lecture 1

7/21/2019 Lecture 1

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S. Turteltaub

Ae 2135-II - 2015

Ae 2135-II

Vibrations

Faculty of Aerospace EngineeringDelft University of Technology

Sergio Turteltaub

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Course setup

AE2135-II Vibrations

Responsible Instructor Sergio Turteltaub

Contact [email protected]

NB2.24

Blackboard

Education Period 2

ECTS 3

Lectures Tuesdays [Theory]Instructions Thursdays [Practice]

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Course setup

AE2135-II Vibrations

Textbook Daniel J. Inman

Special TU Delft edition

ISBN 9781784344726

Exams Regular: Period 2

[22 Jan 2016, 9:00-12:00]

Resit: Period 3

Allowed material during

exam

• Only official formula

sheet

• (Graphical) Calculator

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Course setup

You are here

You need to know this [yesterday!]

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What are vibrations?

Vibrations in solids are repetitive (or quasi-repetitive) motions

of a structure or structural component

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What are vibrations?

Vibrations in solids are repetitive (or quasi-repetitive) motions

of a structure or structural component

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Why do we study vibrations?

• To determine the internal loads on the structure

• To determine the movement (deformation) of thestructure

• To be able to investigate the fatigue lifetime of the

structure

• To be able to investigate possible noise issues withthe structure

Aerospace structures are commonly subjected to vibrationsduring operation. For analysis and design purposes it iscritical to model the response of a structure under free andforced loading conditions.

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Multiple degree of freedom (MDOF)

Definitions

Degree of freedom:

A degree of freedom (DOF) is a (variable) scalar quantity thatis used to specify the location of the system as a function oftime (“coordinates” such as distances, angles, etc.)

Single degree of freedom (SDOF)

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Forced vibration

Definitions

Forced vibration

A system is said to be subjected to forced vibrations if there is atime-varying external force applied to it (called the forcingload) that is independent of the motion of the system

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Forced vibration

Definitions

Free vibration

A system is said to be under free vibrations if it is not subjectedto a forcing load

Free vibration

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Definitions

Dissipation

A system is said to be dissipative if its total energy (potential +kinetic) decays as the system vibrates.

The energy dissipated is converted into another form (e.g.,heat) and transferred out of the system

In this course, dissipation is related to dampers and/orfrictional forces

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Course overview

From simple to more complex systems:

Beginning of the course: Undamped, single degree of freedom

under free vibrations

End of the course: Damped, multiple degrees of freedom underrandom external loading

Topic Simple Complex

Degrees of freedom Single Multiple

Loading “Free” vibrations Forced vibrations

Dissipation Undamped Damped

Forced vibrations Harmonic Random

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Course contents

1. Introduction to vibrations and modelling of structures.

2. Free vibrations in single degree-of-freedom models.

3. Harmonically forced vibrations.

4. Vibrations under general loading types: impulse loading,step loading, arbitrary transient loading.

5. Vibration concepts: eigenfrequency, resonance, critical,undercritical and overcritical damping, transfer function.

6. Multiple degree-of-freedom systems.

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Study goals and learning outcomes

The purpose of this course is to provide an introduction to

vibrations of structures and structural components. At the endof the course, students should be able to analyze basicvibration problems and be able to use this information fordesign purposes.

Learning outcomes1. Represent an actual structure or structural component

using a mass-damper-spring model.

2. Formulate and solve the equation of motion associated tothe mass-damper-spring model.

3. Understand the influence of the main model parameters onthe structural response.

4. Interpret the results of the simulations in terms of globalcases such as resonance and under or over-damped responses.

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Topic 1

Modelling approach

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As an analyst, one must choose a modeling approach:

CompMechLab

Continuous:Flexible components modeled

e.g. via finite elements

Modelling approach

Ae 2135-II

Discrete:Interconnected masses

and/or rigid bodies

Components

Connectors

Support

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Lumped mass/ rigid body discrete models:

• Advantages:

Computational effort: relatively smallMight be solved in closed form in simple cases

Easy to modify for design/parametric analyses

• Limitations:

Might not be able to capture some critical effects

Requires experience to make proper assumptions

Modelling approach

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Basic modelling assumptions


and/or rigid bodies

Components

Connectors

Modelling approach

1. Mass of the system is lumped inthe components only:• Point mass [translation] and/or• Rigid body [translation+rotation]

2. Connectors are assumed massless

3. Components do not deform (theyonly translate and/or rotate)

4. All deformations are lumped in theconnectors

5. Force and/or kinematic conditionsat the supports are known a priori

Support

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Supports

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Kinematic connectors

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Force connectors

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Force connectors

Spring stiffness

Torsional stiffness

Damping / viscosity

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Interconnected masses and/or rigid bodies

Each concentrated mass:In 3D: 3 scalars; in 2D: 2 scalars

• location of massEach rigid body:In 3D: 6 scalars; in 2D: 3 scalars

• location of one point• orientation of body

Degrees of freedom Types of connectors:1. Translational and/or

rotational spring and/or

damping2. Hinges,…

1

2

1. Identify main components2. Introduce connectors

Overview of modelling approach

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Each concentrated mass:In 3D: 3 scalars; in 2D: 2 scalars

• location of massEach rigid body:In 3D: 6 scalars; in 2D: 3 scalars

• location of one point• orientation of body

Degrees of freedom Types of connectors:1. Information about forces

transmitted between

components2. Kinematic constraints

1

2



The size of the problem may bereduced by elimination ofdependent degrees of freedom

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Equation(s) of motion

1

2



Governing equations of motion are obtained from the balance oflinear and/or angular momentum (“Newton’s second law”)

Initial conditions

Initial position and velocity of all components

Boundary conditions

Information about the supports (external connections)

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1

2

All equations are assembled in a system of equations

Equations of motion for each component:

Linear momentum


Free body diagram and kinetic diagramfor each component

Angular momentum (rigid body)

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Discrete force connectorsMassless force connector: sum of forces and moments is zero(formally like equilibrium since inertial term is assumed zero)

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Discretizing continuous force connectors1. Axial stiffness of a (continuous) beam:• Beam of length L, cross section A and Young’s modulus E

subjected to an axial force F at its tip.

