CIV-330 · Lesson 10 of 12 · 20 min

Short columns: axial load and bending

Columns carry the floors above them down to the foundations. Almost every column carries some bending as well as axial load, from unbalanced floor spans, frame action under wind, or simply the unavoidable eccentricity of construction. This lesson develops the strength of a short column, one in which the extra moment from lateral deflection is negligible, under combined axial load and bending. The tool for this is the interaction diagram, built by the same strain-compatibility method used for beams in lesson 4. The example is the ground-floor interior column of the office building.

Types of columns

Most concrete columns are tied columns: longitudinal bars held in position by rectangular ties at regular spacing. Spiral columns have their bars arranged in a circle and enclosed by a continuous helix of closely spaced wire or bar. The spiral confines the core concrete so well that after the cover spalls off the column can carry its load at large deformations, which is why spiral columns get a higher resistance factor and a higher maximum axial load. Composite columns encase a structural steel shape in concrete or fill a steel tube with concrete.

Columns are also classed by slenderness. A short column fails by material failure, crushing of the concrete or yielding of the steel, at the load given by its cross-section. A slender column deflects laterally under load, the axial load acting on that deflection adds moment, and the column fails at a lower load. Most building columns braced against sway are short; lesson 11 covers slender ones.

Axial load capacity

Under concentric axial load, the concrete and steel shorten together. The concrete reaches its capacity at a strain of about 0.002, at which point Grade 400 or Grade 60 steel has just yielded, so the strength is the sum of the two contributions:

\[P_{ro} = \alpha_1\phi_c f'_c\big(A_g - A_{st}\big) + \phi_s f_y A_{st}\]

The factor \(\alpha_1\) reflects the lower strength of concrete in a column compared with a cylinder, caused by the size, the casting position and the sustained nature of the load. No real column is perfectly concentric, so A23.3 caps the factored axial resistance at \(P_{r,max} = 0.80P_{ro}\) for tied columns and \(0.85P_{ro}\) for spiral columns. That cap allows for an accidental eccentricity of roughly 10 per cent of the column depth.

The steel ratio \(\rho_g = A_{st}/A_g\) must be between 1 and 8 per cent. The lower limit gives some resistance to bending and controls creep, which slowly shifts load from the concrete to the steel. The upper limit avoids congestion; in practice 4 per cent is a sensible maximum where bars are lap-spliced, because the laps double the steel at the splice. Rectangular tied columns need at least four bars, one in each corner.

Axial load with bending

When a column carries a moment \(M\) together with an axial load \(P\), the load acts as if it were applied at an eccentricity \(e = M/P\) from the centroid. A small eccentricity puts the whole section in compression and the column fails by crushing. A large eccentricity puts the far face in tension, the far steel yields, and the column behaves more like a beam. Between these lies the balanced condition, in which the concrete reaches its crushing strain at the same moment that the extreme tension steel reaches its yield strain.

For each assumed position of the neutral axis there is one strain profile, one set of bar stresses and one pair of resultants \((P, M)\). Sweeping the neutral axis from far outside the section (pure compression) to near the compression face (tension) traces out a curve: the interaction diagram. Any combination of factored load and moment that falls inside the curve is safe.

\[P_r = C_c + \sum_i \phi_s A_{si} f_{si}, \qquad M_r = C_c\Big(\frac{h}{2} - \frac{a}{2}\Big) + \sum_i \phi_s A_{si} f_{si}\Big(\frac{h}{2} - y_i\Big)\]

Here \(C_c = \alpha_1\phi_c f'_c ab\), \(y_i\) is the depth of bar layer \(i\) from the compression face, \(f_{si} = E_s\varepsilon_{si}\) limited to \(\pm f_y\) and taken positive in compression, and the concrete displaced by compression bars inside the stress block is subtracted. Because the material factors are inside every term, each point is already a factored resistance; A23.3 needs no further reduction.

