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Biaxial column 400x600, 1500 kN

Short answer

For a 400×600 mm biaxially loaded column with 1500 kN axial load, 300 kNm major-axis and 200 kNm minor-axis moments, use 8-32 mm bars (2.68%) and 8 mm ties at 300 mm c/c.

Determine the reinforcement in a biaxially loaded column with the following parameters: Size of column = 400 mm x 600 mm; Factored load, PuP_u = 1500 kN; Factored moment, MuxM_{ux} = 300 kNm; Factored moment, MuyM_{uy} = 200 kNm. Assume M25 concrete and Fe 415 steel.

Data and assumptions

  • Column 400 mm (width bb) × 600 mm (depth DD); Pu=1500P_u = 1500 kN; fck=25f_{ck} = 25 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} = 300 kN·m; about minor axis (y-y) MuyM_{uy} = 200 kN·m.
  • Unsupported length not given; a short column with L=3L = 3 m is assumed. Mux=300M_{ux}=300 kN·m acts about the major axis (depth 600 mm), Muy=200M_{uy}=200 kN·m about the minor axis (depth 400 mm).
  • 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)30006005.00short
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+600/30=26.0e_x = 3000/500 + 600/30 = 26.0 mm (min. 20 mm); adopt 26.0 mm; Mux,min=Pue=39.0M_{ux,min} = P_u e = 39.0 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=30.0M_{uy,min} = 30.0 kN·m

Design moments

  • Design moments: Mux=300.0M_{ux} = 300.0 kN·m, Muy=200.0M_{uy} = 200.0 kN·m (not less than the minimum-eccentricity values).

Longitudinal reinforcement

PufckbD=150000025×400×600=0.250\dfrac{P_u}{f_{ck} b D} = \dfrac{1500000}{25 \times 400 \times 600} = 0.250 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), with effective cover d′=40+8+ϕ/2d' = 40 + 8 + \phi/2.

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}
12-25 φ2.450.101531.6318.30.3361.231.061 ✗
20-20 φ2.620.097550.9331.40.3271.211.021 ✗
8-32 φ2.680.107570.6338.20.3241.210.991 ✓

Check

Puz=0.45fckAc+0.75fyAsc=0.45×25×233566+0.75×415×6434=4630P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 233566 + 0.75 \times 415 \times 6434 = 4630 kN PuPuz=15004630=0.324\dfrac{P_u}{P_{uz}} = \dfrac{1500}{4630} = 0.324, so αn=1.21\alpha_n = 1.21 (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

(MuxMux1)αn+(MuyMuy1)αn=(300.0570.6)1.21+(200.0338.2)1.21=0.991≤1.0\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{300.0}{570.6}\right)^{1.21} + \left(\frac{200.0}{338.2}\right)^{1.21} = 0.991 \le 1.0

The interaction condition is satisfied; the section is safe.

Provide 8-32 mm bars (Asc=6434A_{sc} = 6434 mm², p=2.68%p = 2.68\%), within 0.8% to 4%.

Lateral ties

  • Diameter ≥max⁡(ϕmax/4, 6)=max⁡(32/4,6)=8\ge \max(\phi_{max}/4,\ 6) = \max(32/4, 6) = 8 mm → provide 8 mm.
  • Pitch ≤min⁡\le \min(least lateral dimension, 16ϕmin16\phi_{min}, 300 mm) =min⁡(400,512,300)=300= \min(400, 512, 300) = 300 mm → provide 8 mm ties @ 300 mm c/c.

Detailing

CROSS-SECTION 400 x 600
    +-------------------------+
    |   o        o        o   |
    |                         |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |                         |
    |   o        o        o   |
    +-------------------------+
   o = 32 mm bars, ties 8 mm @ 300 c/c

Answer: 400 × 600 mm column with 8-32 mm longitudinal bars (p=2.68%p = 2.68\%) and 8 mm ties @ 300 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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