Explicit and Implicit Analysis

AIM : solve the given equation, F(u) = u3+9u2+4u in both IMPLICIT and EXPLICIT method.

OBJECTIVE :

• solve the equation, F(u) = u3+9u2+4u using EXPLICIT method.
• solve the equation, F(u) = u3+9u2+4u using IMPLICIT method.
• compare, both the method values.
• plot the graph of the values as comparison and shared it.

PROCEDURE :

1) Brief intro on EXPLICIT method :

EXPLICIT is the method, In which the information at time step (n+1) can be obtained in terms of previous time steps and there is no dependence on the current time step. In this method direct computation of the dependent variables can be made in terms of known quantities.

In EXPLICIT analysis , solution is performed with direct iterative calculations.however, errors accumulate and the answer will diverge over time from the answer you are tring to model but the method is fast and inexpensive computationally.

calculations :

Consider the case of a simple bar in tension. Suppose that the force in the bar is a nonlinear function of the displacement

F(u) = u3+9u2+4u as equation (1)

and it follows that the stiffness of the bar is

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

Clearly it is possible to plot a graph of force, f, versus displacement, u, by using equation (1). 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 delta(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 delta(u) or delta(F),respectively. Use is made of the relationship delta(F) = k*delta(u). The analysis proceeds as follows:

delta(F) = k*delta(u). as equation (3)

initial condition step 1 : at u(0) = 0.0 , `/_ F = 1`

from eq (2) : `k(u) = 3(u^2)+18u+4` `

`k(0) = 3(0^2) + 18(0)+ 4` (:as u(0) = 0.0)

`k(0) = 4`

from eq (3) : `/_ F = kxx /_ u`

`1 = 4 xx /_ u`

`/_ u = 0.25`

now, `u1 = u(0) + /_ u`

`u1 = 0.00+0.25`

`u1 = 0.25`

2nd step condition : u1 = 0.25 , `/_ F = 1`

from eq(2):

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

`k(0.25) = 3xx(0.25^2) + 18xx(0.25)+ 4`

`k(0.25) = 8.6875`

from eq(3):

`/_ F = kxx /_ u```

`/_ u2 = 0.1151`

`u2 = u1 + /_u2`

`u2 = 0.25 + 0.1151`

`u2 = 0.3651`

3rd step condition :

`u2 = 0.3651` , `/_ F = 1`

from eq (2)

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

`k(0.3651) = 3xx(0.3651^2) + 18xx(0.3651)+ 4`

`k(0.3651) = 10.971`

from eq(3) :

`/_ F = k xx /_ u`

`/_ u3 = 0.0911`

`u3 = u2 + /_u3`

`u3 = 0.4562`

now , `F_(ext) = 1 + 1 +1`

and from eq (1)

`f_(i nt 1) = u1^3 + 9xxu1^2+ 4xxu1` `

`f_(i nt1) = (0.25^3) + 9xx(0.25^2)+ 4xx(0.25)`

`f_ (i nt1) = 1.5781`

for (u2)

`f_(i nt 2) = u2^3 + 9xxu2^2+ 4xxu2`

`f_(i nt2) = (0.3651^3) + 9xx(0.3651^2)+ 4xx(0.3651)`

`f_(i nt2) = 2.7087`

for (u3)

`f_(i nt 3) = u3^3 + 9xxu3^2+ 4xxu3`

`f_(i nt3) = (0.4562^3) + 9xx(0.4562^2)+ 4xx(0.4562)`

`f_(i nt3) = 3.7928`

residuals :

for step 1

`R1 = f_(i nt1) - F_(ext1)`

`R1 = 0.5781`

for step 2

`R2 = f_(i nt2) - F_(ext2)`

`R2 = 0.7087`

for step 3

`R3 = f_(i nt3) - F_(ext3)`

`R3 = 0.7928`

TABLE :

step	`/_ F`	`/_ u`	u	`F_(ext)`	`f_(i nt)`	Residual
1	1	0.25	0.25	1	1.5781	0.5781
2	1	0.1151	0.3651	2	2.7087	0.7087
3	1	0.0911	0.4562	3	3.7928	0.7928

2) Brief intro on IMPLICIT analysis method :

IMPLICIT methods are those where the information at time step (n+1) is dependent on previous time levels (n,n-1,..) and the current (n+1). it means that the solution need to be computed by adding another formulation since some terms can not be calculated explicitly at the searched step : this is an indirect calculation.

this methods are more computationally intensive, but tends to have less errors as the simulation progresses.

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^jxx(u^j) + ((dR(u^j))/(du))xxdeltau^(j+1) + ...`

By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for `deltau^(j+1)` the following correction is obtained

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

CALCULATIONS :

Initial step :

u = 0 , k(0) = 4 , `/_ F = 1` , `/_ u = 0.25`

`u1 = u(0) + /_u`

`u1 = 0.25`

`f_(i nt1)= 1.5781`

`F_(ext) = 1`

residual `R0 = f_(i nt1) - F_(ext1)`

`R0 = 0.5781`

as we can see R0>0.01 , now we have to use NEWTON RAPHSON interations method

now ,

u1 = u1'

`deltau' = -[k(u1')]^-1xxR0`

`deltau' = -0.066`

the updated value of `u1'' = u1' + deltau1'`

`u1'' = 0.184`

check for residual again `F_(ext) = 1` and `f_(i nt1) = (u1'')^3+ 9xx(u1'')^2+4xx(u1'')` ; `f_(i nt1) = 1.0469`

residual `R' = f_(i nt1) - F_(ext1)`

`R' = 0.0469` which is greater than 0.01 , again we have to use NEWTON RAPHSON interations

new total correction

u1 = u1'

`deltau'' = deltau' - [k(u1'')]^-1. R'`

`deltau''= -0.0723`

updated value of `u1''' = u1' + deltau''` ; `u1''' = 0.1777`

check residual again

`F_(ext) = 1`

`f_(i nt) = 1.0006`

`R''= 0.0006` < 0.01

hence the final value of u1 = 0.1777

similarly we found out the values of u2= 0.2965 and u3= 0.3912

TABLE

step	`/_ F`	`/_ u`	u	`F_(ext)`	`f_(i nt)`	residual
1	1	0.25	0.1777	1	1.0006	0.0006
2	1	0.1371	0.2965	2	2.0032	0.0032
3	1	0.1041	0.3912	3	3.002	0.002

PLOT GRAPH :

the graph shows the values of force verses displacement by both IMPLICIT and EXPLICIT method.

RESULT AND CONCLUSION :

The results are plotted for comparison. It proofs from the results that the explicit analysis drifts more than an implicit analysis used, which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces.