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3.4  Contact simulation method

• Status method for contact
• Time integration

3.4.1  Status method for contact

The equilibrium condition is given by

(3.48)

with {F} the external force (possibly due to the gap offset), {R}=[NOR]{f_n}+[TAN]{f_t} the contact friction load.

For a static equilibrium, the sliding velocity is naturally the enforced velocity (UNSOLVED PROBLEM should account for convection terms), so the friction force is tangential (e_t=e_θ) µ{f_n}. The projection of the equilibrium equation (3.48) on the observation matrix for relative motion also gives

(3.49)

Combining ft=µ{f_n} and equations (3.48)-(3.48), one obtains the non symmetric equation

(3.50)

which can be solved, while assuming the contact constraint [NOR]^T{u}, to obtain the static equilibrium under sliding contact. This resolution is performed using constraint elimination.

At this stage, some contact points may be found to have negative normal forces, so that the constraint should not have been applied. An iteration of the status method thus removes constraints for points that have negative contact forces and adds them for points with a negative gap. It is assumed and generally found that the procedure converges.

3.4.2  Time integration

The system is then solved using a modified Newton scheme to limit the computation times. If the contact density is non linear, instead of computing at each Newton step a new tangent stiffness matrix, a constant Jacobian having sufficient contracting properties is used for the displacement increments. This Jacobian is formed using a user defined linear contact penalty large enough to allow convergence. The convergence criteria are based on the displacement increment or on the equilibrium equation where the contact is only represented by the contact forces.

For time simulation an outer time incrementation loop based on the Newmark scheme is used as represented in figure 3.10. This method is the basis non linear Newmark scheme in fe_time.

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Figure 3.10: Non-linear Newmark scheme

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Lagrange formulation

The Lagrange formulation keeps the contact forces as unknowns. The system is then solved using Lagrange multipliers. The size of the system is then also governed by the size of the interface in contact. The static equilibrium is solved using

(3.51)

under

(3.52)

Given an initial Lagrange multiplier {λ₀}, the initial displacement {q₀} is

(3.53)

We search for a correction {δ q} such that or {δ λ} such that

If , then δ q_k=0

If , then search

(3.54)

which is equivalent to

(3.55)

We choose the following approximation:

(3.56)