Structural Analysis - Caprani

Structural Analysis IV Chapter 1 – Course Introduction

Dr. C. Caprani 1

Chapter 1 - Course Introduction

1.1 Introduction ......................................................................................................... 2

1.1.1 Background .................................................................................................... 2

1.1.2 Course Aims .................................................................................................. 3

1.1.3 Programme ..................................................................................................... 4

1.1.4 Reading Material ........................................................................................... 5

1.1.5 Website .......................................................................................................... 8

1.2 Syllabus ................................................................................................................. 9

1.2.1 Semester 1 Only ............................................................................................. 9

1.3 Assessment .......................................................................................................... 10

1.3.1 Examination ................................................................................................. 10

1.3.2 Continuous Assessment ............................................................................... 11

Rev. 1


Dr. C. Caprani 2

1.1 Introduction

1.1.1 Background

Within 9 months of starting this course you will be qualified to practice as a structural

engineer. Every single day of your career as a structural engineer, you will be

responsible for the lives of every person that will ever use the structures you design.

But more than that: at a minimum you will also be responsible for:

• The safety of the people who will build your structure;

• The quality of life of future generations – structural engineers are in a unique

position to contribute to limiting the significant carbon emissions of the

construction industry;

• The best economic use of your clients’ money to best achieve their goals;

• The use of your time that best achieves your employer’s goals.

Though mistakes that lead to collapse of a structure are rare, they do happen. Often it

is through an unreasonable faith in a computer analysis that makes this so. With

excellent structural intuition; an ability to properly model the structure with structural

analysis software, and; an ability to check computer output with appropriate hand

calculations, the risk of such collapses can be minimized.

This course builds on your ability to analyse statically indeterminate structures from

the 3rd year course and introduces new ideas and areas of study. We do this so that

you are best equipped to deal with the realities of structural analysis and design.


Dr. C. Caprani 3

1.1.2 Course Aims

Given the background just discussed, the general aims of this course are to provide

students with:

• An improved understanding and intuition of structural behaviour;

• An ability to properly model structures and to check output by hand;

• Knowledge of different types of structures and their behaviour.


Dr. C. Caprani 4

1.1.3 Programme

Teaching

This course is taught in Semester 1 only. It is taught as follows:

• 3 hours lectures per week;

• 2 hours of computer laboratory every two weeks.

Assessment

We asses your performance on this course as follows:

• Submission of laboratory work - 20% of the marks;

• A 3-hour end-of-semester examination - 80% of the marks.

In the unlikely event of changes to the above arrangements, the changes will be

notified to you well in advance of their implementation by your lecturer.


Dr. C. Caprani 5

1.1.4 Reading Material

Reading about projects and new techniques will be a major part of your engineering

career (CPD). You should read as many different versions or explanations of the

same topic or material as you can. This way it is more likely that you will find a

means of explanation that works best for you.

Some good sources for this course are:

General Understanding of Structural Behaviour

• Brohn, D., Understanding Structural Analysis, 4th Edn., New Paradigm

Solutions, 2005.

• Heyman, J., Basic Structural Theory, Cambridge University Press, 2008.

• Jennings, A., Structures: from theory to practice, Spon Press, 2004.

• Ji, T., and Bell, A., Seeing and Touching Structural Concepts, Taylor & Francis,

2008.

• Williams, M.S., and Todd, J.D., Structures: theory & Analysis, Macmillan,

1999.

General Structural Analysis

• Coates, R.C., Coutie, M.G., and Kong, F.K., Structural Analysis, 3rd Edn.,

Chapman & Hall, 1987.

• Ghali, A., Neville, A., Brown, T.G., Structural Analysis: A Unified Classical

and Matrix Approach, 5th Edn., Taylor & Francis, 2003.

• McKenzie, W.M.C., Examples in Structural Analysis, Taylor and Francis,

Abington, 2006.


Dr. C. Caprani 6

Books for Specific Topics

• Charlton, T.M., Analysis of Statically Indeterminate Frameworks, Longmans,

1961.

• Charlton, T.M., Energy Principles in Theory of Structures, Oxford University

Press, 1973.

• Davies, G.A.O., Virtual Work in Structural Analysis, John Wiley & Sons, 1982.

• Dym, C.L., Structural Modeling and Analysis, Cambridge University Press,

2005.

• Guarracino, F. and Walker, A., Energy Methods in Structural Mechanics,

Thomas Telford, 1999.

• Heyman, J., Beams and Framed Structures, 2nd Edn., Pergamon Press, 1974.

• Heyman, J., Elements of the Theory of Structures, Cambridge University Press,

1996.

• Hodge, P.G., Plastic Analysis of Structures, McGraw-Hill, New York, 1959.

• Kong, F.K., Prentis, J.M. and Charlton, T.M., ‘Principle of virtual work for a

general deformable body – a simple proof’, The Structural Engineer, Vol. 61A,

No. 6, 1983.

• Neal, B.G., Structural Theorems and their Applications, Pergamon Press, 1964.

• Rees, D.W.A., Mechanics of Solids and Structures, Imperial College Press,

London, 2000.

• Thompson, F., and Haywood, G.G., Structural Analysis Using Virtual Work,

Chapman and Hall, 1986.


Dr. C. Caprani 7

Structural Dynamics

• Beards, C.F., Structural Vibration Analysis: modelling, analysis and damping of

vibrating structures, Ellis Horwood, Chichester, England, 1983.

• Bhatt, P., Structures, Longman, Harlow, England, 1999.

• Case, J., Chilver, A.H. and Ross, C.T.F., Strength of Materials and Structures,

4th edn., Arnold, London, 1999.

• Clough, R.W. and Penzien, J., Dynamics of Structures, 2nd edn., McGraw-Hill,

New York, 1993.

• Craig, R.R. and Kurdila, A.J., Fundamentals of Structural Dynamics, 2nd End.,

Wiley, New York, 2006.

• Irvine, M., Structural Dynamics for the Practising Engineer, Allen & Unwin,

London, 1986.

• Kreyszig, E., Advanced Engineering Mathematics, 7th edn., Wiley, 1993.

• Smith, J.W., Vibration of Structures – Applications in civil engineering design,

Chapman and Hall, London, 1988.


Dr. C. Caprani 8

1.1.5 Website

The course will be supported through the lecturer’s website:

www.colincaprani.com – go to the Structural Engineering section of the site.

On the site there are two main resources:

• Lecture notes: most of the lecture notes will be available in PDF format for

download from the website. Class handouts will still be the main source of

material.

• Discussion Forum: to facilitate students studying on their own, or maybe when

home for the weekend, there is a forum through which you can liaise with

others. Feel free to ask questions and to answer them. Though the forum will be

facilitated by your lecturer, there is no guarantee that a question will receive an

answer. This is primarily a way to encourage student-to-student remote learning.

Some other resources that may prove useful will be links to sites with good material

and the provision of some software (with absolutely no guarantees!).

The website support for the course is only meant to help, so please:

• Do not abuse either the facility or the facilitator!

• Try to use the site to best help you and your friends.

• Suggest ways to improve the usefulness of the website.

• Do not post inappropriate comment/content – your site access will be removed,

with more serious consequences also possible.

You are required to register for the forum – only registrations in your own name

will be approved. You can change your display name later on.

http://www.colincaprani.com/


Dr. C. Caprani 9

1.2 Syllabus

1.2.1 Semester 1 Only

The topics to be covered in the lectures are as follows:

Virtual Work (Compound Structures)

A Virtual Work analysis is used for structures whose members undergo a

combination of stress resultants, most notably bending and axial force.

Virtual Work (Arches)

Here we use Virtual Work to analyse moments/shears and axial forces in parabolic

and semi-circular arches.

Matrix Stiffness Method

This topic provides an introduction to the basis of modern structural analysis

software. This is a particular case of finite element analysis.

Influence Line Analysis

These are used to determine design loads for members in structures subjected to

moving loads (e.g. bridges) or for repeated analysis of a structure under various

loading scenarios.

Structural Dynamics

This topic covers exact and approximate methods of determining the motion of

structures under dynamic loading situations.


Dr. C. Caprani 10

1.3 Assessment

1.3.1 Examination

The examination will be held at the end of Semester 1. The format is:

Layout

There will be 5 questions and you are to answer 4.

Marking

Each question is worth 25%.

Timing

The exam is 3 hours in duration.

Format

The questions will examine a topic or topics from the lectures. Further information

will be given.

Exam Handout

A handout will be attached to the paper in each exam with relevant information and

formulae. A copy of this will be given to you during Semester 1.

Note: in the event of any changes to these arrangements, they will be notified to you

well in advance.


Dr. C. Caprani 11

1.3.2 Continuous Assessment

General

Continuous Assessment is primarily carried out through laboratory work. These labs

are not the same as traditional labs you may have already done. You will be given

tasks, with a schedule of dates for delivering various aspects of the problem to ensure

an even distribution of workload. You will be given access to the lab to facilitate your

work, not only at scheduled lab times. We hope that this will improve your prospects

to self-direct your learning.

Structural Analysis IV Chapter 2 – Virtual Work: Compound Structures

Dr. C. Caprani 1

Chapter 2 - Virtual Work: Compound Structures

2.1 Introduction ......................................................................................................... 3

2.1.1 Purpose .......................................................................................................... 3

2.2 Virtual Work Development ................................................................................ 4

2.2.1 The Principle of Virtual Work ....................................................................... 4

2.2.2 Virtual Work for Deflections ......................................................................... 8

2.2.3 Virtual Work for Indeterminate Structures.................................................... 9

2.2.4 Virtual Work for Compound Structures ...................................................... 11

2.3 Basic Examples .................................................................................................. 14

2.3.1 Example 1 .................................................................................................... 14

2.3.2 Example 2 .................................................................................................... 21

2.3.3 Example 3 .................................................................................................... 32

2.3.4 Example 4 .................................................................................................... 40

2.3.5 Problems ...................................................................................................... 48

2.4 Past Exam Questions ......................................................................................... 52

2.4.1 Sample Paper 2007 ...................................................................................... 52

2.4.2 Semester 1 Exam 2007 ................................................................................ 53

2.4.3 Semester 1 Exam 2008 ................................................................................ 54

2.4.4 Semester 1 Exam 2009 ................................................................................ 55

2.4.5 Semester 1 Exam 2010 ................................................................................ 56

2.4.6 Semester 1 Exam 2011 ................................................................................ 57

2.5 Appendix – Trigonometric Integrals ............................................................... 58

2.5.1 Useful Identities ........................................................................................... 58


Dr. C. Caprani 2

2.5.2 Basic Results ................................................................................................ 59

2.5.3 Common Integrals ....................................................................................... 60

2.6 Appendix – Volume Integrals ........................................................................... 67

Rev. 1


Dr. C. Caprani 3

2.1 Introduction

2.1.1 Purpose

Previously we only used virtual work to analyse structures whose members primarily

behaved in flexure or in axial forces. Many real structures are comprised of a mixture

of such members. Cable-stay and suspension bridges area good examples: the deck-

level carries load primarily through bending whilst the cable and pylon elements

carry load through axial forces mainly. A simple example is a trussed beam:

Other structures carry load through a mixture of bending, axial force, torsion, etc. Our

knowledge of virtual work to-date is sufficient to analyse such structures.


Dr. C. Caprani 4

2.2 Virtual Work Development

2.2.1 The Principle of Virtual Work

This states that:

A body is in equilibrium if, and only if, the virtual work of all forces acting on

the body is zero.

In this context, the word ‘virtual’ means ‘having the effect of, but not the actual form

of, what is specified’.

There are two ways to define virtual work, as follows.

1. Virtual Displacement:

Virtual work is the work done by the actual forces acting on the body moving

through a virtual displacement.

2. Virtual Force:

Virtual work is the work done by a virtual force acting on the body moving

through the actual displacements.

Virtual Displacements

A virtual displacement is a displacement that is only imagined to occur:

• virtual displacements must be small enough such that the force directions are

maintained.


Dr. C. Caprani 5

• virtual displacements within a body must be geometrically compatible with

the original structure. That is, geometrical constraints (i.e. supports) and

member continuity must be maintained.

Virtual Forces

A virtual force is a force imagined to be applied and is then moved through the actual

deformations of the body, thus causing virtual work.

Virtual forces must form an equilibrium set of their own.

Internal and External Virtual Work

When a structures deforms, work is done both by the applied loads moving through a

displacement, as well as by the increase in strain energy in the structure. Thus when

virtual displacements or forces are causing virtual work, we have:

00I E

E I

WW W

W W

δδ δ

δ δ

=− =

=

where

• Virtual work is denoted Wδ and is zero for a body in equilibrium;

• External virtual work is EWδ , and;

• Internal virtual work is IWδ .

And so the external virtual work must equal the internal virtual work. It is in this

form that the Principle of Virtual Work finds most use.


Dr. C. Caprani 6

Application of Virtual Displacements

For a virtual displacement we have:

0

E I

i i i i

WW W

F y P e

δδ δ

δ δ

==

⋅ = ⋅∑ ∑

In which, for the external virtual work, iF represents an externally applied force (or

moment) and iyδ its virtual displacement. And for the internal virtual work, iP

represents the internal force (or moment) in member i and ieδ its virtual deformation.

The summations reflect the fact that all work done must be accounted for.

Remember in the above, each the displacements must be compatible and the forces

must be in equilibrium, summarized as:

Set of forces in

equilibrium

Set of compatible

displacements


Dr. C. Caprani 7

Application of Virtual Forces

When virtual forces are applied, we have:

0

E I

i i i i

WW W

y F e P

δδ δ

δ δ

==

⋅ = ⋅∑ ∑

And again note that we have an equilibrium set of forces and a compatible set of

displacements:

In this case the displacements are the real displacements that occur when the structure

is in equilibrium and the virtual forces are any set of arbitrary forces that are in

equilibrium.

Set of compatible

displacements

Set of forces in

equilibrium


Dr. C. Caprani 8

2.2.2 Virtual Work for Deflections

Deflections in Beams and Frames

For a beam we proceed as:

1. Write the virtual work equation for bending:

0

E I

i i

WW W

y F M

δδ δ

δ θ δ

==

⋅ = ⋅∑

2. Place a unit load, Fδ , at the point at which deflection is required;

3. Find the real bending moment diagram, xM , since the real curvatures are given

by:

xx

x

MEI

θ =

4. Solve for the virtual bending moment diagram (the virtual force equilibrium

set), Mδ , caused by the virtual unit load.

5. Solve the virtual work equation:

0

1L

xx

My M dxEI

δ ⋅ = ⋅ ∫

6. Note that the integration tables can be used for this step.


Dr. C. Caprani 9

2.2.3 Virtual Work for Indeterminate Structures

General Approach

Using compatibility of displacement, we have:

Final = Primary + Reactant

Next, further break up the reactant structure, using linear superposition:

Reactant = Multiplier × Unit Reactant

We summarize this process as:

0 1M M Mα= +

• M is the force system in the original structure (in this case moments);

• 0M is the primary structure force system;

• 1M is the unit reactant structure force system.

The primary structure can be analysed, as can the unit reactant structure. Thus, the

only unknown is the multiplier, α , for which we use virtual work to calculate.


Dr. C. Caprani 10

Finding the Multiplier

For beams and frames, we have:

( )210 1

0 0

0L L

ii

i i

MM M dx dxEI EI

δδ α⋅= + ⋅∑ ∑∫ ∫

Thus:

( )

0 1

021

0

Li

i

Li

i

M M dxEI

Mdx

EI

δ

αδ

⋅−=∑∫

∑∫


Dr. C. Caprani 11

2.2.4 Virtual Work for Compound Structures

Basis

In the general equation for Virtual Work:

i i i iy F e Pδ δ⋅ = ⋅∑ ∑

We note that the summation on the right hand side is over all forms of real

displacement and virtual force combinations. For example, if a member is in

combined bending and axial force, then we must include the work done by both

effects:

( ) ( ) ( )Axial BendingMemberiW e P e P

PL MP M dxEA EI

δ δ δ

δ δ

= ⋅ + ⋅

= ⋅ + ⋅∫

The total Virtual Work done by any member is:

( )Memberiv

PL M T VW P M dx T VEA EI GJ GA

δ δ δ δ δ= ⋅ + ⋅ + ⋅ + ⋅∫

In which Virtual Work done by axial, bending, torsion, and shear, respectively, is

accounted for. However, most members primarily act through only one of these stress

resultants, and so we commonly have only one term per member. A typical example

is when axial deformation of frame (bending) members is neglected; since the area is

large the contribution to virtual work is small.


Dr. C. Caprani 12

At the level of the structure as a whole, we must account for all such sources of

Virtual Work. For the typical structures we study here, we account for the Virtual

Work done by axial and flexural members separately:

0

E I

i i i i i i

WW W

y F e P M

δδ δ

δ δ θ δ

==

⋅ = ⋅ + ⋅∑ ∑ ∑

In which the first term on the RHS is the internal virtual work done by axial members

and the second term is that done by flexural members.

Again considering only axial and bending members, if a deflection is sought:

0

1

i i i i

Lx

i xi

y F e P M

PL My P M dxEA EI

δ δ θ δ

δ δ

⋅ = ⋅ + ⋅

⋅ = ⋅ + ⋅

∑ ∑

∑ ∑∫

To solve such an indeterminate structure, we have the contributions to Virtual Work:

0 1M M Mα= +

0 1P P Pα= +

for the structure as a whole. Hence we have:


Dr. C. Caprani 13

( ) ( )

( )

1

0

0 1 0 11

0

210 1 0 11 1

0

0

0 1

0

0

E I

i i i i i i

Lx

i xi

Lx x

i x

i

Lxx x

i ii i

WW W

y F e P M

PL MP M dxEA EI

P P L M MP M dx

EA EI

MP L P L M MP P dxEA EA EI

δδ δ

δ δ θ δ

δ δ

α δ αδ δ

δδ δδ α δ α

==

⋅ = ⋅ + ⋅

⋅ = ⋅ + ⋅

+ ⋅ + = ⋅ + ⋅

⋅= ⋅ + ⋅ ⋅ + + ⋅

∑ ∑ ∑

∑ ∑∫

∑ ∑∫

∑ ∑ ∑∫0

L

dxEI∑∫

Hence the multiplier can be found as:

( ) ( )

0 1 0 1

02 21 1

0

Li i i

i i

Li i i

i i

P P L M M dxEA EI

P L Mdx

EA EI

δ δ

αδ δ

⋅ ⋅ ⋅+= −

+

∑ ∑∫

∑ ∑∫

Note the negative sign!

Though these expressions are cumbersome, the ideas and the algebra are both simple.

Integration of Diagrams

We are often faced with the integration of various diagrams when using virtual work

to calculate the deflections, etc. As such diagrams only have a limited number of

shapes, a table of ‘volume’ integrals is used.


Dr. C. Caprani 14

2.3 Basic Examples

2.3.1 Example 1

Problem

For the following structure, find:

(a) The force in the cable BC and the bending moment diagram;

(b) The vertical deflection at D.

Take 3 28 10 kNmEI = × and 316 10 kNEA = × .


Dr. C. Caprani 15

Solution – Part (a)

This is a one degree indeterminate structure and so we must release one redundant.

We could choose many, but the most obvious is the cable, BC. We next analyze the

primary structure for the actual loads, and the unit virtual force placed in lieu of the

redundant:

From the derivation of Virtual Work for indeterminate structures, we have:

( )210 1 0 1

1 1

0 0

0L L

xx xi i

i i

MP L P L M MP P dx dxEA EA EI EI

δδ δδ α δ α ⋅

= ⋅ + ⋅ ⋅ + + ⋅

∑ ∑ ∑ ∑∫ ∫

We evaluate each term separately to simplify the calculations and to minimize

potential calculation error.


Dr. C. Caprani 16

Term 1:

This term is zero since 0P is zero.

Term 2:

Only member BC contributes to this term and so it is:

1

1 1 2 21ii

P L PEA EA EAδ δ ⋅

⋅ = ⋅ =

∑

Term 3:

Here we must integrate the bending moment diagrams. We use the volume integral

for the portion AD of both diagrams. Thus we multiply a triangle by a trapezoid:

( ) ( )( )( )

0 1

0

1 1 40 2 2 4 26

400 3

Lx xM M dxEI EI

EI

δ⋅ = − + −

= −

∑∫

Term 4:

Here we multiply the virtual BMD by itself so it is a triangle by a triangle:

( ) ( )( )( )

21

0

1 1 64 34 4 43

LxM

dxEI EI EI

δ = − − = ∑∫

With all terms evaluated the Virtual Work equation becomes:

2 400 3 64 30 0EA EI EI

α α= + ⋅ − + ⋅


Dr. C. Caprani 17

Which gives:

400 3400

2 64 3 6 64EI

EIEA EI EA

α = =+ +

Given that 3 38 10 16 10 0.5EI EA = × × = , we have:

( )

400 5.976 0.5 64

α = =+

Thus there is a tension (positive answer) in the cable of 5.97 kN, giving the BMD as:

Note that this comes from:

( )( )( )( )

0

0

40 5.97 4 16.1 kN

0 5.97 2 11.9 kNA

D

M M M

M M M

α δ

α δ

= + ⋅ = + − =

= + ⋅ = + − = −


Dr. C. Caprani 18

Solution – Part (b)

Recalling that the only requirement on applying virtual forces to calculate real

displacements is that an equilibrium system results, we can apply a vertical unit force

at D to the primary structure only:

The Virtual Work equation useful for deflection is:

0

1

i i i i

Lx

Dy i xi

y F e P M

PL MP M dxEA EI

δ δ θ δ

δ δ δ

⋅ = ⋅ + ⋅

⋅ = ⋅ + ⋅

∑ ∑

∑ ∑∫

Since 0Pδ = , we need only calculate the term involving the Virtual Work done by

the beam bending. This involves the volume integral of the two diagrams:


Dr. C. Caprani 19

Note that only the portion AD will count as there is no virtual moment on DB. Thus

we have:

However, this shape is not easy to work with, given the table to hand. Therefore we

recall that the real BMD came about as the superposition of two BMD shapes that are

easier to work with, and so we have:

A further benefit of this approach is that an equation of deflection in terms of the

multiplier α is got. This could then be used to determine α for a particular design

requirement, and in turn this could inform the choice of EI EA ratio. Thus:

( )( )( ) ( ) ( )( )( )

0

1 1 12 40 2 2 2 2 4 23 6

160 203

Lx

Dy xM M dxEI

EI

EI

δ δ

α

α

= ⋅

= + ⋅ − + − −

=

∑∫


Dr. C. Caprani 20

Given 5.97α = , we then have:

( ) 33

160 20 5.97 13.9 13.9 10 1.7 mm3 8 10Dy EI EI

δ−

= = = × =×

The positive answer indicates that the deflection is in the direction of the applied

virtual vertical force and so is downwards as expected.

We can also easily work out the deflection at B, since it is the same as the elongation

of the cable:

( )( ) 33

5.97 210 0.75 mm

16 10ByPLEA

δ = = × =×

Draw the deflected shape of the structure.


Dr. C. Caprani 21

2.3.2 Example 2

Problem

For the following structure, find:

(a) The force in the cable CD and the bending moment diagram;

(b) Determine the optimum EA of the cable for maximum efficiency of the beam.



Dr. C. Caprani 22

Solution – Part (a)

Choose the cable CD as the redundant to give:

The equation of Virtual Work relevant is:

( )210 1 0 1

1 1

0 0

0L L

xx xi i

i i



= ⋅ + ⋅ ⋅ + + ⋅

∑ ∑ ∑ ∑∫ ∫

We evaluate each term separately:


Dr. C. Caprani 23

Term 1:

This term is zero since 0P is zero.

Term 2:

Only member CD contributes to this term and so it is:

1

1 1 2 21ii

P L PEA EA EAδ δ ⋅

⋅ = ⋅ =

∑

Term 3:

Here we must integrate the bending moment diagrams. We use the volume integral

for each half of the diagram, and multiply by 2, since we have two such halves.

( )( )( )

0 1

0

2 5 1 10 212

50 3

Lx xM M dxEI EI

EI

δ⋅ = −

= −

∑∫

Term 4:

Here we multiply the virtual BMD by itself:

( ) ( )( )( )

21

0

2 1 4 31 1 23

LxM

dxEI EI EI

δ = − − = ∑∫

Thus the Virtual Work equation becomes:


Dr. C. Caprani 24

2 50 3 4 30 0EA EI EI

α α= + ⋅ − + ⋅

Which gives:

50 350

2 4 3 6 4EI

EIEA EI EA

α = =+ +

Given that 3 38 10 48 10 0.167EI EA = × × = , we have:

( )

50 106 0.167 4

α = =+

Thus there is a tension (positive answer) in the cable of 10 kN, giving:


Dr. C. Caprani 25

As designers, we want to control the flow of forces. In this example we can see that

by changing the ratio EI EA we can control the force in the cable, and the resulting

bending moments. We can plot the cable force and maximum sagging bending

moment against the stiffness ratio to see the behaviour for different relative

stiffnesses:

0

2

4

6

8

10

12

14

0.0001 0.001 0.01 0.1 1 10 100 1000 10000

Ratio EI/EA

Cable Tension (kN)Sagging Moment (kNm)


Dr. C. Caprani 26

Solution – Part (b)

Efficiency of the beam means that the moments are resisted by the smallest possible

beam. Thus the largest moment anywhere in the beam must be made as small as

possible. Therefore the hogging and sagging moments should be equal:

We know that the largest hogging moment will occur at 2L . However, we do not

know where the largest sagging moment will occur. Lastly, we will consider sagging

moments positive and hogging moments negative. Consider the portion of the net

bending moment diagram, ( )M x , from 0 to 2L :

The equations of these bending moments are:


Dr. C. Caprani 27

( )2P

PM x x= −

( ) 2

2 2W

w wLM x x x= − +

Thus:

( ) ( ) ( )

2

2 2 2

W PM x M x M xwL w Px x x

= +

= − −

The moment at 2L is:

( )2

2 2

2

22 2 2 2 2 2

4 8 4

8 4

wL L w L P LM L

wL wL PL

wL PL

= − −

= − −

= −

Which is as we expected. The maximum sagging moment between 0 and 2L is

found at:


Dr. C. Caprani 28

( )

max

max

0

02 2

2 2

dM xdx

wL Pwx

L Pxw

=

− − =

= −

Thus the maximum sagging moment has a value:

( )2

max

2 2 2 2

2 2

2 2 2 2 2 2 2 2 22

4 4 2 4 4 4 4 4

8 4 8

wL L P w L P P L PM xw w w

wL PL w L PL P PL Pw w w

wL PL Pw

= − − − − −

= − − − + − +

= − +

Since we have assigned a sign convention, the sum of the hogging and sagging

moments should be zero, if we are to achieve the optimum BMD. Thus:

( ) ( )max

2 2 2

2 2

22

2 0

08 4 8 8 4

04 2 8

1 08 2 4

M x M L

wL PL P wL PLw

wL PL Pw

L wLP Pw

+ =

− + + − =

− + =

+ − + =

This is a quadratic equation in P and so we solve for P using the usual method:


Dr. C. Caprani 29

( )

2 2

82 4 82 2 2 8

82 2

L L Lw L LP

wwL

± − = = ±

= ±

Since the load in the cable must be less than the total amount of load in the beam, that

is, P wL< , we have:

( )2 2 0.586P wL wL= − =

With this value for P we can determine the hogging and sagging moments:

( )( )2

2

2

2 22

8 42 2 3

8

0.0214

wL LwLM L

wL

wL

−= −

−=

= −

And:

( )

( )

2 2

max

2

2

2

2

8 4 8

2 22 2 38 8

3 2 28

0.0214

wL PL PM xw

wLwL

w

wL

wL

= − +

− − = + −

=

= +


Dr. C. Caprani 30

Lastly, the location of the maximum sagging moment is given by:

( )

( )

max 2 22 2

2 2

2 120.207

L Pxw

wLLw

L

L

= −

−= −

= −

=

For our particular problem, 5 kN/mw = , 4 mL = , giving:

( )0.586 5 4 11.72 kNP = × =

( ) ( )2max 0.0214 5 4 1.71 kNmM x = × =

Thus, as we expected, 10 kNP > , the value obtained from Part (a) of the problem.

Now since, we know P we now also know the required value of the multiplier, α .

Hence, we write the virtual work equations again, but this time keeping Term 2 in

terms of L, since that is what we wish to solve for:

50 11.726 4

1 50 4 0.0446 11.72

EIEA

EIEA

α = =+

∴ = − =

Giving 3 38 10 0.044 180.3 10 kNEA = × = × . This is 3.75 times the original cable area

– a lot of extra material just to change the cable force by 17%. However, there is a


Dr. C. Caprani 31

large saving by reducing the overall moment in the beam from 10 kNm (simply-

supported) or 2.5 kNm (two-span beam) to 1.71 kNm.


Dr. C. Caprani 32

2.3.3 Example 3

Problem

For the following structure:

1. Determine the tension in the cable AB;

2. Draw the bending moment diagram;

3. Determine the vertical deflection at D with and without the cable AB.



Dr. C. Caprani 33

Solution

As is usual, we choose the cable to be the redundant member and split the frame up

as follows:

Primary Structure Redundant Structure

We must examine the BMDs carefully, and identify expressions for the moments

around the arch. However, since we will be using virtual work and integrating one

diagram against another, we immediately see that we are only interested in the

portion of the structure CB. Further, we will use the anti-clockwise angle from

vertical as the basis for our integration.

Primary BMD

Drawing the BMD and identify the relevant distances:


Dr. C. Caprani 34

Hence the expression for 0M is:

( ) ( )0 20 10 2sin 20 1 sinMθ θ θ= + = +

Reactant BMD

This calculation is slightly easier:


Dr. C. Caprani 35

( ) ( )1 1 2 2cos 2 1 cosMθ θ θ= ⋅ − = −

Virtual Work Equation

As before, we have the equation:

( )210 1 0 11 1

0 0

0L L

xx xi i

i i



= ⋅ + ⋅ ⋅ + + ⋅

∑ ∑ ∑ ∑∫ ∫

Term 1 is zero since there are no axial forces in the primary structure. We take each

other term in turn.

Term 2

Since only member AB has axial force:

( )21 2 2Term 2EA EA

= =

Term 3


Dr. C. Caprani 36

Since we want to integrate around the member – an integrand ds - but only have the

moment expressed according to θ , we must change the integration limits by

substituting:

2ds R d dθ θ= ⋅ =

Hence:

( ) ( )

( )( )

( )

20 1

0 0

2

0

2

0

1 2 1 cos 20 1 sin 2

80 1 cos 1 sin

80 1 sin cos cos sin

Lx xM M dx dEI EI

dEI

dEI

π

π

π

δ θ θ θ

θ θ θ

θ θ θ θ θ

⋅= − − +

= − + +

= − − + +

∑∫ ∫

∫

∫

To integrate this expression we refer to the appendix of integrals to get each of the

terms, which then give:

( )

20 1

00

80 1cos sin cos24

80 1 10 1 1 0 1 02 4 4

80 1 11 12 4 4

80 12

Lx xM M dxEI EI

EI

EI

EI

πδ θ θ θ θ

π

π

π

⋅ = − + + −

= − + + − − − − + + − = − + + − + − =

∑∫

Term 4


Dr. C. Caprani 37

Proceeding similarly to Term 3, we have:

( ) ( ) ( )

( )

21 2

0 0

22

0

1 2 1 cos 2 1 cos 2

8 1 2cos cos

LxM

dx dEI EI

dEI

π

π

δθ θ θ

θ θ θ

= − −

= − +

∑∫ ∫

∫

Again we refer to the integrals appendix, and so for Term 4 we then have:

( ) ( )

[ ]

21 22

0 0

2

0

8 1 2cos cos

8 12sin sin 22 4

8 12 0 0 0 02 4 4

8 3 74

LxM

dx dEI EI

EI

EI

EI

π

π

δθ θ θ

θθ θ θ

π π

π

= − +

= − + +

= − + + − − + + − =

∑∫ ∫

Solution

Substituting the calculated values into the virtual work equation gives:

2 80 1 8 3 70 02 4EA EI EIπ πα α− − = + ⋅ + + ⋅

And so:


Dr. C. Caprani 38

80 12

2 8 3 74

EI

EA EI

π

απ

− − =

− +

Simplifying:

20 20

3 7 EIEA

παπ

−=

− +

In this problem, 2EI EA = and so:

20 20 9.68 kN3 5παπ−

= =−

We can examine the effect of different ratios of EI EA on the structure from our

algebraic solution for α . We show this, as well as a point representing the solution

for this particular EI EA ratio on the following graph:

0

2

4

6

8

10

12

14

16

18

20

0.0001 0.001 0.01 0.1 1 10 100 1000 10000

Ratio EI/EA

α F

acto

r


Dr. C. Caprani 39

As can be seen, by choosing a stiffer frame member (increasing EI) or by reducing

the area of the cable, we can reduce the force in the cable (which is just 1 α⋅ ).

However this will have the effect of increasing the moment at A, for example:

Deflections and shear would also be affected.

Draw the final BMD and determine the deflection at D.

0

5

10

15

20

25

30

35

40

45

0.0001 0.001 0.01 0.1 1 10 100 1000 10000

Ratio EI/EA

Ben

ding

Mom

ent a

t A (k

Nm

)


Dr. C. Caprani 40

2.3.4 Example 4

Problem

For the following structure:

1. draw the bending moment diagram;

2. Find the vertical deflection at E.



Dr. C. Caprani 41

Solution

To begin we choose the cable BF as the obvious redundant, yielding:

Virtual Work Equation

The Virtual Work equation is as before:

( )210 1 0 11 1

0 0

0L L

xx xi i

i i



= ⋅ + ⋅ ⋅ + + ⋅

∑ ∑ ∑ ∑∫ ∫

Term 1 is zero since there are no axial forces in the primary structure. As we have

done previously, we take each other term in turn.

Term 2


Dr. C. Caprani 42

Though member AB has axial force, it is primarily a flexural member and so we only

take account of the axial force in the cable BF:

1

1 1 2 2 2 21ii

P L PEA EA EAδ δ

⋅ ⋅ = ⋅ =

∑

Term 3

Since only the portion AB has moment on both diagrams, it is the only section that

requires integration here. Thus:

( )( )( )0 1

0

1 1 220 2200 2 22

Lx xM M dxEI EI EIδ⋅ − = − =

∑∫

Term 3

Similar to Term 3, we have:

( ) ( )( )( )

21

0

1 1 4 32 2 23

LxM

dxEI EI EI

δ = − − = ∑∫

Solution

Substituting the calculated values into the virtual work equation gives:

2 2 220 2 4 30 0EA EI EI

α α= + ⋅ − + ⋅

Thus:


Dr. C. Caprani 43

220 22 2 4 3

EI

EA EI

α =+

And so:

220 242 23

EIEA

α =+

Since:

3

3

120 10 260 10

EIEA

×= =

×

We have:

( )220 2 40.4642 2 2

3

α = = ++

Thus the force in the cable BF is 40.46 kN tension, as assumed.

The bending moment diagram follows from superposition of the two previous

diagrams:


Dr. C. Caprani 44

To find the vertical deflection at E, we must apply a unit vertical load at E. We will

apply a downwards load since we think the deflection is downwards. Therefore we

should get a positive result to confirm our expectation.

We need not apply the unit vertical force to the whole structure, as it is sufficient to

apply it to a statically determinate sub-structure. Thus we apply the force as follows:

For the deflection, we have the following equation:


Dr. C. Caprani 45

0

1

i i i i

Lx

Ey i xi

y F e P M

PL MP M dxEA EI

δ δ θ δ

δ δ δ

⋅ = ⋅ + ⋅

⋅ = ⋅ + ⋅

∑ ∑

∑ ∑∫

However, since 0Pδ = , we only need calculate the second term:

For AB we have:

( )( )( )1 1 1371.2200 142.8 4 22

Bx

xA

M M dxEI EI EI

δ ⋅ = + = ∫

For BC we have:

( )( )( )1 1600200 4 2C

xx

B

M M dxEI EI EI

δ ⋅ = = ∫

For CD, we have the following equations for the bending moments:


Dr. C. Caprani 46

( ) ( )( )100 2sin200sin

M θ θθ

=

= ( ) ( )( )2 1 2sin

2 2sinMδ θ θ

θ= +

= +

Also note that we want to integrate around the member – an integrand ds - but only

have the moment expressed according to θ , we must change the integration limits by

substituting:

2ds R d dθ θ= ⋅ =

Thus we have:

( )( )

( )

2

0

22

0

2 22

0 0

1 200sin 2 2sin 2

800 sin sin

800 sin sin

Dx

xC

M M dx dEI EI

dEI

d dEI

π

π

π π

δ θ θ θ

θ θ θ

θ θ θ θ

⋅ = + ⋅

= +

= +

∫ ∫

∫

∫ ∫

Taking each term in turn:


Dr. C. Caprani 47

[ ] ( )2

2

00

sin cos 0 1 1dπ

πθ θ θ= − = − − − = +∫

( ) ( )22

2 22 2

00

1 1 1 1sin sin 1 0 02 4 4 4 4 4

dππ θ π πθ θ θ − = − = − − − = ∫

Thus:

800 1 200 60014

Dx

xC

M M dxEI EI EI

π πδ − + ⋅ = + = ∫

Thus:

1371.2 1600 200 600 4200Ey EI EI EI EI

πδ += + + = +

Thus we get a downwards deflection as expected. Also, since 3 2120 10 kNmEI = × ,

we have:

3

4200 35 mm120 10Eyδ = = ↓

×


Dr. C. Caprani 48

2.3.5 Problems

Problem 1

For the following structure, find the BMD and the vertical deflection at D. Take 3 28 10 kNmEI = × and 316 10 kNEA = × .

(Ans. 7.8α = for BC, 1.93 mmByδ = ↓ )

Problem 2

For the following structure, find the BMD and the vertical deflection at C. Take 3 28 10 kNmEI = × and 316 10 kNEA = × .

(Ans. 25.7α = for BD, 25 mmCvδ = ↓ )


Dr. C. Caprani 49

Problem 3

For the following structure, find the BMD and the horizontal deflection at C. Take 3 28 10 kNmEI = × and 316 10 kNEA = × .

(Ans. 47.8α = for BD, 44.8 mmCxδ = → )

Problem 4

For the following structure, find the BMD and the vertical deflection at B. Take P =

20 kN, 3 28 10 kNmEI = × and 316 10 kNEA = × .


Dr. C. Caprani 50

(Ans. 14.8α = for CD, 14.7 mmByδ = ↓ )

Problem 5

For the following structure, find the BMD and the vertical deflection at C. Take 3 250 10 kNmEI = × and 320 10 kNEA = × .

(Ans. 100.5α = for BC, 55.6 mmCyδ = ↓)

Problem 6

Analyze the following structure and determine the BMD and the vertical deflection at

D. For ABCD, take 210 kN/mmE = , 4 212 10 mmA = × and 8 436 10 mmI = × , and for

AEBFC take 2200 kN/mmE = and 3 22 10 mmA = × .

