Week - 2 - Explicit and Implicit Analysis

Aim: F(u) = u3+9u2+4u use this equation and solve using both Explicit and Implicit Methods (tolerance of 0.01).

Explicit Analysis:

The explicit analysis is preferred for the following reasons:

• In Explicit analysis no iterations are required.
• It is conditionally stable, Courant-Friedrich-Levy (CFL) criteria.
• It requires low memory & smaller time steps.
• Cost per timestep is low.
• It will calculate the state of system at different time by considering the current time, which results in finding out the solution faster.
• Each increment is founded fast, this results in less computation.
• Explicit method is used for large strain deformation problems, which includes the applications such as: Automotive Crash analysis, Drop test & few dynamic analysis.
• One of the con in explicit analysis is: More increments are needed to match up with the exact solution & it has Less accuracy in the obtained results
• This method is used where the events to be performed are more.

Given equation is F(u)=u³+9u²+4u

Stiffness of the bar is df/du=k(u)=3u²+18u+4

However, if the nonlinear force versus displacement relationship is not known, but the stiffness is known, it is possible to construct the graph of force versus displacement numerically. One method is to use an incremental explicit load control scheme.

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. Use is made of the relationship ΔF = k Δu.

Rearrange the above formula, we get Δu = ΔF/k

Step 1:

u0 = 0, ⇒ k(u0) = 4, ΔF = 1, Δu1 = ΔF/k(u0) = 1/4 = 0.25,

u1 = u0 + Δu1 = 0+0.25=0.25

Hence u₁₌0.25

f_ext =1

f_int = (0.25)³+9(0.25)²+4(0.25) =1.579

Step 2:

u₁₌0.25⇒ k(u1) = 3(0.25)² + 18(0.25)+4 = 8.69, ΔF = 1, Δu2 = ΔF/k(u1) = 1/8.69 =0.115

u2 = u1 + Δu2 = 0.25+0.115 =0.365

Hence u₂=0.365

f_ext =1

f_int = (0.365)³+9(0.365)²+4(0.365) =2.708

Step 3:

u₂=0.365⇒ k(u2) = 3(0.365)²+18(0.365)+4 = 10.72,ΔF = 1, Δu3 = ΔF/k(u2) = 1/10.72 =0.093, u3 = u2+Δu3 =0.365+0.093 =0.458

Hence u₃=0.458

f_ext =1

f_int = (0.458)³+9(0.458)²+4(0.458) =3.808

Finally, from the three steps the following results are obtained.

Fext = ΔF + ΔF + ΔF = 3.0

Explicit Analysis Results

Implicit Analysis:

The implicit analysis is preferred for the following reasons:

• In Implicit analysis iterations are required. & Newton Raphson method is used for the iterations.
• It requires high memory.
• Time is high in implicit analysis.
• Cost per timestep is high.
• Implicit analysis is mainly used for small strain deformation problems.
• In implicit analysis, Acceleration is assumed to vary linearly over the time step(Which could be either positive(or)Negative).
• It will calculate the solution by considering the current & further states of the system, Which helps in attaining the equilibrium between the external & internal forces before moving in to
• More accuracy in the obtained results when compared to the solution obtained by explicit method.
• This method is used where the events to be performed are less.

an incremental force is applied to advance the solution forward at the beginning

of a step. However, internal forces and external forces will not be in equilibrium unless the

stiffness is linear for the given step. Hence, in order to achieve equilibrium, corrections must

be made to the displacement. This is accomplished by using Newton-Raphson iterations to

minimize the residual, R(u) = fint − Fext.

δ u^j+1 = −[k(u^j)]⁻¹R_j .

Given: F(u)=u³+9u²+4u

Stiffness of the bar is df/du=k(u)=3u²+18u+4

Step 1 :

u₀=0 ⇒k(u₀ ) = 3(0)² + 18(0) + 4 = 4, ΔF = 1, Δu1 = ΔF/k(u0) = 1/4 = 0.25,

u1 = u0 + Δu1 = 0+0.25=0.25

Hence u₁₌0.25

Check for Residual; f_ext =1; f_int = (0.25)³+9(0.25)²+4(0.25) =1.579;

R = f_int - f_ext=1.578 – 1 =0.578

R(0) >0.01 Hence Newton-Raphson iterations are necessary.

Iteration 1:

Calculating the new total correction to u_{1 =} u₁⁽⁰⁾

δ u ⁽¹⁾= −[k(u₁⁰ )] ^-1. R = - (8.6875) (0.578) = - 0.0665

Where, K(u₁⁰ ) = 3(0.25)²+18(0.25)+4 = 8.6875

Therefore update value of u is, u₁⁽¹⁾ = u₁⁰ + δ u⁽¹⁾ = 0.25 - 0.0665 ;u = 0.1834

Check the residual again; f_ext =1; f_int = (0.1834)³+9(0.1834)²+4(0.1834) =1.0432;

R = f_int - f_ext=1.0432 – 1 =0.0432 >0.01; Hence Newton-Raphson iterations are necessary

Iteration 2:

