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16-Civ-B2 Advanced Structural Design · May 2015

Question 3 of 7: Three-span continuous welded plate girder

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 98-Civ-B2 Advanced Structural Design, National Exams May 2015 — 3 hours, closed book (textbooks and design handbooks permitted, no notes). Seven design questions of 20 marks each; any five constitute a complete paper, so all seven are solved here. All loads shown on the figures are unfactored. Page 1 supplies the design data used throughout: concrete f′c = 30 MPa, structural steel Fy = 350 MPa, reinforcing steel fy = 400 MPa; for the prestressed member f′ci = 35 MPa, f′c = 50 MPa, n = 6, fpu = 1750 MPa, fpy = 1450 MPa, initial stress 1200 MPa and total losses 240 MPa.

Reference texts. CSA S16:19 Design of Steel Structures; CISC Handbook of Steel Construction, 12th ed.; CSA A23.3:19 Design of Concrete Structures; CSA S6:19 Canadian Highway Bridge Design Code; NBCC 2020 Part 4; Hibbeler, Structural Analysis, 10th ed. (plastic analysis, Ch. 18).

Check — load factors. Page 1 states only that “all loads shown are unfactored” and gives no dead/live split. Throughout this paper the applied loads on the figures are factored by 1.5, and any self weight introduced by the solution itself (the plate girder, the deck, the prestressed beam, the reinforced-concrete frame) by 1.25, consistent with NBCC 2020 Table 4.1.3.2. The mechanisms, section classifications and interaction equations are independent of that choice — only the magnitudes change.

Question 3: Three-span continuous welded plate girder (15 + 5 = 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. A prismatic girder continuous over four supports, pinned at A and on rollers at B, C and D, with three equal 12 m spans and one point load at the mid-point of each span.

QuantityValue
Spans AB, BC, CD12 m each (36 m overall)
Point loads (unfactored)300 kN at 6 m, 200 kN at 18 m, 300 kN at 30 m
Lateral supportat 3 m intervals
SteelFy = 350 MPa, E = 200 GPa, φ = 0.90
Girder self weight (from the section chosen)1.05 kN/m

Find. A welded plate-girder cross-section that satisfies flexure, shear and their interaction at every critical section, and the long-term vertical displacement at the mid-point of the centre span.

ABCD300 kN200 kN300 kN12 m12 m12 mlateral support at 3 m intervals; loads at the mid-point of every span1027 (sag)695 (hog)230factored bending moments, kN·m
Figure 2 — the three-span continuous girder and its factored bending moment diagram. The outer spans carry the larger load, so sagging at their mid-points governs the design.

Approach. Analyse the symmetric continuous beam by the three-moment equation, trial a deep slender-web section, apply the CSA S16 Cl 14.3.4 reduction for web slenderness, check lateral-torsional buckling over the 3 m braced length, check shear with tension-field action and then the Cl 14.6 moment-shear interaction, and finally superpose deflections for part (b).

