**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

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

INPUT PROPERTIES SUMMARY | ||

Parameter | Value | |

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 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.

INPUT PARAMETERS | ||

Parameter | Value | |

Diameter [d] | 50 | mm |

Length [L] | 2000 | |

Density [p] | 7.87 | g/cm^3 |

RESULTS | ||

Parameter | Value | |

Cross section area [A] | 1963.495 | mm^2 |

Mass [M] | 30.905 | kg |

Second moment of area [I_{xx}] |
306796.156 | mm^4 |

Second moment of area [I_{yy}] |
306796.156 | |

Polar second moment of area [J] | 613592.312 | |

Section modulus [S_{xx}] |
12271.847 | mm^3 |

Section modulus [S_{yy}] |
12271.847 | |

Radius of gyration [r_{x}] |
12.5 | mm |

Radius of gyration [r_{y}] |
12.5 | |

CoG distance in x direction [x_{cog}] |
25 | mm |

CoG distance in y direction [y_{cog}] |
25 |

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

INPUT PARAMETERS | ||

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 [n_{d}]** |
5 | |

Modulus of Elasticity [E] | 205 | GPa |

Yield strength [S_{y}] |
345 | MPa |

Compressive yield strength [S_{yc}]* |
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 |

RESULTS | ||

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 [S_{eff}] |
160 | |

Critical load for failure [F_{c}] |
155.18 | kN |

Allowable load for given parameters. This value includes design factor F_{c}/F_{a} =n_{d} = 5 |
||

Allowable load(includes n_{d}) [F_{a}] |
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.

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