Bending Members (Beams)

 

Example:  (continued)

                                   

The moment of inertia of each part about their centroidal axis is:

 

                          Iz1z1 = (1/12)bh3      and     I z2z2 = (1/12)hb3

 

Next use the parallel axis theorem to transfer these moments of inertia to the centroidal

(neutral) axis of the composite section.

 

For part 1:   IzNA1 = inertia of part 1 about the centroidal axis of the composite section is:

 

                          Iz1z1 = (1/12)bh3    +  bh(D  – h/2)2

 

For part 2:   IzNA2 = inertia of part 2 about the centroidal axis of the composite section is:  

 

                         Iz2z2 = (1/12)hb3    +  bh(b/2 + h – D)2

 

Then the moment of inertia of the composite section about its neutral axis is the sum.

 

        IzNA  =  IzNA1  ,  IzNA2  =  = (1/12)bh3    +  bh(D  – h/2)2  +  (1/12)hb3    +  bh(b/2 + h – D)2

 

For the given data:  b = 20 mm, h = 80 mm,  D = 65 mm

                                                                                                                

    IzNA  =  2.906 x 106  mm4  =  moment of inertia of composite section about neutral axis

 

For the top fiber:  σtop  =  MCtop / IzNA    =  10000(35) / 2.906 x 106    =  0.120 N/mm2  = σmax

 

For the bottom fiber:  σbottom  =  MCtop / IzNA   = 10000(65) / 2.906 x 106  =  − 0.224 N/mm2 = σmin

                   

Top fiber has the maximum tensile stress =  0.120 N/mm2       (result)

 

Bottom fiber has the minimum compressive stress =  − 0.224 N/mm2    (result)

 


Return to Notes on Solid Mechanics

Copyright © 2019 Richard C. Coddington
All rights reserved.