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MECHANICS - EXAMPLE

    Example


Shelf Supports and Loading
 

A beam is constructed where one end cannot deflect (pinned joint), but can rotate, and the other end cannot rotate, but can deflect..

Determine the deflection equation by integrating from the loading function.

   
    Solution

   

Starting with the loading function, the deflection can be found by integrating it four times. However, this will require four unique boundary conditions. Those four conditions are

  1. θ(x=L) = v'(x=L) = 0
  2. v(x=0) = 0
  3. M(x=0) = v''(x=0) = 0
  4. V(x=L) = v'''(x=L) = 0
     


Equivalent beam deflections using superposition principle

 

The loading is a constant distributed load, or

     EI v'''' = -w

where 'w' is acting downward. Integration this gives the shear function,

     V(x) = EIv''' = -wx + C1

The 4th boundary condition can be used to determine the integration constant, C1,

     V(x=L) = -wL + C1 = 0
     C1 = wL

Integrating again gives the moment equation,

     M(x) = EIv'' = -wx2/2 + xwL + C2

The 3rd boundary conditions gives,

     0 = -w 02/2 + x 0 L + C2
     C2 = 0

Thus, the final moment equation is

     M(x) = -wx2/2 + xwL

Next, this equation can be integrated to give the rotation equation,

     θ(x) = EIv' = -wx3/6 + wLx2/2 + C3

Using boundary condition 1 gives,

     0 = -wL3/6 + wL3/2 + C3
     C3 = -wL3/3

The final rotation equation is,

     θ(x) = EIv' = -wx3/6 + wLx2/2 - wL3/3

The deflection equation is just the integral of the rotation equation,

     EIv = -wx4/24 + wLx3/6 - wL3x/3 + C4

Applying the last unused boundary conditions, number 2, gives,

     0 = - 0 + 0 - 0 + C4 ==> C4 = 0

The final deflection equation is

     v = -w (x 4/8 - Lx3/2 + xL3) / (3EI)

     
   
 
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