Week - 2 - Explicit and Implicit Analysis

Aim: To Solve the mathematical equation using Explicit and Implicit Method.

f = (u)ᶟ +9(u)² + 4(u)

Time interval problem solving can be broadly classify in two categories

1) Implicit 2) Explicit

The selection of the type of analysis depends on various factor like:

• Computational Characteristics
• Stability
• Accuracy
• Time interval

Implicit Analysis:

In Implicit the solution, will be find by solving an equation that includes current and later stage of the system. In other words, implicit will find the solution of later stage and by add it with present stage we will get the solution of the iteration.

The equation used in Implicit method is

This method is used to find solution for dynamic equation using matrix inversion method. This method is mostly recommended to find the displacement of the system. Even though we can use this method for dynamic method, but it is recommended for static and quais static analysis

This method is inverse of explicit method

Let us consider displacement is zero at force 1

K(u) = 3u² + 18u + 4 will be K(0) = 3(0)² + 18(0) + 4

K(0) = 4

ΔF = K ΔU

ΔU = ΔF/ K

ΔU = 1/4

ΔU = 0.25

U1 = u + ΔU

= 0 + 0.25

= 0.25

External force = F = 1

Internal Force = f = (u)ᶟ +9(u)² + 4(u)

= (0.25)ᶟ +9(0.25)² + 4(0.25)

= 1.5781

Residual force = 1.5781-1

Residual Force = 0.5781

R is greater than 0.001

Therefore we have to use Newton Raphson Method

U1 = U

∂U = [-K(U)]ˆ-1 x R

K(U) = 3u² + 18u + 4

=3(0.25)² + 18(0.25) + 4

= 8.6875

∂U = [-8.6575] ˆ-1 x 0.5781

∂U = -0.0665

Therefore U1 Becomes

U1 = U + ∂U

= 0.25 - 0.0665

= 0.1834

Therefore Internal force will be

Fi = (0.1834)ᶟ +9(0.1834)² + 4(0.1834)

= 1.04

Residual force = 1.04-1

Residual Force = 0.04

R is greater than 0.001

Therefore we have to use Newton Raphson Method

U1 = U

∂U = [-K(U)]ˆ-1 x R

K(U) = 3u² + 18u + 4

=3(0.1834)² + 18(0.1834) + 4

= 7.4021

∂U = [-7.4021] ˆ-1 x 0.04

∂U = -0.0054

Therefore U1 Becomes

U1 = U + ∂U

= 0.1834 – 0.0054

= 0.1779

Therefore Internal force will be

Fi = (0.1779)ᶟ +9(0.1779)² + 4(0.1779)

= 1.0020

Residual force = 1.0020-1

Residual Force = 0.002

R is greater than 0.001

Therefore we have to use Newton Raphson Method

U1 = U

∂U = [-K(U)]ˆ-1 x R

K(U) = 3u² + 18u + 4

=3(0.1779)² + 18(0.1779) + 4

= 7.2971

∂U = [-7.2971] ˆ-1 x 0.002

∂U = -0.0002

Therefore U1 Becomes

U1 = U + ∂U

= 0.1779 – 0.0002

= 0.1776

Therefore Internal force will be

Fi = (0.1776)ᶟ +9(0.1776)² + 4(0.1776)

= 0.9998

Residual force = 0.9998-1

Residual Force = - 0.0001

R is Smaller than 0.001

Therefore For iteration 1

U1 = 0.1776

Iteration 2

U = U1

K(U) = 3u² + 18u + 4

=3(0.1776)² + 18(0.1776) + 4

= 7.2914

ΔF = K ΔU

ΔU = ΔF/ K

ΔU = 1/7.4806

ΔU = 0.13714

U1 = u + ΔU

= 0.1776 + 0.1336

= 0.3147

External Force = F ext = 2

Internal Force = Fi = (u)ᶟ +9(u)² + 4(u)

= (0.3112)ᶟ +9(0.3112)² + 4(0.3112)

= 2.1812

Residual force = 2.1812 -2

Residual Force = 0.1812

R is greater than 0.001

Therefore we have to use Newton Raphson Method

U1 = U

∂U = [-K(U)]ˆ-1 x R

K(U) = 3u² + 18u + 4

=3(0.3112)² + 18(0.3112) + 4

= 9.9617

∂U = [-9.9617] ˆ-1 x 0.1812

∂U = -0.01818

Therefore U1 Becomes

U1 = U + ∂U

= 0.3112 – 0.01818

= 0.2965

Therefore Internal force will be

Fi = (0.2965)ᶟ +9(0.2965)² + 4(0.2965)

= 2.0003

Residual force = 2.003 -2

Residual Force = 0.003

R is greater than 0.001

Therefore we have to use Newton Raphson Method

U1 = U

∂U = [-K(U)]ˆ-1 x R

K(U) = 3u² + 18u + 4

=3(0.2965)² + 18(0.2965) + 4

= 9.6007

∂U = [-9.6007] ˆ-1 x 0.003

∂U = -0.00031

Therefore U1 Becomes

U1 = U + ∂U

= 0.2965 – 0.00031

= 0.2961

Therefore Internal force will be

Fi = (0.2961)ᶟ +9(0.2961)² + 4(0.2961)

= 1.9994

Residual force = 1.9994 -2

Residual Force = -0.0006

R is less than 0.001

Therefore, for iteration 2

U2 = 0.2961

Iteration 3:

