EGB375 · Design of Concrete Structures
Slender Columns — Slenderness & Moment Magnification
A short column fails by crushing at its section capacity, but a slender column bows sideways first, so the axial load acts at a deflection and adds a second-order (P–δ) moment. EGB375 Design of Concrete Structures at Queensland University of Technology treats this as the second half of column design: Chapter 10 gives the section strength (the interaction diagram), and this chapter asks whether the member is slim enough that buckling erodes it. The AS3600-2018 route is a fixed chain — radius of gyration r = 0.3D, effective length Le = kLu, the slenderness ratio Le/r against a limit, an Euler-type critical load Nc, and a moment magnifier that produces the design moment Mmax = δM2*.
What this chapter covers
- 01Why slender columns buckle: the P-delta second-order moment
- 02Radius of gyration r = 0.3D for a rectangular section (and why 0.289D rounds up)
- 03Effective length Le = kLu and the braced/unbraced k values (0.70, 0.85, 1.0; 1.2-2.2 sway)
- 04The slenderness ratio Le/r and evaluating it about each axis
- 05The short/slender limit: braced max{25, alpha_c (38 - f'c/15)(1 + M1*/M2*)}; unbraced < 22
- 06The single- vs double-curvature sign convention on M1*/M2* (negative vs positive)
- 07The Euler-type critical buckling load Nc = pi^2[182 do phiMub]/[Le^2(1 + betad)]
- 08The braced magnifier delta_b = km/(1 - N*/Nc) and km = 0.6 - 0.4(M1*/M2*) with its 0.4 floor
- 09The sway magnifier delta_s and the design moment Mmax = delta M2*
- 10Feeding Mmax back into the Chapter 10 interaction diagram to confirm adequacy
Worked example: classifying a braced column and finding the magnified design moment
- +1Radius of gyration. r = 0.3D = 0.3×350 = 105 mm.
- +1Effective length. Le = kLu = 0.85×4500 = 3825 mm.
- +1Slenderness ratio. Le/r = 3825/105 = 36.4.
- +2Short/slender limit. Single curvature so M1*/M2* = −48/80 = −0.6. With N* = 1485 kN (next step), alpha_c = √(2.25 − 2.5×1485/(0.6×5200)) = 1.03. Second term = alpha_c×(38 − 40/15)×(1 − 0.6) = 1.03×35.33×0.4 = 14.6. Limit = max{25, 14.6} = 25. Since Le/r = 36.4 > 25, the column is slender and the moment must be magnified.
- +1Axial load and sustained ratio. N* = 1.2×800 + 1.5×350 = 960 + 525 = 1485 kN. betad = NG/(NG+NQ) = 800/1150 = 0.696.
- +3Critical buckling load. Nc = pi^2[182·do·phiMub]/[Le^2(1+betad)] = pi^2×182×310×(160×10^6)/[3825^2×1.696] = 8.910×10^13 / (1.4631×10^7×1.696) = 8.910×10^13 / 2.481×10^7 = 3591 kN (keep phiMub in N·mm so Nc comes out in N).
- +1Factor km. km = 0.6 − 0.4(M1*/M2*) = 0.6 − 0.4(−0.6) = 0.6 + 0.24 = 0.84 (within the 0.4-1.0 range).
- +1Magnifier. delta_b = km/(1 − N*/Nc) = 0.84/(1 − 1485/3591) = 0.84/0.586 = 1.43 (≥ 1.0).
- +1Design moment. Mmax = delta_b·M2* = 1.43×80 = 114.6 kN·m. Plot (N*, Mmax) = (1485 kN, 114.6 kN·m) on the phi-reduced interaction diagram to confirm adequacy.
Key terms
- Slender column
- A column slim enough that lateral buckling reduces its strength: for a braced member, one whose slenderness ratio Le/r exceeds the limiting slenderness. It must be designed for a moment magnified by the P-delta effect, unlike a short column which is designed straight from its section (interaction) capacity.
- Radius of gyration r
- A measure of section stiffness, r = sqrt(I/A) in mm. For a rectangular column bending about the axis of dimension D, I = bD^3/12 and A = bD give r = D/sqrt(12) = 0.289D, which AS3600 rounds to r = 0.3D. D is the dimension in the plane of bending.
- Effective length Le
- The buckling length Le = kLu, where Lu is the clear (unsupported) height and k the effective-length factor. Braced frames: k = 0.70 (fixed-fixed), 0.85 (one end pinned), 1.0 (pin-pin). Unbraced (sway) frames: k = 1.2-2.2.
