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Biaxial short square column, 800 kN

Short answer

For 400x400 mm column, Pu=800 kN with 80 mm eccentricity in both directions, provide 8-16 mm bars (Asc=1608 mm2, p=1.01%) and 8 mm ties at 250 mm c/c; interaction ratio is 0.821.

Determine the longitudinal and transverse reinforcements to be provided in a biaxially loaded short square shaped RCC column with following data: Size of column = 400 x 400 mm. Ultimate factored axial load = 800 kN. Inclusive of live load at an eccentricity of 80 mm in both X and Y direction. Use concrete grade of M20 and steel grade of Fe415.

Data and assumptions

  • Column 400 mm (width bb) × 400 mm (depth DD); Pu=800P_u = 800 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} = 64 kN·m; about minor axis (y-y) MuyM_{uy} = 64 kN·m.
  • Moments from eccentricity: Mux=Muy=Pue=800×0.08=64M_{ux} = M_{uy} = P_u e = 800 \times 0.08 = 64 kN·m.
  • Column is short; unsupported length 3 m assumed for the minimum-eccentricity check.
  • 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)30004007.50short
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+400/30=19.3e_x = 3000/500 + 400/30 = 19.3 mm (min. 20 mm); adopt 20.0 mm; Mux,min=Pue=16.0M_{ux,min} = P_u e = 16.0 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=16.0M_{uy,min} = 16.0 kN·m

Design moments

  • Design moments: Mux=64.0M_{ux} = 64.0 kN·m, Muy=64.0M_{uy} = 64.0 kN·m (not less than the minimum-eccentricity values).

Longitudinal reinforcement

PufckbD=80000020×400×400=0.250\dfrac{P_u}{f_{ck} b D} = \dfrac{800000}{20 \times 400 \times 400} = 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}
8-16 φ1.010.140123.2123.20.4151.360.821 ✓

Check

Puz=0.45fckAc+0.75fyAsc=0.45×20×158392+0.75×415×1608=1926P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 158392 + 0.75 \times 415 \times 1608 = 1926 kN PuPuz=8001926=0.415\dfrac{P_u}{P_{uz}} = \dfrac{800}{1926} = 0.415, so αn=1.36\alpha_n = 1.36 (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

(MuxMux1)αn+(MuyMuy1)αn=(64.0123.2)1.36+(64.0123.2)1.36=0.821≤1.0\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{64.0}{123.2}\right)^{1.36} + \left(\frac{64.0}{123.2}\right)^{1.36} = 0.821 \le 1.0

The interaction condition is satisfied; the section is safe.

Provide 8-16 mm bars (Asc=1608A_{sc} = 1608 mm², p=1.01%p = 1.01\%), 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 400
    +-------------------------+
    |                         |
    |   o        o        o   |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |   o        o        o   |
    |                         |
    +-------------------------+
   o = 16 mm bars, ties 8 mm @ 250 c/c

Answer: 400 × 400 mm column with 8-16 mm longitudinal bars (p=1.01%p = 1.01\%) and 8 mm ties @ 250 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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