Week - 2 - Explicit and Implicit Analysis
Aim:
Solve this equation, F(u) = u3+9u2+4u using both Explicit and Implicit Methods with a tolerance of 0.01.
Theory:
The basic equation of motion solved by transient dynamic analysis is given below
[M] {A} +[C] {V}+[K]{u}={F(t)}
[M]= Mass matrix [C]=Dambing matrix, [K]= Stiffness matrix, {F(t)}=Load vector
{A}= Nodal Acceleration, {V}= Nodal velocity vector, {u}= Nodal displacement vector.
- Explicit
At each node point, internal and external forces are added, and nodal acceleration is calculated by dividing nodal mass. The solution is advanced by incorporating this time acceleration. The Courant condition limits the maximum time step size, resulting in an algorithm that typically requires a large number of relatively inexpensive time steps.
Explicit analysis offers a faster solution in events where there is a dynamic equilibrium or otherwise:
Sum of all forces = mass x acceleration
- Implicit
To obtain a displacement increment, a global stiffness matrix is computed, inverted, and applied to the nodal out-of-balance force. The advantage of this approach is that the user can choose the size of the time step. The disadvantage is the significant computational effort required to create, store, and factorise the stiffness matrix. As a result, implicit simulations typically involve a small number of expensive time steps.
The implicit method should be used when the effects of strain rates are minimal and the events are much slower. Once the growth of stress as a function of strain can be established, these can be analyzed using implicit methods. In this case, one can consider a static equilibrium such that:
Sum of all forces = 0
|
IMPLICIT METHOD |
EXPLICIT METHOD |
|
|
Deformation |
small |
Large |
|
What cause large deformation |
Force and dynamic |
Velocity, acceleration with mass and inertia |
|
K matrix inversion |
yes |
No |
|
Time step size |
Large |
Small |
|
What limits the time step |
Accuracy |
Stability |
|
Cost per time step |
large |
Small |
|
What dominates the cost |
Matrix inversion |
Element process |
|
Parallel process |
Possible |
Not possible |
- Explicit Method solving
F= external force applied
k= Stiffness
u= Displacement
∆F = k∆u.
F(u) = u3 +9u2 +4u (1)
K(u) = dF/du = 3u2 +18u + 4 (2)
Step1 :
Take u = 0 in equation (2) =>
3u2 +18u + 4 => k(0) = 3 x 0 + 18 x 0 + 4 = 4
= k(0) = ¼ = 0.25
u= 0.25
Finternal = (0.25)3 + 9 x (0.25)2 +4 x 0.25 = 1.578
Step2
Take u1 = 0.25 in equation (2)
3u2 +18u + 4 => k(u1) = 3 x 0.252 + 18 x 0.25 + 4 = 8.6875
= /k(u1) = 1/8.6875 = 0.1151
U 2= u = 0.3651
Finternal = (0.3651)3 + 9 x (0.3651)2 +4 x 0.3651 = 2.708
Step3
Take u 2 = 0.3651 in equation (2)
3u2 +18u + 4 => k(u2 ) = 3 x 0.36512 + 18 x 0.365 + 4 = 10.9716
= /k(u2 ) = 1/10.9716 = 0.09114
U = u2+ = 0.46
To determine whether the analysis is equilibrium
Fexternal = = 3
Finternal = (0.46)3 + 9 x (0.46)2 +4 x 0.46 = 3.841
Since, Fexternal ≠ Finternal So they are not in equilibrium.
| step | ∆F | ∆u | u | Fexternal | Finternal | Finternal-Fexternal=R |
| 1 | 1 | 0.2500 | 0.25 | 1 | 1.578 | 0.578 |
| 2 | 1 | 0.1151 | 0.3651 | 2 | 2.708 | 0.708 |
| 3 | 1 | 0.09114 | 0.46 | 3 | 3.841 | 0.841 |
- Implicit Method solving
F(u) = u3 +9u2 +4u (1)
K(u) = dF/du = 3u2 +18u + 4 (2)
Step1
Take u = 0 in (2)
(2) => k(u0) = 3 x 0 + 18 x 0 + 4 = 4
= /k(u0) = ¼ = 0.25
u1= u0+ = 0.25
Residual R
Finternal = (0.25)3 + 9 x (0.25)2 +4 x 0.25 = 1.578
R0 = 1.578 – 1 = 0.578 > 102
So, Newton-Raphson iterations is necessary
The correction u1 = u1(0)
u(1) = -[k(u1(0)-1 x R0
= -[(3 x 0.252 ) + (18 x 0.250) +4]-1 x 0.578
=-0.0665
U1(1) = u1(0) + u(1)
= 0.25 – 0.0665 = 0.1835
Residual R1
Finternal = (0.1835)3 + 9 x (0.1835)2 +4 x 0.1835 = 1.0432
R0 = 1.0432 – 1 = 0.432 > 102
So, Newton-Raphson iterations is necessary
The correction u1 = u1(1)
u(2) = u(2) -[k(u1(1)-1 x R1
=(-0.0665) -(8.6875-1 )x 0.0432
=-0.0714
U1(2) = u1(1) + u(2)
= 0.25 – 0.0714 = 0.1786
Residual R1
Finternal = (0.1786)3 + 9 x (0.1786)2 +4 x 0.1786 = 1.0071
R0 = 1.0071 – 1 = 0.0071 < 10 xss=removed>2
No residual needed
U1 = 0.1786
Similarly, u2 = 2966, u3 = 0.3911
|
Step |
∆F |
∆u |
u |
Fexternal |
Finternal |
Finternal-Fexternal=R |
|
1 |
1 |
0.25 |
0.1786 |
1 |
1.0071 |
0.0071 |
|
2 |
1 |
0.1374 |
0.2966 |
2 |
2.0042 |
0.0042 |
|
3 |
1 |
0.1041 |
0.3911 |
3 |
3.008 |
0.008 |
Conclusion:
Using explicit and implicit method solved the equation F(u) = u3+9u2+4u. implicit method is more accurate than the explicit but the time consumption is more in implicit method.