Linear beams: extension to dynamics and ﬁnite … · Linear beams: extension to dynamics and...

Elementi di Meccanica ComputazionaleCorso di Laurea in Ingegneria Civile

Pavia, 2014

Linear beams: extension to dynamics and finite element

formulation

Ferdinando Auricchio 1 2

1DICAR – Dipartimento di Ingegneria Civile e Architettura, Universita di Pavia, Italy2IMATI – Istituto di Matematica Applicata e Tecnologie Informatiche, CNR, Italy

March 13, 2014

F.Auricchio (UNIPV) Linear beam March 13, 2014 1 / 44

Some references I

A. Chopra, “Dynamics of Structures: Theory and Applications to EarthquakeEngineering (2nd Edition) ”, Prentice Hall, 2000


Beam: extension to dynamics I

GOAL: extend beam formulation to include dynamics

1 Weak formulation of the problem

Differential form → Integro-differential form

2 Introduction of approximation fields

Integro-differential form → Algebraic-differential form

3 Introduction of time-marching scheme

Algebraic-differential form → Algebraic form

4 Solution of algebraic problem


Beam: extension to dynamics at the continuous level I

Strong form for static conditions

◦ Assume v = v(x) with x longitudinal abscissa◦ Distinct strong equation

v ′ = θ

χ = θ′

M = EIχ

M′ − S = 0

S ′ − q = 0

◦ Compact strong form

EIv IV − q = 0


Beam: extension to dynamics at the continuous level II

Strong form for dynamic conditions

◦ Assume v = v(x , t) with t indicating time◦ All previous equations valid except for last one (transverse equilibrium) which should

be modified as follow:S ′ − qdyn = 0

setting now

qdyn = qstat − ρv v

where⋆ qdyn : equivalent load under dynamical conditions⋆ qstat : externally applied load (static load)⋆ ρv : transverse mass density (per unit length) given by

ρv =

∫

h/2

−h/2

ρbdy = ρA

with ρ mass density per unit volume, b cross-section depth and A cross-section area

◦ Compact strong form

EIv IV − qdyn = 0 or EIv IV + ρv v − qstat = 0

◮ Approach known as d’Alembert principle


Beam: dynamical weak form I

Weak form

◦ Assume to work with weight function w = w(x)⋆ Note: weight function independent of time !!!!⋆ Note: integral form to be valid for all times !!!!

◦ Multiply differential form by w and integrate over domain [a,b]

∫ b

a

[

w(

EIv IV + ρv v − qstat)]

dx = 0

◦ Integrate twice by parts the first term

∫ b

a

[

w ′′EIv ′′]

dx +

∫ b

a

[wρv v ] dx −

∫ b

a

[wq] dx + Sw |ba − Mw ′∣

∣

b

a= 0


Beam: dynamical weak form II

◮ In compact form, weak form can be written as

f = f int + f inert − f ext = 0

with

f int =

∫ b

a

[

w ′′EIv ′′]

dx

f inert =

∫ b

a

[wρv v ] dx

f ext =

∫ b

a

[wq] dx − Sw |ba + Mw ′∣

∣

b

a

◮ Interesting to observe that

f = f int(v) + f inert(v)− f ext(v) = 0

thereforef (v , v) = 0



Approximated weak form

◦ Introduce an approximation for weight function and unknown function

w(x) =n

∑

i=1

Ni (x)wi

v(x , t) =n

∑

j=1

Nj(x)vj (t)

⋆ Ni : shape functions depending only on spatial coordinate x⋆ wi : arbitrary parameters independent of time t⋆ vi : unknown parameters depending on time t

◦ As a consequence

v(x , t) =

n∑

j=1

Nj (x)¨vj (t)

◦ In matrix form

w(x) = N(x)w

v(x , t) = N(x)v(t)

v(x , t) = N(x)¨v(t)

as well as

{

w′′(x) = B(x)w

v′′(x , t) = B(x)v(t)


