Week - 2 - Explicit and Implicit Analysis

Explicit and Implicit Analysis Hand calculation

 

OBJECTIVE:  Solving the given equation F(u) = u³+9u²+4u using both Explicit and Implicit Methods (with the tolerance of 10^-2.

Explicit: A function in which the dependent variable and independent variable are separated on opposite side of equality are commonly known as Explicit functions.

e.g. y = f(x)

It is a method used to compute the state of certain system at a time different from the current time. It can also be said as a direct computation of the dependent variables can be made in terms of existing quantities.

Explicit analysis provides a faster solution in events which have dynamic equilibrium. The main advantage of explicit method is to solve for acceleration. The explicit methods fast and inexpensive computationally. Each increment is calculated extremely fast. Accuracy is not maintained. Time step has to be extremely small.

 

Implicit:  A function in which the dependent variable and independent variables are not separated on opposite sides of the equality are commonly known as implicit function. It can be function Ф (x, y, z) = 0 where x, y are independent variables and z is dependent variable.

e.g.

1. x³+y³-3axy = 0
2. (x-a)² + (y-b)² = r²
3. Ax²+2hxy+by²+2gx+2fy+c = 0

 

It is a method of solving to find the unknown using matrix inversions. When a dependent variable is calculated by some set of equations, and to have the final solution an iterative technique or a matrix is required then it’s called an implicit method.

 

Implicit methods take lot of time to solve dynamic and nonlinear problems. Increments are big and due to this it is very slow. Accuracy is more compared to explicit methods. Responsiveness of the model with respect to time may differ to that of an actual model. parameters are dependent on each other at the same time.

 

 

 

Explicit Method

The given equation

F(u) = u³+9u²+4u

Now we can write the stiffness equation of as

                              --------------------------------(1)

 

We can represent the stiffness as

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

 

 

Now with the help of equation (1) and (2) we will compute the problem using incremental explicit load control scheme.

from (1) and (2) we get,

 

ΔF = KΔU ------------------------------- (3)

 

 

Step 1:

 

for the first step lets us consider an initial condition where, ΔF = 1, u (0) = 0 substituting in equation (2),

 

K (0) = 3(0)² + 18(0) + 4 = 4

Then we can use the values in equation (3)

ΔU₁ =  =  = 0.25

u₁ = u (0) + ΔU₁

       u₁ = 0 + 0.25 = 0.25

u₁ = 0.25

 

 

 

Step 2:

 

Now we have to consider u = u₁ then we can write from equation 2

 

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

 

k(u₁) = 3(0.25)2 + 18(0.25) + 4           --------------------- (As u = u₁)

k(u₁) = 8.6875

ΔU₂ =  =  = 0.1151

u₂ = u₁ + ΔU₂ = 0.25 + 0.1151

u₂= 0.3651

Step 3:

 

Now we have to consider u = u₂ then we can write from equation 2

 

K(u₂) = 3(0.365)² + 18(0.365) + 4 K(u₂) --------------------- (As u = u₂)

 

k(u₂) = 10.96

 

ΔU₃ =  =  = 0.091

u₃ = u₂ + ΔU₃

              = 0.3651 + 0.091

u₃= 0.456

 

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

                                                          Fext = ΔF + ΔF + ΔF

 

                                                          Fext = 1 + 1 + 1

 

                                                          Fext = 3

 

Now we have,

F(u) = u³+9u²+4u

 

f₁ = (u₁³) + (9u₁²) + 4u₁ = (0.25)³ + (9 * (0.25)²) + (4 * 0.25) = 1.578

 

f₂ = (u₂³) + (9u₂²) + 4u₂ = (0.365)³ + (9 *(0.365)²) + (4 * 0.36) = 2.707

 

f₃ = (u₃³) + (9u₃²) + 4u₃ = (0.456)³ + (9 *(0.456)²) + (4 * 0.45) = 3.791

 

 

 

It is evident from the results that external and internal forces are NOT in equilibrium because

Fext ≠ fint. Table 1 summarizes the main results for the explicit analysis.

 

Step i	ΔFi	ΔUi	ui	(Fext)i	(Fint)i	(Fint)I - (Fext)I = R
1	1	0.25	0.25	1	1.578	0.578
2	1	0.1151	0.3651	2	2.707	0.707
3	1	0.091	0.456	3	3.791	0.791

 

 

 

 

 

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) = f_int − F_ext. Expanding the residual as a Taylor series about

the current displacement uj gives

= + + +… = + k ()+…

By neglecting higher order terms in the series, setting R^j+1 = 0 and solving for the

following correction is obtained

 = − [k ()] ^-1

 

Notice that the iteration variable is j and it may take several iterations for the norm of the

residual, ||R|| to be reduced below the chosen tolerance criteria (for the example below a tolerance

of 10^-2 is chosen). The implicit analysis proceeds as follows (results are approximate,

they are more precise if done in a computer to machine precision):

In implicit analysis, at the end of each step Newton-Raphson iteration is used to bring equilibrium and then proceeded to further steps.

