statics and dynamics

11

Instructor: Dr. Samuel O. Osisanya, P. E.

Associate Professor

Mewbourne School of Petroleum & Geological

Engineering

Course Title

PE 2113 Statics and Dynamics

Fall 2010

2

Petroleum Engineering Curriculum

Basic Science/General

Engineering

Fundamentals of PE -

Geology, Rock & Fluid

Properties

Applications &

Economics

Drilling

Engineering

Reservoir

EngineeringProduction

Engineering

23

PREAMBLE

Objectives

Course Content

Ground Rules

Grading System

Caring Isnt Coddling

Learning is Your Responsibility

4

Text Books

R. C. Hibbeler, Engineering Mechanics-

STATICS, 12th Edition, Pearson-

Prentice Hall,

35

Objectives

To make you all the best engineers of the 21st century

The main objective of this course is to provide students with a clear and thorough

presentation of the theory and application of

engineering mechanics. The concept will be

to apply the principles of mechanics first to a

particle then to a rigid-body subjected to a

coplanar system of forces.

COURSE OUTCOMES1. Provide students with a clear and thorough presentation

of the theory and application of engineering mechanics.

2. The concept will be to apply the principles of mechanics

first to a particle then to a rigid-body subjected to a

coplanar system of forces.

3. Draw free-body diagram which is particularly necessary

for solving mechanics problems.(e)

4. Apply the concept and understanding to solve problems

thereby improving their problem-solving skills.(a, e)

5. To continue to emphasize the importance of units either

SI or US Customary (FPS) (a)

Note:

(a) An ability to apply knowledge of math, science and

engineering

(e) An ability to identify, formulate, and solve engineering

problems

4COURSE EVALUATIONCourse evaluations are an indirect tool to evaluate attainment of

student outcomes. In order to increase the on-line course

evaluation survey response rate, you must do the following:

Go to the link http:/eval.ou.edu

Mewbourne School of Petroleum and Geological Engineering utilizes student ratings as one of the bases for evaluation in the

teaching effectiveness of each of its faculty members. The results

of these ratings are important data used by the faculty members to

improve their own teaching effectiveness, and programs use the

data to assess achievement of a set of learning outcomes. The

original request for the use of these forms came from students,

and it is students who eventually benefit most from their use.

Please take this task seriously, evaluate courses on-line and

respond as honesty and precisely as possible, both to the machine

scored items and to the open-ended questions. We appreciate your feedback.

What is ABET

Accreditation Board for Engineering and Technology, Inc.

MPGE is an accredited program and is reviewed by ABET every six years.

*Upcoming review July 2011.

Accreditation is based on a continuous improvement process of our program

outcomes and objectives.

5A an ability to apply knowledge of mathematics, science, and engineering

Ban ability to design and conduct experiments, as well as to analyze and

interpret data

C

an ability to design an system, component, or process to meet desired needs

(within realistic constraints such as economic, environmental, social,

political, ethical, health and safety, manufacturability, and sustainability)

D an ability to function on multi-disciplinary teams

E an ability to identify, formulate, and solve engineering problems

F an understanding of professional and ethical responsibility

G an ability to communicate effectively

Hthe broad education necessary to understand the impact of engineering

solutions in a global, economic, environmental, and societal context

I a recognition of the need for, and ability to engage in life-long learning

J a knowledge of contemporary issues

Kan ability to use the techniques, skills, and modern engineering tools

necessary for engineering practice

MPGE Program Outcomes

We must demonstrate that our students have:

MPGE

Program Objectives

Area of Technical Skills

Our Alumni will have successful professional careers in petroleum engineering.

Area of Business Acumen

Our Alumni will be emerging or established leaders among their peers demonstrated by leading projects or teams and creating business

value.

Area of Continuous Learning

Our Alumni will be engaged in activities of life-long learning in petroleum industry through professional development, advanced

degrees, certifications, and/or continuing education and training.

Area of Service

Our Alumni will be emerging leaders in service to the profession.

6Program Objectives

Mapping A K Outcomes to Objs.

Mapping Courses

to A- K Outcomes

Data Collection

Course Data

ABET Questionnaire

Exit Interviews

External Constituents

Recruiters

Intern Supervisors

Advisory Board

Alumni

CDRP

Annual Assessment

ABET Certification

Self Study

Evaluation

Semester Annual 2-3 years09/09/09

FLOW DIAGRAM FOR ABET PROCESS

Whats in it for you?

Your academic performance in the PE program indicate where improvements

are needed

Your participation in course evaluations indicate where improvements are

possible

Your participation in surveys after graduation indicate how well the program

has met its objectives

7GROUND RULES

Attendance is required

Be prompt as late arrivals disorganize the class

Must follow adds and drops policies

Adhere to homework rules & format

Adhere to test rules including academic dishonesty

OTHERS

Read ahead of class

Time management/how do you learn

Be the best, neat, and honest professional drilling engineer

Recipe for an F

No, I dont grade on a curve . Its too complicated

Caring Isnt Coddling

Term report (essay) evaluation

815

Engineers, learning ultimately is your responsibility. Today and in the future,

your only security is what you know, what you can do, and how fast you

can learn - Dr. K.K. Millheim - JPT 04/92

Your future is in your hands. While mentors will supply guidance, it is your responsibility to make learning a key priority.

How do you learn?

commitment, set goals, time management,

take notes, study everyday

16

COURSE CONTENT1. General Principles

Statics

2. Force Vectors

3. Equilibrium of a Particle

4. Force System Resultants

5. Equilibrium of a Rigid-Body

6. Moment of Inertia

Dynamics

7. Kinetics of a Particle: Force and Acceleration

8. Kinetics of a Particle: Work and Energy

9. Kinetics of a Particle: Impulse-Momentum.

917

Chapter 1 General Principles

18

Objectives

10

19

Mechanics Mechanics deals with the state of rest or motion of

bodies subjected to action of forces.

