第一讲结构的阻尼

第一讲，结构的阻尼结构动力学基本方程：

MX

CX

KX

0(P(t))1

阻尼的分类(Category

of

damping)2

m

m2

m(

/

D)

2

m,其中，

m

ln(

)1

xst

0

p02

x0

,

xst

0

3,半功率带宽法

确定粘弹性阻尼的试验方法粘弹性阻尼（与速度成正比

反向）的测量方法1,自由振动衰减法xnxn

m

3

2,共振峰处放大系数法

k

n

4,等效阻尼法Reference:

Damping

Characteristics

of

a

Footbridge

-

Mysteries

and

Truths

R.

Cantieni,A.

Bajric,

R.

Brincker,

IMAC-XXXIV,

2016;

Paper

no.

487.ED

d(x)dx

f

(cx)xdt

粘弹性阻尼的物理意义单自由度体系(SDOF)稳态振动情况下:

4

TD0

TD0fd(x)dx

p0x0

sin(

)

mx

cx

kx

f

(t)

p0sin(ωt)粘弹性阻尼一周耗散的能量：外力输入的能量：

TD

0

TD

0x

x0sin(ωt

φ)2

c

x02

c

x0ES

fs(x)dx

(kx)xdt

粘弹性阻尼的物理意义

5

mx

cx

kx

f

(t)

p0sin(ωt)势能

(应变能:

strain

energy)：

TD

0x

x0sin(ωt

φ)

2

0

k(x

0sin(

t

))(

x

0cos(

t

))dt

0动能

(kinetic

energy)：

TD

0

0

x

1粘弹性阻尼耗散的能量

6x

x0sin(ωt

φ)22

2

2

mx

cx

kx

f

(t)

p0sin(ωt)

fd(x)

cx

cωx0cos(ωt

φ)

cω

x0

x0

sin2(ωt

φ)

cω

x0

x2(t)

2

2

f

x0

cωx0

A

πcωx0

kx

c

x0

x0

sin

(

t

)

x

f

kx

1粘弹性阻尼耗散的能量

72

x

x0sin(ωt

φ)Aπcωx0(

plastic

deformation

)2

mx

cx

kx

f

(t)

p0sin(ωt)

试验中测量的总抗力：

fs(x)

fd(x)

kx

cx

2

2

kx

c

x0

x2(t)

2

2

x0

cωx0

讨论：ω

0,

A

0

有塑性变形ED

fDdu

(cu

)u

dt

cu2dt

2

/

0

2

/

0

试验方法

-

等效阻尼法等效阻尼法为最普通的确定结构阻尼的方法，它利用一个循环内的力和位移曲线（滞回曲线）。

滞回曲线所围的面积为结构耗散

的能量：

2

/

0

1应变能(Strain

energy)：

2u0fDEs0

nED

4

eq.

1

ED4

Es0

1

1

ED4

/

n

Es0

eq.

8Reference:

Estimating

Effective

Viscous

Damping

and

Restoring

Force

in

ReinforcedConcrete

Buildings,

Hesam,

A.

Irfanoglu,

T.J.

Hacker,

IMAC-XXXIV,

2016;

Paper

no.

501.E

π92

粘滞阻尼

(Hysteretic

damping,

rate-independent

damping试验发现：1，周期加载耗散能量不与频率相关；

2，结构各阶阻尼比相当，差别不大。

ED

πcωx0xηk

ω

fD

22

HDωx0ηk

ω

πηkx0

2πηES0

结构外阻尼产生的原因

(EXTERNAL

MECHANISMS

OF

DAMPING)1.

Acoustic

radiation

damping,

whereby

the

vibrational

response

couples

with

the

surrounding

fluid

medium,

leading

to

sound

radiation

from

the

structure.2.

Fluid

pumping,

in

which

the

vibration

of

structure

surfaces

forces

the

fluid

medium

within

which

the

structure

is

immersed

to

pass

cyclically

through

narrow

paths

or

leaks

between

different

zones

of

the

system

or

between

the

system

and

the

exterior,

thereby

dissipating

energy.3.

Coulomb

friction

damping,

in

which

adjacent

touching

parts

of

the

machine

or

structure

slide

cyclically

relative

to

one

another,

on

a

macroscopic

or

amicroscopic

scale,

dissipating

energy,

and

etc.10

结构内阻尼产生的原因

（INTERNAL

MECHANISMS

OF

DAMPING）1.

