Week - 2 - Explicit and Implicit Analysis

Objective: To solve the non-linear problem using explicit and implicit method of approach with a force on a bar under tension and displacement being a non-linear displacement.
Introduction
This article presents an example nonlinear static problem. The problem is solved by both an explicit and implicit analysis. In both cases an incremental load control scheme is used. The example as presented is certainly contrived. To a beginner some aspects of this exercise may seem strange. However, the example problem is constructed to illustrate a variety of concepts that arise in a typical nonlinear finite element analysis. It is helpful to think of this exercise as providing some introductory tools that are useful to know for future problems.

Consider the case of a simple bar in tension as shown in fig. Suppose that the force in the bar is a nonlinear function of the displacement
f (u) = u3+9u2+4u ………………………..(1)

From (1) it follows that the stiffness of the bar is
K (u) = dF/du = 3u2+18u+4

Explicit Analysis:
Clearly it is possible to plot a graph of force, f, versus displacement, u, by using equation (1) (see Figure 2). However, if the nonlinear force versus displacement relationship is not known, but the stiffness is known, it is possible to construct the graph of force versus displacement numerically. One method is to use an incremental explicit load control scheme. This is illustrated next. For this example 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 external force applied to the bar and k for stiffness. Incremental displacements or incremental externally applied forces are represented as ∆u or ∆F, respectively. Use is made of the relationship ∆F = k∆u. The analysis proceeds as follows
∆F = k ∆ u
Step-1
uo= 0.0, → k (u) = 4, ∆F = 1,
∆ u1 = ∆F/ k (u0) = 0.25,
u1 = u0 + ∆u1 = 0.0+ 0.25 = 0.25.
F external = ∆F = 1
F internal = u3+9u2+4u = 0.253 +9 x (0.25)2 +4 x (0.25) = 1.5781

Step-2
u1= 0.25, k (u1) =8.688, ∆F =1
∆ u2 = ∆F/ k (u2) = 0.1151,
u2 = u1 + ∆u2 = 0.25+ 0.1151 = 0.3651
F external = ∆F+∆F = 1+1 =2
F internal = u3+9u2+4u = (0.3651)3 +9 x (0.3651)2 +4 x (0.3651) = 2.708
Step -3
u2 = 0.3651, k (u1) = 10.9717, ∆F =1,
∆ u3 = ∆F/ k (u2) = 0.09114,
u3 =u2 +∆u3 = 0.3651 + 0.09114 = 0.4562
F external = ∆F+∆F+∆F = 1+1+1 =2
F internal = u3+9u2+4u = (0.4562)3 +9 x (0.4562)2 +4 x (0.4562) = 3.7928
Here F external ≠ F internal
Step i ∆F i ∆ u i u i (F ext)i (F int)i Fint − Fext = R
1 1.0 0.25 0.25 1 1.5781 0.5781
2 1.0 0.1151 0.3651 2 2.708 0.708
3 1.0 0.09114 0.4562 3 3.7928 0.7928

IMPLICIT ANALYSIS
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, in order 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 u j gives
R j+1 = R j (u j) + dR(u j ) du δuj+1 + ... = R j + k(u j )δuj+1 + ...
By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the following correction is obtained δuj+1 = −[k(u j )]−1R 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 analysis proceeds as follows (results are approximate, they are more precise if done in a computer to machine precision):

Step-1
uo= 0.0, → k (u) = 4, ∆F = 1,
∆ u1 = ∆F/ k (u0) = 0.25,
u1 = u0 + ∆u1 = 0.0+ 0.25 = 0.25
Checking the Residual |R|
F external = ∆F = 1
F internal = u3+9u2+4u = 0.253 +9 x (0.25)2 +4 x (0.25) = 1.5781
R (0) = Fint − Fext =0.5781
| R (0)| > 10-2----- Newton Raphson iteration is necessary.

