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PhD_Wallyson

Mr. WALLYSON SILVA

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5 years, 20 days
University of Miskolc
Mechanical Engineering

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I am a Brazilian PhD student in Mechanical Engineering at the University of Miskolc. My research is about: - The vibration of machine tools - Particle Impact Damper in boring bars to internal turning operation - turning hardened materials - passive damping in turning operation - Chatter in Machining process - Mathematical modelling of particle damping systems - Mathematical modelling of boring bar fixation

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These are replies submitted by PhD_Wallyson

Eigenvalues_3_span_beam...

PhD_Wallyson 35
July 13 2020
0 0

@Rouben Rostamian  

Hi Professor

Just to put you in the context: during an internal turning operation, the overhang (ratio Length/Diameter of the tool [L/D]) is really important to guarantee the stability of the process (minimal vibration as possible). Having said that, it is desirable to increases the overhang to do deep holes, because of this the ratio L/D varies depending on the necessity and consequently the natural frequency of the tool will change.

As you can see in the attached Picture_A and B, I am trying to find the Eigenvalues when the overhang (ratio L/D) changes.

Consider:

L1 = ratio L/D (changeable);
L2 = Fixation of the tool (content)
L3 = the remaining part of the tool out of the fixation (changeable)

Tool length is content = L1 + L2 + L3

13_07_2020_Eigenvalues_3_span_beam.mw13_07_2020_Eigenvalues_3_span_beam.mw

 

 

Calculate the natural frequency...

PhD_Wallyson 35
July 13 2020
0 0

@Rouben Rostamian  

Hi Professor, 

Considering the same reference used before (attached):

I am trying to calculate the frequencies demonstrated on page 69 (Table 3.12), using the parameters of Table 2.1 and 2.2, also the dimensions of the beam in Figure 2.12.

As you can see in the program attached when I change lengths (L1, L2 and L3) to meters the mode shape (mode 3) is deformed, even though the natural frequencies is correct, according to the Table 3.12. 

How can I correct this deformation in the mode shape when I consider the length in meters?

 

https://www.diva-portal.org/smash/get/diva2:837061/FULLTEXT01.pdf


 

Modal analysis of a 3-span beam

Here we formulate and solve for the modal shapes of transverse vibration of a 3-span

free-pinned-pinned-free beam.  The graphs of the first three modes agree with those

in Figure 3.22 on page 70 of the 2007 article of
Henrik Åkesson, Tatiana Smirnova, Thomas Lagö, and Lars Håkansson.

 

Note: Their boundary conditions, given in (5.1) on page 88, don't look correct to me,

so here I do it my way.

 

Rouben Rostamian

2020-07-08

> 

restart;
> 

#Digits := 15:
> 

with(LinearAlgebra):
> 

pde := rho*A*diff(u(x,t),t,t) + EI*diff(u(x,t),x$4) = 0;

rho*A*(diff(diff(u(x, t), t), t))+EI*(diff(diff(diff(diff(u(x, t), x), x), x), x)) = 0

 (1)

To find the fundamental modes, look for a time-harmonic solution:

> 

eval(pde, u(x,t)=X(x)*cos(omega*t));
ode_tmp := isolate(%, diff(X(x),x$4));

-rho*A*X(x)*omega^2*cos(omega*t)+EI*(diff(diff(diff(diff(X(x), x), x), x), x))*cos(omega*t) = 0

  

diff(diff(diff(diff(X(x), x), x), x), x) = rho*A*X(x)*omega^2/EI

 (2)

So our task has been reduced to solving the ODE above.  The coefficient on the right-hand side is

unnecessarily messy, so we name it mu^4.  In the subsequence calculations we will find mu.  Then we

can find "omega," if needed, from the relationship below:

> 

coeff(rhs(ode_tmp), X(x)) = mu^4;
omega_vs_mu := isolate(%, omega^2);

rho*A*omega^2/EI = mu^4

  

omega^2 = mu^4*EI/(rho*A)

 (3)
> 

ode := eval(ode_tmp, omega_vs_mu);

diff(diff(diff(diff(X(x), x), x), x), x) = X(x)*mu^4

 (4)

Solve the ODE:

> 

dsolve(ode);

X(x) = _C1*exp(-mu*x)+_C2*exp(mu*x)+_C3*sin(mu*x)+_C4*cos(mu*x)

 (5)

Maple presents the solution in terms of exponential functions.  We prefer hyperbolic

functions instead. We let X__1(x), X__2(x), X__3(x) be the beam's deflection in each of the

three spans.

