16-Civ-B2 Advanced Structural Design · December 2013
Question 6 of 7: Steel beam-column AB
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.
welded I, web $544 \times 16$, flanges $350 \times 28$
Find. A section satisfying the three S16 Clause 13.8.2 interaction checks — cross-sectional strength, overall member strength and lateral–torsional buckling — under the forces the frame carries at collapse.
Section selected for the beam-column.
Approach. Take the member forces from the Question 5 collapse state, compute the compressive resistance about both axes, the lateral–torsional moment resistance over the 10 m unbraced length, and evaluate the three interaction expressions.
Section properties from the plates. For $d = 600$, $b = 350$, $t_f = 28$, $t_w = 16\ \text{mm}$,$$A = 28\,304\ \text{mm}^2, \quad I_x = 1819\times10^6, \quad I_y = 200.3\times10^6\ \text{mm}^4, \quad Z_x = 6.79\times10^6\ \text{mm}^3$$so $r_x = 253.5$ and $r_y = 84.1\ \text{mm}$, and the member weighs $222\ \text{kg/m}$. The flange ratio $(350-16)/2/28 = 5.96$ is Class 1; the web limit under combined load, $\frac{1100}{\sqrt{F_y}}\left(1 - 0.39\frac{C_f}{\phi C_y}\right) = 55.7$, comfortably exceeds the actual $544/16 = 34$.
Compressive resistance. With $n = 1.34$ and $\lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^2E}}$,$$\lambda_y = \frac{10\,000}{84.1}(0.013316) = 1.583 \Rightarrow C_{r,y} = \phi AF_y(1+\lambda^{2n})^{-1/n} = 2938.7\ \text{kN}$$$$\lambda_x = \frac{2.3(10\,000)}{253.5}(0.013316) = 1.208 \Rightarrow C_{r,x} = 4296.7\ \text{kN}$$The weak axis governs even though its effective length factor is only 1.0, because $r_y$ is a third of $r_x$. Both greatly exceed $C_f = 1195\ \text{kN}$, so pure compression is not the issue — the interaction with bending is.
Lateral–torsional moment resistance. The column bends in single curvature from zero at the pin to $M_f$ at B, so $\kappa = 0$ and $\omega_2 = 1.75$. With $J = 5.86\times10^6\ \text{mm}^4$ and $C_w = I_y(d-t_f)^2/4$,$$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^2I_yC_w} = 3063.0\ \text{kN}\cdot\text{m}$$Since $M_u > 0.67M_p = 1592\ \text{kN}\cdot\text{m}$ the member is in the inelastic range, so$$\boxed{M_r = 1.15\phi M_p\left(1 - \frac{0.28M_p}{M_u}\right) = 1925.2\ \text{kN}\cdot\text{m} \le \phi M_p = 2138.6\ \text{kN}\cdot\text{m}}$$
Interaction checks (S16 Cl 13.8.2). The Euler load about the axis of bending is $C_e = \pi^2EI_x/L^2 = 35\,908\ \text{kN}$; in an unbraced frame Clause 13.8.4 sets $U_{1x} = 1.0$ because the sway amplification is already carried in the analysis. The three checks are$$\text{(a) cross-section: } \frac{C_f}{\phi AF_y} + \frac{0.85U_{1x}M_f}{\phi M_p} = 0.134 + 0.201 = 0.335$$$$\text{(b) overall member: } \frac{C_f}{C_{r,y}} + \frac{0.85U_{1x}M_f}{M_r} = 0.407 + 0.224 = 0.630$$$$\text{(c) lateral-torsional: } \frac{C_f}{C_{r,x}} + \frac{0.85U_{1x}M_f}{M_r} = 0.278 + 0.224 = 0.502$$All three are below 1.0, with the overall-member check governing at 0.63.
Consistency with Question 5. Because the frame was analysed on the assumption that every member shares one $M_p$, the section chosen here must not undermine that mechanism. It does not: $\phi M_p = 2138.6\ \text{kN}\cdot\text{m}$ exceeds the $1706.25\ \text{kN}\cdot\text{m}$ required for the combined mechanism, so no hinge can form in AB before the assumed hinges at mid-span and C. The section is heavier than strength alone demands — the governing check is at 0.63, not near 1.0 — because a 10 m column braced only at its ends is controlled by out-of-plane stability, and $r_y$ is what buys the resistance.
Practical note on the unbraced length. If wall girts or a secondary frame can brace the column at mid-height, $K_yL$ halves, $\lambda_y$ drops to 0.79, $C_{r,y}$ rises to roughly $6100\ \text{kN}$, and a section some 25 % lighter satisfies every check. Where such bracing exists it should be used and stated; where it does not, the 600 mm deep section above is the honest answer to the question as posed.
