Week - 2 - Explicit and Implicit Analysis
AIM
Knowledge of Explicit and Implicit method of analysis
OBJECTIVE
Know about Explicit and Implicit method of analysis
Solve the below equation in both method.
F(u) = u3+9u2+4u
THEORY
Explicit method: Calculate the sate of system at particular time with the help of previous state of system.
Yn+1 = Yn + Δt.Y’n
Implicit method: Calculate the state of system at particular time with solving both previous state and current state.
Yn+1 = Yn + Δt.Y’n+1
EXPLICIT METHOD
Since it is evident to plot a graph between the force and displacement. However if the non linear force vs displacement relationship is not known but stiffness is known then it is possible to construct the graph of force vs displacement. One of the method is to use an incremental explicit load control scheme. For this three equal load increments ΔF = 1 unit each is used. Final equation is ΔF = k Δu
F(u) = u3+9u2+4u -----------------------------(1)
ΔF = k Δu -----------------------------(2)
Here K is the stiffness of material
F’(u) = K(u) = 3u2+18u+4 -----------------------------(3)
Step 1:
Put u = 0 in equation 1 and 3
F (0) = 0 & K(0) = 4
Sine the K is known , lets put in equation 2
Since ΔF = kΔu
Δu = Δf/k = 1/4 = 0.25
Now the increment of displacement is known
- U1 = u0+Δu = 0+0.25 = 0.25
Finally, the displacement at point 1 is known
F internal = (0.25)3+9 X (0.25)2 + 4 X 0.25 = 0.015625 + 0.5625 + 1 = 1.578125
Step 2:
Since we know the u1 value for finding u2 put u1 value in equation 1
K(u1) = 3 X (0.25)2 + 18 X 0.25 + 4
= 0.1875 + 4.5 + 4
= 8.6875
Here we find the stiffness value at 2nd stage.
Now Δu = ΔF/k(u1) = 1/8.6875 = 0.1151
U2 = u1 +Δu = 0.25 + 0.1151 = 0.3651
Here the displacement at point 2 is 0.3651
F internal = (0.36)3+9 X (0.36)2 + 4 X 0.36 = 0.046656 + 1.1664 + 1.44 = 2.653056
Step 3:
Put the u2 = 0.3651 in equation 1
K(u2) = 3 X (0.3651)2 + 18 X 0.3651 + 4
= 0.3998 + 5.47 + 4
= 9.8698
Here also the stiffness is known
Now Δu = ΔF/k(u1) = 1/9.8698 = 0.1013
U3 = u2 +Δu = 0.3651 + 0.1013 = 0.4664
From the above 3 steps
F external = ΔF + ΔF + ΔF = 3
F internal = u3+9u2+4u = (0.4664)3 + 9 X (0.4664)2 + 4 X 0.4664 = 0.1014 + 1.9577 + 1.8656 = 3.9247
|
Step |
ΔF |
Δ u i |
ui |
F external |
F internal |
R = F internal - F external |
|
1 |
1 |
0.25 |
0.25 |
1 |
1.578125 |
0.57 |
|
2 |
1 |
0.11 |
0.36 |
2 |
2.653056 |
0.65 |
|
3 |
1 |
0.10 |
0.46 |
3 |
3.9247 |
0.92 |
From the above results it is evident that internal and external force are not in equilibrium.
IMPLICIT METHOD
An implicit analysis is the same as the explicit, except that at the end of each step Newton Raphson methos is 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.
Newton Raphson method: x = x0-[F(x0)/F’(x0)]
Expanding the residual as a Taylor series about the current displacement u j gives R j+1 = R j (u j ) + dR(u j ) du δuj+1 + ... = R j + k(u j )δuj+1 + ...
By neglecting higher order terms in the series, setting Rj+1 = 0 and solving for δuj+1 the following correction is obtained δuj+1 = −[k(u j )]−1R j .
Step 1:
Put u = 0 in equation 1 and 3
F (0) = 0 & K(0) = 4
Sine the K is known , lets put in equation 2
Since ΔF = kΔu
Δu = Δf/k = 1/4 = 0.25
Now the increment of displacement is known
- U1 = u0+Δu = 0+0.25 = 0.25
Finally the displacement at point 1 is known
Now, Check the residual
F external = ∆F = 1
F internal = (0.25)3+9 X (0.25)2 + 4 X 0.25 = 0.015625 + 0.5625 + 1 = 1.578125
R0 = F internal − F external = 0.57
R0 > 0.001
⇒ Newton-Raphson iterations are necessary.
