CNST-H-418 : Examples of exam questions (answers) !1. Explain the principles of a Newton-Raphson solution procedure for solving a non linear

(scalar) equation. (1.3) p.7->9 Explain how this scheme can be used to solve a discretised equilibrium problem for a structure with a non linear response. (1.3) p.10->12 !

• Principle: Find a new approximation based on an initial value and the slope of the function at this point (=derivative) • Iterative procedure • Quadratic local convergence, if the derivative is right (for proper convergence) • Potential shortcomings: vanishing derivatives, deadlocks between particular points

For structural computations:

!2. Explain what are the various sources of nonlinearity in a mechanical non linear problem.

For each of these sources, give an example of application in which it is present. (2.2) p.4-6 + Exam June 2012

2 possible sources of non linearities: • Material non linearity (related to the stress-strain relationship) e.g. elastoplastic yielding of

reinforcing bars in RC • Geometrical non linearity (large strains, large displacement and contact) e.g. instability in truss

domes and pole vaulting (large displ), rolling of metal (large strains), lack of consistency=change in boundary conditions as a result of loading (contact)

3. What are the main principles to use to define a strain measure when geometrical nonlinearities need to be considered ? (3.1) p.2->8 Are there several types of geometrical nonlinearities ? Explain your answer. What are the consequences of geometrical nonlinearities on the principle of virtual works ? (3.1)p.2 !

Main principles 4 criterions for 1D strain measure: !!!!!!!

• Just a choice • Not allowed to decrease or be constant after a certain value • Avoid to lead to Delandsheer’s course • If small displacement, leads to the standard definition of strain epsilon=(l-l0)/l0

Assumptions:

!• Ejk=Green-Lagrange deformation tensor (initial coordinates) • Ajk=Euler-Almansi deformation tensor (final coordinates) • 1st two terms (in brackets) in order of 1% • Last term in order of 0,01% if strain is small (1%) (=>negligible) !

2 main informations: • These functions all satisfy the four criteria mentioned above •These strain measures are indeed equivalent when the stretch tends to 1. This corresponds (in the simplified 1D case) to the fact that the Green and Euler tensors tend towards the infinitesimal strain tensor when the geometrically non linear terms become negligible.

3D generalisation: - Deformation gradient F

In a multiaxial setting, the strain measures are more complex.

Since the strain measures (f.i. the Green and Euler strain tensors) are constructed based on an inspection of the kinematics (how the scalar product of infinitesimal vectors evolves as a function of the deformation), it is necessary to start from the definition of the transformation between the initial and current (deformed) configurations. Non symmetric tensor as it links vectors in one configuration (the initial one) to vectors in the other

(the deformed one). It contains all informations about deformation but also about rigid body rotation (F = R.U, where R=rigid body rotation and U=3D strecth).

- Objectivity The strain measure CANNOT be based on a quantity that includes the rigid body rotation.

For a rigid body rotation, F represents the rotation matrix. In that case, a strain measure that would be based on a function of the components of F would not vanish, which is unacceptable for a rigid body movement.

A strain measure is defined as ‘objective’ when it is insensitive to the rigid body rotation.

To obtain such an objective measure, one should extract the part of F that represents the rotation (R), and base the strain definition on the remaining part, i.e. the stretch U .

Crucial point is to remember that the usual infinitesimal strain tensor (the symmetric gradient of the total displacements) is NOT objective.

Several types of geometrical non linearities

See Q2

Consequences on virtual work principle

In cases in which non linear terms in the strain tensors cannot be neglected anymore, it is necessary to write the weak form selecting the system in which the various quantities are expressed. On needs to account for the fact that the deformed configuration is not the same as the initial configuration.

• The strains can be defined by differentiating displacements with respect to the initial of current coordinates

• The definition of stress can be based on the current configuration (true stress) or in the initial configuration (stress definitions will be given in the sequel)

These two measures CANNOT be chosen independently, and their product has to be integrated on the proper volume (initial or deformed configuration) in order to correctly represent the internal work in the principle of virtual work.

!4. What are the assumptions to fulfill for a mechanical problem to be geometrically linear ?

Precise clearly the different assumptions and their physical interpretation. (2.2) p.5 + (1.2) p.7 + (3.1) p.9 !