(Ans. 109.3α = for BF, 54.4 mmCyδ = ↓)


Dr. C. Caprani 51

Problem 7

Analyze the following structure. For all members, take 210 kN/mmE = , for ABC, 4 26 10 mmA = × and 7 4125 10 mmI = × ; for all other members 21000 mmA = .

(Ans. 72.5α = for DE)


Dr. C. Caprani 52

2.4 Past Exam Questions

2.4.1 Sample Paper 2007

3. For the rigidly jointed frame shown in Fig. Q3, using Virtual Work:

(i) Determine the bending moment moments due to the loads as shown; (15 marks)

(ii) Draw the bending moment diagram, showing all important values;

(4 marks)

(iii) Determine the reactions at A and E; (3 marks)

(iv) Draw the deflected shape of the frame.

(3 marks) Neglect axial effects in the flexural members. Take the following values: I for the frame = 150×106 mm4; Area of the stay EB = 100 mm2; Take E = 200 kN/mm2 for all members.

FIG. Q3


Dr. C. Caprani 53

2.4.2 Semester 1 Exam 2007

3. For the rigidly jointed frame shown in Fig. Q3, using Virtual Work:

(i) Determine the bending moment moments due to the loads as shown; (15 marks)

(ii) Draw the bending moment diagram, showing all important values;

(4 marks)

(iii) Determine the reactions at A and E; (3 marks)

(iv) Draw the deflected shape of the frame.

(3 marks) Neglect axial effects in the flexural members. Take the following values: I for the frame = 150×106 mm4; Area of the stay EF = 200 mm2; Take E = 200 kN/mm2 for all members.

Ans. 35.0α = .

FIG. Q3


Dr. C. Caprani 54


QUESTION 3

For the frame shown in Fig. Q3, using Virtual Work:

(i) Determine the force in the tie; (ii) Draw the bending moment diagram, showing all important values; (iii) Determine the deflection at C; (iv) Determine an area of the tie such that the bending moments in the beam are minimized; (v) For this new area of tie, determine the deflection at C; (vi) Draw the deflected shape of the structure.

(25 marks)

Note:

Neglect axial effects in the flexural members and take the following values:

• For the frame, 6 4600 10 mmI = × ; • For the tie, 2300 mmA = ; • For all members, 2200 kN/mmE = .

Ans. 21.24α = ; 4.1 mmCyδ = ↓ ; 22160 mmA = ; 2.0 mmCyδ = ↓

FIG. Q3


Dr. C. Caprani 55


QUESTION 3

For the frame shown in Fig. Q3, using Virtual Work:

(i) Determine the axial forces in the members; (ii) Draw the bending moment diagram, showing all important values; (iii) Determine the reactions; (iv) Determine the vertical deflection at D; (v) Draw the deflected shape of the structure.

(25 marks)

Note:

Neglect axial effects in the flexural members and take the following values:

• For the beam ABCD, 6 4600 10 mmI = × ; • For members BF and CE, 2300 mmA = ; • For all members, 2200 kN/mmE = .

Ans. 113.7α = (for CE); 55 mmDyδ = ↓

FIG. Q3


Dr. C. Caprani 56

2.4.5 Semester 1 Exam 2010 QUESTION 3 For the frame shown in Fig. Q3, using Virtual Work:

(i) Draw the bending moment diagram, showing all important values;

(ii) Determine the horizontal displacement at C;

(iii) Determine the vertical deflection at C;

(iv) Draw the deflected shape of the structure.

(25 marks) Note: Neglect axial effects in the flexural members and take the following values:

• For the beam ABC, 3 25 10 kNmEI = × ; • For member BD, 2200 kN/mmE = and 2200 mmA = ; • The following integral results may assist in your solution:

sin cosdθ θ θ= −∫ 1cos sin cos 24

dθ θ θ θ= −∫ 2 1sin sin 22 4

d θθ θ θ= −∫

Ans. 37.1α = (for BD); 104 mmCxδ = ← 83 mmCyδ = ↓

FIG. Q3


Dr. C. Caprani 57

2.4.6 Semester 1 Exam 2011 QUESTION 3 For the frame shown in Fig. Q3, using Virtual Work: (i) Draw the bending moment diagram, showing all important values; (ii) Draw the axial force diagram; (iii) Determine the vertical deflection at D; (iv) Draw the deflected shape of the structure.

(25 marks) Note: Neglect axial effects in the flexural members and take the following values:

• For the member ABCD, 3 25 10 kNmEI = × ; • For members BF and CE, 2200 kN/mmE = and 2200 mmA = ; • The following integral result may assist in your solution:

2 1sin sin 2

2 4d θθ θ θ= −∫

Ans. 48.63α = (for BF); 108.4 mmDyδ = ↓

FIG. Q3


Dr. C. Caprani 58

2.5 Appendix – Trigonometric Integrals

2.5.1 Useful Identities

In the following derivations, use is made of the trigonometric identities:

1cos sin sin 22

θ θ θ= (1)

( )2 1cos 1 cos22

θ θ= + (2)

( )2 1sin 1 cos22

θ θ= − (3)

Integration by parts is also used:

u dx ux x du C= − +∫ ∫ (4)


Dr. C. Caprani 59

2.5.2 Basic Results

Neglecting the constant of integration, some useful results are:

cos sindθ θ θ=∫ (5)

sin cosdθ θ θ= −∫ (6)

1sin cosa d aa

θ θ θ= −∫ (7)

1cos sina d aa

θ θ θ=∫ (8)


Dr. C. Caprani 60

2.5.3 Common Integrals

The more involved integrals commonly appearing in structural analysis problems are:

cos sin dθ θ θ∫

Using identity (1) gives:

1cos sin sin 22

d dθ θ θ θ θ=∫ ∫

Next using (7), we have:

1 1 1sin 2 cos22 2 2

1 cos24

dθ θ θ

θ

= −

= −

∫

And so:

1cos sin cos24

dθ θ θ θ= −∫ (9)


Dr. C. Caprani 61

2cos dθ θ∫

Using (2), we have:

( )2 1cos 1 cos2

21 1 cos22

d d

d d

θ θ θ θ

θ θ θ

= +

= +

∫ ∫

∫ ∫

Next using (8):

1 1 11 cos2 sin 22 2 2

1 sin 22 4

d dθ θ θ θ θ

θ θ

+ = +

= +

∫ ∫

And so:

2 1cos sin 22 4

d θθ θ θ= +∫ (10)


Dr. C. Caprani 62

2sin dθ θ∫

Using (3), we have:

( )2 1sin 1 cos2

21 1 cos22

d d

d d

θ θ θ θ

θ θ θ

= −

= −

∫ ∫

∫ ∫

Next using (8):

1 1 11 cos2 sin 22 2 2

1 sin 22 4

d dθ θ θ θ θ

θ θ

− = −

= −

∫ ∫

And so:

2 1sin sin 22 4

d θθ θ θ= −∫ (11)


Dr. C. Caprani 63

cos dθ θ θ∫

Using integration by parts write:

cos d u dxθ θ θ =∫ ∫

Where:

cosu dx dθ θ θ= =

To give:

du dθ=

And

cos

sin

dx d

x

θ θ

θ

=

=∫ ∫

Which uses (5). Thus, from (4), we have:

cos sin sin

u dx ux x du

d dθ θ θ θ θ θ θ

= −

= −∫ ∫

∫ ∫

And so, using (6) we have:

cos sin cosdθ θ θ θ θ θ= +∫ (12)


Dr. C. Caprani 64

sin dθ θ θ∫

Using integration by parts write:

sin d u dxθ θ θ =∫ ∫

Where:

sinu dx dθ θ θ= =

To give:

du dθ=

And

sin

cos

dx d

x

θ θ

θ

=

= −∫ ∫

Which uses (6). Thus, from (4), we have:

( ) ( )sin cos cos

u dx ux x du

d dθ θ θ θ θ θ θ

= −

= − − −∫ ∫

∫ ∫

And so, using (5) we have:

sin cos sindθ θ θ θ θ θ= − +∫ (13)


Dr. C. Caprani 65

( )cos A dθ θ−∫

Using integration by substitution, we write u A θ= − to give:

1duddu dθ

θ

= −

= −

Thus:

( ) ( )cos cosA d u duθ θ− = −∫ ∫

And since, using (5):

cos sinu du u− = −∫

We have:

( ) ( )cos sinA d Aθ θ θ− = − −∫ (14)


Dr. C. Caprani 66

( )sin A dθ θ−∫

Using integration by substitution, we write u A θ= − to give:

1duddu dθ

θ

= −

= −

Thus:

( ) ( )sin sinA d u duθ θ− = −∫ ∫

And since, using (6):

( )sin cosu du u− = − −∫

We have:

( ) ( )sin cosA d Aθ θ θ− = −∫ (15)


Dr. C. Caprani 67

2.6 Appendix – Volume Integrals

13

jkl 16

jkl ( )1 21 26

j j kl+ 12

jkl

16

jkl 13

jkl ( )1 21 26

j j kl+ 12

jkl

( )1 2

1 26

j k k l+ ( )1 21 26

j k k l+ ( )

( )

1 1 2

2 1 2

1 26

2

j k k

j k k l

+ +

+

( )1 212

j k k l+

12

jkl 12

jkl ( )1 212

j j kl+ jkl

( )1

6jk l a+ ( )1

6jk l b+

( )

( )

1

2

16

j l b

j l a k

+ +

+

12

jkl

512

jkl 14

jkl ( )1 21 3 5

12j j kl+ 2

3jkl

14

jkl 512

jkl ( )1 21 5 3

12j j kl+ 2

3jkl

14

jkl 112

jkl ( )1 21 3

12j j kl+ 1

3jkl

112

jkl 14

jkl ( )1 21 3

12j j kl+ 1

3jkl

13

jkl 13

jkl ( )1 213

j j kl+ 23

jkl

1 2

1 2

Structural Analysis IV Chapter 3 – Virtual Work: Advanced Examples

Dr. C. Caprani 1

Chapter 3 - Virtual Work: Advanced Examples

3.1 Introduction ......................................................................................................... 2

3.1.1 General ........................................................................................................... 2

3.2 Ring Beam Examples .......................................................................................... 3

3.2.1 Example 1 ...................................................................................................... 3

3.2.2 Example 2 ...................................................................................................... 8

3.2.3 Example 3 .................................................................................................... 15

3.2.4 Example 4 .................................................................................................... 23

3.2.5 Example 5 .................................................................................................... 32

3.2.6 Review of Examples 1 – 5 ........................................................................... 53

3.3 Grid Examples ................................................................................................... 64

3.3.1 Example 1 .................................................................................................... 64

3.3.2 Example 2 .................................................................................................... 70

3.3.3 Example 3 .................................................................................................... 79

Rev. 1


Dr. C. Caprani 2

3.1 Introduction

3.1.1 General

To further illustrate the virtual work method applied to more complex structures, the

following sets of examples are given. The examples build upon each other to

illustrate how the analysis of a complex structure can be broken down.


Dr. C. Caprani 3

3.2 Ring Beam Examples

3.2.1 Example 1

Problem

For the quarter-circle beam shown, which has flexural and torsional rigidities of EI

and GJ respectively, show that the deflection at A due to the point load, P, at A is:

3 3 3 8

4 4Ay

PR PREI GJ

π πδ − = ⋅ +


Dr. C. Caprani 4

Solution

The point load will cause both bending and torsion in the beam member. Therefore

both effects must be accounted for in the deflection calculations. Shear effects are

ignored.

Drawing a plan view of the structure, we can identify the perpendicular distance of

the force, P, from the section of consideration, which we locate by the angle θ from

the y-axis:

The bending moment at C is P times the perpendicular distance AC , called m. The

torsion at C is the force times the transverse perpendicular distance CD , called t.

Using the triangle ODA, we have:


Dr. C. Caprani 5

sin sin

cos cos

m m RROD

OD RR

θ θ

θ θ

= ∴ =

= ∴ =

The distance CD , or t, is R OD− , thus:

( )cos

1 cos

t R ODR RR

θθ

= −

= −

= −

Thus the bending moment at point C is:

( )sin

M PmPR

θθ

=

= (1)

The torsion at C is:

( )

( )1 cos

T Pt

PR

θ

θ

=

= − (2)

Using virtual work, we have:

0

E I

Ay

WW W

M TF M ds T dsEI GJ

δδ δ

δ δ δ δ

==

⋅ = ⋅ + ⋅∫ ∫

(3)


Dr. C. Caprani 6

This equation represents the virtual work done by the application of a virtual force,

Fδ , in the vertical direction at A, with its internal equilibrium virtual moments and

torques, Mδ and Tδ and so is the equilibrium system. The compatible

displacements system is that of the actual deformations of the structure, externally at

A, and internally by the curvatures and twists, M EI and T GJ .

Taking the virtual force, 1Fδ = , and since it is applied at the same location and

direction as the actual force P, we have, from equations (1) and (2):

( ) sinM Rδ θ θ= (4)

( ) ( )1 cosT Rδ θ θ= − (5)

Thus, the virtual work equation, (3), becomes:

[ ][ ] ( ) ( )

2 2

0 0

1 11

1 1sin sin 1 cos 1 cos

Ay M M ds T T dsEI GJ

PR R Rd PR R RdEI GJ

π π

δ δ δ

θ θ θ θ θ θ

⋅ = ⋅ + ⋅

= + − −

∫ ∫

∫ ∫ (6)

In which we have related the curve distance, ds , to the arc distance, ds Rdθ= , which

allows us to integrate round the angle rather than along the curve. Multiplying out:

( )2 23 3

22

0 0

sin 1 cosAyPR PRd dEI GJ

π π

δ θ θ θ θ= + −∫ ∫ (7)

Considering the first term, from the integrals’ appendix, we have:


Dr. C. Caprani 7

( )

222

00

1sin sin 22 4

1 0 0 04 4

4

dππ θθ θ θ

π

π

= −

= − ⋅ − −

=

∫

(8)

The second term is:

( ) ( )

2 22 2

0 0

2 2 22

0 0 0

1 cos 1 2cos cos

1 2 cos cos

d d

d d d

π π

π π π

θ θ θ θ θ

θ θ θ θ θ

− = − +

= − +

∫ ∫

∫ ∫ ∫ (9)

Thus, from the integrals in the appendix:

( ) [ ] [ ]

( ) ( ) ( ) ( )

222 22

0 000

11 cos 2 sin sin 22 4

10 2 1 0 0 0 02 4 4

22 43 8

4

dππ

π π θθ θ θ θ θ

π π

π π

π

− = − + +

= − − − + + ⋅ − +

= − +

−=

∫

(10)

Substituting these results back into equation (7) gives the desired result:

3 3 3 8

4 4AyPR PREI GJ

π πδ − = +

(11)


Dr. C. Caprani 8

3.2.2 Example 2

Problem


and GJ respectively, show that the deflection at A due to the uniformly distributed

load, w, shown is:

( )24 4 212 8Ay

wR wREI GJ

πδ

−= ⋅ + ⋅


Dr. C. Caprani 9

Solution

The UDL will cause both bending and torsion in the beam member and both effects

must be accounted for. Again, shear effects are ignored.

Drawing a plan view of the structure, we must identify the moment and torsion at

some point C, as defined by the angle θ from the y-axis, caused by the elemental

load at E, located at φ from the y-axis. The load is given by:

Force UDL length

w dsw R dφ

= ×= ⋅= ⋅

(12)


Dr. C. Caprani 10

The bending moment at C is the load at E times the perpendicular distance DE ,

labelled m. The torsion at C is the force times the transverse perpendicular distance

CD , labelled t. Using the triangle ODE, we have:

( ) ( )

( ) ( )

sin sin

cos cos

m m RROD

OD RR

θ φ θ φ

θ φ θ φ

− = ∴ = −

− = ∴ = −

The distance t is thus:

( )( )

cos

1 cos

t R OD

R R

R

θ φ

θ φ

= −

= − −

= − −

The differential bending moment at point C, caused by the elemental load at E is

thus:

( )[ ][ ] ( )

( )2

Force Distance

sin

sin

dM

wRd m

wRd R

wR d

θ

φ

φ θ φ

θ φ φ

= ×

= ×

= − = −

Integrating to find the total moment at C caused by the UDL from A to C around the

angle 0 to θ gives:


Dr. C. Caprani 11

( ) ( )

( )

( )

2

0

2

0

sin

sin

M dM

wR d

wR d

φ θ

φ

φ θ

φ

θ θ

θ φ φ

θ φ φ

=

=

=

=

=

= −

= −

∫

∫

∫

In this integral θ is a constant and only φ is considered a variable. Using the identity

from the integral table gives:

( ) ( )

( )

2

0

2

cos

cos0 cos

M wR

wR

φ θ

φθ θ φ

θ

=

== −

= −

And so:

( ) ( )2 1 cosM wRθ θ= − (13)

Along similar lines, the torsion at C caused by the load at E is:

( ) [ ][ ] ( ){ }

( )2

1 cos

1 cos

dT wRd t

wRd R

wR d

θ φ

φ θ φ

θ φ φ

= ×

= − −

= − −

And integrating for the total torsion at C:


Dr. C. Caprani 12

( ) ( )

( )

( )

( )

2

0

2

0

2

0 0

1 cos

1 cos

1 cos

T dT

wR d

wR d

wR d d

φ θ

φ

φ θ

φ

φ θ φ θ

φ φ

θ θ

θ φ φ

θ φ φ

φ θ φ φ

=

=

=

=

= =

= =

=

= − −

= − −

= − −

∫

∫

∫

∫ ∫

Using the integral identity for ( )cos θ φ− gives:

( ) [ ] ( ){ }

[ ]{ }

2

0 0

2

sin

sin 0 sin

T wR

wR

φ θφ θ

φ φθ φ θ φ

θ θ

==

= == − − −

= + −

And so the total torsion at C is:

( ) ( )2 sinT wRθ θ θ= − (14)

To determine the deflection at A, we apply a virtual force, Fδ , in the vertical

direction at A. Along with its internal equilibrium virtual moments and torques, Mδ

and Tδ and this set forms the equilibrium system. The compatible displacements

system is that of the actual deformations of the structure, externally at A, and

internally by the curvatures and twists, M EI and T GJ . Therefore, using virtual

work, we have:

0

E I

Ay

WW W

M TF M ds T dsEI GJ

δδ δ

δ δ δ δ

==

⋅ = ⋅ + ⋅∫ ∫

(15)


Dr. C. Caprani 13

Taking the virtual force, 1Fδ = , and using the equation for moment and torque at

any angle θ from Example 1, we have:

( ) sinM Rδ θ θ= (16)

( ) ( )1 cosT Rδ θ θ= − (17)

Thus, the virtual work equation, (15), using equations (13) and (14), becomes:

( ) [ ]

( ) ( )

22

0

22

0

1 11

1 1 cos sin

1 sin 1 cos

Ay M M ds T T dsEI GJ

wR R RdEI

wR R RdGJ

π

π

δ δ δ

θ θ θ

θ θ θ θ

⋅ = ⋅ + ⋅

= −

+ − −

∫ ∫

∫

∫

(18)

In which we have related the curve distance, ds , to the arc distance, ds Rdθ=

allowing us to integrate round the angle rather than along the curve. Multiplying out:

( )

( )

24

0

24

0

sin sin cos

sin cos cos sin

AywR dEI

wR dGJ

π

π

δ θ θ θ θ

θ θ θ θ θ θ θ

= −

+ − − +

∫

∫ (19)

Using the respective integrals from the appendix yields:


Dr. C. Caprani 14

( )

( ) ( )

24

0

24 2

0

4

4 2

4

4 2

1cos cos24

1cos sin cos cos22 4

1 10 14 4

1 10 1 0 1 0 1 0 18 2 4 4

12

1 18 2 4 4

AywREI

wRGJ

wREI

wRGJ

wREI

wRGJ

π

π

δ θ θ

θ θ θ θ θ θ

π π

π π

= − +

+ + − + −

= − − − − + + + − ⋅ + − − − + − + −

=

+ − + +

Writing the second term as a common fraction:

4 4 21 4 4

2 8AywR wREI GJ

π πδ − +

= ⋅ +

And then factorising, gives the required deflection at A:

( )224 4 21

2 8AywR wREI GJ

πδ

−= ⋅ + ⋅ (20)


Dr. C. Caprani 15

3.2.3 Example 3

Problem


and GJ respectively, show that the vertical reaction at A due to the uniformly

distributed load, w, shown is:

( )( )

24 22 2 3 8AV wR

β πβπ π

+ −=

+ −

where GJEI

β = .


Dr. C. Caprani 16

Solution

This problem can be solved using two apparently different methods, but which are

equivalent. Indeed, examining how they are equivalent leads to insights that make

more difficult problems easier, as we shall see in subsequent problems. For both

approaches we will make use of the results obtained thus far:

• Deflection at A due to UDL:

( )24 4 212 8Ay

wR wREI GJ

πδ

−= ⋅ + ⋅ (21)

• Deflection at A due to point load at A:

3 3 3 8

4 4Ay

PR PREI GJ

π πδ − = ⋅ +

(22)

Using Compatibility of Displacement

The basic approach, which does not require virtual work, is to use compatibility of

displacement in conjunction with superposition. If we imagine the support at A

removed, we will have a downwards deflection at A caused by the UDL, which

equation (21) gives us as:

( )24 40 21

2 8Ay

wR wREI GJ

πδ

−= ⋅ + ⋅ (23)

As illustrated in the following diagram.


Dr. C. Caprani 17

Since in the original structure we will have a support at A we know there is actually

no displacement at A. The vertical reaction associated with the support at A, called V,

must therefore be such that it causes an exactly equal and opposite deflection, VAyδ , to

that of the UDL, 0Ayδ , so that we are left with no deflection at A:

0 0VAy Ayδ δ+ = (24)

Of course we don’t yet know the value of V, but from equation (22), we know the

deflection caused by a unit load placed in lieu of V:

3 3

1 1 1 3 84 4Ay

R REI GJ

π πδ ⋅ ⋅ − = ⋅ +

(25)


Dr. C. Caprani 18

This is shown in the following diagram:

Using superposition, we know that the deflection caused by the reaction, V, is V times

the deflection caused by a unit load:

1VAy AyVδ δ= ⋅ (26)

Thus equation (24) becomes:

0 1 0Ay AyVδ δ+ ⋅ = (27)

Which we can solve for V:


Dr. C. Caprani 19

0

1Ay

Ay

Vδδ

= − (28)

If we take downwards deflections to be positive, we then have, from equations(23),

(25), and (28):

( )24 4

3 3

212 8

1 1 3 84 4

wR wREI GJ

VR R

EI GJ

π

π π

−⋅ + ⋅

= − ⋅ ⋅ − − ⋅ +

(29)

The two negative signs cancel, leaving us with a positive value for V indicating that it

is in the same direction as the unit load, and so is upwards as expected. Introducing

GJEI

β = and doing some algebra on equation (29) gives:

( )

( )

( ) ( )

( )( )

12

12

12

2

21 1 1 1 1 3 82 8 4 4

21 1 1 3 82 8 4 4

4 2 3 88 4

4 2 88 2 2 3 8

V wREI EI EI EI

wR

wR

wR

π π πβ β

π π πβ β

β π βπ πβ β

β π ββ βπ π

−

−

−

− − = ⋅ + ⋅ × ⋅ + − − = + ⋅ × + + − + −

= × + −

= × + −

And so we finally have the required reaction at A as:


Dr. C. Caprani 20

( )( )

24 22 2 3 8AV wR

β πβπ π

+ −= + −

(30)

Using Virtual Work

To calculate the reaction at A using virtual work, we use the following:

• Equilibrium system: the external and internal virtual forces corresponding to a

unit virtual force applied in lieu of the required reaction;

• Compatible system: the real external and internal displacements of the original

structure subject to the real applied loads.

Thus the virtual work equations are:

0

E I

Ay

WW W

F M ds T ds

δδ δ

δ δ κ δ φ δ

==

⋅ = ⋅ + ⋅∫ ∫ (31)

At this point we introduce some points:

• The real external deflection at A is zero: 0Ayδ = ;

• The virtual force, 1Fδ = ;

• The real curvatures can be expressed using the real bending moments, MEI

κ = ;

• The real twists are expressed from the torque, TGJ

φ = .

These combine to give, from equation (31):

0 0

0 1L LM TM ds T ds

EI GJδ δ ⋅ = ⋅ + ⋅ ∫ ∫ (32)


Dr. C. Caprani 21

Next, we use superposition to express the real internal ‘forces’ as those due to the real

loading applied to the primary structure plus a multiplier times those due to the unit

virtual load applied in lieu of the reaction:

0 1 0 1M M M T T Tα α= + = + (33)

Notice that 1M Mδ = and 1T Tδ = , but they are still written with separate notation to

keep the ideas clear. Thus equation (32) becomes:

( ) ( )0 1 0 1

0 0

0 1 0 1

0 0 0 0

0

0

L L

L L L L

M M T TM ds T ds

EI GJ

M M T TM ds M ds T ds T dsEI EI GJ GJ

α αδ δ

δ α δ δ α δ

+ += ⋅ + ⋅

= ⋅ + ⋅ ⋅ + ⋅ + ⋅ ⋅

∫ ∫

∫ ∫ ∫ ∫

(34)

And so finally:

0 0

0 01 1

0 0

L L

L L

M TM ds T dsEI GJ

M TM ds T dsEI GJ

δ δα

δ δ

⋅ + ⋅

= −

⋅ + ⋅

∫ ∫

∫ ∫ (35)

At this point we must note the similarity between equations (35) and (28). From

equation (3), it is clear that the numerator in equation (35) is the deflection at A of the

primary structure subject to the real loads. Further, from equation (15), the

denominator in equation (35) is the deflection at A due to a unit (virtual) load at A.

Neglecting signs, and generalizing somewhat, we can arrive at an ‘empirical’

equation for the calculation of redundants:


Dr. C. Caprani 22

of primary structure along due to actual loads line of action of redundant due to unit redundant

δαδ

=

(36)

Using this form we will quickly be able to determine the solutions to further ring-

beam problems.

The solution for α follows directly from the previous examples:

• The numerator is determined as per Example 1;

• The denominator is determined as per Example 2, with 1P = .

Of course, these two steps give the results of equations (23) and (25) which were

used in equation (28) to obtain equation (29), and leading to the solution, equation

(30).

From this it can be seen that compatibility of displacement and virtual work are

equivalent ways of looking at the problem. Also it is apparent that the virtual work

framework inherently calculates the displacements required in a compatibility

analysis. Lastly, equation (36) provides a means for quickly calculating the redundant

for other arrangements of the structure from the existing solutions, as will be seen in

the next example.


Dr. C. Caprani 23

3.2.4 Example 4

Problem

For the structure shown, the quarter-circle beam has flexural and torsional rigidities

of EI and GJ respectively and the cable has axial rigidity EA, show that the tension in

the cable due to the uniformly distributed load, w, shown is:

( ) ( )1

2

34 2 2 2 3 8 8 LT wRR

ββ π πβ πγ

− = + − + − + ⋅

where GJEI

β = and EAEI

γ = .


Dr. C. Caprani 24


Dr. C. Caprani 25

Solution

For this solution, we will use the insights gained from Example 3, in particular

equation (36). We will then verify this approach using the usual application of virtual

work. We will be choosing the cable as the redundant throughout.

Empirical Form

Repeating our ‘empirical’ equation here:


δαδ

=

(37)

We see that we already know the numerator: the deflection at A in the primary

structure, along the line of the redundant (vertical, since the cable is vertical), due to

the actual loads on the structure is just the deflection of Example 1:

( )24 40 21

2 8Ay

wR wREI GJ

πδ

−= ⋅ + ⋅ (38)

This is shown below:


Dr. C. Caprani 26

Next we need to identify the deflection of the primary structure due to a unit

redundant, as shown below:

The components that make up this deflection are:

• Deflection of curved beam caused by unit load (bending and torsion);

• Deflection of the cable AC caused by the unit tension.

The first of these is simply the unit deflection of Example 3, equation (25):

( )3 3

1 1 1 3 8beam4 4Ay

R REI GJ

π πδ ⋅ ⋅ − = ⋅ +

(39)

The second of these is not intuitive, but does feature in the virtual work equations, as

we shall see. The elongation of the cable due to a unit tension is:


Dr. C. Caprani 27

( )1 1cableAy

LEA

δ ⋅= (40)

Thus the total deflection along the line of the redundant, of the primary structure, due

to a unit redundant is:

( ) ( )1 1 1

3 3

beam cable

1 1 3 8 14 4

Ay Ay Ay

R R LEI GJ EA

δ δ δ

π π

= +

⋅ ⋅ − ⋅ = ⋅ + +

(41)

Both sets of deflections (equations (39) and (41)) are figuratively summarized as:

And by making 0 1Ay AyTδ δ= , where T is the tension in the cable, we obtain our

compatibility equation for the redundant. Thus, from equations (37), (38) and (41) we

have:


Dr. C. Caprani 28

( )24 4

3 3

212 8

1 1 3 8 14 4

wR wREI GJ

TR R L

EI GJ EA

π

π π

−⋅ + ⋅

= ⋅ ⋅ − ⋅ ⋅ + +

(42)

Setting GJEI

β = and EAEI

γ = , and performing some algebra gives:

( )

( ) ( )

( ) ( )

12

3

12

3

1

2 3

21 1 1 1 1 3 82 8 4 4

4 2 3 88 4

82 2 3 84 28 8

LT wREI EI EI EI R EI

LwRR

LRwR

π π πβ β γ

β π βπ πβ β γ

ββπ πβ π γβ β

−

−

−

− − = ⋅ + ⋅ ⋅ + + + − + −

= +

+ − + + − =

(43)

Which finally gives the required tension as:

( ) ( )1

2

34 2 2 2 3 8 8 LT wRR

ββ π πβ πγ

− = + − + − + ⋅

(44)

Comparing this result to the previous result, equation (30), for a pinned support at A,

we can see that the only difference is the term related to the cable: 38 LR

βγ⋅ . Thus the

‘reaction’ (or tension in the cable) at A depends on the relative stiffnesses of the beam

and cable (through the 3R

EI,

3RGJ

and LEA

terms inherent through γ and β ). This

dependence on relative stiffness is to be expected.


Dr. C. Caprani 29

Formal Virtual Work Approach

Without the use of the insight that equation (37) gives, the more formal application of

virtual work will, of course, yield the same result. To calculate the tension in the

cable using virtual work, we use the following:

• Equilibrium system: the external and internal virtual forces corresponding to a

unit virtual force applied in lieu of the redundant;

• Compatible system: the real external and internal displacements of the original

structure subject to the real applied loads.

Thus the virtual work equations are:

0

E I

Ay

WW W

F M ds T ds e P

δδ δ

δ δ κ δ φ δ δ

==

⋅ = ⋅ + ⋅ + ⋅∑∫ ∫ (45)

In this equation we have accounted for all the major sources of displacement (and

thus virtual work). At this point we acknowledge:

• There is no external virtual force applied, only an internal tension, thus 0Fδ = ;

• The real curvatures and twists are expressed using the real bending moments and

torques as MEI

κ = and TGJ

φ = respectively;

• The elongation of the cable is the only source of axial displacement and is

written in terms of the real tension in the cable, P, as PLeEA

= .

These combine to give, from equation (45):

0 0

0L L

AyM T PLM ds T ds PEI GJ EA

δ δ δ δ ⋅ = ⋅ + ⋅ + ⋅ ∫ ∫ (46)

As was done in Example 3, using superposition, we write:


Dr. C. Caprani 30

0 1 0 1 0 1M M M T T T P P Pα α α= + = + = + (47)

However, we know that there is no tension in the cable in the primary structure, since

it is the cable that is the redundant and is thus removed, hence 0 0P = . Using this and

equation (47) in equation (46) gives:

( ) ( ) ( )0 1 0 1 1

0 0

0L LM M T T P L

M ds T ds PEI GJ EAα α α

δ δ δ + +

= ⋅ + ⋅ + ⋅ ∫ ∫ (48)

Hence:

0 1

0 0

0 1

0 0

1

0L L

L L

M MM ds M dsEI EI

T TT ds T dsGJ GJ

P L PEA

δ α δ

δ α δ

α δ

= ⋅ + ⋅ ⋅

+ ⋅ + ⋅ ⋅

+ ⋅ ⋅

∫ ∫

∫ ∫ (49)

And so finally:

0 0

0 01 1 1

0 0

L L

L L

M TM ds T dsEI GJ

M T P LM ds T ds PEI GJ EA

δ δα

δ δ δ

⋅ + ⋅

= −

⋅ + ⋅ + ⋅

∫ ∫

∫ ∫ (50)

Equation (50) matches equation (35) except for the term relating to the cable. Thus

the other four terms are evaluated exactly as per Example 3. The cable term,


Dr. C. Caprani 31

1P L PEA

δ⋅ , is easily found once it is recognized that 1 1P Pδ= = as was the case for

the moment and torsion in Example 3. With all the terms thus evaluated, equation

(50) becomes the same as equation (42) and the solution progresses as before.

The virtual work approach yields the same solution, but without the added insight of

the source of each of the terms in equation (50) represented by equation (37).


Dr. C. Caprani 32

3.2.5 Example 5

Problem

For the structure shown, the quarter-circle beam has the properties:

• torsional rigidity of GJ;

• flexural rigidity about the local y-y axis YEI ;

• flexural rigidity about the local z-z axis ZEI .

The cable has axial rigidity EA. Show that the tension in the cable due to the

uniformly distributed load, w, shown is:

( ) ( )12

2

4 2 1 1 8 21 3 82

T wRR

β ππ π

λ β γβ

− + − = + + − +

where Y

GJEI

β = , Y

EAEI

γ = and Z

Y

EIEI

λ = .


Dr. C. Caprani 33


Dr. C. Caprani 34

Solution

We will carry out this solution using both the empirical and virtual work approaches

as was done for Example 4. However, it is in this example that the empirical

approach will lead to savings in effort over the virtual work approach, as will be seen.

Empirical Form

Repeating our empirical equation:


δαδ

=

(51)

We first examine the numerator with the following y-z axis elevation of the primary

structure loaded with the actual loads:

Noting that it is the deflection along the line of the redundant that is of interest, we

can draw the following:


Dr. C. Caprani 35

The deflection Azδ , which is the distance 'AA is known from Example 2 to be:

( )24 4 212 8Az

wR wREI GJ

πδ

−= ⋅ + ⋅ (52)

It is the deflection ''AA that is of interest here. Since the triangle A-A’-A’’ is a 1-1-

2 triangle, we have:

, 4 2Az

A π

δδ = (53)

And so the numerator is thus:

( )24 40 2

2 2 8 2A

wR wRGJEI

πδ

−= + ⋅ (54)


Dr. C. Caprani 36

To determine the denominator of equation (51) we must apply a unit load in lieu of

the redundant (the cable) and determine the deflection in the direction of the cable.

Firstly we will consider the beam. We can determine the deflection in the z- and y-

axes separately and combine, by examining the deflections that the components of the

unit load cause:

To find the deflection that a force of 12

causes in the z- and y-axes directions, we

will instead find the deflections that unit loads cause in these directions, and then

divide by 2 .

Since we are now calculating deflections in two orthogonal planes of bending, we

must consider the different flexural rigidities the beam will have in these two


Dr. C. Caprani 37

directions: YEI for the horizontal plane of bending (vertical loads), and ZEI for loads

in the x-y plane, as shown in the figure:

First, consider the deflection at A in the z-direction, caused by a unit load in the z-

direction, as shown in the following diagram. This is the same as the deflection

calculated in Example 1 and used in later examples:

3 3

1 1 1 3 84 4Az

Y

R REI GJ

π πδ ⋅ ⋅ − = ⋅ +

(55)


Dr. C. Caprani 38

Considering the deflection at A in the y-direction next, we see from the following

diagram that we do not have this result to hand, and so must calculate it:


Dr. C. Caprani 39

Looking at the elevation of the x-y plane, we have:

The lever arm, m, is:

sinm R θ= (56)

Thus the moment at point C is:

( ) 1 1 sinM m Rθ θ= ⋅ = ⋅ (57)

Using virtual work:


Dr. C. Caprani 40

0

1E I

Ay

WW W

M ds

δδ δ

δ κ δ

==

⋅ = ⋅∫ (58)

In which we note that there is no torsion term, as the unit load in the x-y plane does

not cause torsion in the structure. Using ZM EIκ = and ds Rdθ= :

2

0

1 AyZ

M M RdEI

π

δ δ θ⋅ = ∫ (59)

Since sinM M Rδ θ= = , and assuming the beam is prismatic, we have:

3 2

2

0

1 sinAyz

R dEI

π

δ θ θ⋅ = ∫ (60)

This is the same as the first term in equation (7) and so immediately we obtain the

solution as that of the first term of equation (11):

3

1

4Ayz

REI

πδ = ⋅ (61)

In other words, the bending deflection at A in the x-y plane is the same as that in the

z-y plane. This is apparent given that the lever arm is the same in both cases.

However, the overall deflections are not the same due to the presence of torsion in the

z-y plane.

Now that we have the deflections in the two orthogonal planes due to the units loads,

we can determine the deflections in these planes due to the load 12

:


Dr. C. Caprani 41

3

1 2 1 1 3 84 42Az

Y

REI GJ

π πδ − = ⋅ +

(62)

3

1 2 142Ay

z

REI

πδ

= ⋅

(63)

The deflection along the line of action of the redundant is what is of interest:

Looking at the contributions of each of these deflections along the line of action of

the redundant:

From this we have:


Dr. C. Caprani 42

1 2

3

3

12

1 1 1 3 84 42 2

1 1 3 82 4 4

Az Az

Y

Y

AE

REI GJ

REI GJ

δ δ

π π

π π

= ⋅

− = ⋅ ⋅ +

− = ⋅ +

(64)

1 2

3

3

12

1 142 2

12 4

Ay Ay

z

z

AD

REI

REI

δ δ

π

π

=

= ⋅ ⋅

= ⋅

(65)

Thus the total deflection along the line of action of the redundant is:

1, 4

3 31 1 3 8 12 4 4 2 4

A Az Ay

Y z

AE AD

R REI GJ EI

πδ δ δ

π π π

= +

− = ⋅ + + ⋅

(66)

This gives, finally:

3

1, 4

1 1 1 3 82 4 4A

Y z

REI EI GJπ

π πδ − = + +

(67)

To complete the denominator of equation (51), we must include the deflection that

the cable undergoes due to the unit tension that is the redundant:


Dr. C. Caprani 43

1

2

LeEAREA

⋅=

= (68)

The relationship between R and L is due to the geometry of the problem – the cable is

at an angle of 45°.

