16-Civ-A2 Elementary Structural Design · December 2018
Question 6 of 7: B3 — Square reinforced concrete column CD
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2018 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). 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 (1.5 on the specified live/imposed loads, 1.25 on self weight) 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, Clause 13.8 interaction, Clause 13.5 effective widths).
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 (glulam selection tables).
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
National Building Code of Canada, Table 4.1.3.2 (load combinations).
Check — assumptions declared under Note 1 of the paper. (i) In A2 the single dimension h is the wall thickness of the whole fabricated panel (top plate and both stems), which is how the figure dimensions it. (ii) In A3 the bolted joint at the column is a simple (shear-only) connection and the steel tie is a two-force member, which is what makes the stub beam determinate. (iii) In B1/B3 the left column A–B is taken as 6 m high, level with the roller at D, as Figure B1 draws it; the beam self weight is included, the column self weight is not (it does not change any design action at C). (iv) In B2 and C1 the beam is drawn with a tapered soffit, but part (a) asks for a uniform cross-section, so a prismatic member is designed.
6.0 m, monolithic with the beam at C, roller (free to rotate) at D
Factored axial force from Question B1
$N_f = D_y = 453.6$ kN
Factored moment at C from Question B1
$M_f = 360$ kN·m
Horizontal load position
3.0 m above D
Materials
$f_c' = 35$ MPa, $f_y = 400$ MPa
Find. A square section for CD with its longitudinal and tie reinforcement.
Approach. Take the factored actions straight from the B1 frame analysis, form the eccentricity $e = M_f/N_f$, and find a square section whose factored interaction diagram passes outside the point $(N_f, M_f)$. Because the eccentricity is very large, the section must be designed by strain compatibility — the pure-axial formula does not apply.
Design actions along the column. The roller at D supplies no horizontal reaction, so below the 80 kN load the column carries no shear and no moment; above it the shear is constant at $H_f = 120$ kN and the moment grows linearly to
$$M_f=120(3.0)=360\ \text{kN}\cdot\text{m}\ \text{at C},\qquad N_f=D_y=453.6\ \text{kN}$$
The critical section is therefore the head at C, where the moment is greatest and the axial force is least — the worst possible combination for a column.
Eccentricity.
$$e=\frac{M_f}{N_f}=\frac{360\times10^{6}}{453.6\times10^{3}}=\boxed{794\ \text{mm}}$$
For any plausible square section this is well over the depth itself, so CD is not a column carrying a small moment — it is a flexural member carrying a modest axial force. The compression-controlled formula $P_{r,\max}=0.80\left[\alpha_1\phi_cf_c'(A_g-A_{st})+\phi_sf_yA_{st}\right]$, which for the section chosen below returns 5071 kN, is meaningless at this eccentricity and must not be used.
Trial section. Try 500 mm x 500 mm with 8-30M (8 x 700 = 5600 mm2) arranged three per face, 40 mm cover and 10M ties, giving $d' = 40+11.3+14.9 = 66.2$ mm to the bar centres. The steel ratio is
$$\rho=\frac{5600}{500\times500}=2.24\%$$
which sits between the A23.3 Clause 10.9.1 limits of 1 % and 8 % (and below the 4 % at which bar congestion at lap splices becomes a problem).
Strain-compatibility solution at the required eccentricity. Assume the extreme fibre reaches $\varepsilon_{cu}=0.0035$, take an equivalent stress block $\alpha_1\phi_cf_c'$ over a depth $a=\beta_1c$, and give every bar layer the stress $f_s = E_s\varepsilon_{cu}(c-d_i)/c$ capped at $\pm f_y$. Solving for the neutral-axis depth $c$ that yields $M_r/N_r = e$:
$$c=143\ \text{mm}\quad\Longrightarrow\quad N_r=579.4\ \text{kN},\qquad M_r=459.8\ \text{kN}\cdot\text{m}$$
The resistance point lies on the same radial line as the demand, so the comparison is direct:
$$\frac{N_f}{N_r}=\frac{453.6}{579.4}=\boxed{0.78}$$
The section is adequate with 22 % reserve. Since $c = 143$ mm is far less than the balanced depth of roughly 300 mm, the failure mode is tension-controlled — the steel yields first and the column gives warning.
Ties. Clause 7.6.5 sets the tie spacing at the least of 16 longitudinal bar diameters, 48 tie diameters and the least column dimension:
$$16(29.9)=478\ \text{mm},\qquad 48(11.3)=542\ \text{mm},\qquad 500\ \text{mm}$$
Provide 10M ties at 400 mm, closed, with intermediate cross-ties engaging the mid-face bars so that no bar is more than 150 mm clear from a laterally supported one.
Below the 80 kN load. Over the lower 3.0 m the column is in pure axial compression at 453.6 kN, which the same section carries with enormous reserve. The section and the bars are nevertheless run full height for constructability, and the ties are closed at 400 mm throughout.
Check — the "short column" instruction. The question directs that the column be assumed short, and it is designed that way above. A rigorous check does not support the assumption: with a roller at D and no other lateral restraint the frame can sway, so $k = 2.0$ and
$$\frac{k\ell_u}{r}=\frac{2.0(6000)}{0.3(500)}=80\ \gg\ 22$$
the A23.3 Clause 10.15.2 threshold for a member in single curvature. A rigorous design would apply a moment magnifier, and increasing the section to 600 mm square (which carries $N_r = 803$ kN at the same eccentricity, a utilisation of 0.57) would absorb it comfortably. The 500 mm square answer follows the instruction as printed; the 600 mm square is what would be built.
