Properties of section,I beam.

What is x on the diagram?
x is the centre of gravity for Iyy.

Question:
Calculate the value of I beam :-
    a)moment of inertia/centroid/centre of gravity?
    b)Ixx of a beam.

Here the solution:   
 
         a) moment of inertia.
                i)web.!
                      Use this formula :  1/2 x (thickness) x (length^3)
                      
                                                    =1/2 x (12mm) x (400mm^3)
                                                    = 1 x 64000000mm^3
                                                    =64000000mm^4
                                   Simplified= 64x10^6 mm^4
           
                ii)Flag.!
                      Use this formula : 1/2 x (length) x (thickness^3)
                             
                                                    =1/2 x (200mm) x (24mm^3)
                                                    =50/3 mm x (13824mm^3)
                                                    =23400mm^4
                                                    =0.23x10^6 mm^4

      
          b)Ixx of a beam.

              Ixx=(inertia) + (length x thickness x centroid^2)
          
              Ixx=(0.23 x 10^6) + (200 x 24 x 212^2)
              Ixx=230000mm^4 + 215731200mm^4
              Ixx=215961200mm^4
              Ixx=215.9612 x 10^6  mm^4




Are we done?Not yet.!!


We need to combine Ixx and Iyy.Why don't we calculate Iyy.It is because it have been given in the diagram.


Combine= (Ixx Flag) + (Ixx Web)

               =(215.9612 x 10^6 mm^4) + (64 x 10^6mm^4)
               =495.92 x 10^6 mm^4

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