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Fixed-end column 4 m, 1000 kN

Short answer

Design a 300×350 mm short column with 14-16 mm bars (2.68%) and 8 mm ties at 250 mm c/c for a 4 m fixed–fixed column under 1000 kN axial load and 100 kNm uniaxial moment.

A column of 4 m length with both ends fixed and effectively held in position is subjected to a design axial load of 1000 kN and factored bending moment of 100 kN-m. Design the rectangular column with its longitudinal and transverse reinforcements.

Data and assumptions

  • Column 300 mm (width bb) × 350 mm (depth DD); Pu=1000P_u = 1000 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.
  • Moment about major axis: MuxM_{ux} = 100 kN·m.
  • Both ends fixed and held in position: le=0.65L=2.6l_e = 0.65L = 2.6 m. M25 concrete and Fe415 steel are assumed (not stated). The 100 kN·m moment acts about the major axis. Section size is selected by trial.
  • 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)26003507.43short
y-y (about minor)26003008.67short

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+350/30=19.7e_x = 4000/500 + 350/30 = 19.7 mm (min. 20 mm); adopt 20.0 mm; Mux,min=Pue=20.0M_{ux,min} = P_u e = 20.0 kN·m
  • About y-y: ey=4000/500+300/30=18.0e_y = 4000/500 + 300/30 = 18.0 mm (min. 20 mm); adopt 20.0 mm; Muy,min=20.0M_{uy,min} = 20.0 kN·m

Design moments

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

Longitudinal reinforcement

PufckbD=100000025×300×350=0.381\dfrac{P_u}{f_{ck} b D} = \dfrac{1000000}{25 \times 300 \times 350} = 0.381 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} (kN·m)MuxM_{ux} requiredRatio
12-16 φ2.300.16096.0100.01.042 not ok
8-20 φ2.390.16699.2100.01.008 not ok
14-16 φ2.680.160103.3100.00.968 ok

Check

Mux=100.0M_{ux} = 100.0 kN·m ≤Mux1=103.3\le M_{ux1} = 103.3 kN·m. Minimum eccentricity about the minor axis: Puey=20.0P_u e_y = 20.0 kN·m ≤Muy1=88.5\le M_{uy1} = 88.5 kN·m. Both are satisfied; the section is safe.

Provide 14-16 mm bars (Asc=2815A_{sc} = 2815 mm², p=2.68%p = 2.68\%), 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⁡(300,256,300)=256= \min(300, 256, 300) = 256 mm → provide 8 mm ties @ 250 mm c/c.

Detailing

CROSS-SECTION 300 x 350
    +-------------------------+
    |                         |
    |    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: 300 × 350 mm column with 14-16 mm longitudinal bars (p=2.68%p = 2.68\%) and 8 mm ties @ 250 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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