Week - 2 - Explicit and Implicit Analysis

EXPLICIT AND IMPLICIT ANALYSIS

EXPLICIT ANALYSIS are generally used for the problem which has dynamic and inertia effect.It is most helpful in solving high deformation time-dependent problem like Crash,Drop,blast,Impact.etc..It is Non-equilibrium that means sum of all Forces=Mass x Acceleration.

IMPLICIT ANALYSIS are used for static problem where no inertia effect in which equilibrium problem where sum of all forces are zero.In this method of analysis all types of Non-linearity can be defined.Implicit approach is useful in problems in which time dependency of the solution is not an important factor.(eg). Static,Vibration,Oscillation,Structural,Harmonic etc..

DIFFERENCE BETWEEN EXPLICIT & IMPLICIT:

IMPLICIT	EXPLICIT
More accurate	Less accurate
Highly time Consuming for Large Models	Computationally efficient for large models with relatively short dynamics response time
Implicit means we cannot discriminate	Explicit means we can discriminate

Explicit doesn't need any iteration scheme.

Implicit need iteration scheme,where we cannot solve directly.

Both Implicit & Explicit schemes are used for solving dynamic events.Implicit scheme is incremental as well as iterative process(ie) it satisfy the equilibrium conditions at every time step and very unconditionally stable due to which we also choose very large time steps,while in Explicit scheme is only incremental process not iterative(ie) it doesn't satisfy equilibrium conditions at every time step which means if any error occurs during the tme step it will keep on magnifying so it is only conditionally stable.

CALCULATION:

GIVEN

F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit method

EXPLICIT METHOD:

F(u) = u³+9u²+4u

For stiffness,

K(u)=df/du=3u²+18u+4

The variables used are u for displacement, f for internal force in the bar, F for external force applied to the bar and k for stiffness. Incremental displacements or incremental externally applied forces are represented as Δu or ΔF, respectively.

From the above, the relationship made is ΔF = kΔu.

STEP 1:

For initial conditions where, ΔF=1

u₀= 0, => k(u₀) =3(0) ²+18(0) +4

k(u₀) = 4

Δu₁= ΔF/k(u0) = 1/4

Δu₁=0.25

u₁=u₀+Δu₁=0+0.25

u₁=0.25

STEP 2:

k(u₁) = 3(0.25) ²+18(0.25) +4

= 8.687

Δu₂= ΔF/k(u₁) = 1/8.687

Δu₂= 0.115

u₂=u₁+Δu₁=0.25+0.115

u₂= 0.365

STEP 3:

k(u₂) = 3(0.365) ²+18(0.365) +4

= 0.399+6.57+4

k(u₂) = 10.969

Δu₃= 1/10.969

Δu₃= 0.091

u₃= u₂+Δu₃= 0.365+0.091

u₃= 0.456

From the above calculation,

F_ext= ΔF + ΔF + ΔF= 1+1+1=3

f₁, f₂, f₃ are F_int (i.e., Internal forces)

f₁=u₁³+9u₁²+4u₁ = 0.25³+9(0.25) ²+4(0.25)

= 0.015+0.562+1

f₁= 1.577

f₂= u₂³+9u₂²+4u₂ = (0.365) ³+9(0.365) ²+4(0.365)

= 0.048+1.199+1.46

f₂= 2.707

f₃= u₃³+9u₃²+4u₃ = (0.456) ³+9(0.456) ²+4(0.456)

= 0.094+1.871+1.824

f₃= 3.789

The table below summarizes the main results for the explicit analysis

STEP	ΔF	ΔU	U	Fext	Fint	Fint- Fext =R
1	1	0.25	0.25	1	1.577	0.577
2	1	0.115	0.365	2	2.707	0.707
3	1	0.091	0.456	3	3.789	0.789

GRAPH PLOT:

IMPLICIT METHOD:

STEP 1:

Given tolerance is 0.01

u₀=0.0 => k(u₀) = 4, ΔF= 1, Δu₁= ΔF/ k(u₀) =0.25,

u₁= u₀ + Δu₁= 0.25

Checking the residual,

F_ext = ΔF = 1

F_int = u₁³+9u₁²+4u₁ = (0.25) ³+9(0.25) ²+4(0.25)

