16-Civ-B2 Advanced Structural Design · December 2013
Question 1 of 7: Welded steel plate girder — flexure, shear and interaction
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2013 — 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; CISC Handbook of Steel Construction; CSA A23.3:19 Design of Concrete Structures (Clause 18 for prestressed concrete, Clause 17 of S16 for composite members); 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 printed loads are unfactored; it gives no dead/live split, so no code load combination can be assembled uniquely. Every load printed on Figures 1–4 is therefore factored by 1.5, and any self weight the solver itself introduces (the concrete frame, the prestressed girder, the steel sections) by 1.25, consistent with NBCC 2020 Table 4.1.3.2. State this assumption on the answer paper as page-1 Note 1 invites. The choice scales the magnitudes only — the collapse mechanisms, section classifications and interaction equations below are unaffected.
Check: section properties. Rolled-shape properties cannot be reproduced from first principles, so every steel member selected here is a welded three-plate I-section whose area, $I$, $Z$, $J$ and $C_w$ are computed from its plate dimensions and are therefore fully checkable. A candidate with the CISC Handbook would instead take the lightest rolled W-shape whose $Z_x$ (or $M_r$, $C_r$) exceeds the value derived in each question; the design logic is identical.
Find. Web and flange plate sizes, plus a transverse-stiffener spacing, such that the factored moment resistance, the factored shear resistance and the S16 moment–shear interaction are all satisfied at the governing sections.
Figure 1 as analysed: two 12 m spans, a fixed end at C, and the resulting factored bending-moment diagram.
Approach. Solve the two-span beam (one degree of static indeterminacy beyond the fixed end — two redundants in total) by the three-moment equation, trial-size a slender-web girder, then check Clause 14.3.4 for flexure, Clause 13.4.1.1 for shear and Clause 14.6 for their interaction.
Part (a) — support moments from the three-moment equation. With A simply supported, C fixed (modelled as a zero-length imaginary span beyond C) and a central point load on each 12 m span, the load term for span $i$ is $3P_fL^2/8$. The two equations are$$2M_B(L+L) + M_C L = -2\left(\tfrac{3P_fL^2}{8}\right), \qquad M_B L + 2M_C L = -\tfrac{3P_fL^2}{8}$$Substituting $P_f = 750\ \text{kN}$ and $L = 12\ \text{m}$ gives $48M_B + 12M_C = -81\,000$ and $12M_B + 24M_C = -40\,500$, whence$$\boxed{M_B = 1446.4\ \text{kN}\cdot\text{m}, \qquad M_C = 964.3\ \text{kN}\cdot\text{m}\quad(\text{both hogging})}$$
Confirm the result independently by slope-deflection. Writing $M_{ij} = \frac{2EI}{L}(2\theta_i + \theta_j) + \text{FEM}_{ij}$ with $\text{FEM} = P_fL/8 = 1125\ \text{kN}\cdot\text{m}$, imposing $M_{AB} = 0$ and $M_{BA} + M_{BC} = 0$ reproduces $1446.4$ and $964.3\ \text{kN}\cdot\text{m}$ exactly. Two independent methods agreeing is the cheapest guard against a sign slip in an indeterminate beam.
