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1 SD 421 Dynamics of Structures

Lecture 1&2

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Structural dynamics-Introduction

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Page 1: Lecture 1&2

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SD 421

Dynamics of Structures

Page 2: Lecture 1&2

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1.0 Introduction1.1 Objective

The objective of structural dynamics is to predict the behavior of structures under dynamic loads

This is done with a help of a mathematical model for the structural system.

For problems involving complex material properties, loading and boundary conditions, the engineer has to introduce assumptions and idealizations, which will make the problem manageable.

A mathematical model is a symbolic designation for the substitute idealized system including all assumptions imposed on the physical problem.

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1.0 Introduction1.2 Significance

There are cases whereby a structure has to be designed to resist dynamic loads such as gusty winds, earthquakes or other dynamic disturbances. under those circumstances it is important to

determine :

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1.0 Introduction1.2 Significance

The natural frequencies of vibration of the structure the purpose of estimating the likelihood of resonance due to the dynamic disturbances

The maximum displacements, stresses and acceleration the structure may experience when it is subjected to the dynamic disturbances.

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1.0 Introduction1.3 Degrees of Freedom

Degrees of Freedom is the number of independent coordinates or measurements required to define completely the configuration of the structure at any instant

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1.0 Introduction1.3 Examples: One Degrees of Freedom Systems

Simple Pendulum. Configuration is defined fully by angle

L

o(t)

M

o(t)

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1.0 Introduction1.3 Examples: One Degrees of Freedom Systems

Configuration is defined fully by displacement x(t)

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1.0 Introduction1.3 Examples: Two Degrees of Freedom Systems

Configuration is fully defined by the displacements x1(t) and x2(t)

Therefore the system has 2 degrees of freedom

m2

m1

m2

m2

m1 m1h2

h2

h 1 h 1

x1x1

x2x2

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1.0 Introduction1.3 Degrees of Freedom (cont.)

In general a continuous structure has an infinite number of degrees of freedom.

Nevertheless the process of idealization permits the reduction in the number of degrees of freedom to a discrete number and in some cases to just a single degree of freedom .

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1.0 IntroductionExamples of structural systems that can be modeled as one degree of freedom systems

Mass of the columns can be neglected

Mass of the beam can be neglected

F(t)

F(t)

m

m

X(t)

X(t)

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1.0 Introduction1.4. Dynamic Models

One degree of freedom systems may be described conveniently by one of the following models

m

m

kkc

c

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1.0 Introduction1.4. Dynamic Models

Each model has following elements:1) A mass element (m) representing the mass

and inertial characteristics of the structure2) A spring element (k) representing the

stiffness characteristics of the structure. (The stiffness of the structure describes the restoring force and the potential energy characteristics of the structural system.)

3) A damping element (c) representing the frictional characteristics and energy loses of the structure

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2.0 Free Vibrations of Un-damped Single

Degree of Freedom Oscillator2.1 Equation of Motion

Fig. 2.1 Shows un-damped single degree of freedom system, in which the mass has been displaced and then suddenly released

Position of static equilibrium Position of the mass t sec. After being

displaced and released suddenly

Figure 2.1 Un-damped Single Degree of freedom system

kk

m

m

X(t) X

(t)

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.1 Derivation Using Newton’s 2nd Law of Motion

Fig. 2.2 Shows a free body diagram of the oscillator displaced in the positive direction and acted upon by the spring force F = k∙x

k∙x

Positive Direction of motion

Fig. 2.2: Free body Diagram of the mass

m

k

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.1 Derivation Using Newton’s 2nd Law of Motion

From Newton’s 2nd Law of Motion:

F = m∙a

i.e.

Thus [2.1]

Equation [2.1] is the equation of motion of afreely vibrating one degree of freedom oscillator

xmxk masstheofonaccelerati

dtdxxwhereby

2

2

0 xkxm

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.1 Derivation Using D’Alembert’s Principle

It is difficult to use Newton’s 2nd Law of motion to derive the equations of motions of complex structural systems.

In those cases D’Alembert’s principle, in conjunction with the principle of virtual work is widely used

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.1 Derivation Using D’Alembert’s Principle

D’Alembert’s Principle states that a system may be set in a state of dynamic equilibrium by adding to the external forces a fictitious force which is commonly known as inertia force.