> 

X[1] := x -> a[1]*sin(mu*x) + b[1]*cos(mu*x) + c[1]*sinh(mu*x) + d[1]*cosh(mu*x);
X[2] := x -> a[2]*sin(mu*x) + b[2]*cos(mu*x) + c[2]*sinh(mu*x) + d[2]*cosh(mu*x);
X[3] := x -> a[3]*sin(mu*x) + b[3]*cos(mu*x) + c[3]*sinh(mu*x) + d[3]*cosh(mu*x);

proc (x) options operator, arrow; a[1]*sin(mu*x)+b[1]*cos(mu*x)+c[1]*sinh(mu*x)+d[1]*cosh(mu*x) end proc

  

proc (x) options operator, arrow; a[2]*sin(mu*x)+b[2]*cos(mu*x)+c[2]*sinh(mu*x)+d[2]*cosh(mu*x) end proc

  

proc (x) options operator, arrow; a[3]*sin(mu*x)+b[3]*cos(mu*x)+c[3]*sinh(mu*x)+d[3]*cosh(mu*x) end proc

 (6)

so here is the overall deflection:

> 

modal_shape := piecewise(x<L[1], X[1](x), x<L[1]+L[2], X[2](x), X[3](x));

modal_shape := piecewise(x < L[1], a[1]*sin(mu*x)+b[1]*cos(mu*x)+c[1]*sinh(mu*x)+d[1]*cosh(mu*x), x < L[1]+L[2], a[2]*sin(mu*x)+b[2]*cos(mu*x)+c[2]*sinh(mu*x)+d[2]*cosh(mu*x), a[3]*sin(mu*x)+b[3]*cos(mu*x)+c[3]*sinh(mu*x)+d[3]*cosh(mu*x))

 (7)

We have three spans, there are two boundary conditions at the end of each span,

and therefore we have 12 boundary conditions altogether.  These are almost,

but not quite, the same as those given on page 88 of the paper cited above.

> 

bcs :=
(D@@2)(X[1])(0) = 0,   # zero moment at x=0
(D@@3)(X[1])(0) = 0,   # zero shear  at x=0

X[1](L[1]) = 0,        # zero displacement at x=L[1]
X[2](L[1]) = 0,        # zero displacement at x=L[1]
D(X[1])(L[1]) = D(X[2])(L[1]),            # continuous slope at x=L[1]
(D@@2)(X[1])(L[1]) = (D@@2)(X[2])(L[1]),  # no external moment at x=L[1]

X[2](L[1]+L[2]) = 0,        # zero displacement at x=L[1]+L[2]
X[3](L[1]+L[2]) = 0,        # zero displacement at x=L[1]+L[2]
D(X[2])(L[1]+L[2]) = D(X[3])(L[1]+L[2]),            # continuous slope at x=L[1]+L[2]
(D@@2)(X[2])(L[1]+L[2]) = (D@@2)(X[3])(L[1]+L[2]),  # no external moment at x=L[1]+L[2]

(D@@2)(X[3])(L[1]+L[2]+L[3]) = 0,   # zero moment at x=L[1]+L[2]+L[3]
(D@@3)(X[3])(L[1]+L[2]+L[3]) = 0:   # zero shear  at x=L[1]+L[2]+L[3]

The 12 conditions given above constitute a system of 12 linear, homogeneous equations in the

12 variables "`a__1`,`b__1`. ..., `c__3`, `d__3`."  Here we construct the coefficient M of the system's matrix.  The
system's right-hand side, which is a vector of zeros, is captured in the vector MM which is not

used in the subsequent calculations.

