Week - 2 - Explicit and Implicit Analysis

OBJECTIVE: -

        To demonstrated in the PDF, F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods with tolerance of 10^-2.

Explicit method: -

It is use in crash-worthiness, drop test and other such dynamic analysis where deformation is very large. It is very suitable compare to implicit method for complex problem with number of components and elements.

u(n+1) = u(n)+Δt × f(u(n))

Force function: -   F(u)=u3+9u2+4u

Stiffness: -     k(u)=f′(u)=3u2+18u+4

Implicit method: -

The implicit scheme that checks convergence after each time step and then only moves in to the calculation of variables. The convergence is brought about by Newton-Raphson iterations.

For the convergence the F_ext should be equal to F or the residual R=F−F_ext=0

After each time step, the residual need to be brought within the tolerance value. (Here we assume tolerance=0.01) This is done by using the Newton-Rapshon iteration to calculate the correction value until required tolerance reached.

 

PROGRAM: -

Explicit method: -

<!-- [if gte vml 1]> <![endif]--><!-- [if !vml]--><!--[endif]-->

 Summary of Explicit analysis results:  

Step i	▲ Fi	▲ui	ui	(Fe)i	(fi)i	fi - Fe = R
1	1	0.25	0.25	1	1.5781	0.5781
2	1	0.115	0.365	2	2.7076	0.7076
3	1	0.0911	0.4561	3	3.7898	0.7898

 

Implicit method: -

 

  

 

 Summary of Explicit analysis results:  

Step i	▲ Fi	ui	(Fe)i	(fi)i	fi - Fe = R
1	1	0.1777	1	0.9984	-0.0016
2	1	0.2965	2	2.0032	0.0032
3	1	0.3911	3	3.0008	0.0008

 

Exact Solution by solving third order polynominal:

F(u)=u³+9u²+4u

u³+9u²+4u−F(u)=0

Step i	(Fe) i	ui
1	1	0.17762
2	2	0.29616
3	3	0.39103

 

Plot Results:

 

From the above plots, we obtained that the Implicit method gives a better solution comparing with the Explicit method. The implicit method will be equal to the exact solution due to Newton Raphson iterations used to enforce the equilibrium of external & internal forces.