Numerical Solution of Explicit and Implicit Analysis

G DRIVE LINK: https://drive.google.com/drive/u/0/folders/1S18pT4BTUAOJ-_CcfmpdcvvKG5dY2Guq

AIM

• To sloving the given one dimensional quadratic polynominal fuction in both implicit and explicit method and compraing and determine the most feasible sloution.To find that we consider 1d bar element which constraint one end and apply the other end in F the bar under tensile load as shown image

Explicit defination

• It can be computed forward in time using quantities from the previous time step and it has the dynamic analysis during the analysis we can calculate the internal effect structure and there is no force equilibrium between the internal structure and out side
• Each time stepin explicit method is very fastbut the method is stable onlty the when time step is very small ie mile seconds and Explicit Method simulations are conditionally stable. It calculates the system status at a future time from the currently known system times

Implicit defination

• An Implicit FEM analysis is the same as Explicit with the addition that after each increment the analysis does Newton-Raphson iterations to enforce equilibrium of the internal structure forces with the externally applied loads. The equilibirium is usually enforced to some user specified tolerance
• where as the implicit is a time depedent and the force equlibrium ie. the sum of resultant force is equal to the applie d force
• Implicit uses Newton-Raphson iterations to enforce equilibrium and the method is that during the Newton-Raphson iterations one must update and reconstruct the stiffness matrix for each iteration

Numerical solution for given equation

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

The stiffness matrix differentiating (df/du) we get the below eqn

K(u)=3u²+18u+4 -------------------->2

We know that general governing eqn

F=[K]*[u]

∆f=k*∆u -------------------------->3 (in dynamic condition)

where

F=Force K=Stiffness matrix u=Nodal displacement

SOLVING THE EQUATION BY USING THE EXPLICIT METHOD

let as consider and apply eqn=2 ∆F=1 & u0=0

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

∴∆F=k*∆u1

∆u1=∆F/k*u0

∆u1 =1/4 =>0.25

similarly ∆F=1 and u1=0.25 For load step 2

k(u₁) =3(.25)²+18(0.25)+4

k(u₁) =8.6875

∴∆F=k*∆u1

∆u1=∆F/k*u1

∆u1 =1/8.6875 =>0.115

To find the changing the displacement

∆u=u2-u1

u2 =∆u1+∆u2

u2 =0.25+0.115

u₂=0.365

similarly ∆F=1 and u2=0.365 For load step 3

k(u₃)=3(0.365)²+18(0.365)+4

k(u₃)=10.96

∴∆F=k*∆u3

∆u3=∆F/k*u3

∆u3 =1/10.96 =>0.0912

To find the changing the displacement

∆u3=u3-u2

u3 =∆u3+u2

u3 =0.0912+0.365

u3=0.4562

Now we find the displacement change in similarly we find the internal force

wkt

f=u³+9u²+4u

u₁=0.25

f₁=0.25³+9(0.25)²+4(0.25)

f1=1.578

similarly f2=2.707 & f3=3.792

Load step	∆F	ui	Fexternal	Finternal	R= Fint- Fext
1	1	0.25	1	1.578	0.578
2	1	0.115	2	2.707	0.707
3	1	0.365	3	3.792	0.792

It is evident from the results that external and internal forces are NOT in equilibrium because fext≠f intarnal Table summarizes the main results for the explicit analysis.

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
• Basically, an incremental force is applied to advance the solution forward at the beginning of a step
• However, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given step. Hence, in order to achieve equilibrium, corrections must be made to the displacement.
• This is accomplished by using Newton-Raphson iterations to minimize the residual, R(u) = fint − Fext. Expanding the residual as a Taylor series about the current displacement u j gives

• Rj+1=Rj(uj)+dR(uj)/d*{δu}j+1+1+.....

To removing the higher order we get the below eqn

• Rj+1=Rj+k{uj}{δu}j

where R_j+1=0

• 0= Rj+k{uj}{δu}j+1
• -k{uj}{δu}j+1=Rj
• {δu}j+1={−k(uj)}^-1 *Rj---------------------->4

SOLVING THE EQUATION BY USING THE IMPLICIT METHOD

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

The stiffness matrix differentiating (df/du) we get the below eqn

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

∆f=k*∆u-------------------------->3 (in dynamic condition)

where

F=Force K=Stiffness matrix u=Nodal displacement

let as consider and apply eqn=2 ∆F=1 and u0=0

• k(u0)=3(0)2+18(0)+4

• k(u0)=4

• ∴∆F=k*∆u1
• ∆u1=∆F/k*u1
• ∆u1 =1/4 =>0.25

• u₁=0.25

Apply the u1 value in eqn 1

• f(u₁)=u³+9u²+4u

• =0.25³+9*0.25²+4*0.25

• =1.578

Apply the u1 value in eqn 2

• K(u₁)=3u²+18u+4

• =3(0.25)²+18(0.25)+4

• K(u₁)=8.6875

R_f = F_int-Fext

=1.578-1

=0.578

Apply the ku₁ and R_f value in eqn 4

{δu}j+1={-k(uj)}^-1 *Rj

{δu}j+1={-8.68}^-1 *0.578

{δu}1 =-0.0665

For u1 =∆u1+{δu}1

u1=0.25+{-0.0665}

u₁=0.1835

Newton raphson method ittration-1

To find the internal force put the U₁ value in eqn 1

• f(u)=u³+9u²+4u----------------->1

• f_i(u₁)=0.1835³+9(0.1835)²+4(0.1835)