> 

M, MM := LinearAlgebra:-GenerateMatrix({bcs},
  [ a[1], b[1], c[1], d[1], a[2], b[2], c[2], d[2], a[3], b[3], c[3], d[3] ]):

Here is the 12x12 matrix M:

> 

M;

_rtable[18446744074701413662]

 (8)

We set the determinant of M to zero in order to get a nontrivial solution to our homogeneous system.

That yields a transcendental equation for the unknown mu.

> 

Determinant(M):
expand(%):
simplify(%):
characteristic_equation := %/(8*mu^16);  # get rid of the unnecessary factor of 8*mu^16

(((-2*sinh(mu*L[2])*sin(mu*L[2])*cos(mu*L[3])+sin(mu*L[3])*(cosh(mu*L[2])*sin(mu*L[2])-sinh(mu*L[2])*cos(mu*L[2])))*cos(mu*L[1])+sin(mu*L[1])*((cosh(mu*L[2])*sin(mu*L[2])-sinh(mu*L[2])*cos(mu*L[2]))*cos(mu*L[3])+sin(mu*L[3])*(cosh(mu*L[2])*cos(mu*L[2])-1)))*cosh(mu*L[3])+(-sinh(mu*L[3])*(cosh(mu*L[2])*sin(mu*L[2])-sinh(mu*L[2])*cos(mu*L[2]))*cos(mu*L[3])-2*sinh(mu*L[2])*sin(mu*L[2]))*cos(mu*L[1])-(sinh(mu*L[3])*(cosh(mu*L[2])*cos(mu*L[2])-1)*cos(mu*L[3])-cosh(mu*L[2])*sin(mu*L[2])+sinh(mu*L[2])*cos(mu*L[2]))*sin(mu*L[1]))*cosh(mu*L[1])+(-sinh(mu*L[1])*((cosh(mu*L[2])*sin(mu*L[2])-sinh(mu*L[2])*cos(mu*L[2]))*cos(mu*L[3])+sin(mu*L[3])*(cosh(mu*L[2])*cos(mu*L[2])-1))*cos(mu*L[1])-2*sinh(mu*L[2])*sin(mu*L[2])*cos(mu*L[3])+sin(mu*L[3])*(cosh(mu*L[2])*sin(mu*L[2])-sinh(mu*L[2])*cos(mu*L[2])))*cosh(mu*L[3])+(sinh(mu*L[3])*(cosh(mu*L[2])*cos(mu*L[2])-1)*cos(mu*L[3])-cosh(mu*L[2])*sin(mu*L[2])+sinh(mu*L[2])*cos(mu*L[2]))*sinh(mu*L[1])*cos(mu*L[1])-sinh(mu*L[3])*(cosh(mu*L[2])*sin(mu*L[2])-sinh(mu*L[2])*cos(mu*L[2]))*cos(mu*L[3])-2*sinh(mu*L[2])*sin(mu*L[2])

 (9)

We are ready now for some number crunching.  For the span lengths we take the values

given in the caption of Figure 2.12 on page 28.

> 

span_lengths := L[1]=0.035, L[2]=0.050, L[3]=0.215;

L[1] = 0.35e-1, L[2] = 0.50e-1, L[3] = .215

 (10)

Plug the span lengths into the characteristic equation

> 

characteristic_equation_numeric := eval(characteristic_equation, {span_lengths});

