm centres. Then $k_v = 5.34 + 4/(a/h)^2 = 9.34$ and $F_{cri} = 94.6$ MPa. Because $h/w = 133.3 \gt 621\sqrt{k_v/F_y} = 101$, the post-buckling tension field is available (Cl 13.4.1.1): $$f_t = \frac{0.50F_y - 0.866F_{cri}}{\sqrt{1 + (a/h)^2}} = 65.8\ \text{MPa}, \qquad F_s = 94.6 + 65.8 = 160.4\ \text{MPa},$$ $$V_r = \phi A_w F_s = 0.90(4800)(160.4)/10^3 = \boxed{693\ \text{kN}} \;\gt\; 555.6\ \text{kN}\quad\checkmark$$
8. Moment–shear interaction. The prop is the one section where both maxima act together, and because the shear resistance relies on tension field action Cl 14.6 applies: $$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727\!\left(\frac{666.7}{1078}\right) + 0.455\!\left(\frac{555.6}{693}\right) = 0.449 + 0.365 = \boxed{0.814} \;\le\; 1.0\quad\checkmark$$
9. Bearing stiffeners. The prop takes 1111 kN. A pair of $120 \times 16$ plates plus an effective web strip $25w$ gives $A = 4740$ mm$^2$ and $r = 47.1$ mm; as a column of length $0.75h = 600$ mm, $C_r = 1466$ kN $\gt 1111$ kN. Their outstand ratio $b/t = 7.5$ clears the Cl 11.2 limit $200/\sqrt{F_y} = 10.7$. Use the same pair at A, at D and under each 750 kN load.