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I'm having difficulties with Neumann boundary conditions in Navier-Cauchy equations (a.k.a. the elastostatic equations). The trouble is that if I rotate a body then Neumann boundary condition should be satisfied with zero force.

In math language: if deformation is given by

$$u_i ~=~ a_{ij}x_j - x_i.$$

Where $a_{ij}$ is rotational matrix. Then this

$$\mu n_j ( u_{i,j} + u_{j,i}) + \lambda n_i u_{k,k} ~=~ 0 $$

(Neumann boundary condition) should hold everywhere and for any vector $n_i$ (basically it doesn't matter how the body looks like).

But if I substitute for $u_i$ I get

$$2 \mu n_j(a_{ij} - \delta_{ij}) + \lambda n_i ( a_{jj} -3 ), $$

which is not zero. Because first term rotates with $n$ and the rest two just scale $n$. So I cannot get a zero for every $n$.

Can someone see what am I doing wrong? I would be most grateful for any help.

Tom

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1 Answer 1

As far as I understand the question.

The expression $(u_{i,j}+u_{j,i})$ is basically a strain tensor.
And it works only for small deformations, so your rotation cannot be too large,

Lets select the z axis along the axis of rotation. Then we can write the matrix explicitly and try to see what happens when we consider only linear terms in the rotation angle : $$a_{ij} = \left(\begin{array}{ccc} \cos\theta&-\sin\theta&0\\ \sin\theta& \cos\theta&0\\ 0& 0&1 \end{array}\right)\simeq \left(\begin{array}{ccc} 1&-\theta&0\\ \theta& 1&0\\ 0& 0&1 \end{array}\right)+O(\theta^2)$$ So, we've got that $a_{jj}-3=0$, and that diagonal terms in $(a_{ij}-\delta_{ij})$ are equal to zero.

Therefore the part that scales $n_i$ vanishes.

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doesn't the small deformation condition mean that it works as long as Hook's law holds? –  Tom Apr 15 '12 at 10:41
    
@Tom I always thought that Hook's law holds as long as deformations are small enough... –  Kostya Apr 15 '12 at 10:46

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