ipsum

Structural Health Monitoring

(CVEN9840)

Lecture 3

Dr Mehri Makki Alamdari

School of Civil and Environmental Engineering

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2CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Structural Dynamics

Single Degree of Freedom (SDOF)

3CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Mechanical vibrations theory and application to structural dynamics By Michel Geradin, Daniel Rixen

This is a comprehensively updated new edition of the popular textbook. It presents the theory of vibrations in the context of structural analysis. With revised, coherent and uniform notation, «Mechanical Vibrations: Theory and Application to Structural Dynamics, Third Edition» is a must-have textbook for graduate students working with vibration in mechanical, aerospace and civil engineering, and is also an excellent reference for researchers and industry practitioners.

Recommended Textbook

Dynamics of Structures: Theory and Applications to Earthquake Engineering By Anil K. Chopra

This second edition includes many topics encompassing the theory of structural dynamics and the application of this theory regarding earthquake analysis, response, and design of structures. Covers the inelastic design spectrum to structural design; energy dissipation devices; Eurocode; theory of dynamic response of structures; structural dynamics theory; and more. Ideal for readers interested in Dynamics of Structures and Earthquake Engineering.

4CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Recommended Textbook

Fundamentals of Mechanical Vibrations By S. Graham Kelly

This text is designed for undergraduate courses in vibrations taught in departments of engineering. It provides detailed explanations of fundamental aspects of vibrations such as the derivation of differential equations, covers physical interpretation of phenomena using energy methods, and includes chapters on vibration control and nonlinear vibrations. Detailed coverage of the equivalent systems method allows modelling of all one-degree-of-freedom linear systems by a mass-spring-dashpot system.

Intended primarily for teaching dynamics of structures to advanced undergraduates and graduate students in civil engineering departments, this text is the solutions manual to Dynamics of Structures, 2nd edition, which should provide an effective reference for researchers and practicing engineers. The main text aims to present state-of-the-art methods for assessing the seismic performance of structure/foundation systems and includes information on earthquake engineering, taken from case examples.

Dynamics of Structures By Ray W. Clough, Joseph Penzien

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Structural Dynamics

Bridge

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Structural Dynamics

Structural dynamics is the analysis of structures subjected to time varying forces or motions.

Examples of structures which are subjected to dynamic loading:

High speed ground transportation vehicle moving along its guideway Bridge structure subject to ongoing traffic Offshore drilling platform Jet plane flying through a thunderstorm Treadmill High-rise building subject to wind

7CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Structural Dynamics

Bridge in China shakes like waves after being hit by strong winds

Ref: https://www.youtube.com/watch?v=VGjrZDNk_EQ

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Static Loads vs. Dynamic Loads

There are generally two types of loading:

Static loadings Æ does not have a time varying magnitude

Dynamic loadings Æ have time varying magnitudes

¾ The major distinction between static and dynamic loadings is the effect of inertia force.

Note that most of the loadings on a structure are rarely truly static.

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CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Static Loads vs. Dynamic Loads

¾ Examples of such loading are dead load, creep effect, support settlement, and temperature change.

¾ If the loading is applied on a structure slowly, i.e. the shortest period of the loading is much longer than the fundamental period of the structure, the loading is considered as a static loading. In such case, the inertia effect can be disregarded.

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Thermal cycling load

Period of cyclic temperature loading = 24 hrs Fundamental frequency of a bridge = 2Hz Fundamental period of a bridge = 0.5 second

24 hr >>0.5 second! Æ Thermal loading can be considered as a

static loading.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Dynamic Loads

Dynamic loadings may be classified based on the nature of the time history of the loading as: � Deterministic load: ¾ Time variation of the load can be defined Æ for example: a periodic load.

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� Nondeterministic or stochastic or random load: ¾ Time variation of the load is not completely known.

¾ The load can be specified in a statistical sense such as mean value, standard deviation, auto correlation function, power spectrum, density function, etc.

¾ Examples of nondeterministic load are wind pressure on building and water waves.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Deterministic Loads

Periodic Loads: ¾ Repetitive loads which exhibit same time variation successively for a large number of cycles. ¾ Periodic loads can be harmonic or non-harmonic loads. ¾ Examples of harmonic excitation are rotating machines, turbine power plant, etc.

Non-Periodic Loads: ¾ Non-repetitive loads which can be short in durations or long in durations.

Harmonic Loading Non-Harmonic Loading Non-Periodic Impact Loading

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CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Essential Characteristics of a Dynamic Problem

The response of the structure depends on dynamic characteristics of the structure (such as natural frequency, damping ratio) and the characteristics of the loading (loading frequency).

The principle distinction between static and dynamic problems is the consideration of the inertial effects (mass × acceleration) and damping effects (dissipation in the system).

The essential characteristics of a dynamic problem are:

¾ The time varying nature of the loading ¾ Inertial effects Æ an example of inertia effect is shown: ¾ Damping effects

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CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Dynamic Analysis

Dynamic loadings vary with time Æ dynamic analysis inevitably requires much more computational efforts than a static analysis.

Major steps involved in dynamic analysis: Construction of an analytical model or a mathematical model Determination of the solution to the model (response of the system)

¾ Generally, analytical models fall into two categories: discrete models and continuous models.

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14CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Single Degree of Freedom System Æ SDOF

9 A single degree of freedom (SDOF) system is a system having one independent spatial coordinates (translation or rotation) which needs to be defined at any instant to be able to predict its dynamic behaviour.

9 The system may compose of a mass, a spring, and a dashpot.

9 This is the most simplified model used in structural dynamic analysis.

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� Most structures and or structural components can be simplified as a SDOF system.

SDOF System

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� The spring provides stiffness to the system.

� The dashpot or a viscous damper dissipates vibrational energy of the system.

