Week - 2 - Explicit and Implicit Analysis
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
It is a method used to compute the state of certain systems 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.
The explicit analysis provides a faster solution in events which have dynamic equilibrium. the main air of explicit method is to solve for acceleration. explicit method is fast and inexpensive computationally. Each increment is calculated extremely fast. Accuracy is not maintained. Timestep has to be extremely small.
Implicit analysis
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 its called an implicit method.
Implicit methods take a 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. The 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.
The implicit method is 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 is established, these can be analyzed using implicit methods. Since the implicit method cannot be carried out for each kind of differential operator, it is sometimes advisable to make use of the so-called operator splitting method, which means that the differential operator is rewritten as the sum of two complementary operators. A simple qualitative model will help to illustrate how this works. Let Q be a quantity whose value Qn+1 we want to compute at time t=(n+1)it, in terms of its value at time t=ndt, i.e., Qn+1=Qn+dtS, where S is the rate of change in Q.
In an explicit numerical method, S would be evaluated in terms of known quantities at the previous time step n. An implicit method, in contrast, would evaluate some or all of the terms in S in terms of unknown quantities at the new time step n+1. Since new quantities appear on both the left and right sides of the Q-equation, it is said to be an implicit definition of the new n+1 values. Usually, a matrix or iterative solution must be used to compute the new quantities.
The explicit method should be used when the strain rates/velocity is over 10 units/second or 10 m/s respectively. 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 role. Once the acceleration is calculated at the nth step, the velocity at n+1/2 step and displacement at the n+1 step are calculated accordingly. In these calculations, the scheme is not unconditionally stable and thus smaller time steps are required.
The speed of computing is exactly the difference between implicit and explicit methods.
Aim: To solve f(u)=u3+9u2+4uf(u)=u3+9u2+4u using implicit and explicit methods.
Explicit method:
`F(u)=u3+9u2+4uF(u)=u3+9u2+4u` .(1)
now the stifness can be represented as
`K(u)=dF(u)du=3u2+18u+4K(u)=dF(u)du=3u2+18u+4` .(2)
now with the help of equation (1) and (2) we will compute the problem usng incremental explicit load control scheme.
from (1) and (2) we get,
`ΔF=KΔUΔF=KΔU` .(3`
step 1:
for the first step lets us consider an initial conditon where,
`ΔF=1ΔF=1`
`u(0)=0u(0)=0`
substituing in equation (2),
`K(0)=3(0)2+18(0)+4=4K(0)=3(0)2+18(0)+4=4`
`K(0)=4K(0)=4`
then (3) implies,
`ΔU1=14=025ΔU1=14=025`
`u1=u(0)+ΔU1u1=u(0)+ΔU1`
`u1=0.25u1=0.25`
step 2:
Now let us consider,
`u=u1u=u1,then`
`k(u1)=3(0.25)2+18(0.25)+4k(u1)=3(0.25)2+18(0.25)+4`
`k(u1)=8.687k(u1)=8.687`
`ΔU2=ΔFK(u1)=18.687ΔU2=ΔFK(u1)=18.687`
`ΔU2=0.115ΔU2=0.115`
`u2=u1+ΔU2=0.25+0.115u2=u1+ΔU2=0.25+0.115`
`u2=0.365u2=0.365`
Step 3:
Let us consider u=u2u=u2,then
`K(u2)=3(0.365)2+18(0.365)+4K(u2)=3(0.365)2+18(0.365)+4`
`K(u2)=10.96K(u2)=10.96`
`ΔU3=ΔFK(u2)=110.96=0.091ΔU3=ΔFK(u2)=110.96=0.091`
