Predicting Beam Buckling: 5x5 vs. 6x6 with 1x1 Square Hole

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The discussion revolves around predicting which beam will buckle first: a 5x5 square cross-section or a 6x6 square cross-section with a 1x1 square hole. Key factors influencing buckling include the unsupported length, end conditions, and load eccentricities, not just the cross-sectional area. Euler's formula for critical buckling load emphasizes the importance of the second moment of area (I) in determining buckling behavior. Calculations indicate that the 6x6 beam, despite the hole, has a greater second moment of area, suggesting it may buckle after the 5x5 beam. Ultimately, the 5x5 beam is expected to buckle first based on these calculations.
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Homework Statement


Hi, guys!

I am trying to work out with beam will buckle first, either a beam with a square cross section, 5x5 or a beam with a square cross section, 6x6 with a 1x1 square removed from the middle of the section.

Homework Equations

The Attempt at a Solution



Well I'm not really sure as they have the same dimensions in a way, 5 for the 5x5 and 6-1 for the 6x6. But I am think that as a hollow cylinder is stronger than a full cylinder, the 6x6 with a hole in the middle will buckle second.

Any help would be greatly appreciated!
 
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You really can't say what will buckle first, just going by the cross section of the member. The unsupported length of the member, the end conditions, any eccentricities in the applied load all influence calculating the critical load beyond which buckling can be expected.
 
SteamKing said:
You really can't say what will buckle first, just going by the cross section of the member. The unsupported length of the member, the end conditions, any eccentricities in the applied load all influence calculating the critical load beyond which buckling can be expected.

Hi SteamKing,

Yeah I'm not given much expect those are the two cross sections of long and slender bars which are loaded in compression with a pin ended support. They are both subjected to the same force and are the same length.
 
GBA13 said:
Hi SteamKing,

Yeah I'm not given much expect those are the two cross sections of long and slender bars which are loaded in compression with a pin ended support. They are both subjected to the same force and are the same length.

The critical buckling load is given by Euler's formula:

F = π2E I / (K L2), where for pinned-pinned columns, K = 1.0

L = unsupported span
E = Young's modulus for the column material
I = second moment of area of the cross section

In calculating I, you must find the axis about which the second moment of area is a minimum. For sections which have axes of symmetry, the second moment calculated about one of those axes will have a minimum value, and this value must be used in Euler's formula.

You should be able to calculate the second moments for the two beam cross sections given in the OP and decide which figure gives the smallest second moment value.
 
correct me if I'm wrong but the 6x6 even removing the 1x1 core would outweigh the 5x5 increasing its total downward force on the span.
 
dragoneyes001 said:
correct me if I'm wrong but the 6x6 even removing the 1x1 core would outweigh the 5x5 increasing its total downward force on the span.

It's not a matter of weight, insofar as buckling is concerned. It's how the material in the cross-section is distributed around the centroid of the section.
 
So I calculated the second moment of areas and found that 6x6 one has a much greater I than the 5x5. So I think that from the formula that the 5x5 would bend first? Thanks!
 
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