# 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. 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.