Mathematica use in an Engineering Environment-an example-Rev.0/May 2012

Part1

byMircea Dinulescu, Sergiu Ciocirlan, Laurentiu Moruz April 2012

© Mircea Dinulescu/2012

IntroductionMathematica permits symbolic as well as numerical/graphical analysis of problems.In this article we will demonstrate first the symbolic analyses and afterwards usethese symbolic results to make numerical/graphical analyses.To illustrate this, we have chosen Chapter 2.3 ,Theoretical Modelling by equivalent 1-DOF (degrees-of-freedom, autor) mass-spring systems, from Reference 1.The problem consists in replacing a beam of stiffness EI(x) and with N nodal massesdistributed at random along the axis of the beam with a simple mass-spring modelwhich has same own frequency and same strain energy at a given displacement as theoriginal beam structure (see Figure 1). The beam is fixed at x=0 and free at x=L. Theequivalent mass will be located at the point with maximum displacement in the mainmode of vibration, i.e. at the free end. The effect of an axial force Q applied to thenode x=L will be studied for determining its effect on the equivalent stiffness of thespring and the critical point of instability due to compressive force (see Figure 2).This point of instability will be further compared with the classical Euler formulae.To simplify the presentation we will assume EI=constant. Same calculation would bevalid for EI(x) but the results would be more complex.

m1

m2

m3

Q Y HtLpHL, tL

x

yHtL

LL

y

∆y

∆x∆L

Figure 1. Cantilever column with flexural rigidity EI(x), mass m(x) and damping coefficient c(x) subjected to a dynamic load p(x,t) at a distance x above the base.

K

M

PHtL

C

Figure 2. Equivalent mas-spring sys-tem with stiffness K, mass M, damping coefficient C and dynamic load P(t).

Import modules and create special symbolsNeeds@"aGeneralFunctions`"D;Notation`AutoLoadNotationPalette = False;Needs@"Notation`"D;

SymbolizeB KE , WorkingForm ® TraditionalFormF;

SymbolizeB Y

, WorkingForm ® TraditionalFormF;

SymbolizeB Ωn , WorkingForm ® TraditionalFormF;

SymbolizeB x..

, WorkingForm ® TraditionalFormF;

SymbolizeB M1 , WorkingForm ® TraditionalFormF;

SymbolizeB fn , WorkingForm ® TraditionalFormF;

Symbolic analysis of the problemThe key concept. The key concept consists in finding a function j(x) satisfying thefollowing conditions: - (i) it is addimensional e.g. j(x) = I1 3 L4M Ix4 - 4 L x3 + 6 L2 x2M - (ii) it has the value 1 at the point of maximum displacement; in the case of a can-tiliver beam or column j(L) = 1

- (iii) preserves the shape of deformation line at different maximum displacements,i.e. y(x,t) = j(x) Y(t)

- (iv) the dependence of y(x,t) of x being limited to j(x) the mathematical analysis isgreatly simplified - (v) the function j(x) must be determined by using reasonable simplifyingassumptions.

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© Mircea Dinulescu/2012

The key concept. The key concept consists in finding a function j(x) satisfying thefollowing conditions: - (i) it is addimensional e.g. j(x) = I1 3 L4M Ix4 - 4 L x3 + 6 L2 x2M - (ii) it has the value 1 at the point of maximum displacement; in the case of a can-tiliver beam or column j(L) = 1

- (iii) preserves the shape of deformation line at different maximum displacements,i.e. y(x,t) = j(x) Y(t)

- (iv) the dependence of y(x,t) of x being limited to j(x) the mathematical analysis isgreatly simplified - (v) the function j(x) must be determined by using reasonable simplifyingassumptions.Determination of the equivalent mass. We put the condition that the kinetic energy of the mass-spring system is equal to thekinetic energy of the column i.e.:

kin01 = 1 2 M Y 2

;

int01 = à0

L

1 2 Y 2

m@xD j@xD2â x ;

sum01 = â

i=1

N 1

2mi j@xiD2 Y

2;

Print@kin01 TraditionalForm, " = ",int01 + sum01 TraditionalForm, Spacer@121D, "H1L"D

M Y 2

2= â

i=1

N 1

2Y 2

mi jHxiL2+ à

0

L 1

2Y 2

mHxL jHxL2â x H1L

Simplifyng this equation considering that Y

is constant along x we obtain formula H2L :

eqn01 = FullSimplifyAkin01 aExtractConstants@int01D +

aExtractConstants@sum01D, Y

¹ 0E;