Thus the denominator of equation (51) is finally:

3

1, 4 2

1 1 1 3 8 2 22 4 4A

Y z

REI EI GJ R EAπ

π πδ − = + + +

(69)

The solution for the tension in the cable becomes, from equations (51), (54) and (69):

( )2

4

3

2

21 12 2 8 2

1 1 1 3 8 2 22 4 4Y z

wRGJEI

TR

EI EI GJ R EA

π

π π

−+ ⋅

= − + + +

(70)

Using Y

GJEI

β = , Y

EAEI

γ = and Z

Y

EIEI

λ = , we have:

( )2

1

2

21 12 2 8 2

1 1 1 3 8 28 8

YY

Y Y Y Y

T wREIEI

EI EI EI R EI

πβ

π πλ β γ

−

−= + ⋅

− × + + +

(71)

Continuing the algebra:


Dr. C. Caprani 44

( )

( ) ( )

12

2

12

2

21 1 1 1 3 8 218 82 2 8 2

4 2 1 1 8 21 3 88 8 88 2

T wRR

wRR

π π πβ λ β γ

β π π πλ β γβ

−

−

− − = + ⋅ + + +

+ − = + + − +

(72)

Which finally gives the desired result:

( ) ( )12

2

4 2 1 1 8 21 3 82

T wRR

β ππ π

λ β γβ

− + − = + + − +

(73)

Formal Virtual Work Approach

In the empirical approach carried out above there were some steps that are not

obvious. Within a formal application of virtual work we will see how the results of

the empirical approach are obtained ‘naturally’.

Following the methodology of the formal virtual work approach of Example 4, we

can immediately jump to equation (46):

0 0

0L L

AyM T PLM ds T ds PEI GJ EA

δ δ δ δ ⋅ = ⋅ + ⋅ + ⋅ ∫ ∫ (74)

For the next step we need to recognize that the unit redundant causes bending about

both axes of bending and so the first term in equation (74) must become:

0 0 0

L L LY Z

Y ZY Z

M M MM ds M ds M dsEI EI EI

δ δ δ ⋅ = ⋅ + ⋅

∫ ∫ ∫ (75)


Dr. C. Caprani 45

In which the notation YM and ZM indicate the final bending moments of the actual

structure about the Y-Y and Z-Z axes of bending respectively. Again we use

superposition for the moments, torques and axial forces:

0 1

0 1

0 1

0 1

Y Y Y

Z Z Z

M M MM M M

T T TP P P

αααα

= +

= +

= +

= +

(76)

We do not require more torsion terms since there is only torsion in the z-y plane. With

equations (75) and (76), equation (74) becomes:

( ) ( )

( ) ( )

0 1 0 1

0 0

0 1 0 1

0

0L L

Y Y Z ZY Z

Y Z

L

M M M MM ds M ds

EI EI

T T P P LT ds P

GJ EA

α αδ δ

α αδ δ

+ += ⋅ + ⋅

+ +

+ ⋅ + ⋅

∫ ∫

∫

(77)

Multiplying out gives:

0 1

0 0

0 1

0 0

0 1

0 0

0 1

0L L

Y YY Y

Y Y

L LZ Z

Z ZZ Z

L L

M MM ds M dsEI EI

M MM ds M dsEI EI

T TT ds T dsGJ GJ

P L P LP PEA EA

δ α δ

δ α δ

δ α δ

δ α δ

= ⋅ + ⋅ ⋅

+ ⋅ + ⋅ ⋅

+ ⋅ + ⋅ ⋅

+ ⋅ + ⋅ ⋅

∫ ∫

∫ ∫

∫ ∫

(78)

At this point we recognize that some of the terms are zero:


Dr. C. Caprani 46

• There is no axial force in the primary structure since the cable is ‘cut’, and so 0 0P = ;

• There is no bending in the x-y plane (about the z-z axis of the beam) in the

primary structure as the loading is purely vertical, thus 0 0ZM = .

Including these points, and solving for α gives:

0 0

0 01 1 1 1

0 0 0

L LY

YY

L L LY Z

Y ZY Z

M TM ds T dsEI GJ

M M T P LM ds M ds T ds PEI EI GJ EA

δ δα

δ δ δ δ

⋅ + ⋅

= −

⋅ + ⋅ + ⋅ + ⋅

∫ ∫

∫ ∫ ∫ (79)

We will next examine this expression term-by-term.

0

0

LY

YY

M M dsEI

δ⋅∫

For this term, 0YM are the moments caused by the UDL about the y-y axis of bending,

as per equation (13):

( ) ( )0 2 1 cosYM wRθ θ= − (80)

YMδ are the moments about the same axis caused by the unit redundant. Since this

redundant acts at an angle of 45° to the plane of interest, these moments are caused

by its vertical component of 12

. From equation (4), we thus have:

( ) 1 sin2YM Rδ θ θ= − (81)


Dr. C. Caprani 47

Notice that we have taken it that downwards loading causes positive bending

moments. Thus we have:

( )

( )

02

0 0

23

0

1 11 cos sin2

sin sin cos2

L LY

YY Y

Y

M M ds wR R dsEI EI

wR RdEI

π

δ θ θ

θ θ θ θ

⋅ = − −

= − −

∫ ∫

∫ (82)

In which we have used the relation ds Rdθ= . From the integral appendix we thus

have:

[ ]

( ) ( ) ( ) ( )

20 42

000

3

1cos cos242

10 1 1 142

LY

YY Y

Y

M wRM dsEI EI

wREI

ππδ θ θ

⋅ = − − − −

= − − − + − −

∫ (83)

And so finally:

0 4

0 2 2

LY

YY Y

M wRM dsEI EI

δ⋅ = −∫ (84)

0

0

L T T dsGJ

δ⋅∫

The torsion caused by the UDL in the primary structure is the same as that from

equation (14):

( ) ( )0 2 sinT wRθ θ θ= − (85)


Dr. C. Caprani 48

Similarly to the bending term, the torsion caused by the unit redundant is 12

that of

the unit load of equation (17):

( ) ( )1 1 cos2

T Rδ θ θ= − − (86)

Again note that we take the downwards loads as causing positive torsion. Noting

ds Rdθ= we thus have:

( ) ( )

( )( )

202

0 0

24

0

1 1sin 1 cos2

sin 1 cos2

L T T ds wR R RdGJ GJ

wR dGJ

π

π

δ θ θ θ θ

θ θ θ θ

⋅ = − − −

= − − −

∫ ∫

∫ (87)

This integral is exactly that of the second term in equation (19). Hence we can take its

result from equation (20) to give:

( )220 4

0

282

L T wRT dsGJ GJ

πδ

−⋅ = − ⋅∫ (88)

1

0

LY

YY

M M dsEI

δ⋅∫

For this term we recognize that 1Y YM Mδ= and are the moments caused by the 1

2

component of the unit redundant in the vertical direction and are thus given by

equation (1):


Dr. C. Caprani 49

( )1 1 sin2Y YM M Rδ θ θ= = (89)

Hence this term becomes:

1 2

0 0

3 22

0

1 1 1sin sin2 2

sin2

LY

YY Y

Y

M M ds R R RdEI EI

R dEI

π

π

δ θ θ θ

θ θ

⋅ =

=

∫ ∫

∫ (90)

From the integral tables we thus have:

( )

21 3

0 0

3

1 sin 22 2 4

1 0 0 02 4 4

LY

YY Y

Y

M RM dsEI EI

REI

πθδ θ

π

⋅ = −

= − ⋅ − −

∫ (91)

And so we finally have:

1 3

0 8

LY

YY Y

M RM dsEI EI

πδ⋅ = ⋅∫ (92)

1

0

LZ

ZZ

M M dsEI

δ⋅∫

Again we recognize that 1Z ZM Mδ= and are the moments caused by the 1

2

component of the unit redundant in the x-y plane and are thus given by equation (57).

Hence this term becomes:


Dr. C. Caprani 50

1 2

0 0

1 1 1sin sin2 2

LY

YY Y

M M ds R R RdEI EI

π

δ θ θ θ ⋅ = ∫ ∫ (93)

This is the same as equation (90) except for the different flexural rigidity, and so the

solution is got from equation (92) to be:

1 3

0 8

LZ

ZZ Z

M RM dsEI EI

πδ⋅ = ⋅∫ (94)

1

0

L T T dsGJ

δ⋅∫

Once again note that 1T Tδ= and are the torques caused by the 12

vertical

component of the unit redundant. From equation (2), then we have:

( )1 1 1 cos2

T T Rδ θ= = − (95)

Thus:

( ) ( )

( )

21

0 0

232

0

1 1 11 cos 1 cos2 2

1 cos2

L T T ds R R RdGJ GJ

R dGJ

π

π

δ θ θ θ

θ θ

⋅ = − −

= −

∫ ∫

∫ (96)

This integral is that of equation (9) and so the solution is:


Dr. C. Caprani 51

1 3

0

3 88

L T RT dsGJ GJ

πδ − ⋅ = ∫ (97)

1P L P

EAδ⋅

Lastly then, since 1 1P Pδ= = and 2L R= , this term is easily calculated to be:

1 2P L RP

EA EAδ⋅ = (98)

With the values for all terms now worked out, we substitute these values into

equation (79) to determine the cable tension:

( )224 4

3 3 3

282 2 2

3 8 28 8 8

Y

Y Z

wR wREI GJ

R R R REI EI GJ EA

π

απ π π

− − − ⋅ = −

− ⋅ + ⋅ + +

(99)

Cancelling the negatives and re-arranging gives:

( )2

4

3

2

21 12 2 8 2

1 1 1 3 8 2 22 4 4

Y

Y z

wRGJEI

TR

EI EI GJ R EA

π

π π

−+ ⋅

= − + + +

(100)

And this is the same as equation (70) and so the solution can proceed as before to

obtain the tension in the cable as per equation (73).


Dr. C. Caprani 52

Comparison of the virtual work with the empirical form illustrates the interpretation

of each of the terms in the virtual work equation that is inherent in the empirical view

of such problems.


Dr. C. Caprani 53

3.2.6 Review of Examples 1 – 5

Example 1

For a radius of 2 m and a point load of 10 kN, the bending and torsion moment

diagrams are:

Using the equations derived in Example 1, the Matlab script for this is:

function RingBeam_Ex1 % Example 1 R = 2; % m P = 10; % kN theta = 0:(pi/2)/50:pi/2; M = P*R*sin(theta); T = P*R*(1-cos(theta)); hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off;

0 10 20 30 40 50 60 70 80 900

5

10

15

20

Mom

ent (

kNm

)

Degrees from Y-axis

BendingTorsion


Dr. C. Caprani 54

Example 2

For a radius of 2 m and a UDL of 10 kN/m, the bending and torsion moment

diagrams are:


function RingBeam_Ex2 % Example 2 R = 2; % m w = 10; % kN/m theta = 0:(pi/2)/50:pi/2; M = w*R^2*(1-cos(theta)); T = w*R^2*(theta-sin(theta)); hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off;

0 10 20 30 40 50 60 70 80 900

5

10

15

20

25

30

35

40

Mom

ent (

kNm

)

Degrees from Y-axis

BendingTorsion


Dr. C. Caprani 55

Example 3

For the parameters given below, the bending and torsion moment diagrams are:


function [M T alpha] = RingBeam_Ex3(beta) % Example 3 R = 2; % m w = 10; % kN/m I = 2.7e7; % mm4 J = 5.4e7; % mm4 E = 205; % kN/mm2 v = 0.30; % Poisson's Ratio G = E/(2*(1+v)); % Shear modulus EI = E*I/1e6; % kNm2 GJ = G*J/1e6; % kNm2 if nargin < 1 beta = GJ/EI; % Torsion stiffness ratio end alpha = w*R*(4*beta+(pi-2)^2)/(2*beta*pi+2*(3*pi-8)); theta = 0:(pi/2)/50:pi/2;

0 10 20 30 40 50 60 70 80 90-10

-5

0

5

10

15

20

X: 90Y: 17.19

X: 59.4Y: -4.157

X: 90Y: 0.02678M

omen

t (kN

m)

Degrees from Y-axis

X: 28.8Y: -6.039

BendingTorsion


Dr. C. Caprani 56

M0 = w*R^2*(1-cos(theta)); T0 = w*R^2*(theta-sin(theta)); M1 = -R*sin(theta); T1 = -R*(1-cos(theta)); M = M0 + alpha.*M1; T = T0 + alpha.*T1; if nargin < 1 hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off; end

The vertical reaction at A is found to be 11.043 kN. Note that the torsion is

(essentially) zero at support B. Other relevant values for bending moment and torsion

are given in the graph.

By changing β , we can examine the effect of the relative stiffnesses on the vertical

reaction at A, and consequently the bending moments and torsions. In the following

plot, the reaction at A and the maximum and minimum bending and torsion moments

are given for a range of β values.

Very small values of β reflect little torsional rigidity and so the structure movements

will be dominated by bending solely. Conversely, large values of β reflect structures

with small bending stiffness in comparison to torsional stiffness. At either extreme

the variables converge to asymptotes of extreme behaviour. For 0.1 10β≤ ≤ the

variables are sensitive to the relative stiffnesses. Of course, this reflects the normal

range of values for β .


Dr. C. Caprani 57

The Matlab code to produce this figure is:

% Variation with Beta beta = logspace(-3,3); n = length(beta); for i = 1:n [M T alpha] = RingBeam_Ex3(beta(i)); Eff(i,1) = alpha; Eff(i,2) = max(M); Eff(i,3) = min(M); Eff(i,4) = max(T); Eff(i,5) = min(T); end hold on; plot(beta,Eff(:,1),'b:'); plot(beta,Eff(:,2),'k-','LineWidth',2); plot(beta,Eff(:,3),'k-'); plot(beta,Eff(:,4),'r--','LineWidth',2); plot(beta,Eff(:,5),'r--'); hold off; set(gca,'xscale','log'); legend('Va','Max M','Min M','Max T','Min T','Location','NO',... 'Orientation','horizontal'); xlabel('Beta'); ylabel('Load Effect (kN & kNm)');

10-3

10-2

10-1

100

101

102

103

-10

-5

0

5

10

15

20

25

Beta

Load

Effe

ct (k

N &

kN

m)

Va Max M Min M Max T Min T


Dr. C. Caprani 58

Example 4

For a 20 mm diameter cable, and for the other parameters given below, the bending

and torsion moment diagrams are:

The values in the graph should be compared to those of Example 3, where the support

was rigid. The Matlab script, using Example 4’s equations, for this problem is:

function [M T alpha] = RingBeam_Ex4(gamma,beta) % Example 4 R = 2; % m - radius of beam L = 2; % m - length of cable w = 10; % kN/m - UDL A = 314; % mm2 - area of cable I = 2.7e7; % mm4 J = 5.4e7; % mm4 E = 205; % kN/mm2 v = 0.30; % Poisson's Ratio G = E/(2*(1+v)); % Shear modulus EA = E*A; % kN - axial stiffness EI = E*I/1e6; % kNm2 GJ = G*J/1e6; % kNm2 if nargin < 2 beta = GJ/EI; % Torsion stiffness ratio end if nargin < 1 gamma = EA/EI; % Axial stiffness ratio end

0 10 20 30 40 50 60 70 80 90-10

-5

0

5

10

15

20

X: 90Y: 17.58

X: 59.4Y: -3.967X: 28.8

Y: -5.853

X: 90Y: 0.4128

Mom

ent (

kNm

)

Degrees from Y-axis

BendingTorsion


Dr. C. Caprani 59

alpha = w*R*(4*beta+(pi-2)^2)/(2*beta*pi+2*(3*pi-8)+8*(beta/gamma)*(L/R^3)); theta = 0:(pi/2)/50:pi/2; M0 = w*R^2*(1-cos(theta)); T0 = w*R^2*(theta-sin(theta)); M1 = -R*sin(theta); T1 = -R*(1-cos(theta)); M = M0 + alpha.*M1; T = T0 + alpha.*T1; if nargin < 1 hold on; plot(theta.*180/pi,M,'k-'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('Bending','Torsion','location','NW'); hold off; end

Whist keeping the β constant, we can examine the effect of varying the cable

stiffness on the behaviour of the structure, by varying γ . Again we plot the reaction

at A and the maximum and minimum bending and torsion moments for the range of

γ values.

For small γ , the cable has little stiffness and so the primary behaviour will be that of

Example 1, where the beam was a pure cantilever. Conversely for high γ , the cable is

very stiff and so the beam behaves as in Example 3, where there was a pinned support

at A. Compare the maximum (hogging) bending moments for these two cases with

the graph. Lastly, for 0.01 3γ≤ ≤ , the cable and beam interact and the variables are

sensitive to the exact ratio of stiffness. Typical values in practice are towards the

lower end of this region.


Dr. C. Caprani 60

The Matlab code for this plot is:

% Variation with Gamma gamma = logspace(-3,3); n = length(gamma); for i = 1:n [M T alpha] = RingBeam_Ex4(gamma(i)); Eff(i,1) = alpha; Eff(i,2) = max(M); Eff(i,3) = min(M); Eff(i,4) = max(T); Eff(i,5) = min(T); end hold on; plot(gamma,Eff(:,1),'b:'); plot(gamma,Eff(:,2),'k-','LineWidth',2); plot(gamma,Eff(:,3),'k-'); plot(gamma,Eff(:,4),'r--','LineWidth',2); plot(gamma,Eff(:,5),'r--'); hold off; set(gca,'xscale','log'); legend('T','Max M','Min M','Max T','Min T','Location','NO',... 'Orientation','horizontal'); xlabel('Gamma'); ylabel('Load Effect (kN & kNm)');

10-3

10-2

10-1

100

101

102

103

-10

0

10

20

30

40

Gamma

Load

Effe

ct (k

N &

kN

m)

T Max M Min M Max T Min T


Dr. C. Caprani 61

Example 5

Again we consider a 20 mm diameter cable, and a doubly symmetric section, that is

Y ZEI EI= . For the parameters below the bending and torsion moment diagrams are:

The values in the graph should be compared to those of Example 4, where the cable

was vertical. The Matlab script, using Example 5’s equations, for this problem is:

function [My T alpha] = RingBeam_Ex5(lamda,gamma,beta) % Example 5 R = 2; % m - radius of beam w = 10; % kN/m - UDL A = 314; % mm2 - area of cable Iy = 2.7e7; % mm4 Iz = 2.7e7; % mm4 J = 5.4e7; % mm4 E = 205; % kN/mm2 v = 0.30; % Poisson's Ratio G = E/(2*(1+v)); % Shear modulus EA = E*A; % kN - axial stiffness EIy = E*Iy/1e6; % kNm2 EIz = E*Iz/1e6; % kNm2 GJ = G*J/1e6; % kNm2 if nargin < 3 beta = GJ/EIy; % Torsion stiffness ratio end if nargin < 2 gamma = EA/EIy; % Axial stiffness ratio end if nargin < 1

0 10 20 30 40 50 60 70 80 90-20

-10

0

10

20

30

X: 90Y: -19.22

X: 52.2Y: -2.605

X: 90Y: 3.61

X: 90Y: 20.78

X: 25.2Y: -4.378

Mom

ent (

kNm

)

Degrees from Y-axis

YY BendingZZ BendingTorsion


Dr. C. Caprani 62

lamda = EIy/EIz; % Bending stiffness ratio end numerator = (4*beta+(pi-2)^2)/(beta*sqrt(2)); denominator = (pi*(1+1/lamda)+(3*pi-8)/beta+8*sqrt(2)/(gamma*R^2)); alpha = w*R*numerator/denominator; theta = 0:(pi/2)/50:pi/2; M0y = w*R^2*(1-cos(theta)); M0z = 0; T0 = w*R^2*(theta-sin(theta)); M1y = -R*sin(theta); M1z = -R*sin(theta); T1 = -R*(1-cos(theta)); My = M0y + alpha.*M1y; Mz = M0z + alpha.*M1z; T = T0 + alpha.*T1; if nargin < 1 hold on; plot(theta.*180/pi,My,'k'); plot(theta.*180/pi,Mz,'k:'); plot(theta.*180/pi,T,'r--'); ylabel('Moment (kNm)'); xlabel('Degrees from Y-axis'); legend('YY Bending','ZZ Bending','Torsion','location','NW'); hold off; end

Keep all parameters constant, but varying the ratio of the bending rigidities by

changing λ , the output variables are as shown below. For low λ (a tall slender

beam) the beam behaves as a cantilever. Thus the cable requires some transverse

bending stiffness to be mobilized. With high λ (a wide flat beam) the beam behaves

as if supported at A with a vertical roller. Only vertical movement takes place, and the

effect of the cable is solely its vertical stiffness at A. Usually 0.1 2λ≤ ≤ which means

that the output variables are usually quite sensitive to the input parameters.


Dr. C. Caprani 63

The Matlab code to produce this graph is:

% Variation with Lamda lamda = logspace(-3,3); n = length(lamda); for i = 1:n [My T alpha] = RingBeam_Ex5(lamda(i)); Eff(i,1) = alpha; Eff(i,2) = max(My); Eff(i,3) = min(My); Eff(i,4) = max(T); Eff(i,5) = min(T); end hold on; plot(lamda,Eff(:,1),'b:'); plot(lamda,Eff(:,2),'k-','LineWidth',2); plot(lamda,Eff(:,3),'k-'); plot(lamda,Eff(:,4),'r--','LineWidth',2); plot(lamda,Eff(:,5),'r--'); hold off; set(gca,'xscale','log'); legend('T','Max My','Min My','Max T','Min T','Location','NO',... 'Orientation','horizontal'); xlabel('Lamda'); ylabel('Load Effect (kN & kNm)');

10-3

10-2

10-1

100

101

102

103

-20

-10

0

10

20

30

40

Lamda

Load

Effe

ct (k

N &

kN

m)

T Max My Min My Max T Min T


Dr. C. Caprani 64

3.3 Grid Examples

3.3.1 Example 1

Problem

For the grid structure shown, which has flexural and torsional rigidities of EI and GJ

respectively, show that the vertical reaction at C is given by:

12 3CV P

β

= +

Where

EIGJ

β =


Dr. C. Caprani 65

Solution

Using virtual work, we have:

0

0

E I

WW W

M TM ds T dsEI GJ

δδ δ

δ δ

==

= ⋅ + ⋅∫ ∫

(101)

Choosing the vertical reaction at C as the redundant gives the following diagrams:

And the free bending moment diagram is:


Dr. C. Caprani 66

But the superposition gives:

0 1M M Mα= + (102)

0 1T T Tα= + (103)

Substituting, we get:

( ) ( )0 1 0 10M M T T

M ds T dsEI GJα α

δ δ+ +

= ⋅ + ⋅∫ ∫ (104)

2 2

0 1 1 0 1 1 0M M M T T Tds ds ds dsEI EI GJ GJ

α α+ + + =∫ ∫ ∫ ∫ (105)

2 2

0 1 1 0 1 1 0M M M T T Tds ds ds dsEI EI GJ GJ

α α+ + + =∫ ∫ ∫ ∫ (106)

Taking the beam to be prismatic, and EIGJ

β = gives:

2 20 1 1 0 1 1 0M M ds M ds T T ds T dsα β αβ+ + + =∫ ∫ ∫ ∫ (107)


Dr. C. Caprani 67

From which:

0 1 0 1

2 21 1

M M ds T T ds

M ds T ds

βα

β

+ = − +

∫ ∫∫ ∫

(108)

From the various diagrams and volume integrals tables, the terms evaluate to:

( )( )( )

( )

( )( )( )

( )( )( )

3

0 1

0 1

2 31

2 31

13 30 0

1 223 3

PLM M ds L PL L

T T ds

M ds L L L L

T ds L L L L

β β

β β β

= − = −

= =

= =

= =

∫∫

∫

∫

(109)

Substituting gives:

( )

3

3 3

3

3 23

03

23

1 13

PL

L L

PLL

αβ

β

− + = − +

= ⋅ ⋅+

(110)

Which yields:

12 3CV Pα

β

≡ = + (111)


Dr. C. Caprani 68

Numerical Example

Using a 200 × 400 mm deep rectangular concrete section, gives the following:

3 4 3 41.067 10 m 0.732 10 mI J= × = ×

The material model used is for a 50N concrete with:

230 kN/mm 0.2E ν= =

Using the elastic relation, we have:

( ) ( )

66 230 10 12.5 10 kN/m

2 1 2 1 0.2EGν

×= = = ×

+ +

From the model, LUSAS gives: 0.809 kNCV = . Other results follow.

Deflected Shape


Dr. C. Caprani 69

Bending Moment Diagram

Torsion Moment Diagram

Shear Force Diagram


Dr. C. Caprani 70

3.3.2 Example 2

Problem


respectively, show that the reactions at C are given by:

4 4 4 28 5 8 5C CV P M PLβ ββ β

+ += = + +

Where

EIGJ

β =

(Note that the support symbol at C indicates a moment and vertical support at C, but

no torsional restraint.)


Dr. C. Caprani 71

Solution

The general virtual work equations are:

0

0

E I

WW W

M TM ds T dsEI GJ

δδ δ

δ δ

==

= ⋅ + ⋅∫ ∫

(112)

We choose the moment and vertical restraints at C as the redundants. The vertical

redundant gives the same diagrams as before:

And, for the moment restraint, we apply a unit moment:

Which yields the following:


Dr. C. Caprani 72

Again the free bending moment diagram is:

Since there are two redundants, there are two possible equilibrium sets to use as the

virtual moments and torques. Thus there are two equations that can be used:

1 10 M TM ds T dsEI GJ

= ⋅ + ⋅∫ ∫ (113)


= ⋅ + ⋅∫ ∫ (114)


Dr. C. Caprani 73

Superposition gives:

0 1 1 2 2M M M Mα α= + + (115)

0 1 1 2 2T T T Tα α= + + (116)

Substituting, we get from equation (113):

( ) ( )0 1 1 2 2 0 1 1 2 21 10

M M M T T TM ds T ds

EI GJα α α α+ + + +

= ⋅ + ⋅∫ ∫ (117)

20 1 1 2 1

1 2

20 1 1 2 1

1 2 0

M M M M Mds ds dsEI EI EI

T T T T Tds ds dsGJ GJ GJ

α α

α α

+ +

+ + + =

∫ ∫ ∫

∫ ∫ ∫ (118)


β = gives:

2

0 1 1 1 2 2 1

20 1 1 1 1 2 1 0

M M ds M ds M M ds

T T ds T ds T T ds

α α

β α β α β

+ +

+ + + =

∫ ∫ ∫∫ ∫ ∫

(119)

Similarly, substituting equations (115) and (116) into equation (114) gives:

2

0 2 1 1 2 2 2

20 2 1 1 2 2 2 0

M M ds M M ds M ds

T T ds TT ds T ds

α α

β α β α β

+ +

+ + + =

∫ ∫ ∫∫ ∫ ∫

(120)

We can write equations (119) and (120) in matrix form for clarity:


Dr. C. Caprani 74

0 1 0 1

0 2 0 2

2 21 1 2 1 2 1 1

2 221 2 1 2 2 2

0

M M ds T T ds

M M ds T T ds

M ds T ds M M ds T T ds

M M ds TT ds M ds T ds

β

β

β β ααβ β

+ + +

+ + = + +

∫ ∫∫ ∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(121)

Evaluating the integrals for the first equation gives:

3

0 1 0 1

32 2 3

1 1

2 22 1 2 1

03

23

12

PLM M ds T T ds

LM ds T ds L

M M ds L T T ds L

β

β β

β β

−= =

= =

= − = −

∫ ∫

∫ ∫

∫ ∫

(122)

And for the second:

0 2 0 2

2 21 2 1 2

2 22 2

0 0

12

M M ds T T ds

M M ds L TT ds L

M ds L T ds L

β

β β

β β

= =

= − = −

= =

∫ ∫

∫ ∫∫ ∫

(123)

Substituting these into equation (121), we have:

( )

3 23

1

22

23 2

030 1

2

L LPL

L L

β βαα

β β

1 + − + − + = 1 − + +

(124)

Giving:


Dr. C. Caprani 75

( )

13 2 3

1

2 2

23 2

301

2

L L PL

L L

β βαα

β β

− 1 + − + = 1 − + +

(125)

Inverting the matrix gives:

( ) ( )

( ) ( )

33 2

1

22

12 61 1 21

35 8 6 4 01 2 2 3

PLL L

L L

β βαα β

β β

+ + = + + +

(126)

Thus:

( )

( )

( )( )

3

31

32

2

12 13 4 11

2 1 25 8 5 86 1 23

PLL P

LPLL

ββαβα β β

β

+ + = = ++ + +

(127)

Thus, since 1

2

C

C

VM

αα

≡

, we have:

4 4 4 28 5 8 5C CV P M PLβ ββ β

+ += = + +

(128)

And this is the requested result.


Dr. C. Caprani 76

Some useful Matlab symbolic computation script appropriate to this problem is:

syms beta L P A = [ L^3*(2/3+beta) -L^2*(0.5+beta); -L^2*(0.5+beta) L*(1+beta)]; A0 = [P*L^3/3; 0]; invA = inv(A); invA = simplify(invA); disp(simplify(det(A))); disp(invA); alpha = invA*A0; alpha = simplify(alpha);


Dr. C. Caprani 77

Numerical Example

For the numerical model previously considered, for these support conditions, LUSAS

gives us:

5.45 kN 14.5 kNmC CV M= =

Deflected Shape

Shear Force Diagram


Dr. C. Caprani 78




Dr. C. Caprani 79

3.3.3 Example 3

Problem


respectively, show that the reactions at C are given by:

( )( ) ( )2 1 1

2 4 1 4 1C C C

P PL PLV M Tββ β

+= = ⋅ = ⋅

+ +

Where

EIGJ

β =


Dr. C. Caprani 80

Solution

The general virtual work equations are:

0

0

E I

WW W

M TM ds T dsEI GJ

δδ δ

δ δ

==

= ⋅ + ⋅∫ ∫

(129)

We choose the moment, vertical, and torsional restraints at C as the redundants. The

vertical and moment redundants give (as before):

Applying the unit torsional moment gives:


Dr. C. Caprani 81

Again the free bending moment diagram is:

Since there are three redundants, there are three possible equilibrium sets to use. Thus

we have the following three equations:


Dr. C. Caprani 82


= ⋅ + ⋅∫ ∫ (130)


= ⋅ + ⋅∫ ∫ (131)


= ⋅ + ⋅∫ ∫ (132)

Superposition of the structures gives:

0 1 1 2 2 3 3M M M M Mα α α= + + + (133)

0 1 1 2 2 3 3T T T T Tα α α= + + + (134)

Substituting, we get from equation (113):

( ) ( )0 1 1 2 2 3 3 0 1 1 2 2 3 31 10

M M M M T T T TM ds T ds

EI GJα α α α α α+ + + + + +

= ⋅ + ⋅∫ ∫ (135)

20 1 1 2 1 3 1

1 2 3

20 1 1 2 1 3 1

1 2 3 0

M M M M M M Mds ds ds dsEI EI EI EI

T T T T T T Tds ds ds dsGJ GJ GJ GJ

α α α

α α α

+ + +

+ + + + =

∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ (136)


β = gives:

2

0 1 1 1 2 2 1 3 3 1

20 1 1 1 2 2 1 3 3 1 0

M M ds M ds M M ds M M ds

T T ds T ds T T ds T T ds

α α α

β α β α β α β

+ + +

+ + + + =

∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(137)

Similarly, substituting equations (115) and (116) into equations (114) and (132)

gives:


Dr. C. Caprani 83

2

0 2 1 1 2 2 2 3 3 2

20 2 1 1 2 2 2 3 3 2 0

M M ds M M ds M ds M M ds

T T ds TT ds T ds T T ds

α α α


+ + +

+ + + + =

∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(138)

2

0 3 1 1 3 2 2 3 3 3

20 3 1 1 3 2 2 3 3 3 0

M M ds M M ds M M ds M ds

T T ds TT ds T T ds T ds

α α α


+ + +

+ + + + =

∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(139)

We can write equations (119), (120), and (139) in matrix form for clarity:

{ } [ ]{ } { } [ ]{ } { }β β0 0M + δM α + T + δT α = 0 (140)

Or more concisely:

{ } [ ]{ } { }+ =0A δA α 0 (141)

In which { }0A is the ‘free’ actions vector:

{ } { } { }0 1 0 1

0 2 0 2

0 3 0 3

M M ds T T ds

M M ds T T ds

M M ds T T ds

β

β β

β

+ = = +

+

∫ ∫∫ ∫∫ ∫

0 0 0A M + T (142)

And [ ]δA is the virtual actions matrix:

[ ] [ ] [ ]2 2

1 1 2 1 2 1 1 3 1 3

2 21 2 1 2 2 2 2 3 2 3

2 21 3 1 3 2 3 2 3 3 3

M ds T ds M M ds T T ds M M ds TT ds

M M ds TT ds M ds T ds M M ds T T ds

M M ds TT ds M M ds T T ds M ds T ds

β

β β β

β β β

β β β

=

+ + + = + + + + + +

∫ ∫ ∫ ∫ ∫ ∫∫ ∫ ∫ ∫ ∫ ∫∫ ∫ ∫ ∫ ∫ ∫

δA δM + δT

(143)


Dr. C. Caprani 84

And { }α is the redundant multipliers vector:

{ }1

2

3

ααα

=

α (144)

Evaluating the free actions vector integrals gives:

3

0 1 0 1

0 2 0 2

2

0 3 0 3

03

0 0

02

PLM M ds T T ds

M M ds T T ds

PLM M ds T T ds

β

β

β

−= =

= =

= =

∫ ∫∫ ∫

∫ ∫

(145)

The virtual moment and torsion integrals are (noting that the matrices are

symmetrical):

3 2 22

1 2 1 1 3

22 2 3

23

23 2 2

0

L L LM ds M M ds M M ds

M ds L M M ds

M ds L

= = − = −

= =

=

∫ ∫ ∫∫ ∫

∫

(146)

2 3 21 2 1 1 3

22 2 3

23

0

0

T ds L T T ds L TT ds

T ds L T T ds

T ds L

= = − =

= =

=

∫ ∫ ∫∫ ∫

∫

(147)

Substituting these integral results into equation (141) gives:


Dr. C. Caprani 85

3 2 23 23

122

22

2 3

23 2 2

30 0 0

2

02 2

L L LL LPL

L L L LPL L L L

β βα

β β αα

β

+ − − − −

+ − − + = − +

(148)

( )

( )

23 2 3

12

22

32

2 13 2 2

31 1 0 02

0 1 22

LL L PL

L LPL

L L

β βα

β β αα

β

+ − + − − + + =

− − +

(149)

Inverting the matrix gives:

( )( ) ( )( )

( )( ) ( )( )

3 2 2

1 2

2 2

3

2

6 1 3 2 1 3 14 1 4 1 4 1

3 2 1 1 12 20 5 3 2 14 1 2 4 1 1 2 4 1 1

3 1 3 2 1 1 8 54 1 2 4 1 1 2 4 1 1

L L L

L L L

L L L

β ββ β β

αβ β β βαβ β β β β

αβ β

β β β β β

+ + + + + + + + + = + + + + +

+ + + + + + +

3

2

30

2

PL

PL

−

(150)

Thus:


Dr. C. Caprani 86

( )

( )

3 2

3 2

1 3 2

2 2

3 3 2

2

6 313 2

1 3 3 2 12 14 1 3 2 2 1

3 1 8 53 2 2 1

PL PLL L

PL PLL L

PL PLL L

βα

βα ββ β

αββ

+ − + = + − + +

+ − +

(151)

Simplifying, we get:

( )( )

( )

1

2

3

22 1

4 11

4 1

P

PL

PL

αβ

αβ

α

β

+ = ⋅ +

⋅

+

(152)

Since the redundants chosen are the reactions required, the problem is solved.

Some useful Matlab symbolic computation script appropriate to this problem is:

syms beta L P A = [ L^3*(beta+2/3) -L^2*(beta+0.5) -L^2/2; -L^2*(beta+0.5) L*(beta+1) 0; -L^2/2 0 L*(beta+1)]; A0 = [P*L^3/3; 0; -P*L^2/2]; invA = inv(A); invA = simplify(invA); disp(simplify(det(A))); disp(invA); alpha = invA*A0; alpha = simplify(alpha);


Dr. C. Caprani 87

Numerical Example

For the numerical model previously considered, for these support conditions, LUSAS

gives us:

5.0 kN 13.3 kNm 1.67 kNmC C CV M T= = =

Deflected Shape

Shear Force Diagram


Dr. C. Caprani 88



Structural Analysis IV Chapter 4 – Matrix Stiffness Method

Dr. C. Caprani 1

Chapter 4 - Matrix Stiffness Method

4.1 Introduction ......................................................................................................... 3

4.1.1 Background .................................................................................................... 3

4.1.2 Basic Concepts............................................................................................... 4

4.1.3 Computer Programs to Support Learning...................................................... 6

4.2 Basic Approach .................................................................................................. 10

4.2.1 Individual Element ...................................................................................... 10

4.2.2 Assemblies of Elements ............................................................................... 12

4.2.3 Example 1 .................................................................................................... 14

4.2.4 General Methodology .................................................................................. 20

4.2.5 Member contribution to global stiffness matrix .......................................... 22

4.2.6 Interpretation of Stiffness Matrix ................................................................ 27

4.2.7 Restricting a Matrix – Imposing Restraints ................................................. 29

4.3 Plane Trusses ..................................................................................................... 32

4.3.1 Introduction .................................................................................................. 32

4.3.2 Truss Element Stiffness Matrix ................................................................... 35

4.3.3 Element Forces ............................................................................................ 40

4.3.4 Example 2: Basic Truss ............................................................................... 43

4.3.5 Example 3: Adding Members ...................................................................... 52

4.3.6 Example 4: Using Symmetry ....................................................................... 56

4.3.7 Self-Strained Structures ............................................................................... 59

4.3.8 Example 5 – Truss with Differential Temperature ...................................... 63

4.3.9 Example 6 – Truss with Loads & Self Strains ............................................ 69


Dr. C. Caprani 2

4.3.10 Problems ................................................................................................... 74

4.4 Beams .................................................................................................................. 77

4.4.1 Beam Element Stiffness Matrix ................................................................... 77

4.4.2 Beam Element Loading ............................................................................... 82

4.4.3 Example 7 – Simple Two-Span Beam ......................................................... 84

4.4.4 Example 8 – Non-Prismatic Beam .............................................................. 88

4.4.5 Problems ...................................................................................................... 92

4.5 Plane Frames ...................................................................................................... 95

4.5.1 Plane Frame Element Stiffness Matrix ........................................................ 95

4.5.2 Example 9 – Simple Plane Frame ............................................................. 104

4.5.3 Example 10 –Plane Frame Using Symmetry............................................. 109

4.5.4 Problems .................................................................................................... 115

4.6 Appendix .......................................................................................................... 120

4.6.1 Plane Truss Element Stiffness Matrix in Global Coordinates ................... 120

4.6.2 Coordinate Transformations ...................................................................... 129

4.6.3 Past Exam Questions ................................................................................. 137

4.7 References ........................................................................................................ 148

Rev. 1


Dr. C. Caprani 3

4.1 Introduction

4.1.1 Background

The matrix stiffness method is the basis of almost all commercial structural analysis

programs. It is a specific case of the more general finite element method, and was in

part responsible for the development of the finite element method. An understanding

of the underlying theory, limitations and means of application of the method is

therefore essential so that the user of analysis software is not just operating a ‘black

box’. Such users must be able to understand any errors in the modelling of structures

which usually come as obtuse warnings such as ‘zero pivot’ or ‘determinant zero:

structure unstable: aborting’. Understanding the basics presented herein should

hopefully lead to more fruitful use of the available software.


