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Final Min Shear from Rho

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civst500

Structural
Joined
Nov 5, 2015
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thread607-464356

R13, in the calculation you provided with a Rho of 0.0111 the mininmum spiral area = (0.011)(240) = 2.4 in^2 ???? that seems huge. this would required #10 spirals? what am I missing?

Im getting a similar crazy result for a 30 inch column, 3" clear with fc' = 5000, fy = 60,0000. rho min = 0.02, As = 9 in^2!!!!!

This would make sense for the plan area of the spiral but not the cross sectional.
 
The design example is jacked up. They used "cover" instead of "core". The area of the cover, in plan, would be Ag-Ac = 330 in^2 - 240 in^2 = 90 in^2. However, that's still the plan area of the cover, not the cross section of the concrete corresponding to the cross sectional area of the reinforcing. That cross section is a vertical cut, so the concrete area under consideration has a height equal to the pitch of the spiral and a width that goes across the diameter of the column.

I can't completely reconcile the rho with the area of concrete, but it's in the ballpark.

In the example calculation, the result is a #3 spiral at a 1.606" pitch. Maybe follow their calculations all the way through and see if you get a more reasonable answer.
 
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