Editing Linear elasticity (section)

====Displacement formulation====
In this case, the displacements are prescribed everywhere in the boundary. In this approach, the strains and stresses are eliminated from the formulation, leaving the displacements as the unknowns to be solved for in the governing equations.
First, the strain-displacement equations are substituted into the constitutive equations (Hooke's law), eliminating the strains as unknowns:
<math display="block">\sigma_{ij} = \lambda \delta_{ij} \varepsilon_{kk}+2\mu\varepsilon_{ij}
= \lambda\delta_{ij}u_{k,k}+\mu\left(u_{i,j}+u_{j,i}\right).
</math>
Differentiating (assuming <math>\lambda</math> and <math>\mu</math> are spatially uniform) yields:
<math display="block">\sigma_{ij,j} = \lambda u_{k,ki}+\mu\left(u_{i,jj}+u_{j,ij}\right).</math>
Substituting into the equilibrium equation yields:
<math display="block">\lambda u_{k,ki}+\mu\left(u_{i,jj} + u_{j,ij}\right) + F_i = 0</math>
or (replacing double (dummy) (=summation) indices k,k by j,j and interchanging indices, ij to, ji after the, by virtue of [[Symmetry of second derivatives|Schwarz' theorem]])
<math display="block">\mu u_{i,jj} + (\mu+\lambda) u_{j,ji} + F_i = 0</math>
where <math>\lambda</math> and <math>\mu</math> are [[Lamé parameters]].
In this way, the only unknowns left are the displacements, hence the name for this formulation. The governing equations obtained in this manner are called the ''elastostatic equations'', the special case of the steady '''Navier–Cauchy equations''' given below.

{{math proof
| title = Derivation of steady Navier–Cauchy equations in Engineering notation
| proof = First, the <math>x</math>-direction will be considered. Substituting the strain-displacement equations into the equilibrium equation in the <math>x</math>-direction we have
<math display="block">\sigma_x = 2 \mu \varepsilon_x + \lambda(\varepsilon_x + \varepsilon_y +\varepsilon_z) = 2 \mu \frac{\partial u_x}{\partial x} + \lambda \left(\frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z}\right)</math>
<math display="block">\tau_{xy} = \mu\gamma_{xy} = \mu\left(\frac{\partial u_x}{\partial y} + \frac{\partial u_y}{\partial x}\right)</math>
<math display="block">\tau_{xz} = \mu\gamma_{zx} = \mu\left(\frac{\partial u_z}{\partial x} + \frac{\partial u_x}{\partial z}\right)</math>

Then substituting these equations into the equilibrium equation in the <math>x\,\!</math>-direction we have
<math display="block">\frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} + F_x = 0</math>
<math display="block">\frac{\partial}{\partial x}\left( 2\mu\frac{\partial u_x}{\partial x}+ \lambda \left(\frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y}+ \frac{\partial u_z}{\partial z}\right)\right) + \mu\frac{\partial}{\partial y} \left(\frac{\partial u_x}{\partial y}+ \frac{\partial u_y}{\partial x}\right)+ \mu\frac{\partial}{\partial z} \left(\frac{\partial u_z}{\partial x} + \frac{\partial u_x}{\partial z}\right) +F_x=0</math>

Using the assumption that <math>\mu</math> and <math>\lambda</math> are constant we can rearrange and get:
<math display="block">\left(\lambda+\mu\right)\frac{\partial}{\partial x} \left(\frac{\partial u_x}{\partial x} +\frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z}\right)+\mu \left(\frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2}+ \frac{\partial^2 u_x}{\partial z^2}\right) + F_x= 0</math>

Following the same procedure for the <math>y\,\!</math>-direction and <math>z\,\!</math>-direction we have
<math display="block">\left(\lambda + \mu\right) \frac{\partial}{\partial y} \left(\frac{\partial u_x}{\partial x} +\frac{\partial u_y}{\partial y} +\frac{\partial u_z}{\partial z}\right)+\mu\left(\frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} + \frac{\partial^2 u_y}{\partial z^2}\right) + F_y = 0</math>
<math display="block">\left(\lambda+\mu\right) \frac{\partial}{\partial z} \left(\frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z}\right)+ \mu \left(\frac{\partial^2 u_z}{\partial x^2} + \frac{\partial^2 u_z}{\partial y^2} + \frac{\partial^2 u_z}{\partial z^2}\right) + F_z=0</math>

These last 3 equations are the steady Navier–Cauchy equations, which can be also expressed in vector notation as
<math display="block">(\lambda+\mu) \nabla(\nabla \cdot \mathbf{u}) + \mu \nabla^2\mathbf{u} + \mathbf{F} = \boldsymbol{0}</math>
}}

Once the displacement field has been calculated, the displacements can be replaced into the strain-displacement equations to solve for strains, which later are used in the constitutive equations to solve for stresses.

===== The biharmonic equation =====
The elastostatic equation may be written:
<math display="block">(\alpha^2-\beta^2) u_{j,ij} + \beta^2 u_{i,mm} = -F_i.</math>

Taking the [[divergence]] of both sides of the elastostatic equation and assuming the body forces has zero divergence (homogeneous in domain) (<math>F_{i,i}=0\,\!</math>) we have
<math display="block">(\alpha^2-\beta^2) u_{j,iij} + \beta^2u_{i,imm} = 0.</math>

Noting that summed indices need not match, and that the partial derivatives commute, the two differential terms are seen to be the same and we have: <math display="block">\alpha^2 u_{j,iij} = 0</math> from which we conclude that: <math display="block">u_{j,iij} = 0.</math>

Taking the [[Laplacian]] of both sides of the elastostatic equation, and assuming in addition <math>F_{i,kk}=0\,\!</math>, we have
<math display="block">(\alpha^2-\beta^2) u_{j,kkij} + \beta^2u_{i,kkmm} = 0.</math>

From the divergence equation, the first term on the left is zero (Note: again, the summed indices need not match) and we have:
<math display="block">\beta^2 u_{i,kkmm} = 0</math>
from which we conclude that:
<math display="block">u_{i,kkmm} = 0</math>
or, in coordinate free notation <math>\nabla^4 \mathbf{u} = 0</math> which is just the [[biharmonic equation]] in <math>\mathbf{u}\,\!</math>.