Beam: extension to dynamics at the continuous level II

Assume from now only trivial boundary conditions

Neglect to explicitly indicate spatial and time dependence

Plugging approximations into weak form

∫ b

a

[(Bw)EI (Bv)] dx +

∫ b

a

[

(Nw) ρv(

N¨v)]

dx −

∫ b

a

[(Nw) q] dx = 0

which can be rewritten as

wT

{∫ b

a

[

BTEI (Bv)

]

dx +

∫ b

a

[

NTρv

(

N¨v)]

dx −

∫ b

a

[

NTq]

dx

}

= 0

Recalling the arbitrariness of w, we have

∫ b

a

[

BTEI (Bv)

]

dx +

∫ b

a

[

NTρv

(

N¨v)]

dx −

∫ b

a

[

NTq]

dx = 0


Beam: extension to dynamics at the continuous level IIIIn compact form dynamical equilibrium can be written as

f = fint + f

inert− f

ext = 0

where

fint =

∫ b

a

[

BTEI (Bv)

]

dx

finert =

∫ b

a

[

NTρv

(

N¨v)]

dx

fext =

∫ b

a

[

NTq]

dx

In a more standard form dynamical equilibrium can be written as

Kv +M¨v = fext

with

K =

∫ b

a

[

BTEIB

]

dx stiffness matrix

M =

∫ b

a

[

NTρvN

]

dx “consistent” mass matrix


Beam: extension to dynamics at the continuous level IV

The terminology consistent is introduced to stress that the mass matrix is obtainedusing a consistent variational formulation

The terminology consistent is often used in contrast with the terminology lumpedwhich is in general non-consistent and diagonal!!

Using previously introduced shape functions

M = ρv

13

35l

11

210l2

9

70l −

13

420l2

11

210l2

1

105l3

13

420l2 −

1

140l3

9

70l

13

420l2

13

35l −

11

210l2

−13

420l2 −

1

140l3 −

11

210l2

1

105l3

� Exercise. Using a symbolic code verify expression for the Euler-Bernoulli beam mass matrix

� Exercise. Compute the determinants of the element mass matrix and of the element stiffness matrix.

Comment on the fact that the element mass matrix is non-singular while the element stiffness matrix is

singular.



Extend dynamical strong equilibrium to include also a linear viscous term

EIvIV + ρv v + µv − q

stat = 0

Corresponding weak form produces the following system

M¨v+ C ˙v+ Kv = fext (1)

with

M =

∫ b

a

[

NTρvN

]

dx mass matrix

C =

∫ b

a

[

NTµN

]

dx damping matrix

K =

∫ b

a

[

BTEIB

]

dx stiffness matrix

In structural analysis often C is assumed to be in the form

C = αM+ βK

with α and β to be determined experimentally (Rayleigh damping)


Dynamic problem: problem classification I

As a result of semi-discretization in space, time-dependent problem is reduced to aset of ordinary differential equations

In the following discuss three different possible situations:

◦ Determination of free-response (i.e. fext = 0)

◦ Determination of steady-state periodic response (i.e. fext periodic in time)◦ Determination of transient response (i.e. fext arbitrary in time)


Dynamic problem: free-response / eigenvalue problem I

From now on consider always discrete problems (i.e. simplify notation)

If no damping or forcing terms exist, dynamical problem 1 reduces as

Mu+ Ku = 0 (2)

A general solution can be written in the form

u = u exp(iωt)

with the corresponding real part representing an harmonic response since

exp(iωt) = cos(ωt) + i sin(ωt)

Back-substitution into equation 2 returns(

−ω2M+ K

)

u = 0 (3)

i.e. a general linear eigenvalue or characteristic value problem and for non-zerosolutions the determinant of the above coefficient matrix must be zero

det[

−ω2M+ K

]

= 0 (4)


