16-Civ-A2 Elementary Structural Design · December 2014

Question 2 of 7: A2 — Maximum factored load on an eccentrically loaded column

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

Notes on this paper

Paper format. National Examinations, December 2014 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored, unless otherwise specified”, but the paper never splits the figure loads into dead and live components. Throughout Parts A and B the applied figure loads are therefore treated as live load, factored by 1.5, and self-weight of concrete members as dead load, factored by 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). In Part C the paper does name the components, so 1.25D + 1.5S is used directly. If an examiner intended a different split the method is unchanged — only the numerical factor moves.

Check — figure dimensions. Every figure on this paper is a hand sketch on page 3; all geometry below was read from the printed figures. Figure A1 is not drawn to scale horizontally (its printed 600 and 300 dimensions are authoritative, not the drawn proportions). In Figure B1 the dimension “1 m × 1 m” runs from the section centreline to the outer face, so each cell is 1 m square and the box is 2000 mm wide × 1000 mm deep; the “600 × 600” dimension line brackets the void, giving 200 mm walls and a 400 mm central web. That reading is confirmed independently by the “65 typical” cover note: 65 + 65 cover plus two 15M tie legs plus one 30M bar needs exactly 200 mm of wall.

Question 2: A2 — Maximum factored load on an eccentrically loaded column (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. The Question A1 section (A_g = 55 000 mm², F_y = 350 MPa), used as a 6 m column hinged top and bottom, carrying P_f at point N on the x–x axis, 500 mm from the centroid O.

Quantity	Value
Height, L	6000 mm
Effective length factor, K	1.0 (hinged both ends)
Eccentricity, e	500 mm, along x–x
Elastic modulus, E	200 000 MPa
Shear modulus, G	77 000 MPa

Find. The greatest factored axial load P_f the member can carry, satisfying every S16 Clause 13.8.2 combined-load check.

Approach. Compute both second moments of area and radii of gyration; take the compressive resistance from the weaker axis and the moment resistance about the bending axis (including lateral–torsional buckling); then apply the three S16 13.8.2 interaction checks with $M_f = P_f e$ and solve each for P_f.

1. Second moments of area. About the horizontal axis the flanges dominate through their offset and the webs contribute their own depth: $$I_x = 2\left[\frac{600(25)^3}{12} + (600)(25)(137.5)^2\right] + 4\,\frac{25(250)^3}{12} = 699.0\times10^{6}\ \text{mm}^4$$ About the vertical axis the flanges act as deep plates in their own plane and the webs sit far from the axis: $$I_y = 2\,\frac{25(600)^3}{12} + \sum_{i=1}^{4}\left[\frac{250(25)^3}{12} + (250)(25)\,d_i^{\,2}\right] = 2049.3\times10^{6}\ \text{mm}^4$$ with $d_i = 287.5,\ 95.83,\ 95.83,\ 287.5$ mm.
2. Radii of gyration, and hence the governing slenderness. $$r_x=\sqrt{\frac{699.0\times10^{6}}{55\,000}}=112.7\ \text{mm} \qquad r_y=\sqrt{\frac{2049.3\times10^{6}}{55\,000}}=193.0\ \text{mm}$$ Buckling is therefore controlled by the x–x axis: $$\frac{KL}{r_x}=\frac{1.0(6000)}{112.7}=53.2$$
3. Compressive resistance (S16 13.3.1). With $n = 1.34$ for fabricated sections, $$\lambda=\frac{KL}{r_x}\sqrt{\frac{F_y}{\pi^2 E}}=53.2\sqrt{\frac{350}{\pi^2(200\,000)}}=0.708$$ $$C_r=\phi A F_y\left(1+\lambda^{2n}\right)^{-1/n}=0.90(55\,000)(350)\left(1+0.708^{2.68}\right)^{-1/1.34}$$ $$\boxed{C_r = 13\,496\ \text{kN}}$$ For the cross-sectional-strength check the same expression with $\lambda = 0$ gives $\phi A F_y = 17\,325$ kN.
4. Moment resistance about y–y, allowing for lateral–torsional buckling. The assembly is made only of thin rectangles, so warping is negligible and $$J=\sum\frac{bt^3}{3}=2\,\frac{600(25)^3}{3}+4\,\frac{250(25)^3}{3}=11.46\times10^{6}\ \text{mm}^4$$ With equal end eccentricities the moment is uniform, $\omega_2 = 1.0$, and $$M_u=\frac{\omega_2\pi}{L}\sqrt{E I_x G J}=\frac{\pi}{6000}\sqrt{(200\,000)(699.0\times10^{6})(77\,000)(11.46\times10^{6})}=5815\ \text{kN}\cdot\text{m}$$ Since $M_u > 0.67M_p$ (with $M_p = Z_yF_y = 3252$ kN·m), the inelastic branch of S16 13.6 governs: $$M_r=1.15\phi M_p\left(1-\frac{0.28M_p}{M_u}\right)=1.15(0.90)(3252)\left(1-\frac{0.28(3252)}{5815}\right)$$ $$\boxed{M_{ry} = 2839\ \text{kN}\cdot\text{m}} \le \phi M_p = 2927\ \text{kN}\cdot\text{m}$$
5. Moment amplification. The Euler load about the bending axis is $$C_{ey}=\frac{\pi^2 E I_y}{(KL)^2}=\frac{\pi^2(200\,000)(2049.3\times10^{6})}{6000^2}=112\,366\ \text{kN}$$ and with equal end moments in single curvature $\omega_1 = 0.6-0.4\kappa = 1.0$, giving $$U_{1y}=\frac{\omega_1}{1-C_f/C_{ey}} \ge 1.0$$
6. Apply the three interaction checks (S16 13.8.2). The section is a plate assembly, not an I-shape, so the coefficient $\beta$ is 1.0 in every term. With $M_{fy}=0.500P_f$ (in kN·m for P_f in kN), each check has the form $$\frac{P_f}{C_r}+\frac{U_{1y}(0.500 P_f)}{M_{ry}} \le 1.0$$ Solving each in turn:
7. Collect the governing case. Check (a) uses $C_r = \phi AF_y = 17\,325$ kN with $M_r = \phi M_p = 2927$ kN·m and yields $P_f = 4251$ kN. Check (b) substitutes the buckling resistance $C_r = 13\,496$ kN and gives $P_f = 3981$ kN. Check (c) keeps that $C_r$ but uses the lateral–torsional value $M_r = 2839$ kN·m, and is the most severe: $$\boxed{P_{f,\max} = 3898\ \text{kN}}$$ At this load $C_f/C_{ey} = 0.0347$, so $U_{1y} = 1.036$ — the second-order amplification is real but modest, because the 6 m member is very stiff about its bending axis.

The moment term does the damage here. At the governing load the axial ratio $P_f/C_r$ is only 0.289, while the amplified moment term contributes 0.711 — almost three-quarters of the interaction budget — because a 500 mm eccentricity on a 600 mm wide section places the load well outside the middle third. The member is, in every practical sense, a beam-column dominated by flexure.

Quantity	Value
Ix / Iy	699.0 / 2049.3 × 106 mm4
rx / ry	112.7 / 193.0 mm
Governing KL/r	53.2 (about x–x)
Compressive resistance, Cr	13 496 kN
Moment resistance, Mry (LTB)	2839 kN·m
Check (a) cross-sectional strength	Pf = 4251 kN
Check (b) overall member strength	Pf = 3981 kN
Check (c) lateral–torsional buckling	Pf = 3898 kN ← governs
Maximum factored load	Pf = 3898 kN

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