Inverted Tripod as a cantilevered column

[slickdeals]

Folks,
I am analyzing an inverted tripod as a column (you gotta love the architects). The column is significantly large.

However, this column is also subjected to moments in both directions. In one direction, this column is a cantilever, while in the other direction, it is part of a moment frame.

How would one design the three legs of the tripod? How does one calculate the effective length factors?

It is a cantilever in one direction, so K=2.0? Is that a conservative way of estimating? Are there accurate ways to determine the K factor for analysis?

 

[Lion06]

First, how are the three legs attached at the top? I am picturing three legs springing from the ground that get farther away as it goes up.
In addition to looking at the column as a whole, you will need to make sure that none of the three legs buckle before the global section buckles.

Also, if there are shears and moments, what provisions are being made to transfer shear between the three to make it act compositely? Are you even attempting to make this happen?

 

[slickdeals]

The 3 legs form a cradle to support a box girder. One of the legs stops short and the other 2 legs extend higher. A thick diaphragm will connect across the 2 extending columns, and the box girder will bear on the lower column.
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The legs go up about 40'. Yes, the idea is to make a diaphragm on the top connecting the three legs together. This helps a better load distribution (almost keeps the load on the centroid of the tripod).

 

[kslee1000]

For true cantilever (tip are free to translate and rotate), the theoretical k value is 2.0, but 2.1 is suggested. Please read nto steel design text books to find methods to determine proper K values when framing is involved.

Other than that, I think the design can be carried out by any of the conventional methods (computer modelling/analysis is more practical though).

Watch out for OT of the footing.

 

[slickdeals]

I must mention that this is a rather huge concrete column. I see it the way StrEIT sees it. I have 3 columns on a tall pedestal which forms a tall single column.

 

[kslee1000]

I could be missing something, but still don't see any significant problems (other than time consuming).

I evision a big old (non-composite) bold tree, it has a huge trunk with 3 diverging braches that supporting some structures above, which is/are linked to the branches by eith moment connections or else. Analytically, I would first assume the branches are fixed at the top of the trunk, then each branch will get its own share of loads, derived from typical frame analysis, or static method. Now the loads will pass down to the top of the trunk, calculate the resultant forces as usual using static method, here you go. (Provide aboundant/closely spaced shear reinforcement at the root of the branches, and the top of the trunk, for which stress induced spliting is very likely to occur)

Still, watch out for global stability, especially in areas with high wind, seismic events. Settlement could be a big problem too.

 

[4ntunes]

The coefficient "k" which defines the length of buckling of the bar can be obtained through the relationship between the rigidity of the bars that join the column. First you need to assess whether the structure is moved or not. Then, simply relate the stiffness of the bars with their length using the following formula: G = (IV1/L1 + Iv2/L2 +...)/( Ih3/l3 + Ih4/l4 +...). Where "I" is the inertia of the bar seen and "l" the length. On the value of G can enter the chart in the IACS, the section that discusses design of columns, and get the value of "k". If you model this structure in a program of structural analysis such as SAP2000, get active efforts can easily tailor your column. But, as observed by you, the coefficient "k" is important and needs to be assessed in this case really well.
Sorry for my English mistakes.

 

[4ntunes]

I have not seen your third post and so I thought it was a steel column.Disregard my comments above.
Following the description of the structure you made, I understand that two of the tripod bars belong to the same plan and are linked directly to the upper beam, while the third is a cantilever and it make part of another plan. With this description, I understand that only two legs of the tripod are receiving the load of the beam and therefore should be designed to flexo compression. In that case you will design a column in "Y" form. If you put more details, perhaps I can help better.

4ntunes

 

[kslee1000]

4ntunes has very good grab on this topic. Listen on.

 

[4ntunes]

kslee1000,

Sorry, but I dont understand what you mean with the word "grab". Could you explain? Thanks.

 

[kslee1000]

"Handle", or "understanding". Anytime, English is not my native language too.

 

[rb1957]

grab = grasp ?

 

[kslee1000]

rb: Where you want to "fry=fly" us to?

 

[slickdeals]

Please see attached sketch for the column.
The box girder provides considerable rotational restraint to the tripod column at the top and I think a fixed-free case may be conservative.

In the direction parallel to the span, the system acts like a moment frame. I am looking for some guidance regarding the computation of the "k" value for use in concrete frame design.

 

[BAretired]

The box girder is free to move laterally at the top, so it offers no restraint to the tripod column in the direction normal to the girder. Fixed/free is the correct assumption for the direction normal to the girder unless the girder is laterally restrained by supports not shown on your sketch.

In the other direction, it is a frame which can be resolved by ordinary methods of structural analysis.

BA

 

[ishvaaag]

You can summarily dispatch the evaluation of K by using some kind of "advanced" design. You model all linear members in segments modeling properly the connections. Oblige a P-Delta calculation with the reduced stiffness consistent with the kind of load and its level. A satisfactory structure for such analysis (small and acceptable deformations, acceptable stresses) is stable against the loads at the laterally deformed state, hence you can assume amobility, i.e., laterally prevented displacements at the nodes of the dividing segments at such instance. For such condition and using the length of every segment for the checks, K from non-sway cantilever and always less than 1 for the in-member lateral buckling P-small delta checks. Use as well reduced stiffeness as above.

In short, if you use P-Delta for your analysis, you may check with the length of the dividing segments and K=1, or if wanting more precision, K from non-sway chart. I wouldn't use less than 4 segments or so per straight part except if short. The method is also nice in that the length for the buckling checks entirely coincides with that of the segments used in the model.

 

[dik]

You have an inverted 'tetrahedron' where the edges are the compression members. K should be 1 or less. You have to resolve the forces and account for the forces at the connection of the various elements. The system functions if all points of support are pins.

 

[slickdeals]

@Dik,
Are you suggesting that each piece of the branch can be analyzed as a pin-ended member with K=1.0? What happens for the trunk?

I am having a hard time visualizing the buckling in this column. I must mention though that this column is huge (it has an area of about 40000 in2 at the base and about 22000 in2 at the point where it separates into 3 legs).

 

[dik]

The trunk portion would then be considered as being rigidly connected to the 'point' and the overall effect would be treated as a regular column, albeit, with a variable cross-section to accommodate the 'tetrahedron'.

 

[slickdeals]

If I understand correctly, I would consider the trunk to be a column with an effective length K of 2 and the branches with a effective length of 1.0 and every member modified with an unbraced length factor such as to have to have a total unbraced height of 135'