07-Str-A2 · December 2014

Question 6 of 7: B3 — Design of column BC

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

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Str-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 whose OCR text is unusable; all geometry below was read from the printed figure of the original PDF. 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 6: B3 — Design of column BC (12 + 8 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. Column BC from the Question B2 frame: 6 m tall, pinned at C, rigidly joined to beam AB at B, with the 60 kN horizontal load applied 3 m below B. The column is to be treated as short.

Find. A square cross-section and its longitudinal and transverse reinforcement.

Approach. Take the axial force and moment at the column head from the B2 frame solution, form the eccentricity, then check a trial square section by strain compatibility at the actual factored axial load rather than with a pure-axial formula.

Actions at the head of the column. The column carries the 400 kN load applied directly at B plus the end shear delivered by beam AB. From the factored B2 analysis that shear is 310.5 kN, so $$P_f = 1.5(400) + 310.5 = \boxed{910.5\ \text{kN}}$$ The joint moment computed in B2 passes undiminished into the column head: $$M_f = 1.5(180) = \boxed{270\ \text{kN}\cdot\text{m}}$$
Moment distribution down the column. Below B the moment stays constant at 270 kN·m over the upper 3 m (there is no load along that stretch), then falls linearly to zero at the pin at C. The design section is therefore anywhere in the upper 3 m, where the full moment acts together with the full axial load.
Eccentricity, and what it implies. $$e = \frac{M_f}{P_f}=\frac{270\times10^{3}}{910.5}=297\ \text{mm}$$ This is very large — comparable with the whole section depth — so the member is flexure-dominated. It must be designed by strain compatibility on the interaction diagram; the pure-axial expression $P_{r,\max}=0.80[\alpha_1\phi_cf'_c(A_g-A_{st})+\phi_sf_yA_{st}]$ is simply the wrong tool here and would badly overstate capacity.
Trial section. Take a 450 × 450 mm square with 8-25M bars — three in each of the two faces perpendicular to the bending plane and two at mid-depth — and 10M ties: $$\rho = \frac{8(500)}{450^2}=0.0198 = 1.98\ \%$$ which sits between the A23.3 limits of 1 % and 8 %. With 40 mm cover and 10M ties, $$d = 450-40-11.3-\tfrac{25.2}{2}=386.1\ \text{mm}$$
Solve the section at the actual axial load. Finding the neutral-axis depth that makes the internal axial force equal 910.5 kN gives $$c = 160.0\ \text{mm}, \qquad a = \beta_1c = 141.2\ \text{mm}$$ The tension-face bars are then at a strain of $$\varepsilon_t = 0.0035\,\frac{d-c}{c}=0.0035\,\frac{386.1-160.0}{160.0}=0.00495 \; > \; \frac{f_y}{E_s}=0.002$$ so they yield — the section is on the tension-controlled side of the balance point, exactly as the large eccentricity predicted.
Moment resistance at that axial load. Summing the moments of the concrete block and all three bar layers about the section centreline: $$\boxed{M_r = 342\ \text{kN}\cdot\text{m}} \; > \; M_f = 270\ \text{kN}\cdot\text{m} \quad\checkmark$$ The utilisation is $270/342 = 0.79$, a sensible design point — worked hard but not marginal.
Shear in the column. The factored horizontal load is $V_f = 1.5(60) = 90$ kN, carried by the lower 3 m. With $d_v=\max(0.9d,0.72h)=347.5$ mm, $$V_c=0.65(1.0)(0.18)\sqrt{35}(450)(347.5)=108\ \text{kN} \; > \; 90\ \text{kN} \quad\checkmark$$ No designed shear reinforcement is needed; the compression ties already required for the longitudinal bars are sufficient.
Ties. A23.3 7.6.5.2 limits the tie spacing to the least of 16 bar diameters, 48 tie diameters and the least column dimension: $$s \le \min\left[16(25.2),\ 48(11.3),\ 450\right] = \min(403,\ 542,\ 450) = 403\ \text{mm}$$ $$\Rightarrow\ \boxed{\text{10M ties at 400 mm}}$$ Reduce this to 150 mm within one section depth of the joint at B and above the pin at C, where the bars are most heavily stressed and confinement matters most.

Check — the “short column” instruction. The question directs us to assume the column is short, and the design above follows that instruction. It is worth recording what the slenderness actually is: for the adopted 450 mm square, $r = 0.3h = 135$ mm and, taking a sway-free $k = 0.8$, $k\ell_u/r = 35.6$ against the A23.3 10.15.2 threshold of $34-12(M_1/M_2) = 34$. Strictly the member is just past the short-column boundary, and because the roller at A gives the frame no horizontal restraint at B, a rigorous effective-length factor would exceed 2.0. A designer working outside the exam’s stated assumption should either apply the moment magnifier or increase the section to 500 mm square, which restores $k\ell_u/r = 32$ and makes the short-column idealisation genuinely valid.

Quantity	Value
Factored axial load, P_f	910.5 kN
Factored moment, M_f	270 kN·m
Eccentricity, e	297 mm (flexure-dominated)
Section	450 × 450 mm square
Longitudinal steel	8-25M (ρ = 1.98 %)
Neutral-axis depth at P_f	c = 160.0 mm, ε_t = 0.00495 (yields)
Moment resistance, M_r	342 kN·m > 270 ✓ (utilisation 0.79)
Shear: V_f / V_c	90 kN / 108 kN ✓
Ties	10M at 400 mm (150 mm at the joint zones)

Topic: Design of a short reinforced concrete column under combined axial load and large-eccentricity bending.

Key relations: $e = M_f/P_f$; strain compatibility with $\varepsilon_{cu}=0.0035$ and $f_s=E_s\varepsilon_s \le f_y$; equilibrium $P_r=C_c+\sum\phi_sA_{si}f_{si}$ and $M_r=\sum F_iz_i$; $\rho$ between 0.01 and 0.08; tie spacing $\le\min(16d_b,48d_{tie},h)$; slenderness threshold $34-12(M_1/M_2)$.

Why this works: A column’s capacity is not a single number but a curve — the interaction diagram — because the moment it can carry depends on how much axial load is present. At high axial load the whole section is in compression and moment capacity is small; below the balance point, adding axial load actually increases the moment capacity by delaying tension yielding. Since $P_f$ here is far below the balance load, the section behaves essentially as a doubly reinforced beam with a helpful axial thrust, and the tension steel yields well before the concrete crushes. That is why the design must be run at the actual $P_f$ rather than checked against a pure-axial capacity.

Common pitfall: Designing this member with $P_{r,\max}=0.80[\ldots]$, the concentrically loaded expression. At $e = 297$ mm that formula has no physical meaning and will pass a section that fails in flexure. The related error is forgetting that the beam’s hogging moment enters the column head through the monolithic joint — taking $M_f$ from the 60 kN load alone would give a third of the true value.

Source: CSA A23.3, Design of Concrete Structures, Clauses 10.9, 10.10, 10.15 and 7.6.5; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Ch. 11.

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