 

 

Step 1:

consider u= 0, then, F(u) = u³+9u²+4u becomes

differentiating above eq

3u²+18u+4

K (0) = 3(0)² + 18(0) + 4

K (0) = 4

ΔU₁ =  =  = 0.25

ΔU₁ = 0.25

 

u₁ = u (0) + ΔU₁

       u₁ = 0 + 0.25 = 0.25

u₁ = 0.25

f₁ = (u₁³) + (9u₁²) + 4u₁ = (0.25)³ + (9 * (0.25)²) + (4 * 0.25) = 1.578

 

Check the Residual (F_int)_I - (F_ext)_I = R, As

 

 R⁰= 1.578 -1 = 0.578

 

 

R⁰ which is greater than 0.01 i.e. (0.578 >0.01), therefore Newton Raphson model has to be followed

 

                                                          U1 = U1⁰

 

 

3u²+18u+4        

 

K (1) = 3(0.25)² + 18(0.25) + 4

                                                     K (1) = 8.687

 

 = − [k ()⁰] ^-1  

                                       = - [8.687]^-1*0.578

                                                                 

                                               = - 0.066

                                                            

                                                     u¹ = u₁⁰ +

 

u¹ = 0.25-0.066 = 0.184

          

                                                             u¹ = 0.184

 

f₂ = (u¹)³ + (9 u¹)² + 4 u¹ = (0.184)³ + (9 * (0.184)²) + (4 * 0.184) = 1.046

 

 

R₁ = f_i − F_ext = 1.0469 − 1 = 0.0469

 

which is greater than 0.01 i.e. (0.04>0.01),

 

Step 2:

 

 

U1 = U1⁰

 

K (2) = 3(0.184)² + 18(0.184) + 4

                                                     K (2) = 7.413

              = − [k ()¹] ^-1

 

                                        =  - 0.066 – [7.413]^-1*0.0469

                = -0.0723

 

 

                                                        u₁²= u₁⁰+ ∂u²  = 0.25

 

                      = 0.25 − 0.0713

   

    u₁² = 0.178

The updated value of u₁ is 0.178, then the function will be,

 

f₁ = (u₁³) + (9u₁²) + 4u₁ = (0.178)³ + (9 * (0.178)²) + (4 * 0.178) = 1.007

 

therefore,

R¹= 1.007 -1 = 0.007

 

 

 

The R value is less than 0.01, therefore, the final value of u₁ is 0.178

 

Step 3:

 

K (u₁) = 9(0.178)² + 18(0.178) + 4

                                                   K (u₁) = 7.489

 

ΔU₂ =  =  = 0.133

 

u₂ = u₁ + ΔU₂

        u₂ = 0.178 + 0.133

 

u₂ = 0.311

 

 Here, f_ext = 2

 

f₂ = (u₂³) + (9u₂²) + 4u₂ = (0.311)³ + (9 * (0.311)²) + (4 * 0.311) = 2.144

 

 

Residual (F_int)_I - (F_ext)_I = R, As

 

 R⁰= 2.144 -2 = 0.144, which is greater than 0.01 i.e. (0.144>0.01) from Newton Raphson model the updated u₂ is

 

u₂ = 0.296

 

 

Step 4:

 

K (u₂) = 9(0.296)² + 18(0.296) + 4

                                                   K (u₂) = 10.116

 

ΔU₃ =  =  = 0.098

u₃ = u₂ + ΔU₃

       u₃ = 0.296+0.098

                                                       

                                                         u₃ = 0.394

 

 here, f_ext = 1

 

f₃ = (u₃³) + (9u₃²) + 4u₂ = (0.394)³ + (9 * (0.394)²) + (4 * 0.394) = 3.034

 

Residual (F_int)_I - (F_ext)_I = R, As

 

 R⁰= 3.034 -3 = 0.034, which is greater than 0.01 i.e. (0.034>0.01) from Newton Raphson model the updated u₂ is

u₃ = 0.391

 

 

 

 

Step i	ΔFi	ΔUi	ui	(Fext)i	(Fint)i	(Fint)I - (Fext)I = R
1	1	0.25	0.178	1	1.007	0.007
2	1	0.113	0.296	2	2.004	0.004
3	1	0.098	0.391	3	3.001	0.001

 

 

 

 

 

 

It is evident from the plot that the explicit analysis drifts from the exact solution. If 20 increments are used the explicit analysis more closely follows the exact solution. However, even with more increments the explicit analysis still drifts from the exact solution. The drift from the exact solution illustrates the lack of equilibrium between the internal and external forces. To correct for this problem an implicit analysis is required. The results for the implicit analysis are shown in above graphs. The Newton-Raphson iterations correct the incremental steps so that they land on the exact solution according to the specified tolerance.

 

 

CONCLUSION:

 

1. A nonlinear analysis is illustrated using an incremental explicit and implicit procedure.

The results are plotted for comparison. It is evident from the results that the explicit analysis drifts from the true solution. To overcome this problem an implicit analysis is used, which includes Newton-Raphson iterations to enforce equilibrium between internal and external forces.

 

2. Implicit methods require an extra computation, and they can be much harder to implement. But when it comes to the matters of accuracy, it takes much less computational time for an implicit system with larger time step than an explicit system having an impractically small-time step

 

 

3. 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 can’t be used for problems having higher strain rate. For the particular problem Implicit method is the best way to compute and find the solution.

 

4. From this, we can say that the usage of 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.

 

 

 

problem Implicit method is the best way to compute and 􀂦nd the solution.

From this, we can say that the usage of 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.

 

 

KINDLY FIND THE ATTACHED FILES. EQUATIONS MAY NOT BE CORRECTLY WRITTEN OVER HERE.