结构变形后有很多机制以热的形式耗散能量，都与物

质微观结构内部的原子或者分子重建，或者热效应有

关。在特定的条件下，对某种材料（金属，合金等）

只有其中的一种或者两种机制起主导作用。大多数金

属和合金在多数情况下阻尼很小，个别合金由于晶格

结构特殊阻尼很大。2.

结构的阻尼很难确定的原因还在于结构内部的接头和

接触面，以及部件连接处和支撑。3.

由于这些原因，通常几乎不可能或者显然很不容易，

准确地确定和控制一个结构的初始阻尼水平，具体何种机制产生也很难区分。11常用材料的阻尼钢结构的阻尼系数为多少与其工作状态有关，一般试验确定。12

规范的规定1.建筑结构抗震规范

GB

50011-2001的8.2.2条也有说

明,全文如下:钢结构在多遇地震下的阻尼比,对不超过

12层的钢结构可采用0.035,对超过12层的钢结构可采

用0.02;在罕遇地震下的分析,阻尼比可采用0.05.2.在日本道路公团2005年版的设计要领中,关于阻尼做如

下规定:

对于直接承受动荷载的桥梁上部结构，一般

不希望其工作在弹塑性阶段，阻尼系数取0.02~0.03。对于在大地震时可能工作在弹塑性阶段的钢下部结构，当其在弹性域范围内工作时，阻尼系数取0.03~0.05，当其工作在弹塑性域，且采用等价线性化模型解析时，阻尼系数取0.1~0.2。13a

2

1/

瑞利(Rayleigh)阻尼瑞利(Rayleigh)阻尼，指定两个固有频率处的阻尼，

其它处的阻尼也随之确定。c

a0m

a1k

a02

n

1

n

2

n

m

1

1/

m

m

a0

n

n

n

a1

优点：1，能够有效解耦；2，计算简单。缺点：两个参数决定所有阻尼，且其余各阻尼相关。c

a1kc

a0m14

,K

610

1瑞利(Rayleigh)阻尼习题

1：

1

1

1

2

1

2

400

400

400

1

386

m

已知结构为瑞利阻尼，给定第一阶和第二阶阻尼比为5%,求第三阶的阻尼比？1516John

Strutt,

3rd

Baron

Rayleigh

John

William

Strutt,

3rd

Baron

Rayleigh,

OM

(Nov.12,1842

–

June,30,

1919)

was

an

English

physicist

who,

with

William

Ramsay,

discovered

the

element

argon(Ar-18

氩气),

an

achievement

for

which

he

earned

the

Nobel

Prize

for

Physics

in

1904.

He

also

discovered

the

phenomenon

now

called

Rayleigh

scattering,

explaining

why

the

sky

is

blue,

and

predicted

the

existence

of

the

surface

waves

now

known

as

Rayleigh

waves.

Rayleigh's

textbook,

The

Theory

of

Sound,

is

still

referred

to

by

acoustic

engineers

today.

Rayleigh

ratio

(瑞雷商)；Rayleigh-Ritz

method

(瑞雷-瑞兹法)；

c

m

M

T(m

M

1)

I

Tc

(m

M

)C(M

m)

m(

n2

n

n

T

1C

T

1T(M

1

Tm)

I

1

(M

1

Tm)

1

(m

M

1)Mn

N

1

1

T习题

2：质量和刚度阵如习题

1，给定第一阶和第二阶阻

n

1尼比为5%,用叠加法求阻尼矩阵，和第三阶阻尼比？提示：先计算阻尼矩阵、对角化后再计算阻尼比。

模态阻尼矩阵的叠加法

(superposition

of

modal

damping

matrix)可以指定前n阶模态的阻尼比

Tc

C,Cn

n2

nMn

n)m

17教师联系方式：姓名：李东升地址：4号试验楼213室Tel:84708402Email:

dsli@答疑时间：周五

10:00

–

11:00课件下载地址：/s/1bntqAhL密码：0xrm

18c

m

al[m

1k]l两边左乘：

nkm

n

al

n

2l

1

两边左乘：

n(km

)继续左乘：

n(km

)l

0时：

T

nc0

n

n

T(a0m)

n

a0Mn

l

1时：

c

n

(a1k)

n

a1

Mnl

2时：

c

n

(a2km

k)

n

a2

n

n

a2

n

n

k

MCn

al

n

2lMnrrN

1

l

0rn

n

N

1

l

0

T

1

T

1

2

T

1

lcl左乘：I

mm

1

N

1第n阶模态的模态阻尼为：

Cn

Tnc

n

al

Tncl

n

l

0T

T

2n

1

n

nn

2

n

nT

T

1

2

T

4

Caughey

阻尼(扩展的瑞利阻尼)Caughey

阻尼可用于指定多于两个模态的阻尼比，是瑞利阻尼的扩展。19

Cn

1

N

1

k

r

2m

r

2Mn

n

2

l

0

Tn[km

1k]

r

2

Tnk

r

0,n

r.

