Concrete Beam Design - Shear in a Concrete Beam Design

This is a column failing in shear
A concrete column failing in shear

The following Concrete Shear Examples are shown here:

Shear in Concrete (complex method)[1]

V_c = {1.9\sqrt {f'_c} + 2500ρ_w{\left(V_ud\over M_u\right)b_wd } \le { 3.5\sqrt {f'_c} b_wd} }

where:

Vc = The shear in the concrete
f'c = The compression strength of the concrete (e.g. 4 ksi, 6 ksi)
ρw = The steel reinforcement ratio for the web \left(ρ_w = {A_s\over b_wd}\right)
Vu = The ultimate shear acting in the member
Mu = The ultimate moment acting in the member
bw = the width of the concrete beams web
d = the depth of the concrete beam (note: that this is not the same as the height of the beam)

Shear in Concrete (simple method)[1]

V_c = { 2\sqrt {f'_c} b_wd}

ACI 318.11.8 allows a 10% increase in the shear for ribs of floor joist construction (As long as the ribs are at least 4" wide, have a depth no more than 3.5 times the width of the rib, and have a clear spacing less than 30")

Shear Capacity in Steel Stirrups (Vs) :

V_s = {A_vf_{yt}d\over s } \le {8\sqrt {f'_c} b_wd}

where:

Vs = The shear in the steel stirrups
Av = the area of the vertical stirrups (note that it should be 2*Abar since they run on both faces)
fyt = The yield strength of the steel (e.g. 40 ksi, 60 ksi)
s = the spacing of the steel stirrups

If {ÝV_c\over2} \ge V_u then it is not necessary to consider Vs. [ACI 11.5.6.1]

This assumes that:

  • h < 10"
  • d ≤ 2.5 tf
  • d ≤ tw / 2

Shear Checks:

  • Check to ensure Vu < Vu-max

V_{u-max} = {10\sqrt {f'_c} b_wd}

References

  1. American Concrete Institute, "ACI 318", 2005

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