Print@"M = ", M = M . Solve@eqn01, MD@@1, 1DD TraditionalForm, Spacer@249D, "H2L"D

M = â

i=1

N

mi jHxiL2+ à

0

L

mHxL jHxL2â x H2L

Determination of the equivalent stiffness. We put the condition that the strainenergy stored in the spring equals the strain energy stored in the column i.e.:

int02 = à0

L

1 2 E I Ij ''@xD Y

M2

â x;

strain01 = 1 2 KE Y 2

;Print@strain01 TraditionalForm, " = ",

int02 TraditionalForm, Spacer@283D, "H3L"D

1

2Y 2

KE = à0

L 1

2E I Y

2j

¢¢HxL2â x H3L

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© Mircea Dinulescu/2012

Simplifyng the constant Y

we obtain formula H4L :

eqn02 = FullSimplifyB1 2 KE Y 2

aExtractConstantsBà0

L

1 2 E I Ij ''@xD Y

M2

â xF, Y

¹ 0F;

Print@"KE = ", KE = KE . Solve@eqn02, KED@@1, 1DD TraditionalForm, Spacer@377D, "H4L"D

KE = E I à0

L

j¢¢HxL2

â x H4L

Now let us apply these symbolic results to a number of practical example, first insymbolic form and then in numerical and graphical form. Determination of the natural frequency, fn. The expressions for M and K

E,

formulae (2) and (4), can be used to determine the natural frequency of the structure.Assuming an harmonic oscillation of the structure, given by the time function: x = X SinHΩn tL we obtain the displacement x(t) and its second derivative x

..(t):

x@t_D := X Sin@Ωn tDPrint@"xHtL = ",

x@tD TraditionalForm, Spacer@397D, "H5L"D

xHtL = X sinHt ΩnL H5L

PrintA"x..

HtL = ",

x ''@tD TraditionalForm, Spacer@356D, "H6L"E

x..

HtL = -X Ωn2 sinHt ΩnL H6L

The second law of motion, KE x(t) = -M x..(t), becomes :

eqn03 = KE x@tD == -M x ''@tD;Print@FullSimplify@eqn03, 8X ¹ 0, Sin@Ωn tD ¹ 0 <D

TraditionalForm, Spacer@75D, "H7L"D

E I à0

L

j¢¢HxL2

â x Ωn2

à0

L

mHxL jHxL2â x + â

i=1

N

mi jHxiL2 H7L

or KE = M Ωn2 :

PrintAKE M Ωn2 TraditionalForm, Spacer@75D, "H8L"E

E I à0

L

j¢¢HxL2

â x Ωn2

à0

L

mHxL jHxL2â x + â

i=1

N

mi jHxiL2 H8L

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© Mircea Dinulescu/2012

The angular frequency, Ωn , in [rad/sec] is:

Ωn = KE M ;

PrintB"Ωn = KE M =", Ωn TraditionalForm,

Spacer@75D, Spacer@45D, "H9L"F

Ωn = KE M =

E I Ù0

Lj¢¢HxL2 â x

Ù0

LmHxL jHxL2 â x + Úi=1

N mi jHxiL2H9L

and the frequency, fn, in [Hz] is:

fn = 1 H2 ΠL KE M ;

PrintB"fn = 1H2ΠL KE M =",

fn TraditionalForm, Spacer@75D, Spacer@46D, "H10L"F

fn = 1H2ΠL KE M =

E I Ù0

Lj¢¢HxL2 âx

Ù0

LmHxL jHxL2 âx+Úi=1

N mi jHxiL2

2 ΠH10L

The formulae (1) through (10) will be used in the next revision to solve a few simplestructures.

References1. Structural Dynamics for Engineers, Hans AntonBuchholdt & Shodja Edin Moos-savi Nejad, ICE Publishing 2012

Now we will show how we can obtain a print - ready article from the abovecalculations.

We make a copy of the above notebook with all the cells caclulated andremove from this copy : -the CellLabels “In[ ...]” and “Out[ ...]” signs: select from menubar Format/OptionInspector in Option Inspector dialog select Cell Options/Cell Labels and uncheckShowCellLabel or more simple just save/close the notebook and reopen. -the cells which are superfluous for the article.Then we export the notebook as a pdf file or in other formats, as needed. Please look at Part 2 of this revision to see the result in pdf format. By comparing Part 1 with Part 2 it will be obvious which cells had been removed.

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© Mircea Dinulescu/2012