Beam Deflection by Double Integration   (continued)

 

For   0  ≤  x  ≤  L/2    The beam deflection starts with double integration of:

                             

               EI d2y1/dx2  =  3wLx/8  ˗  wx2/2

 

There will be two constants of integration.

 

One boundary condition is    y1(0)  =  0

 

 

For   L/2  ≤  x  ≤  L   The beam deflection starts with double integration of:

                     

                EI d2y2/dx2  =  wL2 /8 ˗ wLx/8 

 

 

There will be two constants of integration.

 

One boundary condition is    y2(L)  =  0

 

 

There are two conditions of continuity that must be satisfied.  They are as follows:

 

   At  x = L/2             y1(x)  =  y2(x)     and       dy1/(x)dx  =  dy2 (x)/dx

 

So the two boundary conditions and the two continuity conditions allow the

determination of the four constants of integration.

 

 

Integration gives:

 

For   0  ≤  x  ≤  L/2         EI dy1/dx  =  3wLx2/16  ˗  wx3 / 6  +  C1  

 

and                                   EI y1 (x)  =  wLx3/16  ˗  wx4 / 24  +  C1  x  +  C2 

 

 

For   L/2  ≤  x  ≤  L         EI dy2/dx  =  wL2 x / 8 ˗ wLx2 /16  +  C3      

 

and                                   EI dy2 (x)  =  wL2 x2 / 16 ˗  wLx3 /48  +  C3 x  +  C4     

 

Application of the boundary and continuity conditions yield:

 

C1   =  ˗  9 w L3 / 384,    C2   =  0,    C3   =   ˗  17 w L3 / 384,    and  C4  = w L4 / 384  

 

 

 

The resulting deflections in each section of the beam are:

 

For   0  ≤  x  ≤  L/2             y1 (x)  =  (w/ 384EI)  [ 24 Lx3  ˗ 16 x4   ˗   9L3  x ] 

 

For   L/2  ≤  x  ≤  L             y2 (x)  =   (w/ 384EI)  [ L4  + 24 L2x2  ˗   8L x3  ˗  17 L3 x ]

 


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