Slender column 400x500, single curvature

Data and assumptions

• Column 400 mm (width $b$) × 500 mm (depth $D$); $P_u = 1600$ kN; $f_{ck} = 20$ MPa, $f_y = 415$ MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties.
• Moments: about major axis (x-x) $M_{ux}$ = 180 kN·m; about minor axis (y-y) $M_{uy}$ = 110 kN·m.
• The unsupported length for the minimum eccentricity is taken equal to the larger effective length, 7 m.
• Single curvature: both end moments have the same sign.
• Sizes are given as $b \times D$ with $D \ge b$; moments $M_{ux}$ act about the major axis (depth $D$) and $M_{uy}$ about the minor axis (depth $b$).

Slenderness

Minimum eccentricity (cl. 25.4)

$$e_{min} = \max\left(\frac{L}{500} + \frac{D}{30},\ 20\right)$$

• About x-x: $e_x = 7000/500 + 500/30 = 30.7$ mm (min. 20 mm); adopt 30.7 mm; $M_{ux,min} = P_u e = 49.1$ kN·m
• About y-y: $e_y = 7000/500 + 400/30 = 27.3$ mm (min. 20 mm); adopt 27.3 mm; $M_{uy,min} = 43.7$ kN·m

Design moments

• Major axis: $M_i = 0.6M_2 + 0.4M_1 = 0.6(180) + 0.4(100) = 148.0$ kN·m, not less than $0.4M_2 = 72.0$; adopt 148.0 kN·m (cl. 39.7.1).
• Minor axis: $M_i = 0.6(110) + 0.4(50) = 86.0$ kN·m, not less than $0.4M_2 = 44.0$; adopt 86.0 kN·m. Additional moment due to slenderness: $M_a = \dfrac{P_u D}{2000}\left(\dfrac{l_e}{D}\right)^2$ (78.4 kN·m about x-x, 72.0 kN·m about y-y), reduced by $k = \dfrac{P_{uz}-P_u}{P_{uz}-P_b} \le 1$.
• For the selected steel: $P_{uz} = 3580$ kN, $P_{bx} = 331$ kN, $k_x = 0.61$, $M_{ax} = 47.8$ kN·m; $P_{by} = 247$ kN, $k_y = 0.59$, $M_{ay} = 42.8$ kN·m.
• $P_b$ is much lower than the concrete force alone ($0.36 f_{ck} b x_{u,max} \approx 600$ kN) because, at the balanced strain, the bars in the tension half (the far face and the inner rows) carry a large tensile force that outweighs the compression steel.
• Design moments: $M_{ux} = 148.0 + 47.8 = 195.8$ kN·m, $M_{uy} = 86.0 + 42.8 = 128.8$ kN·m (not less than the larger end moments or the minimum-eccentricity values).

Longitudinal reinforcement

$\dfrac{P_u}{f_{ck} b D} = \dfrac{1600000}{20 \times 400 \times 500} = 0.400$ Bars are chosen so that $p$ lies between 0.8% (minimum) and 4% (practical maximum; IS 456 cl. 26.5.3.1 allows 6%). The uniaxial capacities $M_{ux1}$, $M_{uy1}$ at $P_u$ 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 $d' = 40 + 8 + \phi/2$.

Check

$P_{uz} = 0.45 f_{ck} A_c + 0.75 f_y A_{sc} = 0.45 \times 20 \times 194110 + 0.75 \times 415 \times 5890 = 3580$ kN $\dfrac{P_u}{P_{uz}} = \dfrac{1600}{3580} = 0.447$, so $\alpha_n = 1.41$ (IS 456 cl. 39.6; linear between 1.0 at 0.2 and 2.0 at 0.8).

$$\left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} = \left(\frac{195.8}{325.3}\right)^{1.41} + \left(\frac{128.8}{241.4}\right)^{1.41} = 0.900 \le 1.0$$

The interaction condition is satisfied; the section is safe.

Provide 12-25 mm bars ($A_{sc} = 5890$ mm², $p = 2.95\%$), within 0.8% to 4%.

Lateral ties

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

Detailing

CROSS-SECTION 400 x 500
    +-------------------------+
    |   o     o     o     o   |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |                         |
    |   o                 o   |
    |                         |
    |                         |
    |   o     o     o     o   |
    +-------------------------+
   o = 25 mm bars, ties 8 mm @ 300 c/c

Answer: 400 × 500 mm column with 12-25 mm longitudinal bars ($p = 2.95\%$) and 8 mm ties @ 300 mm c/c.