An interaction diagram with factored moment resistance on the horizontal axis in kN·m and factored axial resistance on the vertical axis in kN. The curve rises from a pure tension point at minus 1360 kN, passes through pure bending at 268 kN·m, bulges out to the balanced point near 431 kN·m and 1984 kN, and is capped at 4177 kN. A red dot at 120 kN·m and 3205 kN lies well inside.
Factored interaction diagram for the 500 × 500 mm column with 8-25M, with the ground-floor load.

The diagram has a recognizable shape. From the capped top, increasing moment lowers the axial capacity in the compression-controlled region. At the balanced point the moment capacity is a maximum (for the factored diagram of a symmetrical section). Below balanced, in the tension-controlled region, a little axial compression actually increases the moment capacity, because it delays yielding of the tension steel. This last fact matters: a column with low axial load and high moment is unconservative to check with the largest axial load, so designers check the combinations with minimum axial load too, such as 0.9D with wind.

The office column

The ground-floor interior column supports two floors and the roof over a 7.5 m × 7.5 m tributary area. From the floor and roof loads and the column self-weight, Owen finds specified loads of \(D = 1700\) kN and \(L = 720\) kN (with the live load reduction the building code allows for large tributary areas). The frame analysis of CIV-321 gives a factored moment of 120 kN·m from unbalanced live load on the adjacent spans. The factored axial load is \(P_f = 1.25(1700) + 1.5(720) = 3205\) kN. Priya has proposed 500 × 500 mm columns with 8-25M bars (three per face), \(f'_c = 30\) MPa.

Worked problem

Key points of the column interaction diagram

Compute the pure axial point, the cap and the balanced point for the 500 × 500 mm column with 8-25M: three bars at \(y = 64\) mm, two at 250 mm, three at 436 mm.

Show solution
  1. \(A_g = 250\,000\) mm², \(A_{st} = 4000\) mm² (\(\rho_g = 1.6\) per cent), \(\alpha_1 = 0.805\), \(\beta_1 = 0.895\).
  2. \[P_{ro} = 0.805(0.65)(30)(246\,000) + 0.85(400)(4000) = 3862 + 1360 = 5222\ \text{kN} \qquad P_{r,max} = 0.80(5222) = 4177\ \text{kN}\]
  3. Balanced: the bottom bars reach \(\varepsilon_y = 0.002\) when the top fibre reaches 0.0035:
  4. \[c_b = \frac{0.0035}{0.0035 + 0.002}(436) = 277\ \text{mm} \qquad a = 0.895(277) = 248\ \text{mm}\]
  5. \[C_c = 0.805(0.65)(30)(248)(500) = 1950\ \text{kN}\]
  6. Top bars: \(\varepsilon = 0.0035(277 - 64)/277 = 0.0027\), yielded; force \(1500(0.85 \times 400 - 15.7) = 487\) kN. Middle bars: \(\varepsilon = 0.00035\), \(f_s = 69\) MPa, force 59 kN. Bottom bars: yielded in tension, \(-510\) kN.
  7. \[P_r = 1950 + 487 + 59 - 510 = 1985\ \text{kN}\]
  8. \[M_r = 1950(250 - 124) + 487(250 - 64) + 510(436 - 250) = 245 + 91 + 95 = 431\ \text{kN·m}\]

Answer: Pure axial 5222 kN, capped at 4177 kN; balanced point (431 kN·m, 1985 kN). The factored load (120 kN·m, 3205 kN) is far inside the diagram.

Computing a full diagram by hand is tedious, so Sofia extends the strain-compatibility routine of lesson 4 to a net axial force and sweeps the neutral axis depth. She also interpolates the moment capacity at the factored axial load.

