16-Civ-B2 Advanced Structural Design · December 2014

Question 1 of 7: Two-span continuous welded plate girder

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

Notes on this paper

Paper format. National Examinations, December 2014 — 98-Civ-B2 Advanced Structural Design, 3 hours, closed book (handbooks and textbooks permitted). Seven design questions of equal value; any five constitute a complete paper.

Reference texts. CSA S16:19 Design of Steel Structures (Clause 13 members, Clause 14 plate girders, Clause 17 composite, Clause 21 plastic design); CISC Handbook of Steel Construction; CSA A23.3:19 Design of Concrete Structures (Clause 10 flexure and columns, Clause 11 shear, Clause 18 prestressed concrete); National Building Code of Canada 2020 for load combinations; Hibbeler, Structural Analysis, for the frame and continuous-beam analysis.

Design data printed on page 1. Concrete $f_c' = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; reinforcing steel $f_y = 400\ \text{MPa}$. Prestressed concrete: $f_{ci}' = 35\ \text{MPa}$ at transfer, $f_c' = 50\ \text{MPa}$, modular ratio $n = 6$, $f_{ult} = 1750\ \text{MPa}$, $f_y = 1450\ \text{MPa}$, $f_{initial} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$. Design in SI. All loads shown are unfactored.

Check — load factors. Page 1 states only that the loads shown are unfactored; it gives no dead/live split. Throughout this paper the figure loads are treated as a single variable action and factored by 1.5, while self weight introduced by the solver (the reinforced-concrete frame, the prestressed girder, the bridge deck) is factored by 1.25, per NBCC 2020 Case 2. Mechanisms, section classifications and interaction ratios are unaffected by this choice — only the magnitudes are. Serviceability checks (the no-tension prestress design of Q2 and the soil bearing of Q5) use unfactored loads, as they must.

Question 1: Two-span 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.

Quantity	Value
Spans (two, continuous over B)	$L = 16\ \text{m}$ each
Point load at mid-point of each span	$P = 500\ \text{kN}$ (unfactored)
Lateral support to compression flange	every 2 m
Steel	$F_y = 350\ \text{MPa}$, $E = 200\,000\ \text{MPa}$
Resistance factor	$\phi = 0.90$

Find. The web and flange plate sizes of a welded girder whose factored moment, shear and combined moment-shear resistances all envelope the factored effects, with the governing section being the one over the interior support B where $M_f$ and $V_f$ coexist at their maxima.

Figure 1.1 — Two equal 16 m spans with a mid-span point load in each; factored bending moment diagram. The hogging moment over B is 20 per cent larger than the sagging peak and coexists with the maximum shear, so B governs the design.

Approach. Analyse the two-span beam by the three-moment equation to obtain the coexisting $M_f$ and $V_f$ at B, then proportion a welded I-section so that the flange is Class 1–2 and the web slenderness stays just inside the Class 3 limit (which avoids the Clause 14.3.4 slender-web moment reduction), and finally check the web for shear without transverse stiffeners and for the Clause 14.6 moment-shear interaction.

