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Biaxial column 400x500, equal steel all sides

Short answer

Provide 14-16 mm bars (1.41%) equally on all sides with 8 mm ties at 250 mm c/c for a 400×500 mm biaxially loaded column under 1200 kN and moments 120 and 100 kNm, with d'/D=0.15.

Determine the reinforcement equal in all sides of a biaxially loaded column with the following parameters. Size of column = 400 mm x 500 mm; Factored load, PuP_u = 1200 kN; Factored moment MuxM_{ux} = 120 kNm; Factored moment MuyM_{uy} = 100 kNm. M20 concrete and Fe 415 steel. d'/D = 0.15 for both axes.

Data and assumptions

  • Column 400 mm (width bb) × 500 mm (depth DD); Pu=1200P_u = 1200 kN; fck=20f_{ck} = 20 MPa, fy=415f_y = 415 MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties.
  • Moments: about major axis (x-x) MuxM_{ux} = 120 kN·m; about minor axis (y-y) MuyM_{uy} = 100 kN·m.
  • Unsupported length not given; a short column with L=3L = 3 m is assumed.
  • d′/D=0.15d'/D = 0.15 in both directions: d′=0.15×500=75d' = 0.15 \times 500 = 75 mm about the major axis and 0.15×400=600.15 \times 400 = 60 mm about the minor axis.
  • Sizes are given as b×Db \times D with D≥bD \ge b; moments MuxM_{ux} act about the major axis (depth DD) and MuyM_{uy} about the minor axis (depth bb).

Slenderness

Axislel_e (mm)Dimension (mm)le/l_e/dimType
x-x (about major)30005006.00short
y-y (about minor)30004007.50short

Both ratios are below 12, so the column is short; no additional moment is required.

Minimum eccentricity (cl. 25.4)

emin=max⁡(L500+D30, 20)e_{min} = \max\left(\frac{L}{500} + \frac{D}{30},\ 20\right)
  • About x-x: ex=3000/500+500/30=22.7e_x = 3000/500 + 500/30 = 22.7 mm (min. 20 mm); adopt 22.7 mm; Mux,min=Pue=27.2M_{ux,min} = P_u e = 27.2 kN·m
  • About y-y: ey=3000/500+400/30=19.3e_y = 3000/500 + 400/30 = 19.3 mm (min. 20 mm); adopt 20.0 mm; Muy,min=24.0M_{uy,min} = 24.0 kN·m

Design moments

  • Design moments: Mux=120.0M_{ux} = 120.0 kN·m, Muy=100.0M_{uy} = 100.0 kN·m (not less than the minimum-eccentricity values).

Longitudinal reinforcement

PufckbD=120000020×400×500=0.300\dfrac{P_u}{f_{ck} b D} = \dfrac{1200000}{20 \times 400 \times 500} = 0.300 Bars are chosen so that pp lies between 0.8% (minimum) and 4% (practical maximum; IS 456 cl. 26.5.3.1 allows 6%). The uniaxial capacities Mux1M_{ux1}, Muy1M_{uy1} at PuP_u are read from the SP-16 interaction charts for rectangular sections; here they are computed by strain compatibility with the IS 456 stress block and the design stress-strain curve of steel (same basis as the charts), using the given d′/D=0.15d'/D = 0.15 about both axes.

Trialpp (%)d′/Dd'/DMux1M_{ux1}Muy1M_{uy1}Pu/PuzP_u/P_{uz}αn\alpha_n(Mux/Mux1)αn+(Muy/Muy1)αn(M_{ux}/M_{ux1})^{\alpha_n}+(M_{uy}/M_{uy1})^{\alpha_n}
12-16 φ1.210.150189.3151.40.4741.461.061 ✗
8-20 φ1.260.150196.1156.90.4691.451.012 ✗
14-16 φ1.410.150198.9168.60.4531.420.963 ✓

Check

Puz=0.45fckAc+0.75fyAsc=0.45×20×197185+0.75×415×2815=2651P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 197185 + 0.75 \times 415 \times 2815 = 2651 kN PuPuz=12002651=0.453\dfrac{P_u}{P_{uz}} = \dfrac{1200}{2651} = 0.453, so αn=1.42\alpha_n = 1.42 (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

(MuxMux1)αn+(MuyMuy1)αn=(120.0198.9)1.42+(100.0168.6)1.42=0.963≤1.0\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{120.0}{198.9}\right)^{1.42} + \left(\frac{100.0}{168.6}\right)^{1.42} = 0.963 \le 1.0

The interaction condition is satisfied; the section is safe.

Provide 14-16 mm bars (Asc=2815A_{sc} = 2815 mm², p=1.41%p = 1.41\%), within 0.8% to 4%.

Lateral ties

  • Diameter ≥max⁡(ϕmax/4, 6)=max⁡(16/4,6)=6\ge \max(\phi_{max}/4,\ 6) = \max(16/4, 6) = 6 mm → provide 8 mm.
  • Pitch ≤min⁡\le \min(least lateral dimension, 16ϕmin16\phi_{min}, 300 mm) =min⁡(400,256,300)=256= \min(400, 256, 300) = 256 mm → provide 8 mm ties @ 250 mm c/c.

Detailing

CROSS-SECTION 400 x 500
    +-------------------------+
    |                         |
    |   o     o     o     o   |
    |                         |
    |   o                 o   |
    |                         |
    |   o                 o   |
    |                         |
    |   o                 o   |
    |                         |
    |   o     o     o     o   |
    |                         |
    +-------------------------+
   o = 16 mm bars, ties 8 mm @ 250 c/c

Answer: 400 × 500 mm column with 14-16 mm longitudinal bars (p=1.41%p = 1.41\%) and 8 mm ties @ 250 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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