(((-2*sinh(0.50e-1*mu)*sin(0.50e-1*mu)*cos(.215*mu)+sin(.215*mu)*(cosh(0.50e-1*mu)*sin(0.50e-1*mu)-sinh(0.50e-1*mu)*cos(0.50e-1*mu)))*cos(0.35e-1*mu)+sin(0.35e-1*mu)*((cosh(0.50e-1*mu)*sin(0.50e-1*mu)-sinh(0.50e-1*mu)*cos(0.50e-1*mu))*cos(.215*mu)+sin(.215*mu)*(cosh(0.50e-1*mu)*cos(0.50e-1*mu)-1)))*cosh(.215*mu)+(-sinh(.215*mu)*(cosh(0.50e-1*mu)*sin(0.50e-1*mu)-sinh(0.50e-1*mu)*cos(0.50e-1*mu))*cos(.215*mu)-2*sinh(0.50e-1*mu)*sin(0.50e-1*mu))*cos(0.35e-1*mu)-(sinh(.215*mu)*(cosh(0.50e-1*mu)*cos(0.50e-1*mu)-1)*cos(.215*mu)-cosh(0.50e-1*mu)*sin(0.50e-1*mu)+sinh(0.50e-1*mu)*cos(0.50e-1*mu))*sin(0.35e-1*mu))*cosh(0.35e-1*mu)+(-sinh(0.35e-1*mu)*((cosh(0.50e-1*mu)*sin(0.50e-1*mu)-sinh(0.50e-1*mu)*cos(0.50e-1*mu))*cos(.215*mu)+sin(.215*mu)*(cosh(0.50e-1*mu)*cos(0.50e-1*mu)-1))*cos(0.35e-1*mu)-2*sinh(0.50e-1*mu)*sin(0.50e-1*mu)*cos(.215*mu)+sin(.215*mu)*(cosh(0.50e-1*mu)*sin(0.50e-1*mu)-sinh(0.50e-1*mu)*cos(0.50e-1*mu)))*cosh(.215*mu)+(sinh(.215*mu)*(cosh(0.50e-1*mu)*cos(0.50e-1*mu)-1)*cos(.215*mu)-cosh(0.50e-1*mu)*sin(0.50e-1*mu)+sinh(0.50e-1*mu)*cos(0.50e-1*mu))*sinh(0.35e-1*mu)*cos(0.35e-1*mu)-sinh(.215*mu)*(cosh(0.50e-1*mu)*sin(0.50e-1*mu)-sinh(0.50e-1*mu)*cos(0.50e-1*mu))*cos(.215*mu)-2*sinh(0.50e-1*mu)*sin(0.50e-1*mu)

 (11)
> 

plot(characteristic_equation_numeric, mu=0..20*Pi);

 

It is difficult to see the roots in the graph above.  Therefore we limit the vertical range:

> 

plot(characteristic_equation_numeric, mu=0..20*Pi, view=-4..4);

 
> 

fsolve(characteristic_equation_numeric, mu = 0 .. 20*Pi, maxsols = 20)

0., 8.148236435, 20.65743153, 34.65175842, 41.30255339, 49.46090966

 (12)

We proceed to find the first few roots.

 

Important!  The roots depends on the choice of the span lengths "`L__1`, `L__2`, `L__3`."

If these are changed, then the initial guesses in the calculations below should be
re-adjusted according to what is seen in the graph above:

> 

mu__1 := fsolve(characteristic_equation_numeric, mu=8.148236435);
mu__2 := fsolve(characteristic_equation_numeric, mu=20.65743153);
mu__3 := fsolve(characteristic_equation_numeric, mu=34.65175842);
mu__4 := fsolve(characteristic_equation_numeric, mu=41.30255339);

8.148236435

  

20.65743153

  

34.65175842

  

41.30255340

 (13)

We write a proc that calculates and plots the modal shape for any of the mu values

calculated above.  Note that the shapes don't depend on the physical properties

of the beam, that is, the coefficients rho, E, I, A of the original PDE are immaterial.

 

If the mu values shown above were calculated perfectly, then the determinant of M

would be exactly zero.  In reality, the values of mu are only approximations, and

therefore the determinant of M is not exactly zero and depending on the values

of the span lengths, it may be quite far from zero.  The proc prints out the value

of the determinant.  If it is not small (like less that 10^(-5)),  then increase the value

of Digits and try again.

> 

plot_mode := proc(mu)
  local coeff_vals, total_length, my_M, my_mode_shape;
  total_length := add(rhs~({span_lengths}));
  my_M := eval(M, {span_lengths, :-mu=mu});
  printf("Determinant is %g.  Increase Digists if not near zero!\n", Determinant(my_M));
  NullSpace(my_M)[1]:
  <a[1], b[1], c[1], d[1], a[2], b[2], c[2], d[2], a[3], b[3], c[3], d[3]> =~ %:
  coeff_vals := convert(%, set)[];
  #print(coeff_vals);
  my_mode_shape := eval(modal_shape, {:-mu=mu, coeff_vals, span_lengths}):
  eval(my_mode_shape, x=total_length);
  # the (-%) below normalizes the graph so that its value is -1 at the right end
  plot(my_mode_shape/(-%), x=0..total_length);
end proc:
> 

p1 := plot_mode(mu__1);

Determinant is -1.88827e+006.  Increase Digists if not near zero!

