LannyBudd
Structural
- Mar 12, 2012
- 23
I am trying to check the Allowable Shear (Fv) due to the perpendicular wind loading on a 4’-0” tall masonry parapet using the ACI 530-11/IBC 2012. When I performed this calculation using ACI 530-08/IBC 2009; when shear reinforcement is not provided to resist the calculated shear (fv) per Section 2.3.5.2.2 of ACI 530-08, I would use Equation 2-24 (Fv = (f’m)^0.5) not to exceed 50 psi. ACI 530-11 does not have this allowable and I am not sure if I am doing the calc correctly. If I am reading ACI 530-11 correctly, and if I have no shear reinforcement, then all I do is plug and chug into ACI 530-11 equation 2-28.
I have 8” CMU blocks, f’m=1500 psi, d=3.81”, partially grouted with #5 at 48” O.C. At the parapet level, I get a moment of 3200 ft-lb/bar and shear of 1600 lb/bar. The calculated shear area is 79.2 in^2 for a calculated shear of fv=20.2 psi. Per ACI 530-08 Eq 2-24, Fv=(f’m)^0.5 = 38.7 psi. Per ACI 530-11, I am calculating M/(Vd)>1, therefore per ACI 530-11 Eq 2-28, Fvm=43.6psi.
Is what I calculated for Fvm per ACI 530-11 correct? It seems wrong because it’s more than double for the older code.
Another question that I have is if this was not a parapet and if I have a compressive load on the wall, then per ACI 530-11 Eq 2-28, the Fvm is increased by 0.25(P/An).
I have 8” CMU blocks, f’m=1500 psi, d=3.81”, partially grouted with #5 at 48” O.C. At the parapet level, I get a moment of 3200 ft-lb/bar and shear of 1600 lb/bar. The calculated shear area is 79.2 in^2 for a calculated shear of fv=20.2 psi. Per ACI 530-08 Eq 2-24, Fv=(f’m)^0.5 = 38.7 psi. Per ACI 530-11, I am calculating M/(Vd)>1, therefore per ACI 530-11 Eq 2-28, Fvm=43.6psi.
Is what I calculated for Fvm per ACI 530-11 correct? It seems wrong because it’s more than double for the older code.
Another question that I have is if this was not a parapet and if I have a compressive load on the wall, then per ACI 530-11 Eq 2-28, the Fvm is increased by 0.25(P/An).