Week - 2 - Explicit and Implicit Analysis

week - 2 - Explicit and Implicit Analysis:

objectives: To explain about the explicit and implicit method and to solve the basic problem by using explicit and implicit method to understand the working of this schemes in analysis .

Explicit and Implicit Analysis:

• Explicit and implicit are the methods to solve the dynamic problems.even it can solve the static problem but we have some limitation and disadvantage.
• Analysis can be divided into two:

1. Static(Load is constant,deformation maybe linear or it can also be non-linear,time dependent or it may also be time independent)
2. Dynamic(Load may constant or may not also be constant,deformation maybe linear or it can also be non-linear,time independent)

• Explicit and implicit are comes under dynamic analysis,but it can also solve static
• if we take the dynamic governing equation M[X'']+C[X']+K[X]=F(t)

here the [x'']-acceleration

[x;]-velocity

[x]-displacement

F- force

t-time

M-Mass matrix

C- damping matrix

K- stiffness matrix

• When we solve static problem using implicit scheme then the governing equation is reduced to M[X'']+C[X']+K[X]=F(t) same thing for the dynamic analysis,but one thing to keep in mind is whether it is linear or non-linear,because when solving static problem the dynamic govering equation M[X'']+C[X']+K[X]=F(t) in this the mass matrix we become null matrix(every value become zero) and the damping matrix also become zero and finally it reduced to equation K[X]=F(t).
• For linear K[X]=F(t) for non-linear K(x)[X]=F(t).
• Mostly they will use implict method to solve static and quasi static types of problem and explicit scheme is used to solve dynamic cases.
• We can solve static problem by using explcit analysis but the time it would take take to form stiffness matrix will take some time and the solving will not converged at each steps.
• lets take a brief diffrence between the explcit and implicit scheme:
  

EXPLICIT	IMPLICIT
In implicit scheme the unknown[x] find out by the inversion of stiffness matrix	In explicit scheme the unknown[x''] can be found out by the inversion of mass matrix.
for each increment(steps): stiffness matrix will be updated next increment of load is defined newton raphson enforce equilibrum	for each increment: stiffness matrix will be updated next increment of load is defined direct solution
For implicit newmark b method is used	for explicit central differenvce method is used there are two types: inclassical central diffrence method half step central diffrence method.
After solving the govering equationM[X'']+C[X']+K[X]=F(t) we get the value of[x],other value like[x''] and [x'] are found out by the numerical integration method. In implicit new mark b is the numerical integration method.	After solving the govering equationM[X'']+C[X']+K[X]=F(t) we get the value of[x''] other value like[x'] and [x] are found out by the numerical integration method. In explicit central diffrence method is the numerical intrgration method
implict schemes are unconditional statble means we can give any time increment(△t) we want that does not after the solution will converged for it.	in explicit we have to take care of criticsl time step(△tcr) △t>△tcr △tcr=le/c #le-element length,c-wave propagation c=√E/P #E-youngs modulas,P-density means it depends on both material and the element.
more accurate results	less accurate
extremely time consuming for large models	computationally efficient for large model with relatively short dynamic response.
no mathematical time limit for solution	time taken by the explicit analysis is depends on the wave proppgation speed and charaterictics length of the elements if mesh size is so small then the problem would take so much of time to solve problem than the static time.
implicit method require more computer storgae because if we give more time increment the each time the stiffness matrix will be updated and store it so ut take s much storage.	less storage is required than the implicit
effect of strain rate are minimum.	strain rate are significant effect.
easy and moderate contact comdition,means if the propbelm the rigid body motion sometime the implicit will fail.	solve extreme discontinous event or process and can solve complec contact condition.
application:low frequency response problem like static cases,vibration,oscillation.	application:wave propgation type of propbelm ike car crash,drop impact test etc..

• let understand this concepts by taking a simple problem,a bar in tension with one end is fixed ,the given equation is

F(u)=u³+9u²+4u---------(1)

• Diffrentiate the above equation to find out the stiffness

K(u)=3u²+18u+4---------(2)

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For explicit:

For solving this problem the external load is given in a increment at each step

△F-is the external force

△u-incremental displacement. And we will get a incremental displacement at each step for incremental load

F_int-internal force

k-stiffness, the stiffness matrix will be updated at the each steps

u-displacement, displacement is the output at the each step

step1:

• The initial displacement is zero when force is not applied on the body so u₀=0
• In this step the stiffness matrix should be update so substitute the intial displacement in the stiffness equation

k(u₀) = 3u²+18u+4

k(0)=3(0)²+18(0)+4

k(0)=4

the incremental load is △f=1

as we solving a static problem the governing equation is K[X]=F(t) by applying the know value to the governing equation we can find the incremental displacement in this step

x-->△u

substitute the k and △F=1 in that equation we get ‘△u’