  

 
> 

p2 := plot_mode(mu__2);

Determinant is -6.39035e+017.  Increase Digists if not near zero!

  

 
> 

p3 := plot_mode(mu__3);

Determinant is -2.99856e+025.  Increase Digists if not near zero!

  

 

Here are the first three modal plotted together.  This agrees with Figure 3.22(a) in the cited article.

> 

plots:-display([p1,p2,p3], color=["Red","Green","Blue"],
   legend=[mode1, mode2, mode3], axes=boxed);

 
> 

``
> 

``
> 

E := 2.05*10^11

0.2050000000e12

 (14)
> 

rho := 7860

7860

 (15)
> 

I_x := 1.1385*10^(-7)

0.1138500000e-6

 (16)
> 

A_x := 1.1933*10^(-3)

0.1193300000e-2

 (17)

 

``

> 

CM := sqrt(E*I_x/(rho*A_x))

49.88355021

 (18)
> 

w_1 := mu__1^2*CM

3311.956311

 (19)
> 

w_2 := mu__2^2*CM

21286.78131

 (20)
> 

w_3 := mu__3^2*CM

59897.39167

 (21)
> 

f_1 := w_1/(2*Pi)

527.1142180

 (22)
> 

f_2 := w_2/(2*Pi)

3387.896468

 (23)
> 

f_3 := w_3/(2*Pi)

9532.965960

 (24)
> 

``
> 

 


 

Download 9_07_2020_Mode_shapes_matrix_12x12_Figure_3.22_artigo_2_v2.mw

9_07_2020_Mode_shapes_matrix_12x12_Figure_3.22_artigo_2_v2.mw

@vv you are right! it works now hah...

PhD_Wallyson 35
July 09 2020
0 0

@vv you are right! it works now hahahaha
 

Thanks...

PhD_Wallyson 35
July 08 2020
0 0

@tomleslie 
Hi 

Really thanks for your answer. I am a beginner PhD student who should learn too much.
I have learnt many lessons from you after reading your feedback. 

Considering your help, I will try to plot the 3 first modes of vibration as I mentioned in Figure 3.22 (a) (page 70 in the pdf file from the link below). 

 



God Bless you.

Pdf File: https://www.diva-portal.org/sm...

PhD_Wallyson 35
July 08 2020
0 0

Hi Everyone!

I would like your help again.

Considering a Free-Pinned-Pinned-Free beam (page 88 in the pdf file from the link below). In case of a matrix 12x12 how could I find the coefficients (a1, a2, a3, a4, b1, b2, b3, b4, c1, c2, c3, c4) using MAPLE in order to plot the mode shapes of the Figure 3.22 (a) (page 70 in the pdf file from the link below)? in case of a matrix 16x16 and  20x20, the procedure is the same?

I tried to plot the mode shapes but I failed because I believe they should be similar to Figure 3.22 (a) (page 70 in the pdf file from the link below)

 