  1. Part (a) — factored loads. The applied loads are factored by 1.5 and the girder’s own weight by 1.25: $$P_1 = P_3 = 450\ \text{kN},\qquad P_2 = 300\ \text{kN},\qquad w = 1.25(1.05) = 1.31\ \text{kN/m}$$
  2. Analyse by the three-moment equation. By symmetry $M_A = M_D = 0$ and $M_B = M_C = M$. Applying the equation to supports A–B–C with a central point load and a uniform load on each span, $$2M(L+L) + ML = -\left[\frac{3L^2(P_1+P_2)}{8} + \frac{wL^3}{2}\right]$$ $$60M = -\left[\frac{3(144)(750)}{8} + \frac{1.31(1728)}{2}\right] \;\Rightarrow\; \boxed{M_B = M_C = -694.8\ \text{kN}\cdot\text{m}}$$
  3. Span moments and shears. Adding the free bending moment to the linear support-moment line, $$M_{mid,AB} = \frac{P_1L}{4} + \frac{wL^2}{8} + \frac{M_B}{2} = 1350 + 23.6 - 347.4 = \boxed{1027.3\ \text{kN}\cdot\text{m}}$$ $$M_{mid,BC} = \frac{P_2L}{4} + \frac{wL^2}{8} + M_B = 900 + 23.6 - 694.8 = 230.0\ \text{kN}\cdot\text{m}$$ The end shears follow as $V_A = 175.3$ kN and, just left of B, $V = P_1/2 + wL/2 - M_B/L = 291.1$ kN, which is the design shear.
  4. Trial section. A span-to-depth ratio near 12 suits a welded girder, so try a 1000 × 8 mm web with 200 × 14 mm flanges, overall depth 1028 mm. From first principles $$I_x = \frac{b d^3 - (b-w)h^3}{12} = 2.106\times10^9\ \text{mm}^4,\qquad S_x = \frac{2I_x}{d} = 4.098\times10^6\ \text{mm}^3$$ with $A = 13\,600\ \text{mm}^2$ (self weight 1.05 kN/m, as assumed), $I_y = 1.871\times10^7$, $J = 5.389\times10^5$ and $C_w = 4.809\times10^{12}\ \text{mm}^6$. The flange $b/2t = 7.14 < 145/\sqrt{350} = 7.75$ is Class 1, but the web at $h/w = 125$ is slender, which is normal for a plate girder and is handled explicitly by Cl 14.3.4.
  5. Flexural resistance with the slender-web reduction. CSA S16 Cl 14.3.4 reduces the resistance once $h/w$ exceeds $1900/\sqrt{M_f/(\phi S)}$; here that limit is $1900/\sqrt{1027.3\times10^6/(0.9\times4.098\times10^6)} = 113.8 < 125$, so $$M_r' = \phi S F_y\left[1 - 0.0005\frac{A_w}{A_f}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_f/\phi S}}\right)\right] = 1290.6\left[1 - 0.0005(2.857)(11.2)\right] = 1270.2\ \text{kN}\cdot\text{m}$$
  6. Lateral-torsional buckling over the 3 m braced length. Taking $\omega_2 = 1.0$ conservatively, $$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^2 I_yC_w} = 2121\ \text{kN}\cdot\text{m} > 0.67M_y = 961\ \text{kN}\cdot\text{m}$$ so the inelastic branch applies, $M_r = 1.15\phi M_y\left(1 - 0.28M_y/M_u\right) = 1203.3\ \text{kN}\cdot\text{m}$. Combining this with the web reduction gives a governing $$\boxed{M_r = 1184\ \text{kN}\cdot\text{m} \ge M_f = 1027.3\ \text{kN}\cdot\text{m}} \qquad(\text{ratio }0.87)$$ and at the interior supports $694.8/1184 = 0.59$.
  7. Shear with tension-field action. Transverse stiffeners at $a = 1500$ mm give $a/h = 1.5$ and $$k_v = 5.34 + \frac{4}{(a/h)^2} = 7.118,\qquad 621\sqrt{k_v/F_y} = 88.6 < \frac{h}{w} = 125$$ so the web buckles before it yields and the post-buckling tension field is mobilised: $$F_{cri} = \frac{180\,000k_v}{(h/w)^2} = 82.0\ \text{MPa},\qquad f_t = \frac{0.50F_y - 0.866F_{cri}}{\sqrt{1+(a/h)^2}} = 57.7\ \text{MPa}$$ $$V_r = \phi A_w(F_{cri}+f_t) = 0.9(8000)(139.7)/10^3 = 1006\ \text{kN} \gg 291.1\ \text{kN}$$
  8. Moment-shear interaction at B. Where tension-field action is relied on, Cl 14.6 requires $$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727\left(\frac{694.8}{1184}\right) + 0.455\left(\frac{291.1}{1006}\right) = 0.427 + 0.132 = \boxed{0.558 \le 1.0}$$ Flexure, shear and their interaction are therefore all satisfied. Bearing stiffeners are provided at all four supports and under each point load; intermediate stiffeners at 1500 mm are carried through the end panels of each span, where the shear is highest.
    9. 2001000 web14web 1000 × 8flanges 2 – 200 × 14overall depth 1028 mmtransverse stiffenersa = 1500 (a/h = 1.5)
    Plate-girder cross-section and stiffener layout — 1000 × 8 mm web, 2 – 200 × 14 mm flanges, transverse stiffeners at 1500 mm.
  9. Part (b) — service actions on the centre span. Under the unfactored loads the same analysis gives $M_B = M_C = -465.1$ kN·m, with $P_2 = 200$ kN at the mid-point of BC and $w = 1.05$ kN/m throughout.
  10. Superpose the three contributions. With $EI = 200\,000 \times 2.106\times10^9 = 4.212\times10^{14}\ \text{N}\cdot\text{mm}^2$ and $L = 12\,000$ mm, $$\delta = \underbrace{\frac{P_2L^3}{48EI}}_{17.09\ \text{mm}\downarrow} + \underbrace{\frac{5wL^4}{384EI}}_{0.67\ \text{mm}\downarrow} - \underbrace{\frac{|M_B|L^2}{8EI}}_{19.87\ \text{mm}\uparrow}$$ $$\boxed{\delta = -2.11\ \text{mm}, \text{ i.e. } 2.1\ \text{mm}\ \textbf{upward}}$$
  11. Interpret the long-term value. Structural steel does not creep at service temperature, so the long-term displacement equals the elastic displacement under the sustained portion of the load; if the three loads are permanent, the value above is the long-term value. The centre span rises rather than sags because the heavier outer spans hog the girder over B and C by more than the 200 kN load can push it down — the negative-moment term is 20 mm against 18 mm from the loads. The movement is 1/5700 of the span and of no practical consequence, but its sign matters: any cladding, bearing or drainage detail at the centre of BC must accommodate an upward, not a downward, movement.
ResultValue
Support moments MB = MC (factored)−694.8 kN·m
Mid-span moment, spans AB and CD (factored)1027.3 kN·m
Design shear just left of B (factored)291.1 kN
Girder sectionweb 1000 × 8, flanges 2 – 200 × 14 (depth 1028 mm)
Governing flexural resistance1184 kN·m (ratio 0.87)
Shear resistance with a/h = 1.51006 kN (ratio 0.29)
Cl 14.6 M–V interaction at B0.558
Long-term displacement at mid-point of BC2.1 mm upward