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Effective depth "d" for shear ACI 318-19

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CDLD

CDLD

Structural
May 20, 2020
214
Hello,

I am making a concrete column spreadsheet and am unsure of the effective depth, "d", for shear.

As you may know, in the 19 version of ACI, the formula for concrete shear strength was changed and is now dependent on the reinforcing ratio pw.

In the commentary R22.5.5.1, it states" "the value of As to be used in the calculation of pw may be taken as the sum of the areas of longitudinal bars located more than two thirds of the overall member depth away from the extreme compression fiber".

If As is calculted using all rebar below 2/3 of the member depth (i.e., including some sidebars), should "d" be located at the centroid of that reinforcing?

Typically (i.e., before ACI 318-19), "d" for column shear calculations is taken as column depth minus cover, stirrups and half bar.
 
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Technically, yes. See some previous discussion here:
Note that 22.5.2 gives you a floor of 80% diameter if it's a circular column, or prestressed. (AASHTO says 72% of the member height for any section)

There's an argument to be made that while the letter of the law would be the physical centroid of all reinforcement below 2/3 depth, a more rational approach would be the location of the resultant tensile force (accounting for variations in strain across the cross section, where side bars may not be at yield).

If you're making a standardized spreadsheet, you probably won't want to take advantage of these. But it appears some extra margin exists in the real world.
 
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Thanks for the reply.

Celt83's link recommends 0.8H for design.
I suppose I'll make the spreadsheet code compliant for now, and take "d" as the centroid of all bars below 2/3 depth.
 
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It seems the code's recommendations aren't actually that punishing.
Worst case, you get the average between 2/3*h and your most remote "d" (i.e., the bottom layer of rebar).

This winds up being very close to 0.8h anyways.
 
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Not sure if anyone will benefit from this, but the reduction for square columns with equally distributed reinforcement is quite small.

I calculated d at the centroid of all bars below 2/3 depth and got 20.6", which is only an inch higher than the bottom layer of bars.

Obviously the reduction would be greater for a rectangular column, but it doesn't seem unreasonable.
Screenshot_2024-01-31_154833_jwavwc.png


d = 20.6 / 21.5 = 0.95 ~ about a 5% reduction.
 
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