Composite Beams

 

*Example:  (continued)

                  

4.      Next calculate the moment of inertia of the entire “transformed” section

about its neutral axis, y.  Use the parallel axis theorem.

 

   Iyy  =    (1/12) b1h13 + b1h1( c  ˗ h1/2)2  +  (1/12) b2h23  +  (b2h2) ( h1 + h2 /2 ˗ c )2   

 

Iyy  =  (1/12)(80)(0.53) + (80)(0.5)(1.315 – 0.25)2  + (1/12)(4)(5.53) + (4)(5.5)(3.25 – 1.315)2

 

So for the “transformed” section    Iyy  =  184 in4  

 

Now the maximum bending stress can be calculated using  σ  =  Mc/I .  So we need to calculate

the maximum bending moment by constructing the shear and bending moment diagrams

as shown below.

                    

 

The bending stresses in terms of the base material (wood) are as follows:

 

   Let                c1  =  c ,         c2  =  h1 + h2 ˗ c ,           c3  =  c  ˗  h1

 

  At the top of the composite beam  σ  =  Mc2/Iyy =  – (4950)(12)(6-1.315) / 184  = – 1513 psi

  At the bottom of the composite beam σ  =  Mc1/Iyy =  (4950)(12)(1.315) / 184 =  424 psi

  The maximum tensile bending stress in the wood is: 

                             σ  =  Mc3/Iyy =  (4950)(12)(1.315 – 0.5)/184  =  263 psi

 

  The maximum tensile bending stress in the steel is   424(Es/Ew)  =  424(20)  =  8480 psi

 

                                                                                                                           (result)

 

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