The critical buckling load of a column is important for the design and analysis purposes of a structural column to ensure it is structurally sound and safe.Â The critical buckling load of a column can be considered the load at which a column fails and is a function of the height (unbraced length of the column), the way the column is braced at support locations, the moment of inertia of the columnâ€™s cross section and the strength of the column material.

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Here we will consider a column that is slender and narrow with a pinned connection at its base and itâ€™s top and conforms to the qualities of an idealized column in Euler buckling.Â A column that is pinned at both ends is simply one in which at both ends of the column there is a connection that allows some rotation, but no vertical or horizontal movement under loading.Â For a pinned-pinned column with no internal bracing the critical buckling loads can be determined for the following situations:

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Mode 1 Buckling: This occurs when the column deflects and bends in one orientation creating a bow shape.Â This is the most common mode case that is considered.

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P_{CR,PP,1} = [(Ï€^{2})EI] / l_{C}^{2 }

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Mode 2 Buckling: This occurs when the column deflects and bends in two orientations in its top and bottom segments creating what appears as a single cycle of a sine curve.

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P_{CR,PP,2} = [4(Ï€^{2})EI] / l_{C}^{2 }

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Where E is the Modulus of Elasticity, I is the cross sectional moment of inertia and l_{C} is the unbraced or unsupported length of the column.

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When we have the situation where the column is braced internally, we take this in to account in our formulation as this is a major factor when determining our critical buckling load.Â For example if a pinned-pinned column is braced at Â¼ points, this means that the column is effectively Â¼ of the total length and l_{C }is Â¼ the value of what it would have been if otherwise unbraced.

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It should be noted that for a pinned-pinned column or any other column for that matter, we must look at the two principal axes about which the deflection and buckling occurs, the x and y axis.Â When determining the critical buckling load for the column, unless the column cross section and internal bracing is symmetric about each axis, a separate calculation must be performed with respect to each of these axes and the minimum value taken as the true buckling load.Â

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We will see an example of how to perform this calculation.

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There is a 20 ft column that is pinned at both ends and is a W24 x 94.Â In the direction of weak axis buckling (y-axis), I_{y} = 109 in^{4} and the column is braced at 10 ft.Â In the direction of strong axis buckling (x-axis), I_{x} = 2,700 in^{4 }and is not braced internally.Â The Elastic modulus of the steel is 29,000 ksi.Â Determine the critical buckling load of this column.Â Assume this column buckles in mode 1.

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Weak Axis Buckling:

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P_{CR,PP,1,y} = [(Ï€^{2})EI] / l_{C}^{2 }

E = 29,000 ksi

I_{y} = 109 in^{4}

l_{C} = 10 ftÂ = 120 in.

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P_{CR,PP,1,y} = [(Ï€^{2})( 29,000 ksi)( 109 in^{4})] / (120 in.)^{2 }

P_{CR,PP,1,y }= 2,167 kips = 2,167,000 lbs

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Strong Axis Buckling:

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P_{CR,PP,1,x} = [(Ï€^{2})EI] / l_{C}^{2 }

E = 29,000 ksi

I_{x} = 2,700 in^{4}

l_{C} = 20 ftÂ = 240 in.

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P_{CR,PP,1,y} = [(Ï€^{2})( 29,000 ksi)( 2,700 in^{4})] / (240 in.)^{2 }

P_{CR,PP,1,y }= 13,417 kips = 13,417,000 lbs

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So the weak axis buckling controls and the critical buckling load of the column is 2,167 kips.

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REFERENCE:

Gere, James M. (2004).Â Mechanics of Materials (6^{th} ed.).Â USA: Brooks/Cole â€“ Thomson Learning.