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RE: Rigid frame deflections (again)

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Kathleen:

Look at the two hinged frame.  It lends itself to easier discussion (there is no K to fool around with).

Examine the problem first strictly from mathematics.
The L is in the numerator.  The larger the L; the larger the computed lateral drift.

Now from the intuitive engineering approach.
The analysis is independent of gravity loads.  The tendency is to view the dead weight of the beam as a counteracting moment on the column.  This would give rise to the impression that the longer the beam; the greater the counteracting moment; and the supposition that the lateral drift should be less for the longer beam.  But this formula is independent of the dead weight of the beam.  If the frame were weightless, it becomes easier to visualize.  Assume that you had a weightless frame with a length of 200 feet.  The 200 ft frame would not be nearly as stiff as the same frame with a 7 foot length.  A beam of that length becomes a noodle for lateral stiffness.

I hope that this analogy adds a little insight.

Regards,
Harold Sprague
Krawinkler, Luth & Assoc.
4412 W. Eisenhower Blvd.
Loveland, CO 80537
Voice: 970 667-2426
Fax: 970 667-2493
Email: hsprague(--nospam--at)klaalov.com


> ----------
> From: 	Kathleen  A. O'Brien
> Sent: 	Wednesday, January 21, 1998 4:35 PM
> To: 	SEAOC Forum
> Subject: 	Rigid frame deflections (again)
> 
> Guys and Dolls:
> 
> Now that I have the Steel Interchange sheet from Modern Steel
> Construction
> (4/93) which shows the quick hand-calc for rigid frame defelction, I
> have
> started using it.
> 
> HOWEVER, I have noticed something very odd. Extremeley odd.
> 
> The formulas for pinned and fixed bases follow (and please excuse the
> crummy formatting):
>         
>         drift = P*H(squared)/6E  *(H/Icol + L/2Ibm)
>         where    P = load
>                 H= height of frame
>                 L=span of frame
>                 Icol= I of column
>                 Ibm= I of beam
>                 E=29000ksi
> 
> For fixed-at-base frames:
> 
>         drfit=P*H(cubed)/12EIcol * (3K+2/6K+1)
> 
>         where   K= Ibm*H/Icol*L
>                 everything else as above
> 
> Now here is the strange part. I have two frames along line D in my
> project.
> One has a 17 foot span and the other a 7 foot span (L). Both of  them
> are
> 10.5 feet high (H).  And  I am assuming:
> 
> P = 1 kip for each frame  (just getting relative rigidities at this
> point).
> Relative rigidity R is the inverse of the drift; i.e. 1/drift
> 
> For both fixed and pinned at base calcs, my 7 foot long frame is
> stiffer
> than my 17 foot one.
> 
> I have hunted HARD for errors in my calcs and I can't find any. I have
> done
> the calc both by hand and with a spreadsheet and I still get the same
> answer: the tall skinny frame is stiffer than the short fat one.
> 
> I do not buy this. Has anyone else had this problem? Is it me? Is it
> the
> calcualtion??
> 
> Kate O'Brien
> Simi Valley, CA

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