instead of a simple n*n square matrix as for the heat transfer problem. As we have axisymmetry, our 3D problem is reduced to a 2D one, but there are still some 3D “remains”. We have indeed an important difference between our problem and stressplane strain usual problems: there are four strains instead of three because, even though the displacement uθ is identically zero, the strain εθ is not zero because it is caused by radial displacement [247]:
{}
r {} = = z rz
{ } ∂ ur ∂r ur r ∂ uz ∂z
∂ ur ∂ u z ∂z ∂r
In finite element formulation, the strain vector, which is the derivative of the displacement, is defined using the B derivative matrix: {ε} = [B] {u} Stiffness matrix Since we have four strain components, the strain interpolation matrix B is now a 4x6 array. Hence we may write the matrix relating strains and displacements as follows:
[
∂N1 ∂r N1 [ B]= r 0 ∂N1 ∂z
0 0 ∂N1 ∂z ∂N1 ∂r
∂ N2 ∂r N2 r 0 ∂ N2 ∂z
∂N3 ∂r N3 r
0 0 ∂N2 ∂z ∂N2 ∂r
0 ∂N3 ∂z
0 0 ∂N3 ∂z ∂N3 ∂r
]
giving:
[
b1 2 N 1 1 [ B]= r 2 0 c1
0 0 c1 b1
b2 2 N 2 r 0 c2
0 0 c2 b2
b3 2 N 3 r 0 c3
0 0 c3 b3
27
]
[248]