PROFIS ENGINEERING
Reference Figure 6.2 [ 51 ]. It illustrates a hypothetical anchorage consisting of six anchors. Shear load is being modeled in the x- direction( V ua, x) and in the y- direction( V ua, y). The x- and y- edges are fixed. For this example, assume V ua, x = 6000 lb and V ua, y = 3000 lb.
Since shear load acts towards two edges, shear concrete breakout must be considered for each edge. ACI 318 provisions require V ua, x to be checked versus the calculated design concrete breakout strength in shear for the x- edge( ϕV cbg, x), and V ua, y to be checked versus the calculated design concrete breakout strength in shear for the y- edge( ϕV cbg, y). The anchorage design with respect to concrete breakout in shear is satisfied if:
MAX( V ua, x / ϕV cbg, x);( V ua, y / ϕV cbg, y) < 1.0
The resultant load( V ua, res) of V ua, x and V ua, y acts an oblique angle with respect to each edge:
V ua, res =
2 2 V ua, x + V ua, y = 6708lb.
The line of action for V ua, x coincides with the centroid of the anchors; however, V ua, y acts 6 in in the x- direction from the centroid of the anchors. This creates torsion on the anchors such that V ua, res will be eccentric to the centroid of the anchors. Shear concrete breakout calculations must consider any shear eccentricity via the ψ ec, V modification factor. PROFIS Engineering calculations assume V ua, res acts towards either fixed edge in lieu of V ua, x acting towards the x- edge and V ua, y acting towards the y- edge. PROFIS Engineering shear concrete breakout calculations are predicated on:
MAX( V ua, res / ϕV cbg, x);( V ua, res / ϕV cbg, y) < 1.0
For the application illustrated in Figure 6.2 [ 51 ], Shear Load Evaluation – Concrete Breakout at y- Edge( page 52) explains PROFIS Engineering calculations for shear eccentricity when calculating concrete breakout in shear for the y- edge( ϕV cbg, y).
Shear Load Evaluation – Concrete Breakout at y- Edge
Figure 6.2.1a.
Figure 6.2.1b.
V ua, y = factored shear load acting in the y direction towards the-y edge = 3000 lb
V ua, y, n = factored shear load in-y direction acting on anchor n V ua, y, n =( V ua, y / 6 anchors) = 500 lb / anchor
M tor = factored torsion moment =( V ua, y)( 6 in) = 18,000 in-lb F tor, n = factored load( from torsion) acting on anchor n F tor, n, x = factored load( from torsion) acting on anchor n in the x direction F tor, n, y = factored load( from torsion) acting on anchor n in the y direction
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