In the dynamic analysis of a SDOF system subjected to a time varying loading, an equation of motion of the system must be formulated and solved. There are four methods in the formulation of the equation of motion:

Newton’s second law of motion

The D Ale be principle

The e e g e h d

The i ci le f i al di lace e

SDOF System

where k is the

stiffness of the

system.

17CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Equation of Motion (SDOF)

� Force-displacement relation

Consider a SDOF system consists of a linear elastic spring subjected to an externally applied force along the degree of freedom (DOF) .

The relationship between the applied force and the resulting displacement is,

where is the stiffness of the system.

Dampin g force:

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� Damping force � Damping is the process by which free vibration steadily diminishes in amplitude.

� In reality, the energy of the vibrating system is dissipated by various mechanisms such as friction at steel connections, opening and closing of microcracks in concrete or actual mechanical dampers.

� Although in reality, the damping in actual structures is much more complex, it has been proven that, for simplification, a linear viscous damper or a dashpot provides satisfactory results in the SDOF analysis. For a system consists of a dashpot, the damping force is related to the velocity by

Equation of Motion (SDOF)

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Equation of Motion (SDOF)

Ne ec d la f i a e ha if he e l a f ce a b d i zero then the body will move with an acceleration which is proportional to the resultant force acting on the body and in the direction of the force.

Equation of motion (E.o.M.) governing the displacement of a SDOF structure:

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Example

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Write the equation of motion of the beam shown assuming that the beam mass is negligible and there is no damping.

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Solution

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Example

Write the horizontal equation of motion of the steel frame (E=200GPa) shown assuming the column mass is negligible, the beams have infinite stiffness and there is no damping. Given the moments of inertia of W200x36=34.4x106 mm4 and W250x49=70.6x106 mm4.

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Solution

M=L

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Solution

L/2 L/(2EI)

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Solution

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Combining Springs

¾ Springs are usually only available in limited stiffness values. Combing them allows other values to be obtained.

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Example

Write an equation of motion of the systems shown.

(a) (b)

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Solution

(a)

Springs are in parallel Æ Equivalent k=k1+k2

mu cu+(k1+k2)u=p(t)

(b) Springs are in series Æ Equivalent k= 𝒌𝟏𝒌𝟐 𝒌𝟏 𝒌𝟐

mu cu+( 1 2 1 2

)u=p(t)

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Solve SDOF- Free Vibration

𝑚𝑢 𝑐𝑢 𝑘𝑢 𝟎

𝑢 𝐶𝑒 𝑢 𝜆𝐶𝑒 𝑢 𝜆2𝐶𝑒 𝑚𝜆

2 𝑐𝜆 𝑘 0 𝜆 𝑐

2𝑚 𝑐 2𝑚

2 𝑘 𝑚

𝜆1 𝑐

2𝑚 𝑐 2𝑚

2 𝑘 𝑚 𝑢 𝐶1𝑒 𝐶2𝑒𝜆2

𝑐 2𝑚

𝑐 2𝑚

2 𝑘 𝑚

𝜔𝑛 𝑘 𝑚 wn is called the un-damped natural frequency (rad/s)

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

𝑓𝑛 𝜔𝑛/(2𝜋 𝐻𝑧 𝐻𝑒𝑟𝑡𝑧

30CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

SDOF- Free Vibration

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Example

Calculate the displacement and acceleration of a SDOF system with 10 mm/s at a frequency of 23 Hz.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

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Solution

Recall:

𝑥 𝑡 𝑋𝑠𝑖𝑛𝜔𝑡 → 𝑥 𝑡 𝜔𝑋𝑐𝑜𝑠𝜔𝑡 → 𝑥 𝑡 𝜔2𝑋𝑠𝑖𝑛𝜔𝑡

𝑓 = 23 Hz Æ 𝜔=2𝜋𝑓 46𝜋 𝑟𝑎𝑑/𝑠𝑒𝑐

𝑥= 𝜔𝑋= 10 ×10-3 m/s Æ 𝑋=0.01/(46 𝜋) m 𝑥= 𝜔2𝑋 = - 0.01(46 𝜋) m/s2

33CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Determine the natural frequency of the simple spring-mass system shown below by measuring the static deflection.

The undamped natural frequency of a simple SDOF system may be determined from a static-displacement measurement.

Equilibrium Position

Example

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Solution

𝜔2 𝑘 𝑚

The equilibrium of the mass as it hangs on the spring is expressed by

𝐹 0 → 𝑊 𝑓𝑠 0

𝑓𝑠 𝑘𝑢𝑠𝑡 𝑊 𝑚𝑔 Æ 𝑚𝑔 𝑚𝜔2𝑢𝑠𝑡 0 → 𝜔

2 𝑔/𝑢𝑠𝑡

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¾ An example of arch bridge¾ Golden Gate Bridge, San Francisco

T a e e: T 18.2s, f 0.06Hz Ve ical: T 10.9s, f 0.09Hz T i al: T 4.943s, f 0.02Hz

1 1 24 1: .0Mode f Hz= 2 3 08 2 : .Mode f Hz=

3 3 15 3 : .Mode f Hz= 4 3 67 4 : .Mode f Hz=

5 3 72 5 : .Mode f Hz= 1 4 86 6 : .Mode f Hz=

7 5 16 7 : .Mode f Hz= 1 6 03 8 : .Mode f Hz=

9 7 02 9 : .Mode f Hz= 1010 : 8.1 Mode f Hz=

Example

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

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The nature of the two roots 𝜆1 and 𝜆2 depends upon whether the quantity under the square root is positive, exactly zero, or negative. When it is exactly zero, this particular value of c is called the critical value, ccrit, given by:

� The critical value of damping is so called because it marks the boundary between oscillatory and non-oscillatory behaviour.

The non-dimensional viscous damping coefficient, 𝜉 , is defined by:

ξ 𝑐

𝑐

It is sometimes expressed in percentage terms, for example ξ 0.02 may be efe ed a 2% i c da i g .

