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Buckling Analysis of a Column

Title

Euler Buckling Analysis of a pin-ended column

Description A column with both ends pinned has to be checked for buckling instability

i) Find out the buckling mode shapes,

ii) Find the critical buckling compressive load on the column

Assume Column to be an I-section i.e. a Universal Column UC 203x203x60. The

material of the column is S355 as per Eurocode. Height of the column, L= 5 m.

Vertical Load acting on the column, P= 1000 kN.

Structural geometry and analysis model

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Finite Element Modelling:

Analysis Type: 2-D static analysis (X-Z plane)

Step 1: Go to File>New Project and then go to File>Save to save the project with any name

Step 2: Go to Model>Structure Type to set the analysis mode to 2D (X-Z plane)

Unit System: kN,m

Step 3: Go to Tools>Unit System and change the units to kN and m.

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Geometry generation:

Step 4: Go to Model>Structure Wizard>Column and enter Distance as 0.50 m and Repeat

times = 10. Press Add. This will create a column of height 5 m which is composed of 10

finite elements. A column should be divided into several finite elements before performing

the buckling analysis in the software. Select pin as the boundary condition. Enter Material

and Section ID as 1. Click OK. Switch to the Front view by clicking on as shown below.

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Material: S355 (Eurocode). E=2.1x108 kN/m2

Step 7: Go to Model>Properties>Material>Add. Select Steel in the Type of Design and

select Standard as EN(05)S. Select DB as S355. Click OK and Close.

Section Property: Section for the column is UC 203x203x60

Step 8: Go to Model>Properties>Section>Add. Select DB/User tab and select I-section

type. Enter Section DB as BS04-93. Enter Section types as UC 203x203x60. You can take a

note of the section properties i.e. Moment of inertia, Iyy= 6.13x10-5 m4 by clicking on Show

Calculation results. Click OK and Close.

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You can see the column section with the section shape in the model view now. Also take a

look at the works tree to see the information related to the model

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Boundary Condition: Pinned Supports at top and bottom of column. From the Column

Wizard in Step 4, we automatically defined a pinned support at the base of the column. Now

we will define a pinned support at the top.

Step 9: Use select single to select or highlight the topmost node as shown in figure below.

Go to Model>Boundaries>Supports and check on Dx and Apply.

Load Case:

Concentrated Loads:

-Vertically downward load of 1000 kN is applied on the top of column applied in (-Z) dir.

Step 10: Go to Load>Static Load Cases and define static load case ‘P’. Select Load type as

User defined for both of them. Click Close after adding the load case.

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Step 11: Go to Load>Element Beam Loads and select load case P. Select Load type as

Concentrated Force. Select the topmost element using and enter x=1 (marks the end of

the element or end of column) and enter Force Value= -1000 kN in the Global Z direction.

You can display the values from the Works Tree by Right clicking on Element Beam Loads

and click Display.

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Buckling Analysis:

Step 12: Go to Analysis>Buckling Analysis Control. Enter Number modes=5. Check the

Consider Axial Forces only box. Select the Load case P for the buckling load case and mark

it as variable. Click OK.

Step 13: Go to Analysis>Perform Analysis

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Results

Buckling Mode Shapes:

Step 14: Click on Results>Buckling Mode Shapes. Select Mode 1,2 and 3 one by one.

Check on Undeformed, Legend and Contour and click on Apply to see different mode shapes

graphically. You can also view multiple buckling mode shapes at the same time by clicking

on Multi Modes option.

Critical load factor for Mode 1 is 5.417 which means that the maximum compressive force

for buckling is Pcr=1000x5.417=5417 kN.

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Critical load factor for Mode 2 is 19.69 which means that the maximum compressive force

for 2nd mode buckling is Pcr=1000x19.69=19690 kN.

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Critical load factor for Mode 3 is 38.43 which means that the maximum compressive force

for 3rd mode buckling is Pcr=1000x38.43=38430 kN.

Multi Modes:

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Hand Calculations:

Euler Critical Buckling Load for Pinned ended columns:

𝑷𝒄𝒓 =𝝅𝟐𝑬𝑰

𝑳𝟐

For nth buckling mode, Euler Critical Buckling Load:

𝑷𝒄𝒓,𝒏 = 𝒏𝟐𝝅𝟐𝑬𝑰

𝑳𝟐

The higher level harmonics are only possible if columns are restrained at the appropriate

levels, e.g. mid-height point in the case of the 2nd harmonic and the third-height points in

the case of the 3rd harmonic.

In our case we do not have any intermediate restraints, hence we will focus on the first mode.

For 1st mode:

𝑷𝒄𝒓 =𝝅𝟐 × 𝟐. 𝟏 × 𝟏𝟎𝟖 × 𝟔. 𝟏𝟑 × 𝟏𝟎−𝟓

𝟓. 𝟎𝟐= 𝟓𝟎𝟖𝟐 𝒌𝑵

Comparison of Results Unit : kN,m

Mode Number Theoretical Midas Gen

Critical Buckling Load 1 5082 5417

More accurate results can be obtained in finite element analysis if the number of elements

are increased in the analysis. Generally theoretical formulas give lesser (or conservative)

values for critical buckling loads. Hence finite element analysis is very commonly used to

check critical buckling.

Reference William M.C. McKenzie, “Examples in Structural Analysis”, 1st Edition, Taylor &

Francis 2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN, 2006, 5.2.1 Example 5.3,

Page 383.