We will discuss how to calculate the concrete breakout strength according to the ACI 318 code in this part of the concrete anchor foundation bolt design calculation example. Please refer part-1 of the calculation to understand the design problem statement and the calculations for the anchor bolt materials strength.

If you refer the Fig.1 and the Fig.2 of the problem statement then it will clear to you that the distance between two surrounding anchors (in our case it is 6 inch) is less than three times the effective embedment depth (in our case it is 10 inch). So, as per ACI codes, the anchor arrangement can be termed as group and the ACI formula for calculating the concrete breakout strength for the anchor group is:

*φN _{cbg}= (A_{Nc}/A_{Nco})ψ_{ec,N} ψ_{ed,N} ψ_{c,N} ψ_{cp,N}N_{b}……………………………..D-5*

* *

*Where, *

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

*A _{Nc} –Total projected area for the group of anchors of the failure surface as approximated by a rectangle with edges bounded by 1.5h_{ef} (1.5*10 = 15 inch in our case) and the free edges of the concrete from the centerline of the anchors. Confused? You will be clearer while calculating it little later.*

*h _{ef }– Effective embedment depth of anchor (in our case it is 10 inch).*

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

*1.5h*

_{ef}(1.5*10 = 15 inch in our case) from the centerline of the anchors at all the sides.*ψ _{ec,N} – Modification factor for the anchor group loaded eccentrically in tension, in our case there is no eccentric loading , so ψ_{ec,N}=1.*

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

*Ψ _{c,N} – Modification factor based on the presence and absence of crack in the concrete, we will calculate it later.*

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

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

** **

**Calculation of ***A _{Nc}*

Before going to the calculation please refer the plan view of the anchor arrangement as below:

A_{NC }=Area of the square shown in dotted line

= (15+6+5)*(15+6+5) = 676 Square Inch

**Calculation of ***A _{Nco}*

Before going to the calculation please refer the plan view of the anchor arrangement as below:

A_{NC }=Area of the square shown in dotted line

= (15+15)*(15+15) = 900 Square Inch

**Calculation of ψ_{ed,N}**

If, **C _{a,min}** (minimum distances between two anchors, 6 inch in our case) is less than 1.5 times the

*h*_{ef}*(1.5*10=15 inch in our case) then*

ψ_{ed,N} = 0.7+(0.3*C_{a,min})/(1.5*h_{ef})

= 0.7+(0.3*6)/(1.5*10)

= 0.82

**Calculation of ψ_{c,N}**

According to the code the value of ** ψ_{c,N} **=1 where concrete crack is likely to occur.

**Calculation of N _{b}**

*N _{b}= k_{c }λ h_{ef}^{1.5} *

*f*

_{c}’ …………………D-7*Where,*

*K _{c} – a factor which has value of 24 for cast in concrete.*

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

*h _{ef }– Effective embedment depth of anchor (in our case it is 10 inch).*

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

So, by putting the value in D-7 we get:

N_{b }=24*1*10^1.5*4000^0.5

= 48000 lb

Now, by putting all the values in **D-5 **we get the Concrete breakout strength for the anchor group

**φN _{cbg }**=

*(A*

_{Nc}/A_{Nco})ψ_{ec,N}ψ_{ed,N}ψ_{c,N}ψ_{cp,N}N_{b}= 0.70*(676/900)*1*0.82*1*1*48000

= 20694.6131 **lb**

** **

So, finally we got the concrete break out strength of the anchor group in tension as 20694.6131 **lb. **In the next part (Part-3) of the foundation anchor bolt design calculation example, we will discuss about calculating the concrete pull out force in tension.

according to the commentary, ca,min is the minimum edge distance, not the minimum distance between bolts as you stated above.

Ashe is correct ca,min =5″ and f’c =3500 psi as per problem statement also eqn D-7 is missing sq rt symbol for f’c . calculation correctly shows this.

Hi, Thanks for you work, Is the strength reduction factor equal to 0.75 or 0.7. It is taken to be 0.75 in step 1.