Dr. C. Caprani 4

4.1.2 Basic Concepts

Node

The more general name for a connection between adjacent members is termed a node.

For trusses and frames the terms joint and node are interchangeable. For more

complex structures (e.g. plates), they are not.

Element

For trusses and frames element means the same as member. For more complex

structures this is not the case.

Degree of Freedom

The number of possible directions that displacements or forces at a node can exist in

is termed a degree of freedom (dof). Some examples are:

• Plane truss: has 2 degrees of freedom at each node: translation/forces in the x and y

directions.

• Beams: have 2 degrees of freedom per node: vertical displacement/forces and

rotation/moment.

• Plane Frame: has 3 degrees of freedom at each node: the translations/forces similar

to a plane truss and in addition, the rotation or moment at the joint.

• Space Truss: a truss in three dimensions has 3 degrees of freedom: translation or

forces along each axis in space.

• Space Frame: has 6 degrees of freedom at each node: translation/forces along each

axis, and rotation/moments about each axis.


Dr. C. Caprani 5

Thus a plane truss with 10 joints has 20 degrees of freedom. A plane frame with two

members will have three joints (one common to both members) and thus 9 degrees of

freedom in total.

Local and Global

Forces, displacements and stiffness matrices are often derived and defined for an axis

system local to the member. However there will exist an overall, or global, axis

system for the structure as a whole. We must therefore transform forces,

displacements etc from the local coordinate system into the global coordinate system.


Dr. C. Caprani 6

4.1.3 Computer Programs to Support Learning

Matlab Truss Analysis Program

Description

To support the ideas developed here we will introduce some Matlab scripts at each

point to demonstrate how the theory described can be implemented for computer

calculation. This collection of scripts will build into a program that can analyse pin-

jointed trusses. The scripts will only demonstrate the calculations process, and do not

have any graphical user interface facilities. This keeps the calculation process

unencumbered by extra code. (In fact probably 90+% of code in commercial

programs is for the graphical user interface and not for the actual calculations

process.) Of course, this is not to say that graphical displays of results are

unimportant; gross mistakes in data entry can sometimes only be found with careful

examination of the graphical display of the input data.

The scripts that are developed in these notes are written to explain the underlying

concepts, and not to illustrate best programming practice. The code could actually be

a lot more efficient computationally, but this would be at cost to the clarity of

calculation. In fact, a full finite element analysis program can be implemented in

under 50 lines (Alberty et al, 1999)!

It is necessary to use a scripting language like Matlab, rather than a spreadsheet

program (like MS Excel) since the number of members and member connectivity can

change from structure to structure.

The program will be able to analyse plane pin-jointed-trusses subject to nodal loads

only. It will not deal with member prestress, support stiffness or lack of fits: it is quite

rudimentary on purpose.


Dr. C. Caprani 7

Use

To use the program, download it from the course website (www.colincaprani.com).

Extract the files to a folder and change the current Matlab directory to that folder.

After preparing the data (as will be explained later), execute the following statement

at the command line:

>> [D F R] = AnalyzeTruss(nData,eData)

This assumes that the nodal data is stored in the matrix nData, and the element data

matrix is stored in eData – these names are arbitrary. Entering the required data into

Matlab will also be explained later.



Dr. C. Caprani 8

TrussMaster

To provide a bridge between the obvious workings of the Matlab program and the

more ‘black box’ nature of commercial programs, TrussMaster has been developed.

This program can analyse plane trusses of any size. It has a front end that illustrates a

fairly rudimentary commercial program interface, coupled with a back end (and some

dialogs) that expose the calculations the program carries out. In this manner it is easy

to link the hand calculations of the examples with the computer output, strengthening

the link between theory and practice of the method.

TrussMaster is available on the college computers. The help file can be downloaded

from the course website (www.colincaprani.com). It will be used in some labs.



Dr. C. Caprani 9

LinPro

LinPro is very useful as a study aid for this topic: for example, right click on a

member and select “Stiffness Matrix” to see the stiffness matrix for any member. The

latest version (2.7.3) has a very useful “Study Mode”, which exposes the structure

and member stiffness matrices to the user. A user familiar with the underlying theory

can then use the program for more advanced purposes, such as spring supports, for

example.

You can download LinPro from www.line.co.ba.

http://www.line.co.ba/


Dr. C. Caprani 10

4.2 Basic Approach

4.2.1 Individual Element

We consider here the most basic form of stiffness analysis. We represent a structural

member by a spring which has a node (or connection) at each end. We also consider

that it can only move in the x-direction. Thus it only has 1 DOF per node. At each of

its nodes, it can have a force and a displacement (again both in the x-direction):

Notice that we have drawn the force and displacement vector arrows in the positive x-

direction. Matrix analysis requires us to be very strict in our sign conventions.

Using the basic relationship that force is equal to stiffness times displacement, we can

determine the force at node 1 as:

( )1 net displacement at 1F k=

Thus:

( )1 1 2 1 2F k u u ku ku= − = − (4.2.1)

Similarly for node 2:


Dr. C. Caprani 11

( )2 2 1 1 2F k u u ku ku= − = − + (4.2.2)

We can write equations (4.2.1) and (4.2.2) in matrix form to get the element stiffness

matrix for a 1-DOF axial element:

1 1

2 2

F uk kF uk k

− = −

(4.2.3)

And using matrix notation, we write:

{ } [ ]{ }e e=F k u (4.2.4)

Here:

• { }eF is the element force vector;

• [ ]k is the element stiffness matrix;

• { }eu is the element displacement vector.

It should be clear that the element stiffness matrix is of crucial importance – it links

nodal forces to nodal displacements; it encapsulates how the element behaves under

load.

The derivation of the element stiffness matrix for different types of elements is

probably the most awkward part of the matrix stiffness method. However, this does

not pose as a major disadvantage since we only have a few types of elements to

derive, and once derived they are readily available for use in any problem.


Dr. C. Caprani 12

4.2.2 Assemblies of Elements

Real structures are made up of assemblies of elements, thus we must determine how

to connect the stiffness matrices of individual elements to form an overall (or global)

stiffness matrix for the structure.

Consider the following simple structure:

Note that the individual elements have different stiffnesses, 1k and 2k . Thus we can

write the force displacement relationships for both elements as:

1 1 1 1

2 1 1 2

F k k uF k k u

− = −

(4.2.5)

2 22 2

3 32 2

F uk kF uk k

− = −

(4.2.6)

We can expand these equations so that they encompass all the nodes in the structure:

1 1 1 1

2 1 1 2

3 3

00

0 0 0

F k k uF k k uF u

− = −

(4.2.7)

1 1

2 2 2 2

3 2 2 3

0 0 000

F uF k k uF k k u

= − −

(4.2.8)


Dr. C. Caprani 13

We can add equations (4.2.7) and (4.2.8) to determine the total of both the forces and

displacements at each node in the structure:

1 1 1 1

2 1 1 2 2 2

3 2 2 3

0

0

F k k uF k k k k uF k k u

− = − + − −

(4.2.9)

As can be seen from this equation, by adding, we have the total stiffness at each node,

with contributions as appropriate by each member. In particular node 2, where the

members meet, has total stiffness 1 2k k+ . We can re-write this equation as:

{ } [ ]{ }=F K u (4.2.10)

In which:

• { }F is the force vector for the structure;

• [ ]K is the global stiffness matrix for the structure;

• { }u is the displacement vector for the structure.


Dr. C. Caprani 14

4.2.3 Example 1

Problem

The following axially-loaded structure has loads applied as shown:

The individual member properties are:

Member Length (m) Area (mm2) Material, E (kN/mm2)

1 0.28 400 70

2 0.1 200 100

3 0.1 70 200

Find the displacements of the connections and the forces in each member.


Dr. C. Caprani 15

Solution

Our first step is to model the structure with elements and nodes, as shown:

Calculate the spring stiffnesses for each member:

31

1

70 400 100 10 kN/m0.28

EAkL

⋅ = = = ×

(4.2.11)

32

2

100 200 200 10 kN/m0.1

EAkL

⋅ = = = ×

(4.2.12)

33

3

200 70 140 10 kN/m0.1

EAkL

⋅ = = = ×

(4.2.13)

Next we calculate the individual element stiffness matrices:

1 13

2 2

100 10010

100 100F uF u

− = −

(4.2.14)

2 23

3 3

200 20010

200 200F uF u

− = −

(4.2.15)


Dr. C. Caprani 16

3 33

4 4

140 14010

140 140F uF u

− = −

(4.2.16)

We expand and add the element stiffness matrices to get:

( )( )

1 1

2 23

3 3

4 4

100 100 0 0100 100 200 200 0

100 200 200 140 1400 0 140 140

F uF uF uF u

− − + − = − + − −

(4.2.17)

Notice how each member contributes to the global stiffness matrix:

Node 1 Node 2 Node 3 Node 4

Node 1 0 0

Node 2 0

Node 3 0

Node 4 0 0

Notice also that where the member stiffness matrices overlap in the global stiffness

matrix that the components (or entries) are added. Also notice that zeros are entered

where there is no connection between nodes, e.g. node 1 to node 3.


Dr. C. Caprani 17

We cannot yet solve equation (4.2.17) as we have not introduced the restraints of the

structure: the supports at nodes 1 and 4. We must modify equation (4.2.17) in such a

way that we will obtain the known results for the displacements at nodes 1 and 4.

Thus:

( )( )

1

22 3

33

4

0 1 0 0 00 100 200 200 0

100 200 200 140 0

0 0 0 0 1

uuFuFu

+ − = − +

(4.2.18)

What we have done here is to ‘restrict’ the matrix: we have introduced a 1 on the

diagonal of the node number, and set all other entries on the corresponding row and

column to zero. We have entered the known displacement as the corresponding entry

in force vector (zero). Thus when we now solve we will obtain 1 4 0u u= = .

For the remaining two equations, we have:

2 23

3 3

300 20010

200 340F uF u

− = −

(4.2.19)

And so:

( )( ) ( )( )

2 33

3

340 200 50 31 1 1 10 m200 300 100 2010 300 340 200 200 62

0.048 mm

0.322

uu

−− = ⋅ = × − − −

=

(4.2.20)

To find the forces in the bars, we can now use the member stiffness matrices, since

we know the end displacements:


Dr. C. Caprani 18

Member 1

1 3 3

2

100 100 0 4.810 10

100 100 0.048 4.8FF

−− − = × = −

(4.2.21)

Thus Member 1 has a tension of 4.8 kN, since the directions of the member forces are

interpreted by our sign convention:

Also note that it is in equilibrium (as we might expect).

Member 2

2 3 3

3

200 200 0.048 54.810 10

200 200 0.322 54.8FF

−− − = × = −

(4.2.22)

Member 2 thus has tension of 54.8 kN.

Member 3

3 3 3

4

140 140 0.322 45.0810 10

140 140 0 45.08FF

−− = × = − −

(4.2.23)

Thus Member 3 has a compression of 45.08 kN applied to it.


Dr. C. Caprani 19

Problem

Find the displacements of the connections and the forces in each member for the

following structure:

Ans. 0.22 mm, 2.11 mm


Dr. C. Caprani 20

4.2.4 General Methodology

Steps

The general steps in Matrix Stiffness Method are:

1. Calculate the member stiffness matrices

2. Assemble the global stiffness matrix

3. Restrict the global stiffness matrix and force vector

4. Solve for the unknown displacements

5. Determine member forces from the known displacements and member stiffness

matrices

6. Determine the reactions knowing member end forces.


Dr. C. Caprani 21

Matlab Program - Implementation

These steps are implemented in the Matlab Program as follows:

function [D F R] = AnalyzeTruss(nData,eData) % This function analyzes the truss defined by nData and eData: % nData = [x, y, xLoad, yLoad, xRestraint, yRestraint] % eData = [iNode, jNode, E, A]; kg = AssembleTrussK(nData, eData); % Assemble global stiffness matrix fv = AssembleForceVector(nData); % And the force vector [kgr fv] = Restrict(kg, fv, nData); % Impose restraints D = fv/kgr; % Solve for displacements F = ElementForces(nData,eData,D); % Get the element forces R = D*kg; % Get the reactions

The output from the function AnalyzeTruss is:

• D: vector of nodal deflections;

• F: vector of element forces;

• R: vector of nodal forces (indicating the reactions and applied loads).

The input data required (nData and eData) will be explained later.


Dr. C. Caprani 22

4.2.5 Member contribution to global stiffness matrix

Consider a member, ij, which links node i to node j. Its member stiffness matrix will

be:

Node i Node j

Node i k11ij k12ij

Node j k21ij k22ij

Its entries must then contribute to the corresponding entries in the global stiffness

matrix:

… Node i … Node j …

… … … … … …

Node i … k11ij … k12ij …

… … … … … …

Node j … k21ij … k22ij …

… … … … … …

If we now consider another member, jl, which links node j to node l. Its member

stiffness matrix will be:


Dr. C. Caprani 23

Node j Node l

Node j k11jl k12jl

Node l k21jl k22jl

And now the global stiffness matrix becomes:

… Node i … Node j … Node l …

… … … … … … … …

Node i … k11ij … k12ij … … …

… … … … … … … …

Node j … k21ij … k22ij + k11jl

… k12jl …

… … … … … … … …

Node l … … k21lj … k22jl …

… … … … … … … …

In the above, the identifiers k11 etc are sub-matrices of dimension:

ndof × ndof

where ndof refers to the number of degrees of freedom that each node has.


Dr. C. Caprani 24

Matlab Program – Element Contribution

Considering trusses, we have 2 degrees of freedom (DOFs) per node, the x direction

and the y direction. Thus, for a truss with nn number of nodes, there are 2nn DOFs in

total. The x-DOF for any node i is thus located at 2i-1 and the y-DOF at 2i.

Consider a truss member connecting nodes i and j. To add the 4×4 truss element

stiffness matrix into the truss global stiffness matrix, we see that each row adds into

the following matrix columns:

2i-1 2i 2j-1 2j

The rows in the global stiffness matrix corresponding to the rows of the element

stiffness matrix are:

1. Row 1: Adds to row 2i-1 of the global stiffness matrix;

2. Row 2: Adds to row 2i;

3. Row 3: adds to row 2j-1;

4. Row 4: adds to row 2j.

Note of course that the column and row entries occur in the same order.

These rules are implemented for our Truss Analysis Program as follows:

function kg = AddElement(iEle,eData,ke,kg) % This function adds member iEle stiffness matrix ke to the global % stiffness matrix kg. % What nodes does the element connect to? iNode = eData(iEle,1); jNode = eData(iEle,2); % The DOFs in kg to enter the properties into DOFs = [2*iNode-1 2*iNode 2*jNode-1 2*jNode]; % For each row of ke for i = 1:4 % Add the row to the correct entries in kg kg(DOFs(i),DOFs) = kg(DOFs(i),DOFs) + ke(i,:); end


Dr. C. Caprani 25

Matlab Program – Global Stiffness Matrix Assembly

The function that assembles the truss global stiffness matrix for the truss is as

follows:

function kg = AssembleTrussK(nData, eData) % This function assembles the global stiffness matrix for a truss from the % joint and member data matrices % How many nodes and elements are there? [ne ~] = size(eData); [nn ~] = size(nData); % Set up a blank global stiffness matrix kg = zeros(2*nn,2*nn); % For each element for i = 1:ne E = eData(i,3); % Get its E and A A = eData(i,4); [L c s] = TrussElementGeom(i,nData,eData); % Geometric Properties ke = TrussElementK(E,A,L,c,s); % Stiffness matrix kg = AddElement(i,eData,ke,kg); % Enter it into kg end

Note that we have not yet covered the calculation of the truss element stiffness

matrix. However, the point here is to see that each element stiffness matrix is

calculated and then added to the global stiffness matrix.


Dr. C. Caprani 26

Matlab Program – Force Vector

Examine again the overall equation (4.2.10) to be solved:

{ } [ ]{ }=F K u

We now have the global stiffness matrix, we aim to calculate the deflections thus we

need to have a force vector representing the applied nodal loads. Again remember

that each node as two DOFs (x- and y-loads). The code for the force vector is thus:

function f = AssembleForceVector(nData) % This function assembles the force vector % How may nodes are there? [nn ~] = size(nData); % Set up a blank force vector f = zeros(1,2*nn); % For each node for i = 1:nn f(2*i - 1) = nData(i, 3); % x-load into x-DOF f(2*i) = nData(i, 4); % y-load into y-DOF end


Dr. C. Caprani 27

4.2.6 Interpretation of Stiffness Matrix

It is useful to understand what each term in a stiffness matrix represents. If we

consider a simple example structure:

We saw that the global stiffness matrix for this is:

11 12 13 1 1

21 22 23 1 1 2 2

31 32 33 2 2

0

0

K K K k kK K K k k k kK K K k k

− = = − + − −

K

If we imagine that all nodes are fixed against displacement except for node 2, then we

have the following:


Dr. C. Caprani 28

From our general equation:

1 11 12 13 12

2 21 22 23 22

3 31 32 33 32

010

F K K K KF K K K KF K K K K

= =

(4.2.24)

Thus:

1 12 1

2 22 1 2

3 32 2

F K kF K k kF K k

− = = + −

(4.2.25)

These forces are illustrated in the above diagram, along with a free-body diagram of

node 2.

Thus we see that each column in a stiffness matrix represents the forces required to

maintain equilibrium when the column’s DOF has been given a unit displacement.

This provides a very useful way to derive member stiffness matrices.


Dr. C. Caprani 29

4.2.7 Restricting a Matrix – Imposing Restraints

In Example 1 we solved the structure by applying the known supports into the global

stiffness matrix. We did this because otherwise the system is unsolvable; technically

the determinant of the stiffness matrix is zero. This mathematically represents the fact

that until we apply boundary conditions, the structure is floating in space.

To impose known displacements (i.e. supports) on the structure equations we modify

the global stiffness matrix and the force vector so that we get back the zero

displacement result we know.

Considering our two-element example again, if node 1 is supported, 1 0u = . Consider

the system equation:

1 11 12 13 1

2 21 22 23 2

3 31 32 33 3

F K K K uF K K K uF K K K u

=

(4.2.26)

Therefore to obtain 1 0u = from this, we change K and F as follows:

1

2 22 23 2

3 32 33 3

0 1 0 000

uF K K uF K K u

=

(4.2.27)

Now when we solve for 1u we will get the answer we want: 1 0u = . In fact, since we

now do not need this first equation, we could just consider the remaining equations:

2 22 23 2

3 32 33 3

F K K uF K K u

=

(4.2.28)


Dr. C. Caprani 30

And these are perfectly solvable.

Thus to summarize:

To impose a support condition at degree of freedom i:

1. Make the force vector element of DOF i zero;

2. Make the i column and row entries of the stiffness matrix all zero;

3. Make the diagonal entry ( ),i i of the stiffness matrix 1.


Dr. C. Caprani 31

Matlab Program – Imposing Restraints

To implement these rules for our Truss Analysis Program, we will first create of

vector which tells us whether or not a DOF is restrained. This vector will have a zero

if the DOF is not restrained, and a 1 if it is.

Once we have this vector of restraints, we can go through each DOF and modify the

force vector and global stiffness matrix as described before. The implementation of

this is as follows:

function [kg f] = Restrict(kg, f, nData) % This function imposes the restraints on the global stiffness matrix and % the force vector % How may nodes are there? [nn ~] = size(nData); % Store each restrained DOF in a vector RestrainedDOFs = zeros(2*nn,1); % For each node, store if there is a restraint for i = 1:nn % x-direction if nData(i,5) ~= 0 % if there is a non-zero entry (i.e. supported) RestrainedDOFs(2*i-1) = 1; end % y-direction if nData(i,6) ~= 0 % if there is a support RestrainedDOFs(2*i) = 1; end end % for each DOF for i = 1:2*nn if RestrainedDOFs(i) == 1 % if it is restrained f(i) = 0; % Ensure force zero at this DOF kg(i,:) = 0; % make entire row zero kg(:,i) = 0; % make entire column zero kg(i,i) = 1; % put 1 on the diagonal end end


Dr. C. Caprani 32

4.3 Plane Trusses

4.3.1 Introduction

Trusses are assemblies of members whose actions can be linked directly to that of the

simple spring studied already:

EAkL

= (4.3.1)

There is one main difference, however: truss members may be oriented at any angle

in the xy coordinate system (Cartesian) plane:

Thus we must account for the coordinate transformations from the local member axis

system to the global axis system.


Dr. C. Caprani 33

Matlab Program – Data Preparation

In the following sections we will put the final pieces of code together for our Truss

Analysis Program. At this point we must identify what information is required as

input to the program, and in what format it will be delivered.

The node data is stored in a matrix nData. Each node of the truss is represented by a

row of data. In the row, we put the following information in consecutive order in

columns:

1. x-coordinate;

2. y-coordinate;

3. x-load: 0 or the value of load;

4. y-load: 0 or the value of load;

5. x-restraint: 0 if unrestrained, any other number if restrained;

6. y-restraint: 0 if unrestrained, any other number if restrained.

The element data is stored in a matrix called eData. Each element has a row of data

and for each element the information stored in the columns in order is:

1. i-Node number: the node number at the start of the element;

2. j-Node number: the other node the element connects to;

3. E: the Modulus of Elasticity of the element material;

4. A: the element area;

We will prepare input data matrices in the above formats for some of the examples

that follow so that the concepts are clear. In doing so we keep the units consistent:

• Dimensions are in m;

• Forces in kN

• Elastic modulus is in kN/mm2;

• Area is mm2.


Dr. C. Caprani 34

Matlab Program – Data Entry

To enter the required data, one way is:

1. Create a new variable in the workspace (click on New Variable);

2. Name it eData for example;

3. Double click on the new variable to open the Matlab Variable Editor;

4. Enter the necessary input data (can paste in from MS Excel, or type in);

5. Repeat for the nodal data.


Dr. C. Caprani 35

4.3.2 Truss Element Stiffness Matrix

For many element types it is very difficult to express the element stiffness matrix in

global coordinates. However, this is not so for truss elements. Firstly we note that the

local axis system element stiffness matrix is given by equation (4.2.3):

[ ] 1 11 1

k kk

k k− −

= = − − k (4.3.2)

Next, introducing equation (4.3.1), we have:

[ ] 1 11 1

EAL

− = −

k (4.3.3)

However, this equation was written for a 1-dimensional element. Expanding this to a

two-dimensional axis system is straightforward since there are no y-axis values:

[ ]

1 0 1 00 0 0 01 0 1 0

0 0 0 0

i

i

j

j

xyEAxLy

←− ← =

← − ←

k (4.3.4)

Next, using the general element stiffness transformation equation (See the Appendix):

[ ] [ ] [ ][ ]Tk = T k T (4.3.5)

And noting the transformation matrix for a plane truss element from the Appendix:


Dr. C. Caprani 36

cos sin 0 0sin cos 0 00 0 cos sin0 0 sin cos

P

P

α αα α

α αα α

− = −

T 0T =

0 T (4.3.6)

We have:

[ ]

1cos sin 0 0 1 0 1 0sin cos 0 0 0 0 0 00 0 cos sin 1 0 1 00 0 sin cos 0 0 0 0


EAL

α αα α

α αα α

α αα α

α αα α

−−

− = ⋅ − −

− −

k

(4.3.7)

Carrying out the multiplication gives:

2 2

2 2

2 2

2 2

cos cos sin cos cos sincos sin sin cos sin sin


EAL

α α α α α αα α α α α α

α α α α α αα α α α α α

− − − − = − − − −

k (4.3.8)

If we examine the nodal sub-matrices and write cosc α≡ , sins α≡ :

[ ]

2 2

2 2

2 2

2 2

c cs c cscs s cs sEA

L c cs c cscs s cs s

− − − − = − − − −

k (4.3.9)


Dr. C. Caprani 37

Labelling the nodal sub-matrices as:

[ ] =

k11 k12k

k21 k22 (4.3.10)

Then we see that the sub-matrices are of dimension 2 × 2 (No. DOF × No. DOF) and

are:

2

2

c csEAL cs s

=

k11 (4.3.11)

And also note:

k11 = k22 = -k12 = -k21 (4.3.12)

Therefore, we need only evaluate a single nodal sub-matrix (k11) in order to find the

total element stiffness matrix in global coordinates.


Dr. C. Caprani 38

Matlab Program – Element Stiffness Matrix

Calculating the element stiffness matrix for our Truss Analysis Program is easy. The

only complexity is extracting the relevant data from the input node and element data

matrices. Rather than try determine the angle that the truss member is at (remember

we only have the nodal coordinates), we can calculate cosα and sinα directly (e.g.

adjacent/hypotenuse). Further, the element length can be found using Pythagoras,

given the nodal coordinates. These element properties are found in the script below:

function [L c s] = TrussElementGeom(iEle,nData,eData); % This function returns the element length % What nodes does the element connect to? iNode = eData(iEle,1); jNode = eData(iEle,2); % What are the coordinates of these nodes? iNodeX = nData(iNode,1); iNodeY = nData(iNode,2); jNodeX = nData(jNode,1); jNodeY = nData(jNode,2); % Use Pythagoras to work out the member length L = sqrt((jNodeX - iNodeX)^ 2 + (jNodeY - iNodeY)^ 2); % Cos is adjacent over hyp, sin is opp over hyp c = (jNodeX - iNodeX)/L; s = (jNodeY - iNodeY)/L;

The E and A values for each element are directly found from the input data element

matrix as follows:

E = eData(i,3); % Get its E and A A = eData(i,4);

Thus, with all the relevant data assembled, we can calculate the truss element

stiffness matrix. In the following Matlab function, note that we make use of the fact

that each nodal sub-matrix can be determined from the nodal sub-matrix k11 :


Dr. C. Caprani 39

function k = TrussElementK(E,A,L,c,s) % This function returns the stiffness matrix for a truss element k11 = [ c^2 c*s; c*s s^2]; k = (E*A/L) * [ k11 -k11; -k11 k11];


Dr. C. Caprani 40

4.3.3 Element Forces

The forces applied to a member’s ends are got from the element equation:

{ } [ ]{ }e e=F k u (4.3.13)

Expanding this in terms of nodal equations we have:

i i

j j

=

F δk11 k12F δk21 k22

(4.3.14)

Thus we know:

j i j= ⋅ + ⋅F k21 δ k22 δ (4.3.15)

From which we could determine the member’s axial force. However, for truss

members, we can determine a simple expression to use if we consider the change in

length in terms of the member end displacements:

x jx ixL δ δ∆ = − (4.3.16)

y jy iyL δ δ∆ = − (4.3.17)

And using the coordinate transforms idea:

cos sinx yL L Lα α∆ = ∆ + ∆ (4.3.18)

Also we know that the member force is related to the member elongation by:


Dr. C. Caprani 41

EAF LL

= ⋅∆ (4.3.19)

Thus we have:

cos sinx y

EAF L LL

α α = ⋅ ∆ + ∆ (4.3.20)

And introducing equations (4.3.16) and (4.3.17) gives:

[ ]cos sin jx ix

jy iy

EAFL

δ δα α

δ δ−

= ⋅ − (4.3.21)

A positive result from this means tension and negative compression.


Dr. C. Caprani 42

Matlab Program – Element Force

Once the element nodal deflections are known, the element forces are found as

described above. Most of the programming effort is dedicated to extracting the nodal

deflections that are relevant for the particular member under consideration:

function F = TrussElementForce(nData, eData, d, iEle) % This function returns the element force for iEle given the global % displacement vector, d, and the node and element data matrices. % What nodes does the element connect to? iNode = eData(iEle,1); jNode = eData(iEle,2); % Get the element properties E = eData(iEle,3); % Get its E and A A = eData(iEle,4); [L c s] = TrussElementGeom(iEle,nData,eData); % Geometric Properties dix = d(2*iNode-1); % x-displacement at node i diy = d(2*iNode); % y-displacement at node i djx = d(2*jNode-1); % x-displacement at node j djy = d(2*jNode); % y-displacement at node j F = (E*A/L) * (c*(djx-dix) + s*(djy-diy));

Note also that the way the program is written assumes that tension is positive and

compression is negative. We also want to return all of the element forces, so we use

the function just described to calculate all the truss elements’ forces:

function F = ElementForces(nData,eData,d) % This function returns a vector of the element forces % How many elements are there? [ne ~] = size(eData); % Set up a blank element force vector F = zeros(ne,1); % For each element for i = 1:ne % Get its force and enter into vector F(i) = TrussElementForce(nData, eData, d, i); end


Dr. C. Caprani 43

4.3.4 Example 2: Basic Truss

Problem

Analyse the following truss using the stiffness matrix method.

Note that:

• 2200 kN/mmE = ;

• The reference area is 2100mmA = .


Dr. C. Caprani 44

Solution

STEP 1: Determine the member stiffness matrices:

Member 12

The angle this member makes to the global axis system and the relevant values are:

21 1cos cos4522

c cα≡ = = ⇒ =

21 1 1sin sin 452 22

s s csα≡ = = ⇒ = ⇒ =

Therefore:

2

12 212

0.5 0.5200 100 20.5 0.510 2

c csEAL cs s

⋅ = = k11

Thus:

312

0.5 0.510

0.5 0.5

=

k11 (4.3.22)

Notice that the matrix is symmetrical as it should be.


Dr. C. Caprani 45

Member 23

The angle this member makes to the global axis system and the relevant values are:

21 1cos cos31522

c cα≡ = = ⇒ =

21 1 1sin sin3152 22

s s csα≡ = = − ⇒ = ⇒ = −

Therefore:

2

23 223

0.5 0.5200 100 20.5 0.510 2

c csEAL cs s

− ⋅ = = − k11

Thus:

323

0.5 0.510

0.5 0.5−

= − k11 (4.3.23)

Again the matrix is symmetrical.


Dr. C. Caprani 46

STEP 2: Assemble the global stiffness matrix

For 3 nodes, the unrestricted global stiffness matrix will look as follows:

11 12 13

21 22 23

31 32 33

Node 1 Node 2 Node 3

← = ← ←

K K KK K K K

K K K (4.3.24)

Note that each of the sub-matrices is a 2×2 matrix, e.g.:

11 12

21 22

Node 1 Node 1

xx xy

yx yy

k k xk k y

← = ←

11K (4.3.25)

The member stiffness nodal sub-matrices contribute to the global stiffness nodal sub-

matrices as follows:

11 12 13 12 12

21 22 23 12 12 23 23

31 32 33 23 23

= =

K K K k11 k12 0K K K K k21 k22 + k11 k12

K K K 0 k21 k22 (4.3.26)

Expanding this out and filling in the relevant entries from equations (4.3.22) and

(4.3.23) whilst using equation (4.3.12) gives:


Dr. C. Caprani 47

3

0.5 0.5 0.5 0.5 0 00.5 0.5 0.5 0.5 0 00.5 0.5 1 0 0.5 0.5

100.5 0.5 0 1 0.5 0.50 0 0.5 0.5 0.5 0.50 0 0.5 0.5 0.5 0.5

− − − − − − −

= − − − − −

− −

K (4.3.27)

STEP 3: Write the solution equation in full

{ } [ ]{ }F = K δ (4.3.28)

Thus, keeping the nodal sub-matrices identifiable for clarity:

1 1

1 1

23

2

3 3

3 3

0.5 0.5 0.5 0.5 0 00.5 0.5 0.5 0.5 0 0

0 0.5 0.5 1 0 0.5 0.510

100 0.5 0.5 0 1 0.5 0.50 0 0.5 0.5 0.5 0.50 0 0.5 0.5 0.5 0.5

x x

y y

x

y

x x

y y

RR

RR

δδδδδδ

− − − − − − − = − − − − − −

− −

(4.3.29)

In which we have noted:

• 1xR is the reaction at node 1 in the x-direction (and similarly for the others);

• The force at node 2 is 0 in the x-direction and -100 kN (downwards) in the y-

direction.


Dr. C. Caprani 48

STEP 4: Restrict the equation.

Now we impose the boundary conditions on the problem. We know:

• 1 1 0x yδ δ= = since node 1 is pinned;

• 3 3 0x yδ δ= = again, since node 3 is pinned.

Thus equation (4.3.29) becomes:

1

1

23

2

3

3

0 1 0 0 0 0 00 0 1 0 0 0 00 0 0 1 0 0 0

10100 0 0 0 1 0 00 0 0 0 0 1 00 0 0 0 0 0 1

x

y

x

y

x

y

δδδδδδ

= −

(4.3.30)

Since both DOFs are restricted for nodes 1 and 3, we can thus write the remaining

equations for node 2:

23

2

0 1 010

100 0 1x

y

δδ

= − (4.3.31)

STEP 5: Solve the system

The y-direction is thus the only active equation:

32100 10 yδ− = (4.3.32)

Thus:

2 0.1 m 100 mmyδ = − = ↓ (4.3.33)


Dr. C. Caprani 49

STEP 6: Determine the member forces

For truss member’s we outlined a simple method encompassed in equation (4.3.21).

In applying this to Member 12 we note:

• 1 1 0x yδ δ= = since it is a support;

• 2 0xδ = by solution;

• 2 0.1yδ = − again by solution.

Thus:

[ ]cos sin jx ix

jy iy

EAFL

δ δα α

δ δ−

= ⋅ −

3 0 01 1 10010 50 2 kN0.1 02 2 2

F− = = − = − − −

(4.3.34)

And so Member 12 is in compression, as may be expected. For Member 23 we

similarly have:

( )3

0 01 1 10010 50 2 kN0 0.12 2 2

F− = − = − = − − −

(4.3.35)

And again Member 23 is in compression. Further, since the structure is symmetrical

and is symmetrically loaded, it makes sense that Member’s 12 and 23 have the same

force.

STEP 7: Determine the reactions

To determine the remaining unknown forces we can use the basic equation now that

all displacements are known:


Dr. C. Caprani 50

1

1

3

3

3

0.5 0.5 0.5 0.5 0 0 00.5 0.5 0.5 0.5 0 0 0

0 0.5 0.5 1 0 0.5 0.5 010

100 0.5 0.5 0 1 0.5 0.5 0.10 0 0.5 0.5 0.5 0.5 00 0 0.5 0.5 0.5 0.5 0

x

y

x

y

RR

RR

− − − − − − − = − − − − − − −

− −

(4.3.36)

Thus we have:

[ ]1

00.5 0.5 50kN

0.1xR = − − = + −

(4.3.37)

[ ]1

00.5 0.5 50kN

0.1yR = − − = + −

(4.3.38)

[ ]3

00.5 0.5 50kN

0.1xR = − = − −

(4.3.39)

[ ]3

00.5 0.5 50kN

0.1yR = − = + −

(4.3.40)

Again note that the sign indicates the direction along the global coordinate system.

We can now plot the full solution:


Dr. C. Caprani 51

Matlab Program – First Use

All necessary functions have been explained. The main function is given on page 21.

This also gives the single line of code that finds the reactions. The input data for the

example truss just given is:

Node Data Element Data

x y Fx Fy Rx Ry Node i Node j E A

0 0 0 0 1 1 1 2 200 70.71

10 10 0 -100 0 0 2 3 200 70.71

20 0 0 0 1 1

And the results from the program are:

Node DOF D R Element F 1 x 0 50 1 -70.71 y 0 50 2 -70.71 2 x 0 0

y -0.100 -100 3 x 0 -50 y 0 50

These results, of course, correspond to those found by hand.

The importance of the graphical display of the results should also be noted: there

could have been clear mistakes made in the preparation of the input data that would

not reveal themselves unless the physical interpretation of the results is appreciate by

drawing the deflected shape, the member forces, and the directions of the reactions.


Dr. C. Caprani 52

4.3.5 Example 3: Adding Members

Problem

Analyse the truss of Example 2 but with the following member 14 added:

Solution

With the addition of node 4 we now know that the nodal sub-matrices global stiffness

equation will be 4×4 with the fully expanded matrix being 16×16. Rather than

determine every entry in this, let’s restrict it now and only determine the values we

will actually use. Since nodes 1, 3 and 4 are pinned, all their DOFs are fully restricted

out. The restricted equation thus becomes:

{ } [ ]{ }2 2=F K22 δ (4.3.41)

Next we must identify the contributions from each member:

• We already know the contributions of Members 12 and 23 from Example 2.

• The contribution of Member 24 is to nodes 2 and 4. Since node 4 is restricted, we

only have the contribution 24k11 to K22 .

Thus K22 becomes:


Dr. C. Caprani 53

12 23 24= + +K22 k22 k11 k11 (4.3.42)

Next determine 24k11 : this member makes an angle of 270° to the global axis system

giving:

2cos cos270 0 0c cα≡ = = ⇒ =

2sin sin 270 1 1 0s s csα≡ = = − ⇒ = ⇒ =

Therefore:

2

324 2

24

0 0 0 0200 100 2 100 1 0 110

c csEAL cs s

⋅ = = = × k11

Thus:

324

0 010

0 2

=

k11 (4.3.43)

Hence the global restricted stiffness matrix becomes:

3 3 31 0 0 0 1 010 10 10

0 1 0 2 0 3

= + =

K22 (4.3.44)

Writing the restricted equation, we have:

23

2

0 1 010

100 0 3x

y

δδ

= − (4.3.45)


Dr. C. Caprani 54

From which we find the only equation

( )32100 10 3 yδ− = (4.3.46)

Thus:

2 0.033 m 33.3 mmyδ = − = ↓ (4.3.47)

The member forces are:

312

0 01 110 23.6kN0.033 02 2

F− = = − − −

(4.3.48)

( )3

23

0 01 110 23.6kN0 0.0332 2

F− = − = − − −

(4.3.49)

[ ]324

0 010 0 2 66.6kN

0.033 0F

− = = − − −

(4.3.50)

Thus we have the following solution:


Dr. C. Caprani 55

Matlab Program – Input/Output

The input data for this example is:

Node Data Element Data

x y Fx Fy Rx Ry Node i Node j E A

0 0 0 0 1 1 1 2 200 70.71

10 10 0 -100 0 0 2 3 200 70.71

20 0 0 0 1 1 2 4 200 100

10 0 0 0 1 1

The results are:

Node DOF D R Element F 1 x 0 16.66 1 -23.57 y 0 16.66 2 -23.57 2 x 0 0 3 66.66

y -0.033 -100 3 x 0 -16.66 y 0 16.66 4 x 0 0 y 0 66.66


Dr. C. Caprani 56

4.3.6 Example 4: Using Symmetry

Problem

Analyse the truss of Example 3 taking advantage of any symmetry:

Solution

Looking at the structure it is clear that by splitting the structure down the middle

along member 24 that we will have two equal halves:

Notice that we have changed the following:


Dr. C. Caprani 57

• The load is halved since it is now equally shared amongst two halves;

• Similarly the area of member 24 is halved.

We now analyse this new truss as usual. However, we can make use of some previous

results. For Member 12:

312

0.5 0.510

0.5 0.5

=

k11 (4.3.51)

And for Member 24

3 324

0 0 0 01 10 100 2 0 12

= =

k11 (4.3.52)

Since the area is halved from that of Example 3, its stiffnesses are halved.

