Design of column CD: 07-Str-A2, May 2014, Q5

Question B2: Design of column CD (20 marks: 4 + 4 + 12)

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.

Quantity	Value
Column CD	6 m, rigid joint at C, fixed base at D, assumed short
Loads at and on the column	300 kN down at C; 125 kN delivered by beam BC; 80 kN horizontal at mid-height
Materials	$f_c' = 35$ MPa, $f_y = 400$ MPa, $\phi_c = 0.65$, $\phi_s = 0.85$
Load factor	$\alpha_L = 1.5$ on all applied loads

Find. Cross-sectional dimensions, longitudinal steel and ties for CD, checked against the combined axial force and bending moment over its full height.

Approach. Transfer the beam actions across the rigid joint at C to get the column head force and moment, walk down the column past the 80 kN load to get the base moment, factor everything, then find the neutral-axis depth at which a trial section carries the factored axial load and read off the moment it can carry there.

Axial force in the column. Cutting just below C and taking the whole beam plus the joint as a free body, $$N = 250 + 300 - A_y = 250 + 300 - 125 = 425\ \text{kN (compression)}$$ constant over the height, since nothing else is applied vertically.
Moment at the head of the column. Joint C is monolithic, so the beam’s end moment passes straight into the column: $$M_C = |A_y(6.0) - 250(4.5)| = |750 - 1125| = 375\ \text{kN}\cdot\text{m}$$ with tension on the face away from A, continuing round the corner from the tension in the top of the beam.
Moment down the column. Above the 80 kN load the column carries no shear, so the moment is constant at 375 kN·m from C down to mid-height. Below it the shear is 80 kN and the moment grows linearly: $$M_D = 375 + 80(3.0) = 615\ \text{kN}\cdot\text{m}$$ $$\boxed{N = 425\ \text{kN},\quad M_C = 375,\quad M_D = 615\ \text{kN}\cdot\text{m}, \quad V = 80\ \text{kN}\ \text{(specified)}}$$ The two effects add rather than cancel — the 80 kN pushes the column towards A, putting tension on the same face the beam does — which is worth confirming, because the alternative sign would have made the head, not the base, the critical section.
Factored design actions. $$P_f = 1.5(425) = 637.5\ \text{kN},\qquad M_f = 1.5(615) = 922.5\ \text{kN}\cdot\text{m},\qquad V_f = 1.5(80) = 120\ \text{kN}$$ The eccentricity is $e = M_f/P_f = 1.45$ m, far beyond the section depth: this is a flexure-dominated member that happens to carry some compression, and the axial load will help its moment capacity because the section is well below the balance point.
Trial section. Take $b = 400$ mm by $h = 900$ mm with the 900 mm in the plane of bending, matching the beam width for a clean joint. Use symmetric reinforcement, as columns must be: 4–30M in each of the two faces perpendicular to bending ($A_s = A_s' = 2800$ mm$^2$, total 5600 mm$^2$), with $d = 830$ mm and $d' = 70$ mm. The steel ratio is $$\rho = \frac{5600}{400(900)} = 0.0156 = 1.56\ \text{per cent}$$ within the A23.3 Cl.10.9.1 range of 1 to 8 per cent.
Locate the neutral axis for $P_f$. With strain compatibility (concrete crushing at 0.0035) the two steel stresses are $f_s' = 0.0035(c - d')E_s/c$ and $f_s = 0.0035(d - c)E_s/c$, each capped at $f_y$, and axial equilibrium requires $$P_r = \alpha_1\phi_c f_c' b \beta_1 c + \phi_s A_s' f_s' - \phi_s A_s f_s = P_f$$ Solving iteratively gives $c = 129.1$ mm, at which $f_s' = 320$ MPa (compression steel not yet yielded) and $f_s = 400$ MPa (tension steel at yield), with $a = \beta_1 c = 113.9$ mm.
Moment resistance at that axial load. Taking moments about the plastic centroid at mid-depth, $$M_r = C_c\left(\frac{h}{2} - \frac{a}{2}\right) + \phi_s A_s' f_s'\left(\frac{h}{2} - d'\right) + \phi_s A_s f_y\left(\frac{h}{2} - d'\right)$$ $$M_r = 826.8(0.393) + 762.6(0.380) + 952.0(0.380)$$ $$\boxed{M_r = 976\ \text{kN}\cdot\text{m}\ \text{at}\ P_f = 637.5\ \text{kN} \;>\; M_f = 922.5\ \text{kN}\cdot\text{m}\quad (\text{utilisation } 0.95)}$$
Pure-axial and short-column checks. The maximum axial resistance $$P_{r,\max} = 0.80\left[\alpha_1\phi_c f_c'(A_g - A_{st}) + \phi_s f_y A_{st}\right] = 6667\ \text{kN}$$ so the section uses under a tenth of its axial capacity — confirming the member is governed by moment. The question states that the column may be assumed short, so no slenderness magnification (A23.3 Cl.10.15) is applied; that assumption is worth recording explicitly, because with $h = 900$ mm and a 6 m height a real check would be needed for the unbraced case.
Shear and ties. With $d_v = \max(0.9(830), 0.72(900)) = 747$ mm, $$V_c = 0.65(0.18)\sqrt{35}(400)(747) \times 10^{-3} = 206.8\ \text{kN} \;>\; V_f = 120\ \text{kN}$$ so the ties are governed by detailing, not strength. A23.3 Cl.7.6.5 caps their spacing at the least of 16 longitudinal bar diameters (478 mm), 48 tie diameters (542 mm) and the least column dimension (400 mm). Provide 10M ties at 400 mm, which also supply $A_v = 200 > A_{v,\min} = 142$ mm$^2$; tighten to 200 mm over the top and bottom metre where the moment gradient is steepest and the bars lap.

