Week 3---Equilibrium+FBD

EQUILIBRIUM/EQUILIBRIUM/EQUILIBRIUM/ EQUILIBRIUM/ TYPES OFTYPES OFTYPES OF TYPES OF

SUPPORTS/SUPPORTS/SUPPORTS/ SUPPORTS/ FREE BODY FREE BODY DIAGRAMDIAGRAMDIAGRAMDIAGRAM

EQUATIONS OFEQUATIONS OFEQUATIONS OF EQUATIONS OF EQUILIBRIUMEQUILIBRIUMEQUILIBRIUMEQUILIBRIUM

ddandand

FREE BODY DIAGRAMSFREE BODY DIAGRAMSare the most important topics are the most important topics

you will learn inyou will learn inyou will learn in you will learn in ENGINEERING MECHANICSENGINEERING MECHANICSENGINEERING MECHANICSENGINEERING MECHANICS

Tutorial problems:Tutorial problems:

Chapter 5Chapter 5

EQUILIBRIUM OF A PARTICLE IN 2EQUILIBRIUM OF A PARTICLE IN 2--DDEQUILIBRIUM OF A PARTICLE IN 2EQUILIBRIUM OF A PARTICLE IN 2 DDToday’s ObjectivesToday’s Objectives::

Students will be able to :Students will be able to :

a) Draw a free body diagram (FBD), and,a) Draw a free body diagram (FBD), and,

b) Apply equations of equilibrium to solve a 2b) Apply equations of equilibrium to solve a 2--D problem.D problem.

QUIZQUIZQQ1) When a particle is in equilibrium, the sum of forces acting 1) When a particle is in equilibrium, the sum of forces acting

on it equals (Choose the most appropriate answer)on it equals (Choose the most appropriate answer)on it equals ___ . (Choose the most appropriate answer)on it equals ___ . (Choose the most appropriate answer)

A) a constant B) a positive number C) zero A) a constant B) a positive number C) zero D) a negative number E) an integerD) a negative number E) an integerD) a negative number E) an integer. D) a negative number E) an integer.

2) For a frictionless pulley and cable tensions in the cable2) For a frictionless pulley and cable tensions in the cable2) For a frictionless pulley and cable, tensions in the cable 2) For a frictionless pulley and cable, tensions in the cable (T(T11 and Tand T22) are related as _____ .) are related as _____ .

A) TA) T > T> TA) TA) T11 > T> T22

B) TB) T11 = T= T22

C) TC) T11 < T< T22

D) TD) T = T= T sinsin θθD) TD) T11 = T= T22 sin sin θθ

APPLICATIONSAPPLICATIONS

For a spool ofFor a spool ofFor a spool of For a spool of given weight, given weight, what are thewhat are thewhat are the what are the forces in cables AB forces in cables AB

d AC ?d AC ?and AC ?and AC ?

APPLICATIONSAPPLICATIONSAPPLICATIONSAPPLICATIONS

For a given cableFor a given cableFor a given cable For a given cable strength, what is strength, what is th ith ithe maximum the maximum weight that can weight that can g a ag a abe lifted ? be lifted ?

EQUILIBRIUM OF PARTICLE IN 2EQUILIBRIUM OF PARTICLE IN 2 DDEQUILIBRIUM OF PARTICLE IN 2EQUILIBRIUM OF PARTICLE IN 2--DD

This is an example of a 2This is an example of a 2 D orD orThis is an example of a 2This is an example of a 2--D or D or coplanar force system. coplanar force system.

If h h l bl i iIf h h l bl i iIf the whole assembly is in If the whole assembly is in equilibrium, then particle A is equilibrium, then particle A is

also in equilibriumalso in equilibriumalso in equilibrium.also in equilibrium.

To determine the tensions in To determine the tensions in the cables for a given weight the cables for a given weight

of the engine we need toof the engine we need toof the engine, we need to of the engine, we need to learn how to draw a free body learn how to draw a free body diagram and apply equations diagram and apply equations d ag a a d app y equa o sd ag a a d app y equa o s

of equilibrium.of equilibrium.

THE WHAT, WHY AND HOW OF A THE WHAT, WHY AND HOW OF A ( )( )FREE BODY DIAGRAM (FBD) FREE BODY DIAGRAM (FBD)

Free Body DiagramsFree Body Diagrams are one of the most are one of the most important things for you to know how to draw and use.important things for you to know how to draw and use.

What ?What ? -- It is a drawing that It is a drawing that h ll t l fh ll t l fshows all external forces shows all external forces acting on the particle.acting on the particle.

Why ?Why ? -- It helps you write It helps you write the equations ofthe equations ofthe equations of the equations of

equilibrium used to solve equilibrium used to solve for the unknowns (usually for the unknowns (usually ( y( y

forces or angles). forces or angles).