In restricting we note that the only possible displacement is node 2 in the y-direction.

However, we will keep using the node 2 sub-matrices until the last moment:

3 3 30.5 0 0 0 0.5 010 10 10

0 0.5 0 1 0 1.5

= + =

K22 (4.3.53)

Thus:

23

2

0 0.5 010

50 0 1.5x

y

δδ

= − (4.3.54)

And now imposing the boundary condition 2 0xδ = :


Dr. C. Caprani 58

{ } [ ]{ }3250 10 1.5 yδ− = (4.3.55)

From which we solve for the displacement:

{ } [ ]{ }3

2

2

50 10 1.5

0.033 m 33.3 mmy

y

δ

δ

− =

= − = ↓ (4.3.56)

This (of course) is the same result we obtained in Example 3. For the member forces

we have:

312

0 01 110 23.6kN0.033 02 2

F− = = − − −

(4.3.57)

[ ] ( )3

24

0 010 0 1 33.3kN

0 0.033F

− = − = − − −

(4.3.58)

Member 12 has the same force as per Example 3 as is expected.

It might appear that Member 24 has an erroneous force result. It must be remembered

that this is the force in the half-member (brought about since we are using symmetry).

Therefore the force in the full member is 2 33.3 66.6 kN× = as per Example 3.


Dr. C. Caprani 59

4.3.7 Self-Strained Structures

Introduction

A self-strained structure is one where strains are induced by sources other than

externally applied loads. The two main examples are temperature difference and lack

of fit of a member. For example consider the effect if member 13 in the following

structure was too long and had to be ‘squeezed’ into place:

It should be intuitively obvious that to ‘squeeze’ the member into place a

compressive force was required to shorten it to the required length:


Dr. C. Caprani 60

Once the member has been put in place, the source of the ‘squeezing’ is removed.

Since the member wants to spring back to its original length, it pushes on its joints:

In this way members 12 and 14 will now go into tension whilst member 13 will

remain in compression, but a smaller compression than when it was ‘squeezed’ into

place since joint 1 will deflect to the right some amount.

In a similar way to lack of fit, examined above, if member 13 had been subject to a

temperature increase it would try to elongate. However this elongation is restrained

by the other members inducing them into tension and member 13 into some

compression.


Dr. C. Caprani 61

Lack of Fit

We consider a member with original length of OL that is required to be of length

Req'dL . Thus a change in length of L∆ must be applied:

Req'd OL L L= + ∆ (4.3.59)

Thus:

• L∆ is positive: the member is too short and must be lengthened to get into place;

• L∆ is negative, it is too long and must be shortened to get into place.

Thus we must apply a force to the member that will cause a change in length of L∆ .

From basic mechanics:

OFLLEA

∆ = (4.3.60)

Thus the force required is:

O

LF EAL∆

= ⋅ (4.3.61)

From the above sign convention for L∆ :

• F is positive when the member must be put into tension to get it in place;

• F is negative when the member must be put into compression to get it in place.

Lastly, remember to apply the member force in opposite direction to the member’s

nodes.


Dr. C. Caprani 62

Temperature Change

We consider a member that is subject to a differential (i.e. different to the rest of the

structure) temperature change of T∆ degrees Celsius. Also we must know the

coefficient of linear thermal expansion, α , for the material. This is the change in

length, per unit length, per unit change in temperature:

O

L CL

α ∆≡

(4.3.62)

Thus the thermal strain induced in the member is:

T Tε α= ∆ (4.3.63)

And so the change in length is:

OL L Tα∆ = ∆ (4.3.64)

Also, since Eσ ε= , we find the force in the member:

T T TF A EAσ ε= = (4.3.65)

So finally, from equation (4.3.63), the force required to suppress the temperature

change is:

TF EA Tα= ∆ (4.3.66)

Once again, apply this force in the opposite direction to the member’s nodes.


Dr. C. Caprani 63

4.3.8 Example 5 – Truss with Differential Temperature

Problem

Member 13 of the following truss is subject to a temperature change of +100 °C.

Calculate the deflections of node 1 and the final forces in the members.

Take: 5 12 10 Cα − −= × ; 42 10 kNEA = × ; the area of member 12 as 2A; the area of

member 13 as A; and, the area of member 14 as A√2.


Dr. C. Caprani 64

Solution

First we must recognize that there are two stages to the actions in the members:

• Stage I: all displacements are suppressed and only the temperature force in

member 13 is allowed for;

• Stage II: displacements are allowed and the actions of the temperature force in

member 13 upon the rest of the structure are analyzed for.

The final result is then the summation of these two stages:

The force induced in member 13 when displacements are suppressed is:

( )( )( )4 52 10 2 10 100

40kN

TF EA Tα−

= ∆

= × × +

=

(4.3.67)


Dr. C. Caprani 65

Stage I

All displacements are suppressed. Thus:

1 10; 0x yδ δ= = (4.3.68)

12 13 140; ; 0I I ITF F F F= = = (4.3.69)

Stage II

Displacements are allowed occur and thus we must analyse the truss. Using the

matrix stiffness method, and recognizing that only joint 1 can displace, we have:

{ } [ ]{ }1 1=F K11 δ (4.3.70)

Also, since we cleverly chose the node numbers, the member contributions are just:

12 13 14= + +K11 k11 k11 k11 (4.3.71)

• Member 12:

21 1cos cos1202 4

c cα≡ = = − ⇒ =

23 3 3sin sin1202 4 4

s s csα≡ = = ⇒ = ⇒ = −

( )2 4

12 223

1 32 10 2 4 4

2.0 3 34 4

c csEAL cs s

− × = = −

k11


Dr. C. Caprani 66

• Member 13:

2cos cos180 1 1c cα≡ = = − ⇒ = 2sin sin180 0 0 0s s csα≡ = = ⇒ = ⇒ =

2 4

13 223

1 02 100 01.0

c csEAL cs s

× = = k11

• Member 14:

21 1cos cos22522

c cα≡ = = − ⇒ =

21 1 1sin sin 2252 22

s s csα −≡ = = ⇒ = ⇒ =

( )42

14 223

2 10 2 0.5 0.50.5 0.51.0 2

c csEAL cs s

× = = k11

Thus from equation (4.3.71) we have:

4 7 2 32 10

4 2 3 5

−×=

− K11 (4.3.72)

From the diagram for Stage II, we can see that the force applied to joint 1 is acting to

the right and so is positive. Thus:


Dr. C. Caprani 67

4

1

1

40 7 2 32 100 4 2 3 5

x

y

δδ

− ×=

− (4.3.73)

Solve this to get:

( )1

241

3

5 3 2 404 102 10 3 2 77 5 2 3

1.1510 m

0.06

x

y

δδ

−

− = × − ⋅ − −

= −

(4.3.74)

Using equation (4.3.21) we can now find the member forces for Stage II:

( ) ( )( )

43

12

0 1.152 10 2 1 3 10 12.54 kN0 0.062.0 2 2

IIF − − ×

= ⋅ − = + − − (4.3.75)

[ ] ( )( )

43

13

0 1.152 10 1 0 10 23.0 kN0 0.061.0

IIF − − ×= ⋅ − = + − −

(4.3.76)

( ) ( )

( )

4

314

2 10 2 0 1.151 1 10 15.4 kN0 0.061.0 2 2 2

IIF −× − = ⋅ − − = + − −

(4.3.77)


Dr. C. Caprani 68

Final

The final member forces are the superposition of Stage I and Stage II forces:

12 12 12 0 12.54 12.54 kNI IIF F F= + = + = (4.3.78)

13 13 13 40 23.0 17.0 kNI IIF F F= + = − + = − (4.3.79)

14 14 14 0 15.4 15.4 kNI IIF F F= + = + = (4.3.80)

Thus the final result is:


Dr. C. Caprani 69

4.3.9 Example 6 – Truss with Loads & Self Strains

Problem

Analyse the same truss as Example 5, allowing for the following additional load

sources:

• 80 kN acting horizontally to the left at node 1;

• 100 kN acting vertically downwards at node 1;

• Member 14 is 5√2 mm too short upon arrival on site.

All as shown below:


Dr. C. Caprani 70

Solution

Again we will separate the actions into Stage I and Stage II scenarios.

Stage I

Displacements are suppressed and as a result the only sources of forces are self-

straining forces:

The forces and displacements for Stage I are thus:

40kNTF = as before,

( )3

4 5 2 102 2 10 100 2 kN2L

LF EAL

−

∆

∆ ×= = × = (4.3.81)

1 10; 0x yδ δ= = (4.3.82)

12 13 140; 40; 100 2I I IF F F= = − = (4.3.83)


Dr. C. Caprani 71

Stage II

In this stage displacements are allowed and the forces in the self-strained members

are now applied to the joints, in addition to any external loads. Thus we have:

Clearly we need to resolve the forces at node 1 into net vertical and horizontal forces:

Since the members have not changed from Example 5, we can use the same stiffness

matrix. Therefore we have :

4

1

1

140 7 2 32 10200 4 2 3 5

x

y

δδ

− − ×= − −

(4.3.84)


Dr. C. Caprani 72

Solve this:

( )1

241

3

5 3 2 1404 12002 10 3 2 77 5 2 3

3.710 m

7.8

x

y

δδ

−

− − = −× − ⋅ − −

− = −

(4.3.85)

Using equation (4.3.21) we can find the member forces for Stage II:

( ) ( )( )

43

12

0 3.72 10 2 1 3 10 98.1 kN0 7.82.0 2 2

IIF − − − ×

= ⋅ − = + − − (4.3.86)

[ ] ( )( )

43

13

0 3.72 10 1 0 10 74.0 kN0 7.81.0

IIF − − − ×= ⋅ − = − − −

(4.3.87)

( ) ( )

( )

4

314

2 10 2 0 3.71 1 10 162.6 kN0 7.81.0 2 2 2

IIF −× − − = ⋅ − − = − − −

(4.3.88)


Dr. C. Caprani 73

Final

As before, the final member forces are the Stage I and Stage II forces:

12 12 12 0 98.1 98.1 kNI IIF F F= + = + = (4.3.89)

13 13 13 40 74.0 114.0 kNI IIF F F= + = − − = − (4.3.90)

14 14 14 100 2 162.6 21.6 kNI IIF F F= + = − = − (4.3.91)

Thus the final result is:


Dr. C. Caprani 74

4.3.10 Problems

Problem 1

Determine the displacements of joint 1 and the member forces for the following truss.

Take 42 10 kNEA = × .

Problem 2

Determine the displacements of joint 1 and the member forces for the following truss.

Take 42 10 kNEA = × , the area of both members is A√2.

Ans. 1 5 mmxδ = + , 1 0yδ =


Dr. C. Caprani 75

Problem 3

Using any pertinent results from Problem 2, determine the area of member 14 such

that the horizontal displacement of node 1 is half what is was prior to the installation

of member 14. Determine also the force in member 14. Take 42 10 kNEA = × ,

Ans. 14A A= , 14 50 kNF = −

Problem 4

Determine the displacements of the joints and the member forces for the following

truss. Take 42 10 kNEA = × , the area of all members is A√2, except for member 24

which has an area of A.


Dr. C. Caprani 76

Problem 5


truss. Take 42 10 kNEA = × , the area of the members is as shown.

Problem 6


truss. Take 42 10 kNEA = × , the area of all members is A, except for member 14

which has an area of A√2.


Dr. C. Caprani 77

4.4 Beams

4.4.1 Beam Element Stiffness Matrix

To derive the beam element stiffness matrix, we recall some results obtained

previously, summarized here:

Next we must adopt strict local element sign convention and node identification:


Dr. C. Caprani 78

Anti-clockwise moments and rotations (i.e. from the x-axis to the y-axis) are positive

and upwards forces are positive.

Thus for a vertical displacement of ∆ at node i, now labelled iyδ , we have the

following ‘force’ vector:

3

2

3

2

12

6

12

6

iy

iiy

jy

j

EIL

F EIM LF EI

LMEIL

δ

=

−

(4.4.1)

Similarly, applying the same deflection, but at node j, jyδ , gives:

3

2

3

2

12

6

12

6

iy

ijy

jy

j

EIL

F EIM LF EI

LMEIL

δ

− − =

−

(4.4.2)

Next, applying a rotation to node i, iθ , gives:


Dr. C. Caprani 79

2

2

6

4

6

2

iy

ii

jy

j

EIL

F EIM LF EI

LMEIL

θ

=

−

(4.4.3)

And a rotation to node j, jθ , gives:

2

2

6

2

6

4

iy

ij

jy

j

EIL

F EIM LF EI

LMEIL

θ

=

−

(4.4.4)

Since all of these displacement could happen together, using superposition we thus

have the total force vector as:

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

iy

iiy i jy

jy

j

EI EI EI EIL L L L

F EI EI EI EIM L L L LF EI EI EI EI

L L L LMEI EI EI EIL L L L

δ θ δ

− − = + + +

− − −

−

jθ

(4.4.5)

Writing this as a matrix equation, we have:


Dr. C. Caprani 80

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

iy iy

i i

jy jy

j j

EI EI EI EIL L L L

F EI EI EI EIM L L L LF EI EI EI EI

L L L LMEI EI EI EIL L L L

δθδθ

− − =

− − −

−

(4.4.6)

This is in the typical form:

{ } [ ]{ }e e=F k u (4.4.7)

And so the beam element stiffness matrix is given by:

[ ]

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

EI EI EI EIL L L LEI EI EI EIL L L L


− −

= − − −

−

k (4.4.8)

Next we note a special case where the vertical displacements of the beam nodes are

prevented and only rotations of the beam ends is allowed. In this case, all terms

relating to the translation DOFs are removed giving us the reduced stiffness matrix

for a beam on rigid vertical supports:


Dr. C. Caprani 81

[ ]4 2

2 4

EI EIL LEI EIL L

=

k (4.4.9)

As we did for trusses, we will often write these equations in terms of nodal sub-

matrices as:

i i

j j

=

F δk11 k12F δk21 k22

(4.4.10)


Dr. C. Caprani 82

4.4.2 Beam Element Loading

Applied Loads

Beam loads are different to truss loads since they can be located anywhere along the

element, not only at the nodes – termed intermodal loading Beams can also have

loads applied to the nodes – nodal loading. We deal with these two kinds of loads as

follows:

• Nodal loads: apply the load to the joint as usual;

• Inter-nodal loads: apply the equivalent concentrated loads to the joints (these are

just fixed end moment reactions to the load, with the direction reversed).

If a member’s nodes are locked against rotation, the member end forces due to inter-

nodal loading will just be the fixed end moment and force reaction vector we are

familiar with { }FF . If a member also displaces, the total member end forces are:

{ } { } [ ]{ }Tot= +FF F k δ (4.4.11)

Thus the general stiffness equation becomes:

{ } [ ]{ }=F K δ (4.4.12)

Where { }F is now the vector of net nodal loads:

Net Nodal Load Nodal Load Fixed End Reactions= −∑ (4.4.13)


Dr. C. Caprani 83

Lastly, we must note that inter-nodal loads on adjacent members will result in

multiple loads on a node. Thus we must take the algebraic sum of the forces/moments

on each node in our analysis, bearing in mind the sign convention.

As an example, the equivalent nodal loads for a UDL applied to a beam element are:

Member End Forces

After the deformations of the beam are known, we can use the element stiffness

matrices to recover the end forces/moments on each element due to both

deformations and the inter-nodal loading directly from equation (4.4.11).


Dr. C. Caprani 84

4.4.3 Example 7 – Simple Two-Span Beam

Problem

For the following beam, find the rotations of joints 2 and 3 and the bending moment

diagram. Take 3 26 10 kNmEI = × .

Solution

First we write the general equation in terms of nodal sub-matrices:

1 11 12 13 1

2 21 22 23 2

3 31 32 33 3

=

F K K K δF K K K δF K K K δ

(4.4.14)

Next we note that the only possible displacements are the rotations of joints 2 and 3.

Thus we can restrict the equation by eliminating joint 1 as follows:

1 11 12 13 1

2 21 22 23 2

3 31 32 33 3

=


(4.4.15)

To give:


Dr. C. Caprani 85

2 22 23 2

3 32 33 3

=

F K K δF K K δ

(4.4.16)

Since this beam is on rigid vertical supports, we can use the beam stiffness matrix

given by equation (4.4.9). Thus we are left with two equations:

2 22 23 2

3 32 33 3

M k kM k k

θθ

=

(4.4.17)

The member contributions to each of these terms are:

• 22 Term 22 of Member 12 + Term 11 of Member 23k = ;

• 23 Term 12 of Member 23k = ;


• 33 Term 22 of Member 23k =

Thus, for Member 12 we have:

[ ]( ) ( )

( ) ( )3 3

12

4 6 2 64 24 26 610 10

2 4 2 42 6 4 66 6

EI EIL LEI EIL L

= = =

k (4.4.18)

And for Member 23:

[ ]( ) ( )

( ) ( )3 3

23

4 6 2 64 26 610 102 42 6 4 6

6 6

= =

k (4.4.19)


Dr. C. Caprani 86

Thus the global stiffness equation is:

2 23

3 3

8 210

2 4MM

θθ

=

(4.4.20)

To find the moments to apply to the nodes, we determine the fixed-end moments

caused by the loads on each members. Only Member 23 has load, and its fixed end

moments are:

Our sign convention is anti-clockwise positive. Thus the moments to apply to the

joints become (refer to equation (4.4.13)):

23

3

30 8 210

30 2 4θθ

− = − (4.4.21)

Solving the equation:

( )

2 33

3

4 2 30 90 141 1 10 rads2 8 30 150 1410 8 4 2 2

θθ

−− − − = = −⋅ − ⋅

(4.4.22)

Since we know that anti-clockwise is positive, we can draw the displaced shape (in

mrads):


Dr. C. Caprani 87

Using the member stiffness matrices we can recover the bending moments at the end

of each member, now that the rotations are known, from equation (4.4.11):

1 3 3

2

0 4 2 0 12.910 10 kNm

0 2 4 90 14 25.7MM

− − = + = − −

(4.4.23)

2 3 3

3

30 4 2 90 14 25.710 10 kNm

30 2 4 150 14 0MM

− − + = + = −

(4.4.24)

Thus the final BMD can be drawn as:


Dr. C. Caprani 88

4.4.4 Example 8 – Non-Prismatic Beam

Problem

For the following beam, find the vertical deflection of joint 2 and the bending

moment diagram. Take 3 212 10 kNmEI = × .

Solution

First we write the general equation in terms of nodal sub-matrices:

1 11 12 13 1

2 21 22 23 2

3 31 32 33 3

=


(4.4.25)

Next we note that the only possible displacements are those of joint 2. Thus we can

restrict the equation to:

{ } [ ]{ }2 22 2=F K δ (4.4.26)

The member contributions to 22K are:

• Sub-matrix k22 of member 12;

• Sub-matrix k11 of member 23.


Dr. C. Caprani 89

That is:

[ ] [ ] [ ]22 12 23= +K k22 k11 (4.4.27)

For member 12, we have, from equation (4.4.8):

3 2 3 2

3 312

2 2

12 6 12 12 6 122.25 4.54 410 10

6 4 6 12 4 12 4.5 124 4

EI EIL LEI EIL L

⋅ ⋅ − − − = = = ⋅ ⋅ − − −

k22 (4.4.28)

And for member 23:

3 2 3 2

3 323

2 2

12 6 12 24 6 244.5 94 410 10

6 4 6 24 4 24 9 244 4

EI EIL LEI EIL L

⋅ ⋅

= = = ⋅ ⋅

k11 (4.4.29)

Since the load is a directly applied nodal load we can now write equation (4.4.26),

using equations (4.4.27), (4.4.28), and (4.4.29), as:

2 23

2 2

100 6.75 4.510

0 4.5 36y yF

Mδθ

− = =

(4.4.30)

Solving:

( )

2 3 3

2

36 4.5 100 16.16110 104.5 6.75 0 2.026.75 36 4.5 4.5

yδθ

− −− − − = = −⋅ − ⋅

(4.4.31)


Dr. C. Caprani 90

Thus we have a downwards (negative) displacement of 16.2 mm and an

anticlockwise rotation of 2.02 mrads at joint 2, as shown:

Next we recover the element end forces. For member 12, from equation (4.4.6) we

have:

1

1 3 3

2

2

2.25 4.5 2.25 4.5 0 45.54.5 12 4.5 6 0 84.8

10 102.25 4.5 2.25 4.5 16.16 45.54.5 6 4.5 12 2.02 97.0

y

y

FMFM

−

− − = = − − − − − −

(4.4.32)

And for member 23:

2

2 3 3

3

3

4.5 9 4.5 9 16.16 54.59 24 9 12 2.02 97.0

10 104.5 9 4.5 9 0 54.59 12 9 24 0 121.2

y

y

FMFM

−

− − − − − = = − − − − −

(4.4.33)

Thus the member end forces are:


Dr. C. Caprani 91

As can be seen, the load is split between the two members in a way that depends on

their relative stiffness.

The total solution is thus:


Dr. C. Caprani 92

4.4.5 Problems

Problem 1

Determine the bending moment diagram and rotation of joint 2. Take 3 210 10 kNmEI = × .

Problem 2

Determine the bending moment diagram and the rotations of joints 1 and 2. Take 3 220 10 kNmEI = × .


Dr. C. Caprani 93

Problem 3

Determine the bending moment diagram and the displacements of joints 2 and 3.

Take 3 220 10 kNmEI = × .

Problem 4

Determine the bending moment diagram and the vertical displacement under the 100

kN point load. Take 3 210 10 kNmEI = × .


Dr. C. Caprani 94

Problem 5

Determine the bending moment diagram and the rotations of joints 2 and 3. Take 3 220 10 kNmEI = × .

Problem 6

Determine the bending moment diagram and the rotations of all joints. Take 3 240 10 kNmEI = × . You may use Excel or Matlab to perform some of the numerical

calculations. Check your member stiffness and global stiffness matrices with LinPro,

and your final results. Identify and explain discrepancies. Verify with LUSAS.


Dr. C. Caprani 95

4.5 Plane Frames

4.5.1 Plane Frame Element Stiffness Matrix

A plane frame element is similar to a beam element except for some differences:

• The presence of axial forces;

• The member may be oriented at any angle in the global axis system;

• The inter-nodal loads may be applied in the local or global coordinates.

These points are illustrated in the following:

Lastly, an easy way to deal with inter-nodal point loads ( GP , LP ) is to introduce a

node under the point load (splitting the member in two), then it is no longer inter-

nodal and so no transformations or equivalent load analysis is required. The downside

to this is that the number of equations increases (which is only really a problem for

analysis by hand).


Dr. C. Caprani 96

Axial Forces

To include axial forces, we can simply expand the beam element stiffness matrix to

allow for the extra degree of freedom of x-displacement at each node in the member

local coordinates. Thus expanding equation (4.4.8) to allow for the extra DOFs gives:

[ ]

11 14

3 2 3 2

2 2

41 44

3 2 3 2

2 2

0 0 0 012 6 12 60 0

6 4 6 20 0

0 0 0 012 6 12 60 0

6 2 6 40 0

X XEI EI EI EI

L L L LEI EI EI EIL L L L

X XEI EI EI EI

L L L LEI EI EI EIL L L L

−

− = − − −

−

k (4.5.1)

However, these terms that account for axial force are simply those of a plane truss

element in its local coordinate system:

[ ] 1 11 1

EAL

− = −

k (4.5.2)



Dr. C. Caprani 97

[ ]

3 2 3 2

2 2

3 2 3 2

2 2

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0 0 0

12 6 12 60 0

6 2 6 40 0

EA EAL L


EA EAL L


− − − = − − − − −

k (4.5.3)

This is the stiffness matrix for a plane frame element in its local coordinate system

and can also be written in terms of nodal sub-matrices as:

[ ] =

k11 k12k

k21 k22 (4.5.4)

Where the nodal sub-matrices are as delineated in equation (4.5.3).

Note that if axial forces are neglected, we can just use the regular beam element

stiffness matrix instead, though coordinate transformation may be required.


Dr. C. Caprani 98

Transformation to Global Coordinates

From the Appendix, the plane frame element stiffness matrix in global coordinates is:

[ ] [ ][ ]Te K = T k T (4.5.5)

As a consequence, note that we do not need to perform the transformation when:

1. The member local axis and global axis system coincide;

2. The only unrestrained DOFs are rotations/moments.

Again from the Appendix, the transformation matrix for a plane frame element is:

cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

α αα α

α αα α

−

= −

T (4.5.6)


Dr. C. Caprani 99

Inter-nodal Loads

In plane frames, loads can be applied in the global axis system, or the local axis

system. For example, if we consider a member representing a roof beam, we can have

the following laods:

• Case 1: Gravity loads representing the weight of the roof itself;

• Case 2: Horizontal loads representing a horizontal wind;

• Case 3: Net pressure loads caused by outside wind and inside pressures.

Case 1 Case 2 Case 3

Most structural analysis software will allow you to choose the axis system of your

loads. However, in order to deal with these loads for simple hand analysis we must

know how it works and so we consider each case separately.

In the following the member local axis system has a prime (e.g. x’) and the global

axis system does not (e.g. x).


Dr. C. Caprani 100

Case 1: Vertically Applied Loads

In this case we can consider an equivalent beam which is the projection of the load

onto a horizontal beam of length XL :

Since the resulting nodal forces and moments are in the global axis system no further

work is required.


Dr. C. Caprani 101

Case 2: Horizontally Applied Loads:

Similarly to vertically applied loads, we can consider the horizontal projection of load

onto an equivalent member of length YL .

Again the resulting nodal loads are in the global axis system and do not require any

modification.


Dr. C. Caprani 102

Case 3: Loads Applied in Local Member Axis System

In this case there is no need for an equivalent beam and the fixed-fixed reactions are

worked out as normal:

However, there is a complication here since the reactions are now not all in the global

axis system. Thus the forces (not moments) must be transformed from the local axis

to the global axis system. Thus there is a simple case:

If axial forces are neglected, only moments are relevant and so no transformations are

required.

For generality though we can use the transformations given in the Appendix:

{ } [ ] { }'TF = T F (4.5.7)

Writing this out in full for clarity, we have:


Dr. C. Caprani 103

'

'

'

'

cos sin 0 0 0 0sin cos 0 0 0 0

0 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

ij ijix ixij ij

iy iyij iji iij ijjx jxij ijjy jyij ijj j

F FF FM MF FF FM M

α αα α

α αα α

− = −

(4.5.8)


Dr. C. Caprani 104

4.5.2 Example 9 – Simple Plane Frame

Problem

For the following frame, determine the rotation of the joints and the bending moment

diagram. Neglect axial deformations. Take 3 21 10 kNmEI = × .

Solution

The fact that we can neglect axial deformation makes this problem much simpler. As

a consequence, the only possible displacements are the rotations of joints 1 and 2.

Since node 3 is fully restricted out, we have the following partially-restricted set of

equations in terms of nodal sub-matrices:

=

1 1

2 2

F δK11 K12F δK21 K22

(4.5.9)

If we expand this further, we will be able to restrict out all but the rotational DOFs:


Dr. C. Caprani 105

1 33 36 1

2 63 66 2

M k k

M k k

θ

θ

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅

∑

∑

(4.5.10)





• 66 Term 66 of Member 12 Term 33 of Member 23k = + .

• Member 12:

Looking at equation (4.5.3):

3

3

12

4 4 10Term 33 4 101

EIL

⋅ = = = ×

(4.5.11)

3

3

12

2 2 10Term 36 2 101

EIL

⋅ = = = ×

(4.5.12)

3

3

12

2 2 10Term 63 2 101

EIL

⋅ = = = ×

(4.5.13)

3

3

12

4 4 10Term 66 4 101

EIL

⋅ = = = ×

(4.5.14)

• Member 23:

Again, from equation (4.5.3):


Dr. C. Caprani 106

3

3

23

4 4 10Term 33 4 101

EIL

⋅ = = = ×

(4.5.15)

Thus the system equation becomes:

1 13

2 2

4 210

2 8MM

θθ

=

∑∑

(4.5.16)

Next we must find the net moments applied to each node. There are no directly

applied nodal moment loads, so the ‘force’ vector is, from equation (4.4.13):

{ } { }= − FF F (4.5.17)

• Member 12 Moments:

2 2121

2 2122

12 1 1 kNm12 12

12 1 1 kNm12 12

wLM

wLM

⋅= = = +

⋅= − = − = −

(4.5.18)

• Member 23 Moments:


Dr. C. Caprani 107

232

233

16 1 2 kNm8 8

16 1 2 kNm8 8

PLM

PLM

⋅= = = +

⋅= − = − = −

(4.5.19)

Thus the net nodal loads become:

{ } { } 1

2

1 1 kNm

1 2 1MM

+ − = − = − = − = − + −

∑∑FF F (4.5.20)

And so equation (4.5.16) is thus:

13

2

1 4 210

1 2 8θθ

− = −

(4.5.21)

Which is solved to get:

( )

1 33

2

8 2 1 3 141 1 10 rads2 4 1 1 1410 4 8 2 2

θθ

−− − − = = × − − −⋅ − ⋅

(4.5.22)

The negative results indicate both rotations are clockwise.


Dr. C. Caprani 108

Lastly, we must find the member end forces. Since we only need to draw the bending

moment diagram so we need only consider the terms of the member stiffness matrix

relating to the moments/rotations (similar to equation (4.4.9)). Also, we must account

for the equivalent nodal loads as per equation (4.4.11):

• Member 12:

121 3 3122

1 4 2 3 14 010 10 kNm

1 2 4 1 14 12 7MM

−+ − = + = − − −

(4.5.23)

• Member 23:

232 3 3233

2 4 2 1 14 12 710 10 kNm

2 2 4 0 17 7MM

−+ − + = + = − −

(4.5.24)


Dr. C. Caprani 109

4.5.3 Example 10 –Plane Frame Using Symmetry

Problem

For the following frame, determine the rotation of the joints, the displacement under

the 8 kN point load and the bending moment diagram. Neglect axial deformations.

Take 3 21 10 kNmEI = × .

Solution

Again, the fact that we can neglect axial deformation makes this problem much

simpler. Since the structure is symmetrical and it is symmetrically loaded, it will not

sway. Further, because of this symmetry, we can adopt the following model for

analysis:


Dr. C. Caprani 110

Notice two things from this model:

• we have renumber the joints – there is no need to retain the old numbering system;

• The remaining DOFs are 2θ and 3 yδ - we can restrict all other DOFs. Thus in terms

of nodal sub-matrices we immediately have:

2 2

3 3

=

F δK11 K12F δK21 K22

(4.5.25)

And expanding this further, we restrict out all other restrained DOFs:

2 66 68 2

3 86 88 3y y

M k k

F k k

θ

δ

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

∑

∑

(4.5.26)


• 66 Term 66 of Member 12 Term 33 of Member 23k = + ;



• 66 Term 55 of Member 23k = .

Transformation of the member stiffness matrices is not required. Member 12 only has

a rotational DOF and Member 23’s local member coordinate system is parallel to the

global axis coordinate system.


Dr. C. Caprani 111

• Member 12:

From equation (4.5.3):

3

3

12

4 4 10Term 66 4 101

EIL

⋅ = = = ×

(4.5.27)

• Member 23:

Again, from equation (4.5.3):

3

3

23

4 4 10Term 33 4 101

EIL

⋅ = = = ×

(4.5.28)

3

32 2

23

6 6 10Term 35 6 101

EIL

⋅ = − = − = − ×

(4.5.29)

3

32 2

23

6 6 10Term 53 6 101

EIL

⋅ = − = − = − ×

(4.5.30)

3

33 3

23

12 12 10Term 55 12 101

EIL

⋅ = = = ×

(4.5.31)

Thus the system equation becomes:

22 3

33

8 610

6 12 yy

MF

θδ

− = −

∑∑

(4.5.32)

The 4 kN point load is directly applied to node 3 so this causes no difficulty. The

equivalent nodal loads for the UDL are:


Dr. C. Caprani 112

2 2121

2 2122

12 1 1 kNm12 12

12 1 1 kNm12 12

wLM

wLM

⋅= = = +

⋅= − = − = −

(4.5.33)

Notice that we do not need to find the vertical reaction forces as there is no sway of

the frame and we are neglecting axial deformation.

The nodal load vector, from equation (4.4.13) is thus:

{ } { } { }0 1 14 0 4

− + = − = − = − −

∑ FF F F (4.5.34)

And so equation (4.5.32) is thus:

23

3

1 8 610

4 6 12 y

θδ

+ − = − −

(4.5.35)

Which is solved to get:

( )( )

2 33

3

12 6 1 0.2 rads1 1 106 8 4 0.43 m10 8 12 6 6y

θδ

−+ − = = × − −⋅ − − −

(4.5.36)


Dr. C. Caprani 113

The negative results indicate the rotation is clockwise and the displacement

downwards, as may be expected:

Lastly then we find the bending moments. For member 12 only the terms relating to

bending moments are relevant.

• Member 12:

121 3 3122

1 4 2 0 0.610 10 kNm

1 2 4 0.2 1.8MM

−+ + = + = − − −

(4.5.37)

However, for member 23, the downwards deflection also causes moments and so the

relevant DOFs are rotation of node i and vertical movement of node j (as calculated

earlier). It is easier to see this if we write the member equation in full:

• Member 23:

232 3 3

233

4 6 0.210 10

0.432 6

M

M

−

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ −

= × ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ −

⋅ ⋅ ⋅ − ⋅ ⋅

(4.5.38)


Dr. C. Caprani 114

Thus:

232233

1.8 kNm

2.2MM

+ = +

(4.5.39)

And so the BMD is:


Dr. C. Caprani 115

4.5.4 Problems

Problem 1

Determine the bending moment diagram and the rotation of joint 2. Take 3 210 10 kNmEI = × and neglect axial deformations.

(Ans. 2 5 6 mradsθ = )

Problem 2

Determine the bending moment diagram and the rotations of joints 1 and 2. Take 3 220 10 kNmEI = × and neglect axial deformations.


Dr. C. Caprani 116

Problem 3

Determine the bending moment diagram. Take 3 220 10 kNmEI = × and neglect axial

deformations.

Problem 4

Determine the bending moment diagram, the rotation of joint 2, and the horizontal

displacements of joints 2 and 3. Take 3 210 10 kNmEI = × and neglect axial

deformations.

(Ans. 2 211.33 mrads; 44.0mmxθ δ= − = )


Dr. C. Caprani 117

Problem 5

Determine the bending moment diagram. Take 3 220 10 kNmEI = × and neglect axial

deformations.

Problem 6

Determine the bending moment diagram and the vertical displacement of joint 3.

Take 3 240 10 kNmEI = × and neglect axial deformations.


Dr. C. Caprani 118

Problem 7

Determine the bending moment diagram, the rotation of joint 2, and the vertical

displacement under the 80 kN point load. Take 3 210 10 kNmEI = × and neglect axial

deformations.

(Ans. 2 1.071 mrads; 1.93mmyθ δ= − = − )

Problem 8

For the frame of Problem 1, determine the bending moment diagram and the rotation

and vertical displacement of joint 2 if member 24 has 310 10 kNEA = × . Neglect axial

deformation in the other members.

(Ans. 2 0.833 mrads; 0.01mmyθ δ= = − )


Dr. C. Caprani 119

Problem 9

Determine the bending moment diagram for the prismatic portal frame. Take 3 220 10 kNmEI = × and neglect axial deformations. You may use Excel or Matlab to

perform some of the numerical calculations. Check your member stiffness and global

stiffness matrices with LinPro, and your final results. Identify and explain

discrepancies. Verify with LUSAS.


Dr. C. Caprani 120

4.6 Appendix

4.6.1 Plane Truss Element Stiffness Matrix in Global Coordinates

Compatibility Conditions

Firstly we indentify the conditions of compatibility of a truss element nodal

deflections and the member elongation. We use the following notation for the

deflections at each node of the truss:

If we now consider the deflected position of the truss member, we have:


Dr. C. Caprani 121

Obviously the change in length of the truss will be related to the difference between

the nodal deflections. Hence, we define the changes in movements such that an

elongation gives positive changes:

x jx ix y jy iyδ δ δ δ δ δ∆ = − ∆ = −

Moving the deflected position of node i back to its original location gives:

Looking more closely at the triangle of displacements at node j, and remembering

that we are assuming small deflections—which in this case means the deflected

position of the member is still at a rotation of θ . Hence we have:


Dr. C. Caprani 122

And so the elongation is given by:

( ) ( )

cos sin

cos sinx y

jx ix jy iy

e δ θ δ θ

δ δ θ δ δ θ

= ∆ + ∆

= − + − (4.6.1)

Now multiply out and re-order to get:

cos sin cos sinix iy jx jye δ θ δ θ δ θ δ θ= − + − + + (4.6.2)

If we define a direction vector, α , and a displacement vector, δ , as:

cossin

cossin

ix

iy

jx

jy

δθδθδθδθ

− − = =

α δ (4.6.3)

Then, from (4.6.2) and (4.6.3), we can say:


Dr. C. Caprani 123

te = α δ (4.6.4)

Thus we have related the end displacements to the elongation of the member which

therefore maintain compatibility of displacement.


Dr. C. Caprani 124

Virtual Work for Element Forces

Looking at the forces acting on the nodes of the bar element, we have:

This is a force system in equilibrium—the external nodal loading is in equilibrium

with the internal bar force, N. If we consider a pattern of compatible displacements

such as the following:


Dr. C. Caprani 125

We can apply virtual work to this:

0

E I

WW Wδδ δ

==

And we have:

Substituting in our notations for the bar element:

ix ix iy iy jx jx jy jyeN F F F Fδ δ δ δ= + + + (4.6.5)

If we define the force vector, F , as:

ix

iy

jx

jy

FFFF

=

F (4.6.6)

Then we can write (4.6.5) as:

t eN=F δ (4.6.7)

Set of forces in

equilibrium

Set of compatible

displacements


Dr. C. Caprani 126

If we use (4.6.4) we how have:

t t N=F δ α δ (4.6.8)

Post-multiply both sides by 1−δ , and noting that N is a scalar, gives:

t t N=F α

N=F α (4.6.9)

Expanding this out gives:

cossin

cossin

ix

iy

jx

jy

F NF NF NF N

θθθθ

− − =

(4.6.10)

Which are the equations of equilibrium of the bar element:


Dr. C. Caprani 127

Relating Forces to Displacements

Lastly, in order to relate the end forces to the element nodal displacements, we note

from the constitutive law:

EAN eL

= ⋅ (4.6.11)

And so from (4.6.9) we have:

EA eL

=F α (4.6.12)

And using equation (4.6.4) gives:

tEAL

=F α α δ (4.6.13)

Hence the term tEAL

α α relates force to displacement and is called the stiffness

matrix, k , which is evaluated by multiplying out terms:

[ ]

cossin

cos sin cos sincossin

tEAL

EAL

θθ

θ θ θ θθθ

=

− − = − −

k α α

(4.6.14)

And multiplying this out gives:


Dr. C. Caprani 128

2 2

2 2

2 2

2 2



EAL

θ θ θ θ θ θθ θ θ θ θ θ

θ θ θ θ θ θθ θ θ θ θ θ

− − − − = − − − −

k (4.6.15)

And for clarity, we write out the final equation in matrix form and in full:

=F kδ (4.6.16)

2 2

2 2

2 2

2 2



ix ix

iy iy

jx jx

jy jy

FF EAF LF

δθ θ θ θ θ θδθ θ θ θ θ θδθ θ θ θ θ θδθ θ θ θ θ θ

− − − − = − − − −

(4.6.17)

So for example, the stiffness that relates a horizontal force at node j to the horizontal

displacement at node j is:

2cosjx jx

EAFL

θ δ =

And other relationships can be found similarly.


