B2. Design of column CD (4 + 4 + 12 marks): 16-Civ-A2 Elementary Structural Design, May 2014, Q5

Question 5 — B2. Design of column CD (4 + 4 + 12 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.

Figure B1 — determinate reinforced concrete frame: roller at A, internal hinge at B, rigid joint at C, fixed base at D. Loads shown are unfactored.

Given.

Quantity	Value
Column	CD, 6.0 m high, fixed at D, rigidly connected to beam ABC at C
Horizontal load	80 kN unfactored, at mid-height (3 m above D)
Actions from Question B1	beam end shear 233.1 kN, beam end moment 653.6 kN·m
Joint load	300 kN unfactored applied directly at C
Materials	$f_c' = 35$ MPa, $f_y = 400$ MPa; column assumed short (no magnification)

Find. Cross-sectional dimensions, longitudinal steel and ties for column CD.

Approach. Assemble the axial force and the moment diagram down the column from the beam-end actions plus the mid-height horizontal load, design at the base where the moment peaks, and check the chosen symmetric section by strain compatibility at the actual axial load.

Axial load on the column. The column carries the 450 kN factored load applied at the joint, the 233.1 kN beam end shear, and its own weight. Trying a 500 mm × 800 mm section, $$P_f = 450 + 233.1 + 1.25(24)(0.5)(0.8)(6.0) = 755.1\ \text{kN}$$
Moment diagram down the column. The rigid joint at C delivers the beam's hogging moment of 653.6 kN·m into the column top. Below that, the 120 kN factored horizontal load at mid-height adds a further $120(3.0)$ kN·m by the time the base is reached: $$M_{top} = 653.6\ \text{kN}\cdot\text{m}, \qquad M_{base} = 653.6 + 120(3.0) = 1\,013.6\ \text{kN}\cdot\text{m}$$ $$\boxed{P_f = 755.1\ \text{kN}, \qquad M_f = 1\,013.6\ \text{kN}\cdot\text{m at D}}$$ The eccentricity is $M_f/P_f = 1\,342$ mm — far outside the section — so this member is overwhelmingly flexural, with the axial load acting as a modest bonus rather than the dominant action.
Set up the strain-compatibility check. For a symmetrically reinforced section with $A_s$ in each face at $d' = 66.3$ mm from the near face, $$P_r = \alpha_1\phi_cf_c'b a + \phi_sA_s'f_s' - \phi_sA_sf_s, \qquad M_r = \alpha_1\phi_cf_c'ba\!\left(\frac{h}{2}-\frac{a}{2}\right) + \phi_sA_s'f_s'\!\left(\frac{h}{2}-d'\right) + \phi_sA_sf_s\!\left(d-\frac{h}{2}\right)$$ with $a = \beta_1 c$ and the steel stresses from $\varepsilon_{cu} = 0.0035$ linear strain, capped at $f_y$.
Solve for the neutral axis at the applied axial load. Trying 10–30M in total (5 per face, $A_s = 3\,500$ mm$^2$ each face, $\rho = 1.75$ %), the value of $c$ that makes $P_r = 755.1$ kN is $c = 123.0$ mm. At that neutral-axis depth the tension steel is well past yield and the compression steel is not, and the corresponding moment resistance is $$\boxed{M_r = 1\,058\ \text{kN}\cdot\text{m} > M_f = 1\,013.6\ \text{kN}\cdot\text{m}}$$ a utilisation of 0.96 — efficient without being tight.
Check the code limits on the reinforcement. The 1 % minimum for a column requires $0.01(500)(800) = 4\,000$ mm$^2$; the 7 000 mm$^2$ provided sits comfortably between that and the 4 % maximum. Five 30M bars per face fit easily across the 500 mm width. Ties must be at least 10M (for 30M longitudinals) at a spacing not exceeding $\min(16d_b, 48d_{tie}, b) = \min(478, 542, 500) = 478$ mm, so use 10M ties at 400 mm, tightened to 200 mm over a distance $h$ either side of the joint at C and above the base at D where the moment gradient is steepest.
Check shear on the column. The column shear is the 120 kN horizontal load. With $d_v = 660.4$ mm, $V_c = 0.65(0.18)(5.916)(500)(660.4) = 229$ kN, which alone exceeds the demand, so the tie spacing above is governed by confinement and not by shear.

Because the eccentricity is so large, the section behaves essentially as a doubly reinforced beam: the concrete compression block is only 109 mm deep, and almost the whole resistance comes from the steel couple. That is also why symmetric reinforcement is the right choice here even though the moment always acts in one direction — the wind-type horizontal load can reverse, and a symmetric column is detailed once and works either way.

Result	Value
Factored axial load	755.1 kN
Moment at top C / base D	653.6 / 1 013.6 kN·m
Eccentricity $M/P$	1 342 mm
Section	500 mm × 800 mm
Longitudinal steel	10–30M (5 per face), $\rho = 1.75$ %
Neutral axis at $P_f$	$c = 123$ mm
$M_r$ at $P_f$	1 058 kN·m (utilisation 0.96)
Ties	10M at 400 mm, reduced to 200 mm at both ends

Topic. Short reinforced concrete column under combined axial load and large moment.

Key relations. Strain compatibility with $\varepsilon_{cu} = 0.0035$ and $a = \beta_1c$; $P_r$ and $M_r$ from horizontal and moment equilibrium of the internal forces; $0.01 \le \rho \le 0.04$; tie spacing $\le \min(16d_b, 48d_{tie}, \text{least dimension})$.

Why this works. A column's capacity is not a single number but an interaction curve, and the design question is always whether the point $(P_f, M_f)$ falls inside it. Rather than plotting the whole curve, it is quicker to solve for the one neutral-axis depth that reproduces the applied axial load and then read off the moment that accompanies it — a single point on the curve, which is the only point that matters.

Common pitfall. Missing the moment carried up from the beam. The rigid joint at C transfers the full beam hogging moment into the column head, and forgetting it leaves only the $120(3.0) = 360$ kN·m from the horizontal load — about a third of the true base moment. A second pitfall is treating a large-eccentricity column with the pure-axial formula $P_{r,max} = 0.80[\alpha_1\phi_cf_c'(A_g - A_{st}) + \phi_sf_yA_{st}]$, which here would suggest almost 8 000 kN of capacity and is entirely irrelevant.

Source. CSA A23.3, Clauses 10.1.2–10.1.7 (strain compatibility), 10.9.1–10.9.2 (steel limits) and 7.6.5 (ties).

Question 5 of 7: B2. Design of column CD (4 + 4 + 12 marks)

Question 5 — B2. Design of column CD (4 + 4 + 12 marks)