## 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
- s
_{max}, s_{min}, and A_{v-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 V _{u}:**

This is the ultimate factored shear that the concrete beam is experiencing.

**2) Calculate V _{c}:**

The equations for this have been provided in the Concrete Beam Design section.

**3) Find when {øV_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 V _{s-reqd}:**

*Where:*

_{s-reqd}= The required shear strength that the steel must provide.

**5) Find the spacing (s):**

*Where:*

_{v}= Assumed based on what type of rebar? you are using. Note that A

_{v}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.

f

_{yt}= 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)

### s_{max}, s_{min}, and A_{v-min}

- s
_{max}- s
_{max}= min [24" or d/2]`for`

V_s \le 4\sqrt{f'_c}b_wd - s
_{max}= min [12" or d/4]`for`

V_s \ge 4\sqrt{f'_c}b_wd

- s

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

- s
_{min}= 3" - A
_{v-min}

*Where:*

_{v-min}= the minimum shear steel required if {øV_c\over2} \le V_u

f'

_{c}= The compression strength of the concrete (e.g. 4 ksi, 6 ksi)

b

_{w}= the width of the concrete beams web

f

_{yt}= the strength of the rebar (e.g. 40 ksi, 60 ksi)

### Situations where no stirrups will be necessary:

If {øV_c\over2} \ge V_u then it is not necessary to consider V_{s}. [ACI 11.5.6.1]

- h < 10"
- d ≤ 2.5 t
_{f} - d ≤ t
_{w}/ 2