Interaction diagram points for the office column and a check of the ground-floor load.
# Interaction diagram for a tied column, CSA A23.3 (factored resistances)
fc, fy, Es, ecu = 30.0, 400.0, 200000.0, 0.0035
phic, phis = 0.65, 0.85
b = h = 500.0
layers = [(3 * 500.0, 64.0), (2 * 500.0, 250.0), (3 * 500.0, 436.0)]   # 8-25M
a1 = max(0.85 - 0.0015 * fc, 0.67); b1 = max(0.97 - 0.0025 * fc, 0.67)
Ag = b * h; Ast = sum(A for A, _ in layers)
Pro = a1 * phic * fc * (Ag - Ast) + phis * fy * Ast
Prmax = 0.80 * Pro

def point(c):
    a = min(b1 * c, h)
    Cc = a1 * phic * fc * a * b
    P = Cc; M = Cc * (h / 2 - a / 2)
    for As, y in layers:
        es = ecu * (c - y) / c
        fs = max(-fy, min(fy, Es * es))
        f = phis * As * fs - (a1 * phic * fc * As if y < a else 0)
        P += f; M += f * (h / 2 - y)
    return P / 1e3, M / 1e6

print(f"rho = {Ast/Ag*100:.2f}%, Pro = {Pro/1e3:.0f} kN, Pr,max = {Prmax/1e3:.0f} kN")
cb = ecu / (ecu + fy / Es) * 436
print("    c (mm)    Pr (kN)   Mr (kN.m)")
pts = []
for c in (600, 500, 400, 350, cb, 200, 150, 100, 60):
    P, M = point(c)
    pts.append((P, M))
    tag = "  balanced" if c == cb else ""
    print(f"{c:9.0f} {min(P, Prmax/1e3):10.0f} {M:10.0f}{tag}")
# pure bending: find c with P = 0 by bisection
lo, hi = 1.0, 250.0
for _ in range(60):
    c = (lo + hi) / 2
    lo, hi = (lo, c) if point(c)[0] > 0 else (c, hi)
print(f"pure bending: c = {c:.0f} mm, Mr = {point(c)[1]:.0f} kN.m")
# check the ground-floor interior column: Pf = 3205 kN, Mf = 120 kN.m
Pf, Mf = 3205, 120
pts.sort()
for (P1, M1), (P2, M2) in zip(pts, pts[1:]):
    if P1 <= Pf <= P2:
        Mcap = M1 + (M2 - M1) * (Pf - P1) / (P2 - P1)
print(f"at Pf = {Pf} kN the diagram gives Mr = {Mcap:.0f} kN.m vs Mf = {Mf} -> {'OK' if Mf <= Mcap and Pf <= Prmax/1e3 else 'not OK'}")
Output
rho = 1.60%, Pro = 5222 kN, Pr,max = 4177 kN
    c (mm)    Pr (kN)   Mr (kN.m)
      600       4177         49
      500       4177        166
      400       3423        305
      350       2880        361
      277       1984        431  balanced
      200       1233        411
      150        690        378
      100        150        294
       60       -488        178
pure bending: c = 89 mm, Mr = 268 kN.m
at Pf = 3205 kN the diagram gives Mr = 327 kN.m vs Mf = 120 -> OK

At \(P_f = 3205\) kN the column resists 327 kN·m against a demand of 120 kN·m, so the section is controlled by axial load rather than moment. Priya considers reducing the bars to 8-20M, but keeps 8-25M because the same column size continues to the roof, where moments from the roof frame are larger relative to the axial load, and because a uniform bar size simplifies splices.

The table of points also shows why the interaction diagram matters even for a lightly loaded column. Near pure bending the capacity is lowest; adding axial load up to the balanced point increases it. If the roof column carries only a small axial load with a large wind moment, its capacity comes from the lower part of the curve, and that is where it may govern.

Biaxial bending

Corner columns and columns in two-way frames carry moments about both axes. The exact solution needs a three-dimensional interaction surface. A simple, widely used approximation is the Bresler reciprocal load equation, which estimates the capacity under biaxial bending from the uniaxial capacities:

\[\frac{1}{P_r} = \frac{1}{P_{rx}} + \frac{1}{P_{ry}} - \frac{1}{P_{ro}}\]

where \(P_{rx}\) and \(P_{ry}\) (or \(P_{nx}\), \(P_{ny}\)) are the axial capacities at the actual eccentricity about each axis alone. The equation is accurate when axial load is significant, above about a tenth of the pure axial capacity. For lower loads, the section is checked as a beam in biaxial bending, or a computer program integrates the stresses over the section for an inclined neutral axis.

