Shear Stirrup Spacing (for concrete beams)

Shear stirrups are used to resist shear cracking in concrete beams along the 45 angle failure plane.

The following sections will be addressed in this section:

  • Design for Shear Stirrups
  • smax, smin, and Av-min
  • Situations where no stirrups will be necessary
  • An example of shear stirrup design
  • Common design practices for stirrups

Design for Shear Stirrups:

1) Calculate Vu:
This is the ultimate factored shear that the concrete beam is experiencing.

2) Calculate Vc:
The equations for this have been provided in the Concrete Beam Design section.

3) Find when {V_c\over2} \ge V_u :
This region of the concrete beam will not require any stirrups, [ACI 11.5.6.1][1].

Where:

4) Find Vs-reqd:

V_{s-reqd} = {V_u\over }-V_c
Where:

Vs-reqd = The required shear strength that the steel must provide.

5) Find the spacing (s):

s = {A_vf_{yt}d\over V_{s-reqd}}
Where:

Av = Assumed based on what type of rebar? you are using. Note that Av is not the area of one bar but 2* the area of one bar since you would have to shear through both sides of the stirrup for failure to occur.
fyt = the strength of the rebar (e.g. 40 ksi, 60 ksi)
d = the depth of the concrete beam (note: that this is not the same as the height of the beam)

smax, smin, and Av-min

  • smax
    • smax = min [24" or d/2] for V_s \le 4\sqrt{f'_c}b_wd
    • smax = min [12" or d/4] for V_s \ge 4\sqrt{f'_c}b_wd

This is to ensure that for high shear zones the stirrup spacing is tighter than in low shear zones.

  • smin = 3"
  • Av-min

A_{v-min} = {0.75\sqrt {f'_c}{b_ws \over f_{yt}} \ge {50b_ws\over f_{yt}}}

Where:

Av-min = the minimum shear steel required if {V_c\over2} \le V_u
f'c = The compression strength of the concrete (e.g. 4 ksi, 6 ksi)
bw = the width of the concrete beams web
fyt = the strength of the rebar (e.g. 40 ksi, 60 ksi)

Situations where no stirrups will be necessary:

If {V_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

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