k-factor with no restraining girder 4

[Lion06]

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.

 

[slickdeals]

If I understand your question correctly, you have restraining girders at level 2, and the column cantilevers up from L2 - L3 with it being restrained from translation in one direction.

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?

 

[Lion06]

yes, but the column is truly pinned at each level, it is not rotationally fixed to anything

 

[271828]

Is it heavily loaded? The only option I see is to say that the bottom level is the girder that restrains this top level. Then you can use the alignment chart with GA not equal to infinity and GB = infinity. This is invalid if the column has significant axial load,though. You should also make the adjustment in the top Note 2 on Page 16.1-242.

How I hate the ELM. Someday, it'll be Good Riddance!

 

[Lion06]

There is no girder at L2, though, it is a true friction-free pin.

 

[271828]

I meant that the column from the ground to L2 is, in effect, your girder at the bottom of the cantilever.

 

[frv]

Maybe you could model this in RISA or some other analysis program that accounts for second order effects. Then back out a k factor to determine whether it seems reasonable. It would certainly be higher than 2.

 

[msquared48]

Wny not just be conservative and design it as pinned-pinned. That's what I am hearing here. It seems you are making the problem more complicated than it really is.

Mike McCann
MMC Engineering

 

[Lion06]

Mike-
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?

 

[msquared48]

OK.

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

 

[BAretired]

Let h1 and h2 be the height from L1 to L2 and L2 to L3 respectively. If you apply a moment at L2, the rotation at L1 will be mL/6EI and the rotation at L2 will be mL/3EI. The rotation is zero at 0.577 h1 from L1 or 0.423h1 from L2.

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

 

[hokie66]

If the column is pinned at L2, how can it cantilever up to L3? Sounds unstable to me.

 

[slickdeals]

I guess this is when you post a 3D screenshot...

 

[Lion06]

hokie-
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.

 

[kslee1000]

This case is rare, never had experience on it, but let me try to clarify something here, assuming the final goal is to find Pe on the column betwee L1 & L2.

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.

 

[BAretired]

The attached sketch shows a similar column which may be a little more intuitive. A hinge support at L1 and roller support at L2 are the only supports. Level L0 has been added to make the structure symmetric. Force P is applied at L0 and L3 as shown.

The effective length of the column in this case is L = 2h2 + h1.

BA

 

[ishvaaag]

I suggest a more modern approach would be to calculate a second order elastic solution, with nodes at the levels, or shorter segments; if the axial loads produce average stress higher than 0.5 Fy correct the stiffness and repeat. In these conditions, since the effects of non-despicable lateral movement (P-Delta) are already accounted for in the analysis itself (shown in the returned forces) you can use a K-factor between NODES from Non-Sway abaqus, always 1 or less.

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.

 

[jike]

EIT:

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!

 

[GalileoG]

BAretired,

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.

 

[BAretired]

Thanks, Clansman.

BA