Shear, Bond, Slabs, Columns and Footings

After flexure, an RCC member must be checked for shear and bond, and the other members of a building (slabs, columns and footings) have their own rules in IS 456.

Shear

• Nominal shear stress: τ_v = V_u/(b d).
• Design shear strength of concrete τ_c depends on the percentage of tension steel and the grade (IS 456 Table 19). τ_v must not exceed τ_c,max (Table 20), otherwise the section must be enlarged.
• If τ_v > τ_c, provide stirrups for the extra shear V_us = V_u − τ_c b d: Vertical stirrups: V_us = 0.87 f_y A_sv d/s_v.
• Minimum shear reinforcement: A_sv/(b s_v) ≥ 0.4/(0.87 f_y). Maximum spacing: 0.75d or 300 mm, whichever is less.

Bond and development length

Bars must extend far enough to develop their stress:

L_d = φ σ_s/(4 τ_bd), with σ_s = 0.87 f_y.

τ_bd for plain bars in tension: 1.2 MPa (M20), 1.4 (M25), 1.5 (M30); increased by 60% for deformed bars and 25% for bars in compression. For Fe 415 deformed bars in M20, L_d ≈ 47φ.

Slabs

• One-way slab: L_y/L_x > 2; bends mainly along the short span. Design as a 1 m wide beam.
• Two-way slab: L_y/L_x ≤ 2; moments from IS 456 coefficient tables (Annex D).
• Minimum steel: 0.12% (HYSD bars) or 0.15% (mild steel) of the gross area. Maximum bar diameter: D/8.

Columns

• Short column if l_eff/D < 12; otherwise slender (additional moments).
• Minimum eccentricity: e_min = l/500 + D/30, at least 20 mm.
• Axially loaded short column (with e_min ≤ 0.05D): P_u = 0.4 f_ck A_c + 0.67 f_y A_sc.
• Longitudinal steel 0.8% to 6% (practically up to 4%); at least 4 bars in a rectangular column, 6 in a circular one. Ties: diameter ≥ φ/4 and 6 mm; pitch ≤ least lateral dimension, 16φ and 300 mm.
• Helical ties increase the capacity by 5%.

Footings

Check an isolated footing for: bearing pressure (area = load/SBC), one-way shear at d from the column face, two-way (punching) shear at d/2 from the face, and bending at the column face.

Worked example

A short column 400 × 400 mm, M25, Fe 415, with 8 bars of 20 mm (A_sc = 2513 mm²).

• A_c = 160,000 − 2513 = 157,487 mm².
• P_u = 0.4 × 25 × 157,487 + 0.67 × 415 × 2513 = 1574.9 + 698.7 = 2273.6 kN.

Practice