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Triangular Element for Plate Bending - Part 1

Summary

Describes the implementation details of a simple triangular element for thin plate bending

Displacement

For bending of thin plate, we only consider the bending action resulted from forces normal to the plate. In-plane forces are not considered. In this case, the strains and stresses are uniquely described by the vertical displacement, w (assume the plate lies horizontally). w can be expressed with the help of Area Coordinates.

w = β1L1 + β2L2 + β3L3
      + β4(L1²L2 + ½L1L2L3) + β5(L1²L3 + ½L1L2L3)
      + β6(L2²L3 + ½L1L2L3) + β7(L2²L1 + ½L1L2L3)
      + β8(L3²L1 + ½L1L2L3) + β9(L3²L2 + ½L1L2L3)

At any point, the displacement vector is

{d} =   w
θx
θy
  =     w
 ∂w/∂y
-∂w/∂x

Remember from Area Coordinates that

L1
L2
L3
= a1
a2
a3
b1
b2
b3
c1
c2
c3
1
x
y

Let
       Lcy = ∂(½L1L2L3)/∂y = ½(c1L2L3 + L1c2L3 + L1L2c3)

   ∂w/∂y = β1c1 + β2c2 + β3c3
            + β4(2L1c1L2 + L12c2 + Lcy) + β5(2L1c1L3 + L12c3 + Lcy)
            + β6(2L2c2L3 + L22c3 + Lcy) + β7(2L2c2L1 + L22c1 + Lcy)
            + β8(2L3c3L1 + L32c1 + Lcy) + β9(2L3c3L2 + L32c2 + Lcy)

Similarly, let

       Lcx = ∂(½L1L2L3)/∂x = ½(b1L2L3 + L1b2L3 + L1L2b3)

   ∂w/∂x = β1b1 + β2b2 + β3b3
            + β4(2L1b1L2 + L12b2 + Lcx) + β5(2L1b1L3 + L12b3 + Lcx)
            + β6(2L2b2L3 + L22b3 + Lcx) + β7(2L2b2L1 + L22b1 + Lcx)
            + β8(2L3b3L1 + L32b1 + Lcx) + β9(2L3b3L2 + L32b2 + Lcx)

Now, let us evaluate displacement vector {d} at each node. The area coordinates,  (L1,L2,L3) at each node are

 P1 = (1, 0, 0)
 P2 = (0, 1, 0)
 P3 = (0, 0, 1)

We immediately conclude that

   Lcy = Lcx = 0 at all nodes

Substitute the values of (L1,L2,L3) into the expressions for w, θx, θy, we have,

 w1 = β1
θx1 = (∂w/∂y)1 =  β1c1 + β2c2 + β3c3 + β4c2 + β5c3
θy1 =-(∂w/∂x)1 = -β1b1 - β2b2 - β3b3 - β4b2 - β5b3

 w2 = β2
θx2 = (∂w/∂y)2 =  β1c1 + β2c2 + β3c3 + β6c3 + β7c1
θy2 =-(∂w/∂x)2 = -β1b1 - β2b2 - β3b3 - β6b3 - β7b1

 w3 = β3
θx3 = (∂w/∂y)3 =  β1c1 + β2c2 + β3c3 + β8c1 + β9c2
θy3 =-(∂w/∂x)3 = -β1b1 - β2b2 - β3b3 - β8b1 - β9b2

Write this into matrix form,

w1
θx1
θy1
w2
θx2
θy2
w3
θx3
θy3
=
100 000 000
c1c2c3c2c30000
-b1-b2-b3-b2-b30000
010000000
c1c2c300c3c100
-b1-b2-b300-b3-b100
001000000
c1c2c30000c1 c2
-b1-b2-b30000-b1 -b2
β1
β2
β3
β4
β5
β6
β7
β8
β9

or        {d} = [C]{β}

Solve the equations, we have

{β} = [CI] {d}   where [CI] = [C]-1


Part 2 - Strain, stress and moment

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