Lecture 27: Structural Dynamics - Beams.mech420/Lecture27.pdf · MECH 420: Finite Element Applications Lecture 27: Structural Dynamics - Beams. Chapter #16: Structural Dynamics and

MECH 420: Finite Element Applications

Lecture 27: Structural Dynamics - Beams.

Chapter #16: Structural Dynamics and Time Dependent Heat Transfer.

Lectures #1-26 have discussed only ‘steady’ systems.There has been no time dependence in any problems.We will investigate beam dynamics and show the additional steps in the numerical solution of a time dependent problem.There are many time dependent problems discussed in Chapter #16.

§16.1 – 1 DOF spring mass system.§16.2, 16.4 & 16.5 – bar element dynamic response.§ 16.7 – truss and plane frame analysis.§ 16.8 – time dependent (unsteady) heat transfer.



§16.6 Beam Element Mass Matrices and Natural Frequencies.We will revisit the formation of the governing DE for the equilibrium of the beam element.Equilibrium now involves an ‘inertial force’ acting on the differential beam element.The approximate displacement field v(x) will be revised to include the time dependence.We will see the generation of the consistent mass matrix through the evaluation of the weighted residuals.We will look at the assembly process for a beam dynamics problem and discuss the computational advantages of a ‘lumped mass approximation.’The solution process will be expanded to include the integration of the state vector (§16.3).Define the phrase initial conditions.


Lecture 27: Structural Dynamics - Beams.Beam Dynamics:

x

y( )xA vdρ ⋅

The differential element is now accelerating in the positive ydirection in reaction to the external forces.



Consider what happens as a beam element moves (vibrates or translates in space).

The profile of our element is defined by node coordinates and node rotations.The nodal values (the state vector d) is blended by the shape function matrix.For the moving beam the profile is fluctuating.In the case shown the slopes at both nodes 1 and 2 and the vertical displacement at node 1 have changed over an interval Δt.



The approximation must be time dependent.

[ ]{ }1 2 3 4 1 1 2 2

ˆˆ ˆ( )

ˆ ˆˆ ˆ T

y y

v x N d

N N N N d dφ φ

= ⋅

=

( )

( )

( )

( )

3 2 31 3

3 2 2 32 3

3 23 3

3 2 24 3

1 ˆ ˆ2 3

1 ˆ ˆ ˆ2

1 ˆ ˆ2 3

1 ˆ ˆ

N x x L LL

N x L x L xLL

N x x LL

N x L x LL

= − +

= − +

= − +

= −

The shape functions are dependent on space. They are set to satisfy element boundary

conditions.So, the node state vector becomes the time dependent quantity.

ˆˆ ˆ ˆ( ) ( ) (,

ˆˆ ˆ ˆ( ) ( ) ( )

ˆˆ ˆ ˆ( ) ( )

)

( )

t

v x N x d t

v x N

v x N x

x d t

d t

= ⋅

=

⋅

⋅

=



A force balance on the differential element gives.

( )ˆ ˆ ˆ( ) ˆ ˆ( )dV w x dx Adx v xρ− − =We can evaluate our approximate inertial term.

( )4 2

4 2

ˆ ˆ ˆ ˆ( , ) ( , ) 0ˆ

v x t v x tEI w Ax t

ρ∂ ∂+ + =

∂ ∂

3

3

ˆ ˆ( )ˆˆ

d v xV EIdx

=

A new term in our residual equations for the beam element. Note that the shape functions have not

changed.

continuum mechanics



Since the shape functions have not changed, the contributions ofthe first two terms to the element equations will not change.We need only evaluate the residual term produced by the acceleration term in the motion equation.We do this in a consistent manner: we try to drive four weighted residuals to zero using Galerkin’s choice of weighting functions.This leads to the phrase ‘consistent mass matrix’.