Mechanics is divided into: Rigid-Body; Deformable-Body and Fluid Mechanics

Rigid-Body: forms the basis for the design and analysis of structural, mechanical, and electrical

devices.

Rigid-Body is divided into Statics and Dynamics

Statics deals with body at rest or moves with constant velocity (i.e. acceleration = 0)

Dynamics deals with the accelerated motion of bodies

20

Historical Development

Statics development principles are based on formulationfrom measurements of geometry and force. Archimedes

(287-212 BC) developed the levers, pulley, inclined

planes.

Dynamics involves time. Galileo (1564 1642 AD)developed pendulums and falling bodies

Newton (1642 1727 AD) the father of mechanics, madethe most significant contributions: laws of motion; laws of

universal gravitational attraction.

Euler, Lagrange, DAlembert and others applied Newtonslaws

11

21

Fundamental Principles and Concepts

The concepts and principles are: basic quantitiesand Newtons laws of motion

Basic quantities (Units) are

Length, Time, Mass, and Force

Length, Time and Mass are basic units

Force is a derived unit

Other basic quantities concepts are idealization,particle, rigid-body, concentrated force

Newtons laws of motion

22

Fundamental Principles and Concepts

Idealization models or idealizations are used inmechanics in order to simplify application of the theory

Particle has a mass, but a size that can be neglected,Earth is smaller than the orbit, i.e. the Earth can be

modeled as a particle. The geometry of a particle (body)

will not be involved in the analysis of the problem

Rigid-body means deformations are relatively small andhence neglected (concept of rigid-body is relative)

Concentrated force: represents the effect of a loadingwhich is assumed to act at a point on a body

12

23

Newtons Laws of MotionThe entire subject of rigid body mechanics is formulated on the basics of Newtons 3 laws of motion; the validity of which is based on experimental observation

First Law a particle at equilibrium ( i.e. at rest) or moving in a straight line with constant velocity will remain in the state

provided the particle is not subjected to an additional force.

Second Law: F = mass x acceleration = m x a

Third Law: the mutual forces of action and reaction

between two particles are equal, opposite and collinear;

i.e. action and reaction are always equal and opposite

24

Newtons Law of Gravitational Attraction Newton postulated a law governing the attraction between any 2

particles, expressed mathematically as follows: F = Gm1m2/r2

F = force, m1, m2 = mass of each of the two particles

G = universal constant of gravitation and according toexperiment G = 66.73 x 10-12 (m3/(kg.s2))

Weight: is the only sizeable gravitational force between a particle(with mass m) located at or near the earth surface and earth (of

mass Me).

Hence, W =GmMe/r2. If g = GMe/r

2 , then W = mg.

Since F = ma, then g is called acceleration due to gravity.

Question: Show that W = mg

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25

Fundamental Units Used in Mechanics

Units are necessary to describe the results of measurements

Dimensions are used to describe physical quantities; independent of units. [L], [M], [T] for length, mass, and time

respectively. This is a powerful tool in analyzing fluid mechanics problems

There are two systems of units:

SI (aka MKS, SI a moderate version of the metric system)

US or British

SI coherent sets of basic (fundamental) units are Meter, Kilogram and Second for length, mass, and time respectively.

US coherent sets of basic (fundamental) units are Feet, Slug, and Second for length, mass, and time respectively.

26

Systems of Units

14

27

Fundamental Units Used in Mechanics

The basic unit of temperature in the SI system is the Kelvin (K) and it is related to oC as follows: K = 273 +oC

The basic unit of temperature in the US system is the Rankine (R) and it is related to oF as follows: R = 460 +oF

The unit of energy in the SI system is the Joule (1 J = 1 N-m)

The unit of energy in the US system is the ft-lbf (1 ft-lbf = 1.356 N-m)

Question: Show that I ft-lbf = 1.356 N-m

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Conversion Factors

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29

Conversion of Units

All engineers, especially the engineers of the 21st Century, must beable to convert between units. If a physical quantity is expressed in any

system, it is simple matter to convert the units from that system to

another. To do this, the basic unit conversion must be known and a

logical unit analysis must be followed.

Mistakes can be minimized if you remember that a conversion factorsimply relates the same physical quantity in two different unit systems.

As an example 1.0 in = 25.4 mm = 2.54 cm describe the same length

quantity.

The mastering of the procedure for units conversions takes a lot ofpractice. Hence, work a lot of problems.

There is a list of unit conversions in your mechanics textbook and manyother textbooks of physics, chemistry, and engineering. Try to develop

these values on your own. Do not memorize them.

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Prefixes

16

31

Numerical Calculations

Numerical calculations involve the following:

Dimensional homogeneity

Significant figures

Rounding off numbers

Finally, the calculations themselves

32

General Procedure for Engineering Analysis

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33

Summary

11


Associate Professor

Mewbourne School of Petroleum &

Geological Engineering

COURSE TITLE

PE 2113 STATICS AND DYNAMICS

FALL 2010

2

Chapter 3 Equilibrium of a Particle

23

Objectives

To introduce the concept of the free-body diagram for a particle

To show how to solve particle equilibrium problems using the

equation of equilibrium

4

Condition for the Equilibrium of a Particle

Recall: Particle has a mass, but a size that can be neglected. The geometry of a particle (body) will not be

involved in the analysis of the problem

When a particle is at rest or moves with constant velocity, it is in equilibrium. This requires that all the

forces acting on the particle form a zero force

resultant.