Dr. C. Caprani 129

4.6.2 Coordinate Transformations

Point Transformation

We consider the transformation of a single point P from one coordinate axis system

xy to another x’y’:

From the diagram, observe:

' coordinate of

' coordinate of

OC x P

PC y P

=

= (4.6.18)

Also:

coordinate of

coordinate of

OB x P

PB y P

=

= (4.6.19)


Dr. C. Caprani 130

Next we can say:

OC OA AC= + (4.6.20)

PC PD CD= − (4.6.21)

Introducing the relevant coordinates:

cos cosOA OA xα α= = (4.6.22)

sin sinAC BD PB yα α= = = (4.6.23)


' cos sinOC x x yα α= = + (4.6.24)

Next we have:

cos cosPD PB yα α= = (4.6.25)

sin sinCD AB OB xα α= = = (4.6.26)


' cos sinPC y y xα α= = − (4.6.27)

Writing equations (4.6.24) and (4.6.27) together:

' cos sinx x yα α= + (4.6.28)


Dr. C. Caprani 131

' sin cosy x yα α= − + (4.6.29)

And now in matrix form gives:

' cos sin' sin cos

x xy y

α αα α

= −

(4.6.30)

Often we write:

cossin

cs

αα

≡≡

(4.6.31)

To give:

''

x c s xy s c y

= −

(4.6.32)

Lastly, if we generically name the two coordinate systems as q and q’, we then have

in matrix form:

{ } [ ]{ }Nq' = T q (4.6.33)

Where [ ]NT is the nodal transformation matrix given by:

cos sinsin cosN

α αα α

= −

T (4.6.34)


Dr. C. Caprani 132

Force/Displacement Transformation

Forces and moments can be oriented in the local member axis system or in the global

structure axis system. In general we will need to transform the forces and

displacements of both nodes, thus we write:

''

i iN

j jN

F FT 0=

F F0 T (4.6.35)

And finally we can write:

{ } [ ]{ }'F = T F (4.6.36)

Where:

[ ] N

N

T 0T =

0 T (4.6.37)

Similarly for deflections:

{ } [ ]{ }'δ = T δ (4.6.38)

A very useful property of the transformation matrix (not derived here) is that it is

orthogonal. This means that its transpose is equal to its inverse:

[ ] [ ] 1T −=T T (4.6.39)


Dr. C. Caprani 133

Thus when either a force or displacement is known for the local axis system, it can be

found in the global axis system as follows:

{ } [ ] { }'TF = T F (4.6.40)

{ } [ ] { }'Tδ = T δ (4.6.41)


Dr. C. Caprani 134

Transformations for Plane Truss Element

For a plane truss member, there will be x and y components of force at each of its

nodes. Using the transformation for a point, we therefore have:

'

'

cos sinsin cos

x x

y y

F FF F

α αα α

= −

(4.6.42)

And so for a truss element, we have directly from equation (4.6.34):

cos sinsin cosN

α αα α

= −

T (4.6.43)

And so, from equation (4.6.37),

[ ]


α αα α

α αα α

− = −

T (4.6.44)

For clarity, we write the transformation out in full:

'

'

'

'


ix ix

iy iy

jx jx

jy jy

F FF FF FF F

α αα α

α αα α

− = −

(4.6.45)


Dr. C. Caprani 135

Transformations for Plane Frame Element

Based on the DOF transformation matrix for a plane truss member (in terms of

forces), we can determine the transformation matrix for a plane frame node quite

easily:

cos sin 0sin cos 00 0 1

ex xe

y ye

F FF FM M

α αα α

= −

(4.6.46)

This is because a moment remains a moment in the plane. So for a single node, and

both nodes, we have, respectively:

{ } [ ]{ }' NF = T F (4.6.47)

''

i iN

j jN

F FT 0=

F F0 T (4.6.48)

Thus, we can now write the final transformation matrix for a plane frame element as:

cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

α αα α

α αα α

−

= −

T (4.6.49)


Dr. C. Caprani 136

Element Stiffness Matrix Transformation

Using the general expression for a single element:

e e eF = K δ (4.6.50)

Regardless of member type or the number of dimensions, we will always have some

coordinate transform from local to global coordinates such that:

eF = TF (4.6.51)

eδ = Tδ (4.6.52)

Hence from equation (4.6.50) we can write:

eTF = K Tδ (4.6.53)

And so the force-displacement relationship in the global axis system is:

1 e− F = T K T δ (4.6.54)

The term in brackets can now be referred to as the element stiffness matrix in global

coordinates. Thus, using equation (4.6.39), we write:

e T eG LK = T K T (4.6.55)


Dr. C. Caprani 137

4.6.3 Past Exam Questions

Summer 2001


Dr. C. Caprani 138

Summer 2002


Dr. C. Caprani 139

Summer 2004


Dr. C. Caprani 140

Summer 2006


Dr. C. Caprani 141

Sample Paper 2006/7 1. (a) Using the stiffness method, determine the displacement of the joints of the pin-jointed truss shown in Fig.

Q1(a), under the load as shown. (10 marks)

(b) Members 15 and 16 are added to the truss of Fig. 1(a) to form the truss shown in Fig. Q1(b). However,

member 16 is found to be 15 mm too long and is forced into place. The same load of 100 kN is again to be applied. Using the stiffness method, determine the displacement of the joints and the force in member 16.

(15 marks) Take EA = 2×104 kN and the cross sectional areas of the members as:

Members 12, 13, and 16: 3A; Diagonal Members 14 and 15: 3√2A.

Ans. (a) 25 kN; 75 kN; 25√2 kN;

(b) 50√2 kN; 156.1 kN; 60.4 kN; -100√2 kN; -50√2 kN

FIG. Q1(a)

FIG. Q1(b)


Dr. C. Caprani 142

Semester 1 2006/7 1. Using the stiffness method, determine the displacement of the joints and the forces in the members of the pin-

jointed truss shown in Fig. Q1, allowing for:

(i) The 100 kN vertical load as shown, and; (ii) A lack of fit of member 12, which was found to be 5 mm too short upon arrival at site, and

which was then forced into place.

Take EA = 2×104 kN and the cross sectional areas of the members as: • Members 12: 3A; • Members 13 and 14: 3√2A.

(25 marks)

Ans. 50 kN; -75√2 kN; -25√2 kN.


Dr. C. Caprani 143

Semester 1 Repeat 2006/7 1. Using the stiffness method, determine the displacement of the joints and the forces in the members of the pin-

jointed truss shown in Fig. Q1, allowing for:

(ii) The 100 kN vertical load as shown, and; (ii) A lack of fit of member 12, which was found to be 10√2 mm too short upon arrival at site, and

which was then forced into place.

Take EA = 2×104 kN and the cross sectional areas of all members as 3√2A. (25 marks)

Ans. 125√2 kN; -50√2 kN; -75√2 kN.

FIG. Q1


Dr. C. Caprani 144

Semester 1 2007/8 QUESTION 1 Using the stiffness method, determine the displacement of the joints and the forces in the members of the pin-jointed truss shown in Fig. Q1, allowing for: (i) The 100 kN load as shown, and; (ii) A lack of fit of member 13, which was found to be 4 mm too short upon arrival at site, and which was then

forced into place; (iii) A temperature rise of 20 ˚C in member 24. Note: Take 3125 10 kNEA = × and the coefficient of thermal expansion -5 -12 10 Cα = × ° .

(25 marks)

Ans. -55.7 kN; +69.7 kN; -55.3 kN.

FIG. Q1


Dr. C. Caprani 145

Semester 1 2008/9 QUESTION 1 Using the stiffness method, for the continuous beam shown in Fig. Q1, do the following:

(i) determine the displacement of the joints; (ii) draw the bending moment diagram;

(iii) determine the reactions.

Note: Take 3 210 10 kNmEI = × .

(25 marks)

Ans. 98.7 kNm; 102.6 kNm; 60.9 kNm.

FIG. Q1


Dr. C. Caprani 146

Semester 1 2009/10 QUESTION 1 Using the stiffness method, for the frame shown in Fig. Q1, do the following:

(i) determine the vertical displacement at the centre of the middle span; (ii) draw the bending moment diagram;

(iii) determine the reactions.

Note: Take 3 210 10 kNmEI = × .

(25 marks)

Ans. -11.88 mm

FIG. Q1


Dr. C. Caprani 147

Semester 1 2010/11 QUESTION 1 Using the stiffness method, for the truss shown in Fig. Q1: (a) Determine:

(i) The displacement of the joints; (ii) The forces in the members; (iii) The deflected shape of the structure.

(15 marks) (b) Determine the lack of fit of member 23, which would result in no horizontal displacement of joint 2 under the

100 kN load shown. (10 marks)

Note: • Take 2200kN/mmE = for all members.

• Area for member 12 is 2400 10 mmA = . • Area for member 23 is 2960mmA = .

• Area for member 13 is 2640 2 mmA = .

Ans. 68.7 kN, 34.6 kN, 49.3 kN; 5.21 mm.

FIG. Q1


Dr. C. Caprani 148

4.7 References • Alberty, J., Carstensen, C. and Funken, S.A. (1999), ‘Remarks around 50 lines of

Matlab: short finite element implementation’, Numerical Algorithms, 20, pp. 117-

137, available at: web address.

• Brown, D.K. (1990), An Introduction to the Finite Element Method using Basic

Programs, 2nd Edn., Taylor and Francis, London.

• Carroll, W.F. (1999), A Primer for Finite Elements in Elastic Structures, John

Wiley & Sons, New York.

• Coates, R.C., Coutie, M.G. and Kong, F.K. (1987), Structural Analysis, Chapman

and Hall.

• Davies, G.A.O. (1982), Virtual Work in Structural Analysis, John Wiley & Sons.

• Desai, C.S. and Abel, J.F. (1972), Introduction to the Finite Element Method: A

Numerical Method for Engineering Analysis, Van Nostrand Reinhold, New York.

• Ghali, A. and Neville, A.M. (1997), Structural Analysis – A unified classical and

matrix approach, 4th edn., E&FN Spon, London.

• McGuire, W., Gallagher, R.H. and Ziemian, R.D. (2000), Matrix Structural

Analysis, 2nd Edn., John Wiley & Sons.

• Meek, J.L. (1991), Computer Methods in Structural Analysis, 2nd Edn., E&FN

Spon.

• Przemieniecki, J.S. (1968), Theory of Matrix Structural Analysis, McGraw-Hill,

New York.

• Sack, R.L. (1989), Matrix Structural Analysis, Waveland Press, Prospect Heights,

Illinois, US.

• Thompson, F., and Haywood, G.G. (1986), Structural Analysis Using Virtual

Work, Chapman and Hall.

• Weaver, W. and Gere, J.M (1990), Matrix Analysis of Framed Structures, 3rd

Edn., Van Nostrand Reinhold, New York.

http://www.math.huberlin.de/~cc/download/public/software/do-cumentation/acf.pdf

Structural Analysis IV Chapter 5 – Structural Dynamics

Dr. C. Caprani 1

Chapter 5 - Structural Dynamics

5.1 Introduction ......................................................................................................... 3

5.1.1 Outline of Structural Dynamics ..................................................................... 3

5.1.2 An Initial Numerical Example ....................................................................... 5

5.1.3 Case Study – Aberfeldy Footbridge, Scotland .............................................. 8

5.1.4 Structural Damping ...................................................................................... 10

5.2 Single Degree-of-Freedom Systems ................................................................. 11

5.2.1 Fundamental Equation of Motion ................................................................ 11

5.2.2 Free Vibration of Undamped Structures...................................................... 16

5.2.3 Computer Implementation & Examples ...................................................... 20

5.2.4 Free Vibration of Damped Structures .......................................................... 26


5.2.6 Estimating Damping in Structures ............................................................... 33

5.2.7 Response of an SDOF System Subject to Harmonic Force ........................ 35


5.2.9 Numerical Integration – Newmark‟s Method ............................................. 47

5.2.10 Computer Implementation & Examples ................................................... 53

5.2.11 Problems ................................................................................................... 59

5.3 Multi-Degree-of-Freedom Systems .................................................................. 63

5.3.1 General Case (based on 2DOF) ................................................................... 63

5.3.2 Free-Undamped Vibration of 2DOF Systems ............................................. 66

5.3.3 Example of a 2DOF System ........................................................................ 68

5.3.4 Case Study – Aberfeldy Footbridge, Scotland ............................................ 73


Dr. C. Caprani 2

5.4 Continuous Structures ...................................................................................... 76

5.4.1 Exact Analysis for Beams ............................................................................ 76

5.4.2 Approximate Analysis – Bolton‟s Method .................................................. 86

5.4.3 Problems ...................................................................................................... 95

5.5 Practical Design Considerations ...................................................................... 97

5.5.1 Human Response to Dynamic Excitation .................................................... 97

5.5.2 Crowd/Pedestrian Dynamic Loading .......................................................... 99

5.5.3 Damping in Structures ............................................................................... 107

5.5.4 Design Rules of Thumb ............................................................................. 109

5.6 Appendix .......................................................................................................... 114

5.6.1 Past Exam Questions ................................................................................. 114

5.6.2 References .................................................................................................. 121

5.6.3 Amplitude Solution to Equation of Motion ............................................... 123

5.6.4 Solutions to Differential Equations ........................................................... 125

5.6.5 Important Formulae ................................................................................... 134

5.6.6 Glossary ..................................................................................................... 139

Rev. 1


Dr. C. Caprani 3

5.1 Introduction

5.1.1 Outline of Structural Dynamics

Modern structures are increasingly slender and have reduced redundant strength due

to improved analysis and design methods. Such structures are increasingly responsive

to the manner in which loading is applied with respect to time and hence the dynamic

behaviour of such structures must be allowed for in design; as well as the usual static

considerations. In this context then, the word dynamic simply means “changes with

time”; be it force, deflection or any other form of load effect.

Examples of dynamics in structures are:

Soldiers breaking step as they cross a bridge to prevent harmonic excitation;

The Tacoma Narrows Bridge footage, failure caused by vortex shedding;

The London Millennium Footbridge: lateral synchronise excitation.

(a) (after Craig 1981)

(b)

Figure 1.1

m

k


Dr. C. Caprani 4

The most basic dynamic system is the mass-spring system. An example is shown in

Figure 1.1(a) along with the structural idealisation of it in Figure 1.1(b). This is

known as a Single Degree-of-Freedom (SDOF) system as there is only one possible

displacement: that of the mass in the vertical direction. SDOF systems are of great

importance as they are relatively easily analysed mathematically, are easy to

understand intuitively, and structures usually dealt with by Structural Engineers can

be modelled approximately using an SDOF model (see Figure 1.2 for example).

Figure 1.2 (after Craig 1981).


Dr. C. Caprani 5

5.1.2 An Initial Numerical Example

If we consider a spring-mass system as shown in Figure 1.3 with the properties m =

10 kg and k = 100 N/m and if give the mass a deflection of 20 mm and then release it

(i.e. set it in motion) we would observe the system oscillating as shown in Figure 1.3.

From this figure we can identify that the time between the masses recurrence at a

particular location is called the period of motion or oscillation or just the period, and

we denote it T; it is the time taken for a single oscillation. The number of oscillations

per second is called the frequency, denoted f, and is measured in Hertz (cycles per

second). Thus we can say:

1

fT

(5.1.1)

We will show (Section 2.b, equation (2.19)) for a spring-mass system that:

1

2

kf

m (5.1.2)

In our system:

1 1000.503 Hz

2 10f

And from equation (5.1.1):

1 11.987 secs

0.503T

f


Dr. C. Caprani 6

We can see from Figure 1.3 that this is indeed the period observed.

Figure 1.3

To reach the deflection of 20 mm just applied, we had to apply a force of 2 N, given

that the spring stiffness is 100 N/m. As noted previously, the rate at which this load is

applied will have an effect of the dynamics of the system. Would you expect the

system to behave the same in the following cases?

If a 2 N weight was dropped onto the mass from a very small height?

If 2 N of sand was slowly added to a weightless bucket attached to the mass?

Assuming a linear increase of load, to the full 2 N load, over periods of 1, 3, 5 and 10

seconds, the deflections of the system are shown in Figure 1.4.

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

pla

ce

me

nt

(mm

)

Time (s)

Period T

m = 10

k = 100


Dr. C. Caprani 7

Figure 1.4

Remembering that the period of vibration of the system is about 2 seconds, we can

see that when the load is applied faster than the period of the system, large dynamic

effects occur. Stated another way, when the frequency of loading (1, 0.3, 0.2 and 0.1

Hz for our sample loading rates) is close to, or above the natural frequency of the

system (0.5 Hz in our case), we can see that the dynamic effects are large.

Conversely, when the frequency of loading is less than the natural frequency of the

system little dynamic effects are noticed – most clearly seen via the 10 second ramp-

up of the load, that is, a 0.1 Hz load.

0

5

10

15

20

25

30

35

40

0 2 4 6 8 10 12 14 16 18 20

Defl

ecti

on

(m

m)

Time (s)

Dynamic Effect of Load Application Duration

1-sec

3-sec

5-sec

10-sec


Dr. C. Caprani 8

5.1.3 Case Study – Aberfeldy Footbridge, Scotland

Aberfeldy footbridge is a glass fibre reinforced polymer (GFRP) cable-stayed bridge

over the River Tay on Aberfeldy golf course in Aberfeldy, Scotland (Figure 1.5). Its

main span is 63 m and its two side spans are 25 m, also, tests have shown that the

natural frequency of this bridge is 1.52 Hz, giving a period of oscillation of 0.658

seconds.

Figure 1.5: Aberfeldy Footbridge

Figure 1.6: Force-time curves for walking: (a) Normal pacing. (b) Fast pacing


Dr. C. Caprani 9

Footbridges are generally quite light structures as the loading consists of pedestrians;

this often results in dynamically lively structures. Pedestrian loading varies as a

person walks; from about 0.65 to 1.3 times the weight of the person over a period of

about 0.35 seconds, that is, a loading frequency of about 2.86 Hz (Figure 1.6). When

we compare this to the natural frequency of Aberfeldy footbridge we can see that

pedestrian loading has a higher frequency than the natural frequency of the bridge –

thus, from our previous discussion we would expect significant dynamic effects to

results from this. Figure 1.7 shows the response of the bridge (at the mid-span) when

a pedestrian crosses the bridge: significant dynamics are apparent.

Figure 1.7: Mid-span deflection (mm) as a function of distance travelled (m).

Design codes generally require the natural frequency for footbridges and other

pedestrian traversed structures to be greater than 5 Hz, that is, a period of 0.2

seconds. The reasons for this are apparent after our discussion: a 0.35 seconds load

application (or 2.8 Hz) is slower than the natural period of vibration of 0.2 seconds (5

Hz) and hence there will not be much dynamic effect resulting; in other words the

loading may be considered to be applied statically.


Dr. C. Caprani 10

5.1.4 Structural Damping

Look again at the frog in Figure 1.1, according to the results obtained so far which

are graphed in Figures 1.3 and 1.4, the frog should oscillate indefinitely. If you have

ever cantilevered a ruler off the edge of a desk and flicked it you would have seen it

vibrate for a time but certainly not indefinitely; buildings do not vibrate indefinitely

after an earthquake; Figure 1.7 shows the vibrations dying down quite soon after the

pedestrian has left the main span of Aberfeldy bridge - clearly there is another action

opposing or “damping” the vibration of structures. Figure 1.8 shows the undamped

response of our model along with the damped response; it can be seen that the

oscillations die out quite rapidly – this depends on the level of damping.

Figure 1.8

Damping occurs in structures due to energy loss mechanisms that exist in the system.

Examples are friction losses at any connection to or in the system and internal energy

losses of the materials due to thermo-elasticity, hysteresis and inter-granular bonds.

The exact nature of damping is difficult to define; fortunately theoretical damping has

been shown to match real structures quite well.

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 2 4 6 8 10 12 14 16 18 20

Dis

pla

cem

en

t (m

m)

Time (s)

Undamped

Damped

m = 10

k = 100


Dr. C. Caprani 11

5.2 Single Degree-of-Freedom Systems

5.2.1 Fundamental Equation of Motion

(a) (b)

Figure 2.1: (a) SDOF system. (b) Free-body diagram of forces

Considering Figure 2.1, the forces resisting the applied loading are considered as:

a force proportional to displacement (the usual static stiffness);

a force proportional to velocity (the damping force);

a force proportional to acceleration (D‟Alambert‟s inertial force).

We can write the following symbolic equation:

applied stiffness damping inertia

F F F F (5.2.1)

Noting that:

stiffness

damping

inertia

F

F

F

ku

cu

mu

(5.2.2)

that is, stiffness × displacement, damping coefficient × velocity and mass ×

acceleration respectively. Note also that u represents displacement from the

equilibrium position and that the dots over u represent the first and second derivatives

m

k

u(t)

c


Dr. C. Caprani 12

with respect to time. Thus, noting that the displacement, velocity and acceleration are

all functions of time, we have the Fundamental Equation of Motion:

mu t cu t ku t F t (5.2.3)

In the case of free vibration, there is no forcing function and so 0F t which gives

equation (5.2.3) as:

0mu t cu t ku t (5.2.4)

We note also that the system will have a state of initial conditions:

00u u (5.2.5)

00u u (5.2.6)

In equation (5.2.4), dividing across by m gives:

( ) ( ) ( ) 0c k

u t u t u tm m

(5.2.7)

We introduce the following notation:

cr

c

c (5.2.8)

2 k

m (5.2.9)


Dr. C. Caprani 13

Or equally,

k

m (5.2.10)

In which

is called the undamped circular natural frequency and its units are radians per

second (rad/s);

is the damping ratio which is the ratio of the damping coefficient, c, to the

critical value of the damping coefficient cr

c .

We will see what these terms physically mean. Also, we will later see (equation

(5.2.18)) that:

2 2cr

c m km (5.2.11)

Equations (5.2.8) and (5.2.11) show us that:

2c

m (5.2.12)

When equations (5.2.9) and (5.2.12) are introduced into equation (5.2.7), we get the

prototype SDOF equation of motion:

22 0u t u t u t (5.2.13)

In considering free vibration only, the general solution to (5.2.13) is of a form

tu Ce (5.2.14)


Dr. C. Caprani 14

When we substitute (5.2.14) and its derivates into (5.2.13) we get:

2 22 0tCe (5.2.15)

For this to be valid for all values of t, tCe cannot be zero. Thus we get the

characteristic equation:

2 22 0 (5.2.16)

the solutions to this equation are the two roots:

2 2 2

1,2

2

2 4 4

2

1

(5.2.17)

Therefore the solution depends on the magnitude of relative to 1. We have:

1 : Sub-critical damping or under-damped;

Oscillatory response only occurs when this is the case – as it is for almost all

structures.

1 : Critical damping;

No oscillatory response occurs.

1 : Super-critical damping or over-damped;

No oscillatory response occurs.

Therefore, when 1 , the coefficient of ( )u t in equation (5.2.13) is, by definition,

the critical damping coefficient. Thus, from equation (5.2.12):


Dr. C. Caprani 15

2 crc

m (5.2.18)

From which we get equation (5.2.11).


Dr. C. Caprani 16

5.2.2 Free Vibration of Undamped Structures

We will examine the case when there is no damping on the SDOF system of Figure

2.1 so 0 in equations (5.2.13), (5.2.16) and (5.2.17) which then become:

2 0u t u t (5.2.19)

respectively, where 1i . From the Appendix we see that the general solution to

this equation is:

cos sinu t A t B t (5.2.20)

where A and B are constants to be obtained from the initial conditions of the system,

equations (5.2.5) and (5.2.6). Thus, at 0t , from equation (5.2.20):

00 cos 0 sin 0u A B u

0

A u (5.2.21)

From equation (5.2.20):

sin cosu t A t B t (5.2.22)

And so:

0

0

0 sin 0 cos 0u A B u

B u

0u

B

(5.2.23)


Dr. C. Caprani 17

Thus equation (5.2.20), after the introduction of equations (5.2.21) and (5.2.23),

becomes:

0

0cos sin

uu t u t t

(5.2.24)

where 0

u and 0

u are the initial displacement and velocity of the system respectively.

Noting that cosine and sine are functions that repeat with period 2 , we see that

1 12t T t (Figure 2.3) and so the undamped natural period of the SDOF

system is:

2

T

(5.2.25)

The natural frequency of the system is got from (1.1), (5.2.25) and (5.2.9):

1 1

2 2

kf

T m

(5.2.26)

and so we have proved (1.2). The importance of this equation is that it shows the

natural frequency of structures to be proportional to km

. This knowledge can aid a

designer in addressing problems with resonance in structures: by changing the

stiffness or mass of the structure, problems with dynamic behaviour can be

addressed.


Dr. C. Caprani 18

Figure 2.2: SDOF free vibration response for (a) 0

20mmu , 0

0u , (b) 0

0u ,

050mm/su , and (c)

020mmu ,

050mm/su .

Figure 2.2 shows the free-vibration response of a spring-mass system for various

initial states of the system. It can be seen from (b) and (c) that when 0

0u the

amplitude of displacement is not that of the initial displacement; this is obviously an

important characteristic to calculate. The cosine addition rule may also be used to

show that equation (5.2.20) can be written in the form:

( ) cosu t C t (5.2.27)

where 2 2C A B and tan BA

. Using A and B as calculated earlier for the

initial conditions, we then have:

( ) cosu t t (5.2.28)

-30

-20

-10

0

10

20

30

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

pla

cem

en

t (m

m)

Time (s)

(a)

(b)

(c)

m = 10

k = 100


Dr. C. Caprani 19

where is the amplitude of displacement and is the phase angle, both given by:

2

2 0

0

uu

(5.2.29)

0

0

tanu

u

(5.2.30)

The phase angle determines the amount by which ( )u t lags behind the function

cos t . Figure 2.3 shows the general case.

Figure 2.3 Undamped free-vibration response.


Dr. C. Caprani 20

5.2.3 Computer Implementation & Examples

Using MS Excel

To illustrate an application we give the spreadsheet used to generate Figure 1.3. This

can be downloaded from the course website.

http://www.colincaprani.com/structural-engineering/courses/structural-analysis-iv/


Dr. C. Caprani 21

The input parameters (shown in red) are:

m – the mass;

k – the stiffness;

delta_t – the time step used in the response plot;

u_0 – the initial displacement, 0

u ;

v_0 – the initial velocity, 0

u .

The properties of the system are then found:

w, using equation (5.2.10);

f, using equation (5.2.26);

T, using equation (5.2.26);

, using equation (5.2.29);

, using equation (5.2.30).

A column vector of times is dragged down, adding delta_t to each previous time

value, and equation (5.2.24) (“Direct Eqn”), and equation (5.2.28) (“Cosine Eqn”) is

used to calculate the response, u t , at each time value. Then the column of u-values

is plotted against the column of t-values to get the plot.


Dr. C. Caprani 22

Using Matlab

Although MS Excel is very helpful since it provides direct access to the numbers in

each equation, as more concepts are introduced, we will need to use loops and create

regularly-used functions. Matlab is ideally suited to these tasks, and so we will begin

to use it also on the simple problems as a means to its introduction.

A script to directly generate Figure 1.3, and calculate the system properties is given

below:

% Script to plot the undamped response of a single degree of freedom system % and to calculate its properties

k = 100; % N/m - stiffness m = 10; % kg - mass delta_t = 0.1; % s - time step u0 = 0.025; % m - initial displacement v0 = 0; % m/s - initial velocity

w = sqrt(k/m); % rad/s - circular natural frequency f = w/(2*pi); % Hz - natural frequency T = 1/f; % s - natural period ro = sqrt(u0^2+(v0/w)^2); % m - amplitude of vibration theta = atan(v0/(u0*w)); % rad - phase angle

t = 0:delta_t:4; u = ro*cos(w*t-theta); plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)');

The results of this script are the system properties are displayed in the workspace

window, and the plot is generated, as shown below:


Dr. C. Caprani 23

Whilst this is quite useful, this script is limited to calculating the particular system of

Figure 1.3. Instead, if we create a function that we can pass particular system

properties to, then we can create this plot for any system we need to. The following

function does this.

Note that we do not calculate f or T since they are not needed to plot the response.

Also note that we have commented the code very well, so it is easier to follow and

understand when we come back to it at a later date.


Dr. C. Caprani 24

function [t u] = sdof_undamped(m,k,u0,v0,duration,plotflag) % This function returns the displacement of an undamped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % u0 - initial displacement, m % v0 - initial velocity, m/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response

Npts = 1000; % compute the response at 1000 points delta_t = duration/(Npts-1);

w = sqrt(k/m); % rad/s - circular natural frequency ro = sqrt(u0^2+(v0/w)^2); % m - amplitude of vibration theta = atan(v0/(u0*w)); % rad - phase angle

t = 0:delta_t:duration; u = ro*cos(w*t-theta);

if(plotflag == 1) plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); end

To execute this function and replicate Figure 1.3, we call the following:

[t u] = sdof_undamped(10,100,0.025,0,4,1);

And get the same plot as before. Now though, we can really benefit from the

function. Let‟s see the effect of an initial velocity on the response, try +0.1 m/s:

[t u] = sdof_undamped(10,100,0.025,0.1,4,1);

Note the argument to the function in bold – this is the +0.1 m/s initial velocity. And

from this call we get the following plot:


Dr. C. Caprani 25

From which we can see that the maximum response is now about 40 mm, rather than

the original 25.

Download the function from the course website and try some other values.

0 0.5 1 1.5 2 2.5 3 3.5 4-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

Time (s)

Dis

pla

cem

ent

(m)



Dr. C. Caprani 26

5.2.4 Free Vibration of Damped Structures

Figure 2.4: Response with critical or super-critical damping

When taking account of damping, we noted previously that there are 3, cases but only

when 1 does an oscillatory response ensue. We will not examine the critical or

super-critical cases. Examples are shown in Figure 2.4.

To begin, when 1 (5.2.17) becomes:

1,2 d

i (5.2.31)

where d

is the damped circular natural frequency given by:

21

d (5.2.32)

which has a corresponding damped period and frequency of:


Dr. C. Caprani 27

2

d

d

T

(5.2.33)

2

d

df

(5.2.34)

The general solution to equation (5.2.14), using Euler‟s formula again, becomes:

( ) cos sint

d du t e A t B t (5.2.35)

and again using the initial conditions we get:

0 0

0( ) cos sint d

d d

d

u uu t e u t t

(5.2.36)

Using the cosine addition rule again we also have:

( ) cost

du t e t (5.2.37)

In which

2

2 0 0

0

d

u uu

(5.2.38)

0 0

0

tand

u u

u

(5.2.39)

Equations (5.2.35) to (5.2.39) correspond to those of the undamped case looked at

previously when 0 .


Dr. C. Caprani 28

Figure 2.5: SDOF free vibration response for:

(a) 0 ; (b) 0.05 ; (c) 0.1 ; and (d) 0.5 .

Figure 2.5 shows the dynamic response of the SDOF model shown. It may be clearly

seen that damping has a large effect on the dynamic response of the system – even for

small values of . We will discuss damping in structures later but damping ratios for

structures are usually in the range 0.5 to 5%. Thus, the damped and undamped

properties of the systems are very similar for these structures.

Figure 2.6 shows the general case of an under-critically damped system.

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 0.5 1 1.5 2 2.5 3 3.5 4

Dis

pla

cem

en

t (m

m)

Time (s)

(a)

(b)

(c)

(d)

m = 10

k = 100

varies


Dr. C. Caprani 29

Figure 2.6: General case of an under-critically damped system.


Dr. C. Caprani 30


Using MS Excel

We can just modify our previous spreadsheet to take account of the revised equations

for the amplitude (equation (5.2.38)), phase angle (equation (5.2.39)) and response

(equation (5.2.37)), as well as the damped properties, to get:


Dr. C. Caprani 31

Using Matlab

Now can just alter our previous function and take account of the revised equations for

the amplitude (equation (5.2.38)), phase angle (equation (5.2.39)) and response

(equation (5.2.37)) to get the following function. This function will (of course) also

work for undamped systems where 0 .

function [t u] = sdof_damped(m,k,xi,u0,v0,duration,plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % xi - damping ratio % u0 - initial displacement, m % v0 - initial velocity, m/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response


w = sqrt(k/m); % rad/s - circular natural frequency wd = w*sqrt(1-xi^2); % rad/s - damped circular frequency ro = sqrt(u0^2+((v0+xi*w*u0)/wd)^2); % m - amplitude of vibration theta = atan((v0+u0*xi*w)/(u0*w)); % rad - phase angle

t = 0:delta_t:duration; u = ro*exp(-xi*w.*t).*cos(w*t-theta);

if(plotflag == 1) plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); end

Let‟s apply this to our simple example again, for 0.1 :

[t u] = sdof_damped(10,100,0.1,0.025,0,4,1);

To get:


Dr. C. Caprani 32

To plot Figure 2.5, we just call out function several times (without plotting it each

time), save the response results and then plot all together:

xi = [0,0.05,0.1,0.5]; for i = 1:length(xi) [t u(i,:)] = sdof_damped(10,100,xi(i),0.025,0,4,0); end plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); legend('Damping: 0%','Damping: 5%','Damping: 10%','Damping: 50%');

0 0.5 1 1.5 2 2.5 3 3.5 4-0.02

-0.01

0

0.01

0.02

0.03

Time (s)

Dis

pla

cem

ent

(m)

0 0.5 1 1.5 2 2.5 3 3.5 4-0.03

-0.02

-0.01

0

0.01

0.02

0.03

Time (s)

Dis

pla

cem

ent

(m)

Damping: 0%

Damping: 5%

Damping: 10%

Damping: 50%


Dr. C. Caprani 33

5.2.6 Estimating Damping in Structures

Examining Figure 2.6, we see that two successive peaks, n

u and n m

u

, m cycles apart,

occur at times nT and n m T respectively. Using equation (5.2.37) we can get the

ratio of these two peaks as:

2

expn

n m d

u m

u

(5.2.40)

where exp xx e . Taking the natural log of both sides we get the logarithmic

decrement of damping, , defined as:

ln 2n

n m d

um

u

(5.2.41)

for low values of damping, normal in structural engineering, we can approximate

this:

2m (5.2.42)

thus,

exp 2 1 2n

n m

ue m m

u

(5.2.43)

and so,


Dr. C. Caprani 34

2

n n m

n m

u u

m u

(5.2.44)

This equation can be used to estimate damping in structures with light damping (

0.2 ) when the amplitudes of peaks m cycles apart is known. A quick way of

doing this, known as the Half-Amplitude Method, is to count the number of peaks it

takes to halve the amplitude, that is 0.5n m n

u u . Then, using (5.2.44) we get:

0.11

m when 0.5

n m nu u

(5.2.45)

Further, if we know the amplitudes of two successive cycles (and so 1m ), we can

find the amplitude after p cycles from two instances of equation (5.2.43):

1

p

n

n p n

n

uu u

u

(5.2.46)


Dr. C. Caprani 35

5.2.7 Response of an SDOF System Subject to Harmonic Force

Figure 2.7: SDOF undamped system subjected to harmonic excitation

So far we have only considered free vibration; the structure has been set vibrating by

an initial displacement for example. We will now consider the case when a time

varying load is applied to the system. We will confine ourselves to the case of

harmonic or sinusoidal loading though there are obviously infinitely many forms that

a time-varying load may take – refer to the references (Appendix) for more.

To begin, we note that the forcing function F t has excitation amplitude of 0

F and

an excitation circular frequency of and so from the fundamental equation of

motion (5.2.3) we have:

0

( ) ( ) ( ) sinmu t cu t ku t F t (5.2.47)

The solution to equation (5.2.47) has two parts:

The complementary solution, similar to (5.2.35), which represents the transient

response of the system which damps out by exp t . The transient response

may be thought of as the vibrations caused by the initial application of the load.

The particular solution, pu t , representing the steady-state harmonic response of

the system to the applied load. This is the response we will be interested in as it

will account for any resonance between the forcing function and the system.

m

k u(t)

c


Dr. C. Caprani 36

The complementary solution to equation (5.2.47) is simply that of the damped free

vibration case studied previously. The particular solution to equation (5.2.47) is

developed in the Appendix and shown to be:

sinp

u t t (5.2.48)

In which

1 2

2 220 1 2F

k

(5.2.49)

2

2tan

1

(5.2.50)

where the phase angle is limited to 0 and the ratio of the applied load

frequency to the natural undamped frequency is:

(5.2.51)

the maximum response of the system will come at sin 1t and dividing

(5.2.48) by the static deflection 0

F k we can get the dynamic amplification factor

(DAF) of the system as:

1 2

2 22DAF 1 2D

(5.2.52)

At resonance, when , we then have:


Dr. C. Caprani 37

1

1

2D

(5.2.53)

Figure 2.8 shows the effect of the frequency ratio on the DAF. Resonance is the

phenomenon that occurs when the forcing frequency coincides with that of the

natural frequency, 1 . It can also be seen that for low values of damping, normal

in structures, very high DAFs occur; for example if 0.02 then the dynamic

amplification factor will be 25. For the case of no damping, the DAF goes to infinity

- theoretically at least; equation (5.2.53).

Figure 2.8: Variation of DAF with damping and frequency ratios.


Dr. C. Caprani 38

The phase angle also helps us understand what is occurring. Plotting equation

(5.2.50) against for a range of damping ratios shows:

Figure 2.9: Variation of phase angle with damping and frequency ratios.

Looking at this then we can see three regions:

1 : the force is slowly varying and is close to zero. This means that the

response (i.e. displacement) is in phase with the force: for example, when the

force acts to the right, the system displaces to the right.

1 : the force is rapidly varying and is close to 180°. This means that the

force is out of phase with the system: for example, when the force acts to the

right, the system is displacing to the left.

1 : the forcing frequency is equal to the natural frequency, we have

resonance and 90 . Thus the displacement attains its peak as the force is

zero.

0 0.5 1 1.5 2 2.5 30

45

90

135

180

Frequency Ratio

Phase A

ngle

(degre

es)

Damping: 0%

Damping: 10%

Damping: 20%

Damping: 50%

Damping: 100%


Dr. C. Caprani 39

We can see these phenomena by plotting the response and forcing fun(5.2.54)ction

together (though with normalized displacements for ease of interpretation), for

different values of . In this example we have used 0.2 . Also, the three phase

angles are 2 0.04, 0.25, 0.46 respectively.