△u=△f/k

=1/4

△u=0.25

now interpolate the two displacement to find the actaul displacement in this step

Now calculate the displacementu₁=u₀+△u

u₁=0+0.25

u₁=0.25.

internal force(F_int)=u₁³+9u₁²+4u₁

=(0.25)³+9(0.25)²+4(0.25)

=1.578

residual=-△F+F_int

=-1+1.578

residual=0.578.

step2:

• now, u₁=0.25 is the input in the step 2 and we have second load increment △f=1

now in this we have to update the stiffness matrix

k(u₁) = 3u²+18u+4

k(0.25)=3(0.25)²+18(0.25)+4

k(0.25)=8.687

now use the governing equation to find the incremental displacement

△u=△f/k

=1/8.687

△u=0.115

The final displacement u₂= u₁+△u

u₂= 0.25+0.115

u₂=0.365.

internal force(F_int)=u₂³+9u₂²+4u₂

=(0.365)³+9(0.365)²+4(0.365)

=2.707

residual=-△F+F_int

=-2+2.707

residual=0.707

steps3:

• Now, u₂=0.365 is the input in the step 3 and we have second load increment △f=1

now in this we have update the stiffness matrix

k(u₂) = 3u²+18u+4

k(0.365)=3(0.365)²+18(0.365)+4

k(0.365)=10.96

now use the governing equation to find the incremental displacement

△u=△f/k

=1/10.96

△u=0.0912.

The final displcement u₃= u₂+△u

U₃= 0.365+0.0912

U₃=0.4562

internal force(F_int)= U₃³+9 U₃²+4 U₃

=(0.4562)³+9(0.4562)²+4(0.4562)

=3.792

residual=-△F+F_int

=-3+3.792

residual=0.792.

step(increment)	△F	Fint	△u	U	△f	residual=-△F+Fint
1	1	1.578	0	0.25.	1	0.578.
2	2	2.707	0.115	0.365	1	0.707
3	3	3.792	0.0912	0.4562	1	0.792.

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implicit:

For solving this problem the external load is given in a increment at each step

△F-is the external force

△u-incremental displacement. And we will get a incremental displacement at each step for incremental load

F_int-internal force

k-stiffness, the stiffness matrix will be updated at the each steps

u-displacement, displacement is the output at the each step

the given tolerance is 0.001

steps1:

• The initial displacement is zero when force is not applied on the body so u₀=0
• In this step the stiffness matrix should be update so substitute the intial displacement in the stiffness equation

k(u₀) = 3u²+18u+4

k(0)=3(0)²+18(0)+4

k(0)=4

the incremental load is △F=1

as we solving a static problem the governing equation is K[X]=F(t) by applying the know value to the governing equation we can find the incremental displacement in this step

x-->△u

substitute the k and △F=1 in that equation we get ‘△u’

△u=△F/k

=1/4

△u=0.25

now interpolate the two displacement to find the actaul displacement in this step

Now calculate the displacementu₁=u₀+△u

u₁=0+0.25

u₁=0.25.

internal force(F_int)=u₁³+9u₁²+4u₁

=(0.25)³+9(0.25)²+4(0.25)

=1.578

residual=-△F+F_int

=-1+1.578

residual=0.578.

F_int is not equal to △F

Newton-Raphson iterations are necessary,

k(u₁⁽⁰⁾) = 3u²+18u+4

=3(0.25)²+18(0.25)+4

=8.68.

|R(0)|> 10−3

δu⁽¹⁾ = −[k(u₁⁽⁰⁾)]⁻¹R(0)

= −[8.68]⁻¹0.578.

δu⁽¹⁾=-0.0665.

u₁⁽¹⁾ = u₁⁽⁰⁾+ δu⁽¹⁾

^{=0.25-0.0665.}

^{u₁(1) =0.1835.}

^{.again check the internal force}

internal force(F_int)=u₁^{(1) 3}+9u₁^{(1) 2}+4u₁⁽¹⁾

=(0.1835)³+9(0.1835)²+4(0.1835)

=1.0071>0.001

residual=-△f+F_int

=-1+ 1.0071

=0.0071<0.001

no residual needed

same thing we have to do for two load increments,u2=0.2966,u3=0.3911.

step(increment)	△F	Fint	△u	U	△f	residual=-△F+Fint
1	1	1.0071	0.25	0.1786	1	0.0071
2	2	2.0042	0.1374	0.2966	1	0.0042
3	3	3.008	0.1041	0.3911	1	0.008

here when u solve the same problem using different scheme you can actual notice the solution convergence.

result:

• briefly explained about the implicit and explicit methods.
• solved the same problem using implicit and explicit method.

learning objectives:

• learned about the explcit and implcit methods