Week - 2 - Explicit and Implicit Analysis

                                                               Explicit and Implicit Analysis - LS DYNA Week-2

Objective: To use the equation F(u) = u3+9u2+4u, and solve using both Explicit and Implicit Methods ( have a tolerence of 0.001).

Theory:

Implicit Time Integration:

Advantages:

• It is Unconditionally stable(no timestep limit).
• Used in static analysis.
• It is relatively inexpensive for long duration analysis.

Disadvantages:

• Often requires a large amount of memory.
• It can cause problems with strongly non-linear models.

Explicit Time Integration:

Advantages:

• Computationally fast.
• Low core / RAM intensive.

Disadvantages:

• Conditionally stable(timestep limit).
• Expensive to conduct long duration analysis.

Applications:

Implicit:

• Linear and non-linear static analysis.
• Low rate dynamic analysis.
• Vibration and Modal analysis.
• Strength and buckling.
• Springback.
• Gravity loading and pre-loading.

Explicit:

• High rate dynamic analysis.
• Impact / penetration problems.
• Car crash.
• Explosives.

1. Explicit scheme:

• If the nonlinear force v/s displacement relationship is not known and the stiffness is known, then it is possible to plot the graph for force v/s displacement numerically. This can be done using an incremental explicit load control scheme.
• Now, the required equation is solved using explicit method:

`F(u)=u^3+9u^2+4u`

• Differentiating force with respect to displacement, we get stiffness(k) of the bar,(since F = kx).

`k(u)=(dF)/(du)=3u^2+18u+4`

• Three equal load increments or steps are needed with `ΔF=1` unit are used to load the tension bar.
• Here, u = displacement,
• f  = internal force in the bar,
• F = external force applied to the bar
• k = stiffness
• Incremental displacements or incremental externally applied forces are referred to as `Δu` or `ΔF`.

`ΔF=k.Δu` This relation is used to find the incremental displacement.

Step1:

`u_0=0.0, ⇒k(u_0 )=4,ΔF=1,Δu_1=(ΔF)/(k(u_0))=0.25,u_1=u_0+Δu_1=0.25`

`F_(external)= ΔF=1`

`F_(Internal)= u^3+9u^2 + 4u=1.5781 (∵u=u_1)`

Step2:

`u_1=0.25,⇒k(u_1)=8.688,ΔF=1`

`Δu_2=(ΔF)/(k.(u_1))=0.1151.`

`u_2=u_1+Δu_2=0.3651.`

`F_(external)=ΔF+ΔF=2`

`F_(Internal)=u^3+9u^2+4u=2.708 (∵u=u_2)`

Step3:

`u_2=0.3651,⇒k(u_2)=10.9717,ΔF=1,Δu_3=(ΔF)/(k(u_2))=0.0911,u_3=u_2+Δu_3=0.4562`

`F_(external)=ΔF+ΔF+ΔF=3`

`F_(Internal)=u^3+9u^2+4u=3.7928 (∵u=u_3)`

•  It is evident from the results that external and internal forces are NOT in equilibrium because `F_(external)≠F_(Internal)`.

   Table:

2. Implicit scheme:

• Implicit analysis is similar to explicit analysis, except that at the end of each step, Newton-Raphson iterations are introduced to enforce equilibrium before moving to next step. Basically, incremental force is applied to advance the solution forward at the beginning of a step. Anyway, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given step.
• Therefore, in order to achieve equilibrium, corrections should be made to the displacement.
• This is acquired by using Newton-Raphson iterations to reduce the residual, `R=F_(Internal)-F_(external)`
• Expanding the residual as a Taylor series about the current displacement `u^j` gives

`R^(j+1)=R^j(u^j)+(dR(u^j))/(du)δu^(j+1)+...=R^j+k(u^j)δu^(j+1)+...`

•  By neglecting higher order terms in above series, i.e. `R^(j+1)=0` and solving for `δu^(j+1)` the following correction can be obtained

`δu^(j+1)=−[k(u^j)]^(−1)⋅R^j`

• The iteration variable here is j and it may take several iterations for the norm of the residual, R to be reduced below the chosen tolerance criteria 0.01.
• The implicit analysis proceeds as follows

Step1:

`u_0=0.0,⇒k(u_0)=4,ΔF=1,`u_1=(ΔF)/(k(u_0))=0.25,u_1=u_0+Δu_1=0.25`

• Checking the residual,

`F_(external)=ΔF=1`

`F_(Internal)=u_1^3+9u_1^2+4u_1=1.578`

`R^((0))=F_(Internal)−F_(external)=0.578`

`R^((0))>10^(−2)⇒`Newton-Raphson iterations are necessary.