How ?How ?

1. Imagine the particle to be isolated or cut free 1. Imagine the particle to be isolated or cut free from its surroundings.from its surroundings.gg

2. Show all the forces that act on the particle.2. Show all the forces that act on the particle.

Active forces:Active forces: They want to move the particle. They want to move the particle. Reactive forces:Reactive forces: They tend to resist the motion.They tend to resist the motion.

3. Identify each force and show all known magnitudes and 3. Identify each force and show all known magnitudes and directions. Show all unknown magnitudes and / or directions as directions. Show all unknown magnitudes and / or directions as ggvariables . variables .

A

FBD at A FBD at A Note : Engine mass = 250 KgNote : Engine mass = 250 Kg

EQUATIONS OF 2EQUATIONS OF 2--D EQUILIBRIUMD EQUILIBRIUM

Since particle A is in equilibrium, Since particle A is in equilibrium, the net force at A is zerothe net force at A is zerothe net force at A is zero.the net force at A is zero.

So So FFABAB + + FFACAC + + FFADAD = 0= 0A

or or ΣΣ F F = 0= 0FBD at A

A

In general, for a particle in equilibrium, In general, for a particle in equilibrium, ΣΣ F F = 0= 0 ororΣΣFF ii ++ ΣΣFF jj 0 00 0 ii + 0+ 0 jj (A t ti )(A t ti )

Or, written in a scalar form,Or, written in a scalar form,

ΣΣFFxx ii + + ΣΣFFyy j j = 0 = 0 = 0 = 0 ii + 0 + 0 jj (A vector equation)(A vector equation)

, ,, ,

ΣΣFFxx = 0 = 0 and and ΣΣ FFyy = 0= 0h l i f ilib i ( f )h l i f ilib i ( f )These are two scalar equations of equilibrium (EofE). These are two scalar equations of equilibrium (EofE).

They can be used to solve for up to They can be used to solve for up to twotwo unknowns.unknowns.

EXAMPLEEXAMPLE

Note : Engine mass = 250 KgNote : Engine mass = 250 Kg FBD at AFBD at A

Write the scalar EofE:Write the scalar EofE:

SPRINGS CABLES AND PULLEYSSPRINGS CABLES AND PULLEYSSPRINGS, CABLES, AND PULLEYSSPRINGS, CABLES, AND PULLEYS

Spring Force =Spring Force =Spring Force = Spring Force = =spring constant x =spring constant x

deformationdeformation

With a frictionless With a frictionless pulley, Tpulley, T11 = T= T22..

deformation,deformation,or F = k x Sor F = k x S

EXAMPLEEXAMPLEEXAMPLEEXAMPLEGiven:Given: Sack A weighs 20 N and Sack A weighs 20 N and

geometry is as shown.geometry is as shown.

Find: Find: Forces in the cables and Forces in the cables and weight of sack B.weight of sack B.

PlanPlan::PlanPlan::

1. Draw a FBD for Point E.1. Draw a FBD for Point E.

2. Apply EofE at Point E to solve for the unknowns 2. Apply EofE at Point E to solve for the unknowns

(T(TEGEG & T& TECEC).).(T(TEGEG & T& TECEC).).

3. Repeat this process at C.3. Repeat this process at C.

EXAMPLEEXAMPLE

CONCEPT QUESTIONSCONCEPT QUESTIONS

1000 N1000 N1000 N1000 N 1000 N1000 N

( A ) ( B ) ( C )( A ) ( B ) ( C )

1) Assuming you know the geometry of the ropes, you cannot 1) Assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system above?determine the forces in the cables in which system above?

A) The weight is too heavy.A) The weight is too heavy.B) The cables are too thinB) The cables are too thinB) The cables are too thin.B) The cables are too thin.C) There are more unknowns than equations.C) There are more unknowns than equations.D) There are too few cables for a 1000 N D) There are too few cables for a 1000 N

2) Why?2) Why?))

weight.weight.

GROUP PROBLEM SOLVINGGROUP PROBLEM SOLVING

600 N

Given:Given: The car is towed at constant The car is towed at constant speed by the 600 N force and the speed by the 600 N force and the angleangle θθ is 25is 25°°angle angle θθ is 25is 25 ..

Find:Find: The forces in the ropes AB andThe forces in the ropes AB andFind: Find: The forces in the ropes AB and The forces in the ropes AB and AC.AC.

Plan:Plan:

1. Draw a FBD for point A.1. Draw a FBD for point A.

2. Apply the EofE to solve for the forces in ropes AB and AC.2. Apply the EofE to solve for the forces in ropes AB and AC.