This condition may be expressed mathematically as F = 0; where F is the vector sum of all the forces

acting on the particle.

35

Condition for the Equilibrium of a Particle

Not only is the equation F = 0 a necessary condition for equilibrium, it is also a sufficient

condition.

This follows from Newtons 2nd Law of motion which says that F = ma.

Thus, it follows that a = 0, which implies that the particle indeed moves with constant velocity or

remains at rest.

6

Free-Body Diagram In order to account for all the forces acting on a

particle, it is necessary to draw a free-body diagram.

The diagram is an outlined shape of the particle that

shows all the forces, listed with their known or

unknown magnitudes and directions.

The free-body diagram is simply a sketch which shows the particle free from its surroundings with all the forces that act on it.

There are two types of connections often encountered in particle equilibrium problems: Springs and Cables and Pulleys

47

Springs

Springs: If a linear elastic spring is used for support, the length of the spring will change in

direction proportional to the force (this follows

from Hookes law of elasticity). This is expressed mathematically as F = ks

k = spring constant or stiffness, (N/m)

s = change in length; if s is positive, then F pull the spring and if s is negative, then F must push on it

8

Cables and Pulleys In this course (except in Chapter 7 Internal Forces),

all cables (or cords) are assumed to have negligible

weight and cannot stretch.

Also, a cable can support only a tensile or pulling force, and this force always acts in the direction of the

cable.

Also, the tension force developed in a continuous cable which passes over a frictionless pulley must

have a constant magnitude to keep the cable in

equilibrium.

59

Procedure for Drawing a Free-Body Diagram Since, we must account for all the forces on the

particle when applying the equation of equilibrium, the importance of first drawing the free-body diagram cannot be overemphasized. The following 3 steps are necessary:

1. Draw outlined shape: imagine the particle to be isolated or cut free from its surroundings

2. Show all forces: forces can be active (tend to set the particle in motion) or can be reactive (tend to prevent motion)

3. Identify each force: known forces must be labeled with their proper magnitude and directions.

10

Example - Free-Body Diagram The sphere on this figure has a mass of 6 kg and is

supported as shown.

Two forces on the

sphere: its

weight & the

force of cord CE

Two forces on the

cord: FCE and FEC. Both are equal and

opposite (Newtons 3rd Law

Three forces on the knot:

FCE, FCBA, and FCD, .Note:

the weight of the sphere

does not directly act on

the knot

611

Coplanar Forces 2-D Force Systems

Recall: In two dimensions, it is easy to find the angle between two lines or the components of a force

parallel or perpendicular to a line. This is by using

trigonometry.

F = 0

Fxi + Fyj = 0

For the above vector equation to be satisfied, the forces x and y components must both be equal to zero. Hence

Fx= 0 Fy= 0

816

3-Dimensional Force Systems

Recall: In 3-D, vector method is employed to solve the problem. The dot product defines a particular

method for multiplying two vectors and is used to solve the 3-D problems.

F = 0

Fxi + Fyj + Fzk= 0

For the above vector equation to be satisfied, the forces x , y, and z components must both be equal to

zero. Hence, Fx= 0 Fy= 0 Fz= 0

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22

Application of Free-Body Diagram

to Petroleum Engineering -

Drilling

12

23

HOISTING SYSTEM

The hoisting system is one of the 6 rig systems. Its function is to provide a means of lowering and raising equipment into or out of the hole

Principal components

Derrick & substructure

Block & tackle pulley arrangements

Drawworks

Major routine operations

Making connection

Making a trip

Slip and cut program

24

Schematic of Block and Tackle

1. Comprises of crown block,

traveling block, and drilling

line

2. Provides a mechanical

advantage, which permits

easier handling of large

loads

3. Generally mechanical

advantage is less than n

(i.e. less than 100%) due

to friction

4. As n increases,

mechanical advantage

decreases tremendously.

11


Associate Professor



COURSE TITLE


FALL 2010

2

Chapter 4 Force System Resultants

23

Objectives

To discuss the concept of the moment of a force and show how to calculate it in two and 3-dimensions

To provide a method for finding the moment of a force about a specific axis

To define the moment of a couple

To present methods for determining the resultants of non-concurrent force systems

To indicate how to reduce a simple distributed loading to a resultant force having a specified location

4

Moment of a Force - Scalar Formulation

The moment of a forceabout a point or axisprovides a measure ofthe tendency of theforce to cause a body torotate about the point oraxis

On the right figure, theforce tends to cause thepipe to turn about the z-axis

35


The tendency of the forceto cause a body to rotate

about the point or axis is

called a TORQUE, but

most often it is called the

MOMENT of a FORCE or

simply the MOMENT (MO)z

The moment axis (z) isperpendicular to x-y plane

6

In figure b on the right, force Fz will notrotate the pipe about the z-axis. Instead it

tends to rotate it about the x-axis.

However, it may not be possible, but

there is a tendency to rotate.

In figure c, no moment is produced aboutpoint O.

The magnitude of the moment Mo = F*dand it is a vector

d = moment arm or perpendiculardistance from the axis at point O to the

line of action of the force.


47

The direction of Mo will be specifiedby using the RIGHT-HAND RULE

The fingers of the right hand arecurled such that they follow the

sense of rotation which would occur if

the force could rotate about point O

The thumb then points along themoment axis so that it gives the

direction and sense of the moment

vector which is upward and

perpendicular to the shaded plane

containing F and d


8

Resultant Moment of a System of Coplanar Forces

The resultant moment MROof the system

of coplanar forces can be determined bysimply adding the moments of all forcesalgebraically since all the moment vectorsare collinear.