`u3=u2+ΔU3=0.365+0.091=0.456u3=u2+ΔU3=0.365+0.091=0.456`
frpm the above 3 steps we can infer that,
`Fext=ΔF+ΔF+ΔF=3Fext=ΔF+ΔF+ΔF=3`
`f1=(u31)+(9u21)+4U1=(0.253)+(9⋅(0.252))+(4⋅0.25)=1.578f1=(u13)+(9u12)+4U1=(0.253)+(9⋅(0.252))+(4⋅0.25)=1.578`
`f2=(u32)+(9u22)+4U2=(0.3653)+(9⋅(0.3652))+(4⋅0.365)=2.707f2=(u23)+(9u22)+4U2=(0.3653)+(9⋅(0.3652))+(4⋅0.365)=2.707`
`f3=(u33)+(9u23)+4U=(0.45613)+(9⋅(0.45612))+(4⋅0.4561)=3.791f3=(u33)+(9u32)+4U=(0.45613)+(9⋅(0.45612))+(4⋅0.4561)=3.791`
| step | ΔFΔF | ΔUΔU | uu | fextfext | fifi | R=fi−fextR=fi-fext |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.578 | 0.578 |
| 2 | 1 | 0.115 | 0.365 | 2 | 2.707 | 0.707 |
| 3 | 1 | 0.091 | 0.456 | 3 | 3.791 | 0.791 |
Implicit Analysis:
In implicit analysis,at the end of each step Newton-Raphson iteration is used to bring equilibrum and then proceded to further steps
Step 1:
conisder u=0u=0,then
`K(0)=9(0)2+18(0)+4K(0)=9(0)2+18(0)+4`
`K(0)=4K(0)=4`
`ΔU=ΔFK(0)=14ΔU=ΔFK(0)=14`
`ΔU=0.25ΔU=0.25`
`u1=u(0)+ΔU=0.25u1=u(0)+ΔU=0.25`
fext=1fext=1
`fi=(u31)+(9u21)+4u1=(0.253)+(9⋅(0.252))+(4⋅0.25)fi=(u13)+(9u12)+4u1=(0.253)+(9⋅(0.252))+(4⋅0.25)`
`f1=1.518f1=1.518`
`Residual R0=fi−Fext=1.578−1R0=fi-Fext=1.578-1`
`R0=0.578R0=0.578,which is greater than 0.01`
therefore Newton Raphson model has to be followed
`U1=U01U1=U10`
`∂U1=[−k(u1)0]−1⋅R0=−0.06∂U1=[-k(u1)0]-1⋅R0=-0.06`
then the value of u1 will be,
`u11=u01+∂u1=0.25−0.06=0.184u11=u10+∂u1=0.25-0.06=0.184`
`R1=fi−Fext=1.0469−1=0.04R1=fi-Fext=1.0469-1=0.04,which is greater than 0.01`
U1=U01U1=U10
`∂U^2=∂U1−[ku11]−1⋅R1∂U^2=∂U1−[ku11]−1⋅R1= -0.06 -0.005 = -0.0713``
`u21=u01+∂u2=0.25−0.0713u12=u10+∂u2=0.25-0.0713`
The updated value of u1 is 0.178
`fi=(u31)+(9u21)+4u1=(0.1783)+(9⋅(0.1782))+(4⋅0.178)fi=(u13)+(9u12)+4u1=(0.1783)+(9⋅(0.1782))+(4⋅0.178)`fi=1.007fi=1.007
`R2=1.007−1=0.007R2=1.007-1=0.007`The R alue is less than 0.01
therefore the final value of u1 is 0.1787
Step 2:
`u=u1=0.178u=u1=0.178`
`K(u1)=9(0.178)2+18(0.178)+4K(u1)=9(0.178)2+18(0.178)+4`
`K(u1)=7.489K(u1)=7.489`
`ΔU2=ΔFK(u1)=17.489ΔU2=ΔFK(u1)=17.489`
`ΔU2=0.133ΔU2=0.133`
`u2=u1+ΔU2=0.311u2=u1+ΔU2=0.311`
fext=1fext=1
`fi=(u32)+(9u22)+4u2=(0.3113)+(9⋅(0.3112))+(4⋅0.311)fi=(u23)+(9u22)+4u2=(0.3113)+(9⋅(0.3112))+(4⋅0.311)`
`f1=2.144f1=2.144`
Residual R0=fi−Fext=2.144−2R0=fi-Fext=2.144-2
`R^0=0.144`,which is greater than 0.01
from Newton Raphson model the updated u2 is
`u2=.296u2=.296`
step 3:
`u=u2=0.296u=u2=0.296`K(u2)=9(0.296)2+18(0.296)+4K(u2)=9(0.296)2+18(0.296)+4
`K(u2)=10.116K(u2)=10.116`
`ΔU3=ΔFK(u2)=110.116ΔU3=ΔFK(u2)=110.116`
`ΔU3=0.098ΔU3=0.098`
`u3=u2+ΔU3=0.394u3=u2+ΔU3=0.394`
fext=1fext=1
`fi=(u33)+(9u23)+4u3=(0.3943)+(9⋅(0.3942))+(4⋅0.394)fi=(u33)+(9u32)+4u3=(0.3943)+(9⋅(0.3942))+(4⋅0.394)`
f1=3.034f1=3.034
`Residual R0=fi−Fext=3.034−3R0=fi-Fext=3.034-3`
`R^0=0.034,whichisgreaterthan0.0,whichisgreaterthan0.0
from Newton Raphson model the updated u3 is
u3=.391u3=.391
| Step | ΔFΔF | ΔUΔU | U | FextFext | fifi | Residual |
| 1 | 1 | .25 | .178 | 1 | 1.007 | .007 |
| 2 | 1 | .113 | .296 | 2 | 2.004 | .004 |
| 3 | 1 | .098 | .391 | 3 | 3.001 | .001 |
Graph (1)
The above graph (1) shows the force values for implicit, exact and explicit values. we can notice that the implicit and exact value curves are almost overlapping because the implicit method provides more accurate values.
Graph (2)
The above graph (2) represents the force value of exact and explicit values.
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.