600 N600 N

3030°°2525°°

AAFBD at point AFBD at point A

3030°°2525°°

FFABABFFACAC

QUIZQUIZQUIZQUIZ

A 4040°°

1. Select the 1. Select the correct FBD ofcorrect FBD of 3030°°

100 N100 N

correct FBD of correct FBD of particle A.particle A.

A)A) A B)B)FF1 1 FF22

3030°°))

100 N100 N

))4040°°

AF FF11 FF22

C)C) 30°A

F

AA

3030°° 4040°°FF11 FF22

D)D)

100 N100 N 100 N100 N

QUIZQUIZQUIZQUIZ

2. Using this FBD of Point C, 2. Using this FBD of Point C, the sum of forces in the xthe sum of forces in the x--di ti (di ti (ΣΣ FF )) ii UU

FF22

direction (direction (ΣΣ FFXX)) is ___ .is ___ . Use Use a sign convention of + a sign convention of + →→ .. 20 N20 N 5050°°

A) FA) F22 sin 50sin 50°° –– 20 = 0 20 = 0 FF

CC

22

B) FB) F22 cos 50cos 50°° –– 20 = 020 = 0

C) FC) F i 50i 50°° FF 00

FF11

C) FC) F22 sin 50sin 50°° –– FF11 = 0 = 0

D) FD) F22 cos 50cos 50°° + 20 = 0+ 20 = 0

EQUILIBRIUM OF A RIGID BODYEQUILIBRIUM OF A RIGID BODYQQ

Today’s ObjectivesToday’s Objectives::

Students will be able toStudents will be able toStudents will be able toStudents will be able to

a) Identify support a) Identify support reactions, and,reactions, and,

b) Draw a free diagramb) Draw a free diagramb) Draw a free diagram.b) Draw a free diagram.

The Equilibrium EquationsThe Equilibrium EquationsThe Equilibrium EquationsThe Equilibrium Equations

•• When an object acted upon by a system of forces When an object acted upon by a system of forces & moments is in equilibrium, the following & moments is in equilibrium, the following conditions are satisfied:conditions are satisfied:conditions are satisfied:conditions are satisfied:

1 The sum of the forces is zero:1 The sum of the forces is zero:1. The sum of the forces is zero:1. The sum of the forces is zero:

ΣΣ FF = = 002. The sum of the moments about any point is 2. The sum of the moments about any point is

zero:zero:zero:zero:

ΣΣ MMany pointany point = = 00

CONDITIONS FOR RIGIDCONDITIONS FOR RIGID--BODY EQUILIBRIUMBODY EQUILIBRIUM

In contrast to the forces on a In contrast to the forces on a particle, the forces on a rigidparticle, the forces on a rigid--body body

t ll t dt ll t dare not usually concurrent and are not usually concurrent and may cause rotation of the body may cause rotation of the body (due to the moments created by(due to the moments created by(due to the moments created by (due to the moments created by

the forces).the forces).Forces on a particleForces on a particle

For a rigid body to be in equilibriumFor a rigid body to be in equilibrium

pp

For a rigid body to be in equilibrium, For a rigid body to be in equilibrium, the net force as well as the net the net force as well as the net

moment about moment about any arbitrary point O any arbitrary point O must be equal to zero.must be equal to zero.

∑∑ FF 00 ∑∑ MM 00∑∑ F F = 0= 0 and and ∑∑ MMO O = 0= 0Forces on a rigid bodyForces on a rigid body

THE PROCESS OF SOLVING RIGID BODY THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMSEQUILIBRIUM PROBLEMSEQUILIBRIUM PROBLEMSEQUILIBRIUM PROBLEMS

Idealised model

FBDmodel

For analyzing an actual physical system first we need toFor analyzing an actual physical system first we need toFor analyzing an actual physical system, first we need to For analyzing an actual physical system, first we need to create an idealized model.create an idealized model.

Then we need to draw aThen we need to draw a freefree body diagrambody diagram showing all theshowing all theThen we need to draw a Then we need to draw a freefree--body diagrambody diagram showing all the showing all the externalexternal (active and reactive) forces.(active and reactive) forces.

Fi ll d tFi ll d t l th ti f ilib il th ti f ilib i ttFinally, we need to Finally, we need to apply the equations of equilibriumapply the equations of equilibrium to to solve for any unknowns.solve for any unknowns.

PROCEDURE FOR DRAWING A PROCEDURE FOR DRAWING A FREE BODY DIAGRAMFREE BODY DIAGRAMFREE BODY DIAGRAMFREE BODY DIAGRAM

FBD

Idealized modelIdealized model Free body diagramFree body diagram

1.1. Draw an outlined shape. Imagine the body to be Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its isolated or cut “free” from its constraints and draw its

outlined shapeoutlined shapeoutlined shape.outlined shape.

