earthquake action is a time-varying the structural responses...

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Chapter 3

Seismic Responses of SDOF and MDOF

2015/10/13

熊海贝，同济大学土木工程学院

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1

Earthquake action is a “time-varying” action.

The structural responses varied with time history,

which are all time-dependant varieties．

Solving these kinds of issues need dynamic analysis

method.

A structure can be simplified as a Single-Degree-Of-

Freedom (SDOF) system or Multiply-Degree-Of-Freedom

(MDOF) system.

The response may be linear response or nonlinear

response which depended on the structural material

stiffness (Nonlinear Material ) and structural deformations

(Nonlinear Structural Element).

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Keywords

SDOF（单自由度）, MDOF（多自由度）

Free Vibration, Forced Vibration（受迫振动）

Harmonious Vibration（谐振）,

Impulsive Vibration（冲击振动）,

Excitation（激励）, Response（反应）,

Time History（时程），Spectrum (反应谱)

Base Shear method（底部剪力法）,

Response Spectrum Method（振型叠加反应谱法）

Time History Method（时程分析法）

2015/10/13 熊海贝，同济大学土木工程学院

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Key Points

How to determine the eq. load (action) ?

How to transform a dynamic action to an equivalent static load?

What’s kind of equivalent condition?

Action equivalent？Response equivalent？

Acceleration / velocity / displacement equivalent ?

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Outline

3.1 Free Vibration of SDOF Systems

3.2 Forced Vibration of SDOF Systems

3.3 Numerical Analysis of Seismic Response of SDOF

3.4 Spectrum of Seismic Response of SDOF

3.5 Response of Nonlinear SDOF Systems （*）

3.6 Free Vibration of MDOF Systems

3.7 Response Specturm Method of MDOF

3.8 Earthquake Action and Responses of MDOF

3.9 Time History Method of MDOF

Dynamics 3.1 Free Vibration of SDOF Systems

3.1.1 Dynamic Model and Equilibrium Equation

3.1.2 Undamped Free Vibration

3.1.3 Damped Free Vibration

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2

How to determine the “Degree-of-freedom”

In mechanics, the degree of freedom (DOF) of a

mechanical system is the number of independent parameters that

define its configuration.

It is the number of parameters that determine the state of a

physical system.

Degree of freedom is a fundamental concept central to the

analysis of systems of bodies in mechanical engineering,

aeronautical engineering, robotics, and structural engineering.

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How to determine the “Degree-of-freedom” for such

kinds of system

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How to determine the “Degree-of-freedom” for such

kinds of system

Adjustable Slippage Element

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Newton’s 2nd Law

“The force acting on a body and causing its movement, is

equal to the rate of change of momentum in the body.”

• Momentum Q, is equal to the product of the body mass by its velocity.

xmdt

dxmmvQ


maxmdt

xdm

dt

dvm

dt

)mv(d

dt

dQF

D’Alambert suggested that Newton’s Law can be written

in a similar way that principle of equilibrium in statics

(F=0):

Inertial force

D’Alambert Principle

0maF


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x

m

k

0 mx

kx

Based on D’Alambert’s principle:

0 kxxm


Ordinate that describes

the mass movement

stiffness

mass

This is Dynamic Equilibrium Equation of undamped free vibration of SDOF

mass

3

Dividing by m and calling :

The solution is:

If, the initial conditions is:

Then:

0

0

0

(0)

(0)

t

x x

v v

2mk

homogeneous linear differential equation with constant coefficients. 常系数齐次线性微分方程

02 xx

tcosBtsinAtx

tcosxtsinv

tx

00


= frequency in radians/second (rad/s)

= frequency in cycles/second (Hz)

= period in seconds (s)

Period T

t

x

0x 0v

2f

fT 12


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But,

movement tends to decrease with time. This

reduction is associated with loss of energy

present in the system. Energy, kinetic or

potential, transforms in other forms of energy,

such as noise, heat, etc.

In dynamic system this loss of energy is known

as, Damping.


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There are several types of damping:

Viscous - Proportional to movement velocity

Coulomb - Caused by friction

Hysteretic - For materials working in the

nonlinear range where loading curve is different from the unloading curve.

