Problem 10.2. Deflection of an orthotropic plate

The deflection function of an orthotropic plate is given:
w(x,y)=Cx2y21xayb2
where C is a constant. Determine the bending and torsional moments of the plate in the functions of the plate’s stiffnesses D11, D22, D12, D66. Derive the load function acting on the plate. Give the possible supports where the deflection function satisfies the boundary conditions.

Solve Problem

Solve

Derive the load function.

Check load, p(x,y)

p(x,y)=D1124Cy2a2+D2224Cx2b2+2(D12+2D66)2C212xa1xa12yb1yb+81xyab

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Steps

Step by step

Step 1.  Perform the necessary partial derivations of the deflection function according to the variables, x and y.

Show partial derivatives

w(x,y)=Cxyx2yaxy2b2wx=2Cxyx2yaxy2by2xyay2b2wx2=2Cy22y3b+y4b6xy2a+6x2y2b+6xy3ab4wx4=24Cy2bwy=2Cxyxy2bx2yax2xybx2a2wy2=2Cx22x3a+x4a6x2yb+6x2y2a+6x3yab4wy4=24Cx2a2wyx=2Cxy26xb6yb+4x3ya3+4xy3b3+9x2y2ab3wy2=2C212xa+12x2a212yb+12y2b2+81x2y2ab

Step 2.  Give the moment functions in the function of the curvatures.

Show moments

Eq.(10-34) and Table 34.
Mx(x,y)=D112wx2D122wy2=D112Cy22y3b+y4b6xy2a+6x2y2b+6xy3abD122Cx22x3a+x4a6x2yb+6x2y2a+6x3yabMy(x,y)=D122wx2D222wy2=D122Cy22y3b+y4b6xy2a+6x2y2b+6xy3abD222Cx22x3a+x4a6x2yb+6x2y2a+6x3yabMxy(x,y)=D662wxx=D662Cxy26xb6yb+4x3ya3+4xy3b3+9x2y2ab

Step 3.  Give the load function with the partial differential equation of the orthotropic plate.

Show load

Eq.(10-37)
p(x,y)=D114wx4+D224wy4+2(D12+2D66)4wx4  p(x,y)=D1124Cy2a2+D2224Cx2b2+2(D12+2D66)2C212xa1xa12yb1yb+81xyab

Step 4.  Check boundary conditions.

Show boundary condition

At edge x = 0:

w(x=0,y)=0wxx=0=0

Thus the edge is built-in.

At edge y = 0:

w(x,y=0)=0wyy=0=0

Thus the edge is built-in.

On the other two edges none of the derivatives are zero, thus the plate must be subjected to line loads and moments or the plate is connected to other structural elements.

Results

Worked out solution

 The necessary partial derivations of the deflection function according to the variables, x and y are the following:

w(x,y)=Cxyx2yaxy2b2wx=2Cxyx2yaxy2by2xyay2b2wx2=2Cy22y3b+y4b6xy2a+6x2y2b+6xy3ab4wx4=24Cy2bwy=2Cxyxy2bx2yax2xybx2a2wy2=2Cx22x3a+x4a6x2yb+6x2y2a+6x3yab4wy4=24Cx2a2wyx=2Cxy26xb6yb+4x3ya3+4xy3b3+9x2y2ab3wy2=2C212xa+12x2a212yb+12y2b2+81x2y2ab

The moments are the functions of the curvatures:

Eq.(10-34) and Table 34.
Mx(x,y)=D112wx2D122wy2=D112Cy22y3b+y4b6xy2a+6x2y2b+6xy3abD122Cx22x3a+x4a6x2yb+6x2y2a+6x3yabMy(x,y)=D122wx2D222wy2=D122Cy22y3b+y4b6xy2a+6x2y2b+6xy3abD222Cx22x3a+x4a6x2yb+6x2y2a+6x3yabMxy(x,y)=D662wxx=D662Cxy26xb6yb+4x3ya3+4xy3b3+9x2y2ab

The load function is given by the partial differential equation of the orthotropic plate.

Eq.(10-37)
p(x,y)=D114wx4+D224wy4+2(D12+2D66)4wx4  p(x,y)=D1124Cy2a2+D2224Cx2b2+2(D12+2D66)2C212xa1xa12yb1yb+81xyab

Boundary conditions at edge x = 0:

w(x=0,y)=0wxx=0=0

Thus the edge is built-in.

Boundary conditions at edge y = 0:

w(x,y=0)=0wyy=0=0

Thus the edge is built-in.

On the other two edges none of the derivatives are zero, thus the plate must be subjected to line loads and moments or the plate is connected to other structural elements.