Calculate the correction to u1 =u1 (0)
δu (1) = −[k(u1 (0) )]−1. R (0) = - (8.6875)-1. (0.578) = - 0.0665
The updated value of u1 (1) = u1 (0) + δu1 (1) =0.25 - 0.0665 =0.1835
Checking the Residual |R|
F external = ∆F = 1
F internal = (u1 (1)) 3+9 (u1 (1)) 2+4(u1 (1)) = 1.0432
R (1) = Fint − Fext =1-1.0432= 0.0432
| R (1)| > 10-2----- Newton Raphson iteration is necessary.
Calculate the new total correction to u1 = u1 (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 (2) = u1 (0) + δ u (2) = 0.25 -0.0723=0.177

Checking the Residual |R|
F external = ∆F = 1
F internal = (u1 (2)) 3+9 (u1 (2)) 2+4(u1 (2)) = 1.0006
R (2) = Fint − Fext =1- 1.0006= 0.0006
| R (2)|< 10-2----- No further iteration is necessary.
Final value of u1 = u1 (2) = 1.077
Step-2
u1 =0.177, k (u1) = 7.293, ∆F = 1,
∆ u2 = ∆F/ k (u1) = 1/7.293 = 0.137, u2 = u1+ ∆ u2 = 0.3148,
Checking the Residual |R|
F external = ∆F+∆F = 1+1= 2
F internal = (u2) 3+9 (u2) 2+4(u2) =2.182
R (0) = Fint − Fext = 2.0 - 2.182= 0.182
| R (0)|>10-2----- Newton Raphson iteration is necessary.

Calculate the correction to u1 =u1 (0)
δu (1) = −[k(u2 (0) )]−1. R (0) = - (9.963)-1. (0.182) = - 0.0182
The Updated value of u2 (1) = u2 (0) + δ u (1) = 0.314 -0.0182=0.296
Checking the Residual |R|
F external = ∆F+∆F = 1+1 =2
F internal = (u2 (1)) 3+9 (u2 (1)) 2+4(u2 (1)) = 2.0042
R (1) = Fint − Fext =2- 2.0042 = 0.0042
| R (1)|< 10-2----- No further iteration is necessary.
Final value of u2 = u2 (1) = 0.2966

Step-3
u2 = 0.296, k (u2) = 9.602, ∆F = 1,
∆ u3 = ∆F/ k (u2) =0.1041,
u3 = u2 +∆ u3 = 0.4007
Checking the Residual |R|
F external = ∆F+∆F+∆F = 1+1+1= 3
F internal = (u3) 3+9 (u3) 2+4(u3 (2)) = 3.112
R (0) = Fint − Fext =3- 3.112= 0.112
| R (0)| >10-2----- Newton Raphson iteration is necessary.

Calculate the correction to u2 =u2 (0)
δu (1) = −[k(u3 (0) )]−1. R (0) = - (11.6942)-1. 0.112 = -0.00958
The Updated value of u3 (1) = u2 (0) + δ u3 (1) = 0.296 -0.00958= 0.391

Checking the Residual |R|
F external = ∆F+∆F+∆F = 1+1+1= 3
F internal = (u3 (1)) 3+9 (u3 (1)) 2+4(u3 (1)) = 3.008
R (1) = Fint − Fext =3- 3.008= 0.112
| R (1)| >10-2----- No further iteration is necessary.
Final value of u3 = u3 (1) = 0.391

Step i ∆F i u i (F ext)i (F int)i Fint − Fext = R
1 1.0 0.177 1 1.006 0.006
2 1.0 0.296 2 2.0042 0.0042
3 1.0 0.391 3 3.008 0.008

Conclusion:
A nonlinear analysis is illustrated using an incremental explicit and implicit procedure.
The results are plotted for comparison. It is evident from the results that the explicit analysis drifts from the true solution.
To overcome this problem an implicit analysis is used, which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces.
The techniques demonstrated are for a load control scheme. For other schemes such as displacement control, arc length control and generalized displacement control the reader is referred to the references given below. Many of the concepts and methods introduced are similar to concepts and methods used in a nonlinear finite element analysis. Therefore, the given example is a useful learning tool for students learning about methods of solution for nonlinear problems
The results of the implicit analysis are summarized in. It is evident from the table that the external forces and internal forces are coming into equilibrium.