sys_GE := {-a1 + a3 = 0, -a2 + a4 = 0, 0.9341161484*a1 + 0.3569692163*a2 + 1.519943120*a3 + 1.819402948*a4 = 0, 0.6669014188*b1 - 0.7451459573*b2 + 5.530777989*b3 + 5.620454178*b4 = 0, 0.9938777922*c1 - 0.1104849953*c2 + 62.17096851*c3 + 62.17901032*c4 = 0, -0.9341161484*a1 - 0.3569692163*a2 + 1.519943120*a3 + 1.819402948*a4 + 0.9341161484*b1 + 0.3569692163*b2 - 1.519943120*b3 - 1.819402948*b4 = 0, -0.3569692163*a1 + 0.9341161484*a2 + 1.819402948*a3 + 1.519943120*a4 + 0.3569692163*b1 - 0.9341161484*b2 - 1.819402948*b3 - 1.519943120*b4 = 0, 0.3569692163*a1 - 0.9341161484*a2 + 1.819402948*a3 + 1.519943120*a4 - 0.3569692163*b1 + 0.9341161484*b2 - 1.819402948*b3 - 1.519943120*b4 = 0, -0.7451459573*b1 - 0.6669014188*b2 + 5.620454178*b3 + 5.530777989*b4 + 0.7451459573*c1 + 0.6669014188*c2 - 5.620454178*c3 - 5.530777989*c4 = 0, -0.6669014188*b1 + 0.7451459573*b2 + 5.530777989*b3 + 5.620454178*b4 + 0.6669014188*c1 - 0.7451459573*c2 - 5.530777989*c3 - 5.620454178*c4 = 0, 0.7451459573*b1 + 0.6669014188*b2 + 5.620454178*b3 + 5.530777989*b4 - 0.7451459573*c1 - 0.6669014188*c2 - 5.620454178*c3 - 5.530777989*c4 = 0}

 

solve(sys_GE, {a1, a2, a3, a4, b1, b2, b3, b4, c1, c2, c3, c4});
        {a1 = 0.1014436637 c4, a2 = -0.1143870369 c4, 

          a3 = 0.1014436637 c4, a4 = -0.1143870369 c4, 

          b1 = 0.07095134140 c4, b2 = -0.1260395712 c4, 

          b3 = 0.1510591164 c4, b4 = -0.1737777468 c4, 

          c1 = 0.2102272829 c4, c2 = -0.2816561313 c4, 

          c3 = -1.003990622 c4, c4 = c4}

26_06_2020_Transcedental_equation_matrix_12x12_artigo_2.mw

https://www.diva-portal.org/smash/get/diva2:837061/FULLTEXT01.pdf

or you can find the figure in this link below as well (Figure 8 (b)):

https://doi.org/10.1016/j.ymssp.2009.05.012

@Rouben Rostamian  Hi you nailed i...

PhD_Wallyson 35
July 07 2020
0 0

@Rouben Rostamian  Hi

you nailed it!

I will also use your codes to help me in my publication in my scientific paper. 
for you and some others here, understand Maple is not a big deal like to me. For that reason, I really like to thank you for such help.

I still need help, and I am performing the question to ask you all help me one more time.
 

I am honoured to have blessed people in my path like you all.
Thanks 

Best answer until now...

PhD_Wallyson 35
July 07 2020
0 0

@acer 

you read my mind!
Best answer until now.

It is the first time I am using that chat. I have some other questions related to my PhD studies, I am a first-year Brazilian student in Mechanical Engineering at the University of Miskolc (Hungary).

I am extremely happy......
GOD BLESS YOU! 

@Rouben Rostamian   First of all, t...

PhD_Wallyson 35
July 07 2020
0 0

@Rouben Rostamian  
First of all, thanks for helping me! I am a Brazilian PhD student that really needs help. God bless you.

you are correct!


the determinant of the matrix of a clamped free beam (uniform section) is the transcendental equation: cos(βL) cosh(βL) = −1.

Could you please continue your way of thinking....Do you have any idea of how could I plot this graph?

I have some paper which has similar explanations:

https://www.sciencedirect.com/science/article/pii/0020740374900083

https://doi.org/10.1016/0020-7403(74)90008-3

Best Regards,
Wallyson Thomas

@Carl Love I want to read the zeros...

