Column Buckling Example of a Solid 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 Buckling Example

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

Parameter Value
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 : 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 3 : Calculate cross-sectional area and second moment of area by using the values summarized in step 1.

Sectional properties of solid circle (shaft)
Parameter Value
Diameter [d] 50 mm
Length [L] 2000
Density [p] 7.87 g/cm^3

Parameter Value
Cross section area [A] 1963.495 mm^2
Mass [M] 30.905 kg
Second moment of area [Ixx] 306796.156 mm^4
Second moment of area [Iyy] 306796.156
Polar second moment of area [J] 613592.312
Section modulus [Sxx] 12271.847 mm^3
Section modulus [Syy] 12271.847
Radius of gyration [rx] 12.5 mm
Radius of gyration [ry] 12.5
CoG distance in x direction [xcog] 25 mm
CoG distance in y direction [ycog] 25

Step 4 : Go to "Column Buckling Calculator" page to calculate maximum allowable load by using parameters given in step 1 and step 3.

Parameter Value
Column length [L] 2 m
Cross-sectional area [A] 1963.495 mm^2
Second moment of area [I] 306796.15 mm^4
Distance to the neutral axis [c] 25 mm
Loading is concentric and e=0
Eccentricity [e] 0 mm
Design factor [nd]** 5  
Modulus of Elasticity [E] 205 GPa
Yield strength [Sy] 345 MPa
Compressive yield strength [Syc]* 345
Expected load shall be entered. For this example, applied force is not known so the force has been taken as 1000 N.
Applied Force [P] 1000 N
End conditions Pinned - Pinned
Parameter Value
Effective length constant [C] * 1
Radius of gyration of column [r] 12.5 mm
Slenderness ratio of column [S] 160
Effective slenderness ratio of column [Seff] 160
Critical load for failure [Fc] 155.18 kN
Allowable load for given parameters. This value includes design factor Fc/Fa =nd = 5
Allowable load(includes nd) [Fa] 31.04 kN
Factor of safety [fos]** 155.18
According to input parameters, column is classified as long column with central loading. Euler column equations are used for the solution.
Column Category Long columns with central loading

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.