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Shear design of a beam with vertical stirrups
Nominal shear stress, design shear strength of concrete, maximum shear stress and stirrup spacing to IS 456. Reference: IS 456:2000 Cl. 40.1-40.4, Tables 19 and 20.
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# Data
V_u := 220 kN
b := 300 mm
d := 500 mm
f_ck := 25 MPa
f_y := 415 MPa
A_st := 1608 mm^2
# Shear stresses
tau_v := V_u/(b*d) = MPa
p_t := 100*A_st/(b*d) =
beta := max(0.8*(f_ck/(1 MPa))/(6.89*p_t), 1) =
tau_c := 0.85*sqrt(0.8*f_ck/(1 MPa))*(sqrt(1 + 5*beta) - 1)/(6*beta)*1 MPa =
# tau_c above is the formula behind Table 19.
tau_c.max := interp(f_ck/(1 MPa), [15, 20, 25, 30, 35, 40], [2.5, 2.8, 3.1, 3.5, 3.7, 4.0])*1 MPa =
tau_v <= tau_c.max =
# Stirrups
V_us := V_u - tau_c*b*d = kN
A_sv := 2*pi/4*(8 mm)^2 = mm^2
s_v := 0.87*f_y*A_sv*d/V_us = mm
s_v.max := min(0.75*d, 300 mm, 0.87*f_y*A_sv/(0.4 MPa*b)) = mm
s := min(s_v, s_v.max) = mmHow to use this example
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More IS 456 sheets on shear and torsion
- Minimum shear reinforcement in beams
- Beam under torsion, shear and bending
- Torsion: equivalent moments for longitudinal steel