• Calculate the correction to `u_1=u_1^((0))`

`δu^((1))=−[k(u_1^((0)))]^(−1)⋅R^((0))=−(8.6875)^(−1)⋅0.578=−0.0665`

• The updated value of `u_1^((1))=u_1^((0))+δu^((1))=0.25−0.0665=0.1835`
• Checking the residual again,

`F_(external)=ΔF=1`

`F_(Internal)=(u_1^((1)))^3+9(u_1^((1)))^2+4u_1^((1))=1.0432`

`R^((1))=F_(Internal)−F_(external)=0.0432`

`R^((1))>10^(−2)⇒`Newton-Raphson iterations are necessary

• Calculate the new total correction to `u_1=u_1^((0))`

`δu^((2))=δu^((1))−[k(u_1^((1)))]^(−1)⋅R^((1))=−0.0665−(7.404)^(−1)⋅0.0432=−0.0723`

• The updated value of `u_1^((2))=u_1^((0))+δu^((2))=0.25−0.0723=0.1777`
• Checking the residual again,

`F_(external)=ΔF=1`

`F_(Internal)=(u_1^((2)))^3+9(u_1^((2)))^2+4u_1^((2))=1.0006`

`R^((2))=F_(Internal)−F_(external)=0.0006`

`R^((2))<10^(−2)⇒`Newton-Raphson iterations are not necessary

• Therefore, the final value is `u_1=u_1^((2))=0.1777`

Step2:

`u_1=0.1777,⇒k(u_1)=7.2933,ΔF=1,Δu_2=(ΔF)/(k(u_1))=0.1371,u_2=u_1+Δu_2=0.3148`

• Checking the residual,

`F_(external)=ΔF+ΔF=2`

`F_(Internal)=u_2^3+9u_2^2+4u_2=2.1822`

`R^((0))=F_(Internal)−F_(external)=0.1822`

`R^((0))>10^(−2)⇒`Newton-Raphson iterations are necessary.

• Calculate the correction to `u_1=u_1^((0))`

`δu^((1))=−[k(u_2^((0)))]^(−1)⋅R^((0))=−(9.9637)^(−1)⋅0.1822=−0.0182`

⇒The updated value of `u_2^((1))=u_2^((0))+δu^((1))=0.3148−0.0182=0.2966`

• Checking the residual again,

`F_(external)=ΔF+ΔF=2`

`F_(Internal)=(u_2^((1)))^3+9(u_2^((1)))^2+4u_2^((1))=2.0042`

`R^((1))=F_(Internal)−F_(external)=0.0042`

`R^((1))<10^(−2)`⇒Newton-Raphson iterations are not necessary

• Therefore, the final value is `u_2=u_2^((1))=0.2966`

Step3:

 `u_2=0.2966,⇒k(u_2)=9.6027,ΔF=1,Δu_3=(ΔF)/(k(u_2))=0.1041,u_3=u_2+Δu_3=0.4007`

• Checking the residual,

`F_(external)=ΔF+ΔF+ΔF=3`

`F_(Internal)=u_3^3+9u_3^2+4u_3=3.1121`

`R^((0))=F_(Internal)−F_(external)=0.1121`

`R^((0))>10^(−2)⇒`Newton-Raphson iterations are necessary.

• Calculate the correction to `u_2=u_2^((0))`

`δu^((2))=−[k(u_3^((0)))]^(−1)⋅R^((0))=−(11.6942)^(−1)⋅0.1121=−0.00958`

⇒The updated value of `u_3^((1))=u_3^((0))+δu^((2))=0.4007−0.00958=0.3911`

• Checking the residual again,

`F_(external)=ΔF+ΔF+ΔF=3`

`F_(Internal)=(u_3^((1)))^3+9(u_3^((1)))^2+4u_3^((1))=3.0008`

 `R^((1))=F_(Internal)−F_(external)=0.0008`

`R^((1))<10^(−2)⇒`Newton-Raphson iterations are not necessary

• Therefore, the final value is `u_3=u_3^((2))=0.3911`

 Table:

Graph:

• We can clearly see that Explicit curve drifts from the exact solution, this is due to lack of equilibrium between internal and external forces.
• Whereas Implicit curve comes close to the exact curve.Here, Newton-Raphson iterations correct the incremental steps so that the solution comes close to exact      curve, according to the specified tolerance. Excellence is achieved within 3 increments and in each step only 2 Newton-Raphson iterations are required.

Conclusion:

Ths the given equation is solved in both implicit and Explicit methods. From the above observations we can say that implicit method wil give better results in non linear static problems when time and computational cost are considered.

Reference: https://people.wallawalla.edu/~louie.yaw/nonlinear/ExIm_analysis.pdf