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

SDOF- Free Vibration

𝑐 2 𝑚𝑘 2𝑚𝜔

𝑐 2𝜉𝑚𝜔

𝜆 𝑐

2𝑚 𝑐 2𝑚

2 𝑘 𝑚

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𝑚𝑢 𝑐𝑢 𝑘𝑢 0

𝜔𝑛 𝑘 𝑚

𝑐 2𝜉𝑚𝜔

𝑢 2𝜉𝜔 𝑢 𝜔2𝑢 0

wD is called the damped natural frequency

𝜔 𝜔 1 𝜉2

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

SDOF- Free Vibration

Replace in E.o.M. and divide by m

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Depending on the value of the damping ratio, different behaviour of the system is expected:

Undamped System 𝜉 =0 Underdamped System 0 < 𝜉 <1 Critical Damped System 𝜉 =1 Overdamped System 𝜉 >1

𝜔 𝜔 1 𝜉2

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

SDOF- Free Vibration

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Undamped System

¾ When the damping ratio of the system is zero (𝜉 =0), the system is called undamped system.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

𝒕 𝟎 → 𝒖 𝟎 𝒖𝟎 t=0 Æ 𝒖 𝟎 𝒖𝟎 t𝒂𝒏 𝝓

𝒖𝟎 𝝎𝒏𝒖𝟎

40CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

The natural frequency of a cantilever beam with a lumped mass at its tip is to be determined dynamically. The mass is deflected by an amount A=1 in. and released. The ensuing motion, shown below, indicates that the damping in the system is very small. Compute the natural frequency in radians per second and in Hertz. What is the period?

Example

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Solution

¾ At point (a), the mass has executed 1 ¼ cycles in approximately 0.4 second. Therefore, the undamped natural frequency in Hertz is:

𝑓𝑛 1.25𝑐𝑦𝑐𝑙𝑒𝑠 0.4𝑠𝑒𝑐

3.125𝐻𝑧

Æ 𝜔𝑛 2𝜋𝑓𝑛 2𝜋(3.125)=19.64 rad/s

¾ The undamped natural period is: Æ 𝑇𝑛

1 = 1 3.125

=0.320 s

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Underdamped System

¾ When the damping ratio of the system is between 0 and unity, i.e. 0 < 𝜉 <1, the system is called underdamped system. The damping coefficient in this case is 0 < c < ccrit .

𝜔 𝜔 1 𝜉2

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

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Critical Damped System

¾ For the system with damping ratio equals to unity (𝜉 =1) which implies that c = ccrit .

The system is called the critical damped system because it will oscillate if 𝜉 is slightly reduced.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

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Overdamped System

¾ For the system with damping ratio greater than unity (𝜉 >1) which implies that c > ccrit.

¾ There will be no oscillation of the system.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

𝒕 𝟎 → 𝒖 𝟎 𝒖𝟎 t=0 Æ 𝒖 𝟎 𝒖𝟎

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nteEnvelop curves

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

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The system shown, has the following properties: m=1 kg; k=10,000 N/m and c=40 N/m/s. Plot the time history of u for the initial conditions: 𝑢 0,𝑢 0.1 at t = 0.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Example

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Solution

→ 𝜔𝑛 = 10000 1

=100rad/s

→ ξ = 2

= 40 2 1 100

=0.2

→ 𝜔𝐷= 𝜔𝑛 1 𝜉2 100 1 0.22=98rad/s

𝑢ℎ(t)=𝜌𝑒 cos(𝜔𝐷t-𝜙

𝜌 𝑢0 2 0 0 2 0.12 0 0.2 100 0.1 2=0.1021m

𝑡𝑎𝑛𝜑 0 0 0

=0 0.2 100 0.1 0.1

=0.2041 Æ𝜑=0.2014 rad

U=0.1021 𝑒 20 cos(98t-0.2014

𝑢 0 0,𝑢 0 0.1

48CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Solution in MATLAB

k=10000; m=1; c=40; u0=0.1; udot0=0; wn=(k/m)^0.5 zeta=c/(2*m*wn) wd=wn*(1-zeta^2)^0.5 ro=(u0^2+((udot0+zeta*wn*u0)/wd)^2)^0.5 phi=atan((udot0+zeta*wn*u0)/(wd*u0)) t=0:0.001:0.2; u=exp(-zeta*wn.*t)*ro.*cos(wd*t-phi); figure plot(t,u,'linewidth',2) xlabel('Time sec') ylabel ('Response m') set(gca,'fontsize',14) grid on grid minor

49CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

A SDOF system has an undamped natural frequency of 5 rad/s and a damping factor of 20%. It is given the initial conditions u0 = 0 and v0 = 20 in/s.

Determine the damped natural frequency and the expression for the motion of the system for t > 0.

Plot the time history of the response for the first three seconds if damping values of 5%, 10%, 20% and 40% are considered.

Example

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Solution

→ 𝜔𝑛 5rad/s → ξ 0.2 → 𝜔𝐷= 𝜔𝑛 1 𝜉2 5 1 0.22=4.8990~4.90 rad/s

𝑢ℎ(t)=𝜌𝑒 cos(𝜔𝐷t-𝜙

𝜌 𝑢0 2 0 0 2 02 20 0.2 5 0

4. 0 2= 20

4. 0 =4.08

𝑡𝑎𝑛𝜑 0 0 0

=20 0 0

Æ𝜑= /2 rad

U=4.08 𝑒 cos(4.90t- /2)= 4.08 𝑒 sin(4.90t)

𝑢 0 20𝑖𝑛/𝑠𝑒𝑐,𝑢 0 0

51CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Solution in MATLAB u0=0; udot0=20; wn=5 figure for zeta=[0.05 0.1 0.2 0.4]; wd=wn*(1-zeta^2)^0.5; ro=(u0^2+((udot0+zeta*wn*u0)/wd)^2)^0.5; phi=atan((udot0+zeta*wn*u0)/(wd*u0)); t=0:0.001:3; u=exp(-zeta*wn.*t)*ro.*cos(wd*t-phi); hold on plot(t,u,'linewidth',2) hold off end; xlabel('Time sec') ylabel ('Response m') set(gca,'fontsize',14) grid on grid minor legend ('\zeta=0.05','\zeta=0.1','\zeta=0.2','\zeta=0.4') box on

52CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

For a SDOF system, the mass and stiffness of the system change between:

2 kg < m < 3 kg k > 200 N/m

As a result, the natural frequency of the system will be in a range of:

8.16 rad/s < n < 10 rad/s

For initial conditions: x0 = 0, v0 < 300 mm/s, choose a damping c so response is always < 25 mm.