Damping

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Fa

Viscous aF

x

F =cxa

u

F

Fy

Hysteretic

aF

uN

N

N

Coulomb

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3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration

x

m

k

0

c

mass

mx

kx

xc

0 kxxcxm

Damping Force

Inertia Force Restoring Force

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D’Alambert’s Principle:

4

Characteristic equation is:

Its roots:

Its solution:

常系数齐次线性微分方程的特征方程

ttBeAetx 21

m

mkcc

2

42

02 kcm


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When the radical is zero, we have Cc：

Define:

We obtain：

And the roots become:

042 mkcc

cc

c

mc 2

12

mkCc 2


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The solution is :

x

t

0x0v

tt BteAetx

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1cc

cWhen:

No mare vibration

critical damped

x

t

0x0v

ttt BetAeetx

11 22

The solution is :

Over damped

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When: 1cc

c

No mare vibration

The imaginary solution is:

Using Euler transformation:

titit BeAeetx

22 11

tsinDtcosCetx t 22 11

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1cc

cWhen: damped

Solving for the initial conditions:

x

t 0x

0v

aT

1

tsin

xvtcosxetx a

a

a

t

00

0

21a

21

22

a

aT

Damped

frequency

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5

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[3.1] A single story concrete factory with total mass concentrated on the roof, is 204 t, and the lateral stiffness of all columns is 8048.6 kN/m. Please determine the building’s natural period.

[Solution]

single story —— SDOF

Concrete factory —— Damping ratio=5%

( if it given by steel frame? , masonry wall?)

Roof mass —— M= 204 t，( if it given by weight? )

Example

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43390204

680482 ..

.

m

k＝＝＝

Undamped natural vibration circular frequency: 28.6＝

27.605.0128.61 22 ＝＝＝

Hz12f ＝＝

s1f1 ＝T

Damped natural vibration circular frequency:

Natural vibration frequency:

Natural vibration period:

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Example

[Solution]

Harmonic vibration :

a train of sinusoidal waves having a given amplitude．

Impulse vibration :

during a short time, the wave amplitude can be simplified as a constant value, or a triangular shape．

Two basic excitation

Harmonic loading

Impulse loading

No. 28

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3.2 Forced Vibration

undamped system subjected to simple harmonic loading

Where

is the amplitude and a is the circular frequency of

the harmonic load．

For a ground acceleration, the acceleration can be

represented as

The equivalent force amplitude as

and the frequency ratio

0F

( ) sin0 0F t mg t

oe 0F mg

/a

No. 29

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3.2.1 Harmonic Vibration

atFkxxm sin0

Equation of motion

Response

sin2 0n n

Fx 2 x t

m

tdtctbtaetx dd

tn sincossincos

Homogeneous solution a and b satisfy i.c.’s

Decays exponentially in time “Transient” response

瞬态反应

Particular solution c and d satisfy F(t)

Does not decay “Steady-state” response

稳态反应


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Steady-State Response

Steady-state response where and

Response is harmonic; lags behind load by

F0/k is static load displacement, so D is dynamic amplification factor

atDk

Ftx sin0

2

1

1

2tan

a

a

2

2 2

n n

1D

1 2

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a a

Dynamic amplification factor, 𝐷

0

1

2

3

0 1 2 3

Frequency ratio, / n

D

= 0.2

= 0.1

= 0.5

= 0.3

Resonance

1D

2

Static

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𝛽 = 𝑎/𝜔

ξ

ξ=0.1

ξ=0.3

ξ=0.5

t t

R(t)

Undamped system

R(t)

Damped system

0 0

Resonance response 共振

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Harmonic excitation

No. 34

The 6th National Structural Competition

3.2.2 Impulsive Vibration

(3-1)

式中：

=惯性力=

=阻尼力=

=弹性恢复力=

=时效力（随着时间而改变的外力）

为总加速度, 和是相对于地面的位移和速度。

Fi d sF F P t 0

No. 35

cxdf

sf

tp

x x

if mx

x

kx

运动平衡方程


冲击荷载作用下的结构的近似响应

No. 36

'

'sint Pd

x t e tm

瞬时冲量

S Pd

初始速度

/0v S m

结构反应 —— 初始速度下的有阻尼自由振动


7

冲击荷载下的自由振动一系列冲击荷载产生的结构振动

..

0

( )t d

t

g

Px

m

＝

地面运动可表示为


A series impulsive excitation —— Base Excitation

( ) ( ) ( ) - ( )gmx t cx t kx t mx t

No. 38

2( ) 2 ( ) ( ) - ( )gx t x t x t x t

=2

c

m

k=

m

P( ) - ( )gt mx t

Integrate of a series impulsive excitation


At moment, τ, the impulsive loading is P(τ), under this action, the responses of a SDOF is ；

''

'

( )sin ( )

t Pdx t e t d

m

..

: ( )= - gif P m x

''

'0

( )sin ( )

t t Px t e t d

m

' ..'

' 0

1- sin ( )

t tgx t e x t d

Duhamel integral


At moment, τ, the impulsive loading is P(τ), under this action, the responses of a SDOF is ；

''

'

( )sin ( )

t Pdx t e t d

m

..

: ( )= - gif P m x

''

'0

( )sin ( )

t t Px t e t d

m

' ..'

