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Biaxial column 400x600, Fe500

Short answer

Short column with Pu=1000 kN, Mux=125, Muy=200 kNm is safe. Provide 12-20 mm bars (Asc=3770 mm2, p=1.57%) and 8 mm ties at 300 mm c/c; biaxial interaction ratio is 0.95.

Determine the reinforcement in short column having 4 m length fixed in both sides with bi-axially loaded having the following parameters: Size of column = 400 mm x 600 mm; Factored load, Pu = 1000 kN; Factored moment Mux = 125 kNm; Factored moment Muy = 200 kNm; M25 concrete and Fe500 steel

Data and assumptions

  • Column 400 mm (width bb) × 600 mm (depth DD); Pu=1000P_u = 1000 kN; fck=25f_{ck} = 25 MPa, fy=500f_y = 500 MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties.
  • Moments: about major axis (x-x) MuxM_{ux} = 125 kN·m; about minor axis (y-y) MuyM_{uy} = 200 kN·m.
  • Both ends fixed: le=0.65L=2.6l_e = 0.65L = 2.6 m. Mux=125M_{ux}=125 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)26006004.33short
y-y (about minor)26004006.50short
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=4000/500+600/30=28.0e_x = 4000/500 + 600/30 = 28.0 mm (min. 20 mm); adopt 28.0 mm; Mux,min=Pue=28.0M_{ux,min} = P_u e = 28.0 kN·m
  • About y-y: ey=4000/500+400/30=21.3e_y = 4000/500 + 400/30 = 21.3 mm (min. 20 mm); adopt 21.3 mm; Muy,min=21.3M_{uy,min} = 21.3 kN·m

Design moments

  • Design moments: Mux=125.0M_{ux} = 125.0 kN·m, Muy=200.0M_{uy} = 200.0 kN·m (not less than the minimum-eccentricity values).

Longitudinal reinforcement

PufckbD=100000025×400×600=0.167\dfrac{P_u}{f_{ck} b D} = \dfrac{1000000}{25 \times 400 \times 600} = 0.167 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}
10-20 φ1.310.097395.3267.70.2601.101.007 ✗
16-16 φ1.340.093418.8256.00.2581.101.028 ✗
12-20 φ1.570.097457.9277.60.2461.080.950 ✓

Check

Puz=0.45fckAc+0.75fyAsc=0.45×25×236230+0.75×500×3770=4071P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 236230 + 0.75 \times 500 \times 3770 = 4071 kN PuPuz=10004071=0.246\dfrac{P_u}{P_{uz}} = \dfrac{1000}{4071} = 0.246, so αn=1.08\alpha_n = 1.08 (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

(MuxMux1)αn+(MuyMuy1)αn=(125.0457.9)1.08+(200.0277.6)1.08=0.950≤1.0\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{125.0}{457.9}\right)^{1.08} + \left(\frac{200.0}{277.6}\right)^{1.08} = 0.950 \le 1.0

The interaction condition is satisfied; the section is safe.

Provide 12-20 mm bars (Asc=3770A_{sc} = 3770 mm², p=1.57%p = 1.57\%), within 0.8% to 4%.

Lateral ties

  • Diameter ≥max⁡(ϕmax/4, 6)=max⁡(20/4,6)=6\ge \max(\phi_{max}/4,\ 6) = \max(20/4, 6) = 6 mm → provide 8 mm.
  • Pitch ≤min⁡\le \min(least lateral dimension, 16ϕmin16\phi_{min}, 300 mm) =min⁡(400,320,300)=300= \min(400, 320, 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     o     o     o   |
    +-------------------------+
   o = 20 mm bars, ties 8 mm @ 300 c/c

Answer: 400 × 600 mm column with 12-20 mm longitudinal bars (p=1.57%p = 1.57\%) and 8 mm ties @ 300 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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