{ } ( )ˆ ˆ4

4ˆ ˆ

ˆ 2

0 02

ˆ 0

ˆ ˆ( ,ˆ ˆ( , ) ˆ ˆ 0ˆ

) ˆx L x L

T T

x x

x LT

x

v x tN EI v x tN A ddx N wt

dxx

xρ=

=

=

=

=

=

⎧ ⎫ ⎧ ⎫∂+ ⋅⎨ ⎬∂

∂⋅ + ⋅ =⎨ ⎬

⎭∂⎩ ⎭ ⎩∫ ∫ ∫


Following the MWR procedure:

( )

( ) { }( )

ˆ 2

2ˆ 0

ˆ

ˆ 0

ˆ

ˆ 0

ˆ ˆ( , ) ˆ

ˆ ˆ

ˆˆ

x LT

ax

x LT

ax

x LT

ax

v x tI N A dxt

I A N Nd dx

I A N N dx d

ρ

ρ

ρ

=

=

=

=

=

=

⎧ ⎫∂= ⋅⎨ ⎬∂⎩ ⎭

= ⋅

⎡ ⎤= ⋅⎢ ⎥⎣ ⎦

∫

∫

∫


The consistent mass matrix.

{ }1 1 2 2ˆ ˆ ˆˆ ˆ

T

y yd d dφ φ=


When we append this inertial term onto the results of our previous weighted residual evaluations:

Our approximate displacement field keeps the element equations linear (we do not see products of state variables).In MECH 330 you learned how to acquire the natural frequencies and mode shapes for a multi-DOF undamped system.

111 12 2 2 2

1 1 1 13

2 2 22 2

2 2 2

ˆˆˆ ˆ 12 6 12 6 156 22 54 13ˆ ˆˆ ˆ 4 6 2 4 13 3

ˆ ˆ ˆ12 6 156 224204 4ˆˆ ˆ

w yyy yw

z z

wy y y

wz z

ddf f L L L Lm m L L L L L LEI AL

L LLf f dL Lm m

φ φρ

φ

⎧ ⎫⎧ ⎫ ⎧ ⎫ − −⎡ ⎤ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥− −⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥+ = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎩ ⎭ ⎩ ⎭i i

2

2

ˆ

ˆyd

φ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

k̂ m̂F




An alternative to the consistent mass matrix is the lumped mass matrix.The lumped mass representation is a heuristic discretization of mass.The consistent mass matrix is formed by using the actual mass distribution within the residual evaluations.The lumped approximation assumes that the beam is very slender and it lumps the mass at node points.There is no inertia associated with changing node slopes.The lumped approximatio is used because it diagonalizes the mass matrix.Lumping is not a mathematical process – it is a idealization that is executed on the element equations.


Half of the beam’s total mass is at node 1.

Half of the beam’s total mass is at node 2.

The fluctuations of the curvilinear profile do not accelerate any mass.




When we apply the lumped mass approximation:

We will see that the lumped mass approximation eases the computational burden of modal analysis and time domain simulation (Lecture #28).

111 12 2

1 1 1 13

2 2 2 222 2 2

2

ˆˆˆ ˆ 12 6 12 6 1 0 0 0ˆ ˆˆ ˆ 4 6 2 0 0 0

ˆ ˆ ˆ12 6 1 02 ˆ4 0ˆˆ ˆ ˆ

w yyy yw

z z

wy y y y

wz z

ddf f L Lm m L L LEI AL

LLf f d dLm m

φ φρ

φ φ

⎧ ⎫⎧ ⎫⎧ ⎫ ⎧ ⎫ −⎡ ⎤ ⎡ ⎤ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪− ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥+ = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥−⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎪⎩i i

⎪

⎪⎪⎪⎭

k̂ m̂F



We know that in the undamped free vibration case, the transverse vibration at any point along the element will be harmonic.

1 1

1 1

22

22

ˆ ( )ˆ ( )ˆ

ˆ ( )ˆ ( )

n

y y

i t

yy

d t Dt

d eDd t

t

ωφ

φ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪Φ⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪Φ⎩ ⎭⎩ ⎭

An assumed harmonic variation of the state variables.

Magnitudes of the 4 vibration signals.( )

( )

2

2

ˆ ˆ ˆˆ0ˆ ˆ

ˆ ˆ

n

n

i tn

i tn

k d m d

k D m D e

k m De

ω

ω

ω

ω

= ⋅ + ⋅

= ⋅ − ⋅

= − ⋅

Must be a singular matrix.The solution to the undamped free vibration problem is the same as the eigen problem for the system dynamic matrix.

Consider free (natural) vibration.



