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{ FREE_PLATE.PDE  

 

 This example considers the bending of a thin rectangular plate under a

 distributed transverse load.

 

 For small displacements, the deflection U is described by the Biharmonic

 equation of plate flexure

       del2(del2(U)) + Q/D  =  0

 where

       Q is the load distribution,

       D = E*h^3/(12*(1-nu^2))

       E is Young's Modulus

       nu is Poisson's ratio

 and   h is the plate thickness.

 

 The boundary conditions to be imposed depend on the way in which the

 plate is mounted.  Here we consider the case of a simply supported

 boundary, for which the correct conditions are

       U = 0

       Mn = 0

 where Mn is the tangential component of the bending moment, which in turn

 is related to the curvature of the plate. An approximation to the second

 boundary condition is then

       del2(U) = 0.

 

 FlexPDE cannot directly solve the fourth order equation, but if we

 define V = del2(U), then the deflection equation becomes

       del2(U) = V

       del2(V) + Q = 0

 with the boundary conditions

       U = 0

       V = 0.

 

 The particular problem addressed here is a plate of 16-gauge steel,

 8 x 11.2 inches, covering a vacuum chamber, with atmospheric pressure

 loading the plate.  The edges are simply supported.  Solutions to this

 problem are readily available, for example in Roark's Formulas for Stress

 & Strain, from which the maximum deflection is Umax =  0.746, as compared

 with the FlexPDE result of 0.750.

 

 (See FIXED_PLATE.PDE for the solution with a clamped edge.)

 

 Note: Care must be exercised when extending this formulation to more complex

   problems.  In particular, in the equation del2(U) = V, V acts as a source

   in the boundary-value equation for U.  Imposing a value boundary condition

   on U does not enforce V = del2(U).

 

}  

 

Title " Plate Bending - simple support "  

 

Select  

   ngrid=10                   { increase initial gridding }  

   cubic           { Use Cubic Basis }  

 

Variables  

    U(0.1)  

    V(0.1)  

 

Definitions  

   xslab = 11.2  

   yslab = 8  

   h = 0.0598 {16 ga}  

   L = 1.0e6  

   E = 29e6  

   Q = 14.7  

   nu = .3  

   D = E*h^3/(12*(1-nu^2))  

 

Plots  

  contour (U) as "Displacement"  

  elevation(U) from (0,yslab/2) to (xslab,yslab/2) as "Displacement"  

  surface(U) as "Displacement"  

 

End