2.2. Show all the external forces and couple moments. Show all the external forces and couple moments. TheseThese typicallytypically include: a) applied loads b) supportinclude: a) applied loads b) supportThese These typicallytypically include: a) applied loads, b) support include: a) applied loads, b) support

reactions, and, c) the weight of the body.reactions, and, c) the weight of the body.

PROCEDURE FOR DRAWING A PROCEDURE FOR DRAWING A FREE BODY DIAGRAMFREE BODY DIAGRAM


33 Label loads and dimensions: All known forces andLabel loads and dimensions: All known forces and


3.3. Label loads and dimensions: All known forces and Label loads and dimensions: All known forces and couple moments should be labeled with their magnitudes couple moments should be labeled with their magnitudes

and directions. For the unknown forces and couple and directions. For the unknown forces and couple l likl lik didimoments, use letters like Amoments, use letters like Axx, A, Ayy, M, MAA, etc.. Indicate any , etc.. Indicate any

necessary dimensions.necessary dimensions.

SUPPORT SUPPORT SUPPORT SUPPORT REACTIONSREACTIONSREACTIONSREACTIONS

SUPPORT REACTIONS IN 2SUPPORT REACTIONS IN 2--DD

support

i t

A few examples are shown above. Other support A few examples are shown above. Other support reactions are given in your textbookreactions are given in your textbook

pin support

As a general rule, if a support prevents As a general rule, if a support prevents translationtranslation of a of a

reactions are given in your textbookreactions are given in your textbook

body in a given direction, then a body in a given direction, then a forceforce is developed on is developed on the body in the opposite direction. Similarly, if the body in the opposite direction. Similarly, if rotationrotationis prevented, a couple is prevented, a couple momentmoment is exerted on the body.is exerted on the body.

QUIZQUIZ11. If a support prevents translation of a body, then the . If a support prevents translation of a body, then the

support exerts a on the bodysupport exerts a on the bodysupport exerts a ___________ on the body.support exerts a ___________ on the body.

A) couple momentA) couple moment

B) fB) fB) forceB) force

C) Both A and B.C) Both A and B.

D) None of the above D) None of the above

22 Internal forces are shown on the freeInternal forces are shown on the free2. 2. Internal forces are _________ shown on the free Internal forces are _________ shown on the free body diagram of a whole body.body diagram of a whole body.

A) alwaysA) alwaysA) alwaysA) always

B) oftenB) often

C) rarelyC) rarelyC) rarelyC) rarely

D) never D) never

APPLICATIONSAPPLICATIONSIdealised model FBD

A 200 kg platform is suspended off an oil rig. A 200 kg platform is suspended off an oil rig.

How do we determine the force reactions at the joints and How do we determine the force reactions at the joints and jjthe forces in the cables?the forces in the cables?

How are the idealized model and the free body diagram usedHow are the idealized model and the free body diagram usedHow are the idealized model and the free body diagram used How are the idealized model and the free body diagram used to do this? Which diagram above is the idealized model?to do this? Which diagram above is the idealized model?


A steel beam is used toA steel beam is used toA steel beam is used to A steel beam is used to support roof joists. How support roof joists. How

can we determine the can we determine the support reactions at A & B?support reactions at A & B?

Again, how can we make use of an idealized model Again, how can we make use of an idealized model g ,g ,and a free body diagram to answer this question?and a free body diagram to answer this question?

QUIZ

1. The beam and the cable (with a frictionless pulley at D) support an 80 kg load at C. In a FBD of only the beam, there are how many unknowns?A) 2 forces and 1 couple momentA) 2 forces and 1 couple momentB) 3 forces and 1 couple momentC) 3 forcesD) 4 forces

22--Dimensional ApplicationsDimensional Applications22--Dimensional ApplicationsDimensional Applications

•• The Pin Support:The Pin Support:–– Figure a: a pin supportFigure a: a pin support

•• a bracket to which an object a bracket to which an object (such as a beam) is attached (such as a beam) is attached by a smooth pin that passes by a smooth pin that passes through the bracket & thethrough the bracket & thethrough the bracket & the through the bracket & the objectobject

Figure b: side viewFigure b: side view–– Figure b: side viewFigure b: side view

22--Dimensional ApplicationsDimensional Applications22 Dimensional ApplicationsDimensional Applications

To understand the reactions that aTo understand the reactions that a pinpin–– To understand the reactions that a To understand the reactions that a pin pin supportsupport can exert:can exert:

Imagine holding the barImagine holding the bar•• Imagine holding the bar Imagine holding the bar attached to the pin support attached to the pin support If t t th b ith t t ti it (iIf t t th b ith t t ti it (i•• If you try to move the bar without rotating it (i.e. If you try to move the bar without rotating it (i.e. translate the bar), the support exerts a reactive translate the bar), the support exerts a reactive force that prevents this movementforce that prevents this movementforce that prevents this movementforce that prevents this movement