+ MRO= F*d

Sign convention

Clockwise is taken as negative

Counterclockwise is taken as positive

59

Operations Damage (makeup & breakout) Bending during tonguing (avoidable)

Observe max TJ height (refer API RP7G) (tongue jaw)

Back up with winch and snatch block if tight breakout

Always use two tongs

When making or breaking connections, the pull on the tong lines exerts a bending force on the pipe, at the top of the slips. If this force is excessive the

pipe will bend, which is seriously bad news. Its also totally avoidable.

There are two cases; tongs at 90 between the arms and at 180. Which is likely to be the worst case (ask the class)? API RP7G gives a formula to use

to work out H, the height of the tool joint center, to avoid bending.

10

Operations Damage (makeup & breakout) Mis-stabbing - Damage to the box seal area. Always

pick up, inspect, restab - NEVER knock it in!

611

Worked Examples

12

713

14

Cross Product

The cross product of two vectors A and Byields a vector C which is written as follows:C = A x B (C equals A cross B)

The magnitude of C is the product of themagnitude of A and B and the sine of angle between their tails

The direction of vector C is perpendicular tothe plane A and B such that C is specified bythe RHR. Thus C can be written as follows:

C = A x B = (ABsin)uc

ABsin defines the scalar

uc = unit vector defines the direction of C

815

Laws of Operation of Cross Product

1. The commutative law is not valid, i.e

A x B B x A, in fact, A x B = - B x A

2. Multiplication by a scalar

a(A x B) = (aA) x B = A x (aB) = (A x B)a

3. The distributive law

A x (B + D) = (A x B) + (A x D)

16

Commutative Law of Cross Product

The commutative law is not valid, i.e

A x B B x A, in fact, A x B = - B x A = -C

917

Cartesian Vector Formulation of Cross Product

This is concerned with the cross productof the pair of Cartesian vectors. For

example

i x j = ijsin()uc, the magnitude is 1x1Sin(90) = 1 and its direction is

determined by the RHR and it shows

that the resultant vector points in the +k

direction, thus, i x j = (1)k. In similar

manner we have the following:

i x j = k i x k = -j i x i = 0

j x k = i j x i = -k j x j = 0

k x i = j k x j = -i k x k = 0

18

Cross Product of Unit Vectors

i x j = k i x k = -j i x i = 0

j x k = i j x i = -k j x j = 0

k x i = j k x j = -i k x k = 0

1. The above results should not be memorized, rather it should be

clearly understood how each is obtained by using the RHR and the

definition of the cross product. A simple scheme shown right above

is helpful for obtaining the same results when the need arises.

2. If the circle is constructed as shown, then crossing two such vectors in a counterclockwise fashion around the circle yields the

positive third unit vector, e.g. k x i = j.

3. Moving clockwise, a negative unit vector is obtained, e.g. i x k = -j

10

19

Cross Product of 2 General Vectors A and B If A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk , it can be shown that

A x B = (AyBz- AzBy)i - (AxBz- AzBx)j + (AxBy- AyBx)k

The above equation may also be written in a more compact determinant form as follows:

To find the cross product of any two Cartesian vectors A and B, it is

necessary to expand a determinant whose first row of elements consists

of the unit vectors i, j, and k and whose second and third rows represent

the x, y, and z components of the two vectors A and B respectively

20

Cross Product of 2 Vectors

The figure on the right showshow to expand a 3 x 3determinant.

The figure shows how the 3minors are generated. Addingthe results and noting that thej element must include theMINUS sign yields theexpanded form of A x B asfollows:

A x B = (AyBz- AzBy)i - (AxBz-AzBx)j + (AxBy- AyBx)k

11

21

Moment of a Force Vector Formulation (Application of Cross Product)

The moment of a force F about a point O or actually about the moment axispassing through O and perpendicular to the plane containing O and F is

expressed as MO = r x F

r = a position vector drawn from O to any point lying on the line of action of F

Magnitude: MO = rF sin = F (r sin)= Fd

Direction: The direction is determined using RHR. From the figure above, thecurl of the fingers like the curl around the moment vector, indicates thesense of rotation caused by the force. Since the cross product is not

commutative, it is important that the proper order of r and F be

maintained (i.e. you cannot write Mo = F x r)

22

Principle of Transmissibility: From the figure onthe left MO = rA x F. However, r can extend from O toany point on the line of action of F. That is MO = rB x F

= rC x F. As a result F has the properties of a sliding

vector and can therefore act at any point along the line

of action and will create the same moment about point

O. This is refer to as the principle of

transmissibility.

Cartesian Vector Formulation of a Moment: If we establish x, y, and z coordinates axes, then the

position vector r and F can be express in Cartesian

form as shown on the left. Clearly

MO = (ryFz- rzFy)i - (rxFz- rzFx)j + (rxFy- ryFx)k

The physical meaning of these moment components becomes evident by studying the figure on the left. For

example, the i component of MO is determined from

the moments of Fx, Fy, Fz about the x-axis. In

particular, note that Fx does not create a moment or

tendency to cause turning about the x axis since the

force is parallel to the x axis

12

23

Resultant Moment of a System of Forces

If a body is acted upon by asystem of forces as shown on

the figure on the left, the

resultant moment of the forces

about point O can be determined

by vector addition resulting from

the successive application of MO= rA x F. This is expressed

symbolically as follows:

MO= (r x F)

24

Note: uF = rB - rC

AB = (14)0.5

13

25

14

27

Principle of Moments Principle of Moments also known as Varignons

theorem (1654 1722) states that the moment of a force about a point is equal to the sum of the moments of the forces components about the point. That is if F = F1 + F2, then

MO = r x F1 + r x F2 = r x (F1 + F2) = r x F

This concept has important application to the solution of problems and proofs of theorem that follow, since it is often easier to determine the moments of a forces components rather than the moment of the force itself.