Assumptions See Q3 assumptions !5. What are the principles to use to define a stress measure when geometrical nonlinearities

need to be considered ? (3.2) p.2->7 Are there several types of geometrical nonlinearities ? Explain your answer. What are the consequences of geometrical nonlinearities on the principle of virtual works ? (3.1) p.2 !

Main principles Configuration dependency !

This 1D situation allows understanding the true stress concept (general stress state much more complex and requires tensorial quantities).

The true stress is defined as the ratio between an elementary force and and the elementary DEFORMED area on which it is applied.

The true stress is the ONLY

stress quantity that is physically realistic in the sense that it is the ONLY stress measure for which the interpretation is immediate, ONLY stress quantity that can be compared to a material strength limit obtained experimentally.

The other stress vectors are quantities used to transfer the expressions of the virtual work between the initial and deformed configurations (using the deformation gradient F).

Cauchy Stress (generalisation of 1D true stress)

Cauchy stress tensor involves elementary forces in the deformed configuration with elementary surfaces in the deformed configuration. Symmetric tensor (both indices related to the deformed system).

In many softwares, it is possible to plot stress maps and to represent scalar equivalent stresses quantities corresponding to material failure or plastic criteria (the equivalent von Mises stress for instance). It is CRUCIAL to post-process such quantities ONLY for the Cauchy stress since it is the only physical stress. Comparing any equivalent stress derived from the other stress tensors would result in wrong decision concerning the material behaviour.

The use of the Cauchy stress is however made complex by the fact that the matrix representing the tensor quickly becomes ‘fully’ populated when rigid body rotations are present, even when simple loading modes (such as uniaxial tension) are considered

Piola-Kirchhoff 1 Stresses (generalisation of 1D nominal stress)

PK1 tensor involves elementary forces in the deformed configuration with surfaces in the initial configuration. NOT symmetric tensor. Difficult to interprete physically.

Piola-Kirchhoff 2 Stresses (generalisation of 1D nominal stress)

PK2 tensor involves elementary forces in the initial configuration with surfaces in the initial configuration. Symmetric tensor (thanks to F-1). Difficult to interprete physically. Allows to interpret the type of loading.

Relation between stress tensors !At a given point of a solid, these various stress tensors are not independent from each other. The knowledge of one of them together with the deformation state (i.e. the deformation gradient F) is sufficient to compute all three stress tensors defined here.

The choice of the stress tensor to be used is important during the practical implementation of a formulation. If a software provides any of these tensors during post-processing, the knowledge of the displacement field allows to determine the others, and in particular the Cauchy tensor for comparisons with material strength.

6. Based on the discretised weak form of a mechanical equilibrium problem, and assuming that only material nonlinearities have to be considered, explain the principles of an incremental-iterative solution procedure (use a figure to help explaining the solution procedure). (4.1) p.6-13->22

Principles: Obtain equilibrium between internal and external forces at the end of each increment (= loading step), using the Newton-Raphson iterative method (= iterations). 1st loading step

1st iteration (predictor) - Try to satisfy at the end of iteration 1: t+Δt{fint}(1) = t+Δt{fext}- Equilibrium state assumed to be known t{fint} = t+Δt{fint}(0) (corresponds to t{q} = t+Δt{q}(0)) - From this equilibrium, express variation of {fint} in function of the displacement variation t+Δt{δq}(1) => Kt(t+Δt{q}(0)). t+Δt{δq}(1) =~ t+Δt{fext} - t+Δt{fint}(0) => t+Δt{δq}(1) - Incremental displacement update: t+Δt{q}(1) = t+Δt{q}(0) + t+Δt{Δq}(1) = t+Δt{q}(0) + t+Δt{δq}(1) - Internal force: t+Δt{q}(1) => t+Δt{ε}(1) => t+Δt{σ}(1) => t+Δt{fint}(1) = integralv([B]t+Δt{σ}(1)dv) - Equilibrium? No => more iterations starting from t+Δt{q}(1) Following iterations (corrections) - Same development until satisfying at the end of iteration i: t+Δt{fint}(i) = t+Δt{fext}- Different structural stiffness matrix (= new slope) at each iteration (except for modified NR method where we keep the same matrix for each iteration)

Next loading step !

