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Element mechanisms

An element is fully integrated if the integration rule used for an element evaluates the element matrices as accurately as analytical integration and ensures that there are no element mechanisms associated with the element.

If the order of numerical integration used to evaluate an element matrix is lower than the full integration rule (reduced integration), the element can exhibit what is commonly termed a mechanism or a spurious zero energy mode. A mechanism is a mode of deformation that causes no stress or strain to be developed as a result.

For example, the 4-noded plane continuum element is fully integrated with a 2x2 (4-point) Gaussian integration rule. If a 1-point rule (centrally located integration point) is used to evaluate the element stiffness matrix, then the element will exhibit one mechanism (two if the symmetrical deformation is taken into account). This mechanism corresponds to the following element deformation

element_mechanism_linear.gif (2141 bytes)

Another example is the 8-noded plane continuum element. This is fully integrated with a 3x3 (9-point) integration rule. If a 2x2 rule is used to evaluate the element stiffness matrix, then the element will exhibit one mechanism (two if the symmetrical deformation is taken into account). This mechanism corresponds to the following element deformation

element_mechanism_quadratic.gif (2710 bytes)

To illustrate the effect of reduced integration on element matrices, consider the evaluation of the stiffness matrix of a 3-noded bar element. If a 2-point integration rule is used a stiffness matrix is generated that is identical to that obtained by integrating analytically and is as follows

T2.jpg (3138 bytes)

This possesses no mechanisms and is fully integrated. If a 1-point integration rule is used, however, a mechanism is exhibited, which can be seen by considering that the stiffness matrix for such an element can be derived as follows

T1.gif (2268 bytes)

Integrating with a 1-point Gauss rule requires the integration point to be located at x=0 with a weight of 2, giving

T3.gif (3934 bytes)

Where the row and column corresponding to the midside node degrees of freedom can be seen to be zero. This means that an axial force applied to the midside node of a 3-noded bar element using a 1-point integration rule will not generate any force in the element. It is for this reason that the default integration order for this element in LUSAS is two.


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