U2 = U3

U3 = 0.2961

K(U) = 3u² + 18u + 4

=3(0.2961)² + 18(0.2961) + 4

= 9.5928

ΔF = K ΔU

ΔU = ΔF/ K

ΔU = 1/9.5928

ΔU = 0.104

U1 = u + ΔU

= 0.2961 + 0.104

= 0.4003

External Force = F ext = 3

Internal Force = Fi = (u)ᶟ +9(u)² + 4(u)

= (0.4003)ᶟ +9(0.4003)² + 4(0.4003)

= 3.107

Residual force = 3.107 - 3

Residual Force = 0.107

Fi = (0.4003)ᶟ +9(0.4003)² + 4(4.003)

= 3.107

Residual force = 3.107-3

Residual Force = 0.107

R is greater than 0.001

Therefore we have to use Newton Raphson Method

U1 = U

∂U = [-K(U)]ˆ-1 x R

K(U) = 3u² + 18u + 4

=3(4.003)² + 18(4.003) + 4

= 11.6861

∂U = [-11.6861] ˆ-1 x 0.107

∂U = -0.0091

Therefore U1 Becomes

U1 = U + ∂U

= 0.4003 – 0.0091

= 0.3912

Fi = (0.3912)ᶟ +9(0.3912)² + 4(0.3912)

= 3.002

Residual force = 3.002-3

Residual Force = 0.002

R is less than 0.001

Therefore, for iteration 3

U3 = 0.3912

Iteration	ΔF	ΔU	U	F	f	R = F - f
1	1	0.1776	0.1776	1	1.0001	0.0001
2	1	0.1185	0.2961	2	2.0006	0.0006
3	1	0.2727	0.3912	3	3.002	0.002

2) Explicit Analysis:

In Explicit method the solution, will be obtained by direct solution at different time interval. one of the advantages of explicit over implicit is that, explicit does not to converge after every iteration. This method is mostly used to find the acceleration of the system. This method is mostly recommended for velocity more than 10 meter per second.

ƒ(u)= uᶟ + 9u² = 4u ……….(1)

Stiffness matrix

K(u) = dF(u) ÷ d(u)

= 3u² + 18u + 4 ………..(2)

From equation 1 and 2 we get

ΔF = K ΔU .…….(3)

Iteration 1:

Let use consider

ΔF = 1

u(0) = 0

To find the value of stiffness put initial condition in equation 2

K = 3(0)² + 18(0) + 4

K = 4

To find deflection put value of K=4 in equation 3

ΔF = K ΔU

1 = 4 ΔU

ΔU = 1/4

ΔU = 0.25

U1 = u + ΔU

= 0 + 0.25

U1 = 0.25

Iteration (2) :

Considering value of u = u1

Then stiffness matrix will be

K(u) = 3u² + 18u + 4

K = 3(0.25)² + 18(0.25) + 4

K = 8.6875

To find deflection put value of K = 8.6875 in equation 3

ΔF = K ΔU

1 = 8.6875 ΔU

ΔU =1/8.6875

ΔU = 0.1151

U2 = u + ΔU

= 0.25 + 0.1151

U2 = 0.3651

Iteration 3 :

Considering value of u = u1

Then stiffness matrix will be

K(u) = 3u² + 18u + 4

K = 3(0.3651)² + 18(0.3651) + 4

K = 10.9716

To find deflection put value of K = 10.9716 in equation 3

ΔF = K ΔU

1 = 10.9716 ΔU

ΔU =1/10.9716

ΔU = 0.09114

U3 = u + ΔU

= 0.3651 + 0.09114

U3 = 0.4562

Iteration 4 :

Considering value of u = u1

Then stiffness matrix will be

K(u) = 3u² + 18u + 4

K = 3(0.4562)² + 18(0.4562) + 4

K = 12.8359

To find deflection put value of K = 12.8359 in equation 3

ΔF = K ΔU

1 = 12.8359 ΔU

ΔU =1/12.8359

ΔU = 0.07790

U4 = u + ΔU

= 0.4562+ 0.07790

U4 = 0.5341

Iteration 5 :

Considering value of u = u1

Then stiffness matrix will be

K(u) = 3u² + 18u + 4

K = 3(0.5341)² + 18(0.5341) + 4

K = 14.4695

To find deflection put value of K = 14.4695 in equation 3

ΔF = K ΔU

1 = 14.4695 ΔU

ΔU =1/14.4695

ΔU = 0.06911

U5 = u + ΔU

= 0.5341+ 0.06911

U5 = 0.6032

Iteration	ΔF	ΔU	U	F
1	1	0.25	0.25	1
2	1	0.1151	0.3651	2
3	1	0.3411	0.4562	3
4	1	0.193	0.5341	4
5	1	0.4102	0.6032	5

Comparison between implicit and explicit methods:

Which Method is better?

• Both solvers have their own areas of advantages.
• Implicit method helps us to select bi time interval, while explicit does restrict the time interval depending upon size, mesh quality and computational capability of the computer used.
• Implicit method is complicated and slow, while explicit is direct and fast method.
• Implicit will given accurate result with small interval of time.

Conclusion:

• Got to know different method to solve dynamic equation.
• Learned the process that occurs at the back end of the system by solving different mathematical equation.
• Got to know major difference between implicit and explicit, and how it affects the solution and when to use it.