Laterally Unsupported Plate Bending Resistance in Strong Direction

[learning2banengineer]

Hi,

In a situation like the sketch attached, how would we determine the thickness of the plate? It does not fall into a particular situation as listed in table 4-2 in S16.
Thanks in advance.

 

[BAretired]

It is not a very good detail.

I would check shear and bending on the plate. For factored flexural compressive resistance, I would use Table 4-4 with effective length = 2L +b where be is the flange width of the supported beam.

If there are any horizontal loads parallel to the supported beam, the plate must be designed for bending and the column for torsion...messy detail.

BA

 

[BAretired]

I meant b, not be is the flange width.

BA

 

[msquared48]

The single knife plate bothers me here. I would go with two plates of the same or greater thickness mounted to the side of the tube section, two thru-bolts for erection only, and field welding of the two plates to the tube. The plates could be shop welded to the WF beam.

The kicker on the WF is a good idea for the torsion to the beam.

Mike McCann
MMC Engineering

 

[JStephen]

The latest AISC specification does include allowable stresses for flat bars bent the hard way.

 

[JoshPlumSE]

As JStephen mentions, section F11 from AISC 13th (or 14th edition) contains these formulas.

Mn = Cb*(1.52-0.274*(L*d/t^2)*Fy/E)*My < Mp
or
Mn = Fcr*Sx = 1.9E*Cb/(L*d/t^2)*Sx

 

[rb1957]

consider the compression side as a flat plate buckling under a direction compression load. for me the allowable works out at something like .4*pi^2*E/(12*(1-v^2))*(t/b)^2 = 0.36*E*(t/b)^2. for b, i'd assume a distance (30t?), get an allowable, get the bending stress at the mid point, get an MS; then again for 1/2 the effective width (the allowable and the average bending stresses will both increase, but does the MS decrease ?)