con4337a2to131007bl_ch3_dynamicanalysis

STRUCTURES AND TALL BUILDINGS (CON4337)

Chapter 3 1

HD in Civil Engineering

│CHAPTER 3│

Introduction to Structural Dynamics

Learning Objectives

Application of lumped mass idealization.

Identification of simple and multi-degree of freedom systems of vibration.

Comparison of free and forced vibrations, resonance and damping effects.

Approximation of fundamental frequencies and mode shape of simple structures for seismic design.


Chapter 3 2


3. Introduction to Structural Dynamics

3.1 Introduction

Dynamic analysis is similar to static analysis except that there is the extra

dimension of time to take into account. Dynamic analysis also involves the

solution of equilibrium equations but with variation in time. The equations

have additional terms proportional to velocity and acceleration.

Figure 1 – Lumped Mass Model for Dynamic Analysis


Chapter 3 3


Fortunately, in many design problems, it is not necessary to obtain the member

forces and stresses at all instants of time. When structures vibrate, they tend to

adopt smoother deflected shapes than under static loads and the corresponding

stresses are more evenly distributed. These considerations make it possible to

model structures in less fine detail than for static analysis and still obtain

accurate results.

Figure 2 – Earthquake-Induced Motions in a Mutli-storey Building


Chapter 3 4


Source of Excitation

The most common types of dynamic loading on structures are: wind,

earthquakes, human forces, moving vehicles, industrial machinery, blasting and

pile driving. It is standard practice to use equivalent static horizontal forces

when designing buildings for wind and earthquake resistance. This is the

simplest way of obtaining the dimensions of the structural members. Dynamic

calculations may follow to check and perhaps modify the design.

Human forces, in the form of crowd loading, are almost always treated as static

distributed loads. But observations at a pop concert or football ground can

demonstrate the highly dynamic nature of the loading. Moving vehicles can be

designed for by adding an allowance for impact to their static weight. This has

proved satisfactory for the design of highway and railway bridges. But the

procedures may not be justified for load moving at very high speeds.

Effects of structural vibrations

Vibration of structures is undesirable for the following reasons:

i) Structural damage,

ii) Cracking and other damage to non-structural elements

iii) Effect on equipment

iv) Human discomfort

v) Fatigue fracture.

With modern forms of construction, it is feasible to design buildings to resist the

forces arising from major earthquakes. The essential requirement is to prevent

total collapse and consequent loss of life. For economic reasons, however, it is

the usual practice to accept some level of damage which may be repaired. For

an important structure such as a nuclear plant, there are considerations other

than structural effects – damage to nuclear equipment can cause dangerous

consequences, not to mention major financial losses due to down-time. Some

industrial plants with delicate manufacturing processes, e.g. microchip

production, also require special vibration control.

Vibrations in a building may also cause significant human discomfort. The

human body is quite sensitive to vibration and amplitudes as low as 0.05 μm

may be detected by the fingertips. In an industrial plant, it may be necessary to

check whether the vibrations caused by industrial machinery exceed human

tolerance limits. Design of tall buildings for wind effects may also have to

consider whether the wind-excited vibrations disturb the occupants to an

unacceptable degree.


Chapter 3 5


Vibration Control in the Design of Structures

There are three steps necessary in the design of structures that are susceptible to

vibration:

a) Identifying the frequency and amplitude of the dynamic loads,

b) Analyzing the structural response to obtain frequencies, dynamic

deflections and stresses,

c) Checking calculated performance against specified criteria to ensure that

there are no adverse effects.

Control over the natural frequency of a structure is possible by increasing the

stiffnesses and/ or reducing the mass and hence keep the frequency of the

structure above the predominant loading frequency.

There are many applications where vibration isolation is required. Delicate

equipment may require mounting on suspensions for protection against

vibrations. Machines are also sometimes mounted on springs to isolate them

from their supports.

3.2 Systems with single degree and multi-degrees of freedom

The most basic vibration system consists of a single lumped mass and a spring.

