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Max Point Load on Steel Angle

Max Point Load on Steel Angle

Max Point Load on Steel Angle

(OP)
Hello All -

I'm doing some studying for the PE exam in only 4 short days.   There is one aspect of a practice problem I'm working on that is not makng sense to me.  The problem is: one leg of a A36 L6x6x3/8x6" steel angle is fastened to a concrete wall with an expansion anchor.  The other leg supports a point load (4" away from the face of the concrete wall). Assume the concrete is sufficient and assume the expansion anchor is sufficient.  Find the max point load P that can be supported.

The ASD solution goes about treating the supporting angle leg as a thin beam.  

S=1/6*6*0.375^2=0.141 ksi
Fb=0.75*36=27 ksi
Mr=0.141*27=3.8 in-K
Pmax=Mr/e=3.8/4=1.0 K (answer)

I follow the solution entirely except for one detail.  When the calculate the Allowable bending stress Fb, they use a factor of 0.75.  Where is this 0.75 coming from?  In my mind, I think the factor should be 0.66?  Any insight on this problem is greatly appreciated!  Thanks!  

RE: Max Point Load on Steel Angle

Is this an old review manual?  The 0.75 is the old green book factor for bending of flat plates?  You also wouldn't use S with the new 13th manual, you would use Z.

Finally, using an e of 4" would be conservative for checking the angle.  You can typically be pretty conservative using e = 4" - the thickness of the angle.   

RE: Max Point Load on Steel Angle

Yes - weak axis (non-LTB failure mode) bending is 0.75Fy
 

RE: Max Point Load on Steel Angle

That's the way I would do it...but then I'm old.  BTW, S is in^3, not ksi.

RE: Max Point Load on Steel Angle

hokie66

That's the way I just did it yesterday.... but I'm just lazy. I didn't feel like looking in the 13th edition manual.

RE: Max Point Load on Steel Angle

I'm relatively young, and that's the way I'd do it. I've always used 0.75 for plate bending and analogies thereof.
 
You'll probably find that your PE test is practical like this problem for the most part.  

RE: Max Point Load on Steel Angle

(OP)
Thanks guys.  This problem is form the latest and greatest edition of the NCEES sample exam (printed in 2010).  

Anyone know where in the steel manual the 0.75 factor is coming from?  I've looked through Chapter F in the specifications and can not find 0.75 listed in any of the cases?

RE: Max Point Load on Steel Angle

I have no idea where the 0.75 is coming from with the 13th edition.  I would think that using S (instead of Z), it would use 0.9Fy, not 0.75 Fy.

I get 0.9 from (1/1.67)*1.5 = (1/omega)*(shape factor=Z/S=1.5 for rectangular section) = (1/1.67)*1.5 = 0.9Fy

RE: Max Point Load on Steel Angle

Lion06 is describing exactly what the link I posted describes.

I'd move on, study for other parts of the test.  

RE: Max Point Load on Steel Angle

You could probably also subtract the thickness and 3/8" (or k (the fillet if I remember correctly)) from the 4 inches according to Blodget and also Salmon.

EIT

RE: Max Point Load on Steel Angle

The 0.75 factor comes from the 9th Edition manual spec. F2.1

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