Week - 2 - Explicit and Implicit Analysis
Table of Contents
1 Introduction
2 Aims
3 Explicit and Implicit Analysis
4 Explicit and Implicit Calculations
5 Conclusion
1 Introduction
LS-DYNA is an advanced general-purpose multiphysics simulation software package developed by the former Livermore Software Technology Corporation (LSTC), which was acquired by Ansys in 2019. While the package continues to contain more and more possibilities for the calculation of many complex real-world problems, its origins and core competency lie in highly nonlinear transient dynamic finite element analysis (FEA) using explicit time integration. LS-DYNA is used by the automobile, aerospace, construction and civil engineering, military, manufacturing, and bioengineering industries.
The report presents an example of a nonlinear static problem. In both cases, an incremental load control scheme is used. The given equation is solved by both an explicit and implicit scheme.
The Problem
Consider the case of a simple bar in tension as shown in Figure 1. Suppose that the force in the bar is a nonlinear function of the displacement.
f(u)= u^3+ 9u^2+ 4u (1)
From (1) it follows that the stiffness of the bar is
k(u)=df/du= 3u^2+ 18u + 4
2 Aims
1) Write a detailed report about implicit and explicit methods.
2) Solve the given Partial Differential Equation (PDF) using both implicit and explicit schemes and plot the graphs for the same.
3 Explicit and Implicit Analysis
For all nonlinear and non-static analysis, an incremental load (also known as displacement steps) are needed. In more simplistic terminology, this means to break down the physics/time relationship to solve a mathematical problem. To achieve this, two groups have been already formed: either time-dependent or time-independent problems. These problems solved commonly using implicit and/or explicit methods.
All these implicit and explicit problems are expressed through mathematical partial differential equations (PDE’s). While today’s computers cannot single-handedly solve PDEs, they are equipped to solve matrix equations. These matrix equations can be linear or nonlinear. In most structural problems, the nonlinear equations fall into 3 categories:
Material Nonlinearity: Where deformations and strains are large (i.e., polymer materials)
Geometric Nonlinearity: Where strains are small, but rotations are large (i.e., thin structures)
Boundary Nonlinearity: Due to the non-linearity of boundary conditions, (i.e., contact problems)
In linear problems, the PDE’s reduce to a matrix equation as:
[K]{x} = {f}
for non-linear static problems as:
[K(x)] {x} = {f}
For dynamic problems, the matrix equations come down to:
[M]{x´´} + [C]{x´} + [K]{x} = {f}
Implicit Analysis:
One method of solving for the unknowns {x} is through matrix inversion is known as an implicit analysis. When the problem is nonlinear, the solution is obtained in several steps and the solution for the current step is based on the solution from the previous step. For large models, inverting the matrix is highly expensive and will require advanced iterative solvers (over standard direct solvers). Sometimes, this is also known as the backward Euler integration scheme. These solutions are unconditionally stable and facilitate larger time steps. Despite this advantage, the implicit methods can be extremely time-consuming when solving dynamic and nonlinear problems.
Explicit Analysis:
Explicit analyses aim to solve for acceleration {x´´}. In most cases, the mass matrix is considered as “lumped” and thus a diagonal matrix. Inversion of a diagonal matrix is straightforward and includes inversion of the terms on the diagonal only. Once the accelerations are calculated at the nth step, the velocity at the n+1/2 step and displacement at the n+1 step are calculated accordingly. In these calculations, the scheme is not unconditionally stable and thus smaller time steps are required. To be more precise, the time step in an explicit finite element analysis must be less than the Courant time step (i.e., the time taken by a sound wave to travel across an element) while implicit analysis has no such limitations.
4 Explicit and Implicit Calculations
Explicit Calculations:
Three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar. The variables used are u for displacement, f for internal force in the bar, F for the external force applied to the bar, and k for stiffness. Incremental displacements or incremental externally applied forces are represented as ∆u or ∆F, respectively. The relationship between applied forces defined as ∆F = k*∆u.
Step 1
u0 = 0.0, ⇒ k(u0) = 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25
Step 2
⇒ k(u1) = 8.6875, ∆F = 1, ∆u2 = ∆F/k(u1) = 0.115, u2 = u1 + ∆u2 = 0.365
Step 3
⇒ k(u2) = 10.969, ∆F = 1, ∆u3 = ∆F/k(u2) = 0.0911, u3 = u2+∆u3 = 0.4561
Finally, from the three steps the following results are obtained.
Fext = ∆F + ∆F + ∆F = 3.0
fint = u3 + 9u2+ 4u = 3.7908
It is evident from the results that external and internal forces are NOT in equilibrium because Fext ≠ fint.
Table 1 summarizes the main results for the explicit analysis.
| Step i | ΔFi | Δui | ui | (Fext)i | (fint)i | fint- Fext = R |
| 1 | 1.0 | 0.25 | 0.25 | 1.0 | 1.578 | 0.578 |
| 2 | 1.0 | 0.115 | 0.365 | 2.0 | 2.708 | 0.708 |
| 3 | 1.0 | 0.0911 | 0.456 | 3.0 | 3.7908 | 0.7908 |
Implicit Calculations:
An implicit analysis is the same as the explicit analysis, except that at the end of each step Newton-Raphson iterations are used to enforce equilibrium before moving to the next step. Basically, an incremental force is applied to advance the solution forward at the beginning of a step. However, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given step. Hence to achieve equilibrium, corrections must be made to the displacement. This is accomplished by using Newton-Raphson iterations to minimize the residual, R(u) = fint − Fext. Expanding the residual as a Taylor series about the current displacement uj gives
R^(j+1)= R^j (u^j )+ (dR u^j/du)*δu 6(j+1)+ … = R^j+ k (u^j )δu^(j+1)+ …
By neglecting higher-order terms in the series, setting Rj+1 = 0, and solving for δuj+1 the following correction is obtained
Δu j+1 = − [k (u j)] −1 R j
Notice that the iteration variable is ‘j’ and it may take several iterations for the norm of the residual, {R}, to be reduced below the chosen tolerance criteria (for the example below a tolerance of 10−2 is chosen). The implicit calculations proceed as follows.
