Week - 2 - Explicit and Implicit Analysis

Objective:

• To solve the given non-linear equation "F(u) = u3+9u2+4u" using both Explicit and Implicit Methods.
• To perform 3 steps for both explicit and implicit Methods & calculate the various parameters such as displacement, stiffness, internal & external forces. Then, To suggest the best method.
• To plot the obtained results for both explicit and implicit methods.

Theory required:

Explicit Analysis:

The explicit analysis is preferred for the following reasons:

• Acceleration is assumed to be constant over the time step.
• It will calculate the state of system at different time by considering the current time, Which results in finding out the solution faster.
• Each increment is founded fast, This results in less computation.
• Explicit method is used for large strain deformation problems, Which includes the applications such as: Automotive Crash analysis, Drop test & few dynamic analysis.
• One of the con in explicit analysis is: More increments are needed to match up with the exact solution.
• Less accuracy in the obtained results when compared to the solution obtained by implicit method.
• This method is used where the events to be performed are more.

Implicit Analysis:

The implicit analysis is preferred for the following reasons:

• In implicit analysis, Acceleration is assumed to vary linearly over the time step(Which could be either positive(or)Negative).
• It will calculate the solution by considering the current & further states of the system, Which helps in attaining the equilibrium between the external & internal forces before moving in to next step with the help of newton-raphson iterations.
• Implicit method is used for small strain deformation problems.
• More accuracy in the obtained results when compared to the solution obtained by explicit method.
• This method is used where the events to be performed are less.

Procedure:

Given non-linear equation: F(u) = `u^3`+`(9*u^2)`+4u ---->eq(1)

Where, F = external force applied to the bar

u = Displacement

f = Internal force in the bar

k = Stiffness

∆u = Incremental displacements

∆F = Incremental externally applied forces

Explicit Method:

For the nonlinear analysis, the force Vs displacement relationship is not known, but the stiffness is known, it is possible to construct the graph of force Vs displacement numerically. One method is, To use an incremental explicit load control scheme. For this example, three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar.

For finding the stiffness, The following relationship: ∆F =`k*∆u` is used.

Then, k = `(dF)/(du)` ------->eq(2)

Therefore, The analysis proceeds as follows:

Step 1: This step is performed by taking the initial conditions as follows:

u0 = 0

=>k(u0) = `(3*u0^2)`+`(18*u0)`+`4` [From eqns(1 &2)]

By substituting the value of u0 in the above equation, We get:

k(u0) = `(3*(0)^2)+(18*(0))+4` =4

Therfore, k(u0) = 4

∆F = 1

Then, ∆u1 = ∆F/k(u0) = `(1)/(4)` = 0.25

u1 = u0 + ∆u1 = 0+0.25 = 0.25 ------->eq(3)

Then internal force(f) can be calculated by substituting the "u1" value in eqn(1):

f1 = `(0.25)^3`+`(9*(0.25)^2)`+`(4*0.25)`

= 1.578125

Therefore, the Internal force for the first increment is 1.578125.

Step 2:

In this step, The externally applied load is incremented by one unit(`Delta`F=1).

From eq(3), The value for displacement for this iteration is u1= 0.25

Then, Stiffness for this step is calculated by: `k(u1)`= `(3*u1^2)`+`(18*u1)`+4

=> k(u1) = `(3*(0.25)^2)`+18(0.25)+4

k(u1) = 8.6875

Now, The Incremental displacements for this step is: `Delta`u2 = ∆F/k(u1) = `1/(8.6875)`

= 0.11512

u2 = u1+`Delta`u2 = 0.25+0.11512 = 0.36512

Internal force, f2 = `(0.36512)^3`+`(9*(0.36512)^2)`+(4*0.36512)

= 2.708

Step 3:

In this step also, The externally applied load is incremented by one unit(`Delta`F=1).

u2 = 0.36512

k(u2) = `(3*u2^2)`+(18*u1)+4

= `(3*(0.3652)^2)`+(18*(0.36512))+4

k(u2) = 10.97

`Delta`u3 = `Delta`F/k(u2) = `1/(10.97)`

`Delta` u3 = 0.09115

u3 = u2+`Delta`u3 = 0.36512+0.09115 = 0.45627

Internal force, f3 = `(0.45627)^3`+`(9*(0.45627)^2)`+(4*0.45627)

= 3.794

Step No	Incremental external force(`Delta`F)	Incremental Displacement(`Delta`u)	Displacement(u)	External force(F)	Internal force(f)	Residual force= f-F
1	1	0.25	0.25	1	1.578125	0.578125
2	1	0.11512	0.36512	2	2.708	0.708
3	1	0.09115	0.45627	3	3.794	0.794

Result:

From the above table, The residual force(difference between internal & external forces are not equal) & more in magnitude. This depicts that, external and internal forces are NOT in equilibrium. To reduce the residual forces, more iterations should be performed. The only advantage of choosing an explicit method for performing the analysis is, The processing time is very less when compared to the implicit method.

The obtained result in explicit methods is plotted & it appears as shown in the below fig.

Implicit Analysis:

Step 1:

In this step, The initial conditions are taken in to consideration as follows:

u0 = 0

`Delta`F=1

k(u0) = 4 ----> Finded in explicit analysis step 1.