PhD_Wallyson 35
July 07 2020
0 0

@Carl Love I want to read the zeros of equation 22 to different a/L. 

in the code below I did just with a/L = 0,6

M := Matrix(8, [[0, 1, 0, 1, 0, 0, 0, 0], [1, 0, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, -sin(x), -cos(x), sinh(x), cosh(x)], [0, 0, 0, 0, -cos(x), sin(x), cosh(x), sinh(x)], [sin(0.6*x), cos(0.6*x), sinh(0.6*x), cosh(0.6*x), 0, 0, 0, 0], [0, 0, 0, 0, sin(0.6*x), cos(0.6*x), sinh(0.6*x), cosh(0.6*x)], [cos(0.6*x), -sin(0.6*x), cosh(0.6*x), sinh(0.6*x), -cos(0.6*x), sin(0.6*x), -cosh(0.6*x), -sinh(0.6*x)], [-sin(0.6*x), -cos(0.6*x), sinh(0.6*x), cosh(0.6*x), sin(0.6*x), cos(0.6*x), -sinh(0.6*x), -cosh(0.6*x)]])

DET := Determinant(M)

restart;
with(plots);


J := plot(((-2.*cos(x)*sinh(x) + 2.*sin(x)*cosh(x))*cos(0.6000000000*x) + (-2. - 2.*sin(x)*sinh(x) - 2.*cos(x)*cosh(x))*sin(0.6000000000*x))*cosh(0.6000000000*x)^3 + ((4.*cos(x)*sinh(x) - 4.*sin(x)*cosh(x))*cos(0.6000000000*x)^2 + ((2.*cos(x)*cosh(x) - 2.*sin(x)*sinh(x) + 2.)*sinh(0.6000000000*x) + 4.*sin(x)*sinh(x)*sin(0.6000000000*x))*cos(0.6000000000*x) + (2.*cos(x)*sinh(x) + 2.*sin(x)*cosh(x))*sin(0.6000000000*x)*sinh(0.6000000000*x) - 4.*sin(x)*cosh(x)*sin(0.6000000000*x)^2)*cosh(0.6000000000*x)^2 + ((-2.*cos(x)*sinh(x) + 2.*sin(x)*cosh(x))*cos(0.6000000000*x)^3 + (4.*sin(x)*sinh(x)*sinh(0.6000000000*x) + (-2. - 2.*sin(x)*sinh(x) - 2.*cos(x)*cosh(x))*sin(0.6000000000*x))*cos(0.6000000000*x)^2 + ((2.*cos(x)*sinh(x) - 2.*sin(x)*cosh(x))*sinh(0.6000000000*x)^2 + (-2.*cos(x)*sinh(x) + 2.*sin(x)*cosh(x))*sin(0.6000000000*x)^2)*cos(0.6000000000*x) + (2.*cos(x)*cosh(x) + 2.*sin(x)*sinh(x) + 2.)*sin(0.6000000000*x)*sinh(0.6000000000*x)^2 + 4.*sin(x)*sinh(x)*sinh(0.6000000000*x)*sin(0.6000000000*x)^2 + (-2. - 2.*sin(x)*sinh(x) - 2.*cos(x)*cosh(x))*sin(0.6000000000*x)^3)*cosh(0.6000000000*x) + (2.*cos(x)*cosh(x) - 2.*sin(x)*sinh(x) + 2.)*sinh(0.6000000000*x)*cos(0.6000000000*x)^3 + (-4.*cos(x)*sinh(x)*sinh(0.6000000000*x)^2 + (2.*cos(x)*sinh(x) + 2.*sin(x)*cosh(x))*sin(0.6000000000*x)*sinh(0.6000000000*x))*cos(0.6000000000*x)^2 + ((-2. + 2.*sin(x)*sinh(x) - 2.*cos(x)*cosh(x))*sinh(0.6000000000*x)^3 - 4.*sin(x)*sinh(x)*sinh(0.6000000000*x)^2*sin(0.6000000000*x) + (2.*cos(x)*cosh(x) - 2.*sin(x)*sinh(x) + 2.)*sin(0.6000000000*x)^2*sinh(0.6000000000*x))*cos(0.6000000000*x) + (-2.*cos(x)*sinh(x) - 2.*sin(x)*cosh(x))*sin(0.6000000000*x)*sinh(0.6000000000*x)^3 + (2.*cos(x)*sinh(x) + 2.*sin(x)*cosh(x))*sin(0.6000000000*x)^3*sinh(0.6000000000*x), x = 0 .. 6*Pi, color = green);


display({J});
 

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