Wall Shear (Out of Plane)
Wall Shear (Out of Plane)
(OP)
I have an 8" thick concrete wall where I am having problems getting the shear to work. The shear is out of plan of the wall. My Vu is greater than 1/2 Phi Vc, however Vu is less than Phi Vc.
Based on what I can determine, it appears I need to provide shear reinforcement. Any ideas? I was thinking of a #3 bar stirrup horizontal around the vertical reinforcing. Is this correct? It apears that 6" oc spacing is the minimum based on a 5.625" d dimension.
Based on what I can determine, it appears I need to provide shear reinforcement. Any ideas? I was thinking of a #3 bar stirrup horizontal around the vertical reinforcing. Is this correct? It apears that 6" oc spacing is the minimum based on a 5.625" d dimension.





RE: Wall Shear (Out of Plane)
If Vu < φVc then you are OK.
RE: Wall Shear (Out of Plane)
RE: Wall Shear (Out of Plane)
RE: Wall Shear (Out of Plane)
I've never used stirrups in a slab or wall.
RE: Wall Shear (Out of Plane)
Strguy11, is this a retaining wall? Must be for you to have such high shear values. If that is the case, I think you need to worry more about the joint at the bottom of the wall than the shear within the wall. I don't believe I have ever actually had out of plane shear control wall design.
RE: Wall Shear (Out of Plane)
One other question that I have is in relation to this but say for an exterior wall, with only reinforcment in the center. Say d=4". For an exterior wall case, the wind loads can be positive or negative. If the moments are greater than the cracking moment Mcr, wouldnt the concrete crack on both side of the rebar (Ext face and int face) Would there really be any concrete left to take the shear our of plane? Or are most people using a double layer of reinforcing for exterior walls, say 8" thick?
RE: Wall Shear (Out of Plane)
Also, don't forget to include P/A in your Mcr calculation. Many times the axial load increases the Mcr.
Per ACI, Mcr = frIg / yt
This is from f = My/I
But with axial load, f = P/A +/- My/I
So by inserting this into the Mcr equation you get:
Mcr = (fr +/- P/A)Ig / yt
Just be sure to watch your sign convention on the P/A by remembering that fr is a tension stress.
RE: Wall Shear (Out of Plane)
They're not exactly labor intensive to place, and depending on your load pattern may be more economical than thickening your wall.
RE: Wall Shear (Out of Plane)
RE: Wall Shear (Out of Plane)
RE: Wall Shear (Out of Plane)
The only hgue draw back I see is that there is a lot of reinforcing steel in your cage
RE: Wall Shear (Out of Plane)
The wall would have to have one mat tied up, placed on one side of the wall. Then the second mat tied up and stood up on the other side. Then you'd have to have your steel workers climb up the mats of steel and tie individual little double hook stirrups all across the wall face at the specified spacing. Very time consuming, difficult to support the workers hanging off the mats of steel., all the while the two mats of steel are not usually supported well until the other form is placed and the mats cross tied to the wall ties. Just a difficult scenario.
RE: Wall Shear (Out of Plane)
RE: Wall Shear (Out of Plane)
RE: Wall Shear (Out of Plane)
RE: Wall Shear (Out of Plane)