To: "'seaint(--nospam--at)seaint.org'" <seaint(--nospam--at)seaint.org>
Subject: RE: effective length factor for sway moment frames? K for frame s wit h partially restrained connections?
From: "Haan, Scott M." <HaanSM(--nospam--at)ci.anchorage.ak.us>
Date: Fri, 30 Mar 2001 09:38:35 -0900
It is a capacity - limit state issue, not an analysis issue. The code used
to say if you limit drift and use a P-Delta then you could use K=1.
Theoretically for a column fixed at the top and the bottom and allowed to
sway the buckled shape would have K=1.
If the beams allow the ends of the columns to rotate then the buckled shape
would have K>1. There are approximation formulas for K so that you would not
have to use a AISC alignment chart:
K=sqrt[(1.6*Ga*Gb+4*(Ga+Gb)+7.5)/(Ga+Gb+7.5)] <==== from Tamboli "steel
design handbook" for unbraced frames. If K>1 then the columns are going to
have a lower strength / capacity and larger columns are required. Sounds
like a conspiracy.
My point was mainly that the code used to allow K=1 and now the code does
not appear to mention it, so for Anchorage at least get your alignment
Scott M Haan P.E.
Plan Review Engineer
Building Safety Division http://www.muni.org/building,
Development Services Department,
Municipality of Anchorage
phone: 907-343-8183 fax: 907-249-7399
> -----Original Message-----
> From: Scott A. Dunham, PE [SMTP:sadunham(--nospam--at)gte.net]
> Sent: Friday, March 30, 2001 7:56 AM
> To: seaint(--nospam--at)seaint.org
> Subject: Re: effective length factor for sway moment frames? K for
> frames wit h partially restrained connections?
> Being from the old school (I did my homework using a slide rule back in
> college), I guess I fail to appreciate that FEA (finite element analysis,
> assume) as the "only" way to get the correct value. Bearing in mind that
> most of the loads that we apply to the structures we design, and that the
> true material properties we use are at best guesses, what's the big deal
> about getting the "exact" number to arrive at a less that exact solution
> the problem at hand. Having spent the bulk of my career working in the
> production side rather that the design side of this business, I can tell
> for a fact that there are a lot of Engineers that can't see the forest for
> the trees.
> Sorry for the sermon, but I think Mark has a valid point. We as Engineers
> need to spend a little more time getting it right rather than getting it
> Scott A. Dunham, PE
> Dunham Engineering Services
> Dothan, AL
> ----- Original Message -----
> From: "Mark Gilligan" <MarkKGilligan(--nospam--at)compuserve.com>
> To: <seaint(--nospam--at)seaint.org>
> Sent: Friday, March 30, 2001 10:00 AM
> Subject: RE: effective length factor for sway moment frames? K for frames
> wit h partially restrained connections?
> The author of the following response implies that because we have computer
> access we should compute the K factor using the computer.
> I would like to suggest that such practices often take longer than quicker
> but less exact methods. On numerous ocassions I have seen engineers spend
> considerable time creating spread sheets to perform a calculation that
> could be done quicker by hand.
> Given limited design fees and a limited schedule how exact should our
> calculations be?
> Similarly how much effort should we spend to reduce construction costs
> when the fee is limited?
> Have computers in many instances actually increased our work load?
> >>>Theoretical results satisfy all the governing conditions. The
> theoretical K
> factor is theoretical exact. They are more reliable than the ones from
> alignment charts. Those alignment charts were introduced before computer.
> For the time being, almost every structural engineer has a computer for
> analysis. Why should we keep applying those inaccurate charts?
> When computers were not available for most structural engineers, engineers
> used some approximate methods, for example, two cycles, to analyze frame
> structures. No one uses two cycles since computer is everywhere. All the
> results are theoretically analyzed on computers. FEA of K factors is a
> Mark Gilligan