I generally use a 45 degree load spread as well, though there are more exact methods out there. Eg. CSA S6-06 specifies a more exact method (don't recall the formula off hand though) and I expect AASHTO does as well. The times I have compared the two, the 45 load spread has been conservative.
gwynn:
Yes, AASHTO does indeed refer to a 45 degree angle.
I feel however that a 45 degree angle is too generous on longer cantilevers. I try to use a 45 degree angle but not to exceed 5 to 8' depending on my cantilever. It is good to set an upper width limit so that you can distribute reinforcing properly.
Consider the attached sketch. The only load acting is a concentrated load at Point 4.
Negative yield lines form at 1-2, 1-3 and 2-5 (shown solid red).
Positive yield lines form at 1-4 and 2-4 (shown dashed red).
m1 and m2 are the negative and positive moment capacities of the slab per unit length in the direction of the cantilever.
m3 and m4 are the negative and positive moment capacities of the slab per unit length perpendicular to the cantilever (i.e. temperature steel).
Consider a deflection of 1 unit at Point 4 under the action of P.
External Work = P*1 = P.
Internal Work = 2(m1+m2)x/a + 2(m3+m4)a/x
E.W. = I.W., so P = 2(m1+m2)x/a + 2(m3+m4)a/x
dP/dx = 2(m1+m2)/a - 2(m3+m4)a/x2
Setting dP/dx = 0 for max. or min. value of x;
x2 = a2(m3+m4)/(m1+m2)
x + a[√]{(m3+m4)/(m1+m2)}
In the case of a steel plate cantilever, m1 = m2 = m3 = m4 = m, so x = a. This would form an angle of 90 degrees between positive yield lines.
For a concrete slab, bottom steel may be absent, in which case m2 and m4 are zero. Also, m1 is unlikely to be the same as m3, so the angle between positive yield lines will be affected by the area of bars in each direction.
Here is my problem with a 45 degree angle. As the cantilever gets longer, the bending strength get stronger by the same amount assuming consistent reinforcement - that is, double the cantilever, double the beam width, double the capacity. To summarize: You need a minimum rebar but cantilever length is irrelevant. Then it just comes down to punching shear at the load point.
Instead, where possible I prefer to treat it as a T-slab with a stem of very small width and depth and use the maximum allowable width. This may be overly conservative, but I don't get a comfortable feeling with the 45 degree angle of indefinite length.
Well, yes...but as the width gets larger I would also usually have other loads on the wide cantilever overlapping as well so it is not 1:1 load to strength.
But I agree with you, jsdpe25684, that there should be a "gut feel" pragmatic limit on how wide you disperse the point load.
Just re-read my last post. The final line:
x + a?{(m3+m4)/(m1+m2)}
should read:
x = a?{(m3+m4)/(m1+m2)}
where distance 'x' is as shown on the sketch. The effective width of slab supporting the point load varies from 0 at the load to 2*x at the support.
In order to justify a spread angle of 45 degrees, x would need to be a*tan22.5 = 0.4142a. The moment capacity normal to the cantilever would need to be 0.172 times that of the primary reinforcement.
A better way to deal with the problem in my opinion, is to provide a slab band along the edge of the cantilever to spread the load as required to satisfy the intended distribution of the cantilever reinforcement.
BA - I don't know if I'm missing something here, but it seems straightforward. Use plate/shell elements to model the cantilever structure, apply a load to the edge, read off the maximum moment/unit width.
If you use plate elements in your analysis, the stiffness per unit width of plate is the same in all directions. This will yield an exact solution for a plate behaving elastically, but the moment requirement normal to the cantilever will likely be much more than one would ordinarily expect as temperature steel in a concrete slab.
I agree that your FEA solution is valid for a slab reinforced identically in each direction top and bottom, but that condition was not specified by the OP and I don't think it conforms with usual practice.
The effective width of slab carrying a point load at the end is dependent on the placement of the reinforcement.
Well it's even more dependent on the pattern of cracking, which is dependent on the load, the concrete tensile strength, the shrinkage and creep strains, and differential temperature stresses. You will need to do separate non-linear analyses for the Serviceability and Ultimate Limit States. If you want to go that degree of detail then a finite element analysis with orthotropic plates is the only practical way of doing it at the Serviceability Limit State.
For normal practical purposes I think a linear FE analysis with isotropic plates combined with standard good practice to ensure ductile behaviour is quite adequate.
The pattern of cracking is primarily controlled by the placement of the reinforcement. Concrete tensile strength is presumed to be zero. Shrinkage, creep and temperature effects are present just as they are in all cast-in-place slabs.
My answer to the original question is that the best way to design a cantilever slab with a point load is the same as for any other slab with (effectively) a 2 way span. Do a finite element analysis and provide the reinforcement required for the output moments in both directions.
I do not have access to a finite element analysis program and, even if I did, I would not know how to account for all the variables inherent in this problem. What is more, I do not believe that you know how to account for these either.
The original question was to determine the width of a cantilever slab carrying a point load at the end. Others have suggested various criteria for determining that width. My position is that the width of slab resisting the point load depends on the arrangement of reinforcement.
You have not answered the original question. You insist that FEA is the only method of solving the problem. Tell that to the many engineers who have been solving similar problems for many years without the advantage of FEA.
