GATE Civil Engineering (CE) 2027 Test: Shear Free Online Test 2026

Permissible shear stress.
In limit state method of design,
τ_c = 0.25√f_ck
in working state method of design,
τ_c = 0.16√f_ck
∴ Ratio = 25/16

Total shear strength = shear resistance of effective concrete area as a function of longitudinal bars + shear resistance of vertical shear stirrups + shear resistance of-inclined shear stirrups.
Tests have shown that inclined bars alone do not provide a satisfactory solution and their contribution is limited to 50% of net shear strength after deducting the contribution of concrete. The remaining shear resistance is provided by vertical stirrups.

If the calculated shear stress ( ζ​_v ) is
(i) less than allowable shear stress (ζ​_c) but more than 0.5ζ​_c, the minimum shear reinforcement in the form of stirrups shall be provided such that,

(ii) more than ζ​_c, shear reinforcement shall be provided in the form of vertical stirrups or bent up bars with stirrups or inclined stirrups to resist
V_us = Vu - τ_cbd. Here  τ_v < τ_c,max
(iii) If  τ_v >  τ_c,max redesign the section.

Torsion constant of a rectangular section of width b and depth d (b < d) may be expressed as,
J = b³d

For T, L and / sections torsion constant,

where b_i and d_i are the dimensions of each of the component rectangles into which the section may be divided.
Torsional shear stress for rectangular section

τ = T/αb²d

For T, L and / sections torsional shear stress may be calculated for each component rectangle by considering them subjected to torsional moment,

Shear stress in slabs is controlled by increasing the depth of slab and not shear reinforcement while in beams shear stirrups are provided to control shear.

As per IS 456:2000 clause 34.4, the permissible shear stress τ_c = 0.62 × √f_ck. Substituting f_ck = 20 MPa gives τ_c = 0.62 × √20 ≈ 0.62 × 4.47 ≈ 2.77 N/mm², which rounds to 2.8 N/mm².