slab design when resisting lateral thrus

huangyhg

slab design when resisting lateral thrust
hello all;
     i'm desingning a propped retaining wall, with the top cast into a concrete suspended floor and the base sitting on a footing with a floor slab "propping" the wall from the side (thereby providing lateral support).  i'm curious about how to design the slab for this reaction.
     please note: after taking into account friction from the base of the footing against the supporting ground, as well as refining this by doing a coulomb pasive pressure triangle, i'm still left with 15kn/m unballanced thrust taken by the slab.  i don't just want to say that this works by inspection; anyone have a design procedure for such a situation?
thanks and regards,
ys
b.eng (carleton)
working in new zealand, thinking of my snow covered home...
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i would check the slab as a deep beam spaning between shear walls. assume the reinforcement in half the slab(or depending on forces only utilise the 1st 1m depth) is the tension reinforcement with a d (eff) which is at average depth for the tension reinforcement.
check as a normail beam.
hello vinny;
     thanks for your reply.  this sounds like a rational approach, however i was more worried about the possibility of the slab failling by "popping up", or buckling out of plane away from the supporting soil.  any thoughts?
for the record, i'm fully open to the possibility that i'm worrying over nothing.  checking as you've suggested coupled with an argument that 2.5% of the thrusting force is easily overcome by gravity on the slab might be the solution.  that and a check for bearing strength of concrete for a 100mm wide bearing surface.
thus:  
1)  deep beam check: 6m deep, 12.4m span
m* = wl^2/8 = 15kn/m(12.4m)^2/8 = 288.3 kn·m
Ømn = 0.85asfy(d-a/2) where a = asfy/alpha·f'c·b
      assuming the bottom 1m strip to be the tension steel:
      d = 5.5m, b = 0.1m
      665 mesh provides 145mm^2/m, fy = 300mpa, f'c = 25mpa
      therefore: a = 20.47mm
Ømn = 203.0kn·m
      assuming 663 mesh (as = 205mm^2/m)
Ømn = 286.8kn·m  
since 288.3/286.8 is only a 0.5% overstress, accept.
2)  2.5% of 15kn = 0.375kn
dead of slab (0.9g) = 0.9*0.1m*23.5kn/m = 2.115kn >> 0.375kn
3)  bearing of concrete floor slab = Ø(0.85f'ca1)
     where a1 = loaded area in mm^2
     (from nzs 3101 part 1:2006, clause 16.3.1)
a1 = 100000mm^2; Ø=0.65;
bearing strength = 1381.25kn/m >> 15kn
after doing the bearing check it's probably something i should just neglect in these circumstances!
thanks again for your help,
ys
b.eng (carleton)
working in new zealand, thinking of my snow covered home...