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- To: <seaint(--nospam--at)seaint.org>
- Subject: Column Reinforcing
- From: "jwknospam" <jwknospam(--nospam--at)comcast.net>
- Date: Thu, 3 Mar 2005 15:47:53 -0600
I’ve got a building where the client wants to add a partial floor below grade. He’s demolishing the first floor slab, removing the soil (soft clay) to bedrock, and re-building a structural slab where the first floor used to be. The end result will be a below-grade parking level.
The structure is a 4-story concrete building with about 31’ bays. The reinforced concrete columns (32” octagons) stop at grade and are supported by 48” sq. unreinforced piers to bedrock. These piers are confined by the soft clay soil. The re-built first floor will be at original grade, so I’m not worried about the hinge.
One of the uses of this structure in the past 60 years was storage for a greeting card company. Anecdotal evidence has every floor covered in huge pallets of paper stacked from floor to ceiling, i.e., MASSIVE live loads. The new use for the structure will be residential (condos), so the live load is reduced to 40 psf in most places.
What I’m worried about is the newly exposed unreinforced pier. Containment reinforcing steel must be added, but how much? Bending isn’t an issue – we’re talking about 4 or 5 columns out of about 50, and there are new reinforced concrete shear walls around the perimeter of the excavated area. The maximum factored vertical load is 860 kips, and the existing pier strength is 5500psi (per several tests). We’re in a low to moderate seismic zone (12% - 6%). Any axial creep occurred decades ago.
Per ACI 318 10.8.4 and 10.9.1, the minimum reinforcing required is 0.5% of the gross area. Adding long bars and wrapping in new concrete raises the area to 56”x56” which requires As=15.68, or (12) #11 bars. This seems quite excessive, and the owner almost fainted when I told him.
My question is, how can I reduce the number of longitudinal bars and still comply with the code? Can I just use the area of the new concrete wrap (4” on a side)? This brings the area to (56”x56” – 48”x48”)*1% = 8.32, or (12) #8 bars. This is much more reasonable, but does it technically meet code?
Any other ideas?
The ties will be (2) #4 hooked stirrups (“U” shapes with hooks on the ends) installed on opposite sides of the column, alternating directions each level.
Thanks in advance for any ideas,
Jason W. Kilgore
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- From: Harold Sprague
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