It is necessary to determine the critical buckling load of a column to ensure that it can safely carry the loads it will experience while in service.Â This is important to protect live safety and the structure functionality.Â Column that have fixed bases with free (unrestrained or unsupported) tops which are also slender are a common occurrence in residential and commercial structures that we see everyday.

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Calculating the critical buckling load of a column with a fixed base and free top is slightly different than that of a column pinned at both ends but both are based on much of the same input.Â To calculate both, you need to the cross sectional area to determine the moment of inertia of the section with respect to the x and y axis, the elastic modulus of the material, and the unbraced length of the column between supports.Â The difference however is that the fixed base, free top column calculation uses an additional factor, K, to determine the effective column length.Â This K factor is equivalent to 2 and is based on the deflected shape of the fixed base free top column and is used as a multiplier times the actual length of this column to determine effective column length in terms of the length of a column that is pinned at both ends.Â For a column pinned at both ends, K = 1, but is often neglected for simplicity.

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The critical buckling load for a column fixed at its base and free at the top is calculated as follows.

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

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In this formulation the elastic modulus is E, the moment of inertia Â of the section is I, K is the factor for the effective column length and l_{C} is the actual length of the column between supports.

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Itâ€™s important to note that in the case where we have a column that does not have the same bracing and unsupported lengths with respect to buckling about the x and y axes and is not symmetric to both then each axis of buckling must be checked to determine the critical buckling load.

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An example will help to illustrate how this calculation is performed.

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Letâ€™s say we have a column that will be fixed at its base and free at the top that is 20 ft long and it is a W18 x 50.Â The moment of inertia with respect to weak axis buckling (y-axis) is I_{y} = 40.1 in^{4 }and with respect to strong axis buckling (x-axis) is I_{x} = 800 in^{4}.Â The steel has an elastic modulus of 29, 000 ksi.Â The column is not braced in any direction.Â Calculate the critical buckling load for the column.

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Since the bracing conditions are similar for each axis of buckling, we can use the minimum moment of inertia to determine the buckling load.

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

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E = 29,000 ksi

I_{y} = 40.1 in^{4}

K = 2

l_{C} = 20 ft = 240 in.

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P_{CR,F} = [(Ï€^{2})(29,000 ksi)(40.1 in^{4})] / [(2)(240 in.)]^{2 }

P_{CR,F} = 49.8 kips = 49,800 lbs

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REFERENCE:

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