Final Min Shear from Rho

[civst500]

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.

 

[BridgeSmith]

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.