Sample Problem : Column Buckling Calculations for a Solid Round Bar

Specify maximum load that can be carried by a round bar with a diameter of 50 mm and length of 2 m. Use design factor of 5 and consider the ends as pinned. The material for the column is selected as ASTM A588 with following properties:

Material properties: Yield Strength: 345 MPa , Modulus of Elasticity: 205 GPa , Density: 7.87 g/cc

Compression analysis of a round column


Step 1 : Write down input parameters (including material properties) which are defined in the sample example.

Parameter Symbol Value Unit
Diameter of column d 50 mm
Length of Column L 2 m
Design factor nd 5 ---
End Condition Pinned-Pinned
Yield Strength (A588 Steel) Sy 345 MPa
Elastic modulus(A588 Steel) E 205 GPa
Density(A204 Steel) ρ 7.87 g/cm3

Step 2 : Go to "Column Buckling and Compression Member Design"  page to calculate maximum shear stress on the shaft.

Step 3 : Cross sectional properties of round column are needed for compression calculations.   To calculate cross sectional properties, go to "Sectional Properties Calculator"  page  and select "Round Solid Bar".

Step 4 : Calculate cross-sectional area and second moment of area by using the values summarized in step 1.


Step 5 : Go back to "Compression Member Design"  page and calculate maximum allowable load by using parameters given in step 1 and step 4.

Allowable load (maximum load) that can be carried by the column is calculated as 31.04 kN. This is the answer of the sample example.


The problem is completely solved with calculators which are summarized as follows.

Calculator Usage
Column Buckling and Compression Member Design To calculate column allowable load for given parameters. In this example, column is classified as long column and Euler equations (Buckling) have been used.
Sectional Properties Calculator To calculate sectional properties of standard profiles. In this example, solid circle has been calculated.