Game Development Reference
In-Depth Information
void buildKe(CMatrix &Ke, //Stiffness matrix (out)
CPoint3d const &x1, //Element vertex (in)
CPoint3d const &x2, //Element vertex (in)
CPoint3d const &x3, //Element vertex (in)
CPoint3d const &x4, //Element vertex (in)
float mu, //Physical parameter (in)
float lambda) //Physical parameter (in)
{ CMatrix Be(6,12),E(6,6),Pe(4,4);
float vol,mr; //Element volume
buildBarycentric(Pe, x1, x2, x3, x4);
vol = 1.0/(Pe.det() 6);
buildBe(Be,Pe);
E.loadZero();
mr=2 mu+lambda;
E(0,0) = mr; E(0,1) = lambda; E(0,2) = lambda;
E(1,0) = lambda; E(1,1) = mr; E(1,2) = lambda;
E(2,0) = lambda; E(2,1) = lambda; E(2,2) = mr;
E(3,3) = mu; E(4,4) = mu; E(5,5) = mu;
Ke= vol Be.transpose() E Be;
}
Listing 10.3. Building matrix
K e .
A range of standard methods exist to solve this system: Gauss elimination,
Cholesky factorization, and conjugate gradient, to name just a few. In Section 10.4,
we briefly describe some of them.
An important requirement to solve the static system in Equation (10.13) and
find the displacements is to fix and keep hold of at least three nodes (or nine
degrees of freedom), otherwise the system is singular, meaning that no unique
solution exists. This can be intuitively explained by the fact that without any
boundary conditions or prescribed positions, the body has no unique position in
space, and because there is no unique position, there is no unique deformation.
Boundary conditions are also used as a manner of applying external loads to the
body. Refer to Section 10.3.5 for more details on this subject.
Note that the system does not depend on time, meaning that if we simulate
it, the object jumps from one configuration to another without taking into ac-
count any previous configuration. The deformed shape of the body is defined
exclusively by the material properties of the object and the applied forces at that
instant. If the applied forces are removed, the object returns instantaneously to its
rest position; there are no dynamics on the system.
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