Week - 2 - Explicit and Implicit Analysis

AIM: Consider the case of a simple bar in tension. Suppose that the force in the bar is a nonlinear function of the displacement.

• FEM is a numerical method that is used to solve differential equation. there are two types of time and displacements will be considered, some cases when the load is applied suddenly like in milliseconds, and in some of the cases load will applied gradually. So these type of cases considered as linear and non-linear case. In linear case where load is applied gradually and in non-linear case load is applied in milliseconds.

EXPLICIT ANALYSIS: Explicit analysis are mainly depends up on the time and displacement variation. when we put a force at the end of the beam. what is the final deformation. If some load is applied at time 't' what happens at time (t+1). To find out this variation we need to differentiate the time and force seperately. To solve a dynamic equilibrium problem when force and acceleration of a system rapidly changes with in a short amount of time we have to use the explicit method, some cases falls under these cases like automotive crash, ballistic events, and hydrodynamic response. Most of the structural element problems are the time independent problems in verey less cases we come across the linear problems where time is dependent.

EXPLANATION:

GIVEN EQUATION : F(u) = u³ + 9u² + 4u

Step 1: u₀ = 0.0 → k (u₀) = 1 → ∆F = 1, → ∆u₁ = ∆F/k (u₀) = 1

→ u₁ = u₀ + ∆u₁ = 1 → u₁ = 1

Step 2: k (u₁) = 1 + 1² = 2 → ∆F = 1, → ∆u₂ = ∆F/k (u₁) = 1/2 = 0.5

→ u₂ = u₁ + ∆u₂ = 1 + 0.5 → u₂ = 1.5

Step 3: k (u₂) = 1 + 1.5² = 3.25 → ∆F = 1, → ∆u₃ = ∆F/k (u₂) = 1/3.25 = 0.307

→ u₃ = u₂ + ∆u₃ = 1.5 + 0.30 → u₃ = 1.8

Step 4: k (u₃) = 1 + 1.8² = 4.24 → ∆F = 1, → ∆u₄ = ∆F/K (u₃) = 1/4.25 = 0.235

→ u₄ = u₃ + ∆u₄ = 1.8 + 0.235 → u₄ = 2.035

Step 5: k (u₄) = 1 + 2.03² = 5.12 → ∆F = 1, → ∆u₅ = ∆F/k (u₄) = 1/5.12 = 0.195

→ u₅ = u₄ + ∆u₅ = 2.03 + 0.19 → u₅ = 2.22

Finally, from the five steps the following results are obtained

F _ext = ∆F + ∆F + ∆F + ∆F + ∆F = 5.0

f _int = F (u) = u³ + 9u² + 4u

= 2.22³ + 9 (2.22)² + 4 (2.22)

= 64.176

Table 1 shows the main results for the explicit analysis

Table 1: summary of explicit analysis result

step	∆Fi	∆ui	ui	(Fext)i	(Fint)i	Fint-Fext =R
1	1.0	1	1	1.0	14	0.013
2	1.0	0.5	1.5	2.0	29.625	0.027
3	1.0	0.307	1.8	3.0	42.192	0.039
4	1.0	0.235	2.03	4.0	53.573	0.049
5	1.0	0.195	2.22	5.0	64.176	0.059

EXPLICIT PLOT:

X = force applied

y = dispalcement occured in bar

IMPLICIT ANALYSIS: In implicit analysis there is a special case where we consider the force to be equilibrium for a instance of time which is also called as quasi-static process . in the quais static-process we consider load and displacement as linear this is a iseal process so to prove this process we use some special itration methods like Raphson-iteration method and there will be some of the tolerance valve for every process where we use the tolerance valve to solve the equation by Raphson-iteration methodIn implicit analysis there is a special case where we consider the force to be equilibrium for a instance of time which is also called as quasi-static process . in the quais static-process we consider load and displacement as linear this is a iseal process so to prove this process we use some special itration methods like Raphson-iteration method and there will be some of the tolerance valve for every process where we use the tolerance valve to solve the equation by Raphson-iteration method.

CONCLUSION: Comparing the both explicit and impicit analysis, in my point of view explicit analysis scheme is a better solver.