Solid objects deform under the action of applied forces: a point in the solid, originally at \((x,y,z)\) will come to \((X,Y,Z)\) after some time; the vector \(\mathbf{u}=(u_1,u_2,u_3) = (X-x, Y-y, Z-z)\) is called the displacement. When the displacement is small and the solid is elastic, Hooke's law gives a relationship between the stress tensor \(\sigma(u)=(\sigma_{ij}(u) )\) and the strain tensor \(\epsilon(u)=\epsilon_{ij}(u)\) $$ \sigma_{ij}(u) = \lambda \delta_{ij} \nabla.\mathbf{u}+ 2\mu\epsilon_{ij}(u), $$ where the Kronecker symbol \(\delta_{ij} = 1\) if \(i=j\), \(0\) otherwise, with $$\epsilon_{ij}(u) = {1\over 2}({\partial u_i\over\partial x_j} + {\partial u_j\over\partial x_i} ), $$ and where \(\lambda, \mu\) are two constants that describe the mechanical properties of the solid, and are themselves related to the better known constants \(E\), Young's modulus, and \(\nu\), Poisson's ratio: $$ \mu = {E\over 2( 1+\nu)}, \quad \lambda = {E\nu\over (1+\nu)(1-2\nu)}. $$

Lamé's system

Let us consider a beam with axis \(Oz\) and with perpendicular section \(\Omega\). The components along \(x\) and \(y\) of the strain \({ u}(x)\) in a section \(\Omega\) subject to forces \({ f}\) perpendicular to the axis are governed by $$ -\mu \Delta { u} - (\mu+\lambda) \nabla (\nabla .{ u})={ f} \hbox{in} \Omega, $$

where \(\lambda ,\mu \) are the Lamé coefficients introduced above.

Remark, we do not used this equation because the associated variationnal form does not give the right boundary condition, we simply use $$ - div( \sigma ) = \mathbf{f} \quad \mbox{in} \Omega $$ where the corresponding variationnal form is: $$ \int_{\Omega} \sigma(u) : \epsilon(\mathbf{v})\;dx - \int_{\Omega} \mathbf{v} f \;dx =0; $$ where \(:\) denote the tensor scalar product, i.e. \( a: b = \sum_{i,j} a_{ij}b_{ij}\).

So the variationnal form can be written as : $$ \int_{\Omega} \lambda \nabla.u \nabla.v + 2 \mu \epsilon(\mathbf{u}):\epsilon(\mathbf{v}) \; dx - \int_{\Omega} \mathbf{v} f \;dx =0; $$

Example Consider elastic plate with the undeformed rectangle shape

\([0,20]\times [-1,1]\). The body force is the gravity force \(\vec f\) and the boundary force \(\vec g\) is zero on lower, upper and right sides.

The left vertical sides of the beam is fixed. The boundary conditions are

\begin{eqnarray*}

\sigma . n &=& g = 0 \hbox{on} \Gamma_1, \Gamma_4, \Gamma_3, u &=& \mathbf{0} \hbox{on} \Gamma_2 \end{eqnarray*}

Here \({ u}=(u,v) \) has two components.\bigskip

The above two equations are strongly coupled by their mixed
derivatives, and thus any iterative solution on each of the
components is risky. One should rather use `FreeFem++`

's system
approach and write:

// file lame.edp mesh Th=square(10,10,[20*x,2*y-1]); fespace Vh(Th,P2); Vh u,v,uu,vv; real sqrt2=sqrt(2.); macro epsilon(u1,u2) [dx(u1),dy(u2),(dy(u1)+dx(u2))/sqrt2] // EOM //the sqrt2 is because we want: epsilon(u1,u2)'* epsilon(v1,v2) $== \epsilon(\bm{u}): \epsilon(\bm{v})$ macro div(u,v) ( dx(u)+dy(v) ) // EOM real E = 21e5, nu = 0.28, mu= E/(2*(1+nu)); real lambda = E*nu/((1+nu)*(1-2*nu)), f = -1; // solve lame([u,v],[uu,vv])= int2d(Th)( lambda*div(u,v)*div(uu,vv) +2.*mu*( epsilon(u,v)'*epsilon(uu,vv) ) ) - int2d(Th)(f*vv) + on(4,u=0,v=0); real coef=100; plot([u,v],wait=1,ps="lamevect.eps",coef=coef); mesh th1 = movemesh(Th, [x+u*coef, y+v*coef]); plot(th1,wait=1,ps="lamedeform.eps"); real dxmin = u[].min; real dymin = v[].min; cout << " - dep. max x = "<< dxmin<< " y=" << dymin << endl; cout << " dep. (20,0) = " << u(20,0) << " " << v(20,0) << endl;

The numerical results are shown on figure [figlame] and the output is:

-- square mesh : nb vertices =121 , nb triangles = 200 , nb boundary edges 40 -- Solve : min -0.00174137 max 0.00174105 min -0.0263154 max 1.47016e-29 - dep. max x = -0.00174137 y=-0.0263154 dep. (20,0) = -1.8096e-07 -0.0263154 times: compile 0.010219s, execution 1.5827s

[hbtp]

[figlame] Solution of Lamé's equations for elasticity for a 2D beam deflected by its own weight and clamped by its left vertical side; result are shown with a amplification factor equal to 100. \em Remark: the size of the arrow is automatically bound, but the color gives the real length