Calculate the correction to u1 = u01
δu11 = −[k(u01 )]−1R0 =-[ 3(u01) 2 + 18 u01 + 4]-1 R0 = 0.1875+4.5+4 = -[8.68]-1 X 0.57= - 0.06566
⇒ The updated value of u11 = u01 + δu11 = 0.25 + (−0.06566) = 0.18434.
Check the residual again,
F external = 1
F internal = (u11)3+9(u11)2+4 u11 = (0.1843)3 + 9 X (0.1843)2 + 4 X 0.1843 = 0.0062 + 0.3056 + 0.7372 = 1.049
R1 = F internal - F external = 1.049 – 1 = 0.049
R1 > 0.001
⇒ another Newton-Raphson iteration is necessary.
Calculate the correction to u1 = u11
δu21 = − [k(u11 )]−1 R1 = =-[ 3(u11) 2 + 18 u11 + 4]-1 R1 = 0.1018 + 3.3174 + 4 = -[ 7.419]-1 X 0.049= - 0.00660
⇒ The updated value of u12 = u11 + δu21 = 0.1843 + (−0.0066) = 0.1777
Check the residual again,
F external = 1
F internal = (u21)3+9(u21)2+4 u21 = (0.1777)3 + 9 X (0.1777)2 + 4 X 0.1777 = 0.0056 + 0.2841 + 0.7108 =1.0005
R2 = F internal - F external = 1.0005 – 1 = 0.0005
R2 < 0.001
Hence no further iteration is required
Therefore, the final value is u1 = u21 = 0.1777.
Step 2:
Since we know the u1 value for finding u2 put u1 value in equation 1
K(u1) = 3 X (0.25)2 + 18 X 0.25 + 4
= 0.1875 + 4.5 + 4
= 8.6875
Here we find the stiffness value at 2nd stage.
Now Δu = ΔF/k(u1) = 1/8.6875 = 0.1151
U2 = u1 +Δu = 0.25 + 0.1151 = 0.3651
Here the displacement at point 2 is 0.3651
F internal = (0.36)3+9 X (0.36)2 + 4 X 0.36 = 0.046656 + 1.1664 + 1.44 = 2.653056
Now, Check the residual
F external = ∆F = 1
F internal = (0.36)3+9 X (0.36)2 + 4 X 0.36 = 0.046656 + 1.1664 + 1.44 = 2.653056
R0 = F internal − F external = 2.653056 – 1 = 1.6530
R0 > 0.001
⇒ Newton-Raphson iterations are necessary.
Calculate the correction to u2 = u02 = 0.3651
δ u12 = −[k(u02 )]−1R0 =-[ 3(u02) 2 + 18 u02 + 4]-1 R0 = 0.2665 + 6.5718 + 4 = -[10.8382]-1 X 0.57= - 0.0525
⇒ The updated value of u12 = u02 + δ u12 = 0.3651+ (−0.0525) = 0.3126.
Check the residual again,
F external = 1
F internal = (u12)3+9(u12)2+4 u12 = (0.3126)3 + 9(0.3126)2 + 4( 0.3126) = 0.0305 + 0.8794 + 1.2504 =2.1603
R1 = F internal - F external = 2.1603– 1 = 1.1603
R1 > 0.001
Calculate the correction to u2 = u12 = 0.3126
δ u22 = −[k(u12 )]−1 R1 =-[ 3(u12) 2 + 18 u12 + 4]-1 R1 = 0.2931 + 5.6268 + 4 = -[ 9.9199]-1 X 1.1603= - 0.1169
⇒ The updated value of u22 = u22 + δ u22 = 0.3126+ (−0.1169) = 0.1957.
