You can count for longitudinal steel to account for Shear friction. As long as there is no net tension on the section, the longitudinal steel that is used for flexure can also be used for shear friction.
See commentary for 11.7.7 (ACI 318-05)
When moment acts on a shear plane, the flexural tension stresses and flexural compression stresses are in equilibrium. There is no change in the resultant compression Avf fy acting across the shear plane and the shear-transfer strength is not changed. It is therefore not necessary to provide additional reinforcement to resist the flexural tension stresses, unless the required flexural tension reinforcement exceeds the amount of shear-transfer reinforcement provided in the flexural tension zone. This has been demonstrated experimentally.
So if I have a bending moment around a corner of a wall that req's As = 0.31 sq.in. of steel, and the shear values exceed the allowable Vc and I need to use Shear Friction, do I not have to "add" additional steel? If Avf = Vu/(phi*Fy*mu) = 0.09, do I not have to use 0.40 sq.in of reinforcing?
blawson,
Can you expand a bit on your wall corner example? Usually the reinforcement used for bending is orthogonal to the shear reinforcement in a bending member since the shear cracks are usually at an inclined angle to the flexural "beam" and thus the shear reinforcement is not aligned with the flexural. Thus there is no conflict.
Just to point this out, too. You can only use shear friction
in certain circumstances. You need to have a
cold joint or a reasonable expectation that a crack will occur along the plane in question - a SHEAR plane. Diagonal tension cracking doesn't count as a shear plane.