15

29

16

31

32

Reduction of a Simple Distributed Loading

In many situation, a very large surface area of a body may be subjected to DISTRIBUTED LOADINGS such as those

created by wind, fluid, or simply the weight of a material

supported over the bodys surface.

The INTENSITY of these loadings at each point on the surface is defined as the pressure p (force per unit area)

which can be measured in lbf/ft2 or Pascal (N/m2)

In this section, we are going to consider the most common case of a distributed pressure loading which is uniform

along the axis of a flat rectangular body upon which the

loading is applied.

17

33

Distributed Loading

This figure shows the most common case of a distributed pressure loading which is uniform along the axis of a flat rectangular body upon which the loading is applied.

The arrows show the direction of the intensity of the pressure

18

35

Distributed Loading

The entire loading on the plate is a system of parallel forces infinite in

number and each acting on a

separate differential area of the

plate. The loading function p=p(x)

Pascal is only a function of x since

the pressure is uniform along the

y-axis.

Multiplying p=p(x) by the width a-m of the plate results in w = [p(x)a]

N/m = w(x) N/m as shown in the

figure above right. w(x) is a

measure of load distribution along

the y-axis which is in the plane of

symmetry of the loading

Consequently, the load-intensity

diagram for w=w(x) is represented

by a system of coplanar parallel

forces shown in 2-D.

36

Distributed Loading

Since, the load-intensity diagram for w=w(x) is

represented by a system of

coplanar parallel forces

shown in 2-D, using the

method of reduction, this

system of forces can be

simplified to a single

resultant force FR and its

location x can be specified

19

37

Magnitude of Resultant Force for Distributed Loading

Since FR = F, the magnitude of FR is equal to the sum of all the forces in the system.

Since dF is acting on an element of length dx and w(x) is a force/unit length, then at the location x, dF = w(x)dx= dA, the area under the curve. That is, the magnitude of dF is determined from the colored differential area dA under the loading curve.

That is, FR = w(x)dx =dA = A= total area under the loading curve

38

Location of the Resultant Force for a Distributed Loading

Applying MRo = MO the location x of the line of action of FR can be

determined by equating the moment

of the force resultant and the force

distribution about point O (the y-

axis).

Since dF produces a moment of xdF = xw(x)dx about O, then for the

entire plate xFR =xw(x)dx

Solving for x gives the following: 1. The equation represents the x-

coordinate for the geometric center or

centroid of the area under the

distributed loading diagram.

2. Hence, the resultant force has a line of

action which passes through the

centroid C of the area defined by the

distributed loading diagram w(x)

20

39

Location of the Resultant Force for a Distributed Loading

Oncex is determined, FR by symmetry passes through point

(x,O) on the surface of the plate

as shown on the right.

In conclusion, the resultant force has a magnitude equal to

the volume under the

distributed load curve p=p(x)

and a line of action which

passes through the centroid

(geometric center) of this

volume.

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42

22

43

11


Associate Professor



COURSE TITLE

PE 2113 STATICS AND DYNAMICSFALL 2010

2

Chapter 5 Equilibrium of a Rigid Body

23

Objectives

To develop the equations of equilibrium for a rigid body

To introduce the concept of the free-body diagram for a rigid body

To show how to solve rigid-body equilibrium problems using the equations of equilibrium

4

Conditions for Rigid-Body Equilibrium The conditions for rigid-body equilibrium are developed by

considering the free-body diagram of an arbitrary particle of the body. Two forces act on this particle.

The internal force, fi and the external force Fi

The internal force is caused by interactions with adjacent particles and external force represents the effects of gravitational, electrical, magnetic, or contact forces between the particles.

If the particles is in equilibrium, then applying Newtons first law we have: Fi + fi = 0

For all the particles, Fi + fi = 0

35

Conditions for Rigid-Body Equilibrium

For all the particles, Fi + fi = 0

Furthermore, fi will be zero since the internal forces between particles within the body will occur

in equal but opposite collinear pairs (Newtons 3rd

Law). Hence, Fi = 0

Using the same approach, considering the moment of the forces acting on the particle, the particle will

be in equilibrium if ri x (Fi + fi)= ri x Fi + ri x fi = 0

For all the body, we have ri x Fi + ri x fi = 0

6

Conditions for Rigid-Body Equilibrium

For all the body, we have ri x Fi + ri x fi = 0

The second term is zero since the sum of the internal forces = 0, and hence the resultant moment of each pair of forces about point O is zero, that is

Mo = ri x Fi = 0

Hence the TWO equations for equilibrium of a rigid body can be summarized as follows:

F = 0

Mo = 0

47

Conditions for Rigid-Body Equilibrium These two equations F = 0 and Mo = 0 require that a rigid

body will remain in equilibrium provided the sum of all the

external forces acting on the body is equal to zero and the sum

of the moments of the external forces about a point is equal to

zero.

These two equations are also sufficient for maintaining equilibrium.

Many types of engineering problems involve symmetric loading and can be solved by projecting all the forces acting on a body

into a single plane.

We are now going to apply these equations to 2-D and 3-D problems. Recall that for 2-D we can always use scalar solution

and for 3-D we generally used the vector formulation

8

Equilibrium in 2-D:Free-Body Diagrams

Successful application of the equations of equilibrium requires a complete specification of all the known and

unknown external forces that act on the body. The best way

to account for these forces is to draw the bodys free-body diagram.

