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FEA Theory Question
- Thread starter jczarne
- Start date
there are six "rigid body" freedoms (or motions) ... the body can translate along or rotate about any of the three orthogonal axes that define the model (X, Y, Z).
"constant strain" may be the same as "rigid body" ... the body won't strain if it was undergoing rigid body motion.
"constant strain" may be the same as "rigid body" ... the body won't strain if it was undergoing rigid body motion.
There are 6 degrees of freedom (3 translation, 3 rotation) in three dimensional (3d) elements like bricks and tetrahedra, but only 3 degrees of freedom (2 translation, 1 rotation) in two dimensional (2D) elements like quadrilaterals and triangles.
The finite element equations that must be solved are sometimes written:
Ku=F, K is NxN matrix called the stiffness matrix, "u" is the solution vector, Nx1 in dimensions, and F is the force vector, Nx1 in dimensions. This is just a quick way of course to represent "N" equations. K11*a1+K12*a2+..etc.=F1, etc.
This number of degrees of freedom (DOF) is exactly the mininum number of DOF in "K" that must be constrained. "K" is singular--in 2D, there are N-3 independent equations. In 3D, there are N-6 independent equations. The "3" and the "6" correspond exactly to the number of DOF in the elements, which depend on whether the elements are 2D like quadrilaterals or 3D like bricks.
The finite element equations that must be solved are sometimes written:
Ku=F, K is NxN matrix called the stiffness matrix, "u" is the solution vector, Nx1 in dimensions, and F is the force vector, Nx1 in dimensions. This is just a quick way of course to represent "N" equations. K11*a1+K12*a2+..etc.=F1, etc.
This number of degrees of freedom (DOF) is exactly the mininum number of DOF in "K" that must be constrained. "K" is singular--in 2D, there are N-3 independent equations. In 3D, there are N-6 independent equations. The "3" and the "6" correspond exactly to the number of DOF in the elements, which depend on whether the elements are 2D like quadrilaterals or 3D like bricks.
feaplastic
Mechanical
Prost,
r u sure, 3d brick elements have 6 dof. Does not it have only 3 translation direction dof. Or, u meant to say shell element and membrane elements.
r u sure, 3d brick elements have 6 dof. Does not it have only 3 translation direction dof. Or, u meant to say shell element and membrane elements.
feaplastic,
Yes Prost is correct. A 3D brick element has all six degrees of freedom, as does any 3D object (unless planes of symmetry have been applied).
However the nodes of that make up the brick element have only translational degrees of freedom and no rotational degrees of freedom, but this of course does not stop the brick itself from rotating !
Yes Prost is correct. A 3D brick element has all six degrees of freedom, as does any 3D object (unless planes of symmetry have been applied).
However the nodes of that make up the brick element have only translational degrees of freedom and no rotational degrees of freedom, but this of course does not stop the brick itself from rotating !
Yes, I misspoke. It's 6 rigid body modes, 3 translational and 3 rotational. Using the definition of 'degrees of freedom' everybody else uses {
}, an 8 noded brick has 8x3=24 DOF, a 20 noded brick has 20x3=60 DOF.
and no, I wasn't talking about shells at all.
the brick does have rotational rigid body modes; you can see that if you have ever constructed a stiffness matrix in a 3D geometry--for N degrees of freedom (or equations) in Ku=F, there are N-6 independent equations, meaning you've got to delete (or constrain) 6 of those N equations to get a non singular matrix K' you can invert to compute "u".
}, an 8 noded brick has 8x3=24 DOF, a 20 noded brick has 20x3=60 DOF. and no, I wasn't talking about shells at all.
the brick does have rotational rigid body modes; you can see that if you have ever constructed a stiffness matrix in a 3D geometry--for N degrees of freedom (or equations) in Ku=F, there are N-6 independent equations, meaning you've got to delete (or constrain) 6 of those N equations to get a non singular matrix K' you can invert to compute "u".
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