Explicit and Implicit Analysis of a bar in Tension

Objective:

To solve numerically and compare the difference between Non-linear static Explicit and Implicit solver methods of a bar in tension.

Question:

As demonstrated in the PDF, F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods (have a tolerence of 10^-2).

Deliverables :

1. Write a detailed report about Implicit and explicit methods.

2. Do the calculations in a spreadsheet, try to plot the graph and compare them (optional).

3. Calculations that are done in a piece of paper and uploaded as an image will not be taken into consideration for grading.

4. Equations can be typed using the insert equation option present in the toolbar. Equations are supposed to be typed and not copy-pasted as images.

5. plots can be uploaded as images, Don't use the google drive to upload the plots you can upload images directly and post it in the answer section.

EXPLICIT ANALYSIS:

An Explicit FEM analysis is time dependent, does the incremental procedure and at the end of each timestep updates the stiffness matrix based on geometry changes and material changes. Then a new stiffness matrix is constructed at the current time step by considering the state or solution of the system at the previous step and the next increment of load (or displacement) is applied to the system. In this type of analysis if the increments are small enough the results will be accurate. One problem with this method is that you do need many small increments for good accuracy and it is time consuming. If the number of increments are not sufficient the solution tends to drift from the correct solution. Moreover, it does not follow a convergence criterion at each increment but rather computes the local variables at that step and moves to the next step.

The explicit method should be used when the strain rates/velocity is over 10 units/second or 10 m/s respectively. These events can be best exampled by extreme scenarios such as an automotive crash, ballistic event etc. In these cases, the material models do not only need to account for the variation of stress with strain but also the strain rate. On this scale, the strain rates play a particularly important contribution.

IMPLICIT ANALYSIS

An Implicit FEM analysis is the same as Explicit with the addition that after each increment the analysis does Newton-Raphson iterations to enforce equilibrium of the internal structure forces with the externally applied loads. This means that each increment has to converge. The equilibirium is usually enforced to some user specified tolerance. So this is the primary difference between the two types of anlysis, Implicit uses Newton-Raphson iterations to enforce equilibrium.

Since convergence is verified at each time increment, those increments can be large. Each time increment computes slowly, since iterations are needed to get converged.

This type of analysis tends to be more accurate and can take somewhat bigger increment steps. Also, this type of analysis can handle problems better such as cyclic loading, snap through, and snap back so long as sophisticated control methods such as arc length control or generalized displacement control are used. One draw back of the method is that during the Newton-Raphson iterations one must update and reconstruct the stiffness matrix for each iteration. This can be computationally costly.

The implicit method should be used when the events are much slower and the effects of strain rates are minimal. Once the growth of stress as a function of strain can be established, these can be analyzed using implicit methods.

To this extent, both implicit and explicit analysis does more or less the same thing. They divide the analysis time into small increments, and then analyze what is going on in the model one increment of time after another.

Problem:

A bar under tension (shown below) is to be solved using Explicit and Implicit approach with the force in the bar being a non-linear function of the displacement.

`f(u)=u^3+9u^2+4 u`

Relationship between Force and Stiffness is `ΔF=k⋅Δu` (Eqn-1)

Force is a function of displacement `f(u)=u^3+9u^2+4 u` (Eqn-2)

Stiffness is the unit force to unit displacement `k(u)=(df)/(du)=3u^2+18u+4` (Eqn-3)

Where,

F- External load

ΔF- Load Increment

f(u)- Internal force

k- Stiffness

u- Displacement

Δu-Displacement Increment

EXPLICIT METHOD:

For this example three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar. The analysis proceeds as follows:

Step 1: At first load step, ΔF = 1, u₀ = 0

From eqn 3, `k(u_0 )=3(0)+18(0)+4 = 4`

From eqn 1, `∆u_1= (∆F)/k= 1/4=0.25`

Therefore, `u_1= u_0+ ∆u_1 = 0+0.25=0.25`

Where, u₀- Initial displacement at no load

u₁- Final displacement after 1st step load

Δu₁- Change in displacement caused due to step load

Step 2: At second load step, ΔF = 1, u₁ = 0.25

From eqn 3,`k(u_1 )=3(0.25)^2+18(0.25)+4 = 8.687`

From eqn 1, `∆u_2= (∆F)/(k(u_1))= 1/8.687=0.115`

Therefore,`u_2= u_1+ ∆u_2 = 0.25+0.115=0.365`

Step 3: At third load step, ΔF = 1, u₂ = 0.365

From eqn 3,`k(u_2 )=3(0.365)^2+18(0.365)+4 = 10.97`

From eqn 1, `∆u_3= (∆F)/(k(u_2))= 1/10.97=0.091`

Therefore, `u_3= u_2+ ∆u_3 = 0.365+0.091=0.456`

Total external force, `F = ΔF+ ΔF+ ΔF =1+1+1 = 3`

F_ext = 3

Internal force, `f=f(u_3 )=f(0.456)`

`f(u_3 )= (0.456)^3+9*(0.456)^2+4×*(0.456) = 3.79`

f_int= 3.79

F_ext ≠ f_int

Hence, the system is not in equilibrium.

Step i	ΔFi	ui­	(Fext)i	(fint)i	fint - Fext = R Residual
1	1	0.25	1	1.578	0.578
2	1	0.365	2	2.707	0.707
3	1	0.456	3	3.79	0.79

IMPLICIT METHOD:

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.

Residual, `R(u)=f_(Int) -F_(ext)`

Upon expanding the above equation as a Taylor series, we get,

`R^(j+1)= R^j (u^j )+ (dR (u^j))/du (δu)^(j+1)+⋯`

Ignoring higher-order terms, we get,

`R^(j+1)= R^j (u^j )+ (dR (u^j))/du (δu)^(j+1) = R^j+k(u^j ) (δu)^(j+1)`

To attain equilibrium, `R^(j+1)=0`

Hence solving for, `(δu)^(j+1)`, we get

`(δu)^(j+1)= -[K(u^j )]^(-1)×R^j`

NOTE: In Implicit method, we have to reduce the Residual below the chosen tolerance criteria, because we cannot eliminate the residual completely using few Newton-Raphson iterations done numerically.

As given, tolerance limit as`10^(-2)`

Step 1: At first load step, ΔF = 1, u₀ = 0

From eqn 3,`k(u_0 )=3(0)+18(0)+4 = 4`

From eqn 1, `∆u_1= (∆F)/k= 1/4=0.25`

Therefore, `u_1= u_0+ ∆u_1 = 0+0.25=0.25`

Internal force, `f(u_1 )= (0.25)^3+9×(0.25)^2+4×(0.25) = 1.578`

Check Residual:

`R(u)=f_(Int)-F_(ext)=1.578-1=0.578`

`R^((0))> 10^(-2)`

Newton-Raphson Iterations are necessary

Newton-Raphson:

Iteration 1:

`u_1= u_1^((0)=0.25`

`δu^((1) )= -[K(u^j )]^(-1)*R^j`

`δu^((1) )= -[8.68]^(-1)*0.578= -0.0665`

`u_1^((1))=u_1^((0))+ δu^((1) )= 0.25+(-0.0665)=0.1835`

`f (u_1^((1)))= 1.0432`

Check Residual:

`R(u)^(1) = f_(Int)-F_(ext)=1.0432-1=0.0432`

`R^((1))> 10^(-2)`

Another Newton-Raphson Iteration is necessary

Iteration 2:

`u_1^((1))=1.0432`

`δu^((2) )= δu^((1) ) -[K(u_1^((1)) )]^(-1)×R^((1))`

`δu^((2) )=-0.0665 -[26.042]^(-1)×0.0432= -0.068159`

`u_1^((2))=u_1^((0))+ δu^((2) )= 0.25+(-0.068159)=0.18184`

`f (u_1^((2)))= 1.0310`

Check Residual:

`R(u)^(2) = f_(Int)-F_(ext)=1.0310-1=0.0310`

`R(u)^(2)> 10^(-2)`

Another Newton-Raphson Iteration is necessary

Iteration 3:

