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Pullout strength formula [D-15/ACI318-11]

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EBASS

Structural
May 20, 2014
3
Does anyone know from where this equation comes Np=8*Abrg*f'c, i cant understand the multiplier 8 in the equation, it is not consequent with the bearing strength calculations of 0,85f'c. If anybody knows, Thanks.
 
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Probably based on testing. I always wondered the same thing.
 
The 0.85f'c typical bearing is for an unconfined cylinder. I don't know exactly where the multiplier magnitude of 8 came from, but it's a recognition of the much higher bearing stress of confined concrete. You are allowed a multiplier for baseplates and such and that is a large area. It makes sense that the smaller the bearing area, the larger the multiplier.
 
I agree with you Lion06, but in the best case we cannot reach a factor multiplying bearing capacity higher than 2, but the effect of confined concrete and the proyection of the bearing stress can be the explanation to this large number. Waiting for the answer. Thanks for your cooperation.
 
It appears to be a different mode of failure? Bearing strength failure is due to cracking, while anchor bolt pullout failure involves localized crushing around the bolt head (source: ACI D commentary).
 
From "Design of Headed Anchor Bolts" by John G. Shipp and Edward R. Haninger,

"The following section establishes limitations for the combined
effects of bolt spacing, embedment depth and edge
distance, such that the heavy hex head on a standard anchor
bolt provides "full anchorage" in concrete equal to the
tensile capacity of the bolt
. Several agencies/authors have
published reports representing their test data and/or recommendations
to account for these variables, (see Refs. 9,
10, 13, 16 and 17)."

I assume the next step would be to go hunt up those references.
 
Thanks guys for all the information, i'm still searching the answer. i'll share with you the correct answer when i find it.
 
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