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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} = wθxθy = w ∂w/∂y-∂w/∂x Remember from Area Coordinates that L1L2L3 = a1a2a3 b1b2b3 c1c2c3  1xy  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 = 1 0 0 0 0 0 0 0 0 c1 c2 c3 c2 c3 0 0 0 0 -b1 -b2 -b3 -b2 -b3 0 0 0 0 0 1 0 0 0 0 0 0 0 c1 c2 c3 0 0 c3 c1 0 0 -b1 -b2 -b3 0 0 -b3 -b1 0 0 0 0 1 0 0 0 0 0 0 c1 c2 c3 0 0 0 0 c1 c2 -b1 -b2 -b3 0 0 0 0 -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