Tn[(km

1)2k]

r

2

Tn[km

1km

1m]

r

0

T[(km

1)l

1k]

r

0

Tcl

r

cl

(km

1)l

1kcl

mm

1km

1km

1

km

1k

m[m

1k]l

1

1

1

N

a0

2

1

1

Cn

a1

2

2

2

n

2Mn

n

N

2M

2

aM

1

2

N

N

l

0

al

n

2l

1

a1

22

23

M1

M

2

M

3

2M

2

2

2

MM

2

N

N

11

12

13

21

1M

2

1

1

a0

aM

1

4

21

2

2

2

Caughey

阻尼的计算

4

2M

2

1

1

4

2M

2

2

2

N

矩阵的逆Explicit

solution20

M

r

1Reference:

T.

K.

Caughey

(JPL,

Caltech)，Classical

Normal

Modes

in

DampedLinear

Dynamic

Systems，Journal

of

Applied

Mechanics，1960;

27(2):269

–

271.1

N

122005;

12:1–2.Thomas

K.

Caughey21Professor

of

Mech.Engr.

Caltech,

passed

awayrather

suddenly

on

7

December

2004,

at

the

ageof

77.A

Scotsman,

while

an

undergraduate

student

inScotland,

he

solved

all

the

problems

in

thefamous1940

textbook

‘Mathematical

Methods

inEngineering’,

by

Theodore

von

Karman

andMaurice

Biot,

a

book

used

in

the

1950’s

as

areference

for

a

graduate

course

at

Caltech.PhD

in

1954

at

Caltech,

assistant

prof.(1954),

prof.(1962).Contributions:

applied

mathematics,

dynamics

and

control

theory.Design

of

the

Caltech

eccentric-mass

vibration

generator

in

theearly

1960’s,

振动台试验的开始。Freudenthal

Medal,

Karman

Prize

by

ASCE

in

2002.Reference:

T.

K.

Caughey

(JPL,

Caltech)，Obituary，Struct.

Control

Health

Monit.Cn

al

n

2lMn1

a

n

2

n

l

01

2

2

1

N

2

4

2

2M

2

a1

2

2

2

aM

1

2

N

N

2M

2

Reference:

J.

Enrique

Luco，A

note

on

classical

damping

matrices，Earthquake

engineering

and

structural

dynamics，2008;

37:615–626.

习用阻尼模型存在的问题常用阻尼的缺点：1，The

Rayleigh

damping

matrix

is

simple

and

banded

but

the

damping

ratios

foronly

two

modes

can

be

specified.2，A

Caughey

series

representation

is

more

general

but

requires

solution

of

apotentially

ill-conditioned

system

of

equations

to

determine

the

coefficients

of

theterms

in

the

series

in

terms

of

the

prescribed

damping

ratios.3，The

superposition

of

modal

damping

matrices

is

also

simple

but

requirescalculation

of

the

mode

shapes

and

leads

to

zero

damping

for

the

modes

for

whichthe

modal

damping

ratios

are

not

specified.

Classical

dampingcm

1k

km

1c,

Tnc

r

2

r

rMr

rsM

1

l

0M

12ll

n44

22

1

1

2

1

1

M

2

a0

2

1

1

N

NVandermonde’s

matrix

,

ill-conditioned2

4

2

4

4

2

4

2

4

2

2

4221

2

2

1

N

2

4

2

2M

2

a1

2

2

2

N

2M

2

aM

1

2

N

N

a1

22

23

M1

M

2

M

3

2M

2

2

2

MM

2

N

N

al

2

lr

r

r

Mr

1

11

12

13

21

1M

2

1

1

a0

aM

1

44

22

1

1

2

阻尼研究的新进展-1

1

1

M

2

a0

2

1

1

N

矩阵的逆2321

M

12

l

0al

nl

1

n

阻尼比由高阶项控制

l

0

al

n

2l

1al

2

lr

r

r

lr

(

1)