This is said to have one degree of freedom. That is to say, there is only one

possible direction of movement of the lumped mass.

It may seem that such as a crude approximation could produce little in the way

of practical information. In fact, they can be very useful for two reasons.

Firstly, skilled engineers can make good predictions of the behavior of real

structures by intelligent choice of the parameters of simple systems. Secondly,

the complicated motions of a real structure can often be resolved into simpler

motions of a number of mass spring systems and dealt with by superposition.

A sound understanding of the vibration of single degree of freedom (SDOF)

systems is also essential groundwork for understanding the dynamic behavior of

more complicated systems.


Chapter 3 6


Figure 3 – Modeling of Structures for Dynamic Analysis

In many cases, it is impractical to include all the features of a structure in a

dynamic analysis. One method of idealizing a structure is to concentrate the

mass at various discrete points around the system. This type of representation

is termed lumped mass idealization.


Chapter 3 7


Figure 4 – Lumped Mass Model for Dynamic Analysis

The complexity of a structure may be defined by the number of dynamic degrees

of freedom describing its motion. When the number of degrees of freedom is

more than one, the system is called a multi-degree of freedom system.


Chapter 3 8


3.3 Damping ratio

The effect of energy dissipating forces that resists the vibratory motion of a

structure is called damping. Damping forces include internal friction and other

effects, which are complex in nature and difficult to determine exactly.

Figure 5 – Effects of Damping on Vibration


Chapter 3 9


In engineering practice, the energy loss mechanism in a structure is usually

modeled by a viscous damper (in the form of a dashpot), in which the damping

force is a linear function of velocity.

Damping causes decay in the vibratory motion. As damping is increased, the

vibrations decay exponentially. In fact, at a certain level of damping, the

vibratory motion ceases altogether – at which the system is called critically

damped.

In most practical structures, damping is very light and is only a small fraction of

the critical damping value. It is convenient to express the damping in a

structure by a dimensionless value called the damping ratio ξ, which is the ratio

of the damping constant for the structure to the corresponding critical damping

value:

Damping ratio ξ = constant damping Critical

system for theconstant Damping

Typical damping ratios for structures

Type of construction

Damping ξ, percent of critical

Steel frame, welded, with all walls of flexible

construction

2

Steel frame, welded or bolted, with stiff cladding

and all internal wall flexible

5

Steel frame, welded or bolted, with concrete shear

walls

7

Concrete frame, with all walls of flexible

construction

5

Concrete frame, with stiff cladding and all

internal wall flexible

7

Concrete frame, with concrete or masonry shear

walls

10

Concrete and/ or masonry shear wall buildings

10

Timber shear wall construction

15


Chapter 3 10


3.4 Dynamic properties of a single degree of freedom system

The properties of a single degree of freedom system that directly affects its

dynamic response are its mass m, stiffness k and damping ratio ξ.

Free Vibration (No damping)

The free vibration response of a SDOF system is shown below. Free vibration

takes place when a structure vibrates on its own after being imparted an initial

displacement and velocity at time zero.

Figure 6 – Free Vibration of Structural System1

The portion a-b-c-d-e of the displacement-time curve is a typical cycle of free

vibration of the system. From its static equilibrium (or undeformed) position at

a, the mass moves to the right, reaching its maximum positive displacement u0

at b, at which time the velocity is zero and the displacement begins to decrease.


Chapter 3 11


The mass returns to its equilibrium position c, continues moving to the left,

reaching its minimum displacement -u0 at d, and then the displacement

decreases again, returning to the equilibrium position at e.

The time required for the undamped system to complete one cycle of free

vibration is called the natural period of vibration T (seconds) of the structure.

It is related to the natural circular frequency of vibration ω (in units of radians

per seconds) and the natural cyclic frequency of vibration f (cycles/sec or Hertz

or Hz) as follows:

Natural Period of vibration T

2 (sec)

Natural Cyclic Frequency f

2T

1 (cps or Hz)

The natural circular frequency ω is a function of the mass and stiffness of the

structure:

Natural Circular Frequency ω m

k (rad/sec)

Thus the free vibration properties ω, T and f depend only on the mass and

stiffness of the structure.