( )( )

( )

2

1 2

1 2

2 1

ˆ ˆ0

ˆˆ ˆ0

ˆˆ0

ˆˆ

ni tn

n

n

n

k m De

m k m D

m k I D

D m k D

ωω

ω

ω

ω

−

−

−

= − ⋅

= − ⋅

= − ⋅

⎡ ⎤= ⎣ ⎦

The ω2 values are the eigenvalues of the dynamic

matrix.

The D vectors are the eigenvectors of the dynamic

matrix.

Dynamic Matrix



Example problem 16.5:



Forming the consistent element equations.

(1) 111

2 2 2 2(1)1 11

(1) 322

22 2(1)22

2

ˆ12 6 -12 6 156 22 54 13

ˆ4 6 2 4 13 312 6 156 22420 ˆ

4 4ˆ

yyy

z

yyy

z

ddf L L L LL L L L L Lm EI AL

dL Lf L dL Lm

φ φρ

φφ

⎧ ⎫⎧ ⎫ ⎧ ⎫ ⎪ ⎪−⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬ ⎨ ⎬− −⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭⎩ ⎭ ⎪ ⎪⎩ ⎭i i

(2) 222

2 2 2 2(2)2 22

(2) 333

32 2(2)33

3

ˆ12 6 -12 6 156 22 54 13

ˆ4 6 2 4 13 312 6 156 22420 ˆ

4 4ˆ

yyy

z

yyy

z

ddf L L L LL L L L L Lm EI AL

dL Lf L dL Lm

φ φρ

φφ

⎧ ⎫⎧ ⎫ ⎧ ⎫ ⎪ ⎪−⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬ ⎨ ⎬− −⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭⎩ ⎭ ⎪ ⎪⎩ ⎭i i

Element #1

Element #2



Assembling and applying the homogeneous boundary conditions.

(1)11

(1) 2 211

(1) (2)22 2

2 2 2 23(1) (2)22 2

(2)33

(2)33

12 6 -12 6 0 06 4 6 2 0 012 6 12 12 6 6 12 6

6 2 6 6 4 4 6 20 0 12 6 12 60 0 6

yy

zz

yy y

zz z

yy

zz

Ff L Lmm L L L LFf f L L L LEI

L L L L L L L Lm Lm mL LFf

mm

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ −⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪+ − − + − + −⎪ ⎪ = =⎨ ⎬ ⎨ ⎬ − + + −+⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ − − −⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭

1

1

2

2

32 2

3

2 2

2 2 2 2

2 6 4

156 22 54 13 0 022 4 13 3 0 054 13 156 156 22 22 54 13

13 3 22 22 4 4 13 34200 0 54

y

y

y

d

d

dL L L L

L LL L L L

L L L LALL L L L L L L L

φ

φ

φ

ρ

⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥−⎣ ⎦ ⎩ ⎭

−−

+ − + −+

− − − + + −

1

1

2

2

32 2

3

13 156 220 0 13 3 22 4

y

y

y

d

d

L L dL L L L

φ

φ

φ

⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪

⎪ ⎪⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪−⎢ ⎥⎪ ⎪⎢ ⎥− − −⎣ ⎦ ⎪ ⎪⎩ ⎭

0

0

0

00

0

0

0

0



The application of the BC’s is what specifies the system nature.The natural frequencies we calculate are now specific to the fixed-fixed supports.

The associated eigen problem is:

The natural frequencies (x2) are given by:

2 22 23

2 2

0 24 0 312 00 0 8 0 8420

y yd dEI ALL LL

ρφ φ

⎧ ⎫⎧ ⎫⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪= +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎪ ⎪⎩ ⎭

( )22 23

24 0 312 00

0 8 0 8420EI AL

L LLρω⎡ ⎤ ⎡ ⎤− =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

2 21 1 2 24 2 4 2

420 5.68 420 20.49 ; 13

EI EI EI EIAL L A AL L A

ω ω ω ωρ ρ ρ ρ

= ∴ = = ∴ =

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Lecture 27: Structural Dynamics - Beams.mech420/Lecture27.pdf · MECH 420: Finite Element Applications Lecture 27: Structural Dynamics - Beams. Chapter #16: Structural Dynamics and