Week - 2 - Explicit and Implicit Analysis

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 the system at different times 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 & a few dynamic analyses.
• One of the cons in the 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 the implicit method. 

 

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 into the next step with the help of Newton-Raphson iterations.
• An implicit method is used for small strain deformation problems.
• More accuracy in the obtained results when compared to the solution obtained by the explicit method.
• This method is used where the events to be performed are fewer.

 

Procedure:

Consider the case of a simple bar in tension as shown in Figure 1. Suppose that the force
in the bar is a nonlinear function of the displacement

                                F(u)=u3+(9.u²)+4u</math>">F(u)=u3+(9.u²)+4u�(�)=�3+(9.�²)+4�   ----eq(1)

 

From (1) it follows that the stiffness of the bar is

                                k(u) = dF / du = 3*u^2 + 18*u  + 4  

                   

Where,

F = external force applied to the bar

u = Displacement

f = Internal force in the bar

k = Stiffness

Δ</math>">ΔΔ u = Incremental displacements

Δ</math>">ΔΔ 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 Δ</math>">ΔΔF = 1 unit each are used to load the tension bar.

For finding the stiffness, The following relationship: Δ</math>">ΔΔF = Δ</math>">ΔΔ u is used.

Then,  k'(u) = dF/du                              ----- eq(2)

Therefore. The analysis proceeds as follows:

Step 1:

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

u0 = 0.0,

⇒ k(u0) = (3.u0^2)+ (18.(0))+4

therefore k(u0)= 4,

∆F = 1,

then ∆u1 = ∆F/k(u0)=14</math>">1414=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 equation 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(Δ</math>">ΔΔF=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</math>">k(u1)=(3.(u1)2)+(18⋅u1)+4�(�1)=(3.(�1)2)+(18⋅�1)+4

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

k(u1) = 8.6875

Now the incremental displacement for this step is : Δu2</math>">Δu2Δ�2=ΔF</math>">ΔFΔ�/K(u1)= 1/8.6875

Δu2</math>">Δu2Δ�2=0.11512

u2= u1 + Δu2</math>">Δu2Δ�2 = 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  ∆F = 1

u2= 0.36512

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

        = (3*(0.3652)^2) + (18*u1) + 4

k(u2)= 10.97

∆u3  = ∆F/K(u2) = 1/10.97

∆u3  =0.09115

u3    = u2 + ∆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 (∆F)	incremental displacement (∆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.

Implicite Analysis : 

Step 1: 

in this step , the initial condition are taken into consideration as follows :

u0= 0      

∆F= 1

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

∆u1 = ∆F / K(u0) = 1/4 =0.25 

f10</math>">f10�10= (0.25)³+(9 *(0.25)²)+(4*0.25)

= 1.578125

Check the residual, 

residual force (R0</math>">R0�0)= internal force(f10</math>">f10�10)- external force (F)

=1.57812-1

=0.57821 , which is greater than 0.01 

if (R0</math>">R0�0) > (10^−2) ⇒ Newton-Raphson iterations are necessary.

Then Newton-Raphson iterations are necessary to get the forces in to equilibrium.

u1= u1⁰

δu11</math>">δu11��11= -[k(u1)]¹. R0</math>">R0�0=-[8.6875]^(-1)*(0.57821)

                  = -0.066

Now update the value of u1 

u1= u10+δu11</math>">u10+δu11�10+��11 =0.25 + (-0.066) =0.184

f11=(0.184)3+(9⋅(0.184)2)+(4⋅0.184)=1.0469</math>">f11=(0.184)3+(9⋅(0.184)2)+(4⋅0.184)=1.0469�11=(0.184)3+(9⋅(0.184)2)+(4⋅0.184)=1.0469

Now the residual is founded again , R1</math>">R1�1 = internal (f11</math>">f11�11) - external force (F)

                                                              = 1.0469-1 =0.0469 which is greater then 0.01 

Newton-Raphson iterations are necessary.

u1= u1⁰

δu12</math>">δu12��12= δu11</math>">δu11��11-[k(u1)]¹. R1</math>">R1�1=(-0.066)-[8.6875]^(-1)*(0.0469)

=-0.066-0.0054

=-0.0711

therefore , δu12</math>">δu12��12= -0.0711

Now the updated u1= u10+δu12</math>">u10+δu12�10+��12

=0.25+(-0.0711) = 0.1789

f12=(0.1789)3+(9⋅(0.1789)2)+(4⋅0.1789)=1.0093</math>">f12=(0.1789)3+(9⋅(0.1789)2)+(4⋅0.1789)=1.0093�12=(0.1789)3+(9⋅(0.1789)2)+(4⋅0.1789)=1.0093

R² = internal force(ƒ1²)-external force (F) = 1.0093-1=0.0093 Which is <0>

Therefore, The final value of u1 = 0.1789

 

step 2:

u2= u1= 0.1789

k(u2)=3⋅(0.1789)2+(18⋅(0.1789))+4</math>">k(u2)=3⋅(0.1789)2+(18⋅(0.1789))+4�(�2)=3⋅(0.1789)2+(18⋅(0.1789))+4

