Week - 2 - Explicit and Implicit Analysis

Explicit and Implicit Analysis:

Aim:

• Write a detailed report about Implicit and explicit methods.
• Solving the partial differential equation using Implicit and explicit methods methods.

Implicit and explicit methods:

For all nonlinear and non static analysis, incremental loads are needed to set up equillibrium betwee. Two methods have been introduced to solve partial differential equations(PDE) related to any probelm in FEA one is time dependent approach where Implicit method is used while the another is time independent approach where Explicit method is used. There are 3 types of non-linearity mainly Geometric non-linearity which occurs due to high deformation or rotation,material non-linearity which occurs due to high change in material parameters with respect to each other, contact non-linearity occurs due to sudden increase or decrease in contact stiffness during the contact of two or more parts of same or different bodies.

During linear problem PDE reduces to matrix equation as per [K][x]=[f] while that for non-linear static problem [K(x)][x]=[f], for non-linear

non-static problem [M][a]+C[v]+[K(x)][x]=[f],

were [M]=Mass matrix,[K]=Stiffness matrix,[x]=Displacement matrix,[v]=velocity matrix,[v]=velocity matrix,C=damping coefficient & [f] is external force matrix.

Calculations:

Given value of internal force F(u) = u^3+9u^2+4u

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

Explicit analysis calculation:

Three equal load increments (or steps) of ∆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 ∆u or ∆F,
respectively. Use is made of the relationship ∆F = k∆u. The analysis proceeds as follows:

Step 1
u0 = 0.0, ⇒ k(u0) =4 , ∆F = 1, ∆u1 = ∆F/k(u0) = 1/4, u1 = u0 + ∆u1 = 0.25
Step 2
⇒ k(u1) =3X0.25^2+18X0.25+4 = 8.68, ∆F = 1, ∆u2 = ∆F/k(u1) = 1/8.68=0.115, u2 = u1 + ∆u2 = 0.3651
Step 3
⇒ k(u2) = 3X0.3651^2+18X0.3651+4=10.972
∆F = 1, ∆u3 = ∆F/k(u2) = 1/10.972=0.091, u3 = u2+∆u3= 0.45625
Finally, from the three steps the following results are obtained.
Fext = ∆F + ∆F + ∆F = 3.0
fint = u3^3+9u3^2+4u3=3.7934

It can be clearly obserevd from the table steps above that external & internal forces are not in equillibrium.

Summary of explicit calculations:

Step i	∆F	ui	∆ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.25	0.25	1	1.578	0.578
2	1	0.3651	0.115	2	2.708	0.708
3	1	0.45625	0.091	3	3.7934	0.7934

Implicit Analysis calculation:

An implicit analysis is slight different than explicit analysis,except at the end of each step, Newton-Raphson iterations is used to enforce equillibrium before moving to the next step.An increamental force is applied to advance the solution forward at the beginning of each step.Internal & external force will not be in equillibrium until the stiffness is linear for the given step.To minimize reasiduals we have R(u)=fint-Fext,

Then again expanding the residuals as Taylor series about displacement(u),

Rj+1 = Rj(uj) + dR(uj)du δuj+1 + ... = Rj + k(uj)δuj+1 + ...

By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the
following correction is obtained
δuj+1 = −[k(uj)]−1Rj. (4)
Notice that the iteration variable is j and it may take several iterations for the norm of the
residual, kRk, to be reduced below the chosen tolerance criteria (for the example below a tolerance of 10−2
is chosen). The implicit analysis proceeds as follows (results are approximate,
they are more precise if done in a computer to machine precision):
Step 1:

u0 = 0.0, ⇒ k(u0) = 4, ∆F = 1, ∆u1 = ∆F/k(u0) = 0.25, u1 = u0 + ∆u1 = 0.25

Check the residual, Fext = ∆F = 1, fint = u1^3 +9 u1^2+4u1=1.578, R(0) = fint − Fext = 0.578

R(0)>1E-02, Newton-Raphson iterations are necessary

Calculate the correction to u1 =u1(0)

δu(1) = −[k(u1(0))]^−1R(0) = −[(3X0.25^2)+(18X0.25)+4]^−1Xo.578 = −0.0665,

updated value of u(1)1 = u(0)1 + δu(1) = 0.25 + (−0.0665) = 0.1835.

Checking the residual again, Fext = 1, fint = 0.1835^3+(9X0.1835)+(4X0.1835) = 1.0432

R1=1.0432-1=0.0432>1E-02, Newton-Raphson iterations are necessary,

Calculating the correction to u1=u1(1), δu(2) = δu(1) − [k(u(1)1)]−1R(1) = −0.0665 − (8.6875)^−1X0.0432 = −0.0714

updated value of u(2)1 = u(0)1 + δu(2) = 0.25 + (−0.0714) = 0.1786,

Checking the residuals again,R2,Fext=1, fint=0.1786^3+(9X0.1786^2)+(4X0.1786)=1.0071,

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

Therfore futhur Iterations of Newton Raphson for step 1 terminated with final value of u(2)1=0.1786 .

Step 2:

k(u1)=(3X0.1786)+(18X0.1786)+4=7.293

∆F=1, ∆u2=∆F/k(u1)=0.1371

u2=u1+∆u2=0.3148

Check residual,Fext=∆F=1,fint=u2^3+9u2^2+4u2=2.1822

R(0)=fint-Fext=0.1822

R(0)>1E-02, Newton-Raphson iterations are necessary

Calculate the correction to u2=u(0)2

δu(1) = −[k(u(0)2)]−1R(0) = −9.9637^−1X(0.1822) = −0.0183
⇒ The updated value of u(1)2 = u(0)2 + δu(1) = 0.2966

Check the residuals again,Fext=∆F=1,fint=u(1)2^3+9u(1)2^2+4u(1)2=2.0042,

R(1)=fint-Fext=0.0042<10^-2

Therfore futhur Iterations of Newton Raphson for step 2 terminated with final value of u2=u2(1)=0.2966

Step 3:

u2=0.2966,k(u2)=9.6027,∆F=1,∆u3=∆F/k(u2)=0.1041,u3=u2+∆u3=0.4007,

Check the residual, Fext=∆F=1,fint=u3^3+9u3^2+4u3=3.1121

R(0)=fint-Fext=0.1121

R(0)>1E-02, Newton-Raphson iterations are necessary,

Calculate the correction to u3=u(0)3,

δu(1) = −[k(u(0)3)]−1R(0) = −(11.6943)^−1X(0.1121) = −0.009585
⇒ The updated value of u(1)3 = u(0)3 + δu(1) = 0.39112

R(1)=fint-Fext=0.00085<10^-2

Therfore futhur Iterations of Newton Raphson for step 3 terminated with final value of u2=u2(1)=0.39112

Summary of implicit calculations:

Step i	∆Fi	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.1786	1	1.0071	0.0071
2	1	0.2966	2	2.00042	0.0042
3	1	0.39112	3	3.00085	0.00085

Conclusion:

Partial differnetial equation is solved using both impilicit & explicit methods and we can observe from the steps above that time taken for solving explicit method is less compared to implicit one.