•• However, you can rotate the bar about the axis of However, you can rotate the bar about the axis of the pinthe pinpp

•• The support cannot exert a couple about the pin The support cannot exert a couple about the pin axis to prevent rotationaxis to prevent rotationpp

22--Dimensional ApplicationsDimensional Applications22 Dimensional ApplicationsDimensional Applications•• Thus, a Thus, a pin supportpin support can’t exert a couple about the can’t exert a couple about the

i i b i f h bj ii i b i f h bj ipin axis but it can exert a force on the object in any pin axis but it can exert a force on the object in any direction, which is usually expressed by direction, which is usually expressed by representing the force in terms of componentsrepresenting the force in terms of components

•• The arrows indicate theThe arrows indicate the

representing the force in terms of componentsrepresenting the force in terms of components

•• The arrows indicate the The arrows indicate the directions of the reactions if directions of the reactions if AAxx & & AAyy are positiveare positiveyy

•• If you determine If you determine AAxx or or AAyy to be to be negative the reaction is in thenegative the reaction is in thenegative, the reaction is in the negative, the reaction is in the direction opposite to that of the direction opposite to that of the arrowarrow

22 Dimensional ApplicationsDimensional Applications22--Dimensional ApplicationsDimensional Applications–– The The pin supportpin support is used to represent any is used to represent any

real support capable of exerting a force in real support capable of exerting a force in any direction but not exerting a coupleany direction but not exerting a couple

•• Used in many common Used in many common devices, particularly devices, particularly , p y, p ythose designed to allow those designed to allow connected parts to connected parts to pprotate relative to each rotate relative to each otherother

22--Dimensional ApplicationsDimensional Applications22 Dimensional ApplicationsDimensional Applications•• The Roller Support:The Roller Support:pppp

–– A pin support mounted on wheelsA pin support mounted on wheelsLike a pin support it cannot exert a coupleLike a pin support it cannot exert a couple–– Like a pin support, it cannot exert a couple Like a pin support, it cannot exert a couple about the axis of the pinabout the axis of the pinSi it f l i th di ti ll lSi it f l i th di ti ll l–– Since it can move freely in the direction parallel Since it can move freely in the direction parallel to the surface on which it rolls, it can’t exert a to the surface on which it rolls, it can’t exert a f ll l t th f b t tf ll l t th f b t tforce parallel to the surface but can exert a force parallel to the surface but can exert a force normal (perpendicular) to this surfaceforce normal (perpendicular) to this surface

22 Dimensional ApplicationsDimensional Applications22--Dimensional ApplicationsDimensional Applications

–– Other commonly used conventions equivalent to Other commonly used conventions equivalent to the roller support:the roller support:

–– The wheels of vehicles & wheels supporting parts The wheels of vehicles & wheels supporting parts of machines are roller supports if the friction forcesof machines are roller supports if the friction forcesof machines are roller supports if the friction forces of machines are roller supports if the friction forces exerted on them are negligible in comparison to exerted on them are negligible in comparison to the normal forcesthe normal forces


–– A plane smooth surface can also be A plane smooth surface can also be modeled by a roller support:modeled by a roller support:

Beams & bridges are sometimes supported Beams & bridges are sometimes supported in this way so that they will be free toin this way so that they will be free toin this way so that they will be free to in this way so that they will be free to undergo thermal expansion & undergo thermal expansion & contractioncontraction


–– These supports are similar to the roller These supports are similar to the roller support in that they cannot exert a couplesupport in that they cannot exert a couplesupport in that they cannot exert a couple support in that they cannot exert a couple & can only exert a force normal to a & can only exert a force normal to a particular direction (friction is neglected)particular direction (friction is neglected)particular direction (friction is neglected)particular direction (friction is neglected)

(a) Pin in a slot (b) Slider in a slot (c) Slider on a shaft


–– In these supports, the supported object is In these supports, the supported object is attached to a pin or slider that can moveattached to a pin or slider that can moveattached to a pin or slider that can move attached to a pin or slider that can move freely in 1 direction but is constrained in freely in 1 direction but is constrained in the perpendicular directionthe perpendicular directionthe perpendicular directionthe perpendicular direction

–– Unlike the roller support, these supports Unlike the roller support, these supports can exert a normal force in either direction can exert a normal force in either direction


•• The Fixed Support:The Fixed Support:–– The fixed support shows the supported objectThe fixed support shows the supported objectThe fixed support shows the supported object The fixed support shows the supported object

literally built into a wall (builtliterally built into a wall (built--in)in)