On the sketch it is necessary to show all the forces and couple moments that the surrounding exert on the body so

that these effects can be accounted for when the equations

of equilibrium are applied.

Hence, a thorough understanding of how to draw a free-body diagram is of primary importance for solving problems

in mechanics

59

Free-Body Diagrams Support Reactions

As a general rule, if a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction.

Likewise, if reaction is prevented, a couple moment is exerted on the body

10

Example - Support Reactions

This figure shows one way a horizontal member such

as a beam is supported a roller or cylinder. The roller prevents the beam from translating in vertical direction

611

Example Support Reactions

This figure shows another way a horizontal member such as a

beam is supported using a PIN. The pin passes through a hole in the beam and two leaves which are fixed to the

ground. The pin can prevent translation in any direction.

Easier to represent this force by 2 components, Fx and Fy.

12

Example Support Reactions

The most restrictive way to support the beam would be to

use a fixed support as shown above. This support will

prevent both translation and rotation of the beam. Hence,

to do this a force and couple moment must be developed

on the beam at its point of connection.

713

14

815

External and Internal Forces

A rigid body is a composition of particles hence both external and internal loadings may act on it.

However, internal forces are not represented on the free-body diagram. Why? These internal forces always occur in

equal but opposite collinear pairs and hence their net effect

on the body is zero.

In automobiles, the connecting parts represent internal forces which would not be included on the free-body

diagram of the automobile.

Particles or bodies outside the boundary exert external forces on the system and hence these forces must be

shown on the free-body diagram

16

Weight and Center of Gravity

A body subjected to gravitational field has a specific weight. This force resultant is called the weight of the body

and the location of its point of application is called center of

gravity. The method used for its calculation will be

developed in Ch.9

In this Ch. 5 however, if the weight of the body is important, this force will be reported in the problem statement.

Also, when the body is uniform or made of homogeneous material, the center of gravity will be located at the bodys geometric center or centroid. If the body is non-

homogeneous (i.e. heterogeneous) or has unusual shape,

then the location of its center of gravity will be given.

917

Idealized Models When an engineer performs a force analysis of any object, he or she

considers a corresponding analytical or idealized model that gives

results that approximate as closely as possible the actual situation.

To do this, careful choices have to be made so that selection of the type of supports, the material behavior, and the objects dimension can be justified.

In complex cases the above process may require developing several different models of the object that must be analyzed but in any case,

this selection process requires both SKILL and EXPERIENCE.

In this course, idealized models of specific objects will be given in some of the examples. It should be realized, however, that each

case represents the reduction of a practical situation using simplified

assumptions.

18

Procedure for Drawing a Free-Body DiagramTo construct a free-body diagram for a rigid body or group of bodies

considered as a single system, the following steps should be performed:

Draw Outlined Shape

Imagine the body to be isolated or cut free from its constraints and connections and draw (sketch) its outlined shape.

Show All Forces and Couple Moments

Identify all the external forces and couple moments that act on the body. Those generally encountered are due to (1) applied loadings, (2) reactions occurring at the

supports or at points of contact with other bodies (see Table 5-1), and (3) the

weight of the body. To account for all these effects, it may help to trace over the

boundary, carefully noting each force or couple moment acting on it.

Identify Each Loading and Give Dimensions

The forces and couple moments that are known should be labeled with their proper magnitudes and direction. Letters are used to represent the magnitudes and

direction angles of forces and couple moments that are unknown. Establish an x, y

coordinate system so that these unknowns, Ax, By, etc., can be identified. Indicate

the dimensions of the body necessary for calculating the moments of forces.

10

19

Important Points No equilibrium problem should be solved without first drawing the free-

body diagram, so as to account for all the forces and couple moments

that act on the body.

If a support prevents translation of a body in a particular direction, the support exerts a force on the body in that direction.

If rotation is prevented, then the support exerts a couple moment on the body.

Study Table 5-1. DO NOT MEMORIZE IT

Internal forces are never shown on the free-body diagram since they occur in equal but opposite collinear pairs and therefore cancel out.

The weight of a body is an external force, and its effect is shown as a single resultant force acting through the bodys center of gravity G.

Couple moments can be placed anywhere on the free-body diagram since they are free vectors. Forces can act at any point along their lines

of action since they are sliding vectors.

20

Example 5.1

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Equations of Equilibrium Recall that the two equations which are both necessary and sufficient

for equilibrium are as follows:

F = 0 Mo = 0

When the body is subjected to a system of forces which all lie on the x-y plane, then the forces can be resolved into their x and y components. Hence, the conditions for equilibrium in 2-D are:

Fx = 0 Fy = 0 Mo = 0

Fx and Fy represent, respectively, the algebraic sums of the x and y components of all the forces acting on the body, and Mo represents the algebraic sum of the couple moments and the moments of all the force components about the axis perpendicular to the x-y plane and passing through the arbitrary point O, which may lie either on or off the body

24

Equations of Equilibrium Procedure for AnalysisCoplanar force equilibrium problems for a rigid body can be solved using the

following procedure.

Free-Body Diagram

Establish the x, y coordinate axes in any suitable orientation. Draw an outlined shape of the body. Show all forces and couple moments acting on the body. Label all the loadings and specify their directions relative to the x, y axes. The sense of a

force or couple moment having an unknown magnitude but known line of action can be

assumed.