Figure 2.10: Steady-state responses to illustrate phase angle.

Note how the force and response are firstly “in sync” ( ~ 0 ), then “halfway out of

sync” ( 90 ) at resonance; and finally, “fully out of sync” ( ~180 ) at high

frequency ratio.

0 0.5 1 1.5 2 2.5 3

-2

0

2

Dis

p.

Ratio

0 0.5 1 1.5 2 2.5 3

-2

0

2

Dis

p.

Ratio

0 0.5 1 1.5 2 2.5 3

-2

0

2

Dis

p.

Ratio

Time Ratio (t/T)

Dynamic Response

Static Response = 0.5; DAF = 1.29

= 0.5; DAF = 2.5

= 2.0; DAF = 0.32


Dr. C. Caprani 40

Maximum Steady-State Displacement

The maximum steady-state displacement occurs when the DAF is a maximum. This

occurs when the denominator of equation (5.2.52) is a minimum:

1 22 22

2 2

2 22

2 2

0

1 2 0

4 1 4 210

2 1 2

1 2 0

d D

d

d

d

The trivial solution to this equation of 0 corresponds to an applied forcing

function that has zero frequency –the static loading effect of the forcing function. The

other solution is:

21 2 (5.2.54)

Which for low values of damping, 0.1 approximately, is very close to unity. The

corresponding maximum DAF is then given by substituting (5.2.54) into equation

(5.2.52) to get:

max 2

1

2 1D

(5.2.55)

Which reduces to equation (5.2.53) for 1 , as it should.


Dr. C. Caprani 41

Measurement of Natural Frequencies

It may be seen from equation (5.2.50) that when 1 , 2 ; this phase

relationship allows the accurate measurements of the natural frequencies of

structures. That is, we change the input frequency in small increments until we can

identify a peak response: the value of at the peak response is then the natural

frequency of the system. Example 2.1 gave the natural frequency based on this type

of test.


Dr. C. Caprani 42


Using MS Excel

Again we modify our previous spreadsheet and include the extra parameters related

to forced response. We‟ve also used some of the equations from the Appendix to

show the transient, steady-sate and total response. Normally however, we are only

interested in the steady-state response, which the total response approaches over time.


Dr. C. Caprani 43

Using Matlab

First let‟s write a little function to return the DAF, since we will use it often:

function D = DAF(beta,xi) % This function returns the DAF, D, associated with the parameters: % beta - the frequency ratio % xi - the damping ratio

D = 1./sqrt((1-beta.^2).^2+(2*xi.*beta).^2);

And another to return the phase angle (always in the region 0 ):

function theta = phase(beta,xi) % This function returns the pahse angle, theta, associated with the % parameters: % beta - the frequency ratio % xi - the damping ratio

theta = atan2((2*xi.*beta),(1-beta.^2)); % refers to complex plane

With these functions, and modifying our previous damped response script, we have:

function [t u] = sdof_forced(m,k,xi,u0,v0,F,Omega,duration,plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % xi - damping ratio % u0 - initial displacement, m % v0 - initial velocity, m/s % F - amplitude of forcing function, N % Omega - frequency of forcing function, rad/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response


w = sqrt(k/m); % rad/s - circular natural frequency wd = w*sqrt(1-xi^2); % rad/s - damped circular frequency

beta = Omega/w; % frequency ratio D = DAF(beta,xi); % dynamic amplification factor ro = F/k*D; % m - amplitude of vibration theta = phase(beta,xi); % rad - phase angle


Dr. C. Caprani 44

% Constants for the transient response Aconst = u0+ro*sin(theta); Bconst = (v0+u0*xi*w-ro*(Omega*cos(theta)-xi*w*sin(theta)))/wd;

t = 0:delta_t:duration; u_transient = exp(-xi*w.*t).*(Aconst*cos(wd*t)+Bconst*sin(wd*t)); u_steady = ro*sin(Omega*t-theta); u = u_transient + u_steady;

if(plotflag == 1) plot(t,u,'k'); hold on; plot(t,u_transient,'k:'); plot(t,u_steady,'k--'); hold off; xlabel('Time (s)'); ylabel('Displacement (m)'); legend('Total Response','Transient','Steady-State'); end

Running this for the same problem as before with 0

10 NF and 15 rad/s gives:

[t u] = sdof_forced(10,100,0.1,0.025,0,20,15,6,1);

As can be seen, the total response quickly approaches the steady-state response.

0 1 2 3 4 5 6-0.06

-0.04

-0.02

0

0.02

0.04

0.06

Time (s)

Dis

pla

cem

ent

(m)

Total Response

Transient

Steady-State


Dr. C. Caprani 45

Next let‟s use our little DAF function to plot something similar to Figure 2.8, but this

time showing the frequency ratio and maximum response from equation (5.2.54):

% Script to plot DAF against Beta for different damping ratios xi = [0.0001,0.1,0.15,0.2,0.3,0.4,0.5,1.0]; beta = 0.01:0.01:3; for i = 1:length(xi) D(i,:) = DAF(beta,xi(i)); end % A new xi vector for the maxima line xi = 0:0.01:1.0; xi(end) = 0.99999; % very close to unity xi(1) = 0.00001; % very close to zero for i = 1:length(xi) betamax(i) = sqrt(1-2*xi(i)^2); Dmax(i) = DAF(betamax(i),xi(i)); end plot(beta,D); hold on; plot(betamax,Dmax,'k--'); xlabel('Frequency Ratio'); ylabel('Dynamic Amplification'); ylim([0 6]); % set y-axis limits since DAF at xi = 0 is enormous legend( 'Damping: 0%','Damping: 10%','Damping: 15%',... 'Damping: 20%','Damping: 30%','Damping: 40%',... 'Damping: 50%', 'Damping: 100%', 'Maxima');

This gives:

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

Frequency Ratio

Dynam

ic A

mplif

ication

Damping: 0%

Damping: 10%

Damping: 15%

Damping: 20%

Damping: 30%

Damping: 40%

Damping: 50%

Damping: 100%

Maxima


Dr. C. Caprani 46

Lastly then, using the phase function we wrote, we can generate Figure 2.9:

% Script to plot phase against Beta for different damping ratios xi = [0.0001,0.1,0.2,0.5,1.0]; beta = 0.01:0.01:3; for i = 1:length(xi) T(i,:) = phase(beta,xi(i))*(180/pi); % in degrees end plot(beta,T); xlabel('Frequency Ratio'); ylabel('Phase Angle (degrees)'); ylim([0 180]); set(gca,'ytick',[0 45 90 135 180]); grid on; legend('Damping: 0%','Damping: 10%','Damping: 20%','Damping: 50%',... 'Damping: 100%','Location','SE');


Dr. C. Caprani 47

5.2.9 Numerical Integration – Newmark’s Method

Introduction

The loading that can be applied to a structure is infinitely variable and closed-form

mathematical solutions can only be achieved for a small number of cases. For

arbitrary excitation we must resort to computational methods, which aim to solve the

basic structural dynamics equation, at the next time-step:

1 1 1 1i i i i

mu cu ku F (5.2.56)

There are three basic time-stepping approaches to the solution of the structural

dynamics equations:

1. Interpolation of the excitation function;

2. Use of finite differences of velocity and acceleration;

3. An assumed variation of acceleration.

We will examine one method from the third category only. However, it is an

important method and is extensible to non-linear systems, as well as multi degree-of-

freedom systems (MDOF).


Dr. C. Caprani 48

Development of Newmark’s Method

In 1959 Newmark proposed a general assumed variation of acceleration method:

1 11

i i i iu u t u t u

(5.2.57)

2 2

1 10.5

i i i i iu u t u t u t u

(5.2.58)

The parameters and define how the acceleration is assumed over the time step,

t . Usual values are 1

2 and

1 1

6 4 . For example:

Constant (average) acceleration is given by: 1

2 and

1

4 ;

Linear variation of acceleration is given by: 1

2 and

1

6 .

The three equations presented thus far (equations (5.2.56), (5.2.57) and (5.2.58)) are

sufficient to solve for the three unknown responses at each time step. However to

avoid iteration, we introduce the incremental form of the equations:

1i i i

u u u

(5.2.59)

1i i i

u u u

(5.2.60)

1i i i

u u u

(5.2.61)

1i i i

F F F

(5.2.62)

Thus, Newmark‟s equations can now be written as:

i i iu t u t u (5.2.63)


Dr. C. Caprani 49

2

2

2i i i i

tu t u u t u

(5.2.64)

Solving equation (5.2.64) for the unknown change in acceleration gives:

2

1 1 1

2i i i i

u u u utt

(5.2.65)

Substituting this into equation (5.2.63) and solving for the unknown increment in

velocity gives:

12

i i i iu u u t u

t

(5.2.66)

Next we use the incremental equation of motion, derived from equation (5.2.56):

i i i i

m u c u k u F (5.2.67)

And introduce equations (5.2.65) and (5.2.66) to get:

2

1 1 1

2

12

i i i

i i i i i

m u u utt

c u u t u k u Ft

(5.2.68)

Collecting terms gives:


Dr. C. Caprani 50

2

1

1 11

2 2

i

i i i

m c k utt

F m c u m t c ut

(5.2.69)

Let‟s introduce the following for ease of representation:

2

1k̂ m c k

tt

(5.2.70)

1 1ˆ 1

2 2i i i i

F F m c u m t c ut

(5.2.71)

Which are an effective stiffness and effective force at time i. Thus equation (5.2.69)

becomes:

ˆ î i

k u F (5.2.72)

Since k̂ and î

F are known from the system properties (m, c, k); the algorithm

properties ( , , t ); and the previous time-step (i

u , i

u ), we can solve equation

(5.2.72) for the displacement increment:

ˆ

î

i

Fu

k

(5.2.73)

Once the displacement increment is known, we can solve for the velocity and

acceleration increments from equations (5.2.66) and (5.2.65) respectively. And once


Dr. C. Caprani 51

all the increments are known we can compute the properties at the current time-step

by just adding to the values at the previous time-step, equations (5.2.59) to (5.2.61).

Newmark‟s method is stable if the time-steps is about 0.1t T of the system.

The coefficients in equation (5.2.71) are constant (once t is), so we can calculate

these at the start as:

1

A m ct

(5.2.74)

1

12 2

B m t c

(5.2.75)

Making equation (5.2.71) become:

î i i i

F F Au Bu (5.2.76)


Dr. C. Caprani 52

Newmark’s Algorithm

1. Select algorithm parameters, , and t ;

2. Initial calculations:

a. Find the initial acceleration:

0 0 0 0

1u F cu ku

m (5.2.77)

b. Calculate the effective stiffness, k̂ from equation (5.2.70);

c. Calculate the coefficients for equation (5.2.71) from equations (5.2.74) and

(5.2.75).

3. For each time step, i, calculate:

î i i i

F F Au Bu (5.2.78)

ˆ

î

i

Fu

k

(5.2.79)

12

i i i iu u u t u

t

(5.2.80)

2

1 1 1

2i i i i

u u u utt

(5.2.81)

1i i i

u u u

(5.2.82)

1i i i

u u u

(5.2.83)

1i i i

u u u

(5.2.84)


Dr. C. Caprani 53


Using MS Excel

Based on our previous spreadsheet, we implement Newmark Integration. Download it

from the course website, and see how the equations and algorithm are implemented.

In the example shown, we‟ve applied a sinusoidal load of 10 N for 0.6 secs to the

system we‟ve been using so far:



Dr. C. Caprani 54

Using Matlab

There are no shortcuts to this one. We must write a completely new function that

implements the Newmark Integration algorithm as we‟ve described it:

function [u ud udd] = newmark_sdof(m, k, xi, t, F, u0, ud0, plotflag) % This function computes the response of a linear damped SDOF system % subject to an arbitrary excitation. The input parameters are: % m - scalar, mass, kg % k - scalar, stiffness, N/m % xi - scalar, damping ratio % t - vector of length N, in equal time steps, s % F - vector of length N, force at each time step, N % u0 - scalar, initial displacement, m % v0 - scalar, initial velocity, m/s % plotflag - 1 or 0: whether or not to plot the response % The output is: % u - vector of length N, displacement response, m % ud - vector of length N, velocity response, m/s % udd - vector of length N, acceleration response, m/s2

% Set the Newmark Integration parameters % gamma = 1/2 always % beta = 1/6 linear acceleration % beta = 1/4 average acceleration gamma = 1/2; beta = 1/6;

N = length(t); % the number of integration steps dt = t(2)-t(1); % the time step w = sqrt(k/m); % rad/s - circular natural frequency c = 2*xi*k/w; % the damping coefficient

% Calulate the effective stiffness keff = k + (gamma/(beta*dt))*c+(1/(beta*dt^2))*m; % Calulate the coefficients A and B Acoeff = (1/(beta*dt))*m+(gamma/beta)*c; Bcoeff = (1/(2*beta))*m + dt*(gamma/(2*beta)-1)*c;

% calulate the change in force at each time step dF = diff(F);

% Set initial state u(1) = u0; ud(1) = ud0; udd(1) = (F(1)-c*ud0-k*u0)/m; % the initial acceleration

for i = 1:(N-1) % N-1 since we already know solution at i = 1 dFeff = dF(i) + Acoeff*ud(i) + Bcoeff*udd(i); dui = dFeff/keff; dudi = (gamma/(beta*dt))*dui-(gamma/beta)*ud(i)+dt*(1-

gamma/(2*beta))*udd(i); duddi = (1/(beta*dt^2))*dui-(1/(beta*dt))*ud(i)-(1/(2*beta))*udd(i); u(i+1) = u(i) + dui; ud(i+1) = ud(i) + dudi;


Dr. C. Caprani 55

udd(i+1) = udd(i) + duddi; end

if(plotflag == 1) subplot(4,1,1) plot(t,F,'k'); xlabel('Time (s)'); ylabel('Force (N)'); subplot(4,1,2) plot(t,u,'k'); xlabel('Time (s)'); ylabel('Displacement (m)'); subplot(4,1,3) plot(t,ud,'k'); xlabel('Time (s)'); ylabel('Velocity (m/s)'); subplot(4,1,4) plot(t,udd,'k'); xlabel('Time (s)'); ylabel('Acceleration (m/s2)'); end

Bear in mind that most of this script is either comments or plotting commands –

Newmark Integration is a fast and small algorithm, with a huge range of applications.

In order to use this function, we must write a small script that sets the problem up and

then calls the newmark_sdof function. The main difficulty is in generating the

forcing function, but it is not that hard:

% script that calls Newmark Integration for sample problem m = 10; k = 100; xi = 0.1; u0 = 0; ud0 = 0; t = 0:0.1:4.0; % set the time vector F = zeros(1,length(t)); % empty F vector % set sinusoidal force of 10 over 0.6 s Famp = 10; Tend = 0.6; i = 1; while t(i) < Tend F(i) = Famp*sin(pi*t(i)/Tend); i = i+1; end

[u ud udd] = newmark_sdof(m, k, xi, t, F, u0, ud0, 1);

This produces the following plot:


Dr. C. Caprani 56

Explosions are often modelled as triangular loadings. Let‟s implement this for our

system:

0 0.5 1 1.5 2 2.5 3 3.5 4-5

0

5

10

Time (s)

Forc

e (

N)

0 0.5 1 1.5 2 2.5 3 3.5 4-0.1

0

0.1

Time (s)

Dis

pla

cem

ent

(m)

0 0.5 1 1.5 2 2.5 3 3.5 4-0.5

0

0.5

Time (s)

Velo

city (

m/s

)

0 0.5 1 1.5 2 2.5 3 3.5 4-1

0

1

Time (s)

Accele

ration (

m/s

2)


Dr. C. Caprani 57

% script that finds explosion response m = 10; k = 100; xi = 0.1; u0 = 0; ud0 = 0; Fmax = 50; % N Tend = 0.2; % s t = 0:0.01:2.0; % set the time vector F = zeros(1,length(t)); % empty F vector % set reducing triangular force i = 1; while t(i) < Tend F(i) = Fmax*(1-t(i)/Tend); i = i+1; end

[u ud udd] = newmark_sdof(m, k, xi, t, F, u0, ud0, 1);

As can be seen from the following plot, even though the explosion only lasts for a

brief period of time, the vibrations will take several periods to dampen out. Also

notice that the acceleration response is the most sensitive – this is the most damaging

to the building, as force is mass times acceleration: the structure thus undergoes

massive forces, possibly leading to damage or failure.


Dr. C. Caprani 58

0 0.5 1 1.5 2 2.5 3 3.5 40

20

40

60

Time (s)

Forc

e (

N)

0 0.5 1 1.5 2 2.5 3 3.5 4-0.2

0

0.2

Time (s)

Dis

pla

cem

ent

(m)

0 0.5 1 1.5 2 2.5 3 3.5 4-0.5

0

0.5

Time (s)

Velo

city (

m/s

)

0 0.5 1 1.5 2 2.5 3 3.5 4-5

0

5

Time (s)

Accele

ration (

m/s

2)


Dr. C. Caprani 59

5.2.11 Problems

Problem 1

A harmonic oscillation test gave the natural frequency of a water tower to be 0.41 Hz.

Given that the mass of the tank is 150 tonnes, what deflection will result if a 50 kN

horizontal load is applied? You may neglect the mass of the tower.

Ans: 50.2 mm

Problem 2

A 3 m high, 8 m wide single-bay single-storey frame is rigidly jointed with a beam of

mass 5,000 kg and columns of negligible mass and stiffness of EIc = 4.5×103 kNm

2.

Calculate the natural frequency in lateral vibration and its period. Find the force

required to deflect the frame 25 mm laterally.

Ans: 4.502 Hz; 0.222 sec; 100 kN

Problem 3

An SDOF system (m = 20 kg, k = 350 N/m) is given an initial displacement of 10 mm

and initial velocity of 100 mm/s. (a) Find the natural frequency; (b) the period of

vibration; (c) the amplitude of vibration; and (d) the time at which the third maximum

peak occurs.

Ans: 0.666 Hz; 1.502 sec; 25.91 mm; 3.285 sec.

Problem 4

For the frame of Problem 2, a jack applied a load of 100 kN and then instantaneously

released. On the first return swing a deflection of 19.44 mm was noted. The period of

motion was measured at 0.223 sec. Assuming that the stiffness of the columns cannot


Dr. C. Caprani 60

change, find (a) the damping ratio; (b) the coefficient of damping; (c) the undamped

frequency and period; and (d) the amplitude after 5 cycles.

Ans: 0.04; 11,367 kg·s/m; 4.488 Hz; 0.2228 sec; 7.11 mm.

Problem 5

From the response time-history of an SDOF system given:


Dr. C. Caprani 61

(a) estimate the damped natural frequency; (b) use the half amplitude method to

calculate the damping ratio; and (c) calculate the undamped natural frequency and

period.

Ans: 4.021 Hz; 0.05; 4.026 Hz; 0.248 sec.

Problem 6

Workers‟ movements on a platform (8 × 6 m high, m = 200 kN) are causing large

dynamic motions. An engineer investigated and found the natural period in sway to

be 0.9 sec. Diagonal remedial ties (E = 200 kN/mm2) are to be installed to reduce the

natural period to 0.3 sec. What tie diameter is required?

Ans: 28.1 mm.

Problem 7

The frame of examples 2.2 and 2.4 has a reciprocating machine put on it. The mass of

this machine is 4 tonnes and is in addition to the mass of the beam. The machine

exerts a periodic force of 8.5 kN at a frequency of 1.75 Hz. (a) What is the steady-

state amplitude of vibration if the damping ratio is 4%? (b) What would the steady-

state amplitude be if the forcing frequency was in resonance with the structure?

Ans: 2.92 mm; 26.56 mm.

Problem 8

An air conditioning unit of mass 1,600 kg is place in the middle (point C) of an 8 m

long simply supported beam (EI = 8×103 kNm

2) of negligible mass. The motor runs

at 300 rpm and produces an unbalanced load of 120 kg. Assuming a damping ratio of

5%, determine the steady-state amplitude and deflection at C. What rpm will result in

resonance and what is the associated deflection?

Ans: 1.41 mm; 22.34 mm; 206.7 rpm; 36.66 mm.


Dr. C. Caprani 62

Problem 9

Determine the response of our example system, with initial velocity of 0.05 m/s,

when acted upon by an impulse of 0.1 s duration and magnitude 10 N at time 1.0 s.

Do this up for a duration of 4 s.

Ans. below

Problem 10

Determine the maximum responses of a water tower which is subjected to a

sinusoidal force of amplitude 445 kN and frequency 30 rad/s over 0.3 secs. The

tower has properties, mass 17.5 t, stiffness 17.5 MN/m and no damping.

Ans. 120 mm, 3.8 m/s, 120.7 m/s2

Problem 11

Determine the maximum response of a system (m = 1.75 t, k = 1.75 MN/m, = 10%)

when subjected to an increasing triangular load which reaches 22.2 kN after 0.1 s.

Ans. 14.6 mm, 0.39 m/s, 15.0 m/s2

0 0.5 1 1.5 2 2.5 3 3.5 4-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

0.025

Time (s)

Dis

pla

cem

ent

(m)


Dr. C. Caprani 63

5.3 Multi-Degree-of-Freedom Systems

5.3.1 General Case (based on 2DOF)

(a)

(b) (c)

Figure 3.1: (a) 2DOF system. (b) and (c) Free-body diagrams of forces

Considering Figure 3.1, we can see that the forces that act on the masses are similar

to those of the SDOF system but for the fact that the springs, dashpots, masses, forces

and deflections may all differ in properties. Also, from the same figure, we can see

the interaction forces between the masses will result from the relative deflection

between the masses; the change in distance between them.

For each mass, 0x

F , hence:

1 1 1 1 1 1 2 1 2 2 1 2 1mu c u k u c u u k u u F (5.3.1)


Dr. C. Caprani 64

2 2 2 2 1 2 2 1 2m u c u u k u u F (5.3.2)

In which we have dropped the time function indicators and allowed u and u to

absorb the directions of the interaction forces. Re-arranging we get:

1 1 1 1 2 2 2 1 1 2 2 2 1

2 2 1 2 2 2 1 2 2 2 2

u m u c c u c u k k u k F

u m u c u c u k u k F

(5.3.3)

This can be written in matrix form:

1 1 1 2 2 1 1 2 2 1 1

2 2 2 2 2 2 2 2 2

0

0

m u c c c u k k k u F

m u c c u k k u F

(5.3.4)

Or another way:

Mu + Cu + Ku = F (5.3.5)

where:

M is the mass matrix (diagonal matrix);

u is the vector of the accelerations for each DOF;

C is the damping matrix (symmetrical matrix);

u is the vector of velocity for each DOF;

K is the stiffness matrix (symmetrical matrix);

u is the vector of displacements for each DOF;

F is the load vector.

Equation (5.3.5) is quite general and reduces to many forms of analysis:


Dr. C. Caprani 65

Free vibration:

Mu + Cu + Ku = 0 (5.3.6)

Undamped free vibration:

Mu + Ku = 0 (5.3.7)

Undamped forced vibration:

Mu + Ku = F (5.3.8)

Static analysis:

Ku = F (5.3.9)

We will restrict our attention to the case of undamped free-vibration – equation

(5.3.7) - as the inclusion of damping requires an increase in mathematical complexity

which would distract from our purpose.


Dr. C. Caprani 66

5.3.2 Free-Undamped Vibration of 2DOF Systems

The solution to (5.3.7) follows the same methodology as for the SDOF case; so

following that method (equation (2.42)), we propose a solution of the form:

sin t u = a (5.3.10)

where a is the vector of amplitudes corresponding to each degree of freedom. From

this we get:

2 2sin t u = a u (5.3.11)

Then, substitution of (5.3.10) and (5.3.11) into (5.3.7) yields:

2 sin sint t Ma + Ka = 0 (5.3.12)

Since the sine term is constant for each term:

2 K M a = 0 (5.3.13)

We note that in a dynamics problem the amplitudes of each DOF will be non-zero,

hence, a 0 in general. In addition we see that the problem is a standard eigenvalues

problem. Hence, by Cramer‟s rule, in order for (5.3.13) to hold the determinant of

2K M must then be zero:

2 0K M = (5.3.14)

For the 2DOF system, we have:


Dr. C. Caprani 67

2 2 2 2

2 1 1 2 2 20k k m k m k K M = (5.3.15)

Expansion of (5.3.15) leads to an equation in 2 called the characteristic polynomial

of the system. The solutions of 2 to this equation are the eigenvalues of 2 K M

. There will be two solutions or roots of the characteristic polynomial in this case and

an n-DOF system has n solutions to its characteristic polynomial. In our case, this

means there are two values of 2 ( 2

1 and 2

2 ) that will satisfy the relationship; thus

there are two frequencies for this system (the lowest will be called the fundamental

frequency). For each 2

n substituted back into (5.3.13), we will get a certain

amplitude vector n

a . This means that each frequency will have its own characteristic

displaced shape of the degrees of freedoms called the mode shape. However, we will

not know the absolute values of the amplitudes as it is a free-vibration problem;

hence we express the mode shapes as a vector of relative amplitudes, nφ , relative to,

normally, the first value in n

a .

As we will see in the following example, the implication of the above is that MDOF

systems vibrate, not just in the fundamental mode, but also in higher harmonics.

From our analysis of SDOF systems it‟s apparent that should any loading coincide

with any of these harmonics, large DAF‟s will result (Section 2.d). Thus, some modes

may be critical design cases depending on the type of harmonic loading as will be

seen later.


Dr. C. Caprani 68

5.3.3 Example of a 2DOF System

The two-storey building shown (Figure

3.2) has very stiff floor slabs relative to

the supporting columns. Calculate the

natural frequencies and mode shapes.

3 24.5 10 kNmc

EI

Figure 3.2: Shear frame problem.

Figure 3.3: 2DOF model of the shear frame.

We will consider the free lateral vibrations of the two-storey shear frame idealised as

in Figure 3.3. The lateral, or shear stiffness of the columns is:

1 2 3

6

3

6

122

2 12 4.5 10

3

4 10 N/m

cEI

k k kh

k

The characteristic polynomial is as given in (5.3.15) so we have:


Dr. C. Caprani 69

6 2 6 2 12

6 4 10 2 12

8 10 5000 4 10 3000 16 10 0

15 10 4.4 10 16 10 0

This is a quadratic equation in 2 and so can be solved using 615 10a ,

104.4 10b and 1216 10c in the usual expression

2

2 4

2

b b ac

a

Hence we get 2

1425.3 and 2

22508 . This may be written:

2425.3

2508n

ω hence 20.6

50.1n

ω rad/s and 3.28

7.972

n

ωf Hz

To solve for the mode shapes, we will use the appropriate form of the equation of

motion, equation (5.3.13): 2 K M a = 0 . First solve for the

2 E K M

matrix and then solve Ea = 0 for the amplitudes n

a . Then, form nφ .

In general, for a 2DOF system, we have:

2

1 2 2 1 1 2 1 22

2

2 2 2 2 2 2

0

0

n

n n

n

k k k m k k m k

k k m k k m

E

For 2

1425.3 :

6

1

5.8735 410

4 2.7241

E


Dr. C. Caprani 70

Hence

16

1 1

2

5.8735 4 010

4 2.7241 0

a

a

E a

Taking either equation, we calculate:

1 2 1 2

1 1

1 2 1 2

5.8735 4 0 0.681 1

4 2.7241 0 0.681 0.681

a a a a

a a a a

φ

Similarly for 2

22508 :

6

2

4.54 410

4 3.524

E

Hence, again taking either equation, we calculate:

1 2 1 2

2 1

1 2 1 2

4.54 4 0 0.881 1

4 3.524 0 0.881 0.881

a a a a

a a a a

φ

The complete solution may be given by the following two matrices which are used in

further analysis for more complicated systems.

2425.3

2508n

ω and 1 1

1.468 1.135

Φ

For our frame, we can sketch these two frequencies and associated mode shapes:

Figure 3.4.


Dr. C. Caprani 71

Figure 3.4: Mode shapes and frequencies of the example frame.

Larger and more complex structures will have many degrees of freedom and hence

many natural frequencies and mode shapes. There are different mode shapes for

different forms of deformation; torsional, lateral and vertical for example. Periodic

loads acting in these directions need to be checked against the fundamental frequency

for the type of deformation; higher harmonics may also be important.

As an example; consider a 2DOF idealisation of a cantilever which assumes stiffness

proportional to the static deflection at 0.5L and L as well as half the cantilever mass

„lumped‟ at the midpoint and one quarter of it lumped at the tip. The mode shapes are

shown in Figure 3.5. In Section 4(a) we will see the exact mode shape for this – it is

clear that the approximation is rough; but, with more DOFs it will approach a better

solution.


Dr. C. Caprani 72

Figure 3.5: Lumped mass, 2DOF idealisation of a cantilever.

Mode 1

Mode 2


Dr. C. Caprani 73

5.3.4 Case Study – Aberfeldy Footbridge, Scotland

Returning to the case study in Section 1, we will look at the results of some research

conducted into the behaviour of this bridge which forms part of the current research

into lateral synchronise excitation discovered on the London Millennium footbridge.

This is taken from a paper by Dr. Paul Archbold, formerly of University College

Dublin.

Mode Mode

Type

Measured

Frequency

(Hz)

Predicted

Frequency (Hz)

1 L1 0.98 1.14 +16%

2 V1 1.52 1.63 +7%

3 V2 1.86 1.94 +4%

4 V3 2.49 2.62 +5%

5 L2 2.73 3.04 +11%

6 V4 3.01 3.11 +3%

7 V5 3.50 3.63 +4%

8 V6 3.91 4.00 +2%

9 T1 3.48 4.17 20%

10 V7 4.40 4.45 +1%

11 V8 4.93 4.90 -1%

12 T2 4.29 5.20 +21%

13 L3 5.72 5.72 +0%

14 T3 5.72 6.07 +19%

Table 1: Modal frequencies Figure 3.6: Undeformed shape

Table 1 gives the first 14 mode and associated frequencies from both direct

measurements of the bridge and from finite-element modelling of it. The type of


Dr. C. Caprani 74

mode is also listed; L is lateral, V is vertical and T is torsional. It can be seen that the

predicted frequencies differ slightly from the measured; however, the modes have

been estimated in the correct sequence and there may be some measurement error.

We can see now that (from Section 1) as a person walks at about 2.8 Hz, there are a

lot of modes that may be excited by this loading. Also note that the overall

fundamental mode is lateral – this was the reason that this bridge has been analysed –

it is similar to the Millennium footbridge in this respect. Figure 1.7 illustrates the

dynamic motion due to a person walking on this bridge – this is probably caused by

the third or fourth mode. Several pertinent mode shapes are given in Figure 3.7.


Dr. C. Caprani 75

Mode 1:

1st Lateral mode

1.14 Hz

Mode 2:

1st Vertical mode

1.63 Hz

Mode 3:

2nd

Vertical mode

1.94 Hz

Mode 9:

1st Torsional mode

4.17 Hz

Figure 3.7: Various Modes of Aberfeldy footbridge.


Dr. C. Caprani 76

5.4 Continuous Structures

5.4.1 Exact Analysis for Beams

General Equation of Motion

Figure 4.1: Basic beam subjected to dynamic loading: (a) beam properties and

coordinates; (b) resultant forces acting on the differential element.

In examining Figure 4.1, as with any continuous structure, it may be seen that any

differential element will have an associated stiffness and deflection – which changes

with time – and hence a different acceleration. Thus, any continuous structure has an

infinite number of degrees of freedom. Discretization into an MDOF structure is

certainly an option and is the basis for finite-element dynamic analyses; the more

DOF‟s used the more accurate the model (Section 3.b). For some basic structures


Dr. C. Caprani 77

though, the exact behaviour can be explicitly calculated. We will limit ourselves to

free-undamped vibration of beams that are thin in comparison to their length. A

general expression can be derived and from this, several usual cases may be

established.

Figure 4.2: Instantaneous dynamic deflected position.

Consider the element A of Figure 4.1(b); , hence:

(5.4.1)

after having cancelled the common shear term. The resultant transverse

inertial force is (mass × acceleration; assuming constant mass):

(5.4.2)

Thus we have, after dividing by the common term:

(5.4.3)

0yF

,

, , 0I

V x tp x t dx dx f x t dx

x

,V x t

2

2

,,I

v x tf x t dx mdx

t

dx

2

2

, ,,

V x t v x tp x t m

x t


Dr. C. Caprani 78

which, with no acceleration, is the usual static relationship between shear force and

applied load. By taking moments about the point A on the element, and dropping

second order and common terms, we get the usual expression:

(5.4.4)

Differentiating this with respect to and substituting into (5.4.3), in addition to the

relationship (which assumes that the beam is of constant stiffness):

(5.4.5)

With free vibration this is:

(5.4.6)

,

,M x t

V x tx

x

2

2vM EI

x

4 2

4 2

, ,,

v x t v x tEI m p x t

x t

4 2

4 2

, ,0

v x t v x tEI m

x t


Dr. C. Caprani 79

General Solution for Free-Undamped Vibration

Examination of equation (5.4.6) yields several aspects:

It is separated into spatial ( ) and temporal ( ) terms and we may assume that the

solution is also;

It is a fourth-order differential in ; hence we will need four spatial boundary

conditions to solve – these will come from the support conditions at each end;

It is a second order differential in and so we will need two temporal initial

conditions to solve – initial deflection and velocity at a point for example.

To begin, assume the solution is of a form of separated variables:

(5.4.7)

where will define the deformed shape of the beam and the amplitude of

vibration. Inserting the assumed solution into (5.4.6) and collecting terms we have:

(5.4.8)

This follows as the terms each side of the equals are functions of and separately

and so must be constant. Hence, each function type (spatial or temporal) is equal to

and so we have:

(5.4.9)

(5.4.10)

x t

x

t

,v x t x Y t

x Y t

4 2

2

4 2

1 1constant

x Y tEI

m x x Y t t

x t

2

4

2

4

xEI m x

x

2 0Y t Y t


Dr. C. Caprani 80

Equation (5.4.10) is the same as for an SDOF system (equation (2.4)) and so the

solution must be of the same form (equation (2.17)):

(5.4.11)

In order to evaluate we will use equation (5.4.9) and we introduce:

2

4 m

EI

(5.4.12)

And assuming a solution of the form , substitution into (5.4.9) gives:

4 4 exp 0s G sx (5.4.13)

There are then four roots for and when each is put into (5.4.13) and added we get:

1 2 3 4exp exp exp expx G i x G i x G x G x (5.4.14)

In which the ‟s may be complex constant numbers, but, by using Euler‟s

expressions for cos, sin, sinh and cosh we get:

1 2 3 4sin cos sinh coshx A x A x A x A x (5.4.15)

where the ‟s are now real constants; three of which may be evaluated through the

boundary conditions; the fourth however is arbitrary and will depend on .

00 cos sin

YY t Y t t

exp( )x G sx

s

G

A


Dr. C. Caprani 81

Simply-supported Beam

Figure 4.3: First three mode shapes and frequency parameters for an s-s beam.

The boundary conditions consist of zero deflection and bending moment at each end:

2

20, 0 and 0, 0

vv t EI t

x

(5.4.16)

2

2, 0 and , 0

vv L t EI L t

x

(5.4.17)

Substituting (5.4.16) into equation (5.4.14) we find . Similarly, (5.4.17)

gives:

1 3

2 2

1 3

sin( ) sinh( ) 0

'' sin( ) sinh( ) 0

L A L A L

L A L A L

(5.4.18)

from which, we get two possibilities:

2 4 0A A


Dr. C. Caprani 82

3

1

0 2 sinh( )

0 sin( )

A L

A L

(5.4.19)

however, since is never zero, 3

A must be, and so the non-trivial solution

must give us:

sin( ) 0L (5.4.20)

which is the frequency equation and is only satisfied when L n . Hence, from

(5.4.12) we get:

2

n

n EI

L m

(5.4.21)

and the corresponding modes shapes are therefore:

1sin

n

n xx A

L

(5.4.22)

where 1

A is arbitrary and normally taken to be unity. We can see that there are an

infinite number of frequencies and mode shapes ( n) as we would expect from

an infinite number of DOFs. The first three mode shapes and frequencies are shown

in Figure 4.3.

sinh( )x

1 0A


Dr. C. Caprani 83

Cantilever Beam

This example is important as it describes the sway behaviour of tall buildings. The

boundary conditions consist of:

0, 0 and 0, 0v

v t tx

(5.4.23)

2 3

2 3, 0 and , 0

v vEI L t EI L t

x x

(5.4.24)

Which represent zero displacement and slope at the support and zero bending

moment and shear at the tip. Substituting (5.4.23) into equation (5.4.14) we get

4 2A A and

3 1A A . Similarly, (5.4.24) gives:

2 2 2 2

1 2 3 4

3 3 3 3

1 2 3 4

'' sin( ) cos( ) sinh( ) cosh( ) 0

''' cos( ) sin( ) cosh( ) sinh( ) 0

L A L A L A L A L

L A L A L A L A L

(5.4.25)

where a prime indicates a derivate of x , and so we find:

1 2

1 2

sin( ) sinh( ) cos( ) cosh( ) 0

cos( ) cosh( ) sin( ) sinh( ) 0

A L L A L L

A L L A L L

(5.4.26)

Solving for 1

A and 2

A we find:

2

1

2

2

cos( ) cosh( )0

sin( ) sinh( ) sin( ) sinh( )

cos( ) cosh( )0

sin( ) sinh( ) sin( ) sinh( )

L LA

L L L L

L LA

L L L L

(5.4.27)


Dr. C. Caprani 84

In order that neither 1

A and 2

A are zero, the expression in the brackets must be zero

and we are left with the frequency equation:

cos( )cosh( ) 1 0L L (5.4.28)

The mode shape is got by expressing 2

A in terms of 1

A :

2 1

sin( ) sinh( )

cos( ) cosh( )

L LA A

L L

(5.4.29)

and the modes shapes are therefore:

1

sin( ) sinh( )

sin( ) sinh( )cosh( ) cos( )

cos( ) cosh( )

n

x x

x A L Lx x

L L

(5.4.30)

where again1

A is arbitrary and normally taken to be unity. We can see from (5.4.28)

that it must be solved numerically for the corresponding values of L The natural

frequencies are then got from (5.4.21) with the substitution of L for n . The first

three mode shapes and frequencies are shown in Figure 4.4.


Dr. C. Caprani 85

Figure 4.4: First three mode shapes and frequency parameters for a cantilever.


Dr. C. Caprani 86

5.4.2 Approximate Analysis – Bolton’s Method

We will now look at a simplified method that requires an understanding of dynamic

behaviour but is very easy to implement. The idea is to represent, through various

manipulations of mass and stiffness, any complex structure as a single SDOF system

which is easily solved via an implementation of equation (1.2):

1

2

E

E

Kf

M (5.4.31)

in which we have equivalent SDOF stiffness and mass terms.

Consider a mass-less cantilever which carries two different masses, Figure 4.5:

Figure 4.5: Equivalent dynamic mass distribution for a cantilever.

