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ha in ACI318 1

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mekafime

Mechanical
Aug 14, 2015
88
Hi everybody,
What's the value of ha for Avc? in fig. RD6.2.1(b) states this.
 
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ha is the thickness of the member to which you are anchoring. It comes into play when the projected breakout area is truncated by the face of the member.
 
You mean what is the thickness of the concrete when I'm anchoring? or the thickness of the motherboard?
because the concrete thickness (it is a cantilever) is 440 mm and the base plate is 12 mm
 
ha is the thickness of concrete. You can refer to next Fig. RD.6.2.4, in which all the distances are shown clearly.

Euphoria is when you learn something new.
 
My problem is ha= 440 mm, Ca1,1= 90 mm, Ca1,2 = 90 mm, Shear force= 80kN for 2 anchor rods 5/8"

Anclaje_1_k9vnwz.png
 
I don't understand the problem. Your ha>1.5ca,1 and ca,2<1.5ca,1. So you're Case 1. ha doesn't appear in that equation. Or did I miss something?
 
phamENG said:
I don't understand the problem. Your ha>1.5ca,1 and ca,2<1.5ca,1. So you're Case 1. ha doesn't appear in that equation. Or did I miss something?

I agree, however I have always thought that ACI does a bad job in this commentary figure. They combine the ha>ca1 condition with the ca2<1.5ca1 condition. In reality, OP's equation isn't really provided in that detail. I think a simple "if ha>ca1, use 1.5ca1" statement would clear it up.
 
Fair point, jdcollins. mekafime - your ca,2 is the space between anchors. Use the the figure in the commentary, but as jdcollins said use 1.5ca,1 rather than ha.
 
Only if you're far enough away from the edge. If you're more than 1.5ca,1 from the edge, then there's no reduction in shear capacity. If you are closer than that, the reduction in shear capacity is proportional to the reduction in projected area (as defined by AVc/AVco)
 
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