Dynamic problem: free-response / eigenvalue problem II

Provided that M and K are n × n symmetric-definite matrices, condition 4 gives npositive roots or eigenvalues ωi with i = 1..n

For each eigenvalue, using equation 3, possible to compute system eigenvectors ornormal modes ui , in general requiring a proper normalization, i.e. s.t.

uTi Mui = 1 for i = 1..n

At this stage also possible to introduce classical modal orthogonality condition

uTi Muj = u

Tj Mui = 0 for i 6= j

as well as to prove the result

uTi Kui = ω

2i


Dynamic problem: free-response / eigenvalue problem III

� Exercise. Using a symbolic code prove that using Euler-Bernoulli shape functions a beam element ischaracterized by the following eigenvalues

ω1 =720

l4, ω2 = ω3 = 0 , ω4 =

8400

l4

and by the following eigenvectors

u1 =

l/6−1l/61

, u2 =

−l

101

, u3 =

1010

, u4 =

−l/121

l/121

� Exercise. Verify normality and orthogonality conditions on the given eigenvectors.

� Exercise. Comment on the dimensionality of the terms appearing in each single eigenvalue as well as on

the shape of each single eigenvector.


Dynamic problem: free-response / eigenvalue problem IV

To preserve methods based on the invertibility of stiffness matrix, possible to use anartifice

Modify problem[

−ω2M+ K

]

u = 0

into[

−(

ω2 + α

)

M+ (K+ αM)]

u = 0

where α is an arbitrary constant of the same order of the typical ω2 sought

The new matrix K+ αM is no more singular


Dynamic problem: damped free-response / eigenvalue problem I

If damping exists but no no forcing terms exist , dynamical problem 1 reduces as

Mu+ Cu+ Ku = 0 (5)

A general solution can again be written in the form

u = u exp(αt)

Back-substitution into equation 5 returns

(

α2M+ αC+ K

)

u = 0 (6)

with α and u possibly complex quantities, with a real part representing a decayingvibration

Equation 6 corresponds again to an eigenvalue problem, which is more complex thanthe one previously considered


Dynamic problem: damped free-response / eigenvalue problem II

Usually the problem is solved by splitting it into two first-order problems, defining

u = v

s.t. we get

[

M 0

0 −M

]{

v

u

}

+

[

C K

M 0

]{

v

u

}

=

{

0

0

}

Setting now

u = u exp(αt) , v = v exp(αt)

and substituting gives the general linear eigenvalue

(

α

[

M 0

0 −M

]

+

[

C K

M 0

]){

v

u

}

=

{

0

0

}

(7)


Dynamic problem: forced periodic response I

If in equation 1 the forcing term is periodical, i.e.

f = f exp(αt)

with α = α1 + iα2, then general solution can be again written as

u = u exp(αt)

which plugged into 1 gives

(

α2M+ αC+K

)

u = f

Obtained problem is no more an eigenvalue problem and it can simply solved as

u = K−1

f

where

K = α2M + αC+ K


Dynamic problem: transient response procedures I

To treat the response of a linear dynamical system to a general force we can adopttwo approaches:

◦ frequency response procedures◦ numerical time-integration

⋆ Start considering frequency response procedures

In general a completely arbitrary forcing function can be represented approximatelyby a Fourier series or in the limit, exactly, as a Fourier integral

Accordingly, response to a completely arbitrary forcing function can be obtainedproperly combining the response to a series of periodic forcing functions

The frequency response technique is readily adapted to problems where the matrix C

is of an arbitrary form


Dynamic problem: modal decomposition analysis I

For free response the general solution is of the form

u =n

∑

i

ui exp(αi t)

where αi are the complex eigenvalues and ui are the complex eigenvectors.