- Slenderness ratio Le/r
- The dimensionless ratio Le/r = kLu/(0.3D) that decides whether a column is short or slender. It is compared with a limiting slenderness and should be evaluated separately about each axis, since both Le and r depend on the axis.
- Limiting slenderness (braced)
- The threshold below which a braced column is short: Le/r <= max{25, alpha_c (38 - f'c/15)(1 + M1*/M2*)}. M1*/M2* is negative for single curvature and positive for double; for an unbraced column the limit is Le/r < 22. alpha_c multiplies the (38 - f'c/15)(1 + M1*/M2*) term, with alpha_c = sqrt(2.25 - 2.5 N*/(0.6 Nuo)) at normal load levels.
- M1*/M2* (curvature sign)
- The ratio of the smaller to the larger factored end moment (M2* is the larger). It is negative when the column is in single curvature (bowed one way) and positive in double curvature (an S-shape). The sign changes both the slenderness limit and the factor km, so getting it wrong is a high-value error.
- Critical buckling load Nc
- An Euler-type buckling load, Nc = pi^2[182 do phiMub]/[Le^2(1 + betad)] in N, where do is the depth to the outer tension bars (mm), phiMub the balanced-point moment capacity from the interaction diagram (N.mm), and betad = G/(G+Q) the sustained-axial-load ratio. It replaces the elastic pi^2 EI/Le^2 with an effective cracked-section rigidity.
- Moment magnifier delta_b
- The braced amplification factor delta_b = km/(1 - N*/Nc) >= 1.0, with km = 0.6 - 0.4(M1*/M2*) bounded 0.4 <= km <= 1.0. As N* approaches Nc the denominator shrinks and delta_b grows, signalling incipient buckling. The design moment is Mmax = delta M2*.
Slender Columns — Slenderness & Moment Magnification FAQ
When is a column short enough to ignore buckling?
When its slenderness ratio Le/r falls below the limiting slenderness. For a braced column that limit is max{25, alpha_c (38 - f'c/15)(1 + M1*/M2*)}, where alpha_c = sqrt(2.25 - 2.5 N*/(0.6 Nuo)) at normal load levels, and for an unbraced (sway) column it is 22. Below the limit the column is short and you design it straight from the Chapter 10 section (interaction) capacity for the actual moment; above it the column is slender and the end moment must be magnified for the P-delta second-order effect.
Why does single curvature make a column more likely to be slender than double curvature?
The sign of M1*/M2* changes the limit. In single curvature the ratio is negative, which lowers the moment-ratio term (it can vanish when M1* = -M2*), so the limit drops toward 25 and the column is more easily slender. In double curvature the ratio is positive, raising the limit, so the same geometry can still be short. Physically, a single-curvature column has its maximum moment and maximum deflection at the same place, which is the more critical arrangement.
Can AI help me with slender columns and moment magnification in EGB375?
Yes, for learning and rehearsal. Sia is an AI tutor that walks the r -> Le -> Le/r -> limit -> Nc -> delta_b workflow with you step by step, checks the M1*/M2* sign convention and the km and delta_b bounds on practice columns, and helps you interrogate why each AS3600 step is there. It supports your study rather than replacing it: it will not sit the closed-book exam for you, and no tool can promise a particular mark or grade. Always confirm code clauses against AS3600 and your unit materials.
Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.
Exam move
Treat the slender-column route as a fixed six-step chain rather than a set of formulas to recognise: radius of gyration r = 0.3D, effective length Le = kLu, slenderness ratio Le/r, the short/slender limit, and only if slender the critical load Nc and the magnifier delta_b that gives Mmax = delta M2*. The single most common way to lose marks is the M1*/M2* sign, so fix it firmly: single curvature is negative, double curvature is positive, and M2* is the larger end moment. Keep phiMub in N.mm when you form Nc so it lands in newtons, clamp km to the 0.4-1.0 range and delta_b to a minimum of 1.0, and always carry the magnified Mmax (not M2*) back into the interaction diagram. Because the final exam is closed-book and paper-based, condense the chain, the k values, the Nc constant of 182 and one fully worked magnification onto your permitted note sheets. The exam is marked out of 100 and worth 50% of the unit; once you confirm its duration on the QUT exam timetable, divide those minutes by 100 to set a consistent minutes-per-mark rate and spend on a slender-column question in proportion to its marks.
