Basic Explicit and Implicit Analysis Study in LS-DYNA
Objective: To solve the required problem using implicit and explicit analysis
Procedure: The equation is given to us as :
f(u) = u^3 + 9u^2 + 4u
As can be seen, force is a non-linear function of displacement
k(u) = `(df)/(du)` = 3u^2 + 18u + 4
Consider `DeltaF = 1`. Since F = ku, `DeltaF = kDeltau`.
Explicit analysis:
At `u_0`, k(`u_0`) = 4, `DeltaF = 1, Deltau = 0.25`, `u_1 = u_0 + Deltau_1` = 0.25
`k(u_1) = 8.6875, Deltau = 1/8.6875 = 0.11510`, `u_2 = u_1 + Deltau_2` = 0.3658
Similarly, `u_3`= 0.4514
In the next step, the total f(u) is found out using the above equation.
f1 = 1.578125 f2 = 2.716434 f3 = 3.731435
| f(u) | F | Residual (R) |
| 1.57825 | 1 | 0.57825 |
| 2.716434 | 2 | 0.2716434 |
| 3.731435 | 3 | 0.731435 |
The next step is to consider implicit analysis for the same problem
Implicit Analysis
`u_1`:
`u_0 = 0, k(u_0) = 4, DeltaF =1, Deltau1 = 0.25, u_1 = 0.25`
F = 1, f(u) = 1.578125, Residual (R) = 0.578125 > 10^9
Hence, we correct u1 = `u_1^(0)`,`deltau_1^(0) = -[k(u_1^0)]^(-1) R` = -1/8.6875 x 0.578125
`deltau_1^(0) = -0.066546`
Hence, `u_1 = 0.25 - 0.066546 = 0.1834`
f(u) = 1.0424, F = 1, R = 0.0425 > 0.01
The procedure is repeated to find the new `u_1`which is 0.1776
`(u) = 0.99987, F = 1, R = 0.0001223 < 0.019`
Hence, the new value of `u_1` = 0.1776
`u_2` :
`k(u_1) = 7.29142, DeltaF = 1, Deltau_2 = 0.13714`
`u_2 = u_1 + Deltau_2 = 0.1776 + 0.1374 = 0.315`
f(u) = 2.184, F = 2, R = 0.184 > 0.01
We correct this `u_2` to `u_2^(0)`
`k(u_2^0) = 9.96765, delta(u_2) = -0.0184, u_2 = u_2^(0)+ delta(u_2) = 0.315 - 0.01845 = 0.2965`
f(u) = 2.00239
R = f(u) - F = 0.00239 < 0.01
Hence, we take `u_2` = 0.2965
`u_3` :
`k(u_2) = 5.8486, delta(u_3) = 1/5.8486 = 0.170981`
`u_3 = u_2 + Delta(u_3) = 0.4676`
f(u) = 3.9404, F = 3, R = 0.9404 > 0.01
`u_3 = u_3^(0)`
`k(u_3^0) = 13.0727, delta(u_3) = -1/13.0727 x 0.9404 = -0.0719`
`u_3 = u_3^0 + deltau = 0.3957`
f(u) = 3.0539, R = 0.0539 > 0.01
Carrying this method out again to get the f(u) under tolerance value, the final `u_3` is 0.39105
f(u) = 3.0002, F = 3, R = 0.000280
Hence, `u_3` = 0.39105
| f(u) | F | R |
| 0.9987 | 1 | 0.0001223 |
| 2.00239 | 2 | 0.00239 |
| 3.0002 | 3 | 0.0002 |
Conclusion: As can be seen from the above values, the implicit analysis gives us results nearer to the exact value than explicit analysis.