7. Give the definition of a material tangent stiffness. Explain how it may be used in the Newton-Raphson solution scheme of a materially non-linear structural problem. (4.1) p.10-11-23 + (5.1) p.2->4

The material tangent stiffness corresponds to the slope of σ-ε curve (in 1D = Tangential Young modulus). It is a NON constant matrix, it depends on the state of the material (=> different at each iteration since the state changes). !!

The material tangent stiffness leads to the definition of the structural tangent stiffness, which is used in the solution procedure of a materially non-linear structural problem (see Q6 for more details). !!!!!!!!!!

8. Assume that you have to code or modify a constitutive law in a nonlinear finite element code. Explain in which part of the code you will have to work and why. For each of the required modifications, explain its use in the global solution scheme of the equilibrium problem. (4.1) p.10-11-23 + (5.1) p.2->4

In case of a modification of the constitutive law, the stages which are surrounded by red boxes have to be modified as the tangent stiffness matrix depends on the material tangent stiffness, which itself depends on the constitutive law.

In the same way, the internal forces computation requires the update of the stresses based on the updated displacements. This stress update operation depends on the constitutive law as well.

The material tangent stiffness corresponds to the slope of σ-ε curve (in 1D = Tangential Young modulus). It is a NON constant matrix, it depends on the state of the material (=> different at each iteration since the state changes). The material tangent stiffness leads to the definition of the structural tangent stiffness, which is used in the solution procedure of a materially non-linear structural problem (see Q6 for more details). !!

9. Define the structural tangent stiffness matrix. Explain its link to the discretised equilibrium of a structure, and its potential use in nonlinear solution procedures. (4.1) p.10-11-23 + (5.1) p.2->4 !

See Q8. !10. What are the main control parameters of a nonlinear calculation met in nonlinear finite

element codes ? (4.1) p.24 Identify them by illustrating them for the case of a Newton-Raphson solution procedure. (4.1) p.15-23

!!11. Explain what is a limit point in the solution procedure of a nonlinear problem, and explain

how their presence may induce computational problems. (4.2) p.2->4 !A limit point is a point on the equilibrium curve with extremal value, above which it is difficult to find an intersection with a hypersurface that is a plan (load or displacement) controlled by an increasing monodically quantity. Every limit point is associated with a type of control (force, displacement, arc-length), which impacts on the type of hypersurface used.

The arc-length control does not fix the external force or the displacement (these 2 will change).

!These limit points may lead to a local divergence when computing the solution. Note that the convergence of the NR method is local and even the limit point is local, the NR method will not often allow to pass it.

!

12. Explain what is a limit point in the solution procedure of a nonlinear problem, and explain in the principles (thus not the equations) which control methodologies allow to pass control limit points in structural response. (4.2) p.2->6 !

See Q11. !13. What is a bifurcation point in the structural response of a structure. Give physical

examples of such a phenomenon. What are the potential impacts of such a phenomenon on the results of a nonlinear computation ? Explain. (8) p.1->3 !

Definition A bifurcation point is a point where at least 2 equilibrium paths exist. Path without perturbation = fundamental path Other paths obtained by perturbation = alternative paths Examples Eulerian buckling of beams under compression Bifurcation point = Buckling limit point Fundamental path = unlimited compression Alternative paths = sinusoidal solutions Elastic instabilities in shells

Potential impacts Since all solutions satisfy the equilibrium equations, the chosen solution may be the wrong solution (i.e. not the real one) and leads to an overestimation of limit loads. The real solution is the one that minimises the supplied work (most critical), it implies to know all the solutions. The fundamental solution is rarely the most critical. !14. How should a bifurcation point be detected in a nonlinear computation ? (8) p.3 !

In the computation, the idea is to detect at each converged step (= equilibrium) the presence of a bifurcation by checking any negative/null eigenvalue which indicates a potential bifurcation. A negative/null eigenvalue means that the structural tangent stiffness matrix Kt is no more positive definite and cannot be inverted in the computation to find the increase of displacement. !

15. What are the main classes of material nonlinear behaviour as presented in the course ?Explain them and give an example for each class. (5.1) p.5-7 + (2.2) p.6

To distinguish plastic behavior (3 curves above), it is necessary to obtain additional informations with respect to a monodically increasing test (only under displacement

control). !!The (very basic) classification of the inelastic behaviours presented above is certainly not the most general one. Its objective is only to identify the most commonly encountered types of effects.