Example

53CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Solution

𝑥(t)=𝜌𝑒 sin(𝜔𝐷t-𝜙

¾ 𝑥 0 300 𝑚𝑚/𝑠,𝑥 0 0

¾ 𝜌 𝑥0 2 0 0 2 0 Æ 𝑥(t)= 0 𝑒 sin(𝜔𝐷t-𝜙

¾ Worst case happens at smallest 𝜔𝐷 ⇒ 𝜔𝑛 = 8.16 rad/s ¾ Worst case happens at maximum of 𝑥0 = 300 mm/s ¾ With 𝜔𝑛 and 𝑥0 fixed at these values, we can find the time instant that corresponds to

maximum response: ¾ Note: since the effect of 𝜙 is only phase shift and our concern here is amplitude, we disregard 𝜙.

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Solution

𝑥(t)= 0 𝑒 sin(𝜔𝐷tSubstitute Tmax into x(t):

It can be further simplified:

~

To keep the max value less then 0.025 m solve: Amax (𝜉 ) = 0.025 Æ 𝜉 = 0.281

U i g he e li i he a ( = 3 kg) → 𝑐 2𝜉𝑚𝜔 2 0.281 3 8.16 14.15 /

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The decay rate of a single-DOF spring mass damper system can be used to estimate the damping using logarithmic decrement method.

Consider the two peaks X1 and X2. These two peaks are at time t1 and t2 respectively, where t2-t1 represents a cycle.

𝑢 𝑡 𝜌𝑒 𝑐𝑜𝑠 𝜔 𝑡 𝜙

𝑢 𝑡1 𝑋1 𝑢 𝑡2 𝑋2

𝑡2 𝑡1 2𝜋 𝜔

𝑋1 𝑋2

𝑒

𝑒 2 𝑒

2 1 𝛿 𝑙𝑛

𝑋1 𝑋2

𝜉 2𝜋 1 𝜉2

𝛿 is called logarithmic decrement.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

¾ Logarithmic Decrement Method

De e e S e Da f e F ee V b a Re e

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If 𝜉 ≪ 1,𝛿 2𝜋𝜉 This shows that the ratio of any two successive amplitudes for an underdamped system, vibrating freely, is constant and is a function of the damping only.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

𝛿 𝑙𝑛 𝑋1 𝑋2

𝜉 2𝜋 1 𝜉2

De e e S e Da f e F ee V b a Re e

57CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

¾ A major advantage of this method is that the requirements for equipment's and instruments are minimal.

¾ The vibration can be initiated by any convenient method such as pushing or pulling and only the relative displacement amplitudes need to be measured.

¾ If the damping is truly of the linear viscous form as assumed, any set of j consecutive cycles will yield the same damping factor.

𝛿 𝑙𝑛 𝑋1 𝑋2

𝜉 2𝜋 1 𝜉2

De e e S e Da f e F ee V b a Re e

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For a viscously damped system, the free vibration response shows a 60% reduction in vibration amplitude after 15 cycles. The critical damping is 50 Ns/m. Find the damping coefficient of the system.

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Example

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j=15 cycles X 15 1 0.6 X 0.4X

15δ ln X

X 15 ln

X 0.4X

0.916 → δ 0.061

ξ 2π 1 ξ2

0.061 → ξ 0.0097 0.97%

X X

X X 1

X 1 X 2

X 2 X 3

. . . X 1 X

X X

e 2 1

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Solution

X X 15

e 15 2

1 δ ξ 2π 1 ξ2

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Find the approximate number of cycles to get half vibration amplitude for damping values of:

𝜉 0.01

𝜉 0.02

𝜉 0.05

𝜉 0.1

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Example

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𝑙𝑛 𝑋

0.5𝑋 𝑗𝛿 2𝜋𝑗𝜉 → 𝑗

ln 2 2𝜋𝜉

0.11 𝜉

𝜉 N be f c cle half a li de 0.01 11 0.02 5.5 0.05 2.2 0.1 1

CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Solution

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Example

A one story building is idealized as a rigid girder supported by weightless columns, as shown below. In order to evaluate the dynamic properties of this structure, a free vibration test is made, in which the roof system (rigid girder) is displaced laterally by a hydraulic jack and then suddenly released. During the jacking operation, it is observed that a force of 20 kips [9,072 kg] is required to displace the girder 0.20 in [0.508 cm]. After the instantaneous release of this initial displacement, the maximum displace- ment on the first return swing is only 0.16 in [0.406 cm] and the period of this displacement cycle is T = 1.40 sec.