For two structures with the same mass, the one with greater stiffness will have a

higher vibration frequency and a shorter vibration period. Similarly, for two

structures with the same stiffness, the one with greater mass will have a lower

vibration frequency.

Damped Vibration

For a damped system under free vibration, the vibration amplitude decreases

with every cycle of vibration.

The period of vibration TD, circular frequency ωD and cyclic frequency fD of the

damped structure are interrelated in the same way as for the undamped structure:

Damped Period of vibration TD

D

2 (sec)

Damped Cyclic Frequency fD

2T

1

D

D (cps or Hz)


Chapter 3 12


Figure 7 – Damped Vibration1

The circular frequency ωD and the vibration period TD are related to the

corresponding undamped values by:

Damped Circular Frequency ωD 21 (rad/sec)

Damped Period of vibration TD 21

T

(rad/sec)

where ξ = damping ratio of the structure

Damping has the effect of lowering the natural circular frequency from ω to ωD

and increasing the natural period from T to TD. For most structures, the

damped values ωD and TD are approximately equal to the undamped values ω

and T respectively.

3.5 Types of Dynamic Loading

Almost any type of structural system may be subjected to one form or another of

dynamic loading during its lifetime. From an analytical standpoint, it is

convenient to divide dynamic loading into two basic categories, periodic and

non-periodic.


Chapter 3 13


A periodic loading repeats the same variation over a large number of cycles.

The simplest periodic loading has a sinusoidal or simple harmonic variation.

An example of periodic type loading is the effect of rotating machinery.

Periodic loading can be more complex, e.g. the forces generated by a ship

propeller. However, any periodic loading can be resolved into a series of

simple harmonic components by means of Fourier Analysis.

Figure 8 – Different Types of Dynamic Loads3

Non-periodic loadings may be either short duration impulsive loadings, e.g. an

explosion or long duration such as earthquake forces.


Chapter 3 14


3.6 Response to Periodic Loading

For structural design purposes, we are especially interested in the maximum

values of response (over time) of a system under dynamic loading since those

values determine the maximum forces in the structure. For periodic type

loading, the maximum displacements of single degree of freedom systems can

be assessed directly from response charts.

Consider the SDOF system shown below, which is subject to a periodic type

loading of P sin(ωp t). The maximum displacements of the structure, umax,

under vibration may be determined by obtaining the corresponding Dynamic

Magnification Factor from the chart below.

Figure 9 – Dynamic Amplification Factor for Different Damping Ratios


Chapter 3 15


where Dynamic Magnification Factor D = staticu

umax

and umax = maximum displacement under vibration

ustatic = displacement produced by a static force P

ωp = frequency of the applied force

ω = natural frequency of the structure

ξ = damping ratio of the structure

Note that the Dynamic Magnification Factor depends only on the frequency

ratio and the damping factor.

It can be observed that as the frequency ratio approaches 1.0, the dynamic

magnification factor becomes very large especially when damping is low (which

is true for most structures). When the applied force frequency and the natural

frequency of the structure almost coincide, the structure is said to resonate and it

will undergo maximum dynamic displacement. When an undamped structure

resonates, the displacement tends to infinity. It is therefore important to ensure,

as far as possible, that the applied force frequency does not approach the natural

frequency of the structure.

For buildings of low to medium height, the structural frequency is generally

different enough from that of wind loading such that the dynamic magnification

factor is practically 1.0. In the design of such structures, wind load can be

treated as equivalent static loading.

Units in dynamic analysis

In dynamic analysis, it is common to express the mass of the structure in terms

of kg. The unit of force is a derived unit and is known as a Newton.

A Newton (N) is defined as a force that will produce an acceleration of 1 m/s2 on

a mass of 1 kg. Since, according to the Newton’s law of motion, force is equal

to the product of mass and acceleration, we have

N = 2s

m kg

The above equation relating N and kg is often used in dynamic analysis.