In particular, the behaviour of a real material will always be a combination of several effects. For instance, cracking in materials like rocks or masonry is accompanied both with the appearance of stiffness degradation and of permanent strains.

Be also aware of not confusing non linearity and non reversibility of the behaviour. A material behaviour may be non linear and remain reversible from a thermodynamic point of view.

Last but not least, pay attention to the fact that a behaviour can be reversible in strain without being reversible thermodynamically.

Non linear elastic: rubber-like materials Elasto-plastic: metals (dislocations: sliding of atomic planes => permanent strains) Elasto-visco-plastic: polymers (depend also on the loading speed) Damaging: cracking concrete (stiffness degradation)

16. What are the main ingredients of a damage model as presented in the course ?What are the main assumptions associated with it, and what are the corresponding limitations of the model (to help yourself, you can use illustrations based on simple physical examples) ? (5.2a) p.2->7 !

Main ingredients - Stiffness degradation: the cracked material does no contribute to stiffness anymore !- Continuum damage

Continuous variable representing the average effect of defaults (= density of cracks). Value from 0 to 1 (1 = material totally damaged). The multiaxial damage (3D case) should be a tensor since the influence of microcraks is different for tangential and normal loading, but it is assumed to be a scalar damage (see limitations) !

- Constitutive law 1D

!!MultiD

- Damage criterion 1D

MultiD

For the multiaxial case, the irreversibility criterion is a bit more complex since it depends on different strains components (aij). Therefore, the criterion is allows to aggregate all the effects of

each component and to get only one value which can be compared to the limit value. !For initially isotropic materials, the equivalent strain is in function of the principal strains. For anisotropic materials, the equivalent strain is in function of the principal strains and its orientations. An alternative is to express it in function of all strain components (not only principal) !Different sources of anisotropy exist (each ingredient): anisotropy in the elastic stiffness, in the initial damage criterion (= strength) and in the evolution law. !!

- Evolution law 1D MultiD

In laboratory, test with concrete under tension will look like the graph on the left. Determination of K => determination of D(K). e.g. 4 points bending test of concrete beam (exponential stiffness degradation, no permanent strain)

Attention! Test made with imposed displacement (not imposed load). !!Limitations - From 1D to multiaxial case: isotropy assumption (scalar damage) => But in reality, as soon as

cracks appear, no isotropy stands anymore. - Theoretically, if we unload the system, no permanent strain should be present. However, in

reality, any failure presents rugosity and therefore a bit of permanent deformation is present. - Different damage behaviour if alterning compression and tension loading. !

!17. What are the main ingredients of a damage model as presented in the course?(5.2a) p.2->7

What are the required experimental tests to identify it ? See Q16. !18. What are the main ingredients of a damage model as presented in the course?(5.2a) p.2->7

How are the specific features of its implementation in a finite element code ? (5.2b) p.2->4 + (5.2a) p.4

See Q16.

!!

19. What are the main ingredients of a plastic law as presented in the course ? (6) p.2->7 What are the main assumptions associated with it, and what are the corresponding limitations of the model (to help yourself, you can use illustrations based on simple physical examples) ? !

Main ingredients - Strain partition:

1D MultiD

!!!!!- Plasticity criterion:

1D

This latter case could not happen, hardening has to occur in order to have an increase of strain. If the stress increases, the boundary of the admissible domain fp adapts and the admissible domain expands (if σy(K) is an increasing function). The yield surface is moving due to plasticity (hardening) => consistency condition MultiD

In multiaxial case, we have to say in which direction plastic loading occurs. Indeed, if we pull in one direction, plastic strains may also occur in the transversal direction (e.g. necking= striction). The plastic potential function is often taken as the yield function, which is then called « associate plasticty »

- Constitutive law: 1D MultiD !!!!

