05-10-2012, 11:58 AM
Finite Element Methods
Finite Element.ppt (Size: 436.5 KB / Downloads: 55)
Beam elements
Beam elements were developed first, as a stiffness matrix [Ke] of a beam can be found exactly for elastic behaviour and small deflections if there are boundary conditions and loading at each end only.
The matrix [Ke] links forces and moments (a vector F) to displacements and rotations (a vector u) at each end (each node).
F = [Ke]u
In 3D 3 force components and 3 moment components act on each end - hence the element matrix is 12 by 12.
Individual terms are listed on the fifth slide. They depend on the length L, the area of cross-section A, the second moments of area, an effective area deforming in shear etc.
Euler-Bernoulli v Timoshenko beam elements
The simplest beam element uses Euler-Bernouilli beam theory.
This neglects deflection transversely due to shear strain.
The same theory is used as in hand calculation
M/(EI) = 1/R integrated twice implies deflection
However, if the length of a beam is short (eg similar to its depth) then this assumption leads to error, so most FE packages use beam elements obtained from Timoshenko beam theory, which includes deflection due to shear strain, as well as that due to bending strain and rigid body motion.
An element may revert to a Euler-Bernoulli beam if the “shear area” (cross-sectional area used to find shear deformation) is not specified by the user.
Terms in a Beam Element Matrix
With Euler-Bernoulli beam theory, the following types of terms arise in a beam element matrix, for bending a beam of length L on a principal axis of the cross-section, with 2nd moment of area I.
Transverse force/deflection relations give ±12EI/L3
Transverse force/rotation or bending moment/displacement relations give ±6EI/L2
Bending moment/rotation relations give 4EI/L or 2EI/L
Twisting a beam gives GIP/L
Axial deformation gives EA/L (A = cross-sectional area0
Transverse shear deformation gives GAS/L
where AS = effective cross-sectional area for shear.
Rotation of coordinates
A beam will not in general be aligned with the “global” xyz axes. To rewrite the stiffness terms in terms of forces/displacements in the global axis directions, a rotation matrix R is found, the individual terms of which are direction cosines.
The rotation R applies to both forces and displacements. The inverse of R is just is transpose. This corresponds to rotating back the other way.
Hence if in local axes aligned with a beam F = K u, then in global coordinates RFG = K RuG or FG = RTKR uG
Even for a single spring, this transformation leads to a 6 by 6 matrix, containing products of direction cosines, as each end of the spring can move in 3 directions, giving 6 equations.
A beam has rotations at each end as
well, giving 12 equations.
Solution Process
Once they are all written in the same coordinate system, element matrices are combined (assembled) to give a large sfiffness matrix for the full structure, by assuming that forces and moments at ends of beams connected (nodes) sum to zero or to the resultant external loading. Eg Fx1 + Fx2 + Fx3 = 0 at the joint below.
The full set of linear equations is solved for the displacements and rotations. This is only possible if some displacements or rotations are “restrained” to prevent rigid body motion.
The forces and moments acting on any one beam can then found from its stiffness matrix times known displacements.
These then imply the stresses within a beam.
More detail later on this …..
Finite Element.ppt (Size: 436.5 KB / Downloads: 55)
Beam elements
Beam elements were developed first, as a stiffness matrix [Ke] of a beam can be found exactly for elastic behaviour and small deflections if there are boundary conditions and loading at each end only.
The matrix [Ke] links forces and moments (a vector F) to displacements and rotations (a vector u) at each end (each node).
F = [Ke]u
In 3D 3 force components and 3 moment components act on each end - hence the element matrix is 12 by 12.
Individual terms are listed on the fifth slide. They depend on the length L, the area of cross-section A, the second moments of area, an effective area deforming in shear etc.
Euler-Bernoulli v Timoshenko beam elements
The simplest beam element uses Euler-Bernouilli beam theory.
This neglects deflection transversely due to shear strain.
The same theory is used as in hand calculation
M/(EI) = 1/R integrated twice implies deflection
However, if the length of a beam is short (eg similar to its depth) then this assumption leads to error, so most FE packages use beam elements obtained from Timoshenko beam theory, which includes deflection due to shear strain, as well as that due to bending strain and rigid body motion.
An element may revert to a Euler-Bernoulli beam if the “shear area” (cross-sectional area used to find shear deformation) is not specified by the user.
Terms in a Beam Element Matrix
With Euler-Bernoulli beam theory, the following types of terms arise in a beam element matrix, for bending a beam of length L on a principal axis of the cross-section, with 2nd moment of area I.
Transverse force/deflection relations give ±12EI/L3
Transverse force/rotation or bending moment/displacement relations give ±6EI/L2
Bending moment/rotation relations give 4EI/L or 2EI/L
Twisting a beam gives GIP/L
Axial deformation gives EA/L (A = cross-sectional area0
Transverse shear deformation gives GAS/L
where AS = effective cross-sectional area for shear.
Rotation of coordinates
A beam will not in general be aligned with the “global” xyz axes. To rewrite the stiffness terms in terms of forces/displacements in the global axis directions, a rotation matrix R is found, the individual terms of which are direction cosines.
The rotation R applies to both forces and displacements. The inverse of R is just is transpose. This corresponds to rotating back the other way.
Hence if in local axes aligned with a beam F = K u, then in global coordinates RFG = K RuG or FG = RTKR uG
Even for a single spring, this transformation leads to a 6 by 6 matrix, containing products of direction cosines, as each end of the spring can move in 3 directions, giving 6 equations.
A beam has rotations at each end as
well, giving 12 equations.
Solution Process
Once they are all written in the same coordinate system, element matrices are combined (assembled) to give a large sfiffness matrix for the full structure, by assuming that forces and moments at ends of beams connected (nodes) sum to zero or to the resultant external loading. Eg Fx1 + Fx2 + Fx3 = 0 at the joint below.
The full set of linear equations is solved for the displacements and rotations. This is only possible if some displacements or rotations are “restrained” to prevent rigid body motion.
The forces and moments acting on any one beam can then found from its stiffness matrix times known displacements.
These then imply the stresses within a beam.
More detail later on this …..