F_int = 1.578

R ⁽⁰⁾ = F_int - F_ext =1.578 -1

R ⁽⁰⁾ = 0.578

R ⁽⁰⁾ > 10^-2 => Newton-Raphson iterations are necessary

Calculate the new total correction to u₁=u₁⁽⁰⁾

δu ⁽¹⁾ = −[k(u₁⁰)]^-1R⁰ = - [3(0.25) ²+18(0.25) +4]^-1 * 0.578

= - (8.687)^-1*0.578 = - (0.115) * 0.578

δu ⁽¹⁾ = -0.06647

updating value u₁¹= u₀ + δu ⁽¹⁾ = 0.1835

Checking again

F_ext = ΔF = 1

F_int = u³+9u²+4u = (0.1835) ³ +9 (0.1835) ² +4(0.1835) = 1.043

R ⁽¹⁾ = F_int - F_ext = 0.043

R ⁽¹⁾ > 10^-2 => so, Newton-Raphson iterations are necessary

Calculate the new total correction to u₁=u₁⁽⁰⁾

δu ⁽²⁾ = δu ⁽¹⁾ − [k(u₁¹)]^-1R¹ = -0.06647 - [3(0.1835) ²+18(0.1835) +4]^-1 * 0.043

δu ⁽²⁾ = -0.07227

u₁²= u₁⁽⁰⁾ + δu ⁽²⁾ = 0.25 -0.07227

u₁² = 0.1777

Checking the residual,

F_ext = ΔF = 1

F_int = u₁³+9u₁²+4u₁ = (0.177) ³+9(0.177) ²+4(0.177)

F_int = 1.0006

R ⁽²⁾ = F_int - F_ext =1.0006 -1

R ⁽²⁾ = 0.0006

R ⁽⁰⁾ < 10^-2 => Newton-Raphson iterations are not necessary

Therefore, the final value is u₁= u₁² = 0.1777

STEP 2:

u₁=0.1777 => k(u₁) = 7.293, ΔF= 1, Δu₂= ΔF/ k(u₁) =0.137,

u₂= u₁ + Δu₂= 0.3148

Checking the residual,

F_ext = ΔF + ΔF = 2

F_int = u₂³+9u₂²+4u₂ = (0.137) ³+9(0.137) ²+4(0.137)

F_int = 2.182

R ⁽⁰⁾ = F_int - F_ext =2.182 - 2

R ⁽⁰⁾ = 0.182

R ⁽⁰⁾ > 10^-2 => Newton-Raphson iterations are necessary

Calculate the new total correction to u₁=u₁⁽⁰⁾

δu ⁽¹⁾ = −[k(u₂⁰)]^-1R⁰ = - [3(0.137) ²+18(0.137) +4]^-1 * 0.182

δu ⁽¹⁾ = -0.0182

updating value u₂¹= u₂⁰ + δu ⁽¹⁾ = 0.3148 -0.0182 = 0.2966

Checking the residual,

F_ext = ΔF+ ΔF = 2

F_int = u³+9u²+4u = 2.004

R ⁽¹⁾ = F_int - F_ext = 2 – 2.004 = 0.004

R ⁽¹⁾ < 10^-2 => Newton-Raphson iterations are not necessary

Therefore, the final value is u₂= u₂¹ = 0.2966

The above procedure has been followed to solve Step 3,

The final value is u₃ = 0.3911

The table below summarizes the main results for the Implicit analysis

STEP	ΔF	ΔU	U	Fext	Fint	Fint- Fext =R
1	1	0.25	0.17	1	1.007	0.0071
2	1	0.137	0.29	2	2.004	0.0042
3	1	0.104	0.39	3	3.000	0.0008

GRAPH PLOT:

COMPARING EXPLICIT AND IMPLICIT:

CONCLUSION:

Hence we have explained about explicit and implicit analysis,calculation has been done separately for both the method.we defines the Graph of Explicit and Implicit method and compared both the result from that we have come to a conclusion that Implict method are accurate but time consuming,Whereas Explicit is not that much accurate.