' 0

1- sin ( )

t tgx t e x t d

Duhamel integral


D’Alambert’s Equilibrium Equation:

)(tPkxxcxm

( )P t( )P t

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Arbitrary excitation （随机激励）

3.2.3 Arbitrary excitation

x

k

0

c

mass

mx

kx

xc

Dynamic Equilibrium Equation

For a dynamic or time-dependant force p (t)，the

equation of static equilibrium becomes one of dynamic

equilibrium:

where a dot represents differentiation with respect to time.

Equation must be satisfied at each instant of time during

the time interval under consideration．

mx cx kx p t

No. 42

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8

Where

=inertia force=

= damping force =

= elastic restoring force = k x

= time-dependent applied force

is the total acceleration of the mass, and ,

x are the velocity and displacement of the mass

relative to the base.

tpfff sdi

cxdf

sf

tp

x x

if mx

No. 43

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Displacement Response under Arbitrary excitation

Duhamel Integral (杜哈密积分)

dtsineFm

tx t

t

2

02

11

1

dt

F( )F t

p( )

d

p(t)

dv( )t

t'=t-

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Base excitation (Earthquakes)

x

y

ground

mass

k c

y

x

ground

mass

k c

gx

gxmkxxcxm 2015/10/13


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3.2.4 Base excitation

And dividing by m

gxmkxxcxm

Then gxxxx 22

dtsinextx t

t

g

2

02

11

1

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Base excitation (Earthquakes) x

ground

mass k

c

gx

3.2.4 Base excitation

Basic Principles

From the moment, (k-1)

Derivate the moment, k

principles：linear acceleration increase

Method: linear acceleration method,

Newmark-β method, and Wilson-θ method.

-1 -1 -1( ), ( ), ( )k k kx t x t x t

( ), ( ), ( )k k kx t x t x t

( ), ( ), ( )x t x t x t

-1 -1 -1( ), ( ), ( )k k kx t x t x t

( ), ( ), ( )k k kx t x t x t

t

-1kt kt

3.3 Numerical Analysis of Seismic Response

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k-1k-1 k k-1

k k-1

t - tx t x t + x t - x t

t - t ＝

k-1 k-1 k-1

t t tk k-1

k-1 k-1t t t

k k-1

x t - x tx t dt x t dt + t - t dt

t - t ＝

2

k-1

k-1 k-1 k-1 k k-1

k k-1

t - t1x t x t + x t t - t + x t - x t

2 t - t ＝

2 3k k-1

k-1 k-1 k-1 k-1 k-1 k-1

k k-1

x t - x t1 1x t x t + x t t - t + x t t - t + t - t

2 6 t - t＝

t

区段[ tk-1, t ]积分


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9

When let, kt t＝

= = = = = - k k k-1 k-1 k k k k k k-1x x t x x t x x t x x t t t t，，，，，

k k-1 k-1 k k-1

k k-1 k-1 k

1x x + x t + x - x t

2

1 1x x + x t + x t

2 2

＝

＝

Then

2 2

k k-1 k-1 k-1 k

1 1x x + x t + x t + x t

3 6 ＝

k-1k-1 k k-1

k k-1

t - tx t x t + x t - x t

t - t ＝


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Let, k-1 k-1 K-1

1x + x t B

2 ＝

2

k-1 k-1 k-1 k-1

1x + x t + x t A

3 ＝

Then, we can get,

k k-1 k

tx B + x

2

＝

（3.20）

2

k k-1 k

tx A + x

6

＝（3.21）


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satisfy the dynamic equilibrium equation,

22 2

k k-1 k k-1 k g k

t tx +2x w B +2x w x +w A +w x - x

2 6

＝

22

t6

t1s

＝Let,

then

k kB As

2

k g k 1 1

1x x 2

＝

（3.22） We can obtain, , use same way to solve all results

( ), ( )k kx t x t


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[Example 3.2] Consider a SDOF system with a mass of 200t,

a lateral stiffness k of 7200 kN/m, and a damping ratio ζ=0.05.

The record of ground motion acceleration is shown below.

Please calculate the displacement, velocity, acceleration,

maximum absolute acceleration, and maximum horizontal

seismic action of the mass using linear acceleration method

(Take the time step as 0.1s and ends at 1.2s).

Table 3-1

2cm/sgx

t 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2

108 207 300 200 95 0 -150 -198 -120 -18 0 0


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[Solution] Natrual frequency

62007200mk ＝＝

091106

61060501

61

22

22

....tts

Conduct iterative calculation according to Eqn. (3.20),

(3.21) and (3.22) (Please refer to the textbook for detail to

obtain the displacement, velocity, acceleration in Table 3.2).


2

max aF S m 544.03 10 200 1088 kN ＝＝＝

Please

Review Dynamics

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earthquake action is a time-varying the structural responses...

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