Week - 2 - Explicit and Implicit Analysis

Aim:- Solve the system of equation in both explicit and implicit methods and compare the result in graphical formate.

Implicit Method:-

• This method is used for static analysis.
• Acceleration is vary linearly with given time step.
• Iteration in allowed in Implicit method.

Explicit Method:-

• This method is used for Dynamic analysis.
• Iteration is not allowed in this method.

Solving equation with Implicit method:-

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

stiffness,k(u){df/du}= 3u^2+18u+4---------------(2)

δu= Displacement

δF= external force

Now,

we know the relation between force,displacement and stiffness."δF=Kδu" tolrance=0.001

Step-1:

From equation(2)

k(u0)=3(0)^2+18(0)+4

k(u0)=4

δu1=δF/k(u0)

δu1= 1/4 =0.25

Then,U1=u0+δu1=0+0.25=0.25

Checking Residual:-

F(external)=δF=1

From equation(1):- F(u)=0.25^3+9*(0.25)^2+4*0.25=1.578

R(Residual)=F(internal)- F(external)

1.578-1=0.578 which is greater than given tolerance value 0.001 So, Newton raphson iterration is necessary.

Calculate the correction to u1=u1(0)

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

Updated ,u1(1)=u1(0)+δu1
=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)

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

u1(2)=u1(1)+δu(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	δf	δui	ui	F(ext)	f(int)	R=f(int)-F(ext)
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

Solving equation with Explicit method:-

Step-1

From equation(2)

k(u0)=3(0)^2+18(0)+4

k(u0)=4

δu1=δF/k(u0)

δu1= 1/4 =0.25

Then,U1=u0+δu1=0+0.25=0.25

Step-2;

u1=0.25

K(u1)=3*0.25^2+18*0.25+4
=8.6875
now,

δu1=δF/u1
=1/8.6875
=0.1151
u2=u1+△u2=0.3651

Step 3:

Take u2=0.3651
K(u1)=3*0.3651^2+18*0.3651+4
=10.9716
now,

δu2=δF/K(u2)
=1/10.9716
=0.09114
u3=u2+δu2=0.46
finally to determine whether the analysis is in equilibrium
F(ext)=δF+δF+δF=3
f(int)=(0.46)^3+(9*0.46^2)+(4*0.46)
=3.841
∴∴ F(ext) is not equal to f(int), they are not in equilibrium

Step1	δf	δu	u	F(ext)	F(int)	R=f(int)-F(ext)
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

Conclusion :

Implicit method takes more time than the explicit method for calculation.