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Short column 35x45 cm, 1800 kN

Short answer

Column is short. Provide 6-32 mm bars on two faces (Asc 4825 mm2, 3.06%) plus one 12 mm nominal bar at mid-depth of each 450 mm face, with 8 mm ties at 300 mm c/c; biaxial check is safe.

Design the longitudinal and transverse reinforcements to be provided for a short column of size 35 cm x 45 cm subjected to the following forces: Factored axial load PuP_u = 1800 kN; Factored moment MuxM_{ux} = 175 kN-m; Factored moment MuyM_{uy} = 105 kN-m. Reinforcements are distributed equally on two sides. Use M25 concrete and Fe500 steel. Unsupported length = 3.1 m

Data and assumptions

  • Column 350 mm (width bb) × 450 mm (depth DD); Pu=1800P_u = 1800 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} = 175 kN·m; about minor axis (y-y) MuyM_{uy} = 105 kN·m.
  • As the question states, the bars are placed equally on two opposite faces: the two 350 mm faces, which are parallel to the major axis x-x and so resist the larger moment MuxM_{ux} most effectively.
  • 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)31004506.89short
y-y (about minor)31003508.86short

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=3100/500+450/30=21.2e_x = 3100/500 + 450/30 = 21.2 mm (min. 20 mm); adopt 21.2 mm; Mux,min=Pue=38.2M_{ux,min} = P_u e = 38.2 kN·m
  • About y-y: ey=3100/500+350/30=17.9e_y = 3100/500 + 350/30 = 17.9 mm (min. 20 mm); adopt 20.0 mm; Muy,min=36.0M_{uy,min} = 36.0 kN·m

Design moments

  • Design moments: Mux=175.0M_{ux} = 175.0 kN·m, Muy=105.0M_{uy} = 105.0 kN·m (not less than the minimum-eccentricity values).

Longitudinal reinforcement

PufckbD=180000025×350×450=0.457\dfrac{P_u}{f_{ck} b D} = \dfrac{1800000}{25 \times 350 \times 450} = 0.457 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}
12-20 φ2.390.129231.9122.00.5731.621.418 ✗
8-25 φ2.490.134238.2131.90.5621.601.303 ✗
6-32 φ3.060.142286.1161.30.5101.520.996 ✓

Check

Puz=0.45fckAc+0.75fyAsc=0.45×25×152675+0.75×500×4825=3527P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 25 \times 152675 + 0.75 \times 500 \times 4825 = 3527 kN PuPuz=18003527=0.510\dfrac{P_u}{P_{uz}} = \dfrac{1800}{3527} = 0.510, so αn=1.52\alpha_n = 1.52 (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

(MuxMux1)αn+(MuyMuy1)αn=(175.0286.1)1.52+(105.0161.3)1.52=0.996≤1.0\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{175.0}{286.1}\right)^{1.52} + \left(\frac{105.0}{161.3}\right)^{1.52} = 0.996 \le 1.0

The interaction condition is satisfied; the section is safe.

Provide 6-32 mm bars (Asc=4825A_{sc} = 4825 mm², p=3.06%p = 3.06\%), within 0.8% to 4%.

The two layers are 450−2×64=322450 - 2 \times 64 = 322 mm apart, which is more than the 300 mm maximum spacing along the periphery (IS 456 cl. 26.5.3.1), so also provide one 12 mm nominal bar at mid-depth of each 450 mm face (not counted in the design).

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⁡(350,512,300)=300= \min(350, 512, 300) = 300 mm → provide 8 mm ties @ 300 mm c/c.

Detailing

CROSS-SECTION 350 x 450
    +-------------------------+
    |                         |
    |    o       o       o    |
    |                         |
    |                         |
    |                         |
    |    x               x    |
    |                         |
    |                         |
    |                         |
    |    o       o       o    |
    |                         |
    +-------------------------+
   o = 32 mm bars, x = 12 mm nominal side bars, ties 8 mm @ 300 c/c

Answer: 350 × 450 mm column with 6-32 mm longitudinal bars (p=3.06%p = 3.06\%), 3 on each 350 mm face, plus one 12 mm nominal bar on each 450 mm face, and 8 mm ties @ 300 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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