• f_i(u₁)=1.0432

∴ R(u) = fint - Fext

• = 1.0432-1
• R(u1) = 0.0432

Newton raphson method ittration-2 u₁=1.0424

{δu}2 = δu 1 {-k(u1)}-1 *R1

WKT δu 1=-0.066 u1=1.0432 apply the u1 value in eqn 2

• K(u₁)=3u²+18u+4 ------------->2

• K(u₁)=3(1.0417)²+18(1.0417)+4

• K(u₁)= 26.042

{δu}21 = δu 1 {-k(u1)}-1 *R1

{δu}2 1= -0.0665-{{26.042}-1*0.0432}

{δu}21= -0.068159

u12 =∆u1+{δu}12

u12 =0.25+(-0.073419)

u₁² = 0.181841

f(u₁²)= 0.181841³+9(0.181841)²+4(0.181841)

f(u₁²)=1.0309

∴ R(u) = f int - Fext

= 1.0309-1

R¹(u) = 0.0309

R¹ >10^-2 it indicate the need to perform newton raphson method

Wkt {δu}21= -0.068159 and u12 = 0.181841 apply the u1 value in eqn 2

K(u₁)=3(0.181841)²+18(0.181841)+4

K(u₁)= 7.3723

{δu}3= δu 2 {-k(u12)}-1 *R(2)

{δu}3= -0.068159-{(7.3723)-1 *0.0310}

{δu}3= -0.07236

u13 =∆u1+{δu}13

u₁³ =0.25+(-0.07236)

u₁³ = 0.17764

f(u₁³)= 0.17764³+9(0.17764)²+4(0.17764)

f(u₁³)=1.000169

f(u₁³)=1.0002

∴ R(u) = fint - Fext

= 1.0002-1

R¹(u₁³) = 0.0002

R¹<10^-2

there for u₁=u₁³=0.17764

In load step two and u₁=0.17764

Apply the value in eqn 2

K(u₁)=3u²+18u+4

=3(0.17764)²+18(0.17764)+4

K(u₁)=7.29218

Apply the value in K(u₁) eqn 3

∆f=k*∆u-------------->3

∆u2=1/7.29218

∆u2=0.13714

u2=u1+∆u2

u₂=0.17764+0.13714

u₂=0.31478

Internal force

f_i(u)=u³+9u²+4u

=0.31478³+9(0.31478)²+4(0.31478)

f_i(u)=2.182

R =f_i-f_ex

R =2.182-2

R =.182

R¹ >10^-2 it indicate the need to perform newton raphson method

Newton raphson method ittration-1

u₂=u₂

_{WKT {δu}j+1={-k(uj)}-1 *Rj}

_{{δu}1={-k(uj)}-1 *Rj}

Apply the u₂=0.31478 in eqn 2

K(u)=3u²+18u+4

=3(0.31478)²+18(0.31478)+4

K(u) =9.9632

Apply the k(u) in above eqn`

{δu}1={-k(uj)}-1 *Rj

={-(9.9632)^-1}*0.182

{δu2}1= -0.018267

u21=u20 +{δu2}1

u₂¹=0.31478+(-0.018267)

u₂¹= 0.2965

Internal force

f_i(u)=(u₂¹)³+9(u₂¹)²+4(u₂¹)

=0.2965³+9(0.2965)²+4(0.2965)

f_i(u)=2.0033

R =f_i-f_ex

R =2.0033-2

R =0.0033

R<10^-2

there for u₂¹= 0.2965

In load step three and u₂=0.2965

Apply the value in eqn 2

K(u₂)=3u²+18u+4

=3(0.2965)²+18(0.2965)+4

K(u₂)=9.60073`

∆f=k*∆u2

∆u2=1/9.60073

∆u2=0.10415

u3=u2+∆u2

u₃=0.2965+0.10415

u₃=0.40065

Internal force

f_i(u)=(u₃)³+(9u₃)²+4u₃

=0.40065³+9(0.40065)²+4(0.40065)

f_i(u) =3.1116

R =f_i-f_ex

R =3.1116-3

R =0.1116

R >10^-2 it indicate the need to perform newton raphson method

Newton raphson method ittration-1

WKT _j+1={-k(u_j)}^-1 *R_j

{δu}1={-k(uj)}-1 *Rj

Apply the u₃=0.40065 in eqn 2

K(u)=3u²+18u+4 .

=3(0.40065 )²+18(0.40065 )+4

K(u) =11.69326

Apply the k(u) in above eqn

{δu3}0={-k(uj)}-1 *Rj

={-(11.69326)^-1}*0.1116

=0.0095439

u3=u3 +{δu3}0

u₃=0.40065 +(-0.0095439)

u₃= 0.39110

Internal force

f_i(u)=(u₃⁰)³+9(u₃⁰)²+4(u₃⁰)

=0.39110³+9(0.39110)²+4(0.39110)

f_i(u)=3.00092

R =f_i-f_ex

R =3.00092-3

R =0.00092

R<10^-2

there for u₃=0.39110

Load step		ui	Fexternal	Finternal	R= Fint- Fext
1	1	0.17764	1	1.0002	0.0002
2	1	0.2965	2	2.0033	0.0033
3	1	0.39110	3	3.00092	0.00092

^RESULT

• ^{which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces. The techniques demonstrated are for a load control scheme}
• ^{Many of the concepts and methods introduced are similar to concepts and methods used in a nonlinear finite element analysis. Therefore, the given example is a useful learning tool for me learning about methods of solution for
  nonlinear problems.}

GRAPH COMPARISION

Consulsion

• As of we seen the graph the implicit values are over laping with the explicit value but in other hand the the explicit vs extact values are is not under the force equiblirium because of the time step the implicit time step very large when we comparing the explicit