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Square column with 80 mm eccentricity

Short answer

For 350x350 mm column, Pu=800 kN with 80 mm eccentricity in both directions, provide 8-20 mm bars (Asc=2513 mm2, p=2.05%) and 8 mm ties at 300 mm c/c; interaction ratio is 0.998.

Design a square shaped reinforced concrete column that has to carry ultimate factored load of 800 kN inclusive of live load, at an eccentricity of 80 mm in both X and Y directions. Use concrete grade M20 and steel grade Fe415.

Data and assumptions

  • Column 350 mm (width bb) × 350 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.
  • Short column assumed; 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)30003508.57short
y-y (about minor)30003508.57short
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+350/30=17.7e_x = 3000/500 + 350/30 = 17.7 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+350/30=17.7e_y = 3000/500 + 350/30 = 17.7 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=80000020×350×350=0.327\dfrac{P_u}{f_{ck} b D} = \dfrac{800000}{20 \times 350 \times 350} = 0.327 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}
10-16 φ1.640.16088.395.10.4681.451.192 ✗
12-16 φ1.970.160102.7102.70.4371.391.035 ✗
8-20 φ2.050.166105.8105.80.4301.380.998 ✓

Check

Puz=0.45fckAc+0.75fyAsc=0.45×20×119987+0.75×415×2513=1862P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 119987 + 0.75 \times 415 \times 2513 = 1862 kN PuPuz=8001862=0.430\dfrac{P_u}{P_{uz}} = \dfrac{800}{1862} = 0.430, so αn=1.38\alpha_n = 1.38 (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

(MuxMux1)αn+(MuyMuy1)αn=(64.0105.8)1.38+(64.0105.8)1.38=0.998≤1.0\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{64.0}{105.8}\right)^{1.38} + \left(\frac{64.0}{105.8}\right)^{1.38} = 0.998 \le 1.0

The interaction condition is satisfied; the section is safe.

Provide 8-20 mm bars (Asc=2513A_{sc} = 2513 mm², p=2.05%p = 2.05\%), 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⁡(350,320,300)=300= \min(350, 320, 300) = 300 mm → provide 8 mm ties @ 300 mm c/c.

Detailing

CROSS-SECTION 350 x 350
    +-------------------------+
    |                         |
    |   o        o        o   |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |   o        o        o   |
    |                         |
    +-------------------------+
   o = 20 mm bars, ties 8 mm @ 300 c/c

Answer: 350 × 350 mm column with 8-20 mm longitudinal bars (p=2.05%p = 2.05\%) and 8 mm ties @ 300 mm c/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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