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Column design, 1350 kN, 3.4 m

Short answer

Short column with l_e=2.21 m; design moments 150 and 100 kNm. Use 8-25 mm bars (A_sc=3927 mm², p=2.45%) with 8 mm ties at 300 mm c/c; biaxial interaction 0.977 ≤ 1, safe.

Design a column member to transfer the factored load of PuP_u = 1350 kN, MuxM_{ux} = 150 kN m and MuyM_{uy} = 100 kNm having length 3.4 m unsupported. Use M25 concrete and Fe 500 steel. Use reinforcement distributed equally on all four sides and assume that the column is fully restrained at both the supports for displacement and rotation.

Data and assumptions

  • Column 400 mm (width bb) × 400 mm (depth DD); Pu=1350P_u = 1350 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} = 150 kN·m; about minor axis (y-y) MuyM_{uy} = 100 kN·m.
  • Both ends fully restrained for displacement and rotation: le=0.65L=0.65×3.4=2.21l_e = 0.65L = 0.65 \times 3.4 = 2.21 m (IS 456 Table 28).
  • Section size is not given and 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)22104005.53short
y-y (about minor)22104005.53short
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=3400/500+400/30=20.1e_x = 3400/500 + 400/30 = 20.1 mm (min. 20 mm); adopt 20.1 mm; Mux,min=Pue=27.2M_{ux,min} = P_u e = 27.2 kN·m
  • About y-y: ey=3400/500+400/30=20.1e_y = 3400/500 + 400/30 = 20.1 mm (min. 20 mm); adopt 20.1 mm; Muy,min=27.2M_{uy,min} = 27.2 kN·m

Design moments

  • Design moments: Mux=150.0M_{ux} = 150.0 kN·m, Muy=100.0M_{uy} = 100.0 kN·m (not less than the minimum-eccentricity values).

Longitudinal reinforcement

PufckbD=135000025×400×400=0.338\dfrac{P_u}{f_{ck} b D} = \dfrac{1350000}{25 \times 400 \times 400} = 0.338 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}
16-16 φ2.010.140184.6184.60.4551.421.162 ✗
12-20 φ2.360.145204.9204.90.4261.381.024 ✗
8-25 φ2.450.151212.9212.90.4181.360.977 ✓

Check

Puz=0.45fckAc+0.75fyAsc=0.45×25×156073+0.75×500×3927=3228P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 156073 + 0.75 \times 500 \times 3927 = 3228 kN PuPuz=13503228=0.418\dfrac{P_u}{P_{uz}} = \dfrac{1350}{3228} = 0.418, 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=(150.0212.9)1.36+(100.0212.9)1.36=0.977≤1.0\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{150.0}{212.9}\right)^{1.36} + \left(\frac{100.0}{212.9}\right)^{1.36} = 0.977 \le 1.0

The interaction condition is satisfied; the section is safe.

Provide 8-25 mm bars (Asc=3927A_{sc} = 3927 mm², p=2.45%p = 2.45\%), within 0.8% to 4%.

Lateral ties

  • Diameter ≥max⁡(ϕmax/4, 6)=max⁡(25/4,6)=6\ge \max(\phi_{max}/4,\ 6) = \max(25/4, 6) = 6 mm → provide 8 mm.
  • Pitch ≤min⁡\le \min(least lateral dimension, 16ϕmin16\phi_{min}, 300 mm) =min⁡(400,400,300)=300= \min(400, 400, 300) = 300 mm → provide 8 mm ties @ 300 mm c/c.

Detailing

CROSS-SECTION 400 x 400
    +-------------------------+
    |                         |
    |   o        o        o   |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |   o        o        o   |
    |                         |
    +-------------------------+
   o = 25 mm bars, ties 8 mm @ 300 c/c

Answer: 400 × 400 mm column with 8-25 mm longitudinal bars (p=2.45%p = 2.45\%) and 8 mm ties @ 300 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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