Step 1
u0 = 0.0, ⇒ k(u0) = 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25
Check the residual, Fext = ∆F = 1, fint = u13 + 9 u12 + 4 u1= 1.578, R (0) = fint − Fext = 0.578
R (0) > 10−2 ⇒ Newton-Raphson iterations are necessary.
Calculate the correction to u1 = u (0)
δu (1) = −[k(u1(0))] −1 R (0) = −8.6875 −1 * 0.578 = −0.0665
The updated value of u1 (1) = u1(0) + δu(1) = 0.25 + (−0.0665) = 0.1835
Check the residual again, Fext = 1, fint = (u1(1))3 + 9 (u1(1))2 + 4 (u1(1)) = 1.0432
R (1) = fint − Fext = 0.0432
R (1) > 10−2 ⇒ another Newton-Raphson iteration is necessary.
Calculate the new total correction to u1 = u (0)
δu(2) = δu(1) − [k(u1(1))] −1 R(1) = -0.0665 – (7.404)-1*0.0432= -0.07233
The updated value of u1 (2) = u1 (0) + δu(2) = 0.25 + (-0.0723) = 0.1777
Check the residual, Fext = 1, fint = (u1(2))3 + 9 (u1(2))2 + 4 (u1(2)) = 1.0006
R (2) = fint − Fext = 0.0006
R (2) < 10>
Therefore, the final value is u1 = u1 (2) = 0.1777
Step 2
k(u1) = 3(0.17772) + 18(0.1777) + 4 = 7.293, ∆F = 1, ∆u2 = ∆F/k(u1) = 0.1371
u2 = u1 + ∆u2 = 0.3148
Check the residual, Fext = ∆F = 1, fint = u23 + 9 u22 + 4 u2 = 2.1822
R (0) = fint − Fext = 0.1822
R (0) > 10−2 ⇒ Newton-Raphson iterations are necessary.
Calculate the correction to u2 = u2(0)
δu (1) = −[k(u2(0))] −1 R (0) = -(9.9637)-1 * 0.1822 = -0.0183
⇒ The updated value of u2 (1) = u2 (0) + δu (1) = 0.2966
Check the residual again, Fext = ∆F = 1, fint = (u2 (1))3 + (9 u2 (1))2 + 4 u2 (1) = 2.0042
R (1) = fint − Fext = 0.0042
R (1) < 10>
Therefore, the final value is u2 = u2(1) = 0.2966
Step 3
U2 = 0.2966, ⇒ k(u2) = 9.6027, ∆F = 1, ∆u3 = ∆F/k(u2) = 0.1041, u3 = u2 + ∆u3 = 0.4007
Check the residual, Fext = ∆F = 1, fint = u33 + 9 u32 + 4 u3 = 3.1121
R (0) = fint − Fext = 0.1121
R (0) > 10−2 ⇒ Newton-Raphson iterations are necessary.
Calculate the correction to u3 = u3(0)
δu (1) = −[k(u3(0))] −1 R (1) = -(11.6943)-1 * 0.1121 = - 0.009585
⇒ The updated value of u3 (1) = u3 (0) + δu (1) = 0.39112
Check the residual again, Fext = ∆F = 1, fint = (u3 (1))3 + (9 u3 (1))2 + 4 u3 (1) = 3.00085
R (1) = fint − Fext = 0.00085
R (1) < 10>
Therefore, the final value is u3 = u3(1) = 0.39112
Table 2: Summary of implicit analysis results
| Step i | ΔFi | ui | (Fext)i | (fint)i | fint - Fext = R |
| 1 | 1.0 | 0.1777 | 1.0 | 1.0006 | 0.0006 |
| 2 | 1.0 | 0.2966 | 2.0 | 2.00042 | 0.0042 |
| 3 | 1.0 | 0.39112 | 3.0 | 3.00085 | 0.00085 |
The results of the implicit analysis are summarized in Table 2. It is evident from the table that the external forces and internal forces are coming into equilibrium. However, the table 3 presents a comparison of the results obtained using explicit, implicit, and exact schemes, respectively.
Table 3: Summarises results of explicit, implicit, and exact schemes.
| Step i | Force (F) | Explicit (u) | Implicit (u) | Exact (u) |
| 0 | 0 | 0 | 0 | 0 |
| 1 | 1.0 | 0.25 | 0.1777 | 0.17762 |
| 2 | 2.0 | 0.365 | 0.2966 | 0.29616 |
| 3 | 3.0 | 0.456 | 0.39112 | 0.39103 |
5 Conclusion
The main aims of the Week 2 challenge were to write a detailed report about implicit and explicit methods and also solve the given Partial Differential Equation (PDF) using both implicit and explicit methods. The results obtained using both schemes were plotted to understand which of the schemes yield better and effective results. The results presented that the explicit analysis drifts from the true solution illustrating the lack of equilibrium between the internal and external forces. To overcome this problem an implicit analysis should be used, which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces.