Time Step Control Limitations

Many of the time step control options influence the solution results. The solution of the nonlinear dynamic response of a finite element system accurate is the numerical model correctly represents the physical model. The critical time step given for finite element system is determined by a theoretical approach in which the highest frequency of the discretized system controls this value. Therefore, the time step limitations are related to the model and cannot be changed without incidence on the quality of results.

Using the DEL option can significantly alter the model, since elements and nodes are removed without replacement. In fact, mass and/or volume is lost. Using either /DT/NODA/CST or /DT/INTER/CST will add mass to the model to allow mathematical solution. The added mass will increase the kinetic energy. This should be checked by the user to see if there is a significant effect. Switching to a small strain option using brick or shell elements also introduces errors as it was seen in Small Strain Formulation.

Generally, in the study of the nonlinear dynamic response of a system, three physical laws have to be respected:
  • Conservation of mass
  • Conservation of energy
  • Conservation of momentum dynamic equilibrium

The last one is generally respected as the equation of motion is resolved at each resolution cycle. However, in the case of adding masses especially when using /DT/NODA/CST option, it is useful to verify the momentum variation. The two other conservation laws are not explicitly satisfied. They should be checked a posteriori after computation to ensure the validity of the numerical model with respect to the physical problem.

Example: Time Step

Explicit Scheme Stability Condition

(1) M X ¨ + K X = 0 (2) X ¨ = M 1 K X (3) X ˙ n + 1 2 = X ˙ n 1 2 + d t X ¨ n (4) X n = X n 1 + d t X ˙ n 1 2
Equation 2 and n MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4saaaa@36C6@ Equation 3 are added:(5) X ˙ n + 1 2 = X ˙ n 1 2 + d t M 1 K X n
The following equation is obtained:(6) X ˙ n 1 2 = X ˙ n 1 2 1 + d t M 1 K X n 1
When Equation 6 is subtracted from Equation 5, and Equation 4 is used:(7) X ˙ n + 1 2 = A X ˙ n 1 2 X n 1 2 1

Where,

A = 2 I + d t 2 M 1 K

For non-divergence of Equation 7:

λ : largest eigen value of A I MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgk HiTiaadMeaaaa@3877@ is smaller than 1

det ( λ I + I + d t 2 M 1 K ) = 0

det ( I ( ( λ + 1 ) / d t 2 ) + M 1 K ) = 0

2 d t 2 > largest eigen value of M 1 K

2 d t 2 < smallest eigen value of 2 K 1 M

Application



Figure 1.

d t 1 = 2 m e l k e l

d t 2 = 2 m e 2 k e 2

d t e = min ( d t 1 , d t 2 )

d t n = m e 1 + m e 2 k e l + k e 2

Interface

Interfaces have stiffness but no mass: d t e = 0

Solution 1: (interface TYPE3, TYPE4, TYPE5)

k int e r f a c e < < k e l e m e n t s

Solution 2: nodal time step (interface TYPE7 and TYPE10)

d t n = 2 m k k = k interface + k elements

Kinematic time step (interface TYPE7):

dt< ( gappene ) 2V

Where,
V
Impact speed