Week - 2 - Explicit and Implicit Analysis

Objective : To solve a non linear Equation with both implicit and Explicit Analysis

Explicit Analysis : It is used to calculate the state of sysytem At a different time from a current time , from the current time Explicit analys is offers a faster solution in the events where there is a dynamic Equillibrium .Explicit Analysis is fast Inexpensive computationally . Each increment is calculated extremly fast .

Sum of all force = mass * acceleration

The Explicit should be used when the strain rate/ velocity is over 10 units/sec , these evenets can be best exampled by extreme scenarious such as automotive crash or meteor impact In this strain rate plays a important role.

Implicit Analysis: An implicit analysis finds a solution by solving an equation that includes both current and later states of the system

the implicit method should be used when the events are minimal . once the growth of the stress as a function of strain can be Established the solution can be found using implicit analysis

Sum of all forces = 0

Implicit method take a lot oif time to solve dynamic and non linear problema, but the accuracy is more compared to Explicit

Procedure :

The Equation of force is f(u) = u3 +9u2 +4u

Equation of stiffness is K = df/du

k = 3u2 +18U +4 ________(1)

To solve this Equation and find load vs displacement effect for simple bar under tension we can use implicit and explicit method

• Implicit method of analysis

solving these equation with 3 equal load increment with ΔΔ F = 1

Step -1

U(0) = 0

k(u) = 4

ΔΔF = 1

ΔΔU1 = ΔΔF/ΔΔk(u)

= 1/4 =0.25

u1 = U0+ΔΔU1 = 0+0.25 = 0.25

F1int = (0.25^3)+9(0.025^2)+4(0.25)

= 1.578125

Step -2

u1 =0.25

K(u1)= 3(0.25^2)+9(0.25^2)+4

=8.6875

ΔΔF=1

ΔΔU2= ΔΔF/ΔΔ(u2)

= 1/8.6875

= 0.1151

U2 =u1=ΔΔU2

= 0.25+0.1151

= 0.3651

f2int = (0.365^3)+9(0.365^2)+4(0.3+651)

= 2.708

Step - 3

U2 = 0.3651

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

= 10.97

ΔΔF=1

ΔΔu3 =ΔΔF/k(u2)

= 0.09114

u3 = u2+

ΔΔu3= 0.365+0.09114

=0.4562

f3int = (0.4652^3)+9(0.4652^2)+4(0.4562)

=3.77928

step	ΔΔF	U	Fext	Fint	residual = Fint-Fext
1	1	0.25	1	1.5781	0.5781
2	1	0.3651	2	2.7087	0.7087
3	1	0.4562	3	3.792	0.792

So from the table , it is noted that the difference in force i.e Residual is considerable , but the processing time is also very less,

To reduce the residual . more no of Iteration must be performed.

Graph for force vs displacement for explicit.

IMPLICIT ANALYSIS :

Step -1

U(0) = 0

K(u) = 4

ΔΔF =1

ΔΔU1 =ΔΔF/`K(u) from k = df/du

= 1/4 = 0.25

fint = (0.1834^3)+9(0.1834^2)+4(0.1834)

= 0.578125

R1 = f1int-f1ext = 0.578125

since R1> 0.01 newton raphson iteration are done

NEWTON RAPHSON METHOD It is a root finding algorithm which produces succesfully better approximation to the roots of a real valued function

I.e x1=x0-f(x0)/f1(x0)

Applying this we ger

ΔΔu1(1)=R1(1)/k(1U1(1)) = -0.066549

U1(2) = u1(1)+ΔΔU1(!)= 0.25-0.06654 =0.1834

F1(2) = 1.04249

R1(2) = 0.04249>0.01 so newton Raphson method is applied

ΔΔU1 (2) = R1(2)/K(u1(2))=0.07228

U1(3) = u1(2)+9(0.177^)+4(0.177)=1.00075

R1(3) = 0.000753

which is less than 0.01 so no need for Newton Raphson method

thus U1 =0.1772

F1 = 1.0007

R1=0.000753

Step -2

u1 = 0.1772

K(u) = 3(0.1772^2)+18(0.1772)+4 = 7.29

ΔΔF =1

ΔΔU2= ΔΔF/K(u) = 0.137

u2 = U1+ΔΔu2 = 0.3148

F2 int = (0.177^3)+9(0.177^2)+4(0.177) = 2.182

R2 = F1int - F1et = 0.1822

since R2 > 0.01 Newton raphson method is done

ΔΔu2(1) = R2(1)/K(u2(1)) = -0.018294

U2(2) = U2(1)+ΔΔu2(1)=0.3148-0.0182=0.2965

F2(2) = 2.0037

R2(2) = 0.00327<0.01 ; so Newton raphson method is no further needed

Thus

U2 =0.2965

F2 = 2.0032

R2 = 0.003276

Step -3

U2 = 0.2965

K(u) = 3(0.2965^2) +18(0.2965)+4 = 9.60073

ΔΔF=1

ΔΔu3 =ΔΔF/k(u) = 0.1046

U3 = U2+ΔΔU3 = 0.4006

F3 int = (0.296^3)+9(0.2965^2)+4(0.2965)

R3 = F1int-F1ext = 0.111

It is greater than 0.01 so newton raphson method is done

ΔΔU3 (1) = R3(1)/K(U3(1)) =-0.009493

U3 (2) = U3(1)+ΔΔU3(1)=0.3911

F3(@) = 3.0008

R3(2) = 0.00855<0.01

thus

u3 = 0.3911

F3 = 3.00855

R3 = 0.00855

Step	ΔΔF	U	Fext	Fint	R=Fint-Fext
1	1	0.1772	1	1.0075	0.000753
2	1	0.2965	2	2.003276	0.0032
3	1	0.3911	3	3.00855	0.0085

So from the table we can say that the Residual is very low and solution is accurate this is due to iteration performed by newton raphson method

But this Iteration Step has taken more time but this process is accurate and slow.

Graph for force vs Displacement for implicit .

Comparision graph force vs displacement for implicit and explicit calculations

Conclusion

For all nonlinear and non-static analyses, the incremental load is needed. Implicit methods require extra computation, and they can be much harder to implement. But when it comes to matters of accuracy, it takes much less computational time for an implicit system with a larger time step than an explicit system having an impractically small time step. explicit methods can be used in extreme scenarios such as an automotive crash, ballistic event, etc, wherein the materials not only account stress with respect to strain but also strain rate. implicit methods cant be used for problems having higher strain rate. for the particular problem, the Implicit method is the best way to compute and find the solution.

From this, we can say that the usage of the implicit or explicit system depends upon the problem that needs to be solved. The decision to use implicit and explicit method directly impacts the speed and accuracy.