For forced response and linear problem, the solution can be again expressed as alinear combination of the modes

u =

n∑

i

uiyi (t) (8)

with yi scalar modal partecipation factors, assumed to be function of time

Position 8 presents no restriction as all the modes are linearly indipendent


Dynamic problem: modal decomposition analysis II

Using position 8 in dynamical equation and premultiplying by the complex conjugateeigenvectors uT

i , the result is simply a set of scalar, indipendent equations

mi y + ci y + kiyi = fi (9)

where

mi = uTi Mui , ci = u

Ti Cui , ki = u

Ti Kui , fi = u

Ti f

since we have

uTi Muj = u

Ti Cuj = u

Ti Kuj = 0 when i 6= j

Each scalar equation 9 can be solved indipendently by elementary procedures

Total motion is obtained by superposition using 8


Dynamic problem: modal decomposition analysis III

Most classical approach is to work with real eigenvalues obtained as solution of

[

−ω2M+ K

]

u = 0

With this position decoupled equations only in the case

uiCuj = 0 if i 6= j

which is a non-trivial condition since eigenvectors in general guarantee orthogonalitywrt M and K and not necessarily wrt C

If using damping matrix in the form of Rayleigh damping, then trivial.


Exercises I

� Exercise. Modify the frame Matlab code to compute also mass matrix, structural eigenvalues andeigenmodes.

� Exercise. For some simple beam problems (i.e. simply supported or cantilever beams), computeeigenvalue and eigenvector solutions in close form and compare with numerical solutions obtained for anincreasing number of elements. Plot convergence diagrams in a log-log-scale.

� Exercise. Choose a structure and compute free response comparing it with a standard commercialsoftware.

� Exercise. Develop mass matrix for a Timoshenko beam model


MDOF numerical solution I

Analytical solutions previously described provide insights into the behaviour patternsof linear systems

◦ Not computationally economical for the solution of transient problems◦ Cannot be extended to nonlinear cases

We now investigate discretization methods directly applicable to time-domain

Since time domain is infinite, refer to a finite time increment ∆t between tn andtn+1 = tn +∆t, obtaining the so-called recurrence relations


MDOF numerical solution II

GOAL: solve equation of motion for a general load condition and possibly forgeneral internal force response

Time interval of interest: [0,T ]

Discretize time interval in N time sub-intervals of size ∆t, such that

t0 = 0, t1 = ∆t, t2 = 2∆t, .... tN = N∆t = T

or assume variable time sub-intervals

Resort to time-marching numerical scheme to compute motion in time, i.e.assuming to know solutions up to tn, compute solution at tn+1

Look for numerical solutions with a bounded error wrt exact solution, i.e.

un+1 ≈ u(tn+1) such that un+1 = u(tn+1) + O(∆tk)

then time integration method is k-order accurate

Second order accurate methods are desirable !!!!


MDOF numerical solution: central difference method I

Equation of motion at time tn

Mun + Cun + Kun = fn (10)

Central difference approximation of first derivative

un =un+1 − un−1

2∆t

Central difference approximation of second derivative

un =un+ 1

2− un− 1

2

∆t=

un+1−un∆t

−un−un−1

∆t

∆t

=un+1 − 2un + un−1

(∆t)2


MDOF numerical solution: central difference method II

Substituting positions 28 and 28 into equation of motion 10, we obtain:

[

1

(∆t)2M+

1

2∆tC

]

un+1 = fn −

[

1

(∆t)2M−

1

2∆tC

]

un−1 −

[

K −2

(∆t)2M

]

un

(11)i.e.

Kun+1 = fn

where

K =

[

1

(∆t)2M+

1

2∆tC

]

fn = pn −

[

1

(∆t)2M−

1

2∆tC

]

un−1 −

[

K−2

(∆t)2M

]

un

Note: explicit method since restoring force based only on quantities evaluated at tn

Easy to extend to non-linear problems


MDOF numerical solution: central difference method III

Starting procedure◮ Required u0, u−1 to solve for u1 !!◮ Compute acceleration at t = 0

u0 = M−1 (f0 − Cu0 − Ku0)