The end deflection of a cantilever loaded at its end by a force P is well known to be

3

3PL

EI and hence the stiffness is 3

3EIL

. Therefore, the frequencies of the two

cantilevers of Figure 4.5 are:

1 3

1

1 3

2

EIf

M x (5.4.32)

3

1 3;

2E

E

EIf

M L (5.4.33)


Dr. C. Caprani 87

And so, if the two frequencies are to be equal, and considering 1

M as the mass of a

small element dx when the mass per metre is m , the corresponding part of E

M is:

3

E

xdM mdx

L

(5.4.34)

and integrating:

3

0

0.25

L

E

xM mdx

L

mL

(5.4.35)

Therefore the cantilever with self-mass uniformly distributed along its length vibrates

at the same frequency as would the mass-less cantilever loaded with a mass one

quarter its actual mass. This answer is not quite correct but is within 5%; it ignores

the fact that every element affects the deflection (and hence vibration) of every other

element. The answer is reasonable for design though.

Figure 4.6: Equivalent dynamic mass distribution for an s-s beam

Similarly for a simply supported beam, we have an expression for the deflection at a

point:


Dr. C. Caprani 88

22

3x

Px L x

EIL

(5.4.36)

and so its stiffness is:

22

3x

EILK

x L x

(5.4.37)

Considering Figure 4.6, we see that, from (5.4.31):

2 32

1

3 48

E

EIL EI

L Mx L x M

(5.4.38)

and as the two frequencies are to be equal:

2

2

40

16

8/15

L

E

L xM x mdx

L

mL

(5.4.39)

which is about half of the self-mass as we might have guessed.

Proceeding in a similar way we can find equivalent spring stiffnesses and masses for

usual forms of beams as given in Table 1. Table 4.1 however, also includes a

refinement of the equivalent masses based on the known dynamic deflected shape

rather than the static deflected shape.


Dr. C. Caprani 89

Table 4.1: Bolton‟s table for equivalent mass, stiffnesses and relative amplitudes.

Figure 4.7: Effective SDOFs: (a) neglecting relative amplitude; (b) including relative

amplitude.

In considering continuous beams, the continuity over the supports requires all the

spans to vibrate at the same frequency for each of its modes. Thus we may consider


Dr. C. Caprani 90

summing the equivalent masses and stiffnesses for each span and this is not a bad

approximation. It is equivalent to the SDOF model of Figure 4.7(a). But, if we

allowed for the relative amplitude between the different spans, we would have the

model of Figure 4.7(b) which would be more accurate – especially when there is a

significant difference in the member stiffnesses and masses: long heavy members will

have larger amplitudes than short stiff light members due to the amount of kinetic

energy stored. Thus, the stiffness and mass of each span must be weighted by its

relative amplitude before summing. Consider the following examples of the beam

shown in Figure 4.8; the exact multipliers are known to be 10.30, 13.32, 17.72, 21.67,

40.45, 46.10, 53.89 and 60.53 for the first eight modes.

Figure 4.8: Continuous beam of Examples 1 to 3.


Dr. C. Caprani 91

Example 1: Ignoring relative amplitude and refined ME

From Table 4.1, and the previous discussion:

3

48 3 101.9E

EIK

L ; and

8 13

15 2E

M mL

,

and applying (5.4.31) we have: 4

110.82

2

EIf

mL

The multiplier in the exact answer is 10.30: an error of 5%.

Example 2: Including relative amplitude and refined ME

From Table 4.1 and the previous discussion, we have:

3 3 3

48 101.93 1 0.4108 185.9

E

EI EI EIK

L L L

3 0.4928 1 0.4299 0.4108 1.655E

M mL mL mL

and applying (5.4.31) we have:

4

110.60

2

EIf

mL

The multiplier in the exact answer is 10.30: a reduced error of 2.9%.


Dr. C. Caprani 92

Example 3: Calculating the frequency of a higher mode

Figure 4.9: Assumed mode shape for which the frequency will be found.

The mode shape for calculation is shown in Figure 4.7. We can assume supports at

the midpoints of each span as they do not displace in this mode shape. Hence we have

seven simply supported half-spans and one cantilever half-span, so from Table 4.1 we

have:

3 3

3

48 101.97 1 0.4108

0.5 0.5

3022.9

7 0.4928 0.5 1 0.4299 0.5 0.4108

1.813

E

E

EI EIK

L L

EI

L

M m L m L

mL

again, applying (5.4.31), we have:

4

140.8

2

EIf

mL

The multiplier in the exact answer is 40.45: and error of 0.9%.


Dr. C. Caprani 93

Mode Shapes and Frequencies

Section 2.d described how the DAF is very large when a force is applied at the

natural frequency of the structure; so for any structure we can say that when it is

vibrating at its natural frequency it has very low stiffness – and in the case of no

damping: zero stiffness. Higher modes will have higher stiffnesses but stiffness may

also be recognised in one form as

1M

EI R (5.4.40)

where R is the radius of curvature and M is bending moment. Therefore, smaller

stiffnesses have a larger R and larger stiffnesses have a smaller R . Similarly then,

lower modes have a larger R and higher modes have a smaller R . This enables us to

distinguish between modes by their frequencies. Noting that a member in single

curvature (i.e. no point of contraflexure) has a larger R than a member in double

curvature (1 point of contraflexure) which in turn has a larger R than a member in

triple curvature (2 points of contraflexure), we can distinguish modes by deflected

shapes. Figures 4.3 and 4.4 illustrate this clearly.


Dr. C. Caprani 94

Figure 4.10: Typical modes and reduced structures.

An important fact may be deduced from Figure 4.10 and the preceding arguments: a

continuous beam of any number of identical spans has the same fundamental

frequency as that of one simply supported span: symmetrical frequencies are

similarly linked. Also, for non-identical spans, symmetry may exist about a support

and so reduced structures may be used to estimate the frequencies of the total

structure; reductions are shown in Figure 4.10(b) and (d) for symmetrical and anti-

symmetrical modes.


Dr. C. Caprani 95

5.4.3 Problems

Problem 1

Calculate the first natural frequency of a simply supported bridge of mass 7 tonnes

with a 3 tonne lorry at its quarter point. It is known that a load of 10 kN causes a 3

mm deflection.

Ans.: 3.95Hz.

Problem 2

Calculate the first natural frequency of a 4 m long cantilever (EI = 4,320 kNm2)

which carries a mass of 500 kg at its centre and has self weight of 1200 kg.

Ans.: 3.76 Hz.

Problem 3

What is the fundamental frequency of a 3-span continuous beam of spans 4, 8 and 5

m with constant EI and m? What is the frequency when EI = 6×103 kNm

2 and m =

150 kg/m?

Ans.: 6.74 Hz.

Problem 4

Calculate the first and second natural frequencies of a two-span continuous beam;

fixed at A and on rollers at B and C. Span AB is 8 m with flexural stiffness of 2EI and

a mass of 1.5m. Span BC is 6 m with flexural stiffness EI and mass m per metre.

What are the frequencies when EI = 4.5×103 kNm

2 and m = 100 kg/m?

Ans.: 9.3 Hz; ? Hz.


Dr. C. Caprani 96

Problem 5

Calculate the first and second natural frequencies of a 4-span continuous beam of

spans 4, 5, 4 and 5 m with constant EI and m? What are the frequencies when EI =

4×103 kNm

2 and m = 120 kg/m? What are the new frequencies when support A is

fixed? Does this make it more or less susceptible to human-induced vibration?

Ans.: ? Hz; ? Hz.


Dr. C. Caprani 97

5.5 Practical Design Considerations

5.5.1 Human Response to Dynamic Excitation

Figure 5.1: Equal sensation contours for vertical vibration

The response of humans to vibrations is a complex phenomenon involving the

variables of the vibrations being experienced as well as the perception of it. It has


Dr. C. Caprani 98

been found that the frequency range between 2 and 30 Hz is particularly

uncomfortable because of resonance with major body parts (Figure 5.2). Sensation

contours for vertical vibrations are shown in Figure 5.1. This graph shows that for a

given frequency, as the amplitude gets larger it becomes more uncomfortable; thus it

is acceleration that is governing the comfort. This is important in the design of tall

buildings which sway due to wind loading: it is the acceleration that causes

discomfort. This may also be realised from car-travel: at constant velocity nothing is

perceptible, but, upon rapid acceleration the motion if perceived ( F ma ).

Figure 5.2: Human body response to vibration

Response graphs like Figure 5.1 have been obtained for each direction of vibration

but vertical motion is more uncomfortable for standing subjects; for the transverse

and longitudinal cases, the difference has the effect of moving the illustrated bands

up a level. Other factors are also important: the duration of exposure; waveform

(which is again linked to acceleration); type of activity; and, psychological factors.

An example is that low frequency exposure can result in motion sickness.


Dr. C. Caprani 99

5.5.2 Crowd/Pedestrian Dynamic Loading

Lightweight Floors

Figure 5.3: Recommended vibration limits for light floors.

Vibration limits for light floors from the 1984 Canadian Standard is shown in Figure

5.2; the peak acceleration is got from:

00.9 2

Ia f

M (5.5.1)

where I is the impulse (the area under the force time graph) and is about 70 Ns and

M is the equivalent mass of the floor which is about 40% of the distributed mass.


Dr. C. Caprani 100

This form of approach is to be complemented by a simple analysis of an equivalent

SDOF system. Also, as seen in Section 1, by keeping the fundamental frequency

above 5 Hz, human loading should not be problematic.


Dr. C. Caprani 101

Crowd Loading

This form of loading occurs in grandstands and similar structures where a large

number of people are densely packed and will be responding to the same stimulus.

Coordinated jumping to the beat of music, for example, can cause a DAF of about

1.97 at about 2.5 Hz. Dancing, however, normally generates frequencies of 2 – 3 Hz.

Once again, by keeping the natural frequency of the structure above about 5 Hz no

undue dynamic effects should be noticed.

In the transverse or longitudinal directions, allowance should also be made due to the

crowd-sway that may accompany some events a value of about 0.3 kN per metre of

seating parallel and 0.15 kN perpendicular to the seating is an approximate method

for design.

Staircases can be subject to considerable dynamic forces as running up or down such

may cause peak loads of up to 4-5 times the persons bodyweight over a period of

about 0.3 seconds – the method for lightweight floors can be applied to this scenario.


Dr. C. Caprani 102

Footbridges

As may be gathered from the Case Studies of the Aberfeldy Bridge, the problem is

complex, however some rough guidelines are possible. Once again controlling the

fundamental frequency is important; the lessons of the London Millennium and the

Tacoma Narrows bridges need to be heeded though: dynamic effects may occur in

any direction or mode that can be excited by any form of loading.

An approximate method for checking foot bridges is the following:

max st

u u K (5.5.2)

where st

u is the static deflection under the weight of a pedestrian at the point of

maximum deflection; K is a configuration factor for the type of structure (given in

Table 5.1); and is the dynamic response factor got again from Figure 5.4. The

maximum acceleration is then got as 2

max maxu u (see equations (2.30) and (3.11) for

example, note: 2 2 f ). This is then compared to a rather simple rule that the

maximum acceleration of footbridge decks should not exceed 0.5 f .

Alternatively, BD 37/01 states:

“For superstructures for which the fundamental natural frequency of vibration

exceeds 5Hz for the unloaded bridge in the vertical direction and 1.5 Hz for the

loaded bridge in the horizontal direction, the vibration serviceability requirement is

deemed to be satisfied.” – Appendix B.1 General.

Adhering to this clause (which is based on the discussion of Section 1‟s Case Study)

is clearly the easiest option.


Dr. C. Caprani 103

Also, note from Figure 5.4 the conservative nature of the damping assumed, which,

from equation (2.35) can be seen to be so based on usual values of damping in

structures.

Table 5.1: Configuration factors for footbridges.

Table 5.2: Values of the logarithmic decrement for different bridge types.


Dr. C. Caprani 104

Figure 5.4: Dynamic response factor for footbridges


Dr. C. Caprani 105

Design Example

A simply-supported footbridge of 18 m span has a total mass of 12.6 tonnes and

flexural stiffness of 3×105 kNm

2. Determine the maximum amplitude of vibration and

vertical acceleration caused by a 0.7 kN pedestrian walking in frequency with the

bridge: the pedestrian has a stride of 0.9 m and produces an effective pulsating force

of 180 N. Assume the damping to be related to 0.05 . Is this a comfortable bridge

for the pedestrian (Figure 5.1)?

The natural frequency of the bridge is, from equations (2.19) and (4.21):

8

2

3 103.17 Hz

2 18 12600/18f

The static deflection is:

3

8

700 180.2835 mm

48 3 10st

u

Table 5.1 gives 1K and Figure 5.4 gives 6.8 and so, by (5.5.2) we have:

max

0.2835 1.0 6.8 1.93 mmu

and so the maximum acceleration is:

22 3 2

max max2 3.17 1.93 10 0.78 m/su u

We compare this to the requirement that:


Dr. C. Caprani 106

max

2

0.5

0.5

0.78 0.89 m/s

u f

f

And so we deem the bridge acceptable. From Figure 5.1, with the amplitude 1.93 mm

and 3.17 Hz frequency, we can see that this pedestrian will feel decidedly

uncomfortable and will probably change pace to avoid this frequency of loading.

The above discussion, in conjunction with Section 2.d reveals why, historically,

soldiers were told to break step when crossing a slender bridge – unfortunately for

some, it is more probable that this knowledge did not come from any detailed

dynamic analysis; rather, bitter experience.


Dr. C. Caprani 107

5.5.3 Damping in Structures

The importance of damping should be obvious by this stage; a slight increase may

significantly reduce the DAF at resonance, equation (2.47). It was alluded to in

Section 1 that the exact nature of damping is not really understood but that it has been

shown that our assumption of linear viscous damping applies to the majority of

structures – a notable exception is soil-structure interaction in which alternative

damping models must be assumed. Table 5.3 gives some typical damping values in

practice. It is notable that the materials themselves have very low damping and thus

most of the damping observed comes from the joints and so can it depend on:

The materials in contact and their surface preparation;

The normal force across the interface;

Any plastic deformation in the joint;

Rubbing or fretting of the joint when it is not tightened.

Table 5.4: Recommended values of damping.

When the vibrations or DAF is unacceptable it is not generally acceptable to detail

joints that will have higher damping than otherwise normal – there are simply too

many variables to consider. Depending on the amount of extra damping needed, one


Dr. C. Caprani 108

could wait for the structure to be built and then measure the damping, retro-fitting

vibration isolation devices as required. Or, if the extra damping required is

significant, the design of a vibration isolation device may be integral to the structure.

The devices that may be installed vary; some are:

Tuned mass dampers (TMDs): a relatively small mass is attached to the primary

system and is „tuned‟ to vibrate at the same frequency but to oppose the primary

system;

Sloshing dampers: A large water tank is used – the sloshing motion opposes the

primary system motion due to inertial effects;

Liquid column dampers: Two columns of liquid, connected at their bases but at

opposite sides of the primary system slosh, in a more controlled manner to oppose

the primary system motion.

These are the approaches taken in many modern buildings, particularly in Japan and

other earthquake zones. The Citicorp building in New York (which is famous for

other reasons also) and the John Hancock building in Boston were among the first to

use TMDs. In the John Hancock building a concrete block of about 300 tonnes

located on the 54th storey sits on a thin film of oil. When the building sways the

inertial effects of the block mean that it moves in the opposite direction to that of the

sway and so opposes the motion (relying heavily on a lack of friction). This is quite a

rudimentary system compared to modern systems which have computer controlled

actuators that take input from accelerometers in the building and move the block an

appropriate amount.


Dr. C. Caprani 109

5.5.4 Design Rules of Thumb

General

The structure should not have any modal frequency close to the frequency of any

form of periodic loading, irrespective of magnitude. This is based upon the large

DAFs that may occur (Section 2.d).

For normal floors of span/depth ratio less than 25 vibration is not generally a

problem. Problematic floors are lightweight with spans of over about 7 m.

Human loading

Most forms of human loading occur at frequencies < 5 Hz (Sections 1 and 5.a) and so

any structure of natural frequency greater than this should not be subject to undue

dynamic excitation.

Machine Loading

By avoiding any of the frequencies that the machine operates at, vibrations may be

minimised. The addition of either more stiffness or mass will change the frequencies

the structure responds to. If the response is still not acceptable vibration isolation

devices may need to be considered (Section 5.c).

Approximate Frequencies

The Bolton Method of Section 4.b is probably the best for those structures outside the

standard cases of Section 4.a. Careful thought on reducing the size of the problem to

an SDOF system usually enables good approximate analysis.

Other methods are:


Dr. C. Caprani 110

Structures with concentrated mass: 1

2

gf

Simplified rule for most structures: 18

f

where is the static deflection and g is the acceleration under gravity.


Dr. C. Caprani 111

Rayleigh Approximation

A method developed by Lord Rayleigh (which is always an upper bound), based on

energy methods, for estimating the lowest natural frequency of transverse beam

vibration is:

22

2

02

12

0

L

L

d yEI dx

dx

y dm

(5.5.3)

This method can be used to estimate the fundamental frequency of MDOF systems.

Considering the frame of Figure 5.5, the fundamental frequency in each direction is

given by:

2

1 2 2

i i i ii i

i i i ii i

Qu mu

g gQu mu

(5.5.4)

where i

u is the static deflection under the dead load of the structure i

Q , acting in the

direction of motion, and g is the acceleration due to gravity. Thus, the first mode is

approximated in shape by the static deflection under dead load. For a building, this

can be applied to each of the X and Y directions to obtain the estimates of the

fundamental sway modes.


Dr. C. Caprani 112

Figure 5.5: Rayleigh approximation for the fundamental sway frequencies of a

building.

Figure 5.6: Rayleigh method for approximating bridge fundamental frequencies.

Likewise for a bridge, by applying the dead load in each of the vertical and horizontal

directions, the fundamental lift and drag modes can be obtained. The torsional mode

can also be approximated by applying the dead load at the appropriate radius of

gyration and determining the resulting rotation angle, Figure 5.6.


Dr. C. Caprani 113

This method is particularly useful when considering the results of a detailed analysis,

such as finite-element. It provides a reasonable approximate check on the output.


Dr. C. Caprani 114

5.6 Appendix

5.6.1 Past Exam Questions

Summer 2005

Question 5

(a) The system shown in Figure Q.5(a) is known to have a static deflection of 32.7 mm for an unknown mass.

1) Find the natural frequency of the system.

(10%)

2) Given that the mass is 10 kg, find the peak displacement when this mass is given an initial

velocity of 500 mm/s and an initial displacement of 25 mm.

(10%)

3) What time does the first positive peak occur?

(10%)

4) What value of damping coefficient is required such that the amplitude after 5 oscillations is 10%

of the first peak?

(10%)

5) What is the peak force in the spring?

(20%)

(b) A cantilever riverside boardwalk has been opened to the public as shown in Figure Q.5(b); however, it was

found that the structure experiences significant human- and traffic-induced vibrations. An harmonic

oscillation test found the natural frequency of the structure to be 2.25 Hz. It is proposed to retro-fit braced

struts at 5m spacings so that the natural period of vibration will be 9 Hz – given E = 200 kN/mm2 and

ignoring buckling effects, what area of strut is required?

(40%)

Ans. (a) 2.756 Hz; 38.2 mm; 0.05 s; 99 kg.s/m;114.5 N; (b) 67.5 mm2.

FIG. Q.5(b)

m

k

FIG. Q.5(a)


Dr. C. Caprani 115

Sample Paper Semester 1 2006/7

5. (a) The single-degree-of-freedom system shown in Fig. Q5(a) is known to have a static deflection of 32.7

mm for an unknown mass.

(i) Find the natural frequency of the system;

(2 marks)

(ii) Given that the mass is 10 kg, find the peak displacement when the mass is given an initial

velocity of 500 mm/s and an initial displacement of 25 mm;

(2 marks)

(iii) At what time does the first positive peak occur?

(2 marks)

(iv) What damping ratio is required such that the amplitude after 5 oscillations is 10% of the first

peak?

(2 marks)

(v) What is the peak force in the spring?

(6 marks)

(b) The beam shown in Fig. Q5(b) is loaded with an air conditioning (AC) unit at its tip. The AC unit

produces an unbalanced force of 100 kg which varies sinusoidally. When the speed of the AC unit is

varied, it is found that the maximum steady-state deflection is 20.91 mm. Determine:

(i) The damping ratio;

(4 marks)

(ii) The maximum deflection when the unit‟s speed is 250 rpm;

(7 marks)

Take the following values:

• EI = 1×106 kNm

2;

• Mass of the unit is 500 kg.

Ans. (a) 2.756 Hz; 38.2 mm; 0.05 s; 99 kg.s/m;114.5 N; (b) ??.

FIG. Q5(a)

FIG. Q5(a)


Dr. C. Caprani 116

Semester 1 2006/7

5. (a) A simply-supported reinforced concrete beam, 300 mm wide × 600 mm deep spans 8 m. Its

fundamental natural frequency is measured to be 6.5 Hz. In your opinion, is the beam cracked or

uncracked?

Use a single degree-of-freedom (SDOF) system to represent the deflection at the centre of the beam.

Assume that 8/15 of the total mass of the beam contributes to the SDOF model. Take the density of

reinforced concrete to be 24 kN/m3 and E = 30 kN/mm

2.

(10 marks)

(b) The beam shown in Fig. Q5(b) is loaded with an air conditioning (AC) unit at its tip. The AC unit

produces an unbalanced force of 200 kg which varies sinusoidally. When the speed of the AC unit is

varied, it is found that the maximum steady-state amplitude of vibration is 34.6 mm. Determine:

(i) The damping ratio;

(5 marks)

(ii) The maximum deflection when the unit‟s speed is 100 rpm;

(10 marks)


• EI = 40×103 kNm

2;

• Mass of the unit is 2000 kg;

• Ignore the mass of the beam.

Ans. (a) Cracked; (b) 5.1%; 41.1 mm.

FIG. Q5(a)

FIG. Q5(b)


Dr. C. Caprani 117

Semester 1 2007/8

QUESTION 5

(a) For the frame shown in Fig. Q5, using a single-degree-of-freedom model, determine:

(i) The natural frequency and period in free vibration;

(ii) An expression for the displacement at time t if member BC is displaced 20 mm and suddenly released

at time t = 1 sec.

(8 marks)

(b) The frame is found to have 5% damping. Using appropriate approximations, what is the percentage change in

deflection, 4 cycles after the frame is released, of the damped behaviour compared to the undamped behaviour?

(10 marks)

(c) A machine is placed on member BC which has an unbalanced force of 500 kg which varies sinusoidally.

Neglecting the mass of the machine, determine:

(i) the maximum displacement when the unit‟s speed is 150 rpm;

(ii) the speed of the machine at resonance;

(iii) the displacement at resonance.

(7 marks)

Note:


EI = 20×103 kNm

2;

M = 20 tonnes;

Consider BC as infinitely rigid.

Ans.(a) 3.93 Hz; 0.254 s; 20cos[24.72(t-1)], t>1; (b) Ratio: 28.4%, change: 71.6%;

(c) 0.67 mm; 236 rpm; 4.01 mm.

FIG. Q5


Dr. C. Caprani 118

Semester 1 2008/9

QUESTION 5

The structure shown in Fig. Q5 supports a scoreboard at a sports centre. The claxton (of total mass M) which sounds

the end of playing periods includes a motor which has an unbalanced mass of 100 kg which varies sinusoidally when

sounded. Using a single-degree-of-freedom model for vibrations in the vertical direction, and neglecting the mass of

the truss members, determine:

(i) the natural frequency and period in free vibration;

(ii) the damping, given that a test showed 5 cycles after a 10 mm initial displacement was imposed, the

amplitude was 5.30 mm;

(iii) the maximum displacement when the unit‟s speed is 1500 rpm;

(iv) the speed of the machine at resonance;

(v) the displacement at resonance.

(25 marks)

Note:


For all truss members: 320 10 kNEA ;

M = 5 tonnes;

Ignore the stiffness and mass of member EF.

Ans. 4.9 Hz; 0.205 s; 2%; 0.008 mm; 293.2 rpm; 5.2 mm.

FIG. Q5


Dr. C. Caprani 119

Semester 1 2009/10

QUESTION 5

(a) A 3 m high, 6 m wide single-bay single-storey frame is rigidly jointed with a beam of mass 2,000 kg and columns

of negligible mass and stiffness of 3 22.7 10 kNmEI . Assuming the beam to be infinitely rigid, calculate the

natural frequency in lateral vibration and its period. Find the force required to deflect the frame 20 mm laterally.

(10 marks)

(b) A spring-mass-damper SDOF system is subject to a harmonically varying force. At resonance, the amplitude of

vibration is found to be 10 mm, and at 0.80 of the resonant frequency, the amplitude is found to be 5.07 mm.

Determine the damping of the system.

(15 marks)

Ans. 5.51 Hz, 48 kN.; 0.1.


Dr. C. Caprani 120

Semester 1 2010/11

QUESTION 5

(a) For the shear frame shown in Fig. Q5(a), ignoring the mass of the columns:

(i) How many modes will this structure have?

(ii) Sketch the mode shapes;

(iii) Indicate the order of the natural frequencies associated with each mode shape (i.e. lowest to highest).

(10 marks)

(b) For the frame shown in Fig. Q5(b), using a single-degree-of-freedom model, determine the natural frequency and

period in free vibration given that EI = 27×103 kNm

2 and M = 24 tonnes. If a machine is placed on member BC

which has an unbalanced force of 500 kg varying sinusoidally, neglecting the mass of the machine, determine:

(i) the maximum displacement when the unit‟s speed is 360 rpm;

(ii) the speed of the machine at resonance;

(iii) the displacement at resonance.

(15 marks)

Ans. 0.34 mm, 426.6 rpm, 1.02 mm.

FIG. Q5(a)

FIG. Q5(b)


Dr. C. Caprani 121

5.6.2 References

The following books/articles were referred to in the writing of these notes;

particularly Clough & Penzien (1993), Smith (1988) and Bolton (1978) - these should

be referred to first for more information. There is also a lot of information and

software available online; the software can especially help intuitive understanding.

The class notes of Mr. R. Mahony (D.I.T.) and Dr. P. Fanning (U.C.D.) were also

used.

Archbold, P., (2002), “Modal Analysis of a GRP Cable-Stayed Bridge”,

Proceedings of the First Symposium of Bridge Engineering Research In Ireland,

Eds. C. McNally & S. Brady, University College Dublin.

Beards, C.F., (1983), Structural Vibration Analysis: modelling, analysis and

damping of vibrating structures, Ellis Horwood, Chichester, England.

Bhatt, P., (1999), Structures, Longman, Harlow, England.

Bolton, A., (1978), “Natural frequencies of structures for designers”, The

Structural Engineer, Vol. 56A, No. 9, pp. 245-253; Discussion: Vol. 57A, No. 6,

p.202, 1979.

Bolton, A., (1969), “The natural frequencies of continuous beams”, The Structural

Engineer, Vol. 47, No. 6, pp.233-240.

Case, J., Chilver, A.H. and Ross, C.T.F., (1999), Strength of Materials and

Structures, 4th edn., Arnold, London.

Chopra, A.K., (2007), Dynamics of Structures – Theory and Applications to

Earthquake Engineering, 3rd edn., Pearson-Prentice Hall, New Jersey.

Clough, R.W. and Penzien, J., (1993), Dynamics of Structures, 2nd edn., McGraw-

Hill, New York.

Cobb, F. (2004), Structural Engineer’s Pocket Book, Elsevier, Oxford.

Craig, R.R., (1981), Structural Dynamics – An introduction to computer methods,

Wiley, New York.


Dr. C. Caprani 122

Ghali, A. and Neville, A.M., (1997), Structural Analysis – A unified classical and

matrix approach, 4th edn., E&FN Spon, London.

Irvine, M., (1986), Structural Dynamics for the Practising Engineer, Allen &

Unwin, London.

Kreyszig, E., (1993), Advanced Engineering Mathematics, 7th edn., Wiley.

Paz, M. and Leigh, W., (2004), Structural Dynamics – Theory and Computation,

5th edn., Springer, New York.

Smith, J.W., (1988), Vibration of Structures – Applications in civil engineering

design, Chapman and Hall, London.


Dr. C. Caprani 123

5.6.3 Amplitude Solution to Equation of Motion

The solution to the equation of motion is found to be in the form:

cos sinu t A t B t (5.5.5)

However, we regularly wish to express it in one of the following forms:

cosu t C t (5.5.6)

cosu t C t (5.5.7)

Where

2 2C A B (5.5.8)

tanA

B (5.5.9)

tanB

A (5.5.10)

To arrive at this result, re-write equation (5.5.5) as:

cos sinA B

u t C t tC C

(5.5.11)

If we consider that A, B and C represent a right-angled triangle with angles and ,

then we can draw the following:


Dr. C. Caprani 124

Thus:

sin cosA

C (5.5.12)

cos sinB

C (5.5.13)

Introducing these into equation (5.5.11) gives two relationships:

sin cos cos sinu t C t t (5.5.14)

cos cos sin sinu t C t t (5.5.15)

And using the well-known trigonometric identities:

sin sin cos cos sinX Y X Y X Y (5.5.16)

cos cos cos sin sinX Y X Y X Y (5.5.17)

Gives the two possible representations, the last of which is the one we adopt:

sinu t C t (5.5.18)

cosu t C t (5.5.19)


Dr. C. Caprani 125

5.6.4 Solutions to Differential Equations

The Homogenous Equation

To find the solution of:

2

2

20

d yk y

dx (5.5.20)

we try xy e (note that this k has nothing to do with stiffness but is the conventional

mathematical notation for this problem). Thus we have:

2

2

2;x xdy d y

e edx dx

Substituting this into (5.5.20) gives:

2 2 0x xe k e

And so we get the characteristic equation by dividing out xe :

2 2 0k

From which:

2k

Or,


Dr. C. Caprani 126

1 2

;ik ik

Where 1i . Since these are both solutions, they are both valid and the expression

for y becomes:

1 2

ikx ikxy Ae A e (5.5.21)

In which 1

A and 2

A are constants to be determined from the initial conditions of the

problem. Introducing Euler‟s equations:

cos sin

cos sin

ikx

ikx

e kx i kx

e kx i kx

(5.5.22)

into (5.5.21) gives us:

1 2cos sin cos siny A kx i kx A kx i kx

Collecting terms:

1 2 1 2cos siny A A kx iA iA kx

Since the coefficients of the trigonometric functions are constants we can just write:

cos siny A kx B kx (5.5.23)


Dr. C. Caprani 127

The Non-homogenous Equation

Starting with equation (5.2.47) (repeated here for convenience):

0

( ) ( ) ( ) sinmu t cu t ku t F t (5.5.24)

We divide by m and introduce equations (5.2.10) and (5.2.12) to get:

2 0( ) 2 ( ) ( ) sin

Fu t u t u t t

m (5.5.25)

At this point, recall that the solution to non-homogenous differential equations is

made up of two parts:

The complimentary solution ( Cu t ): this is the solution to the corresponding

homogenous equation, which we already have (equation (5.5.23));

The particular solution ( Pu t ): particular to the function on the right hand side

of equation (5.5.24), which we must now find.

The final solution is the sum of the complimentary and particular solutions:

C Pu t u t u t (5.5.26)

For the particular solution we try the following:

sin cosP

u t C t D t (5.5.27)

Then we have:

cos sinP

u t C t D t (5.5.28)


Dr. C. Caprani 128

And

2 2sin cosP

u t C t D t (5.5.29)

Substituting equations (5.5.27), (5.5.28) and (5.5.29) into equation (5.5.25) gives:

2 2

2 0

sin cos

2 cos sin

sin cos sin

C t D t

C t D t

FC t D t t

m

(5.5.30)

Collecting sine and cosine terms:

2 2

2 2 0

2 sin

2 cos sin

C D t

FC D t t

m

(5.5.31)

For this to be valid for all t, the sine and cosine terms on both sides of the equation

must be equal. Thus:

2 2 02F

C Dm

(5.5.32)

2 22 0C D (5.5.33)

Next, divide both sides by 2 :


Dr. C. Caprani 129

2

0

2 21 2

FC D

m

(5.5.34)

2

22 1 0C D

(5.5.35)

Introduce the frequency ratio, equation (5.2.51), , and 2k m from

equation (5.2.9) to get:

2 01 2F

C Dk

(5.5.36)

22 1 0C D (5.5.37)

From equation (5.5.37), we have:

21

2C D

(5.5.38)

And using this in equation (5.5.36) gives:

22

01

22

FD

k

(5.5.39)

To get:

2 22

01 2

2

FD

k

(5.5.40)


Dr. C. Caprani 130

And rearrange to get, finally:

0

2 22

2

1 2

FD

k

(5.5.41)

Now using this with equation (5.5.38), we have:

2

0

2 22

1 2

2 1 2

FC

k

(5.5.42)

To get, finally:

2

0

2 22

1

1 2

FC

k

(5.5.43)

Again we use the cosine addition rule:

2 2C D (5.5.44)

tanD

C (5.5.45)

To express the solution as:

( ) sinP

u t t (5.5.46)

So we have, from equations (5.5.44), (5.5.43) and (5.5.41):


Dr. C. Caprani 131

2 22

0

2 22 22 2

1 2

1 2 1 2

F

k

(5.5.47)

This simplifies to:

2 22

0

22 22

1 2

1 2

F

k

(5.5.48)

And finally we have the amplitude of displacement:

1

2 2 220 1 2F

k

(5.5.49)

To obtain the phase angle, we use equation (5.5.45) with equations (5.5.43) and

(5.5.41) again to get:

0

2 22

2

0

2 22

2

1 2tan

1

1 2

F

k

F

k

(5.5.50)

Immediately we see that several terms (and the minus signs) cancel to give:

2

2tan

1

(5.5.51)


Dr. C. Caprani 132

Thus we have the final particular solution of equation (5.5.46) in conjunction with

equations (5.5.49) and (5.5.51).

As referred to previously, the total solution is the sum of the particular and

complimentary solutions, which for us now becomes:

C Pu t u t u t (5.5.52)

cos sin sint

d du t e A t B t t (5.5.53)

Notice here that we used equation (5.2.35) since we have redefined the amplitude and

phase in terms of the forcing function. To determine the unknown constants from the

initial parameters, 0

u and 0

u we differentiate equation (5.5.53) to get:

cos sin cost

d d d du t e B A t A B t t (5.5.54)

Now at 0t , we have from equations (5.5.53) and (5.5.54):

00 sinu u A (5.5.55)

And:

00 cos

du u B A (5.5.56)

Solving for A first from equation (5.5.55) gives:

0

sinA u (5.5.57)


Dr. C. Caprani 133

And introducing this into equation (5.5.56) gives:

0 0sin cos

du B u (5.5.58)

Multiplying out and rearranging gives:

0 0cos sin

du B u (5.5.59)

From which we get:

0 0

cos sin

d

u uB

(5.5.60)

And now we have completely defined the time history of the problem in terms of its

initial parameters.

Remember that:

The complimentary solution ( Cu t ): represents the transient state of the system

which dampens out after a period of time, as may be realized when it is seen that

it is only the complementary response that is affected by the initial state (0

u and

0u ) of the system, in addition to the exponentially reducing term in equation

(5.5.53);

The particular solution ( Pu t ): represents the steady state of the system which

persists as long as the harmonic force is applied, as again may be seen from

equation (5.5.53).


Dr. C. Caprani 134

5.6.5 Important Formulae

SDOF Systems

Fundamental equation of motion ( ) ( ) ( ) ( )mu t cu t ku t F t

Equation of motion for free vibration 2( ) 2 ( ) ( ) 0u t u t u t

Relationship between frequency, circular

frequency, period, stiffness and mass:

Fundamental frequency for an SDOF system.

1 1

2 2

kf

T m

Coefficient of damping 2c

m

Circular frequency 2 k

m

Damping ratio cr

c

c

Critical value of damping 2 2cr

c m km

General solution for free-undamped vibration

( ) cosu t t

2

2 0

0;

uu

0

0

tanu

u

Damped circular frequency, period and

frequency

21d

2;

d

d

T

2

d

df

General solution for free-damped vibrations

( ) cost

du t e t

2

2 0 0

0;

d

u uu

0 0

0

tand

u u

u


Dr. C. Caprani 135

Logarithmic decrement of damping ln 2n

n m d

um

u

Half-amplitude method 0.11

m when 0.5

n m nu u

Amplitude after p-cycles 1

p

n

n p n

n

uu u

u

Equation of motion for forced response

(sinusoidal) 0

( ) ( ) ( ) sinmu t cu t ku t F t

General solution for forced-damped vibration

response and frequency ratio

sinp

u t t

1 2

2 220 1 2 ;F

k

2

2tan

1

Dynamic amplification factor (DAF) 1 2

2 22DAF 1 2D


Dr. C. Caprani 136

MDOF Systems

Fundamental equation of

motion Mu + Cu + Ku = F

Equation of motion for

undamped-free vibration Mu + Ku = 0

General solution and derivates

for free-undamped vibration

sin t u = a

2 2sin t u = a u

Frequency equation 2 K M a = 0

General solution for 2DOF

system

1 1 1 2 2 1

2 2 2 2 2

0 0

0 0

m u k k k u

m u k k u

Determinant of 2DOF system

from Cramer‟s rule 2 2 2 2

2 1 1 2 2 20k k m k m k K M =

Composite matrix 2 E K M

Amplitude equation Ea = 0


Dr. C. Caprani 137

Continuous Structures

Equation of motion

Assumed solution for free-

undamped vibrations

General solution

1 2

3 4

sin cos

sinh cosh

x A x A x

A x A x

Boundary conditions for a simply

supported beam

2

20, 0 and 0, 0

vv t EI t

x

2

2, 0 and , 0

vv L t EI L t

x

Frequencies of a simply

supported beam

2

n

n EI

L m

Mode shape or mode n: (A1 is

normally unity) 1

sinn

n xx A

L

Cantilever beam boundary

conditions

0, 0 and 0, 0v

v t tx

2 3

2 3, 0 and , 0

v vEI L t EI L t

x x

Frequency equation for a

cantilever cos( )cosh( ) 1 0L L

Cantilever mode shapes

1

sin( ) sinh( )

sin( ) sinh( )

cos( ) cosh( )

cosh( ) cos( )

n

x x

L Lx A

L L

x x

Bolton method general equation 1

2

E

E

Kf

M

4 2

4 2

, ,,

v x t v x tEI m p x t

x t

,v x t x Y t


Dr. C. Caprani 138

Practical Design

Peak acceleration under foot-loading

00.9 2

Ia f

M

70 NsI 40% mass per unit areaM

Maximum dynamic deflection max stu u K

Maximum vertical acceleration 2

max maxu u

BD37/01 requirement for vertical acceleration 0.5 f


Dr. C. Caprani 139

5.6.6 Glossary

Structural dynamics introduces many new terms and concepts so it‟s beneficial to

keep track of them in one place. Fill this out as you progress through the notes.

Symbol Name Units

Documents

Structural Analysis - Caprani