Chapter 3 16


Example 1

The structural frame shown in Figure 10 is rigid jointed and fixed to its supports.

The mass of the structure of 5000 kg is concentrated on the beam which is

assumed to be rigid. The columns are assumed weightless and each has

flexural stiffness (EI) of 4.5 x 106 Nm

2 in the plane of the structure. If the

structure is assumed to have a viscous damping ratio of 4%, calculate the

damped and undamped natural frequencies.

Figure 10

Solution

The mass of the equivalent one degree of freedom system is

m = 5000 kg

Note that when the frame is displaced, each column is deflected as shown in

Figure 10 (b) with zero rotations at top and bottom. The forces corresponding to

a unit lateral displacement can be calculated from elementary bending theory and

are as shown. Hence the effective spring stiffness of the structure is twice the

lateral stiffness of each column and is therefore

k = 2 x 3

12

h

EI = 2 x 12 x 4.5 x 10

6 / 3

3 = 4.0 x 10

6 N/m

Hence, the undamped natural frequency is

f

2T

1

m

k

2

1

5000

104

2

1 6x

= 4.502 Hz

The damped natural frequency fD 21 f = 4.502 x 204.01

= 4.498 Hz

which is practically the same.


Chapter 3 17


Example 2

For portal frame ABCD shown in Figure 11, it may be assumed that the

horizontal member is infinitely stiff and that the vertical members have

negligible mass compared with that of the horizontal member. If there is no

damping, determine the natural frequency f and the natural period T.

Figure 11

Solution

The stiffnesses of the members AB and CD are:

kAB = 3

12

L

EI N/mm kCD =

3)2(

)3(12

L

IE = 4.5

3L

EI N/mm

Hence

k = kAB + kCD = 16.5 3L

EI N/mm = 16500

3L

EI N/m

ω m

k (where k is in newtons per metre and m in kilograms)

3

16500

ML

EI

3128

ML

EI radians per second

T =

2 =

EI

ML3

128

2 seconds

f

2T

1

32

128

ML

EI

cycles per second


Chapter 3 18


3.7 Dynamic Properties of a multi-degree of freedom system

The natural frequencies and vibration modes of a structure play a central role in

the analysis of its response to dynamic loading. The frequencies and mode

shapes are determined by the mass matrix M and the stiffness matrix K of the

system.

Figure 12 – Structural Model for Multi-Degree of Freedom System

M =

Nm

m

m

.

.

2

1

K =

NN

N

kk

k

k

kkkk

kkk

.

.

.

)(

)(

3

3322

221

Several properties of the above matrices are useful to bear in mind. First, both

M and K are symmetric matrices. Second, M is a diagonal matrix for a

lumped mass system. Third, the diagonal elements of the K matrix are always

positive.


Chapter 3 19


Computation of Natural Frequencies and Modes of Vibration

Computation of the vibration properties of a multi-degree of freedom system

requires solution of the following matrix equation (which in mathematically

terminology is called an eigen problem):

0X M ω K 2

where X is a column vector of displacements that corresponds to a particular

mode of vibration and is called an eigenvector.

The frequencies can be computed from the following relationship, i.e. the

determinant of the eigen problem equals zero:

0 M ω K 2

For a N-DOF system, the mass and stiffness matrices are of order N. Solution

of the eigen problem leads to the N natural frequencies and eigen vectors: ωn, Xn

(n=1 to N).

It is usual to represent the smallest frequency by ω1. This frequency is called

the fundamental frequency of the system and corresponds to the first mode of

vibration. The remaining frequencies correspond to the second and higher

modes of vibration. The eigen vectors Xn are often re-expressed in

dimensionless form by dividing all the components by one reference component

(usually the largest), a procedure called normalization. The resulting vector is

called the n-th mode shape n:

n =

Nn

n

n

.

.

.

2

1

=

Nn

n

n

X

X

X

X

.

.

.1

2

1

max

in which Xmax is the largest value in Xn.


