Before diving into the concrete anchor bolt design calculation example for calculating anchor breakout strength in shear, please go through the problem statement and part-1, part-2, part-3, part-4 and part-5 of this series. The calculation exercise is carried out according to the ACI 318 appendix D codes.

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**Calculation**

The code gives the following formula for finding out the break out strength in shear for a group of anchor bolts:

*φV _{cbg}= φ(A_{Vc}/A_{Vco})ψ_{ec,V} ψ_{ed,V} ψ_{c,V} ψ_{cp,V}V_{b}……………………………..D-22*

Where,

*Φ – Strength reduction factor for breakout strength and its value for non-reinforcement in tension is 0.70*

*A _{vc} – projected concrete failure area of the group of anchors, for calculation of strength in shear (in inch^{2})*

*as approximated by a rectangle with edges bounded by 1.5*Ca1 (1.5*5 = 7.5 inch in our case) and the free edges of the concrete from the centerline of the anchors. We will calculate it little later.*

*Ca1 – Distance from the center of an anchor shaft to the edge of concrete in one direction (in inch). **Refer Fig.1 below.*

*A _{vco} – Total projected shear failure area approximated by a square bounded by *

*1.5*Ca1 (1.5*5 = 7.5 inch in our case) from the centerline of a anchor at all the sides. We will calculate it little later.*

*ψ _{ec,V} – Modification factor for the anchor group loaded eccentrically in shear, in our case there is no eccentric loading , so ψ_{ec,V}=1.*

*ψ _{ed,V} – Modification factor for edge effect for the anchor group in shear, we will calculate it later.*

*Ψ _{c,V} – Modification factor based on the presence and absence of crack in the concrete, in our case value of it = 1..*

*ψ _{cp,V} – Modification factor for post weld types of anchor. For cast in place anchor ψ_{cp,V} = 1*

*V _{b} – Basic concrete break out strength for a single anchor in shear, we will calculate it later.*

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**Calculation of ***A _{Vc}*

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Refer the above two figure (Fig.1 and Fig.2), it should be clear by now that

A_{VC} = 16.5*(5+6+16.5) = **453.75 inch ^{2}**

**Calculation of ***A _{Vco}*

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By referring the above figure (Fig.3), we can calculate

A_{vco}= 16.5*(16.5+16.5) = **544.5 inch ^{2}**

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**Calculation of ψ_{ed,V}**

For C_{a2} <1.5 Ca1

ψ_{ed,V} = 0.7+(0.3*C_{a2})/(1.5*C_{a1}) = 0.7+ (0.3*5)/(1.5*11) = **0.79**

Where,

C_{a2} – Minimum concrete edge distance from anchor perpendicular to C_{a1}.

**Calculation of V _{b}**

*V _{b}=(7*√d_{a}*[l_{e}/d_{a}]^{0.2} )*λ*√f’_{c}*(C_{a1})^{1.5} ………………………..D-24*

Where,

d_{a} = Diameter of the anchor bolt in inch., in our case it is **0.75 inch**. Remember we have started with the 0.75 inch anchor in the part-1 of this calculation.

l_{e} = Length of the anchor inserted inside the concrete in inch. In our case it is **10 inch**.

*λ = A factor related to the reduced mechanical properties of concrete, for normal weight concrete λ =1.*

*C _{a1 }= we already know it from the Fig.1. Its value is 11 inch.*

*f _{c}’ = Specified compressive strength of the concrete, in our case f_{c}’=4000 psi (refer problem statement of part-1).*

So, by putting all the values, the above equation (D-24) becomes

V_{b} = (7*√0.75*[10/0.75]^{0.2})*1*(√4000)*11^{1.5}

_{ }**= 23482.02 lb**

Now, from the equation D-22, after putting all the values at right hand side,

*φV _{cbg} = φ(A_{Vc}/A_{Vco})ψ_{ec,V} ψ_{ed,V} ψ_{c,V} ψ_{cp,V}V_{b}*

= 0.7*(453.75/544.5)*1*0.79*1*1*23482.02 lb

= **10821.297 lb**

The concrete break out strength in shear for the anchor bolt group according to the ACI codes is calculated as **10821.297 lb.** The next part (Part-7) will talk about Concrete Pullout Strength in Shear.

nyc job

bt i guess there is some prob with the area calculated for shear for group of anchor..

there should be 35 degree leverage to the edge…

Vb=9*1*sqrt 4000* (11)^1.5 =20766 lbs is smaller than 23482lbs as per ACi