`u_1^((2))=0.18184`

`δu^((3) )= δu^((2) ) -[K(u_1^((2)) )]^(-1)×R^((2))`

`δu^((3) )=-0.068159 -[7.3723]^(-1)×0.0310= -0.072364`

`u_1^((3))=u_1^((0))+ δu^((3) )= 0.25+(-0.072364)=0.17764`

`f (u_1^((3) ))= 1.0002`

Check Residual:

`R(u)^(3) = f_(Int)-F_(ext)=1.0002-1=0.0002`

`R(u)^(3) < 10^(-2)`

Therefore,

`u_1=u_1^((3))=0.17764`

Step 2: At second load step, ΔF = 1, u₁ = 0.17764

From eqn 3, `k(u_1 )=3(0.17764)+18(0.17764)+4 = 7.28`

From eqn 1, `∆u_2= (∆F)/k= 1/7.28=0.13736`

Therefore, `u_2= u_1+ ∆u_2=0.177+0.13736=0.31436`

Internal force,

`f(u_2 )= (0.31)^3+9×(0.31)^2+4×(0.31) = 2.1779`

Check Residual:

`R(u)=f_(Int)-F_(ext)=2.1779-2=0.1779`

`R^((0))> 10^(-2)`

Newton-Raphson Iterations are necessary

Newton-Raphson:

Iteration 1:

`u_2= u_2^((0))=0.31436`

`δu^((1) )= -[K(u_j )]^(-1)*R^j`

`δu^((1) )= -[9.9549]^(-1)×0.1779= -0.017871`

`u_2^((1))=u_2^((0))+ δu^((1) )= 0.31436+(-0.017871)=0.29649`

`f (u_2^((1) ))= 2.0032`

Check Residual:

`R(u)=f_(Int)-F_(ext)=2.0032-2=0.0032`

`R^((1))< 10^(-2)`

Therefore, `u_2=0.29649`

Step 3: At third load step, ΔF = 1, u₂ = 0.29649

From eqn 3, `k(u_2 )=3( 0.29649)+18( 0.29649)+4 = 9.6005`

From eqn 1, `∆u_3= (∆F)/k= 1/9.6005=0.10416`

`u_3= u_2+ ∆u_3 = 0.29649+0.10416=0.40065`

Internal force,

`f(u_3 )= (0.40065)^3+9×(0.40065)^2+4×(0.40065) = 3.1116`

Check Residual:

`R(u)=f_(Int)-F_(ext)=3.1116-3=0.1116`

`R^((0))> 10^(-2)`

Newton-Raphson Iterations are necessary

Newton-Raphson:

Iteration 1:

`u_3= u_3^((0))=0.40065`

`δu^((1) )= -[K(u_j )]^(-1)×R_j`

`δu^((1) )= -[11.693]^(-1)×0.1116= -0.009544`

`u_3^((1))=u_3^((0))+ δu^((1) )= 0.40065+(-0.0095440)=0.39111`

`f (u_3^((1) ))= 3.0009`

Check Residual:

`R(u)=f_(Int)-F_(ext)=3.0009-3=0.0009`

`R^((1))< 10^(-2)`

Therefore, `u_3=0.39111`

Step i	ΔFi	ui­	(Fext)i	(fint)i	fint - Fext = R Residual
1	1	0.17764	1	1.002	0.0002
2	1	0.29649	2	2.0032	0.0032
3	1	0.39111	3	3.0009	0.0009

Results & Comparison:

• The difference between internal force and external force is called residual value. The higher deviation in result implies Explicit analysis is a poor solver method.
• 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
• From the above graph, we can clearly see that the results obtained from Implicit method almost overlaps with the exact solution. The small deviation is due to tolerance of 0.01.
• Hence, we can conclude that the Implicit method provides a much more accurate solution when compared to the explicit method.
• This is because the implicit method checks for convergence at each increments and hence minimizes the residual value using the Newton-Raphson method before moving onto the next step.
• The Newton-Raphson iterations correct the incremental steps so that they land on the exact solution according to the specified tolerance.
• So for the given problem Implicit method Analysis is suitable.