Chapter 3 20


Many methods have been developed for numerical solution of the eigen problem

and are available in many textbooks. Usually just the first few vibration modes,

and sometimes just the fundamental mode is sufficient for most dynamic

analysis.

Idealization of multi-storey buildings for horizontal dynamic loading

For the analysis of the response of a multi-storey building to horizontal dynamic

loading (e.g. wind or earthquake), a useful idealization is the shear building or

shear frame model:

i) The mass of the structure is assumed to be lumped at the floor levels.

ii) The floors (i.e. horizontal members) are assumed to be rigid.

iii) Axial deformations in the columns are ignored.

iv) The lateral stiffness of the building is due entirely to the stiffness of the

columns in bending.

In a shear building, rotation of joints is assumed not to occur and the floors can

only move horizontally. Accordingly an N-storey frame will have N degrees of

freedom.

M = mass matrix =

3

2

1

00

00

00

m

m

m

K = stiffness matrix =

33

3322

221

0

)(

0)(

kk

kkkk

kkk

The lateral storey stiffness kj of the j-th storey is the storey shear force required

to cause a unit horizontal displacement for that storey. The values of the lateral

storey stiffnesses k1, k2, … kN can be calculated from the bending stiffness of the

columns. The contribution of each column to the lateral storey stiffness is

given by3

12

L

EI.


Chapter 3 21


Figure 13 – Lateral Stiffness of Column in a Frame

Figure 14 – Idealized Multi-Storey Building


Chapter 3 22


Mass Matrix M =

N

j

m

m

m

m

.

.

.

2

1

Stiffness Matrix K =

NN

N

kk

k

kkkk

kkkk

kkk

.

.

)(

)(

)(

4433

3322

221


Chapter 3 23


Example 3

Find the natural frequencies and mode shapes of the three-storey building shown

in Figure 15. Assume that the building is undamped.

Figure 15

Solution

The building can be idealized as a three-degree of freedom lumped mass system

with the mass matrix given by

M = 105

400

040

004

kg

The stiffness matrix can be obtained by assuming unit displacements at each storey

and then calculating the forces in the structure. Hence

K = 120 x 106

110

132

026

N/m

The natural frequencies can be obtained from the determinant of

0 M ω K 2


Chapter 3 24


K – ω2 M = 120 x 10

6

)1(10

1)3(2

02)6(

where α = 300

2

The determinant of the matrix is

(6 – α) [(3 – α) (1 – α) – 1 x 1] – (-2) x [- 2 x(1 – α) + 0] = 0

(6 – α) [3 – 4α + α2 – 1] + 2 [ 2α – 2 ] = 0

α3 – 10α

2 + 22α – 8 = 0

Solutions of this equation are

α1 = 0.45 α2 = 2.52 α3 = 7.034

thus giving

ω2 =

2110

756

135

and ω =

3

2

1

=

94.45

50.27

62.11

rad/s

The mode shapes can now be calculated. Assuming a normalized shape for each

mode, i.e. 3n = 1, gives

n =

1

2

1

n

n

Thus 0X M ω K 2 becomes

120 x 106

)1(10

1)3(2

02)6(

1

2

1

n

n

=

0

0

0

The mode shapes for the three natural frequencies are

1 =

1

55.0

198.0

; 2 =

1

52.1

874.0

; 3 =

1

03.6

67.11


Chapter 3 25


Example 4

An idealized 2-storey shear building is shown below, with mass lumped at the

floor levels and individual storey stiffnesses. Determine the vibration mode

shapes and the natural circular frequencies of the building.