Based on this information, identify the followings: Undamped frequency of vibration Effective mass of the girder Damping ratio Damped frequency Amplitude after six cycles

63CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Example

Assumption: 𝑇𝐷 2 ~𝑇𝑛

2

→ 𝑇𝑛 2𝜋 𝜔𝑛

1.4sec → 𝜔𝑛 2𝜋 1.4

4.48 𝑟𝑎𝑑 𝑠𝑒𝑐

Effective stiffness: 𝑘𝑒𝑓𝑓 0 2 . 1

0.50 10 1.7519𝑒7

Effective Mass: 𝑚𝑒𝑓𝑓 1. 51 4.4 2

8.7288e5 kg

𝜔𝐷= 𝜔𝑛 1 𝜉2

𝑗 1 , 𝑙𝑛 𝑙𝑛 0.2 0.1

2𝜋𝜉 → 𝜉 1 2

𝑙𝑛 0.2 0.1

0.0355 → 𝜉 3.55%

𝜔𝐷= 𝜔𝑛 1 𝜉2 4.48 1 0.03552 4.447

e 2 e 2 0.0355 3.8126 Æ X =0.2/ 3.8126= 0.0525 ine

64

Forced Vibration: Harmonic Loading (SDOF)

The response of a SDOF system to a harmonic loading is a classical topic in structural dynamics because understanding of the response to harmonic excitation provides an insight to how the structures will respond to other types of forces.

where 𝑃0 is the amplitude of the excitation (N) and 𝜔 is the excitation frequency (rad/s).

𝑃 𝑃0𝑠𝑖𝑛𝜔𝑡

𝑢 2𝜉𝜔 𝑢 𝜔2𝑢 1 𝑚 𝑝 𝑡

𝟏 𝒎 𝑷𝟎𝒔𝒊𝒏𝝎𝒕

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65

𝑢 𝜔2𝑢 1 𝑚 𝑃0𝑠𝑖𝑛𝜔𝑡

𝑢 𝑡 𝐴1𝑠𝑖𝑛𝜔 t+𝐴2𝑐𝑜𝑠𝜔 t

Since the loading is harmonic, the particular solution will be in the form of harmonic function with the same frequency as the loading frequency.

u t 𝑢 𝑡 𝑢 𝑡

𝑢 𝑡 𝐺𝑠𝑖𝑛𝜔𝑡 𝜔2 𝜔2 𝐺𝑠𝑖𝑛𝜔𝑡 1 𝑚 𝑃0𝑠𝑖𝑛𝜔𝑡 → 𝐺

1 𝑚 𝑃0

1 𝜔2 𝜔2

𝐺 𝑃0

𝑘 𝑚𝑘

1 𝜔2 𝜔2

𝑃0 𝑘

1

1 𝜔𝜔 2 β

𝜔 𝜔

u 𝑡 𝐴1𝑠𝑖𝑛𝜔 𝑡 +𝐴2𝑐𝑜𝑠𝜔 𝑡 + 11 𝑠𝑖𝑛𝜔𝑡 CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari

Forced Vibration: Harmonic Loading & Undamped System

¾ Undamped System

Homogeneous solution when 𝑃0=0

𝑢 𝑡 :𝐻𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑢 𝑡 :𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

= ustatic

66

Initial conditions:

u 𝑡0 𝑢0 𝑢 𝑡0 𝑢0

u 𝑡 1

𝑠𝑖𝑛𝜔 t+𝑢0𝑐𝑜𝑠𝜔 t + 1

1 𝑠𝑖𝑛𝜔𝑡

¾ The homogeneous solution or the free response terms are called the transient solution while the particular solution or the forced response term is called the steady stead solution.

¾ The transient vibration will depend on the initial conditions while the steady state vibration depends on the applied force and not the initial conditions.

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u 𝑡 𝐴1𝑠𝑖𝑛𝜔 𝑡 +𝐴2𝑐𝑜𝑠𝜔 𝑡 + 11 𝑠𝑖𝑛𝜔𝑡

Forced Vibration: Harmonic Loading & Undamped System

67

¾ For special case with zero initial conditions (at rest initial condition: 𝑢 0 0,𝑢 0 0), the total response becomes,

u 𝑡 1 1

𝑠𝑖𝑛𝜔𝑡 β𝑠𝑖𝑛𝜔 𝑡

¾ Define 𝑢 , the static displacement or the displacement produced by the

load 𝑃0 applying to the system statically, the Response Ratio 𝑹 𝒕 𝒖 𝒕 𝒖𝒔𝒕

becomes,

𝑅 𝑡 1 1

𝑠𝑖𝑛𝜔𝑡 𝛽𝑠𝑖𝑛𝜔 )

¾ The term 1 1

is called the dynamic magnification factor representing dynamic amplification effect of harmonically applied load.

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Forced Vibration: Harmonic Loading & Undamped System

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Forced Vibration: Harmonic Loading & Undamped System

𝑅 𝑡 1 1

𝑠𝑖𝑛𝜔𝑡 𝛽𝑠𝑖𝑛𝜔 )= MF 𝑠𝑖𝑛𝜔𝑡 𝛽𝑠𝑖𝑛𝜔 ) MF= 𝟏 𝟏 𝜷𝟐

Steady state response component

Transient response component

Final response

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Example

The system shown, has a spring constant of k=10,000 N/m and the mass of m=1kg. If the system is initially at rest, that is 𝑢 0 0,𝑢 0 0, when an excitation of p t 10sin 10𝑡 𝑁 begins, determine an expression for the resulting motion.

Plot time response of the displacement for the first 20 seconds for the excitation frequencies of 𝜔 10, 50, 80, 90 a d 99.9 .

