Week - 2 - Explicit and Implicit Analysis
Week 2 - Explicit and Implicit Analysis
Implicit and Explicit explanation
Implicit and explicit are different codes for solving dynamics problems.
Dynamics problems are car crash, metal forming etc , any model which changes with time is a dynamic problem.
Implicit method works on backward euler method.
U (n+1) = u (n) + dt F(u(n+1)
For each time increment (time step), the global equilibrium (converges) is achieved and then only it moves to the next time step. Once equilibrium is achieved then displacements, stress are calculated.
To achieve equilibrium iterations has to be done to match the internal and external forces, hence time consuming. But the advantage is that time step can be big in this case as in each step equilibrium is achieved.
But in case of explicit the solver assumes equilibrium exists and goes on finding the displacements without any checks or iterations. Time step has to be really small to reduce errors due to forward euler method.
Time step are calculated based on how much time it takes to travel across the smallest element in the model. Critical time step for a 1 d element
= length of the element / sound wave propagation speed through the material
Basically implicit applications includes impact which takes more than 1sec, explicit application include where run times are less than 1s like crash crash, phone drops
Computation time is mostly difference between both
Aim - Solving the non-linear static problem using explicit and implicit methods.
Consider a simple bar in tension
Given that the force developed in the bar internally is given by
Stiffness of the bar becomes
k (u) =( df /du) = 3u2 + 18u + 4
Explicit Analysis
Force step of 1 units applied to the bar. 3 load steps is applied.
Variables used –
u – Displacement
f – Internal force in the bar
F – External force
k – Stiffness
Step 1 :
U0 = 0
k (u0) = 3u2 + 18u + 4 = 4
∆F = 1
∆F = k∆u1
=>
∆u1 = ∆F/k = ¼
New displacement u1 = u0+ ∆u1 = 0 + ¼ = ¼ = 0.25
Step 2:
k(u1) = = 3(0.25)2 + 18 (0.25) + 4 = 8.6875
∆F = 1
∆F =k∆u2
=>
∆u2 = ∆F/k = 1/8.6875 = 0.1151
New displacement u2 = u1+ ∆u2 = 0.25 + 0.1151 = 0.3651
Step 3:
k(u2) = = 3(0.3651)2 + 18 (0.3651) + 4 = 10.9716
∆F = 1
∆F = k∆u3 => ∆u3 = ∆F/k = 1/10.9716= 0.0911
New displacement u3 = u2+ ∆u3 = 0.3651 + 0.0911= 0.4562
Results
Fe = ∆F + ∆F + ∆F = 3.0
f = u3 + 9u2 + 4u = > sub u3 = 4.5429
F is not equal to f internal, hence not in equilibrium
| step | ∆F | ∆u | u | F | f | R = F- f |
| 1 | 1 | 0.25 | 0.25 | 1 | 1.578125 | 0.578125 |
| 2 | 1 | 0.1151 | 0.3651 | 2 | 2.708749 | 0.708749 |
| 3 | 1 | 0.0911 | 0.4562 | 3 | 3.79281 | 0.79281 |
Implicit Analysis
This method is similar to explicit except that next load step calculation is done only after the global equilibrium is done, hence time consuming but more accurate.
To achieve equilibrium R residual has to be close to zero, but to approximate we are given a tolerance of 1e-2.
R = f – F = > u3 + 9u2 + 4u – 1
Newton Raphson method is used to iterate the value to zero.
X = x – f(x)/ dy/dx is iterated till tolerance is achieved.
Writing the residual in terms of Taylor series it becomes
f(x) = f(0)+ f′(0)x+ f′′(0)x2/2! + …..
j is the number of iterations required
Neglecting the higher order terms and correction for displacement becomes, making Rj+1 =0
Solving method is similar to explicit but at the end of each step we are checking the residual.
Step 1:
U0 = 0
k (u0) = 3u2 + 18u + 4 = 4
∆F = 1
∆F = k∆u1
=>
∆u1 = ∆F/k = ¼
New displacement u1 = u0+ ∆u1 = 0 + ¼ = ¼ = 0.25
f = u3 + 9u2 + 4u = 1.578125
Check R(0) = f – F = 0.5781, this values is not less than 0.01, hence iterate
δu(1) = −R(0)/ [k(u 1 ] =- 0.5781/ 8.6875 = - 0.066
Correction u1 = 0.25 + (- 0.066) = 0.184
Checking R (1) = f – F = 0.04693 still not less than 0.01
δu(2) = δu(1) − R(1)/ [k(u 1 ] = -0.066– 0.0469/ 7.413 = -0.0723
Correction u2 = -0.0723+0.25= 0.1776
Check R (2) = f – F = 0.999878-1 = 0.000122, tolerance reached
Final value of displacement is 0.1776
Step 2
U1 = 0.1776
k (u0) = 3u2 + 18u + 4 = 7.72
∆F = 1
∆F =k∆u2
=>
∆u2 = ∆F/k =1/7.72 = 0.1295
New displacement u2 = u1+ ∆u2 = 0.1776 + .1295= 0.3071
f = u3 + 9u2 + 4u = 2.106156
R (0) = f – F = 2.10 -2 = 0.10
δu(1) = −R(0)/ [k(u 2] =-0.10/ 9.810731= - 0.011
Correction u2 = 0.3071-0.011= 0.2961
Checking R (1) = f – F = 0.005 , tolerance reached
Hence Final value of displacement is 0.2961
Step 3
U2 = 0.2961
k (u0) = 3u2 + 18u + 4 = 9.592
∆F = 1
∆F =k∆u3
= f – F = ∆u3 = ∆F/k =1/9.59 = 0.104
New displacement u3 = u2+ ∆u3 = 0.2961 + .104= 0.400
R (0) = f – F = 3.104 – 3 = 0.104
δu(1) = −R(0)/ [k(u 3] =- 0.104/ 11.68 = 0.0089
Correction u3 = 0.4-0.0089= 0.2961 = 0.39109
Checking R (1) = f – F = 0.00074, tolerance reached
Hence the final displacement = 0.39109
| step | ∆F | u | f | R = F- f |
| 1 | 1 | 0.1776 | 0.999878 | 0.000122 |
| 2 | 1 | 0.2961 | 1.999438 | 0.005 |
| 3 | 1 | 0.39109 | 3.00074 | 0.00074 |
Conclusion
We can observe that the results of implicit analysis are more closer towards exact solution than explicit.