Solution

The stiffness matrix can be obtained by assuming unit displacements at each storey

and then calculating the forces in the structure.

kAD = kBE = 3EI/L3 =

34

*3 EI=0.04688 EI

kCF =12EI/L3 =

34

*12 EI=0.1875 EI

kEG = kFH = 12EI/L3 =

33

*12 EI=0.4444 EI

Total stiffness at level 1 = kAD + kBE + kCF = (0.04688*2 + 0.1875) EI

= 0.2813 EI

Total stiffness at level 2 = kEG + kFH = (2*0.4444) EI = 0.8888 EI

7m 7m

3m

4m For all columns

EI = 7.68*108 N/m

2

4000 kg/m

3000 kg/m 3000 kg/m

G H

D E F

A B C


Chapter 3 26


K = EI

8888.08888.0

8888.01701.1

= 7.68 x 108

8888.08888.0

8888.01701.1

= 6.826 x 108

11

13165.1 N/m

Mass Matrix:

M = 104

8.20

02.4 kg = 2.8*10

4*

10

05.1

(iii) The natural 0 M ω K 2

K – ω2 M = 6.826*10

8

11

15.13165.1

where α =8

24

10*826.6

10*8.2

The determinant

[(1.3165 – 1.5α) (1 – α) – (-1) * (-1)] = 0

1.3165 -2.8165 + 1.5α2 -1= 0

1.5α2 – 2.8165α + 0.3165 = 0

Solutions of this equation are

α1 = 0.1200 α2 = 1.7576

thus giving

ω2 =

42848

4.2925 and ω =

2

1

=

0.207

09.54 rad/s

For 1 = 0.1200

6.826*108

0

01

12.011

112.0*5.13165.1

21X

1365.1

11

21X

Hence

1

8799.0

1

11


Chapter 3 27


For 2 = 1.7576

6.826*108

0

01

7576.111

17576.1*5.13165.1

22X

3199.1

11

22X

Hence

1

7576.0

1

12

Revision

Read reference 1 on P.39 – 65, P.401 - 443.

Read reference 2 on P.3 – 33, P.201 – 226.

Main Reference

1. Dynamics of Structures, Theory and Applications to Earthquake Engineering,

3rd

Edition, Anil K. Chopra, Prentice Hall.

2. Structural Dynamics, Theory and Computation, 3rd

Edition, Mario Paz, Chapman & Hall.

3. Dynamics of Structures, 2nd

Edition, Ray W. Clough & Joseph Penzien, McGraw Hill.


Chapter 3 28


│TUTORIAL 3│

Q1. An idealized 3-storey shear building is shown below, with mass lumped at the floor levels and

individual storey stiffnesses.

Determine the vibration mode shapes and the natural circular frequencies of the building.

Ans:


Chapter 3 29


│TUTORIAL 3│

Q2. Compute the natural angular frequency of vibration in sidesway for the frame shown below and

calculate the natural period of vibration. Idealize the frame as one degree of freedom system.

Neglect axial and shear deformations and the weight of the columns.

(Ans: = 7.103 rad/sec, T = 0.885 sec)

If the system has a damping coefficient = 0.1, what are the damped natural circular frequency d

and the natural period of damped vibration Td ?

(Ans: d = 7.067 rad/sec, Td = 0.889 sec)


Chapter 3 30


│TUTORIAL 3│

Q.3 A two-storey undamped building frame is shown in the following figure. By using the lumped mass

idealization and neglecting the axial and shear deformation and self-weight of columns, calculate:

(a) Mass (M) and stiffness (K) matrix of the system;

(b) Natural cyclic frequencies (f) of the building; and

(c) Corresponding mode shapes of vibration.

Q.4 The structural frame of an idealised two-storey shear building is shown as follows. The lumped

masses at floor levels and the relative values of flexural rigidities of the individual columns are

given as shown. Determine:

(a) The mass matrix M of the building.

(b) The stiffness matrix K of the building.

(c) The natural periods and natural cyclic frequencies of the building.

(d) The corresponding mode shapes of vibration normalized with respect to the components of the

second storey.

A B

C D

E

F 3m

3.5m 2EI

2EI

EI

EI

G

EI

4 * 105 kg

EI = 2 * 109 Nm

2

for all columns

H

3 * 105 kg

6 m 8 m

4 m

5 m

For all columns

EI = 6.0*108 Nm

2

2000 kg/m

2000 kg/m 2000 kg/m

A B

C

D E

A

F

G

A

H

A

3 m

con4337a2to131007bl_ch3_dynamicanalysis

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