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Solution

𝜔 = 𝑘/𝑚 10000/1 100

𝜔 10 𝑟𝑎𝑑/sec →β = 10 100

0.1

𝐹0 10𝑁 → = 10

10000 0.001𝑚

u 𝑡 1 1

𝑠𝑖𝑛𝜔𝑡 β𝑠𝑖𝑛𝜔 𝑡 0.001 1 1 0.1

𝑠𝑖𝑛10𝑡 0.1𝑠𝑖𝑛100𝑡

𝑢 𝑡 0.001 𝑠𝑖𝑛10𝑡 0.1𝑠𝑖𝑛100𝑡

71CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Solution in MATLAB k=10000; %N/m m=1; %kg u0=0; %initial displacement udot0=0; %initial velocity p0=10; %applied force amplitude wn=(k/m)^0.5 %natural frequency in rad/sec figure i=0; for wbar=[10 50 80 90 99.9]; %excitation frequencies in rad/sec

i=i+1; beta=wbar/wn t=0:0.001:20; u=p0/k*(1/(1-beta^2))*(sin(wbar*t)-beta*sin(wn*t)); subplot(5,1,i) hold on plot(t,u,'linewidth',2) hold off xlabel('Time sec') ylabel ('Response m') set(gca,'fontsize',20) grid on grid minor title (['wbar=', num2str(wbar)]) box on end;

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Solution in MATLAB

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Resonance Æ Undamped System

¾ A resonant frequency is defined as the forcing frequency at which the largest response amplitude occurs.

¾ The steady state response amplitude of an undamped system tends to increase to infinity as the frequency ratio β approaches unity.

for the case of an undamped system, when β =1 or 𝜔 = 𝜔 , the equation of motion is:

The particular solution is:

The total solution of the equation of motion is:

𝑢 𝑡 𝐴1𝑠𝑖𝑛𝜔 t+𝐴2𝑐𝑜𝑠𝜔 t

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With at rest initial conditions, the solution is

The response ratio is,

At time t +Tn , the response ratio is

Consider the j -1 and the j peaks: 𝑡𝑗 𝑗𝑇𝑛 → 𝑢 𝑗𝑇𝑛 𝑇𝑛)=

1 2 𝑢𝑠𝑡

2 2𝜋 𝑐𝑜𝑠 2 𝑠𝑖𝑛 2 1 2 𝑢𝑠𝑡 2𝜋𝑗 2𝜋

𝑡𝑗 1 𝑗 1 𝑇𝑛 Æ 𝑢 𝑗 1 𝑇𝑛 𝑇𝑛)= 1 2 𝑢𝑠𝑡

2 1 2𝜋 𝑐𝑜𝑠 2 1 𝑠𝑖𝑛 2 1

1 2 𝑢𝑠𝑡 2𝜋𝑗

Resonance Æ Undamped System

𝑅 𝑡 𝒖 𝒕 𝒖𝒔𝒕

= 1 2 𝜔 tcos𝜔 t sin𝜔 t)

|𝑢 𝑡𝑗 |- 𝑢 𝑡𝑗 1 1 2 𝑢𝑠𝑡 2π π𝑢𝑠𝑡

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The resonant response of an undamped system

� Note this is an academic result and should be interpreted appropriately for real structure.

� In reality, as the deformation continues to increase, at some point in time the system would fail if it is brittle.

� On the other hand, the system would yield if it is ductile and its stiffness would decrease which leads to the decrease in the natural frequency.

� The he e a al frequency would no longer be equal to the forcing frequency and the particular solution would no longer valid.

Resonance Æ Undamped System

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Example

The system shown, has a spring constant of k=10,000 N/m and the mass of m=1kg. If the system is initially at rest, that is 𝑢 0 0,𝑢 0 0, when an excitation of p t 10sin 100𝑡 𝑁 begins, plot time response of the displacement for the first 2 seconds and calculate the difference in amplitudes between two consecuitive peaks.

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Solution in MATLAB

clc; clear all; close all; k=10000; %N/m m=1; %kg u0=0; %initial displacement udot0=0; %initial velocity p0=10; %applied force amplitude wn=(k/m)^0.5 %natural frequency in rad/sec wbar=100 %excitation frequency in rad/sec figure beta=wbar/wn t=0:0.001:2; u=-p0/(2*k)*(wn*t.*cos(wn*t)-sin(wn*t)); plot(t,u,'linewidth',2) xlabel('Time sec') ylabel ('Response m') set(gca,'fontsize',20) grid on grid minor title (['wbar=', num2str(wbar)]) box on

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Solution in MATLAB

|𝑢 𝑡𝑗 |- 𝑢 𝑡𝑗 1 1 2 𝑢𝑠𝑡 2π π𝑢𝑠𝑡= π 0 π

10 10000

=0.001π

𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑝𝑙𝑜𝑡:0.0204 0.01726 0.0031𝑚 = 0.001π

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𝑃0=0

Forced Vibration: Harmonic Loading & Viscous Damped System

Replace up in the equation of motion

¾Homogeneous solution

¾Particular solution (with the same frequency as forcing frequency)

80

� The homogeneous solution or the free vibration solution term is called a transient solution because it will damp out very quickly.

� The particular solution terms is called a steady state solution. This steady state response vibrates at the frequency of the applied loading.

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Forced Vibration: Harmonic Loading & Viscous Damped System

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Forced Vibration: Harmonic Loading & Viscous Damped System

𝒖 𝒕 𝒆 𝝃𝝎𝒏𝒕 𝑨𝟏𝐬𝐢𝐧𝝎𝑫𝒕 𝑨𝟐𝒄𝒐𝒔𝝎𝑫𝒕 𝝆𝒔𝒊𝒏 𝝎𝒕 𝝓

¾ A1 and A2 will be determined based on the initial conditions ¾ The homogeneous component will damp out in a few cycles depending on the

damping value

82

Dynamic Amplification Factor

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Dynamic magnification factor depends only on the frequency ratio and the damping ratio of the system and it provides the information of the magnitude of the dynamic response in comparison to the magnitude of the static response of the system under the same magnitude of the applied force.

Dynamic magnification factor D(𝜉,𝛽 ):

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Example

The system shown, has a spring constant k=10,000 N/m and the mass of m=1kg. A damping equivalent to 𝜉=0.2 is considered for the system. If the system is initially at rest, that is 𝑢 0 0,𝑢 0 0, when an excitation p t 10sin 80𝑡 𝑁 begins, determine an expression for the resulting motion.

