Difference between revisions of "MLSM and Solid Mechanics"

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(Equations)
(Steady-state solution)
 
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= Steady-state solution =
 
 
To obtain the steady state solution we have two possible strategies:
 
# Add a (linear) dampening term to the Navier equation, and simulate the dynamic behaviour until the motion ceases: \[\rho \frac{\partial^2 \b{u}}{\partial t^2} + \eta_C \frac{\partial \b{u}}{\partial t} = \mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F}.\]
 
# Directly solve for steady state: \[\mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F} = 0.\]
 
 
Displacement boundary conditions are very simple to implement with both approaches. The desired values are simply prescribed for dynamic simulation with explicit methods, In case of implicit methods where we need to solve a system of equations the prescribed displacements are placed into the right-hand side of the system, and the value 1 is placed in the diagonal of the matrix (for a given node).
 
 
For the traction boundary condition
 
 
\[\begin{bmatrix}
 
\sigma_{xx} & \sigma_{xy} \\
 
\sigma_{yx} & \sigma_{yy}
 
\end{bmatrix} \cdot
 
\begin{bmatrix}
 
n_x \\
 
n_y
 
\end{bmatrix} =
 
\begin{bmatrix}
 
\bar{t}_x \\
 
\bar{t}_y
 
\end{bmatrix}\]
 
 
we first use Hooke's law and the strain-displacement relations to express the traction boundary in terms of displacements:
 
 
\[\begin{bmatrix}
 
(2\mu+\lambda)\partial_x u_x + \lambda \partial_y u_y & \mu(\partial_x u_y + \partial_y u_x) \\
 
\mu(\partial_x u_y + \partial_y u_x) & \lambda \partial_x u_x + (2\mu + \lambda) \partial_y u_y
 
\end{bmatrix} \cdot
 
\begin{bmatrix}
 
n_x \\
 
n_y
 
\end{bmatrix} =
 
\begin{bmatrix}
 
\bar{t}_x \\
 
\bar{t}_y
 
\end{bmatrix}\]
 

Latest revision as of 11:54, 24 February 2018

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