Dear Spyros,

the previous answers do unfortunately mix up locking and hourglass effects. In the following, I will try to elaborate a bit on fully and reduced integrated linear finite elements.

In your question do you refer to a reduced integrated linear element (1 Gauss point) or to the fully integrated one (2x2 Gauss points)? What is a 2-point Gauss integration?

In fully integrated linear finite elements (2x2 Gauss points) such as the 4-node quadrilateral finite element (Q4) you are referring to there is no such thing as an hourglass instability. The hourglass instabilities also known as zero-energy modes are caused by the under-integration of the stiffness matrix by just using 1 Gauss point for a reduced integration technique. This results in a rank-deficiency of the stiffness matrix which causes the existence of deformation states that do not need any energy to come into existence. So a clear no to the question if hourglass modes exist in fully integrated finite elements.

Chitaranjan and Tran are, however, right that fully integrated 4-node quadrilateral elements suffer from locking effects. This can be shear locking which results in too small deformations if the elements experience a bending dominated state. The remedy for this problem consists in a reduced integration which in turn leads to hourglassing if not special hourglassing controls are used. What you basically do is alleviate one shortcoming of the element by committing a variational crime and than you have to alleviated the effects of this crime with yet another numerical trick. In this case, the best recommendation would be to switch to quadratic finite elements (8-node elements, Q8).

Another problem could result from volume locking as these elements cannot handle the incompressibility constraint (nu --> 0.5). A mixed variational formulation could be used to circumvent this problem.

Kind regards,

Sascha

Q4 elements develop will resist even pure bending by developing both normal and shear stresses. This makes it too stiff in bending.

The principal defects of Q4 element is its over stiffness in bending.

It is recommended to not use Q4 element with too large aspect ratios - as it will have infinite stiffness.

Dear Spyros,

the previous answers do unfortunately mix up locking and hourglass effects. In the following, I will try to elaborate a bit on fully and reduced integrated linear finite elements.

In your question do you refer to a reduced integrated linear element (1 Gauss point) or to the fully integrated one (2x2 Gauss points)? What is a 2-point Gauss integration?

In fully integrated linear finite elements (2x2 Gauss points) such as the 4-node quadrilateral finite element (Q4) you are referring to there is no such thing as an hourglass instability. The hourglass instabilities also known as zero-energy modes are caused by the under-integration of the stiffness matrix by just using 1 Gauss point for a reduced integration technique. This results in a rank-deficiency of the stiffness matrix which causes the existence of deformation states that do not need any energy to come into existence. So a clear no to the question if hourglass modes exist in fully integrated finite elements.

Chitaranjan and Tran are, however, right that fully integrated 4-node quadrilateral elements suffer from locking effects. This can be shear locking which results in too small deformations if the elements experience a bending dominated state. The remedy for this problem consists in a reduced integration which in turn leads to hourglassing if not special hourglassing controls are used. What you basically do is alleviate one shortcoming of the element by committing a variational crime and than you have to alleviated the effects of this crime with yet another numerical trick. In this case, the best recommendation would be to switch to quadratic finite elements (8-node elements, Q8).

Another problem could result from volume locking as these elements cannot handle the incompressibility constraint (nu --> 0.5). A mixed variational formulation could be used to circumvent this problem.

Kind regards,

Sascha

Zheng-Kun Liu

Cheers,

Sascha

As Sascha already explained, mixing concepts (hourglassing with locking) is misleading. For instance, the 3-node triangle is free of hourglassing but suffers from locking. Regarding your question, the 4-node quadrilateral element needs to be fully integrated with a 2x2 Gauss rule to make this element free of hourglass instability. This will be clear after reading the following paragraphs.

One can find out the true nature of the hourglassing behavior doing a simple exercise, as follows. The 3-node triangle uniquely determines a linear approximation. Using a normalized system, the basis that span the space of functions of degree 1 is {1,xi,eta}. We say that the 3-node triangle has 0 zero-energy modes because of the following: the 3-node linear elastic triangle has 6 degrees of freedom (6 DOFs); therefore, its stiffness matrix is a 6x6 matrix. Sampling the stiffness matrix at 1 Gauss point results in 3 independent relations (those stemming from the constitutive equation sigma = D*strain). Thus, the stiffness matrix has 6 - 3 = 3 equations that are not independent. However, these three equations are related to the three zero-energy modes corresponding to the three rigid body modes (two translations and one rotation) that need to be fixed. So basically, from the 6 DOFs we need to fix 3 DOFs. After fixing these 3 DOFs, the stiffness matrix has (6-3) - 3 = 0 equations that are not independent, which is equivalent to saying that the 3-node linear elastic triangle has 0 zero-energy modes.

Now, consider the 4-node quadrilateral. The approximation on this element is span by the basis {1,xi,eta,xi*eta}. We already know that the first three components of this basis can be dealt with 1 Gauss point. The remainder monomial (xi*eta) gives rise to 2 additional zero-energy modes (one per DOF). These two modes cannot be related to rigid body modes as these were already taken into account in the analysis of the first three terms of the basis (i.e., {1,xi,eta}). Therefore, these 2 additional zero-energy modes are not physical. We need a way to eliminate these unphysical modes to avoid hourglassing. To this end, 2 more independent relations are needed. At this point, we should make the DOFs exercise for the 4-node quadrilateral: the 4-node linear elastic quadrilateral has 8 DOFs. We know that we have to fix 3 DOFs to eliminate the three rigid body (physical) modes. Of course, we do this. Now, sampling the stiffness matrix at 1 Gauss point gives 3 independent relations leaving (8-3) - 3 = 2 equations that are not independent (we already knew that! no surprise!) So, this reveals that hourglassing is a type of instability associated with a stiffness matrix that is not sufficiently integrated (aka underintegrated). Now, considering that each Gauss point gives 3 independent relations, using a 2x2 Gauss rule (i.e., 4 Gauss points total) results in 4*3 - 3 = 9 additional independent equations, which is enough since we needed at least 2 independent equations. We say that the 4-node quadrilateral element is fully integrated with a 2x2 Gauss rule.

A similar analysis can demonstrate that for the 8-node linear elastic quadrilateral (basis given by {1,xi,eta,xi*xi,xi*eta,eta*eta,xi*xi*eta,xi*eta*eta}), a 3x3 Gauss rule is needed to make this element free of hourglass instability.

In my opinion, hourglass instability can be better understood when viewed through the prism of the Virtual Element Method. A very good reference regarding this topic is the following:

A. Cangiani, G. Manzini, A. Russo and N. Sukumar, "Hourglass stabilization and the virtual element method," Int. J. Numer. Meth. Engng 2015; 102:404–436.

Best!

Zheng-Kun Liu

As you can see in the previous posts, hourglassing is related to stability. Therefore, to investigate the hourglassing you will need a proper stability test. You can find one test in the paper I already mentioned in my previous answer:

A. Cangiani, G. Manzini, A. Russo and N. Sukumar, "Hourglass stabilization and the virtual element method," Int. J. Numer. Meth. Engng 2015; 102: 404–436. (Therein, see the last paragraph in Section 10.1.1)

Alternatively, you can study the stability numerically by performing eigenvalue analyses. See Section 6.2.6 in this paper:

A. Ortiz-Bernardin, A. Russo, N. Sukumar, "Consistent and stable meshfree Galerkin methods using the virtual element decomposition," Int. J. Numer. Meth. Engng 2017; 112 (7): 655-684.

Best!

Patch test