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Solution

𝜔 = 𝑘/𝑚 10000/1 100

𝜔 80 𝑟𝑎𝑑/𝑠𝑒𝑐 →𝛽 = 0 100

0.8

𝑃0 10𝑁 → = 10

10000 0.001𝑚

𝜉=0.2 𝜔𝐷= 𝜔𝑛 1 𝜉2 100 1 0.22 97.9796 rad/sec

𝜌 0 1 1 2 2 2 2

0.001 1 1 0. 2 2 2 0.2 0. 2

0.0021 m

𝑡𝑎𝑛𝜙 2 1 2

2 0.2 0. 1 0. 2

0.8889 → 𝜙= 0.7266 rad

𝑢 𝑡 𝑒 𝐴1𝑠𝑖𝑛𝜔𝐷𝑡 𝐴2𝑐𝑜𝑠𝜔𝐷𝑡 𝜌𝑠𝑖𝑛 𝜔𝑡 𝜙 Æ

𝑢 𝑡 𝑒 20 𝐴1𝑠𝑖𝑛97.98𝑡 𝐴2𝑐𝑜𝑠97.98𝑡 0.0021𝑠𝑖𝑛 80𝑡 0.7266

𝑢 0 𝐴2+ 0.0021𝑠𝑖𝑛 0.7266 0 → 𝐴2= 0.0014 𝑢 𝑡 𝜉𝜔𝑛𝑒 𝐴1𝑠𝑖𝑛𝜔𝐷𝑡 𝐴2𝑐𝑜𝑠𝜔𝐷𝑡 𝑒 𝐴1𝜔𝐷𝑐𝑜𝑠𝜔𝐷𝑡 𝐴2𝜔𝐷𝑠𝑖𝑛𝜔𝐷𝑡 𝜌𝜔𝑐𝑜𝑠 𝜔𝑡 𝜙 𝑢 0 0.2 100 0.0014+97.98𝐴1+0.1661cos(-0.7266)=0 Æ 𝐴1= -9.8125e-04

𝒖 𝒕 𝒆 𝟐𝟎𝒕 𝟎.𝟎𝟎𝟎𝟗𝟖𝒔𝒊𝒏𝟗𝟕.𝟗𝟖𝒕 𝟎.𝟎𝟎𝟏𝟒𝒄𝒐𝒔𝟗𝟕.𝟗𝟖𝒕 𝟎.𝟎𝟎𝟐𝟏𝒔𝒊𝒏 𝟖𝟎𝒕 𝟎.𝟕𝟐𝟔𝟔

85CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari

Solution in MATLAB clc; clear all; close all; k=10000; %N/m m=1; %kg u0=0; %initial displacement udot0=0; %initial velocity p0=10; %applied force amplitude wn=(k/m)^0.5 %natural frequency in rad/sec wbar=80 %excitation frequency in rad/sec beta=wbar/wn zeta= 0.2;% damping ratio wd=wn*(1-zeta^2)^0.5 %damped frequency ro=(p0/k)*(1)/(((1-beta^2)^2+(2*zeta*beta)^2)^0.5) phi=atan(2*zeta*beta/(1-beta^2)) A2=u0-ro*sin(-phi) A1=(zeta*wn*A2-ro*wbar*cos(-phi))/wd figure t=0:0.0001:2; u=exp(-zeta*wn.*t).*(A1*sin(wd*t)+A2*cos(wd*t))+ro*sin(wbar*t-phi); plot(t,u,'linewidth',2) xlabel('Time sec') ylabel ('Response m') set(gca,'fontsize',20) grid on; grid minor; box on t=0; u0=exp(-zeta*wn.*t).*(A1*sin(wd*t)+A2*cos(wd*t))+ro*sin(wbar*t-phi) udot=diff(u); udot0=udot(1)

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Solution in MATLAB

87

The displacement amplitude of an SDOF system due to harmonic force is known for two excitation frequencies. At 𝜔 𝜔𝑛 ,𝑢0 5 mm; at 𝜔 5𝜔𝑛 ,𝑢0 0.02 mm. Estimate the damping ratio of the system.

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Example

𝜌 𝑝0 𝑘

1 1 𝛽2 2 2𝜉𝛽 2

@ 𝜔 𝜔𝑛 : 𝛽=1 Æ 𝜌 0 1 2

0.005

@ 𝜔 5𝜔𝑛 : 𝜌~ 0 1 2 = 0

1 25

= 0.00002

Æ 𝜉= 0.05 (5%)

88

A bridge structure is modelled as a SDOF system with a damping factor of 𝜉. Find the maximum possible Dynamic Magnification Factor of the bridge in terms of 𝜉.

For a bridge with 5% damping determine the maximum possible dynamic magnification factor.

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Example

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Solution

𝐷 1 1 2 2 2 2

→ 1 2

1 𝛽2 2 2𝜉𝛽 2 2 1 𝛽2 2𝛽 2 2𝜉𝛽 2𝜉 0

The first term can not be zero, so the second term must be zero:

2 1 𝛽2 2𝛽 2 2𝜉𝛽 2𝜉 =0

Æ 2𝛽3 2𝛽 4𝜉2𝛽 0

Æ 𝛽𝑝𝑒𝑎𝑘 1 2𝜉2

It can be seen that only when 𝜉 1 2

yields positive real value of 𝛽. Replace 𝛽𝑝𝑒𝑎𝑘 in the original equation of D:

𝐷𝑚𝑎𝑥 1

1 1 2𝜉2 2 2 2𝜉 1 2𝜉2 2

1 2𝜉 1 𝜉2

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Solution in MATLAB

clc; clear all; close all; zeta=0.05; beta=0:0.0001:2; D=((1-beta.^2).^2+(2*zeta.*beta).^2).^-0.5; figure plot(beta,D,'linewidth',2) xlabel('\beta') ylabel('D') grid on grid minor set(gca,'fontsize',20) title ('\ e a=0.05 ) %using the plot

[i,j]=max(D); D_max=i beta_max=beta(j)

%using formula beta_peak=(1-2*zeta^2)^0.5 Dmax=1/((2*zeta)*(1-zeta^2)^0.5)

beta_max = 0.9975 Æ Maximum D is happening slightly less than 1. D_max = 10.0125 Æ Considering dynamic effects, the level of response is approximately 10 times that of static!