Indicate the dimensions of the body necessary for computing the moments of forces.Equations of Equilibrium

Apply the moment equation of equilibrium, about a point (O) that lies at the intersection of the lines of action of two unknown forces. In this way, the moments of these unknowns are

zero about O, and a direct solution for the third unknown can be determined

When applying the force equilibrium equations, orient the x and y axes along lines that will provide the simplest resolution of the forces into their x and y components.

If the solution of the equilibrium equations yields a negative scalar for the force or couple moment magnitude, this indicates that the sense is opposite to that which was assumed in

the free-body diagram.

13

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26

14

27

Two- and Three-Force Members

The solution to some equilibrium problems can be simplified by recognizing members that are

subjected to only 2 or 3 members

2-Force Members: when a member is subjected to no couple moments and forces are applied

only at only two points on the member, the

member is called a two-force member.

3-Force Member: If a member is subjected to only three forces, then it is necessary that the forces

be either concurrent or parallel for the member

to be in equilibrium

28

Constraints for a Rigid Body

To ensure the equilibrium of a rigid body, it is not only necessary to satisfy the

equations of equilibrium, but the body must

also be properly held or constraints by its

supports.

Some bodies may have more supports than are necessary for equilibrium, whereas

others may not have enough or the support

may be arranged in a particular manner that

could cause the body to collapse (i.e. not in

equilibrium).

15

29

Constraints for a Rigid Body -Redundant Constraints When a body has redundant supports, that is, more

supports than are necessary to hold it in equilibrium it

becomes statistically indeterminate.

Statistically indeterminate means that there will be more unknown loadings on the body than equations of

equilibrium available for their solutions.

The additional equations needed to solve indeterminate problems of these types are generally obtained from the

deformation conditions at the points of support. These

equations involve the physical properties of the body

which are studied in subjects dealing with the mechanics

of deformation, such as Mechanics of Materials or Strength of Materials.

30

In this figure, 2-D case, there are five unknowns, that is MA, Ax, Ay,

By and C, for which only three

equilibrium equations can be written

(Fx = 0, Fy = 0, Mo = 0); that

is the problem is statistically

indeterminate.

The additional equations needed to solve indeterminate problems of

these types are generally obtained

from the deformation conditions at

the points of support.

Constraints for a Rigid Body -

Redundant Constraints

16

31

Constraints for a Rigid Body -Improper Constraints

In some cases, there may be many unknowns forces on the body as there are

equations of equilibrium, however,

instability of the body can develop because

of improper constraining by the supports.

For 2-D problems, the axis is perpendicular to the plane of the forces and therefore

appears as a point. Hence, when all the

reacting forces are concurrent at this point,

the body is improperly constrained.

Furthermore it becomes impossible to solve completely for all the unknowns.

Basically, the summation of the moment is

not zero.

32

Summary - Determinacy and Stability

If a body is supported by a minimum number of

constraints to ensure

equilibrium, then it is

statistically determinate.

If it has more constraints than required, then it is

statistically indeterminate.

To properly constrain the body, the reactions must

all be parallel to one

another or concurrent

11


Associate Professor



COURSE TITLE


2

Chapter 9 Center of Gravity and Centroid

23

Objectives

To discuss the concept of the center of gravity, center of mass, and the centroid

To show how to determine the location of the center of gravity and centroid for a system of

discrete particle and a body of arbitrary

shape.

4

Center of Gravity

The center of gravity G is a point which locates the resultant weight of a system of particles.

The weights of the particles comprise of a system of parallel forces which can be replaced by a single (equivalent) resultant weight

having the defined point G of application.

The generalized formulas for determining the center of gravity are given as follows:

35

Center of Gravity

The above equations are easily remembered if it is kept in mind that they simply represent a

balance between the sum of the moments of the

weight of each particle of the system and the

moment of the resultant weight for the system

6

Center of Mass

If the acceleration due to gravity g for every particle in the previous slide is constant, then W = mg and the center of gravity equation becomes the equation above. These are the equations for center of mass.

By comparison, then, the location of the center of gravity coincides with that of center of mass if g is constant.

However, center of gravity is not independent of gravity whereas center of mass is independent of gravity.

47

Center of Gravity for a Rigid-Body

A rigid body is composed of an infinite number of particles and so if the principles used to determine the equation for the CG of discrete

particles are applied to the system of particles composing a rigid body,

it becomes necessary to use integration rather than a discrete

summation of the terms. The resulting equations are given as follows:

8

Center of Mass of a Rigid Body

For the center of mass of a rigid body, the density or mass per unit volume is used.

This density is related to specific weight by

the equation = g.

Substituting g into Eqn. 9.4, and assuming g is constant, the center of mass

is determined.

59

Centroid The centroid C is a point which defines geometric center of an

object. Its location can be determined from formulas similar to

those used to determine the bodys center of gravity or center of mass.

In particular, if the material composing a body is uniform or homogeneous, the density or specific weight will be constant

throughout the body, and therefore this term will factor out of the

integrals and cancel out of Eqn. 9.4.

The resulting formulas define the centroid of the body since they are independent of the body weight and instead depend only on

the body geometry.

Three specific cases are considered: Volume, Area, and Line

10

Centroid

Volume

Area

Length

611

Solving Centroid Problems

In order to use Eqns. 9.4 to 9.7, it is best to choose a coordinate system that

simplifies as much as possible the

equation used to describe the objects boundary.

The terms x, y, and z in the equations refer to the moment arms or coordinates of the center of gravity or centroid of the

differential element used.