–– To understand the reactions:To understand the reactions:•• Imagine holding a bar attached to the fixed Imagine holding a bar attached to the fixed

supportsupport

22--Dimensional ApplicationsDimensional Applications22--Dimensional ApplicationsDimensional Applications•• If you try to translate the bar, the support exerts a If you try to translate the bar, the support exerts a

reactive force that prevents translationreactive force that prevents translationIf t t t t th b th t tIf t t t t th b th t t•• If you try to rotate the bar, the support exerts a If you try to rotate the bar, the support exerts a reactive couple that prevents rotationreactive couple that prevents rotation

•• A fixed support can exert 2 components of force & aA fixed support can exert 2 components of force & a•• A fixed support can exert 2 components of force & a A fixed support can exert 2 components of force & a couplecouple

22--Dimensional ApplicationsDimensional Applications22--Dimensional ApplicationsDimensional Applications•• The term The term MMAA is the couple exerted by the support is the couple exerted by the support

& the curved arrow indicates its direction& the curved arrow indicates its direction

–– Fence posts & lampposts have fixed supportsFence posts & lampposts have fixed supports

–– The attachments of parts connected so that The attachments of parts connected so that ppthey cannot move or rotate relative to each they cannot move or rotate relative to each other, such as the head of a hammer & its other, such as the head of a hammer & its ,,handle, can be modelled as fixed supportshandle, can be modelled as fixed supports

22--Dimensional ApplicationsDimensional Applicationspppp•• Table 5.1 summarizes the support conventions commonly used Table 5.1 summarizes the support conventions commonly used

in 2in 2--D applications:D applications:

EQUATIONSEQUATIONSEQUATIONS EQUATIONS

OFOFOF OF EQUILIBRIUMEQUILIBRIUMEQUILIBRIUM EQUILIBRIUM

IN 2IN 2--DDIN 2IN 2 DD

EQUATIONS OF EQUILIBRIUM IN 2EQUATIONS OF EQUILIBRIUM IN 2--DD

Today’s Objectives:Today’s Objectives:

Students will be able toStudents will be able to

a) Apply equations of a) Apply equations of equilibrium toequilibrium tosolve for unknownssolve for unknownssolve for unknowns, solve for unknowns, and,and,

b) Recognize twob) Recognize two--forceforceb) Recognize twob) Recognize two--force force members.members.

QUIZQUIZ1. The three scalar equations 1. The three scalar equations ∑∑ FFXX = = ∑∑ FFYY = = ∑∑ MMOO = 0, = 0,

are equations of equilibrium in two dimensions.are equations of equilibrium in two dimensions.are ____ equations of equilibrium in two dimensions.are ____ equations of equilibrium in two dimensions.

A) incorrectA) incorrect B) the only correctB) the only correct

C) th t l dC) th t l d D) t ffi i tD) t ffi i tC) the most commonly usedC) the most commonly used D) not sufficientD) not sufficient

2 A i id b d i bj t d t2 A i id b d i bj t d t2. A rigid body is subjected to 2. A rigid body is subjected to forces as shown. This body can forces as shown. This body can be considered as a be considered as a ____________member.member.

A)A) singlesingle--forceforce)) gg

B) twoB) two--forceforce

C)C) threethree forceforceC)C) threethree--forceforce

D) sixD) six--forceforce


For a given load onFor a given load onFor a given load on For a given load on the platform, how the platform, how can we determinecan we determinecan we determine can we determine the forces at the the forces at the

joint A and the forcejoint A and the forcejoint A and the force joint A and the force in the link (cylinder) in the link (cylinder)

BC?BC?BC?BC?


ffA steel beam is used to support roof A steel beam is used to support roof joists. joists. jj

How can we determine the support How can we determine the support i h d f h bi h d f h breactions at each end of the beam?reactions at each end of the beam?

EQUATIONS OF EQUILIBRIUMEQUATIONS OF EQUILIBRIUM

A body is subjected to a system of A body is subjected to a system of forces that lie in the xforces that lie in the x--y planey plane y F3forces that lie in the xforces that lie in the x--y plane. y plane.

When in equilibrium, the net force and When in equilibrium, the net force and F1

3

F4

net moment acting on the body are net moment acting on the body are zero. zero. x

F1

O

This 2This 2--D condition can be represented D condition can be represented by the three scalar equations:by the three scalar equations: F2

∑∑ FFxx = 0 = 0 ∑∑ FFyy = 0 = 0 ∑∑ MMOO = 0= 0Where point O is any arbitrary pointWhere point O is any arbitrary pointWhere point O is any arbitrary point.Where point O is any arbitrary point.

Pl tPl t th t th ti thth t th ti th ttPlease notePlease note that these equations are the ones that these equations are the ones most most commonly usedcommonly used for solving 2for solving 2--D equilibrium problems. D equilibrium problems.