Calculating the new total correction to u_{1 =} u₁⁽¹⁾

δ u ⁽²⁾= δ u 1−[k(u₁¹ )] ^-1. R =- 0.0665 - (7.402) (0.0432) = - 0.0723

Where, K(u₁¹ ) = 3(0.1834)²+18(0.1834)+4 = 7.402

Therefore update value of u is, u₁⁽²⁾ = u₁⁰ + δ u⁽²⁾ = 0.25 - 0.0723 ;u₁= 0.1776

Check the residual again; f_ext =1; f_int = (0.1776)³+9(0.1776)²+4(0.1776) =1.0006;

R = f_int - f_ext=1.0006 – 1 =0.0006 <0.01:

Hence Newton-Raphson iterations are not necessary

Step 2 :

u₁= 0.1776

u₁=0.1776 ⇒k(u₀ ) = 3(0.1776)² + 18(0.1776) + 4 = 7.291, ΔF = 1, Δu1 = ΔF/k(u0) = 1/7.291 = 0.137,

u2 = u1 + Δu2 = 0.1776+0.137=0.3147

Hence u₂₌0.3147

Check for Residual; f_ext =2; f_int = (0.3147)³+9(0.3147)²+4(0.3147) =2.181;

R = f_int - f_ext=2.181 – 2 =0.181

R(0) >0.01 Hence Newton-Raphson iterations are necessary.

Iteration 1:

Calculating the new total correction to u_{2 =} u₂⁽⁰⁾

δ u ⁽¹⁾= −[k(u₂⁰ )] ^-1. R = - (9.9617)^-1 (0.181) = - 0.0182

Where, K(u₁⁰ ) = 3(0.3147)²+18(0.3147)+4 = 9.9617

Therefore update value of u is, u₁⁽¹⁾ = u₁⁰ + δ u⁽¹⁾ = 0.3147 - 0.0182;u₂ = 0.2962

Check the residual again; f_ext =2; f_int = (0.2962)³+9(0.2962)²+4(0.2962) =2.00039;

R = f_int - f_ext=2.00039 – 2 =0.00039 <0.01; Hence Newton-Raphson iterations are not necessary

Step 3 :

U₂= 0.2962

U₂=0.2962 ⇒k(u₁ ) = 3(0.2962)² + 18(0.2962) + 4 = 9.6007, ΔF = 1, Δu1 = ΔF/k(u0) = 1/9.6007 = 0.1041,

U3 = u2+ Δu3= 0.2962+0.1041=0.4006

Hence u₃₌0.4006

Check for Residual; f_ext =3; f_int = (0.4006)³+9(0.4006)²+4(0.4006) =3.116;

R = f_int - f_ext=3.116 – 3 =0.116

R(0) >0.01 Hence Newton-Raphson iterations are necessary.

Iteration 1:

Calculating the new total correction to u_{3 =} u₃⁽⁰⁾

δ u ⁽¹⁾= −[k(u₃⁰ )] ^-1. R = - (11.683)^-1 (0.116) = - 0.0092

Where, K(u₁⁰ ) = 3(0.4006)²+18(0.4006)+4 = 11.683

Therefore update value of u is, u₃⁽¹⁾ = u₃⁰ + δ u⁽¹⁾ = 0.4006- 0.0092;u₃ = 0.3911

Check the residual again; f_ext =3; f_int = (0.3911)³+9(0.3911)²+4(0.3911) =3.0008;

R = f_int - f_ext=3.0008 – 3 =0.0008 <0.01; Hence Newton-Raphson iterations are not necessary

Summary of Implicit Analysis Results

Exact Result:-Using Calculator

THIRD ORDER POLYNOMIAL OF NON LINEAR DISPLACEMENT EQUATION

Take F(u)= u³+9u²+4u ;

u³+9u²+4u -F(u) = 0

F(u₁) = 1 ; u³+9u²+4u -1 = 0 equation is solved by using scientific calculator with polynomial degree 3 and the value of u₁ =0.1176

F(u₂ ) = 2 ; u³+9u²+4u -2 = 0 this equation is solved by using scientific calculator with polynomial degree 3 and the value of u₂ =0.2961

F(u₃ ) = 3 ; u³+9u²+4u -3= 0 this equation is solved by using scientific calculator with polynomial degree 3 and the value of u₃ =0.3910

Comparison Table of Explicit, Implicit and Impact analysis for Force vs Displacement

Conclusion:

1.It is evident that for the given polynomial equation among the three methods like Explicit analysis, Implicit analysis and Exact analysis, Implicit analysis results match with Exact analysis as shown in above figure. The equation F(u) = u3+9u2+4u is solved using both methods (with tolerance of 0.01). It as per the results to get the accurate result Implicit method is best suited to solve this.

2.This is due to Newton Raphson method because it reduces the residues in implicit analysis whereas explicit analysis has no such iterations.

3.But Implicit analysis which is not suitable for Non-linear transient dynamic problem because the curve extends still linear where in that case which is not in real time scenario at that time explicit analysis play a vital role.

Examples:

Implicit analysis Software’s - Abaqus -Strength and durability

Explicit analysis Software’s - Ls Dyna - Crashworthiness Behaviour