◮ Assume that acceleration u0 at t = 0 acts for whole time-interval [t−1, t0]◮ Estimate u−1 backward from t0 to t−1 and initial conditions at t0

u−1 = u0 + (−∆t) u0 +1

2(−∆t)2 u0

i.e.

u−1 = u0 −∆tu0 +1

2(∆t)2 u0


MDOF numerical solution: central difference method IV

Properties of numerical scheme

Consistency with order of accuracy k :◮ gives informations on the relation between numerical approximated scheme and

original differential equation based on time-interval discretization ∆t◮ “Approximated” numerical scheme is said consistent if it reproduces “exact”

differential equation when ∆t → 0

Stability:◮ Local truncation error in numerical approximation does not grow in time

Consistency + stability = convergence

Convergence:◮ Numerical solution approaches exact solution when ∆t goes to zero


SDOF central difference method: consistency I

Consider equationu+ ω

2u = f(t)

Define local truncation error τ as

τ = approx. ODE− exact ODE

In the particular case under investigation

τ =

[

un+1 − 2un + un−1

∆t2+ ωun − fn

]

−[

un + ω2un − fn

]

and after simplyfication

τ =un+1 − 2un + un−1

∆t2− un (12)

Expand displacement in Taylor series

un+1 = un + un∆t +1

2un (∆t)2 +

1

6

...

u (∆t)3 +O(

∆t4)

un−1 = un − un∆t +1

2un (∆t)2 −

1

6

...

u (∆t)3 +O(

∆t4)


SDOF central difference method: consistency II

Substituting Taylor expansion in 12 we obtain local truncation error as:

τ =un+1 − 2un + un−1

∆t2− un = O

(

∆t2)

(13)

Central difference method is consistent with second order accuracy

Damping term does not affect second order accuracy


MDOF central difference method: stability I

We obtain a condition on time step ⇒ numerical approximation of ODE isconditionally stable

∆t

T<

1

π

However, for accuracy a good guideline is

∆t

T<

1

10


MDOF numerical solution: central difference method & nonlinear responseI

Equation of motion at time tn with nonlinear internal force

Mun + Cun + fint(un) = fn

Substituting positions 28 and 28 into equation of motion 35, we obtain:[

1

(∆t)2M+

1

2∆tC

]

un+1 = fn−

[

1

(∆t)2M−

1

2∆tC

]

un−1+

[

2

(∆t)2Mun + f

int(un)

]

(14)i.e.

Kun+1 = fn

where

K =

[

1

(∆t)2M+

1

2∆tC

]

fn = fn −

[

1

(∆t)2M−

1

2∆tC

]

un−1 +

[

2

(∆t)2Mun + f

int(un)

]

Nonlinear internal force can be produced for example by an elasto-plastic spring


MDOF numerical solution: Newmark’s method I

Start from the assumption on the form of acceleration during a time interval [tn, tn+1]

constant acceleration, i.e. u(τ ) =1

2(un + un+1)

linear acceleration, i.e. u(τ ) = un + (un+1 − un)τ

∆t

(15)

with τ ∈ [0,∆t] and ∆t = tn+1 − tn

Integrating twice to obtain velocity and displacement over the time interval we get:

constant acceleration

u(τ ) = un +1

2(un + un+1) τ

u(τ ) = un + unτ +1

4(un + un+1) τ

2

linear acceleration

u(τ ) = un + unτ +1

2(un+1 − un)

τ 2

∆t

u(τ ) = un + unτ +1

2unτ

2 +1

6(un+1 − un)

τ 3

∆t

(16)


MDOF numerical solution: Newmark’s method IIEvaluating at the end of time interval

constant acceleration

un+1 = un +1

2(un+1 + un)∆t

un+1 = un + un∆t +1

4(un+1 + un)∆t

2

linear acceleration

un+1 = un +1

2(un+1 + un)∆t

un+1 = un + un∆t +

(

1

6un+1 +

1

3un

)

