k-factor with no restraining girder
k-factor with no restraining girder
(OP)
I am stumbling on calc'ing a k factor for an unusual column. It goes from ground to L3. It is restrained from translation in both directions at ground, and L2, but cantilevers up to L3 (and is only restrained against translation in one direction at L3. I can't seem to calc a k factor for the column because there are no restraining girders. Even though we would typically assume a pinned connection is a pin, the rotational stiffness of the column from ground to L2 does offer some rotational restraint at L2 for the cantilevered portion above, but because there is no girder restraining at that level, G goes to infinity, and, as a result, k goes to infinity. This can't be the first time someone has dealt with a column like this. How do you handle it?
Before anyone asks, it is completely architecturally driven and I don't have an opportunity to add a translational restraint in the other direction at L3.
Before anyone asks, it is completely architecturally driven and I don't have an opportunity to add a translational restraint in the other direction at L3.






RE: k-factor with no restraining girder
So in one direction, you have a restrained-restrained condition and in the other you have a restrained-free end?
Am I understanding this right?
RE: k-factor with no restraining girder
RE: k-factor with no restraining girder
How I hate the ELM. Someday, it'll be Good Riddance!
RE: k-factor with no restraining girder
RE: k-factor with no restraining girder
RE: k-factor with no restraining girder
RE: k-factor with no restraining girder
Mike McCann
MMC Engineering
RE: k-factor with no restraining girder
pinned-pinned has k=1. I think my k is much larger than 1. Likely larger than 2.
frv-
How will an analysis program let me back out a k?
RE: k-factor with no restraining girder
A thought... put a subsdtantial vertical load on the column with a lateral load, say 10% of the vertical load, at the tip of the cantilever, and see where the inflection point is. Calculate your worst "K" using this approach. I think this can be done in RISA with proper modeling, but use 3D.
Mike McCann
MMC Engineering
RE: k-factor with no restraining girder
The column, in the direction under consideration has an effective length of 2*(0.422h1 + h2).
If h1 = h2 = h, the effective length of the column is 2.845h, so you could say that k = 2.846.
Otherwise, k = 2 + 0.844h1/h2.
BA
RE: k-factor with no restraining girder
RE: k-factor with no restraining girder
RE: k-factor with no restraining girder
It's a continuous column, pinned at it's base, pinned at L2, then cantilevering up to L3. It's the same as a pin-pin beam with a cantilever.
I'll post a sketch asap.
BA-
Excellent analysis. Star for that.
RE: k-factor with no restraining girder
1. For column ends restrained against translation, but without moment restrain: "Zero moment restraint at ends constitutes the weakest situation for compression members when translation of one end relative to the other is prevented...known as the effective length, equals the actual length, i.e., K = 1.0".
"The distance between the points of inflection...is the effective or equivalent pinned length for the column".
(Quotes taken from Salmon & Johnson, 2nd, p.277)
2. If your concern is instability caused by combined axial compression and moment (from cantilever above), a review on Ch. 12 of Salmon & Johnson's book will bring better understanding.
RE: k-factor with no restraining girder
The effective length of the column in this case is L = 2h2 + h1.
BA
RE: k-factor with no restraining girder
That is, you take account of material nonlinearity checking if the correction of the stiffness is required (and making it if required, of course), you account for geometrical (sway) nonlinearity directly by the P-Delta analysis, and you account of geometrical in-member (P-delta) nonlinearity by use of the non-sway K-factor abaqus.
The stiffness correction can be found in the AISC Engineering Journal vol 32-1; you may use also some formulation of the steel tangent modulus, or derive the recommended stiffness reduction from the Fcr LRFD statement. The SSRC publicaiton "Is your structure suitably braced?" has also something about in p. 161, derived form Baker, 1991.
RE: k-factor with no restraining girder
Draw the deflected shape of the member. From the top of the column to where the deflected shape becomes vertical equals X. Then KL = 2X.
This is similar to what we do with a cantilever and a fixed base.
I hope this helps!
RE: k-factor with no restraining girder
That was a great post. Thank you. Star for you.
Clansman
If a builder has built a house for a man and has not made his work sound, and the house which he has built has fallen down and so caused the death of the householder, that builder shall be put to death." Code of Hammurabi, c.2040 B.C.
RE: k-factor with no restraining girder
BA