Practical column design

  • Size columns for axial load first: with 1.5 to 2 per cent steel, a tied column resists a factored axial load of roughly \(0.5f'_c A_g\) at its cap, so \(A_g \approx 2P/f'_c\) is a good first guess; then check moments with the interaction diagram.
  • Keep column sizes constant over several floors and vary the steel and concrete strength, so formwork can be reused. Higher-strength concrete in lower storeys is common.
  • Columns from 300 mm (12 in.) upward are common; very small columns are hard to cast and congested at beam-column joints.
  • Symmetrical reinforcement is used because moments usually reverse under wind and pattern loads.
  • Load combinations with minimum axial load must be checked, because they may govern in the tension-controlled region.

Common mistakes

  • MistakeChecking only the largest axial load combination.

    InsteadIn the tension-controlled region less axial load means less moment capacity; check minimum axial load combinations too.

  • MistakeOmitting the 0.80 cap on axial resistance.

    InsteadTied columns are limited to \(0.80P_{ro}\) for accidental eccentricity.

  • MistakeCounting compression bars inside the stress block at full value without deducting displaced concrete.

    InsteadSubtract the concrete stress times the bar area for bars inside the block.

  • MistakeAssuming all bars yield.

    InsteadBars near the neutral axis have small strains; compute each layer's stress from strain compatibility.

Summary and key terms

  • Tied and spiral columns differ in confinement; spiral columns have higher resistance factors and axial caps.
  • \(P_{ro} = \alpha_1\phi_c f'_c(A_g - A_{st}) + \phi_s f_y A_{st}\), capped at \(0.80P_{ro}\) for tied columns.
  • The interaction diagram is traced by sweeping the neutral axis through strain-compatibility analysis.
  • Below the balanced point, axial compression increases moment capacity, so minimum axial load cases must be checked.
  • Steel ratios are limited to 1 to 8 per cent; 4 per cent is a practical maximum with lap splices.
  • The office column, 500 × 500 mm with 8-25M, resists 327 kN·m at \(P_f = 3205\) kN.
Tied column
A column with longitudinal bars held by individual lateral ties.
Spiral column
A column with a continuous helical transverse bar confining a circular core.
Short column
A column whose strength is governed by its cross-section, with negligible second-order effects.
Interaction diagram
A curve of combinations of axial load and moment that a section can resist.
Balanced point
The point at which concrete crushing and tension steel yielding occur together.
Eccentricity
The ratio M/P, the distance from the centroid at which the axial load effectively acts.
Accidental eccentricity
Unavoidable eccentricity from construction tolerances, covered by the cap on axial load.
Biaxial bending
Bending about both principal axes of a column at once.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. Find \(P_{ro}\) and \(P_{r,max}\) for a 400 × 400 mm tied column, 8-20M, 35 MPa.

    Show answer

    \(\alpha_1 = 0.797\); \(P_{ro} = 0.797(0.65)(35)(157\,600) + 0.85(400)(2400) = 2858 + 816 = 3674\) kN; \(P_{r,max} = 2939\) kN.

  2. What is the balanced neutral axis depth for \(d = 340\) mm with grade 400 steel?

    Show answer

    \(c_b = 0.0035/0.0055 \times 340 = 216\) mm.

  3. Why can a column with a small axial load have less moment capacity than the same column with a moderate axial load?

    Show answer

    Below balanced, axial compression delays tension yielding and adds to the internal couple.

  4. What is the eccentricity of the office column load?

    Show answer

    \(e = 120/3205 = 0.037\) m = 37 mm, 7.5 per cent of the depth.

  5. When is the Bresler reciprocal equation reliable?

    Show answer

    When the axial load is more than about 10 per cent of the pure axial capacity.

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