Week - 2 - Explicit and Implicit Analysis

Aim:

To solve the given eqn using both implicit and explicit methods.

Procedure:

1.) Explicit Analysis

The explicit analysis is a kind of analysis in which the value of a a function at the required time step does not relies on the previous timesteps.

Consider the force in the bar is a nonlinear function of the displacement:

f(u)=u3+9u2+4u ------------->1

The stiffness of the bar is:

k(u)= 3u2+18u+4 --------------->2

step1:

u0 = 0.0

k(u)= 3u2+18u+4

k(0)= 4

ΔF=kΔu =>ΔF=1

∆u1 = ∆F/k(u0) =1/4= 0.25, ∆u1=0.25

u1 = u0 + ∆u1 = 0.25

step2:

k(u1) =3u1^2+18 u+4

k(0.25)= 8.6875

∆F = 1,

∆u2 = ∆F/k(u1) =0.11510

u2 = u1 + ∆u2 = 0.3651

step3:

k(u2) = 3(0.3651)^2 + 18 (0.3651) + 4

K(u2)= 10.9716

∆F = 1,

∆u3 = ∆F/k(u2) = 0.0911

u3 = u2+∆u3 = 0.4562

Finally, from the three steps the following results are obtained.

Fext = ∆F + ∆F + ∆F = 3.0

Fint u1= u(1)3+9u(1)2+4u(1)

=0.25^3 + 0.25Z^2 + 4*0.25

= 1.5781

Fint u2= u(2)3+9u(2)2+4u(2)

= 2.7086

Fint u3= u(3)3+9u(3)2+4u(3)

= 3.7927

Summary of explicit analysis results

Step i	∆Fi	∆ui	ui	(Fext)i	(fint)i	fint − Fext = R
1	1.0	0.25	0.25	1	1.5781	0.5781
2	1.0	0.115	0.3651	2	2.7086	0.7086
3	1.0	0.0911	0.4562	3	3.7927	0.7927

2.) Implicit Analysis

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.

R(u) = fint − Fext.

Expanding the residual as a Taylor series about the current displacement uj gives Rj+1 = Rj(uj) + dR(uj)du δuj+1 + ... = Rj + k(uj)δuj+1 + ...

δuj+1 = −[k(uj)]−1Rj
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

step1:

u0 = 0.0,

k(u0) = 3u2+18u+4 =4

∆F= 1 => ∆F = k ∆u

∆u= ∆F/K

∆u= 0.25

check the residual

Fext = ∆F = 1

fint = u1^3+9u1^2+4u1

= 1.58125

R(0) = fint − Fext = 0.58125

R(0) > 10−2 ⇒ Newton-Raphson iterations are necessary

correction to u1 = u1(0)
δu(1) = −[k(u(0)1)]^−1R(0)

k(0.25)= 3u2+18u+4

= 3*0.1875+4.5+4

= (8.6875)^-1= 0.1151

δu(1) =- 0.1151- 0.5812

= -0.0669

updated value u(1)1 = u(0)1 + δu(1) = 0.1831

Check the residual again, Fext = 1, fint = u3+9u2+4u = 1.04018

R(1) = fint − Fext = 1.04018-1

= 0.04018

R(1) > 10−2 ⇒ another Newton-Raphson iteration is necessary.

new total correction to u1 = u1(0)

δu(2) = δu(1) − [k(u(1)1)]−1R(1) = -0.0669-[k(0.1831)]^-1*0.04018

= -0.0725

updated value of u1(2) == u(0)1 + δu(2) = 0.1775

Fext = 1, fint = u3+9u2+4u

= 1.0001

R(2) = fint − Fext = 0.0001

R(2) < 10−2 ⇒ no further iterations are necessary.

final value is u1 = u1(2) = 0.1775

step 2:

k(u1) = 3u1+18u1+4 =7.28451

∆F= 1 => ∆F = k ∆u

∆u= ∆F/K

∆u2=0.137

check the residual

Fext = ∆F = 1

fint = u2^3+9u2^2+4u2

= 2.18118

R(0) = fint − Fext = 0.58125

R(0) > 10−2 ⇒ Newton-Raphson iterations are necessary

correction to u2 = u2(0)
δu(1) = −[k(u(0))]^−1R(0)

k(0.31477)= 3u2+18u+4

= 0.9441+5.6646+4

= (10.6087)^-1=-0.0942

δu(1) =-0.094*1.1814

= -0.11134

updated value u(2)1 = u(0)2 + δu(1) = 0.20343

Check the residual again, Fext = 1, fint = u3+9u2+4u = 0.86351

R(1) = fint − Fext = 0.86351-1

= -0.1364

R(2) < 10−2 ⇒ no further iterations are necessary.

final value = u2=u2(2)=0.20343

step3:

The above procedure is carried out untill the value of R(3) is lesser than 10^-2(0.01)

The final value obtained is is 0.3001.

Summary of implicit analysis results

Step i	∆Fi	ui	(Fext)i	(fint)i	R
1	1	0.25	1	1.0001	0.8225
2	1	0.1371	2	2.18118	0.18118
3	1	0.1046	3	3.001	0.001

Result:

Thus the given equation is solved by means of both implicit and explicit methods inorder to obtain precise values which can be used to plot graph and compare the values.