- Hardening law: (determines how the yield stress increases with plastic strain) εp has to be related to the way energy is dissipated

1D MultiD

!!Limitations - Potential material anisotropy (constitutive law) - Equivalent plastic strain is generally taken for the hardening law since the strain changes as a

function of the stretch. - Hardening law can be isotropic or kinematic (not material isotropy)

Isotropic: stress envelop in σ1, σ2, σ3 is going to inflate in the same way Kinematic: stress envelop in σ1, σ2, σ3 is going to move (not inflate). This represents the fact that if we pull on a material, the yield surface if going to increase in tension but not necessary in compression. (move to the right = increase only tension yield) !!!!!!!!

20. What are the main ingredients of a plastic constitutive law as presented in the course?(6) p.2->7 How are the specific features of its implementation in a finite element code ? (6) p.7->10

See Q19.

21. What are the main differences between the formulations of a damage and a plastic law from the viewpoint of their assumptions and of their implementation in a computational code ? !

!Damage: explicit problem If ε is known, σ can be directly determined (no local Newton-Raphson procedure) !Plasticity: implicit problem εp has to be known to determine σ but σ is unknown and depends implicitly on εp. It requires the use of local Newton-Raphson procedure at each Gauss point to determine σ. !Of course, stresses have to be updated ONLY incrementally (i.e. not iteratively)! When using a history dependent material behaviour, decisions have to be taken from equilibrated configurations (= when σ has converged). The iteration values are non physical. !22. Give an example of constitutive law for which the constitutive law is independent of the

strain path followed by the material, and an example of constitutive law for which the behaviour is history dependent. Hooke’s law + (5.2b) p.5 Give an example of material for each of these and explain the consequence of the history dependence on the treatment of equations in the solution procedure of a nonlinear structural problem. (5.2b) p.4 !

Dependent: {σ} = [H]({ε}-{εp}) => Steel in plasticity or cracking concrete Independent: {σ} = [H]{ε} => Steel in elastic domain !Of course, stresses have to be updated ONLY incrementally (i.e. not iteratively)! When using a history dependent material behaviour, decisions have to be taken from equilibrated configurations (= when σ has converged). The iteration values are non physical. !23. Define the concept of internal forces by explaining how it is used in the solution procedure

of a nonlinear structural problem. (4.1) p.16-17 See Q6. !24. Define the concept of anisotropy in the material response. Explain where anisotropy may

appear in the constitutive law of a material. (5.2a) p.7 See Q16. !

Damage Plasticity

Stiffness degradation Permanent strains

1D constitutive law: σ = (1-D)Eε 1D constitutive law: σ = E(ε-εp)

Strain threshold (irreversibility criterion)!fd = f(ε) - K

Stress threshold (irreversibility criterion)!fp = σeq

History variable most critical state experienced by material: K

History variable cumulative plastic strain: K

Evolution law: D(K) Evolution law: σy(K)

25. Explain the conceptual problem of using softening constitutive laws in finite element codes and the physical origin of these problems. (7) p.2->7 What are the main solutions to avoid this problem ? (7) p.8->18

Physical origins

This lost cracking width should be a material property! Physically, any failure is localised in

a zone with finite size. !!The area under the curve represents the amount of energy that we have to supply to the material. Its unit is [J/m3], therefore we need to know the cracking zone volume. !!!!

Problem

!

Solutions

Solution 1: Correction of the global dissipation (i.e. modify the fracture energy [J/m3] to provide the good number of J when meshing refinement occurs (= when the number of m3 decreases due to softening problem)) Solution 2: Modify the model to introduce an intrinsic length parameter which leads to the correct cracking volume e.g. with a non local model.

!

!Solution 3: Only the final crack is source of interest, do not care about the cracking width. Thus use a law to describe the crack opening. As a crack is a lign, its « constitutive law » (i.e. cohesive law) is not a relation between stress and strain tensors anymore, but a relation between stress and relative displacement vectors. In that case, all the dissipated energy occurs on one lign and there is no intervention of the discretization (cracking on a zero-thickness element). The area under the stress-relative displacement curve has [N/m] or [J/m²] as units (not J/m³). Of course, it requires a « glue » finite element (i.e. zero-thickness element). There exist 2 possibilities: (i) put it between classical finite elements (implies to know where the crack will start), (ii) add it when running calculations (really complex e.g. re-meshing). !

!26. What are the main ingredients of a model to use to represent failure by cracking in a

quasi-brittle material ? (7) See Q25.