/

(

s

2

r

2),(l

0:M

1;r

1:M)阻尼研究的新进展-2r

r

s1

s

r

r

s

0

1,

m

2

2

2

1

2

1

2

m,(1

m

M

1)

sn

rl

r

Ms

1s

rM

l

1

Mr

11

M

12

n

1

2

2

21

2

2

2

2

2

2

22

21例如：

r

=1,

0

1

1

2

3

M

2

2

3

2

4

2

M

3

4

3

5

M

1

2

3

4

M解析解

见前面矩阵左乘以其逆。基本对称函数24c

m

al[m

k]

,(1)al

2

lr

r

r,(2)[L]

m

lr[m

1k]l,(4)/

(ωs

2

ωr

2),(l

0:M

1;r

1:M)变换符号及积分限：l

M

m

1,1)l

0,m

M

1;[L]

m

(

1)

(m

k)/

(

s

r

2),(5)

阻尼研究的新进展-3

Mr

1

1

lM

1

l

0将(2)代入

(1)：

M

1

M

M

M

1

M

1

l

1

l

l

0

r

1

r

1

l

0

r

1M

1

l

0M

1rσ

Ms

1s

rrM

l

1M

m

1αlr

(

1)2mM

1rm

r

1M

m

1M

1m

0

Ms

1s

r

m

M

l

1.

2)l

M

1,m

0.25[L]

m

(

1)

(m

k)/

(

s

r

2),(5)

(

1)

m(m

k)M

m

1

0(m

k)M

1

1

(m

k)M

2

M

r

2(m

k)

M

r

1I

(m

k

s

2I)c

2

r

r[LMr

1],(3)[LMr

1]

m

(m

1k

s

2I)/

(

s

2

r

2),(6)阻尼研究的新进展-4m

1

r

1

1

1M

1m

0

1

Ms

1s

r2mM

1rm

r

1M

m

1M

1m

0

Ms

1s

rr

r

s1

s

r

r

s

0

1,

m

2

2

2

1

2

1

2

m,(1

m

M

1)

sn

r例如：

r

=1,1

2

2

21

2

2

2

2

2

2

22

2

1

2

3

M

2

2

3

2

4

2

M

3

4

3

5

M

1

2

3

4

M

M

Ms

1

s

1s

r

s

r伟达定理：方程根与系数关系

M

r

1确定阻尼矩阵公式26c

2

r

r[LMr

1],(3)阻尼研究的新进展-5

Mr

1确定阻尼矩阵公式

M

M

s

1

s

1

s

r

s

r1,

The

damping

matrix

[c]

defined

by

Equations

(3)

and

(6)

represents

afactorized

Caughey

series

and

hence

leads

to

classical

normal

modes.2,

The

modal

damping

ratios

for

the

first

M

modes

correspond

exactly

andexplicitly

to

the

prescribed

M

modal

damping

ratios

r.3,Although

the

calculation

of

the

matrix

products

appearing

in

Equation

(6)requires

some

effort,

the

potential

numerical

problems

arising

from

a

numericalsolution

of

the

ill-conditioned

Vandermonde’s

matrix

are

eliminated.4,Finally,

even

though

the

modal

damping

ratios

of

only

M

modes

are

specified,the

damping

ratios

of

the

other

modes

are

non-zero.

271

al

2

lr

r

r

lr

(

1)

/

(

s

2

r

2),(l

0:M

1;r

1:M)

r

r

(

1)

/

(

s

r

2)

r

r

n)

(

s

n

2)/

(

s

r

2),(M

n

N)(

/

阻尼研究的新进展-6指定的M个阻尼比之外的阻尼比的计算方法：l

r

Ms

1s

rM

l

1

Mr

12

M

12

l

0al

nl

1

n

M

M

M

1

2l

1

2l

12

l

0

r

1

r

1

l

0282lM

M

1r

2l

1M

l

1

nr

1

l

0

Ms

1s

r2

2r

1

s

1

s

1M

M

M

s

r

s

r

1

,K

k

c

2

r

r[LMr

1],(3)[LMr

1]

m

(m

1k

s

2I)/

(

s

2

r

2),(6)习题

3：

1

1

1

2

1

1

2

111

1

2

2

1m

m

2

3

M

Ms

1

s

1s

r

s

r设定第一阶，第二阶和第三阶阻尼比为：

1求阻尼矩阵，和第四阶的阻尼比？

M

r

129工程经验：•结构各阶阻尼比相差不多。•阻尼的非严格线性，随结构振动幅度的变化而有所改变，但相差不大。强迫振动和环境振动测试时阻尼比识别结果不同。30不同识别方法阻尼比识别结果不同，原因不明。31Reference:

Damping

Characteristics

of

a

Footbridge

-

Mysteries

and

Truths

R.

