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Design steps for long column, unbraced frame

Short answer

For an unbraced slender column, compute effective lengths from sway-frame charts, check l_e/D > 12, then add additional moment M_a = P_u D/2000 (l_e/D)^2 reduced by k to the larger end moment. Check biaxial interaction using SP-16 uniaxial capacities and provide ties; revise steel if needed.

Write down design steps for the design of RC long column of unbraced frame.

A column is long (slender) when lex/Dl_{ex}/D or ley/bl_{ey}/b exceeds 12. In an unbraced frame (sway frame) the lateral stability depends on the columns themselves, so the effective length is larger than the unsupported length and the additional moments due to slenderness must be added (IS 456 cl. 25.1.2, 39.7).

Design steps

  1. Data: size b×Db \times D, factored loads PuP_u, end moments Mux1,Mux2,Muy1,Muy2M_{ux1}, M_{ux2}, M_{uy1}, M_{uy2}, unsupported length LL, grades of concrete and steel.
  2. Effective length: for a sway frame, find lexl_{ex} and leyl_{ey} from the charts of IS 456 Annex E (Fig. 27 / Fig. 28) using β1,β2\beta_1, \beta_2 (ratio of column stiffness to beam stiffness at both ends), or from Table 28 when end conditions are known.
  3. Slenderness ratio: lex/Dl_{ex}/D and ley/bl_{ey}/b. If either is above 12, the column is slender about that axis; if above 60, redesign (cl. 25.3.1: l/b≤60l/b \le 60).
  4. Minimum eccentricity: emin=L500+D30≥20e_{min} = \dfrac{L}{500} + \dfrac{D}{30} \ge 20 mm for both axes.
  5. Assume a trial steel percentage (say 1% to 3%) and bar arrangement; d′/Dd'/D for cover.
  6. Additional moments due to slenderness for the slender axis: Max=PuD2000(lexD)2,May=Pub2000(leyb)2M_{ax} = \frac{P_u D}{2000}\left(\frac{l_{ex}}{D}\right)^2, \qquad M_{ay} = \frac{P_u b}{2000}\left(\frac{l_{ey}}{b}\right)^2 and reduce by k=Puz−PuPuz−Pb≤1k = \dfrac{P_{uz} - P_u}{P_{uz} - P_b} \le 1, with Puz=0.45fckAc+0.75fyAscP_{uz} = 0.45f_{ck}A_c + 0.75f_yA_{sc}.
  7. Total design moments: in an unbraced column the additional moment is added to the larger end moment, and the end moment is not reduced (for a sway frame there is no reduction for single/double curvature): Mux=Mux,2+MaxM_{ux} = M_{ux,2} + M_{ax}, Muy=Muy,2+MayM_{uy} = M_{uy,2} + M_{ay}, not less than PueminP_u e_{min}.
  8. Check the section: find Mux1,Muy1M_{ux1}, M_{uy1} (uniaxial capacities at PuP_u) from SP-16 charts, and check the biaxial interaction (MuxMux1)αn+(MuyMuy1)αn≤1\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} \le 1 with αn\alpha_n from 1.0 (Pu/Puz≤0.2P_u/P_{uz} \le 0.2) to 2.0 (≥0.8\ge 0.8) (cl. 39.6).
  9. Revise: if not satisfied, increase steel (limit 0.8% to 4%, maximum 6%) or the size, and repeat from step 6.
  10. Lateral ties: diameter ≥max⁡(ϕ/4,6)\ge \max(\phi/4, 6) mm, pitch ≤min⁡(b,16ϕ,300)\le \min(b, 16\phi, 300) mm; in a seismic zone use ductile detailing (IS 13920).
  11. Draw the section and elevation, showing bars, ties and laps.

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