Question 5 of 7: B2 — Maximum load on a laterally loaded concrete column
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018 — 07-Str-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). A candidate submits five solutions — two from Part A, two from Part B and the one question in Part C — all of equal value. All seven questions are solved below, because this set is a study resource rather than an exam script. Page 1 states that all loads shown are unfactored, so every load case is factored here before any resistance is compared against it.
Reference texts.
CSA S16:19, Design of Steel Structures — and CISC, Handbook of Steel Construction, 11th ed. (section tables, Table 2 class limits, Table 7 minimum fillet sizes).
CSA A23.3:19, Design of Concrete Structures — and Cement Association of Canada, Concrete Design Handbook, 4th ed.
CSA O86:19, Engineering Design in Wood — and Canadian Wood Council, Wood Design Manual.
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
National Building Code of Canada 2015, Table 4.1.3.2 (load combinations).
Check — one dimension is missing from the source. Question A2 describes a stub cantilever welded to a column but the paper contains no Figure A2 and never states the cantilever projection. The projection is therefore taken as L = 1.5 m from the column face throughout question A2; every result below is also given in the general form so any other projection can be substituted directly. All other data are read from the printed text and from Figure A3, B1, B2 and B3 on page 3.
Question 5: B2 — Maximum load on a laterally loaded concrete column (8 + 6 + 6 marks)
Given. Square column 700 × 700 mm, 12–25M longitudinal bars (Ast = 6000 mm2, four per face including corners), 15M ties at 200 mm, height L = 6 m, fixed base and pinned head, horizontal load PF at mid-height with axial 10PF at the head, fc′ = 35 MPa, fy = 400 MPa. PF is taken as the factored load, consistent with the notation Pf used in Question A1.
Find. The largest PF for which the factored actions (10PF, Mf) lie on or inside the column's P–M interaction surface.
Figure 5.1 — Elevation, factored bending moment diagram and cross-section. The propped-cantilever action puts the peak moment at the fixed base.
Approach. Analyse the propped cantilever to express Mf in terms of PF, note that the eccentricity is therefore fixed by geometry alone, check slenderness, then march along the interaction diagram by strain compatibility until the ratio M/N equals that eccentricity.
Analysis of the propped cantilever. For a member fixed at one end and propped at the other with a point load at mid-height, the prop reaction is R = PF/4 and the fixed-end moment is
$$M_{\text{base}}=P_{F}\frac{L}{2}-\frac{P_{F}}{4}L=\frac{P_{F}L}{4}=\frac{6P_{F}}{4}=1.5P_{F}\ \ (\text{kN}\cdot\text{m})$$
with a value of 0.75PF at the load point. The base therefore governs.
The eccentricity is a constant. Because both actions scale with the same unknown,
$$e=\frac{M_{f}}{N_{f}}=\frac{1.5P_{F}}{10P_{F}}=0.150\ \text{m}=150\ \text{mm},\qquad \frac{e}{h}=0.214$$
This is the key simplification: the load path on the interaction diagram is a straight radial line of fixed slope, so there is exactly one intersection with the failure surface and the problem inverts directly.
Slenderness. With the head laterally held the frame is braced, so k = 0.8 and r = 0.3h = 210 mm:
$$\frac{kl_{u}}{r}=\frac{0.8(6000)}{210}=22.9\ \lt\ 34-12\!\left(\frac{M_{1}}{M_{2}}\right)=34$$
so by Cl. 10.15.2 the column is nominally short. Because a transverse load acts between the ends, the magnifier is still computed as a check with Cm = 1.0:
$$EI=0.4E_{c}I_{g}=0.4(4500\sqrt{35})\frac{700^{4}}{12}=2.13\times10^{14}\ \text{N}\cdot\text{mm}^{2}$$
$$P_{c}=\frac{\pi^{2}EI}{(kl_{u})^{2}}=\frac{\pi^{2}(2.13\times10^{14})}{4800^{2}}=91\,270\ \text{kN}$$
Iterating with the load found below, δb = 1/(1 − Nf/0.75Pc) = 1.099, so the design eccentricity becomes 150(1.099) = 165 mm.
Bar layout for strain compatibility. With 40 mm cover, 15M ties and 25M bars, d′ = 40 + 16 + 12.6 = 68.6 mm and the four rows sit at 68.6, 256.2, 443.8 and 631.4 mm from the compression face, carrying 4, 2, 2 and 4 bars respectively (12 in total).
March along the interaction diagram. For a trial neutral-axis depth c, take εcu = 0.0035 at the compression face, a = β1c, a concrete force Cc = α1φcfc′ba and bar forces φsAbfs from the linear strain profile (capped at ±fy, and reduced by the displaced concrete for bars inside the block). Bisecting on c until M/N matches the amplified eccentricity gives
$$c=495\ \text{mm},\qquad a=437\ \text{mm},\qquad N_{r}=6180\ \text{kN},\qquad M_{r}=1019\ \text{kN}\cdot\text{m}$$
Since c = 495 mm exceeds the balanced depth cb = 0.636d = 402 mm, the failure is compression-controlled — the extreme layer of bars is still elastic at fs = −193 MPa.
Recover the load. The axial force is 10PF, so
$$\boxed{P_{F,\max}=\frac{6180}{10}=618\ \text{kN}}$$
delivering a factored axial load of 6180 kN and a factored base moment of 1.5(618) = 927 kN·m before amplification.
Check the pure-compression ceiling. The maximum permitted axial resistance of a tied column is
$$P_{r,\max}=0.80\left[\alpha_{1}\phi_{c}f_{c}'(A_{g}-A_{st})+\phi_{s}f_{y}A_{st}\right]=8657\ \text{kN}$$
and Nr = 6180 kN is 71 % of it, which is the expected proportion at e/h ≈ 0.21. The reinforcement ratio ρ = 6000/490 000 = 1.22 % satisfies the 1 %–8 % limits of Cl. 10.9.1.
A designer reading this result should notice that the wind load itself, 618 kN applied as a point load at mid-height, is enormous; the arithmetic is nevertheless consistent, because the accompanying axial load is ten times larger and it is the axial force that consumes most of the capacity.