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Example Centroid ProblemLocate the centroid (x, y) of the exparabolic segment of the shaded area

10

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Example Centroid ProblemLocate the centroid (x, y) of the shaded area

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Associate Professor



COURSE TITLE


2

Chapter 10 Moments of Inertia

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3

Objectives

To develop a method of determining the moment of inertia for an area

To introduce the product of inertia and show how to determine the maximum and

minimum moments of inertia for an area

4

Definition of Moments of Inertia for Area

Centroid is determine in Chapter 9 by considering the first moment of the area about an axis. That is, we evaluate

the integral xdA

Moment of inertia (I) is the integral of the second moment of an area such as x2dA

The moment of inertia for an area is a quantity that relates the normal stress or force per unit area, acting on the transverse cross section of an elastic beam, to the applied external

moment M, which causes bending of the beam.

The theory of mechanics of materials shows that the stress within the beam varies linearly with its distance from an axis

passing through the centroid C of the beams cross-sectional area.

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Practical Application of

Moments of Inertia

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Bottom Hole Assembly (BHA)

Buckling

BHA Design contributing factors;

1. Resistance of the collars to buckling (OD, ID, material).

Resistance is higher with a larger OD or a smaller ID

2. Length between supports (stabilizers). The greater the distance,

the less the resistance to buckling

Downhole conditions contributing factors;

1. Torque high torque will make buckling more likely as more stress is imposed on the collars

2. Hole angle high angle hole gives support to the DC and decreases the likelihood of buckling

3. In gauge or over gauge hole If hole is over gauged, then lateral support is less effective and buckling is more likely

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Critical buckling load of a tubular

2

2Critical buckling load, c

EIP

L

WhereE = Young's Modulus, 30 x 106 psi for steel

I = 2nd moment of area, in4

L = Length between stabilizers, inches.

Be careful of units !

4 4OD ID64

I

8

Critical Buckling - Example

A 90-ft build assembly is to be run; bit - NB stab 90-ft of 8 OD x 3 ID DCs - FG stab. What is the maximum load on the column before buckling of

these drill collars occurs?

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Critical buckling Example 2

A 90-ft build assembly is to be run; bit - NB stab 90-ft of x 9 x 3 ID DCs - FG stab. What is the maximum load on the column

before buckling of these drill collars occurs?

4 44

9 3318 in

64I

2

280,700 lbsc

EIP

L

10

Effect of Length Between Stabilizers

8 x 3 drill collars. I = 197 in4.

With 90-ft between stabilizers, Pc = 50,000 lbs.

What is the effect of reducing length to 60-ft?

Recalculate the Pc for L = 60-ft,the reduced length

2 2 6

22

30 10 197112,518 lbs

60 12c

EIP

L

More resistance to buckling !

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BHA Buckling - Conclusions

1. Drilling BHAs can and do buckle under high WOB.

2. OD has a major influence on the critical load to buckle, Pc.

3. Length between stabilizers has a major influence on Pc.

4. If the BHA buckles dynamically, there are potential

problems;

a. Fatigue damage on the connections.

b. Physical damage to collars and hole at contact points.

c. High torques required to turn the string.

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Moment of Inertia

Considering the area A shown in Figure

10.2 which lies in the

x-y plane, the

moments of inertia are

determined by

integration as shown

in the equations.

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Polar Moment of Inertia The second moment of dA about pole O or z-axis from

Figure 10.2 is referred to as the polar moment of inertia. Polar Moment of Inertia is used to determine the torsional stress in a shaft.

It is defined as dJo = r2dA, where r is the perpendicular

distance from the pole (z-axis) to the element A.

For the entire area the polar moment of inertia is given below. The relationship between Jo and Ix, Iy, is possible since r2 = x2 + y2

14

Parallel-Axis Theorem for an Area If the moment for an area is known about an axis passing through its centroid, which

is often the case, it is convenient to determine the moment of inertia of the area about a corresponding parallel axis using the parallel-axis theorem.

That is, the moment of inertia of an area with respect to any axis not through its centroid is equal to the moment of inertia of that area with respect to its own centroidal axis plus the product of the area and the square of the distance between the two parallel axes. In equation form: Ix = Ix + Ad2. This equation is also called the transfer equation. Transfer may be made only between parallel axes.

2

A similar expression can be written for Iy, i.e.

The form of each of these 3 equations states that the moment of inertia for an area about an axis is

equal to the moment of inertia for the area about a parallel axis passing through the areas centroid

plus the product of the area and the square of the perpendicular distance between the axes

This 2nd integral = 0 since

YdA = YdA andY= 0.

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Radius of Gyration

The radius of gyration of a planar area has units of length and is a quantity that is often

used for the design of columns in structural

mechanics.

Provided the areas and moment of inertia are known, the radius of gyration are determined

from the following formulas:

(10.6)

16

Moment of Inertia for an Area by

Integration

When the boundaries for a planar area are expressed by mathematical functions, Eqs. 10.1 may be integrated

to determine the moment of inertia for the area

If the element of area chosen for integration has a differential size in two directions, a double integration

must be performed to evaluate the moment of inertia.

Most often however, it is easier to perform a single integration by choosing an element having a differential

size or thickness in only one direction

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Orient the element so that the length is parallel to the axis about which the

moment of inertia is computed. This situation occurs when the rectangular

element shown above (a) is used to determine Ix for the area. Hence the

entire area is at a distance y from the x-axis since it has a thickness dy.

Thus Ix = y2dA. To find Iy, the element is oriented as shown in (b) above. The element lies at the same distance x from y axis so that Iy = x2dA

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1. Introduction2. Force Vectors3. Equilibrium of a Particle4. Force System Resultants5. Equilibrium of a Rigid Body6. Center of Gravity and Centroid7. Moment of Inertia

statics and dynamics

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