Week - 2 - Explicit and Implicit Analysis

AIM:

Solve the above given equation in both Implicit and Explicit method and compare the result with exact result

IMPLICIT Method:-

• This method is mostly used in static analysis
• Acceleration will vary linearly within given time step
• Time step is more compared to explicit method
• It needs a numerical solver to invert the stiffness matrix, this costs more for large models
• It Uses Newton-Raphson iterations to enforce equilibrium

EXPLICIT Method:-

• This method of calculation is used in Crash analysis
• Acceleration will be constant within the given time step
• Because we considering the acceleration to be constant the time step step should be very small
• In this method we will inverse Mass diagonal matrix, this makes the solving time less compared to implicit method

Solving equation with Implicit method:

F(u)=`u^3`+9`u^2`+4u -------->(1)
Stiffness K(U)=df/du=3`u^2`+18u+4 ------------->(2)

`triangle`u=incremental displacement,
`triangle`F=Incremental external applied forces

using `triangle`F=K`triangle`u, tolerance 10^-2

STEP 1:-

Take uo=0
(2)----->K(uo)=3(0)^2+18*0+4
=4
now ,`triangle`u1=`triangle`F/K(uo)=1/4=0.25

u1=uo+`triangle`u1=.25

checking residual, R

F(ext)=`triangle`F=1

(1)--->=.25^3+9(.25^2)+(4*.25)
=1.578

RO=1.578-1
=0.578 >10^2

So , newton raphson iterations are necessary

Calculate the correction to u1=u1(0)

`delta`u(1)=-[K(u1(0))]^-1*Ro
=-[(3*.25^2)+(18*.25)+4]^-1*.578
=-.0665

Updated ,u1(1)=u1(0)+`delta`u1
=0.25-0.0665
=0.1835

Checking residual again,R1

F(ext)=1; f(int)=0.185^3+(9*0.185^3)+(4*0.1853)
=1.0432

R1=1.0432-1
=0.0432>10^-2

another newton raphson iteration

Calculate the correction to u1=u1(1)

`delta`u(2)=`delta`u(1)-[K(u1(1))]*R1
=(-0.0655)-(8.6875)^-1*0.0432=-0.0714

u1(2)=u1(1)+`deltau`(2)
=0.25-0.0714
=0.1786

Checking residual again R2

F(ext)=1; f(int)=(1)--->(0.1786)^3+(9*0.1786^2)+(4*0.1786)
=1.0071

R2=1.0071-1
=0.0071<10^-2

Therfore no iteration needed

u1=0.1786

Similarly from the above step calculate for step 2 and step 3

u2=0.2966

u3=0.3911

Step1	`delta`f	`delta`ui	ui	F(ext)i	f(int)i	R=f(int)i-F(ext)i
1	1	.25	0.1786	1	1.0071	0.0071
2	1	0.1375	0.2966	2	2.0042	0.0042
3	1	0.1041	0.3911	3	3.0008	0.0008

IMPLICIT Method:

Step 1:

Take uo=0

(2)--->K(u(0))=3*0^2+18*0+4=4

now `triangle`u1=`triangle`F/K(uo)
=1/4=0.25

u1=u0+`triangle`u1=0.25

Step 2:

u1=0.25
(2)----> K(u1)=3*0.25^2+18*0.25+4
=8.6875
now, `triangle`u1=`triangle`F/u1
=1/8.6875
=0.1151
u2=u1+`triangle`u2=0.3651

Step 3:

Take u2=0.3651
(2)----> K(u1)=3*0.3651^2+18*0.3651+4
=10.9716
now,`triangle`u2=`triangle`F/K(u2)
=1/10.9716
=0.09114
u3=u2+`triangle`u3=0.46
finally to determine whether the analysis is in equilibrium
F(ext)=`triangle`F+`triangle`F+`triangle`F=3
f(int)=(1)-->(0.46)^3+(9*0.46^2)+(4*0.46)
=3.841
`therefore` F(ext) is not equal to f(int), they are not in equilibrium

Step1	`delta`f	`delta`ui	ui	F(ext)i	f(int)i	R=f(int)i-F(ext)i
1	1	.25	0.25	1	1.578	0.578
2	1	0.1151	0.3651	2	2.708	0.708
3	1	0.09114	0.46	3	3.841	0.841

Actual values
Take F(u)=u^3+9u^2+4u=1
we get u=0.1776

F(u)=u^3+9u^2+4u=2
u=0.2962

F(u)=u^3+9u^2+4u=3
u=0.3910

CONCLUSION:

• Comparing the Above results with the exact values we can see that the implicit analysis gives more accurate value
• Implicit method is accurate and suggestale for static analysis
• But we can see the calculation was easy in explicit method
• Explicit analysis is used in the case of non liniearities

CONCLUSION:

• From the above two tables we can see the results