Δu2=ΔF</math>">Δu2=ΔFΔ�2=Δ�/ K(u1) =0.1366

u20=u1+Δu2=0.1789+0.1366=0.3115</math>">u20=u1+Δu2=0.1789+0.1366=0.3115�20=�1+Δ�2=0.1789+0.1366=0.3115

f20=(u20)3+9(u20)2+4(u20)</math>">f20=(u20)3+9(u20)2+4(u20)�20=(�20)3+9(�20)2+4(�20)

f20=(0.3115)3+(9⋅(0.3115)2)+(4⋅0.3115)=2.15</math>">f20=(0.3115)3+(9⋅(0.3115)2)+(4⋅0.3115)=2.15�20=(0.3115)3+(9⋅(0.3115)2)+(4⋅0.3115)=2.15

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

k(u21)=(3⋅(u20)2)+(18⋅(u20))+4=(3⋅(0.3115)2)+(18⋅(0.3115))+4=9.898</math>">k(u21)=(3⋅(u20)2)+(18⋅(u20))+4=(3⋅(0.3115)2)+(18⋅(0.3115))+4=9.898�(�21)=(3⋅(�20)2)+(18⋅(�20))+4=(3⋅(0.3115)2)+(18⋅(0.3115))+4=9.898

δu21=-[k(u2¹)]-1⋅RO</math>">δu21=−[k(u2¹)]−1⋅RO��21=-[�(�2¹)]-1⋅�� = -0.01515`

Therefore,u2¹=u2º+δu21=0.3115-0.01515=0.296</math>">u2¹=u2º+δu21=0.3115−0.01515=0.296�2¹=�2º+��21=0.3115-0.01515=0.296

Now, f21=(u2¹)³+9(u2¹)²+4(u2¹)=2.0032</math>">f21=(u2¹)³+9(u2¹)²+4(u2¹)=2.0032�21=(�2¹)³+9(�2¹)²+4(�2¹)=2.0032

Residual, R¹=f2¹-F=2.0032-2=0.0032<0.01.</math>">R¹=f2¹−F=2.0032−2=0.0032<0.01.�¹=�2¹-�=2.0032-2=0.0032<0>

So, The updated u2 is u2^1 = 0.296

 

step 3:

u3= u2= 0.293

k(u3)=3⋅(0.293)2+(18⋅(0.293))+4</math>">k(u3)=3⋅(0.293)2+(18⋅(0.293))+4�(�3)=3⋅(0.293)2+(18⋅(0.293))+4=9.53

Δu3=ΔF</math>">Δu3=ΔFΔ�3=Δ�/ K(u3) =1/9.53 = 0.1049

u30=u2+Δu3=0.293+0.1049=0.3979</math>">u30=u2+Δu3=0.293+0.1049=0.3979�30=�2+Δ�3=0.293+0.1049=0.3979

f30=(u30)3+9(u30)2+4(u30)</math>">f30=(u30)3+9(u30)2+4(u30)�30=(�30)3+9(�30)2+4(�30)

f30=(0.3979)3+(9⋅(0.3979)2)+(4⋅0.3979)=3.07951</math>">f30=(0.3979)3+(9⋅(0.3979)2)+(4⋅0.3979)=3.07951�30=(0.3979)3+(9⋅(0.3979)2)+(4⋅0.3979)=3.07951

Residual, R^O = f3^0-F = 3.07951-3 = 0.07951 Which is >0.01. So, The Newton's Raphson iterations are used to find u3.

k(u31)=(3⋅(u30)2)+(18⋅(u30))+4=(3⋅(0.3979)2)+(18⋅(0.3979))+4=11.64</math>">k(u31)=(3⋅(u30)2)+(18⋅(u30))+4=(3⋅(0.3979)2)+(18⋅(0.3979))+4=11.64�(�31)=(3⋅(�30)2)+(18⋅(�30))+4=(3⋅(0.3979)2)+(18⋅(0.3979))+4=11.64

δu31=-[k(u3¹)]-1⋅RO</math>">δu31=−[k(u3¹)]−1⋅RO��31=-[�(�3¹)]-1⋅�� = -0.00683

Therefore,u3¹=u3º+δu31=0.3979-0.00689=0.39107</math>">u3¹=u3º+δu31=0.3979−0.00689=0.39107�3¹=�3º+��31=0.3979-0.00689=0.39107

Now, f31=(u3¹)³+9(u3¹)²+4(u3¹)=3.0005</math>">f31=(u3¹)³+9(u3¹)²+4(u3¹)=3.0005�31=(�3¹)³+9(�3¹)²+4(�3¹)=3.0005

Residual, R¹=f2¹-F=3.0005-3=0.0005<0.01.</math>">R¹=f2¹−F=3.0005−3=0.0005<0.01.�¹=�2¹-�=3.0005-3=0.0005<0>

So, The updated u3 is u3^1 = 0.39107

 

step no.	incremental external force (∆F)	incremental displacement (∆u)	displacement (u)	external force (F)	internal force (f)	residual force = f-F
1	1	0.25	0.1789	1	1.0093	0.0093
2	1	0.1366	0.296	2	2.0032	0.0032
3	1	0.1049	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.