Factor the loads and analyse the continuous beam. With a single variable action, $P_f = 1.5 \times 500 = 750\ \text{kN}$. For two equal spans, each carrying a mid-span point load, the three-moment equation $4M_B L = -\tfrac{3}{4}P_f L^2$ gives $$M_B = -\frac{3P_fL}{16} = -\frac{3(750)(16)}{16} = -2250\ \text{kN}\cdot\text{m}$$ Statics of span AB then returns the end reaction and the sagging peak.
Extract the design actions. Taking moments about B for span AB, $$R_A = \frac{P_fL/2 + M_B}{L} = \frac{750(8) - 2250}{16} = 234.4\ \text{kN}$$ so the sagging moment under the load is $M = R_A(8) = 1875\ \text{kN}\cdot\text{m}$ and the shear immediately to the left of B is $V = R_A - P_f = -515.6\ \text{kN}$. By symmetry the interior reaction is $R_B = 2(515.6) = 1031\ \text{kN}$. The design pair at B is therefore $$\boxed{M_f = 2250\ \text{kN}\cdot\text{m}, \qquad V_f = 515.6\ \text{kN}}$$ Both act at the same cross-section, which is why the interaction check of step 7 cannot be skipped.
Choose trial plates from a span/depth rule. A depth near $L/15 = 1070\ \text{mm}$ is economical for a welded girder. Try a web 1100 × 10 mm with flanges 300 × 20 mm, giving an overall depth $d = 1100 + 2(20) = 1140\ \text{mm}$. The gross properties follow from first principles: $$I_x = 2\left[\frac{bt^3}{12} + bt\left(\frac{h+t}{2}\right)^{2}\right] + \frac{wh^3}{12} = 4872.8 \times 10^{6}\ \text{mm}^4$$ $$S_x = \frac{I_x}{d/2} = \frac{4872.8 \times 10^{6}}{570} = 8.549 \times 10^{6}\ \text{mm}^3$$
Classify the flange and the web. The flange outstand ratio is $b/2t = 150/20 = 7.5$, against the Class 1 limit $145/\sqrt{F_y} = 7.75$, so the flange is Class 1. The web ratio is $h/w = 1100/10 = 110$; the Class 3 limit for a girder web in flexure is $$\frac{h}{w} \le \frac{1900}{\sqrt{M_f/(\phi S_x)}} = \frac{1900}{\sqrt{2250\times10^{6}/(0.9 \times 8.549\times10^{6})}} = 111.1$$ Since $110 \le 111.1$ the web is Class 3 and the Clause 14.3.4 slender-web reduction does not apply. Note that a Class 3 web caps the section at its elastic resistance, so nothing is gained by making the flange Class 1 — but it costs nothing either, and it protects the flange against local buckling at the support.
Check flexural resistance. Lateral support at 2 m means the unbraced length is short and lateral-torsional buckling does not reduce the resistance, so $$M_r = \phi S_x F_y = 0.9 \times 8.549\times10^{6} \times 350 = \boxed{2693\ \text{kN}\cdot\text{m}} \;>\; M_f = 2250\ \text{kN}\cdot\text{m}$$ The utilisation is 0.84.
Check shear on an unstiffened web. With no transverse stiffeners the shear buckling coefficient is $k_v = 5.34$. Because $$\frac{h}{w} = 110 \;>\; 621\sqrt{k_v/F_y} = 76.7$$ the web is in the elastic buckling range and carries no tension field, so $$F_{cri} = \frac{180\,000\,k_v}{(h/w)^2} = \frac{180\,000(5.34)}{110^2} = 79.4\ \text{MPa}$$ $$V_r = \phi A_w F_{cri} = 0.9(11\,000)(79.4) = \boxed{786\ \text{kN}} \;>\; V_f = 515.6\ \text{kN}$$ No intermediate stiffeners are required over either span.
Check the combined moment and shear at B. Clause 14.6 governs a girder whose web is designed without a tension field: $$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727\left(\frac{2250}{2693}\right) + 0.455\left(\frac{515.6}{786}\right) = 0.607 + 0.298 = \boxed{0.91 \le 1.0}$$ The section passes with a nine per cent reserve, and this is the check that actually sizes the girder: flexure alone would have allowed a lighter web.
Provide a bearing stiffener at the interior support. The reaction $R_B = 1031\ \text{kN}$ must be carried by a stiffened web strut. Use two plates 140 × 16 mm, one each side, with $b/t = 8.75 \le 200/\sqrt{F_y} = 10.7$. The effective strut area includes $12w$ of web, $$A_{st} = 2(140)(16) + 12(10)(10) = 5680\ \text{mm}^2, \qquad \frac{KL}{r} = \frac{0.75(1100)}{r} = 10.9$$ $$C_r = \phi A_{st} F_y\left(1+\lambda^{2n}\right)^{-1/n} = 1782\ \text{kN} \;>\; R_B = 1031\ \text{kN}$$ The stiffeners are fitted tight to both flanges and welded to the web with 6 mm fillets.

Result	Value
Factored hogging moment at B	2250 kN·m
Factored sagging moment at mid-span	1875 kN·m
Factored shear at B	515.6 kN
Web plate	1100 × 10 mm
Flange plates	300 × 20 mm (both)
Overall depth	1140 mm
Moment resistance $M_r$	2693 kN·m (0.84 utilised)
Shear resistance $V_r$ (unstiffened)	786 kN (0.66 utilised)
Clause 14.6 interaction	0.91 ≤ 1.0
Bearing stiffener at B	2 plates 140 × 16 mm, $C_r = 1782$ kN

Figure 1.2 — Designed welded plate girder cross-section: web 1100 × 10, flanges 300 × 20, overall depth 1140 mm.

Topic. Limit states design of a welded plate girder for coexisting flexure and shear.

Key relations. Three-moment equation $M_B = -3PL/16$ for two equal spans with mid-span point loads; $M_r = \phi S_x F_y$ for a Class 3 web; elastic shear buckling $F_{cri} = 180\,000\,k_v/(h/w)^2$ with $k_v = 5.34$ unstiffened; Clause 14.6 interaction $0.727\,M_f/M_r + 0.455\,V_f/V_r \le 1.0$; Clause 14.3.4 slender-web reduction when $h/w > 1900/\sqrt{M_f/\phi S}$.

Why this works. A plate girder is a deliberately unbalanced section: the flanges are sized for moment and the web for shear, and the designer chooses the web slenderness to sit at a code threshold rather than anywhere in between. Setting $h/w$ just inside the Class 3 limit buys the full elastic moment resistance while keeping the web as thin as the shear check allows, which is where the economy of a welded girder comes from. Over an interior support both actions peak together, so the two independently adequate resistances must still be combined; Clause 14.6 recognises that the flanges carry most of the moment and the web most of the shear by weighting the two ratios unequally rather than summing them directly.

Common pitfall. Two mistakes recur. The first is designing for the sagging moment because it is the one drawn first: for two equal spans the hogging moment at the interior support is 20 per cent larger and it coexists with the peak shear. The second is claiming a tension field on an unstiffened web — a tension field needs transverse stiffeners to anchor it, and without them the resistance is the elastic buckling stress alone, roughly a quarter of the shear yield here.

Source. CSA S16:19 Clauses 13.4 (shear), 14.3 (plate girder flexure) and 14.6 (combined shear and moment); CISC Handbook of Steel Construction, Part 4; Hibbeler, Structural Analysis, Ch. 11 (three-moment equation).

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