∆t2

(17)

Both the constant acceleration case and the linear acceleration case can be put inthe same form in terms of two parameters β and γ defining the integration method:

un+1 = un + [(1− γ) un + γun+1]∆t

un+1 = un + un∆t +

[(

1

2− β

)

un + βun+1

]

∆t2

(18)

such that

constant acceleration γ =1

2and β =

1

4

linear acceleration γ =1

2and β =

1

6

(19)


MDOF numerical solution: Newmark’s method III

It is interesting that both methods can be put in a predictor-corrector

un+1 = [un + (1− γ) un∆t] + [γun+1∆t]

un+1 =

[

un + un∆t +

(

1

2− β

)

un∆t2

]

+[

βun+1∆t2] (20)

which can be rewritten as{

un+1 = ˜un+1 + γ∆tun+1

un+1 = un+1 + β∆t2un+1

(21)

where

˜un+1 = un + (1− γ) un∆t

un+1 = un + un∆t +

(

1

2− β

)

un∆t2

(22)


MDOF numerical solution: Newmark’s method IV

Assume to know solution at tn, want to solve at tn+1

Mun+1 + Cun+1 + Kun+1 = fn+1

Two options:

◦ Express un+1 and un+1 in terms of un+1 and solve for un+1

◦ Express un+1 and un+1 in terms of un+1 and solve for un+1

Second approach is particular interesting in case of nonlinear internal forces


MDOF Newmark’s method in terms of displacements I

From equation 212 we express

un+1 =1

β∆t2(un+1 − un+1) (23)

and using 211un+1 = ˜un+1 + γ∆tun+1

= ˜un+1 +γ

β∆t(un+1 − un+1)

(24)

Plugging back into equilibrium equation we obtain

M

[

1

β∆t2(un+1 − un+1)

]

+ C

[

˜un+1 +γ

β∆t(un+1 − un+1)

]

+Kun+1 = fn+1

i.e.[

1

β∆t2M+

γ

β∆tC+ K

]

un+1 =

[

fn+1 +M1

β∆t2un+1 + C

γ

β∆tun+1

]


MDOF Newmark’s method in terms of displacements II

Previous equation can be rewritten as:

Kun+1 = fn+1

where

K =

[

1

β∆t2M+

γ

β∆tC+ K

]

fn+1 =

[

fn+1 +M1

β∆t2un+1 + C

γ

β∆tun+1

]

and where we recall that

˜un+1 = un + (1− γ) un∆t

un+1 = un + un∆t +

(

1

2− β

)

un∆t2

(25)

Once solved for un+1, possible to compute new velocities and accelerations as

un+1 =1

β∆t2(un+1 − un+1)

un+1 = ˜un+1 + γ∆tun+1


MDOF Newmark’s method in terms of displacements III

Simple starting procedure: only need is to compute acceleration u0 = u(0)

u0 = M−1 (p0 − Cu0 − Ku0)


SDOF Newmark’s method: consistency I

Evaluation of local truncation error of velocity and acceleration shows that:

τ = O(∆tk)

with:

k = 2 for γ =1

2i.e second order

k = 1 for γ 6=1

2i.e first order


Possible exercises I

� Exercise. Modify the frame Matlab code to include some time-integration scheme.

� Exercise. Michael A. Koplow, Abhijit Bhattacharyya, Brian P. Mann “Closed form solutions for thedynamic response of EulerBernoulli beams with step changes in cross section”, Journal of Sound andVibration 295 (2006) 214225

� Exercise. Seon M. Han, Haym Bebaroya and Timothy Wei “Dynamics of transversely vibrating beams

using four engineering theories”, Journal of Sound and Vibration (1999) 225(5), 935-988


Date post:	17-Aug-2018
Category:	Documents
Upload:	dotruc
View:	258 times
Download:	0 times

Linear beams: extension to dynamics and ﬁnite … · Linear beams: extension to dynamics and...

Documents