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Resonance- Damped System

� When the value of damping ratio is small for a damped system, the maximum steady state response amplitude occurs at a frequency ratio slightly less than unity.

� These resonant frequencies can be determined by setting the derivative of the dynamic

magnification factor, with respect to frequency ratio to be zero, i.e.,

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Since the first terms cannot be zero, the second term in the above equation must be zero, i.e.

Only yields positive real value of 𝛽.

¾ Note that although when the frequency of the applied load equals to the natural frequency of the system does not cause maximum response for a damped system, some literature refers such condition as the resonance.

Resonance- Damped System

1 2

1 𝛽2 2 2𝜉𝛽 2 2 1 𝛽2 2𝛽 2 2𝜉𝛽 2𝜉 0

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In such case, the maximum dynamic magnification factor is

Typical values of structural damping are generally less than 20%. Hence, the term in the square root is very small, i.e.,

Resonance- Damped System

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¾ Resonant Amplification Method

Determine the Under-Da ed S e Da

¾ This method is based on measuring the steady state amplitudes of relative displacement response produced by separate harmonic loadings at discrete values over a wide range including the natural frequency.

¾ When this series of test is done, a frequency- response curve, a plot between the steady state amplitudes and the frequency ratio is obtained.

¾ Generally, the peak of the frequency-response curve for a low damped system is quite narrow, it is usually necessary to shorten the intervals of the discrete frequencies in the neighbourhood of the peak to get a good resolution of its shape.

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For small damping ratio in the practical range of interest, the approximation of the maximum dynamic magnification factor is:

From Dmax the damping ratio can then be determined.

Determine the Under-Da ed S e Da

¾ This method requires a simple instrumentation to measure dynamic response amplitudes at discrete values of frequency and fairly simple dynamic loading equipment.

¾ However, obtaining the 𝜌0 may present a problem because the typical harmonic loading system cannot produce a loading at zero frequency.

¾ Close to the peak, we need to use a very short intervals of frequency change to get a good resolution shape.

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Determine the Under-Da ed S e Da

¾ Half-Power (Band-Width) Method

This method requires essentially the same experiment as the resonant amplification method. However, this method is based on the frequencies at which the response amplitude is reduced to rather than the amplitude .

The maximum amplification factor is:

Determine the frequency ratios that produce the dynamic amplification factor of

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For small values of 𝜉 ,

Determine the Under-Da ed S e Da

This method is also called 3 dB method as:

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Example

A structure has been placed on a shake table which can apply harmonic base vibration of specified frequencies and amplitudes. At each excitation frequency, the steady state response is measured and recorded. Below figure is showing the result. From this figure, obtain the natural frequency and damping ratio.

Frequency (Hz)

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Solution

The peak of the frequency response curve occurs at 3.59Hz. Assuming damping is small, the natural frequency will be fn=3.59Hz.

The peak value of response is 12.8. If we draw a horizontal line at 12. 2

9.05 we can get half power points at fa=3.44Hz and fb=3.74Hz . So damping ratio will be:

𝜉 2

3. 4 3.44 2 3.5

=0.042=4.2%

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State-Space Representation for a Single DOF System

x is the state space vector y is the output (measured response) A is the state transition matrix B is the input matrix C is the output matrix u=p is the input

101

Structural or Hysteresis Damping

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¾ Structural damping is an alternative damping model in vibration analysis. ¾ This damping is originated from the hysteresis property of stress strain curves for

metal materials. ¾ The damping force is proportional to the response amplitude with a 90 degrees

phase difference.

he e i i he i agi a i .

Viscous damping model is often not representative when applied to MDOF systems.

𝜂: structural damping loss factor

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Assignment 2 (10% of Final Mark)

A bridge is sitting on top of four identical piers as shown in below. The bridge is subject to a ground motion as a result of an earthquake. Assume the ground displacement is modelled as a harmonic loading ug(t)= U0sin(𝜔t) with U0 being 0.5m. The mass of the bridge deck is Mdeck = 6×106 kg. The piers have a circular cross section with a height of L=10m, a radius of 3m at the base and 1.5m at the top just under the deck. Consider an elastic modulus of E = 20 GPa for the piers.

(1) Plot the time response of the bridge for 10 seconds under the loading if 𝜔 2

where 𝜔 is the natural frequency of the bridge.

(2) Determine the maximum bending stress at the base of the pier if the excitation frequency varies between 𝜔1 2 to 𝜔2 2𝜔 where 𝜔 is the natural frequency of the bridge. (3) As a result of material deterioration, the elastic modulus of the material in all piers has dropped by 10%. Investigate this damage by comparing the responses of the healthy and damaged bridge in time and frequency domain and discuss your results. You can compare displacement, velocity, acceleration and strain time responses. (4) If a strain gauge is installed at the outer surface of the pier in the bottom, compare the time response of strain between the

healthy and damaged bridge if the excitation frequency is 𝜔1 2 . Assumptions: - Consider a SDOF vibration system in horizontal direction. - Assume zero initial conditions. - Damping ratio is 𝜁 10%. - Ignore the mass of the piers. - The stiffness is due to the piers only. - The deck of the bridge is rigid. - The effect of damage is only on pier stiffness.

Due at 5pm 23 October 2020. (A penalty of 10% will apply for each day of late submission for assignments.) Online submission portal will be created on Moodle.