Cantieni,A.

Bajric,

R.

Brincker,

IMAC-XXXIV,

2016;

Paper

no.

487.

经典阻尼模型存在的问题1,

Solution

of

a

potentially

ill-conditioned

Vandermonde

system

of

linearequations2,

Extreme

care

must

be

exercised

to

select

the

number

of

frequenciesand

the

specific

set

of

frequencies

at

which

the

modal

damping

ratios

areprescribed3,

The

resulting

damping

matrix

may

be

a

full

matrix

thus

impacting

thecomputational

effort

and

complicating

the

physical

interpretation

of

thedifferent

elements

of

the

damping

matrix.2M

1

l

0al

nl

12

n

n

References:[1]

J.

Enrique

Luco，A

note

on

classical

damping

matrices，Earthquake

engineeringand

structural

dynamics，2008;

37:615–626.[2]

LETTER

TO

THE

EDITOR,

Earthquake

Engng

Struct.

Dyn.

2008;

37:1801–1804

1:

可以扩展为

指数形式；

2：给出一个简便的多项式方法求系数a;[3]

AUTHOR’SREPLY,

Earthquake

Engng

Struct.

Dyn.

2008;

37:1805–1809

1:

原文为

Lagrange

interpolation

–

Sylvester

formula

32增减后pivot

需要改变等。。。c

(m

M

)C(M

m)

m(

n2

n

n

T经典阻尼模型存在的问题33Mn

n)m

Nn

1叠加法的阻尼

1

1

T

抛弃比例的要求，假定一个振型，则可得

N

n

1

MnReferences:Eduardo

Kausel，Damping

Matrices

Revisited，Journal

of

EngineeringMechanics，2014;

04014055.阻尼矩阵的应用、问题及解决例子科研中的讨论、以及进步Engineering

Mechanics，2009,135(11):1248–1256.

阻尼矩阵的对角近似非对称和复阻尼矩阵出现的场合：1，Nonsymmetric

stiffness

and

damping

matrices

can

arise

when

structures

areactively

controlled,

and

in

areas

like

microdynamics.2,

Damping

matrices

adduced

from

experimental

measurements

often

turn

out

to

benonsymmetric,

and

are

often

used

in

modeling

complex

systems.3,

Matrices

with

complex

entries

can

arise

when

dealing

with

modeling

systemsthat

have

structural

and

viscoelastic

damping.非对称和复阻尼矩阵的工程处理方法：A

common

procedure

in

structural

engineering,

is

to

retain

only

the

diagonalelements

of

C,

zero

out

the

off-diagonal

elements,

and

thereby

obtain

a

newdamping

matrix,

C1=diag(C),

for

which

the

system

is

now

uncoupled.One

might

want

to

know

how

good

this

approximation

of

the

matrixsolely

by

its

diagonal

elements

might

be.

To

measure

“closeness”

of

two

matrices

weshall

use

the

Euclidean

Frobenius

norm.

Reference:

F.E.

Udwadia，A

Note

on

Nonproportional

Damping，Journal

Of34Engineering

Mechanics，2009,135(11):1248–1256.

阻尼矩阵近似的四个有用结果Result

1.

When

all

the

eigenvalues

of

the

matrix

K

are

distinct,

of

all

the

matrices

thatcommute

with

k,

the

matrix,

Ed,

that

comes

“closest”

to

the

matrix

E

in

Euclideannorm,

is

obtained

by

simply

deleting

all

the

off-diagonal

terms

of

E.Result

2.

When

all

the

eigenvalues

of

the

matrix

K

are

not

distinct

but

not

all

equal,

ofall

the

matrices

that

commute

with

,

the

block-diagonal

matrix,

Ed

comes

closestin

Euclidean

norm

to

the

matrix

E.

The

size

of

each

subblock

Ei

along

the

diagonal

ofthe

matrix

Ed

equals

the

multiplicity

of

the

corresponding

eigenvalue

of

K.Result

3.

Let

the

matrix

K

be

hermitian,

and

let

D

and

K

not

commute.

If

theeigenvalues

of

the

matrix

K

are

all

distinct,

then

of

all

the

mat