In the NZSEE the follow equations are given for the in-plane capacity of a masonry wall, whenever rocking occurs:
Vr = (0,9 (α · P + 0,5Pw)Lw)/Heff
While the following equations are given for the toe-crushing capacity:
Vr = ((α · P + 0,5Pw)Lw)/Heff) · (1-(fa/0,7f'm))
With:
- α = factor for boundary conditions (0,5 fixed-free; 1,0 fixed-fixed);
- P = vertical load acting on top of wall;
- Pw = Own weight of the wall;
- Lw = length of the wall;
- Heff = effective height of the wall.
My questions:
- Could anyone explain to me why the own weight (Pw) is multiplied by a factor 0,5 in both formulas?
- And could anyone explain to me why α = 0,5 for fixed-free walls?
Amir Hosein Salmanpour
Dear B. Vries, de,
The factor 0.5 comes from the stabilizing moment arm, i.e. Lw/2. Considering a rigid block with fixed-free boundary conditions, the following equilibrium holds:
(P+Pw)Lw/2=Vr.Hw ---> Vr=(0.5P+0.5Pw)Lw/Hw
for fixed-fixed boundary conditions, the moment equilibrium would be:
P.Lw+Pw.Lw/2=Vr.Hw ---> Vr=(1P+0.5Pw)Lw/Hw
Best,
Amir
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