In Euler's equasion is the critical buckling load the force that will initiate a bend in the column, or is it the force that leads to catastophic failure.

If we start with the initial assumption that there is some lateral force due to system imperfections to initiate the buckling. It seems an infinitely small axial force causes an infinitely small bend, and a moderate axial force causes a moderate bend that is stable and the member returns to its original shape when the force is removed. Does a coulmn "bend" before it "buckles"? Has buckling occured if the column springs back after the load is removed, or have we not yet reached the critical buckling load?

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 Recognitions: Gold Member Homework Help Science Advisor Hi JayPee, welcome to PF. Buckling is sometimes used to describe the catastrophic failure phenomenon, and sometimes the preceding occurrence of large, nonlinear lateral deformation of some part of the structure. (Small, linear lateral deformations, however, do not constitute buckling.) If the load is sustained, buckling will generally lead to failure; however, it's also possible for a material to buckle without failing if the accompanying compression relieves the load. Does this answer your questions?
 I have a small problem with a column buckling experiment I've conducted. The rig consisted of an aluminium member tested in compression using an Instron loading machine. Strain gauges were placed on both sides of the member and a computer recorded values of strain at various compressive loads between 0 and 600N. The properties of the member are listed below: length: 500mm width: 20mm thickness: 4.29mm Young's modulus: 52.4 GPa The member is actually pin-ended, but for all intents and purposes, it can be considered as fixed at both ends because the joints are quite rigid. The Euler buckling load is therefore $$P = \frac{4\pi^2EI}{l^2}$$ as far as i understand. The value of P that I get from this equation is 1015.0 N. Having calculated the theoretical buckling load of the member, I need to check this against the actual buckling load. All I have is Hooke's law ($$P = EA\epsilon$$) and a load of different values for strain. I have used Hooke's law to calculate a value of P for each value of strain, but I don't see how this helps... Basically, I need the critical buckling load, but I don't know how to calculate it from experimental values of strain. Can anyone help? Thanks Rhys

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Isn't the highest value of P before buckling the critical buckling load?

 Well yes, but I only have a very small amount of data, so I can't determine exactly when the member starts to buckle. I know I have to use Hooke's law, but I don't know how to implement it... Is there a graph I can plot that would clearly show when the member buckles?
 Recognitions: Gold Member Homework Help Science Advisor If you don't have any type of data (from the loading machine or the strain gauges) corresponding to the buckling point, I'm not sure how you'd determine the critical buckling load. You could calculate a minimum buckling load, I suppose, from the last data point, but it would be impossible to say how close it is.
 Here's all the data I have: (SG1 and SG2 are the strains recorded on each side of the beam) Compressive Load (kN)..SG1 (µStrain)......SG2 (µStrain) 0..............................-0.04374............-0.63108 0.047........................-8.88919............-11.52479 0.106........................-13.40135...........-33.98217 0.15..........................-17.01924...........-49.0698 0.207.........................-19.19852..........-73.82884 0.245.........................-19.89741..........-89.66715 0.294.........................-19.77105..........-112.4289 0.346.........................-14.35199..........-141.75526 0.401.........................-10.95378..........-168.79844 0.444.........................-12.06381..........-195.79951 0.502..........................15.95878...........-248.28922 0.542..........................75.93387...........-328.41915 0.599..........................219.3245...........-496.27437 Is it possible to determine the buckling load from this?
 How did the geometry of the member change during loading? I would say that buckling occurs at around 0.5 kN because the opposing strains become opposite in sign. One side of the beam experiences compressive strains while the other side experiences tensile strains. Someone with more hands-on experience with strain gauges can probably give you a better answer.

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 Quote by rdt24 Here's all the data I have: (SG1 and SG2 are the strains recorded on each side of the beam) Compressive Load (kN)..SG1 (µStrain)......SG2 (µStrain) 0..............................-0.04374............-0.63108 0.047........................-8.88919............-11.52479 0.106........................-13.40135...........-33.98217 0.15..........................-17.01924...........-49.0698 0.207.........................-19.19852..........-73.82884 0.245.........................-19.89741..........-89.66715 0.294.........................-19.77105..........-112.4289 0.346.........................-14.35199..........-141.75526 0.401.........................-10.95378..........-168.79844 0.444.........................-12.06381..........-195.79951 0.502..........................15.95878...........-248.28922 0.542..........................75.93387...........-328.41915 0.599..........................219.3245...........-496.27437 Is it possible to determine the buckling load from this?
Take a look at the plot of these points and see what you can infer.

CS
Attached Files
 Buckling load.pdf (103.4 KB, 90 views)

 Recognitions: Homework Help Science Advisor rdt24: From the data plotted by stewartcs, I think we can see a transition at 520 N, very close to the value predicted by Brian_C. I don't know how you computed P = 1015 N (in post 3), because your posted formula gives 1089 N. Regardless, it appears your current effective length factor (k = 0.50) might be inaccurate, and should instead perhaps be k = 0.723. Therefore, using k = 0.723, the buckling load becomes Pcr = E*I*[pi/(k*L)]^2 = (52 400)(131.59)[pi/(0.723*500)]^2 = 521 N.

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