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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

##### Solution:

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

INPUT PROPERTIES SUMMARY |

Parameter |
Symbol |
Value |
Unit |

Diameter of column |
d |
50 |
mm |

Length of Column |
L |
2 |
m |

Design factor |
n_{d} |
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/cm^{3} |

## 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.

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Summary

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