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Re: One way slab design with point load.

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My original post read:

>I have a photocopy out of an old concrete design book a senior engineer
> me years ago that has the following simplistic assumption for approximating
> participating width of a one way slab under a point load.   Take TAN^-1(0.6)
> lines from the point load radiating to the supported sides of the slab.  For
> your case with a point load at the center, the participating width would be
> 2*0.6*11 = 13.2 feet.  Obviously, if the load is not at center span, the
> effective width will be smaller at one support than at the other since the
> slab cannot form a symmetric "saucer" shape.  I've always thought this was a
> little too optimistic for my comfort level though, and usually take no more
> than the point load width plus 4x the slab depth in width and provide extra
> reinforcing in this strip as necessary.  I'd be curious to hear how other
> people handle this situation, too >>

Others Responses:

1)  >>Joe, I don't understand the formula, is part of it missing?  
>>Tom VanDorpe, P.E.
>>VCA Engineers

2)  >>We encountered an engineer (old S.E.) who was promoting that approach
a few
>>years ago (trying to justify placing a point load on an existing slab on
>>ground).  We couldn't accept it or rationalize it.
>>Neil Moore, S.E.

3)  >>that too much (four feet wide )in my first respons I said only the
point load
>>width  plus the the slab depth
>>or for max 2x slab depth  not more than that if i find the steel is too much
>>but the other queston you consider punching shear or not and what the area of
>>punching shear?

My Replys:

1)  Sorry Tom, I scratched out this message in a hurry.  The method as I
understand it, is to take two lines radiating from the point load at an
included angle of tan^-1(0.6).  When the lines hit the supported edge of the
slab, you will have drawn a triangle.  The length of the triangle along the
supported edge is the effective width of the slab.  If the point load is 11
feet from the edge of the slab, the (opposite side/adjacent side) = 0.60
lead to the length of "opposite side" = 0.6 * 11 feet = 6.6 feet.  There are
two of these triangles since two lines radiate from the point load to the
supported edge, so the effective width is 2*(6.6 feet) = 13.2 feet.

2)  I don't believe the method is intended for slab-on-grade, but rather for
elevated one-way slabs.  I understand the rationalization for the method as
an admission that the slab will form a saucer shape under a point load, even
if we call the slab a "one-way" slab (good reason to have some transverse
steel in one way slabs!).  The stresses distribution is obviously
complicated, and the method claims to be a simple way of ensuring enough
flexural capacity exists in the slab to allow "saucering" as required.  Hey,
if you want a simple method, it's going to be simple-minded!  The authors
(wish I knew who they were) claim that both tests and analysis will confirm
this rough approximation.  I don't have access to either.

3)  I agree that 13.2 feet is too wide for my comfort level for a 22 foot
span, but I don't want to get **too** conservative by ignoring the fact that
a one-way slab won' just bend like an isolated beam across supports...the
load will distribute itself somewhat.  I simply mention a method that an
older engineer learned when he was in school long ago.  Obviously, theress a
wide range of opinions from like-educated engineers (what's new).  If truly
accurate information is critical, do some testing or FEA, or better yet both!