Solution of a load-displacement non linear relation for static analysis using Explicit and Implicit method in MS Excel using VBA programming

OBJECTIVE:

For a given non linear relation between load and displcement where load is defined as a function of displacement.

F(u) = `u^3` + 9`u^2` + 4u

Solve the above relation for static analysis using Implicit and Explicit scheme.

DESIGN OF EXPERIMENT:

Loading is done in the increements of 1.0 N with initial displacement 0.0 mm

i.e `triangle`F = 1 N

u`circ` = 0.0

THEORY:

In static analysis:-

1. Dead load, i.e load doesnot vary with time for a load step.
2. Equilibrium is achieved in each load step

In explicit analysis:-

1. Stiffnes is calculated for initial deformation.
2. From Stiffness and load step, `triangle`u is calculated.
3. From `triangle`u, deformation for next step is calculated.
4. Internal forces are calculated from expression by substituting calculated u. F(u) = `u^3` + 9`u^2` + 4u
5. Stiffness for next step is calculated from calculated u.
6. Repeat from step 2-6.

In explicit analysis, residual forces exist and equations arenot converged, i.e equilibrium is not achieved at each iteration.

In implicit analysis:-

Implicit anlayis is similar to explicit analysis, but at every step solution is converged by using Newton Raphson method before going to next step.

For a given step, stiffness should be linear for that particular displacement value then only internal forces will be equal to external forces.

In this report, a function is defined using VBA to run iterations in order to achieve convergence at each step.

THEORY:

For solving the non linear static analysis by iterations, equation is:

[F]- [K][u] = [R]

In explicit analysis, Residual need not be 0.

In implicit analysis, residual forces need to be approximated to 0.

For non linear analysis, Stiffness is a function of displcement.

F(u) = `u^3` + 9`u^2` + 4u

K(u) = d F(u)/du = 3`u^2`+ 18u + 4

Stiffness for Explicit and Implicit:

1. In explicit analysis, Stiffness is calculated for the calculated u.
2. In implicit analysis, Stiffness is calculated till stiffnes is linear for the given step. This is done by a convergence method such as Newton Raphson method or modified newton raphson method.

SOLUTION:

EXPLICIT:

LOAD STEP 1:

u`circ` = 0

`triangle`F = 1

K(u) = 3`u^2`+ 18u + 4 = 4

`triangle`u = `triangle`F/K(u) = 1/4 = 0.25

u1 - uo = 0.25

i.e u1 = 0.25

Internal force, l1 = `u^3` + 9`u^2` + 4u

= `0.25^3` + 9`times``0.25^2` + 4`times`0.25

= 1.578125 N

Reaction force, R1 = L1-F1

=1.578125 - 1.000000

=0.578125 N

LOAD STEP 2:

u`circ` = 0.25

`triangle`F = 1

K(u) = 3`u^2`+ 18u + 4

= 3`times``0.25^2` + 18`times`0.25 +4

= 8.6875

`triangle`u = `triangle`F/K(u) = 1/8.6875 = 0.115

u2 - u1 = 0.115

i.e u2 = 0.115 + 0.25

= 0.365

Internal force, l2 = `u^3` + 9`u^2` + 4u

= `0.365^3` + 9`times``0.365^2` + 4`times`0.365

= 2.707 N

Reaction force, R2 = L1-F1

=2.707652 - 2.000000

=0.707652 N

LOAD STEP 3:

u`circ` = 0.365

`triangle`F = 1

K(u) = 3`u^2`+ 18u + 4

= 3`times``0.365^2` + 18`times`0.365 +4

= 10.969675

`triangle`u = `triangle`F/K(u) = 1/10.969675 = 0.091

u3 - u2 = 0.091160

i.e u3 = 0.091160 + 0.365

= 0.456160

Internal force, l3 = `u^3` + 9`u^2` + 4u

= `0.456^3` + 9`times``0.456^2` + 4`times`0.456

= 3.790242 N

Reaction force, R3 = L1-F1

=3.790242 - 3.000000

=0.790242 N

Further iterations are carried out in excel:

IMPLICIT:

LOAD STEP 1:

u`circ` = 0

`triangle`F = 1

K(u) = 3`u^2`+ 18u + 4 = 4

`triangle`u = `triangle`F/K(u) = 1/4 = 0.25

u1 - uo = 0.25

i.e u1 = 0.25

Internal force, l1 = `u^3` + 9`u^2` + 4u

= `0.25^3` + 9`times``0.25^2` + 4`times`0.25

= 1.578125 N

Reaction force, R1 = L1-F1

=1.578125 - 1.000000

=0.578125 N

Since R1>0.01 hence we need to converge.

Using Newton Raphson method:

ITERATION1:

K(0.25) = 8.6875

dU1 = -`[k(u1)]^-1` `times` R1 = -`(8.6875)^-1` `times` 0.578125 = -0.066546

U1 = u1 + dU1 = 0.25 + (-0.66546) = 0.183453

Internal force, L1 = `U1^3` + 9`U1^2` + 4U1

= `0.183^3` + 9`times``0.183^2` + 4`times`0.183

= 1.039529 N

Reaction force, R1 = L1-F1

=1.039529 - 1.000000

=0.039529 N

Since R1>0.01 hence we need to converge.