TWOTWO--FORCE MEMBERSFORCE MEMBERS

The solution to some equilibrium The solution to some equilibrium problems problems can be simplifiedcan be simplified if we if we

i b th ti b th trecognize members that are recognize members that are subjected to forces at only two subjected to forces at only two points (e g at points A and B)points (e g at points A and B)points (e.g., at points A and B).points (e.g., at points A and B).

If we apply the equations of equilibrium to such a member, If we apply the equations of equilibrium to such a member, pp y q q ,pp y q q ,we can quickly determine that we can quickly determine that the resultant forces at A the resultant forces at A

and B must be equal in magnitude and act in the and B must be equal in magnitude and act in the i di i l h li j i i i A d Bi di i l h li j i i i A d Bopposite directions along the line joining points A and Bopposite directions along the line joining points A and B..

EXAMPLE OF TWOEXAMPLE OF TWO--FORCE MEMBERSFORCE MEMBERS

In the cases above, members AB can be considered as twoIn the cases above, members AB can be considered as two--force members, provided that their weight is neglected.force members, provided that their weight is neglected.

This fact simplifies the equilibrium analysis of some rigid This fact simplifies the equilibrium analysis of some rigid bodies since the directions of the resultant forces at A and Bbodies since the directions of the resultant forces at A and Bbodies since the directions of the resultant forces at A and B bodies since the directions of the resultant forces at A and B

are thus known (along the line joining points A and B).are thus known (along the line joining points A and B).

STEPS FOR SOLVING 2STEPS FOR SOLVING 2--D D EQUILIBRIUM PROBLEMSEQUILIBRIUM PROBLEMS

1. If not given, establish a suitable x1. If not given, establish a suitable x -- yy1. If not given, establish a suitable x 1. If not given, establish a suitable x y y coordinate system.coordinate system.

2. Draw a 2. Draw a free body diagramfree body diagram (FBD) of (FBD) of y gy g ( )( )the object under analysis.the object under analysis.

3. Apply the three equations of equilibrium 3. Apply the three equations of equilibrium (E fE) t l f th k(E fE) t l f th k(EofE) to solve for the unknowns.(EofE) to solve for the unknowns.

IMPORTANT NOTESIMPORTANT NOTESO O SO O S

1.1. If we have more unknowns than the number of If we have more unknowns than the number of independent equations, then we have a statically independent equations, then we have a statically indeterminate situation.indeterminate situation.

We cannot solve these problems using just statics.We cannot solve these problems using just statics.

2 The2 The order in which we apply equationsorder in which we apply equations may affect themay affect the2. The 2. The order in which we apply equationsorder in which we apply equations may affect the may affect the simplicity of the solution. For example, if we have two simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal unknown vertical forces and one unknown horizontal force, then solving force, then solving ∑∑ FFXX = O= O first allows us to find the first allows us to find the horizontal unknown quickly.horizontal unknown quickly.

3. If the 3. If the answeranswer for an unknown comes outfor an unknown comes out as negative as negative numbernumber, then the sense (direction) of the unknown force, then the sense (direction) of the unknown forcenumbernumber, then the sense (direction) of the unknown force , then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.is opposite to that assumed when starting the problem.

QUIZQUIZ1. For this beam, how many 1. For this beam, how many

support reactions are theresupport reactions are there F F Fsupport reactions are there support reactions are there and is the problem statically and is the problem statically determinate?determinate?

F F F F

A) (2, Yes)A) (2, Yes) B) (2, No)B) (2, No)C) (3, Yes)C) (3, Yes) D) (3, No)D) (3, No)

2. For the given beam loading: a) 2. For the given beam loading: a) how many support reactions arehow many support reactions are Fixedhow many support reactions are how many support reactions are there, b) is this problem there, b) is this problem statically determinate, and, c) is statically determinate, and, c) is

FFixed support

y , , )y , , )the structure stable?the structure stable?

A) (4, Yes, No)A) (4, Yes, No) B) (4, No, Yes)B) (4, No, Yes)Pin joints

C) (5, Yes, No)C) (5, Yes, No) D) (5, No, Yes)D) (5, No, Yes)

QUIZ1. Which equation of equilibrium

allows you to determine FB right away? 100 Naway?

A) ∑ FX = 0 B) ∑ FY = 0AX A B

100 N

C) ∑ MA = 0

D) Any one of the above.

FBAY

2. A beam is supported by a pin joint and a roller. How many support reactions are there and is the structure stable forand is the structure stable for all types of loadings?