The inertia force is equal to mass multiplied by the acceleration and it is always directed NEGATIVELY with respect to the corresponding coordinate

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.1 Derivation Using D’Alembert’s Principle

Application of the Principle to a single degree of freedom system is illustrated in Fig. 2.3 below

Note that in Fig. 2.3 is the inertia Force

k∙x

Fig. 2.3: Free body Diagram of the mass

m

k

xm Positive Direction of motion

xm

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.1 Derivation Using D’Alembert’s Principle

The requirement that the two forces must be in equilibrium gives the equation of motion as follows:

00 xkxmFx [2.1]

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.2 Solution of the Differential Equation of Motion

The Equation of motion is given by equation

[2.1] i.e.0 xkxm

The equation is of

2nd ORDER,

LINEAR,

HOMOGENEOUS and has

CONSTANT COEFFICIENTS

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.2 Solution of the Differential Equation of Motion

To solve the equation one can start by assuming a trial solution given by

tBxor

tAx

cos

sin

[2.2]

[2.3]

Substituting Eq. [2.2] into Eq. [2.1] gives 02 tAkm sin [2.4]

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.2 Solution of the Differential Equation of Motion

If Eq. [2.4] is to be satisfied at any time (-m∙ω2 + k) must be equal to zero

i.e. -m∙ω2 + k =0 [2.5]

i.e.

The quantity is known as the

NATURAL FREQUENCY of the system

mkor

mk 2

mk

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.2 Solution of the Differential Equation of Motion

Similarly substituting x=B cosωt into gives:

i.e. -m∙ω2 + k =0 i.e.

mkor

mk 2

0 xkxm

02 tAkm cos

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.2 Solution of the Differential Equation of Motion

Since either x=Asinωt or x=Bcosωt is a solution of the differential equation and since this equation is linear, the superposition

of these two is also a solution. Thus the solution of the

equation of motion is given by

x=Asinωt + Bcosωt [2.6] Where A and B are constants of integration which are determined from the initial conditions.

0 xkxm

0 xkxm

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.2 Solution of the Differential Equation of Motion

The velocity is given by

Let xo = Displacement of the mass at time t=0

vo = Velocity of the mass at t=o.

Substituting x(t=0 )=xo into Eq. [2.6] gives

xo=B i.e. B=xo [2.8]

Substituting v(t=0)= vo into Eq. [2.7] gives

vo= Aω i.e. [2.9]

tBtAdtdxxv sincos [2.7]

ovA

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2.0 Equation of Motion Un-damped Single Degree of Freedom Oscillator

2.2 Solution of the Differential Equation of Motion

Finally substitution of A and B from Eqs.[2.8] and [2.9] gives the displacement of the mass as a function of time:

txtvtx

cossin)( 00 [2.10]

Where xo = Displacement of the mass at time t=0 vo = Velocity of the mass at t=o. ω = natural frequency of the system

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2.3 Springs in Parallel or in Series

Sometimes it is necessary to determine the equivalent spring constant for a system in which two or more springs are arranged in parallel or in series.

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2.3.1 Springs in Parallel

Fig. 2.4: Two Springs in Parallel

k1 k2 k1 k2m

m x

F

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2.3.1 Springs in Parallel

Fig. 2.5: Free Body Diagrams

k1

xk2x

kex

F F(a) Actual System (b) Equivalent

System

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2.3.1 Springs in Parallel

Equilibrium requirements for the actual system (fig. 2.5(a)) and of the equivalent system (Fig. 2.5(b)) gives

F = kex= k1x+k2x=(k1+k2)x Thus The Equivalent spring constant is given

by: ke=(k1+k2)In general the equivalent spring constant for n Springs in parallel is given by

n

i

ie kk1

[2.11]

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2.3.2 Springs in Series

k1

k2

k2

k1

m

F

x

m

Let:

x1= Change in length of 1st Spring

x2=Change in Length of 2nd Spring

Displacement of the mass is

given by:

x=x1+x2

Fig. 2.6: Spring In Series

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2.3.2 Springs in Series

k1x1

F F

k2x2

Fig. 2.7: Free Body Diagrams

(a) (b)

Equilibrium Requirements give

2222

kFxorxkF

and

1111

kFxorxkF

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2.3.2 Springs in Series

If ke is the equivalent spring constant the Displacement of the mass is given by

ekFx

2121

kF

kFxxx

21

21

111

11

kkkei

kkF

kF

e

e

..

But

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2.3.2 Springs in Series

In general the equivalent spring constant for n Springs in series is given by

n

iie kk

1

11 [2.12]

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2.4 Frequency and Period

Equation [2.10] shows that the motion of a freely oscillating mass can be expressed by a sine or cosine function of same frequency ω. Since both cosine and sine functions have a period of 2the period T of the motion is given by

ωT= 2

2TOr

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2.4 Frequency and Period

The period is usually expressed in seconds per cycle or simply in seconds with the understanding that it is per cycle.

The value of the reciprocal to the period is the natural frequency f

2

1 T

fi.e.

The natural frequency is usually expressed in HERTZ or CYCLES PER SECOND

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2.4 Frequency and Period

The quantity ω is sometimes referred to as the natural frequency. To distinguish between ω and f, ω is sometimes called CIRCULAR or ANGULAR natural frequency. The circular frequency is is given in RADIANS PER SECOND