# ECCENTRIC COLUMN BUCKLING EXAMPLE OF AN I BEAM

Eccentric Column Buckling Example of an I Beam: A structure with mass of 110 kg will be carried by two I-beams as shown in the figure. I-beams are so located that center of mass of structure remain just at the center between two beams. The height of the I-beam is 1 meter and cross sectional dimensions of I-beam are also given in the figure. If the design factor of the I beam is taken as 3, what is the factor of safety of I- beams ? (Assume that all the mass of structure is concentrated at Cog). The material of the I-beam is ASTM A572 steel with yield strength of 290 MPa and elastic modulus of 140 GPa. I beams are fixed to ground and structure so end conditions are fixed-fixed.

### Solution of Eccentric Column Buckling Example:

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

The force due to the mass of the structure can be calculated by F=m*a. So F=110*9.81=1079.1 N. Since there are two columns, the force per column is 1079.1/2=539.55 N

 INPUT PROPERTIES SUMMARY Parameter Value Width of Column [B] 76 mm Flange Thickness [h] 7.6 mm Web Thickness [b] 4 mm Flange-flange inner face height [H] 119 mm Length of Column [L] 1 m Design factor [nd] 3 --- Eccentricity [e] 2 m Applied Force [P] 539.55 N End Condition Fixed-Fixed Yield Strength (A572 Steel) [Sy] 290 MPa Elastic modulus(A588 Steel) [E] 140 GPa Density(A204 Steel) [ρ] 7.85 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 "I Beam".

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

 INPUT PARAMETERS Parameter Value Flange-flange inner face height [H] 119 mm Width [B] 76 Flange thickness [h] 7.6 Web thickness [b] 4 Length [L] 1000 Density [p] 7.85 g/cm^3
 RESULTS Parameter Value Cross section area [A] 16.312 cm^2 Mass [M] 12.805 kg Second moment of area [Ixx] 519.604 cm^4 Second moment of area [Iyy] 55.667 Section modulus [Sxx] 77.437 cm^3 Section modulus [Syy] 14.649 Radius of gyration [rx] 56.439 mm Radius of gyration [ry] 18.473 CoG distance in x direction [xcog] 38 mm CoG distance in y direction [ycog] 67.1

Cross section area [A] and second moment of area [Ixx and Iyy] values are used in the following step.

Step 4 : Go to ""Column Buckling Calculator" "  page and calculate factor of safeties for x-x and y-y direction by using parameters given in step 1 and step 3.

CALCULATION RESULTS IN X-X DIRECTION

 INPUT PARAMETERS Parameter Value Column length [L] 1 m Cross sectional area and second moment of area values are calculated in the step 3. Cross-sectional area [A] 16.312 cm^2 Second moment of area [I] 55.667 cm^4 Distance from outer fiber to neutral axis in x-x direction. Equals to half of the width 76/2=38 mm. Distance to the neutral axis [c] 38 mm Loading is eccentic and e=2 m. Eccentricity [e] 2 m Design factor [nd]** 3 Modulus of Elasticity [E] 140 GPa Yield strength [Sy] 290 MPa Compressive yield strength [Syc]* 290 Applied force per column as calculated in the step 1. Applied Force [P] 539.55 N End conditions Fixed-Fixed

 RESULTS Parameter Value Effective length constant [C] * 0.65 Radius of gyration of column [r] 18.47 mm Slenderness ratio of column [S] 54.13 Effective slenderness ratio of column [Seff] 35.19 Critical load for failure [Fc] 2114.64 N Allowable force and factor of safety in x-x direction. Allowable load(includes nd) [Fa] 704.88 N Factor of safety [fos]** 3.92 Column Category Struts or short columns with eccentric loading

CALCULATION RESULTS IN Y-Y DIRECTION

 INPUT PARAMETERS Parameter Value Column length [L] 1 m Cross sectional area and second moment of area values are calculated in the step 3. Cross-sectional area [A] 16.312 cm^2 Second moment of area [I] 519.604 cm^4 Distance from outer fiber to neutral axis in y-y direction. Equals to half of the flange-flange outer face 119/2+7.6=67.1 mm. Distance to the neutral axis [c] 67.1 mm Loading is eccentic and e=2 m. Eccentricity [e] 2 m Design factor [nd]** 3 Modulus of Elasticity [E] 140 GPa Yield strength [Sy] 290 MPa Compressive yield strength [Syc]* 290 Applied force per column as calculated in the step 1. Applied Force [P] 539.55 N End conditions Fixed-Fixed

 RESULTS Parameter Value Effective length constant [C] * 0.65 Radius of gyration of column [r] 56.44 mm Slenderness ratio of column [S] 17.72 Effective slenderness ratio of column [Seff] 11.52 Critical load for failure [Fc] 10968.06 N Allowable force and factor of safety in y-y direction. Allowable load(includes nd) [Fa] 3656.02 Factor of safety [fos]** 20.33 Column Category Struts or short columns with eccentric loading

According to calculation results, x-x direction is more critical and factor of safety is calculated as 3.92 (minimum value). This value is larger than design factor(3). As a result, the columns can carry the structure.

### Summary

The problem is completely solved with calculators given below.

 Calculator Usage Column Buckling and Compression Member Design To calculate column factor of safety for given parameters. In this example, column is classified as strut or short column with eccentric loading and stress equations have been used. Sectional Properties Calculator To calculate sectional properties of standard profiles. In this example, I-beam has been calculated.