`Delta`u1 = `(∆F)/(k(u0))` = `(1)/(4)` = 0.25

`f1^0`= `(0.25)^3`+`(9*(0.25)^2)`+`(4*0.25)`

= 1.578125

Residual force(`R^0`) = internal force(`f1^0`)-external force(F)

= 1.578125-1

= 0.57821, Which is Greater than 0.01. If residual is `gt``10^−2`, Then Newton-Raphson iterations are necessary to get the forces in to equilibrium.

u1= `u1^0`

`delta``u1^1`= `−[k(u1)]^(−1)*R^0` = -[8.6875]^(-1)*(0.57821)

= -0.066

Now the updated u1 value is u1 = `u1^(0)`+`delta``u1^(1)`=0.25+(-0.066) = 0.184

`f1^1` = (0.184)^3+(9*(0.184)^2)+(4*0.184) = 1.0469

Now the residual is founded again, `R^1` = internal force(`f1^1`)-external force(F)

= 1.0469-1 = 0.0469 Which is > 0.01

So, The Newton-Raphson iterations are further performed.

Now, u1 = `u1^0`

`delta``u1^2` = `delta``u1^1`-`[k(u1)]^(−1)R(1)` = -0.066-[8.6875]^(-1)*(0.0469) = -0.066-0.0054 = -0.0711

Therefore, `delta``u1^2` = -0.0711

Now the updated u1 value is u1 = `u1^(0)`+`delta``u1^(2)` = 0.25+(-0.0711) = 0.1789

`f1^2` = `(0.1789)^3`+`(9*(0.1789)^2)`+(4*0.1789)

= 1.0093

`R^2` = internal force(`f1^2`)-external force(F) = 1.0093-1 = 0.0093 Which is <0.01.

Therefore, The final value of u1 = 0.1789

Step 2:

u = u1=0.1789

k(u1) = 3*(0.1789)^2+(18*(0.1789))+4

= 7.316

`delta`u2 = `(∆F)/(k(u1))` = `1/(7.316)` = 0.1366

`u2^0`= u1+ `delta`u2 = 0.1789+0.1366 = 0.3115

`f2^0`=`(u2^0)^3`+`9(u2^0)^2`+4(`u2^0`) = (0.3115)^3+(9*(0.3115)^2)+(4*0.3115) =2.15

Residual, R^0 = f2-F = 2.15-2 =0.15 Which is >0.01. So, The Newton's Raphson iterations are used to find u2.

k(u2^1) = (3*(u2^0)^2)+(18*(u2^0))+4 = (3*(0.3115)^2)+(18*(0.3115))+4 = 9.898

`delta`u2^1 = `−[k(`u2^1`)]^(−1)*R^0` = -0.01515

Therefore, `u2^1` = `u2^0`+`delta`u2^1 = 0.3115-0.01515 = 0.296

Now, f2^1 = `(u2^1)^3`+`9(u2^1)^2`+4(`u2^1`) = 2.0032

Residual, `R^1` = `f2^1`-F = 2.0032-2 = 0.0032 <0.01.

So, The updated u2 is u2^1 = 0.296

Step 3:

u=u2 = 0.293

K(u2) = 3*`(0.293)^2`+(18(0.293))+4 = 9.53

`delta`u3^0 = `(∆F)/(k(u2))` = `1/(9.53)` = 0.1049

`u3^0`= u2+`delta`u3 = 0.293+0.1049 = 0.3979

`f3^0` = (`u3^0`)^3+(9*`(u3^0)^2`)+(4*`u3^0`) =`(0.3979)^3`+(9*`(0.3979)^2`)+(4*0.3979) = 3.07951

Residual, R^0 = f3^0-F = 3.07951-3 = 0.07951, Which is >0.01. So, The Newton-Raphson iteration is necessary.

K(u3^1) = (3*(`u3^0`)^2)+(18*(`u3^0`))+4 = `(3*(0.3979)^2)`+(18*0.3979)+4 = 11.64

`delta``u3^1`= -(R^0/(k(u3^1)))=-(0.07951/(11.64)) =-0.00683

u3^1 = u3^0+ `delta``u3^1` = 0.3979-0.00683=0.39107

`f3^1` = (`u3^1`)^3+(9*`(u3^1)^2`)+(4*`u3^1`) = (0.39107)^3+(9*(0.39107)^2)+(4*0.39107) = 3.0005

Residual, R^1 = f3^1-F = 3.0005-3 = 0.0005 <0.01

Therefore, The updated u3 is u3^1 = 0.39107

Step No	Incremental external force(`Delta`F)	Displacement(u)	External force(F)	Internal force(f)	Residual force= f-F
1	1	0.1789	1	1.0093	0.0093
2	1	0.296	2	2.0032	0.0032
3	1	0.39107	3	3.0005	0.0005

Result:

From the above table, The residual force(difference between internal & external forces are not equal) & less in magnitude & tries to attain the equilibrium m between the forces when compared to the results obtained in Explicit analysis. This depicts that, external and internal forces are almost brought to equilibrium by using the newton-Raphson iterations.

The obtained result in implicit methods is plotted & it appears as shown in the below fig.

Conclusion:

Final comparison graph distinguishing the implicit, explicit & exact calculations with force Vs Displacement:

From the above fig, The following aspects are concluded:

• The explicit solution drifts a bit from the exact solution. This is due to the residual force (difference between internal & external forces are not equal) & more in magnitude. In order to attain equilibrium between forces, More iterations are needed.
• But the solution obtained by the implicit method with the newton-Raphson method iterations, Will help the solution to attain the equilibrium between the internal & external forces. Which results in getting the solution much closer to the exact solution.
• As the given equation F(u) = u3+9u2+4u is non-linear, The obtained result in terms of force Vs displacement will be in an exponential form. As shown in above fig.