• The tip displaces a distance δ axially.

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Discretizing continuous force connectors1. Axial stiffness of a (continuous) beam:• Beam of length L, cross section A and Young’s modulus E

subjected to an axial force F at its tip.


• Equivalent discrete

system?

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Discretizing continuous force connectors1. Axial stiffness of a (continuous) beam:• Neglecting the mass of the beam, it exerts an axial force equal

to F at the tip• Equivalent spring should exert the same force

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Discretizing continuous force connectors1. Axial stiffness of a (continuous) beam:• Neglecting the mass of the beam, it exerts an axial force equal

to F at the tip• Equivalent spring should exert the same force

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Discretizing continuous force connectors1. Axial stiffness of a (continuous) beam: Equivalent spring

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Discretizing continuous force connectors2. Axial stiffness of two beams in parallel:• Two beams of length L, cross sections A

1 , A

2 and Young’s

moduli E 1 , E 2 subjected to an axial force F at its tip.


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1 , A

2 and Young’s




system?

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1 , A

2 and Young’s


• Equivalent spring should exert the same force

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1 , A

2 and Young’s



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Discretizing continuous force connectors2. Axial stiffness of two beams in parallel: Equivalent spring

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Discretizing continuous force connectors3. Axial stiffness of two beams in series:• Two beams of lengths L

1 , L

2 , cross section A and Young’s

moduli E 1 , E 2 subjected to an axial force F at its tip. • The tip displaces a distance δ axially. Beams 1 and 2 contract δ

1

and δ2

respectively

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1 , L


moduli E 1 , E 2 subjected to an axial force F at its tip. • The tip displaces a distance δ axially. Beams 1 and 2 contract δ

1

and δ2

respectively


system?

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1 , L




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1 , L




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Discretizing continuous force connectors3. Axial stiffness of two beams in series: Equivalent spring

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Discretizing continuous force connectors4. Torsional stiffness of a (continuous) rod:• Rod of length L, (polar) moment of inertia J and shear modulus

G subjected to a torque T at its tip.

• The rod rotates an angle ϕ.

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G subjected to a torque T at its tip.

• The rod rotates an angle ϕ.

• Equivalent discrete system?

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G subjected to a torque T at its tip.• Equivalent torsional spring should exert the same torque

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G subjected to a torque T at its tip.• Equivalent torsional spring should exert the same torque

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Discretizing continuous force connectors4. Torsional stiffness of a (continuous) rod: equivalent torsionalstiffness

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Discretizing continuous force connectors5. Bending stiffness of a (continuous) beam:• Beam of length L, moment of inertia I and Young’s modulus E

subjected to a transverse force F at its tip.

• The tip displaces a distance δ transversely.

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Discretizing continuous force connectors5. Bending stiffness of a (continuous) beam:• Beam of length L, moment of inertia I and Young’s modulus E

subjected to a transverse force F at its tip.

• The tip displaces a distance δ transversely.

• Equivalent discrete system?

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Discretizing continuous force connectors5. Bending stiffness of a (continuous) beam:• Neglecting the mass of the beam, it exerts a transverse force

equal to F at the tip• Equivalent (transverse) spring should exert the same force

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Discretizing continuous force connectors5. Bending stiffness of a (continuous) beam:• Neglecting the mass of the beam, it exerts a transverse force

equal to F at the tip• Equivalent (transverse) spring should exert the same force

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Discretizing continuous force connectors5. Bending stiffness of a (continuous) beam: Equivalent spring

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Discretizing continuous force connectors6. Bending stiffness with torsional spring• Neglect mass of the beam and torsional spring

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6. Bending stiffness with torsional spring• Neglect mass of the beam and torsional spring

• Equivalent (transverse) spring should exert the same (reaction)force

Discretizing continuous force connectors

f

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6. Bending stiffness with torsional spring• Neglect mass of the beam and torsional spring

• Equivalent (transverse) spring should exert the same (reaction)force• Relation between angle ϕ and force F

Discretizing continuous force connectors

i i i i f

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• Equivalent (transverse) spring should exert the same (reaction)force• Transverse deflection

i i i i f

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• Equivalent (transverse) spring should exert the same (reaction)force

Di i i i f

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Discretizing continuous force connectors6. Bending stiffness with torsional spring

Modelling and solution steps

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Formulation of the problem:


and/or rigid bodies

Components

Connectors


1. Identify the system that needs tobe analyzed

2. Discretize the system: components,internal connectors andinteractions with the environment

3. Introduce coordinates (DOFs) tofully describe the system4. Use free body diagrams and

kinetic diagrams to set up theequation(s) of motion

5. Identify the initial conditions and

the boundary conditions6. Solve the problem7. Test your solutionSupport


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Issues to take into account in order to choose a solutionmethod:


and/or rigid bodies

Components

Connectors


• How many independent degrees-of-freedom does the system have?

• Is the system forced or free tovibrate?

• Is the system undamped or damped?

• How to test the correctness of thesolution? (i.e., how to check that itmakes sense?)

Support

Topic Simple Complex

DOF Single Multiple

Loading “Free” Forced

Dissipation Undamped Damped

Forcing Harmonic Random

A id thi

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Avoid this

A good design is important

lecture 1

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