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checker plate floor loading

checker plate floor loading

checker plate floor loading

Existing 1/4" thick steel checker plate floor is under scrutiny.  I'm having difficulty finding a method to determine the maximum allowable concentrated load.  A typical plate is rectangular and simply supported on four sides, is removable so not fastened to supports.  The load is gravitational so is perpendicular to the plane of the plate.  The plates vary in size but an example would be 3 feet by 4 feet.  Any thoughts?

RE: checker plate floor loading

To truly calculate it, you could do a FEA.

Personally, I'd use the "Floor Plate Bending Capacity" tables in the Steel manual, and calculate the equivalent point load from the moment the uniform load would create.
Page 2-145 in the 9th Ed.
Pages 3-154 & 155 in the 13th Ed.

Hope it helps

RE: checker plate floor loading

I agree with ChipB... Also, in Table 3-18a, you have the moment of intertia per one inch strip. You can calculate your maximum allowable deflection... I'm sure "what it looks like" is more of a concern than the actual yielding anyway.

RE: checker plate floor loading

Roark's formulas for stress and strain. 6th ed. Table 26 Case 1b.

a=4 b =3.  b/a = 1.33   Beta = 0.74  Alpha = 0.0093

Max def = alpha x W x b x b / ( E x t x t x t)

Max stress = (3W /(2 x pi x t x t)) x ((1 + poison ratio) x
ln(2b/pi x r'o) + Beta))

r'o = sqrt (1.6 ro x ro + t x t).
ro = radius of load contact

RE: checker plate floor loading


My 5th Edition Table 25 shows:

alpha=.1478 at a/b=1.2
alpha=.1621 at a/b=1.4

RE: checker plate floor loading

I think to get Roark's formula to work, you'll need dimensions in inches (to match PSI).

If I remember correctly, don't the building codes define what a "concentrated load" is?  It's not a point load, it's something like 2' square (ie, space a person occupies) and that will make a sizable difference in your case.  And that would put you closer to formula 1c rather than 1b in Roark's book.

RE: checker plate floor loading

My calcs show that the method of finding an equivalent concentrated load from the distributed load tables in the steel manual is very conservative in comparison to the result given by Roark's formula from Table 26 case 1b.  The resulting max point load per Roark is about twice that of the first method.  The plate's support on four edges apparently makes the difference.  The first method assumes support on only two edges.  Thank you all for your help!

RE: checker plate floor loading

Side note:  Table 3-18a in Steel Manual gives moment of inertia per one foot of width, not one inch - a minor correction to the contribution by brane23.

RE: checker plate floor loading

ASCE 7-05, 4.3- Concentrated load is over a 2.5' square area.

RE: checker plate floor loading

I have ASCE 7-02 which reads "Unless otherwise specified, the indicated concentration shall be assumed to be uniformly distributed over an area 2.5 ft square and shall be located so as to produce the maximum load effects in the structural members."  My particular situation falls under "otherwise specified".  It is, however, good to know for other cases.

RE: checker plate floor loading

The alpha figures I posted earlier were, in fact, for Table 26, case 1b.  Interpolating, I get alpha=.158, which gives 0.46" deflection under a 1 kip load.

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