A) (3, Yes) B) (3, No)A) (3, Yes) B) (3, No)

C) (4, Yes) D) (4, No)

EQUATIONSEQUATIONSEQUATIONS EQUATIONS

OFOFOF OF EQUILIBRIUMEQUILIBRIUMEQUILIBRIUM EQUILIBRIUM

IN 3IN 3--DDIN 3IN 3 DD

RIGID BODY EQUILIBRIUM IN 3RIGID BODY EQUILIBRIUM IN 3--DDG O QU U 3G O QU U 3

Today’s ObjectiveToday’s Objective::

Students will be able toStudents will be able to

a) Identify supporta) Identify supporta) Identify support a) Identify support reactions in 3reactions in 3--D and draw D and draw a free body diagram, and,a free body diagram, and,y g , ,y g , ,

b) apply the equations of b) apply the equations of equilibrium.equilibrium.equilibrium.equilibrium.

QUIZQUIZ

1. If a support prevents rotation of a body about an 1. If a support prevents rotation of a body about an axis, then the support exerts a ________ on the axis, then the support exerts a ________ on the body about that axis.body about that axis.

A) couple momentA) couple moment B) forceB) force

C) Both A and B.C) Both A and B. D) None of the above.D) None of the above.C) Both A and B.C) Both A and B. D) None of the above.D) None of the above.

2 Wh d i 32 Wh d i 3 D bl l i hD bl l i h2. When doing a 32. When doing a 3--D problem analysis, you have D problem analysis, you have ________ scalar equations of equilibrium.________ scalar equations of equilibrium.

A) 2A) 2 B) 3B) 3 C) 4C) 4

D)D) 55 E) 6E) 6


BallBall--andand--socket joints and journal bearings are often socket joints and journal bearings are often used in mechanical systemsused in mechanical systemsused in mechanical systems. used in mechanical systems.

H d t i th t ti t thH d t i th t ti t thHow can we determine the support reactions at these How can we determine the support reactions at these joints for a given loading?joints for a given loading?

SUPPORT REACTIONS IN 3SUPPORT REACTIONS IN 3--DD

A few examples are shown above. Other support reactions A few examples are shown above. Other support reactions

As a general rule, if a support prevents translation of aAs a general rule, if a support prevents translation of a

p ppp ppare given in your text book (Table 5are given in your text book (Table 5--2).2).

As a general rule, if a support prevents translation of a As a general rule, if a support prevents translation of a body in a given direction, then a reaction force acting in the body in a given direction, then a reaction force acting in the

opposite direction is developed on the body. opposite direction is developed on the body.

Similarly, if rotation is prevented, a couple moment is Similarly, if rotation is prevented, a couple moment is exerted on the body by the support.exerted on the body by the support.

IMPORTANT NOTEIMPORTANT NOTEO OO O

A single bearing or hinge can prevent rotation by providing A single bearing or hinge can prevent rotation by providing a resistive couple moment. a resistive couple moment.

However, it is usually preferred to use two or more However, it is usually preferred to use two or more properly aligned bearings or hinges. properly aligned bearings or hinges.

Thus, in these cases, only force reactions are generated and Thus, in these cases, only force reactions are generated and there are no moment reactions created.there are no moment reactions created.

EQULIBRIUM EQUATIONS IN 3EQULIBRIUM EQUATIONS IN 3--DDQ QQ QAs stated earlier, when a body is in equilibrium, the net force and the As stated earlier, when a body is in equilibrium, the net force and the

net moment equal zero i enet moment equal zero i e ∑∑ FF = 0= 0 andand ∑∑ MM = 0= 0net moment equal zero, i.e., net moment equal zero, i.e., ∑∑ FF = 0= 0 and and ∑∑ MMOO = 0 .= 0 .

These two vector equations can be written as six scalar These two vector equations can be written as six scalar qqequations of equilibrium (EofE). These are equations of equilibrium (EofE). These are

∑∑ FF == ∑∑ FF == ∑∑ FF = 0= 0∑∑ FFXX = = ∑∑ FFYY = = ∑∑ FFZZ = 0 = 0

∑∑MM == ∑∑ MM == ∑∑ MM = 0= 0∑∑MMXX = = ∑∑ MMYY = = ∑∑ MMZZ = 0= 0The moment equations can be determined about any point. The moment equations can be determined about any point. q y pq y p

Usually, choosing Usually, choosing the point where the maximum number of unknown the point where the maximum number of unknown forces are present simplifies the solutionforces are present simplifies the solution. Those forces do not appear . Those forces do not appear p pp p ppppin the moment equation since they pass through the point. Thus, in the moment equation since they pass through the point. Thus, they do not appear in the equation. they do not appear in the equation.


Given:Given: The load is 2The load is 2 kNkN..Given:Given: The load is 2 The load is 2 kNkN..

Find: Find: Reactions at A and B.Reactions at A and B.

PlanPlan::PlanPlan::

1. Draw a FBD.1. Draw a FBD.

2. Apply the2. Apply the EofEEofE to solve for the forces.to solve for the forces.2. Apply the 2. Apply the EofEEofE to solve for the forces.to solve for the forces.

Week 3---Equilibrium+FBD

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