ITERATION2:

K(0.183453) = 7.394467

dU2 = -`[k(U2)]^-1``times`R2 = -`7.394467^-1` `times` 0.039529 = -0.00534575

U2 = U1 + dU2 = 0.183453 + -0.00534575 = 0.178798

Internal force, L2 = `U2^3` + 9`U2^2` + 4U2

= `0.178^3` + 9`times``0.178^2` + 4`times`0.178

= 1.002795 N

Reaction force, R2 = L2-F1

=1.002795 - 1.000000

=0.002795 N

Since R2<0.01 hence, no further iterations are required.

Also, K(0.183453) = 7.394467 is the linear stiffness at that point.

Corrsponding displacement for which solution get converged is u1 = U2 = 0.178798

LOAD STEP 2:

u1 = 0.178798

`triangle`F = 1

K(u) = 3`u^2`+ 18u + 4 = 7.299052

`triangle`u = `triangle`F/K(u) = 1/7.299052 = 0.137004

u2 - u1 = 0.137004

i.e u2 = 0.315802

Internal force, l2 = `u^3` + 9`u^2` + 4u

= `0.315^3` + 9`times``0.315^2` + 4`times`0.315

= 2.184280 N

Reaction force, R2 = L2-F2

=2.184280 - 2.000000

=0.184280 N

Since R1>0.01 hence we need to converge.

Using Newton Raphson method:

ITERATION1:

K(0.315) = 9.967675

dU1 = -`[k(u1)]^-1` `times` R1 = -`(9.967675)^-1` `times` 0.184280 = -0.018487

U1 = u1 + dU1 = 0.315802 + (-0.018487) = 0.297315

Internal force, L1 = `U1^3` + 9`U1^2` + 4U1

= `0.297^3` + 9`times``0.297^2` + 4`times`0.297

= 2.00807 N

Reaction force, R1 = L1-F1

=2.00807 - 2.000000

=0.00807 N

Since R1<0.01 hence we need not to converge.

Also, K(0.315802) = 9.967675 is the linear stiffness at that point.

Corrsponding displacement for which solution get converged is u1 = U1 = 0.297315

LOAD STEP 3:

u1 = 0.297315

`triangle`F = 1

K(u) = 3`u^2`+ 18u + 4 = 9.610627

`triangle`u = `triangle`F/K(u) = 1/9.610627 = 0.104051

u2 - u1 = 0.104051

i.e u2 = 0.401366

Internal force, l2 = `u^3` + 9`u^2` + 4u

= `0.401^3` + 9`times``0.401^2` + 4`times`0.401

= 3.115690 N

Reaction force, R2 = L2-F2

=3.115690 - 3.000000

=0.115690 N

Since R1>0.01 hence we need to converge.

Using Newton Raphson method:

ITERATION1:

K(0.401366) = 11.700403

dU1 = -`[k(u1)]^-1` `times` R1 = -`(11.700403)^-1` `times` 0.115690 = -0.009887

U1 = u1 + dU1 = 0.401366 + (-0.009887) = 0.391478

Internal force, L1 = `U1^3` + 9`U1^2` + 4U1

= `0.391^3` + 9`times``0.391^2` + 4`times`0.391

= 2.999705 N

Reaction force, R1 = L1-F1

=2.999705 - 3.000000

=-0.000295 N

Since R1<0.01 hence we need not to converge.

Also, K(0.401366) = 11.700403 is the linear stiffness at that point.

Corrsponding displacement for which solution get converged is u1 = U1 = 0.391478

Further iterations are carried out using VBA script in excell where Newton raphson method is defined as a function.

Parameters to this function are u from previous step and Net external force for the load step.

Here, a function "Iteration" is defined. It runs the Newton Raphson iteration using while loop until unless residue is less than 0.01.

For each load step, u from last step and net external load is input to get the K(u) and u for which the load step gets converged.

Here, parametrs of function Iteration are u and F

i.e Iteration(u,f) gives K(u) and u for which the solution converges.

EXACT:

From given relation, F(u) = u3 + 9u2 + 4u. we can plot the cureve by sustituting values of u in intervals of 0.1

VISUALISATION AND DISCUSSION:

Load vs deformation curves

The visualisation of load vs deformation curve for three iteration for each case

It is evident from above that Implicit and Exact solutions overlap eachother. However there is deviation associated with explicit solution from exact solution.

The visualisation of load vs deformation curve for three iteration for each case

The solutions almost identical, however still the deviation of explicit from exact in initial iterations.

Residue in each load step

It is evident from above plot that:

• Implicit and exact solutions coincide, thus for static non lienear load step control method, implicit ,method gives better results. It is because in static loading, member attains equilibrium for each load step. Reaction forces are minimized for each load increement and are made 0.
• Explicit solution deviated from exact and gives greater values of u as reaction forces are present within this model.

CONCLUSIONS:

Q. When to use implicit and explicit?

• Whenver there is deformation due to load steps where a successive load step is added only after memeber achieves equilibrium during previous load step stage, in such cases implicit where as in load addition where system doesnot achieve equilibrium in between load step, we need to use explicit.
• In other words, we need to use explicit when transient loading is time dependent, Example: crash of vehicle and when transient loading is gradual such that time can be neglected(quasi static) we need to use implicit method. Example: sheet metal forming
• When stiffnes is linear for every load increement - implicit and when stiffnes isnot linear for a increment of load then explicit.
• For nonlinear transient dynamic loading - explicit.
• For nonlinear transient static loading - implict.

Q. How to decide load step increement value?

Mainly by deducing, minimum load applied and maximum load applied and load steps. More number of steps in between min and max value gives more closer results to exact solution.

SPREADSHEET WITH CALCULATIONS IS ATTACHED.

Week 2.xlsm