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How To Calculate The Critical Buckling Load In A Column Pinned At Both Ends

By Edited Jul 23, 2016 0 0

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.  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 column’s cross section and the strength of the column material.

 

Here we will consider a column that is slender and narrow with a pinned connection at its base and it’s top and conforms to the qualities of an idealized column in Euler buckling.  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.  For a pinned-pinned column with no internal bracing the critical buckling loads can be determined for the following situations:

 

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

 

PCR,PP,1 = [(Ï€2)EI] / lC2

 

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.

 

PCR,PP,2 = [4(Ï€2)EI] / lC2

 

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

 

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.  For example if a pinned-pinned column is braced at ¼ points, this means that the column is effectively ¼ of the total length and lC is ¼ the value of what it would have been if otherwise unbraced.

 

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.  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. 

 

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

 

There is a 20 ft column that is pinned at both ends and is a W24 x 94.  In the direction of weak axis buckling (y-axis), Iy = 109 in4 and the column is braced at 10 ft.  In the direction of strong axis buckling (x-axis), Ix = 2,700 in4 and is not braced internally.  The Elastic modulus of the steel is 29,000 ksi.  Determine the critical buckling load of this column.  Assume this column buckles in mode 1.

 

Weak Axis Buckling:

 

PCR,PP,1,y = [(Ï€2)EI] / lC2

E = 29,000 ksi

Iy = 109 in4

lC = 10 ft  = 120 in.

 

PCR,PP,1,y = [(Ï€2)( 29,000 ksi)( 109 in4)] / (120 in.)2

PCR,PP,1,y = 2,167 kips = 2,167,000 lbs

 

Strong Axis Buckling:

 

PCR,PP,1,x = [(Ï€2)EI] / lC2

E = 29,000 ksi

Ix = 2,700 in4

lC = 20 ft  = 240 in.

 

PCR,PP,1,y = [(Ï€2)( 29,000 ksi)( 2,700 in4)] / (240 in.)2

PCR,PP,1,y = 13,417 kips = 13,417,000 lbs

 

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

 

 

REFERENCE:

Gere, James M. (2004).  Mechanics of Materials (6th ed.).  USA: Brooks/Cole – Thomson Learning.

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