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# RE: One way slab design with point load.

• To: "'seaoc(--nospam--at)seaoc.org'" <seaoc(--nospam--at)seaoc.org>
• Subject: RE: One way slab design with point load.
• From: James Hagensen <JHAGENSEN(--nospam--at)HNTB.com>
• Date: Mon, 29 Sep 1997 19:01:43 -0500

```A.A.S.H.T.O. Standard Specifications for Highway Bridges is a great
reference related to distribution widths for loads to slabs.  I have the
14th edition (1989) which is not the latest edition (I do very little
bridge work).  Section 3.24 deals with distribution of loads and design
of concrete slabs.  It gives a distribution width for wheel loads, E, of
(4 + 0.06S) where S is the span length of the slab in feet.  It also
limits the distribution width to a maximum of 7.0 feet.  This is under
the heading of Main Reinforcement Parallel to Traffic.  I have found in
running the numbers that the condition for Main Reinforcement
Perpendicular to traffic (I don't have the time to explain the details
on how this width is calculated now) is less conservative and will give
you a wider distribution width.  The A.A.S.H.T.O code also addresses
requirements for distribution reinforcement (perpendicular to main
reinforcement) which also has to be met.

Hope this helps.

> -----Original Message-----
> From:	Joe McCormick [SMTP:jmccormick(--nospam--at)proaxis.com]
> Sent:	Monday, September 29, 1997 11:17 AM
> To:	seaoc(--nospam--at)seaoc.org
> Subject:	Re: One way slab design with point load.
>
>
>
> >I have a photocopy out of an old concrete design book a senior
> engineer
> >gave
> > 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
> >>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?
> >>thanks
> >>D.A.
>
>
> 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
> 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
>
> 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!
>
>
>
>
>

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