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steel PL over RC slab

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The question that you ask is a typical problem in Strength of Materials text 
books, at least it was when I went to school and the last time I taught 
Strength (more than 20 years ago).  It was used to introduce students to 
deformation compatibility as an introduction to solving indeterminate 
structures by the method of consistent deformations.

The procedure is to equate deformation of the steel plate *and* the 
deformation of the concrete slab.  In the situation that you describe, the 
load is applied to the steel plate.  An unknown is the force between the 
plate and the slab.

A FBD of the steel plate has the applied load acting down, and an unknown 
distributed load acting up.  A FBD of the concrete slab has the unknown 
distributed load acting down.  The deflection of the steel plate and the 
deflection of the concrete slab has to be equal: deflection of steel plate 
due to applied load - deflection of steel plate due to unknown distributed 
load = deflection of concrete slab due to unknown distributed load.  Solve 
for unknown distributed load.  No FEM or computer or calculator required.  
Slide rule helps (G).


A. Roger Turk, P.E.(Structural)
Tucson, Arizona

Andrew D. Kester wrote:

. > Quick problem, somewhat theoretical:

. > You have a reinforced slab that spans between two beams, in a ribbed slab
. > type of construction. Next place a steel plate continuous over two or more
. > of the spans. The steel plate is not bonded or bolted, just laying on top 
. > of the floor. Next pass a moving load from a wheel over this plate. How 
. > do you calculate the portion of the load carried by the steel plate and 
. > the slab? I don't have any FEA software either. I think there is a 
. > somewhat simple solution involving the various EI's of the two materials. 
. > In my head I see the steel deflecting under the load, carrying some of 
. > the load, then passing the remainder into the slab, until it starts 
. > deflecting, then so and so forth until some type of equilibrium is 
. > reached.

. > Also is there a way to : a) size the plate so that it carries a specific
. > portion of the load
. > b) the slab carries no load at all, the plate carries all of it

. > I think b) is nearly impossible because you would have to have such a 
. > thick plate that it virtually had zero deflection. Noone could transport 
. > such a theoretical plate :)

. > Sorry if this is confusing, I am a bit confused but I am sure one of you
. > will clear it up...

. > Thanks in advance,

. > Andrew Kester, EI
. > Longwood, FL

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