Factored bending-moment diagram for column CD and the adopted section. The moment is constant at 562.5 kN·m over the upper half, where the column carries no shear, and grows to 922.5 kN·m at the fixed base.

Result	Value
Specified axial force in CD	425 kN (compression, constant)
Specified moments: head / base	375 / 615 kN·m
Factored actions	$P_f = 637.5$ kN; $M_f = 922.5$ kN·m; $V_f = 120$ kN
Design eccentricity $M_f/P_f$	1.45 m — flexure dominated
Section adopted	400 mm × 900 mm (900 in the plane of bending)
Longitudinal steel	8–30M, 4 per face; $\rho = 1.56$ per cent
Neutral-axis depth at $P_f$	$c = 129$ mm, $a = 114$ mm
$M_r$ at $P_f$	976 kN·m (utilisation 0.95)
$P_{r,\max}$	6667 kN
Ties	10M at 400 mm, closed; 200 mm over the end metres

Topic. Short reinforced concrete columns under combined axial load and bending — a single point on the P–M interaction diagram.

Key relations. Strain compatibility $f_s = 0.0035(d - c)E_s/c \leq f_y$; axial equilibrium $P_r = \alpha_1\phi_c f_c' b\beta_1 c + \phi_s A_s'f_s' - \phi_s A_s f_s$; moment about the plastic centroid; and $P_{r,\max} = 0.80[\alpha_1\phi_c f_c'(A_g - A_{st}) + \phi_s f_y A_{st}]$.

Why this works. A column section does not have one moment resistance; it has a curve of them, one for each axial load, and the calculation above evaluates that curve at the single point the frame demands. Below the balance point — which is where this column sits, with $c/d = 0.16$ — adding axial compression increases the moment the section can carry, because the extra compression must be balanced by more tension steel force, which is only available if the neutral axis moves down and the lever arm grows. That is why a member with an eccentricity of 1.45 m, nearly twice its depth, is still a perfectly sensible column: it is behaving as a beam with a helpful axial assist. The one-line reason the base rather than the head governs is that the 80 kN lateral load bends the column the same way the beam already does, so the two moments add over the lower half.

Common pitfall. Designing the column for its axial load alone. Here $P_f/P_{r,\max} = 0.10$, so a strength-only approach would suggest a section a fraction of the size, and the member would fail in bending long before it was loaded. The second is placing the reinforcement asymmetrically to suit the computed tension face: columns are detailed symmetrically because the moment reverses under wind and because a mis-oriented cage is the easiest error to make on site.

Source. CSA A23.3:19 Cl.10.9 to 10.11 (columns), Cl.7.6.5 (ties), Cl.10.15 (slenderness, excluded by the question); MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Ch.11.

Question 5 of 7: Design of column CD

Question B2: Design of column CD (20 marks: 4 + 4 + 12)