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
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.
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
CALCULATION RESULTS IN Y-Y DIRECTION
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.
The problem is completely solved with calculators given below.