Check the residual again,
F external = 1
F internal = (u22)3+9(u22)2+4 u22 = (0.1957)3 + 9(0.1957)2 + 4( 0.1957)= 0.0074 + 0.0382 + 0.7828 =0.8284
R1 = F internal - F external = 0.8284– 1 = -0.1716
R1 < 0.001
Hence no further iteration is required
Therefore, the final value is u1 = u22 = - 0.1716
Hence u2 = -0.1716
Step 3:
Put the u2 = 0.3651 in equation 1
K(u2) = 3 X (0.3651)2 + 18 X 0.3651 + 4
= 0.3998 + 5.47 + 4
= 9.8698
Here also the stiffness is known
Now Δu = ΔF/k(u1) = 1/9.8698 = 0.1013
U3 = u2 +Δu = 0.3651 + 0.1013 = 0.4664
From the above 3 steps
F external = ΔF + ΔF + ΔF = 3
F internal = u3+9u2+4u = (0.4664)3 + 9 X (0.4664)2 + 4 X 0.4664 = 0.1014 + 1.9577 + 1.8656 = 3.9247
R = F internal -- F external = 3.9247 – 3 = 0.9247
R > 0.001
⇒ Newton-Raphson iterations are necessary.
Calculate the correction to u3 = u03 = 0.4664
δu13 = −[k(u03 )]−1R0 =-[ 3(u03) 2 + 18 u03 + 4]-1 R0 = 0.6525+8.3952+4 = -[13.0477]-1 X 0.9247= - 0.0708
⇒ The updated value of u31 = u03+ δu13 = 0.4664 + (−0.0708) = 0.3956.
Check the residual again,
F external = 1
F internal = (u13)3+9(u13)2+4 u13 = (0.3956)3 + 9 X (0.3956)2 + 4 X 0.3956= 0.0619 + 1.4084 + 1.5824 = 3.0527
R1 = F internal - F external = 3.0527– 1 = 2.0527
R1 > 0.001
Again iteration required
Calculate the correction to u3 = u13 = 0.3956
δ u23 = −[k(u13 )]−1R1 =-[ 3(u13) 2 + 18 u13 + 4]-1 R1 = 0.4694+7.590+4 = -[12.0594]-1 X 2.0527= - 0.1702
⇒ The updated value of u32 = u13+ δu23 = 0.3956 + (−0.1702) = 0.2254
Check the residual again,
F external = 1
F internal = (u23)3+9(u23)2+4 u23 = (0.2254)3 + 9 X (0.2254)2 + 4 X 0.2254= 0.0114 + 0.4572 + 0.9016 = 1.3702
R2 = F internal - F external = 1.3702– 1 = 0.3702
R2 > 0.001
Again iteration required
Calculate the correction to u3 = u23 = 0.2254
δ u33 = −[k(u23 )]−1R2 =-[ 3(u23) 2 + 18 u23 + 4]-1 R2 = 0.1524+4.0572+4 = -[8.2096]-1 X 0.3702= - 0.0450
⇒ The updated value of u33 = u23+ δu23 = 0.2254 + (−0.0450) = 0.1804
Check the residual again,
F external = 1
F internal = (u33)3+9(u33)2+4 u33 = (0.1804)3 + 9 X (0.1804)2 + 4 X 0.1804= 0.0058 + 0.2928 + 0.7216 = 1.0202
R3 = F internal - F external = 1.0202– 1 = 0.0202
R3 > 0.001
Again iteration required
Calculate the correction to u3 = u33 = 0.1804
δ u43 = −[k(u33 )]−1R3 =-[ 3(u33) 2 + 18 u33 + 4]-1 R3= 0.0976+3.2472+4 = -[7.3448]-1 X 0.3702= - 0.0504
⇒ The updated value of u34 = u33+ δu43 = 0.1804 + (−0.0504) = 0.13
Check the residual again,
F external = 1
F internal = (u43)3+9(u43)2+4 u43 = (0.13)3 + 9 X (0.13)2 + 4 X 0.13= 0.0021 + 0.0169 + 0.52 = 0.539
R4 = F internal - F external = 0.539– 1 = - 0.461
R4 < 0.001
Hence no further iteration is required
Therefore, the final value is u3 = u43 = - 0.461
|
Step |
ΔF |
ui |
F external |
F internal |
R = F internal - F external |
|
1 |
1 |
0.1777 |
1 |
1.0005 |
0.0005 |
|
2 |
1 |
-0.1716 |
2 |
0.8284 |
1.1716 |
|
3 |
1 |
-0.461 |
3 |
0.539 |
2.461 |
CONCLUSION
From the above results it is clear that explicit method analysis deviates from true solution. So to counter this implicit method of analysis is considered. In implicit method of analysis Newton- Raphson iteration is used to enforce equilibrium between internal and external forces.