Week - 2 - Explicit and Implicit Analysis

Explicit and Implicit Methods

Aim: To understand and solve the given equation using both implicit and Explicit Methods.

Explicit and Implicit Methods:

Explicit Method: Explicit method can be used for dynamic and non-linear problems with shorter time steps. Each time step in the explicit method is very fast, but the method is stable only when time step `delta t` is small.

Implicit Method: Implicit method can be used to solve linear problems or problems with higher time increments. Implicit algorithms may not always form Newton-Raphson scheme convergence.

Given equation:

F(u) = `u^3 + 9(u)^2 + 4u` (1)

From (1), the stiffness of the bar will be

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

Explicit method:

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:

Step 1:

u0 = 0 => K(u0) = 4, `triangle F` = 1 => `triangleu1 =(triangleF)/ (k(u0)`= 1/4 = 0.25

Step 2:

k(u1) = `3(1/4)^2+ 18(1/4) + 4` = 8.6875, `triangle F =1`

=> `triangleu2 = (triangleF)/(k(u1)` = 1/8.6875 = 0.115`

u2 = u1 + `triangle u2`= 0.25 +0.1151 = 0.3651

Step 3:

k(u2) = `3(0.3651)^2 + 18 (0.3651) +4`= 10.9716 , `triangle F =1`

=>`triangleu3 = (triangleF)/(k(u2)` = `1/10.9716` = 0.0911

u3 = u2 + `triangle u3` = 0.365+0.0911 = 0.4562

Implicit Method:

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 uj gives

`R^(j+1) = R^j(u^j)+(dR(u^j))/(du)deltau^(j+1 )+.....= R^j +k(u^j)deltau^(j+1)+.....` (3)

By neglecting higher-order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the following correction is obtained

`deltau^(j+1) = -[k(u^j)]^-1 R^j` (4)

Step1

u0 = 0.0, => k(u0) = 1, `triangleF = 1``triangleu1 = (triangleF)/(k(u0) ) = 0.25` , u1 = u0 +`triangleu1` = 0.25

checking the residual ,F(ext) = `triangleF = 1`, f(int) =1.5781, R(0) = f(int)-F(ext) = 0.5781

(Given tolerence = 0.001)

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

Calucating the correction to u1 = u1(0)

`triangleu1 = -[k(u(1) 0) ]^-1R0(0) = (-8.6875)^-1(0.5781) = -0.0665.`

the updated value of u1(1) = u1(0)+`triangleu1` = 0.25+(-0.0665) = 0.1835

Checking the residual, R(1) = f(int)- F(ext) =0.0432

`(1)> `10^-2` =>Newton Raphson iterations are necessary`

calculating the correction to u1 =u1(0)

`triangle u2=triangle u1-[k(u1(1))]^-1R(1)`= -0.0665-(7.404^-1) 0.0432 = -0.0723

=> the updated value of u1(2) = u1(0) + `triangleu2` = 0.25 +(-0.0723) = 0.1777

check the residual, R(2) = f(int)- F(ext) = 0.0006

R(2)<`10^-2`=>Newton Raphson iterations is not necessary`

Thus, the final value of u1 = 0.177

Similarly, u2 = 0.2132 and u3 = 0.3411

Conclusion:

A nonlinear analysis is illustrated using an incremental explicit and implicit procedure. 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. Many of the concepts and methods introduced are similar to concepts and methods used in a nonlinear finite element analysis.