Reactions, shears and the design moment. For span AB, $R_A = P_f/2 - M_B/L = 375 - 120.5 = 254.5\ \text{kN}$, so the sagging moment under the first load is $254.5 \times 6 = 1526.8\ \text{kN}\cdot\text{m}$. The shear just left of B is $V = P_f/2 + M_B/L = 495.5\ \text{kN}$, and at C it is $334.8\ \text{kN}$. The four shears sum to $1500\ \text{kN} = 2P_f$, confirming vertical equilibrium. The girder must therefore carry$$M_f = 1526.8\ \text{kN}\cdot\text{m} \ (\text{at }6\ \text{m}), \qquad V_f = 495.5\ \text{kN}\ (\text{at B, coincident with }M = 1446.4)$$
Trial cross-section. A first estimate $S \ge M_f/(\phi F_y) = 1526.8\times 10^6/(0.9\times350) = 4.85\times10^6\ \text{mm}^3$ suggests a web about $L/12 \approx 1000\ \text{mm}$ deep. Take a web $1000 \times 8$ and flanges $250 \times 18$, giving $d = 1036\ \text{mm}$, $I_x = 2999\times10^6\ \text{mm}^4$ and $S = 5.79\times10^6\ \text{mm}^3$. The flange element ratio is $(250-8)/2/18 = 6.72 < 145/\sqrt{350} = 7.75$, so the flange is Class 1. The web ratio $h/w = 1000/8 = 125$ makes this a genuine plate girder rather than a rolled beam.
Part (a) — flexural resistance with a slender web (S16 Cl 14.3.4). The reduction applies once $h/w > 1900/\sqrt{M_f/(\phi S)}$. Here $M_f/(\phi S) = 1526.8\times10^6/(0.9\times5.79\times10^6) = 293\ \text{MPa}$, so the limit is $1900/\sqrt{293} = 111.0 < 125$ and the web is slender. With $A_w/A_f = 8000/4500 = 1.778$,$$M_r = \phi S F_y\left[1 - 0.0005\frac{A_w}{A_f}\left(\frac{h}{w} - \frac{1900}{\sqrt{M_f/\phi S}}\right)\right] = 1800.8\ \text{kN}\cdot\text{m}$$which exceeds $M_f = 1526.8\ \text{kN}\cdot\text{m}$ by 18 %. Lateral–torsional buckling does not govern: the flanges are braced every 2 m, and Clause 14.3.4 already caps the girder at its elastic resistance.
Part (b) — shear resistance (S16 Cl 13.4.1.1). With $h/w = 125$, an unstiffened web would be far below the demand, so transverse stiffeners are placed at $a = 1500\ \text{mm}$, giving $a/h = 1.5$ and$$k_v = 5.34 + \frac{4}{(a/h)^2} = 7.118$$Since $h/w = 125 > 621\sqrt{k_v/F_y} = 88.6$, the web is in the elastic-buckling branch: $F_{cre} = 180\,000\,k_v/(h/w)^2 = 82.0\ \text{MPa}$ and $k_a = 1/\sqrt{1+(a/h)^2} = 0.5547$, so tension-field action contributes$$F_s = F_{cre} + k_a\left(0.15F_y - 0.4F_{cre}\right) = 92.9\ \text{MPa}, \qquad \boxed{V_r = \phi A_w F_s = 669.1\ \text{kN} > V_f = 495.5\ \text{kN}}$$
Part (c) — moment–shear interaction (S16 Cl 14.6). At B the girder carries the largest shear and a near-maximum moment simultaneously. Because $V_f = 495.5 > 0.60V_r = 401.4\ \text{kN}$ and $M_f = 1446.4 > 0.75M_r = 1350.6\ \text{kN}\cdot\text{m}$, the interaction equation must be satisfied:$$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727(0.803) + 0.455(0.741) = \boxed{0.921 \le 1.0}$$The section passes with 8 % reserve, which is the correct place to be: the panel at an interior support of a plate girder is meant to be worked hard in combined action.
Self weight and detailing. The three plates give $A = 17\,000\ \text{mm}^2$, i.e. $1.31\ \text{kN/m}$; factored by 1.25 over a 12 m span this adds only $29.5\ \text{kN}\cdot\text{m}$, well inside the 274 $\text{kN}\cdot\text{m}$ flexural reserve. Complete the design with bearing stiffeners over A, B and C and beneath each 750 kN load, a stiffened end panel at A (no tension field can develop in an end panel), and continuous fillet welds sized for the flange–web horizontal shear flow $q = V_fA_f\bar{y}/I_x$.
