Engineering Mechanics: Statics Chapter 5: Equilibrium of a Rigid Body.

Engineering Mechanics: StaticsEngineering Mechanics: Statics

Chapter 5: Equilibrium of a Rigid

Body

Chapter 5: Equilibrium of a Rigid

Body

Chapter ObjectivesChapter 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.

Chapter OutlineChapter Outline

Conditions for Rigid EquilibriumFree-Body DiagramsEquations of EquilibriumTwo and Three-Force MembersEquilibrium in Three DimensionsEquations of EquilibriumConstraints for a Rigid Body

5.1 Conditions for Rigid-Body Equilibrium


Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity

Two types of forces that act on it, the resultant internal force and the resultant external force

Resultant internal force fi is caused by interactions with adjacent particles



Resultant external force Fi represents the effects of gravitational, electrical, magnetic, or contact forces between the ith particle and adjacent bodies or particles not included within the body

Particle in equilibrium, apply Newton’s first law,

Fi + fi = 0



When equation of equilibrium is applied to each of the other particles of the body, similar equations will result

Adding all these equations vectorially,

∑Fi + ∑fi = 0 Summation of internal forces =

0 since internal forces between particles in the body occur in equal but opposite collinear pairs (Newton’s third law)



Only sum of external forces will remain

Let ∑Fi = ∑F, ∑F = 0 Consider moment of the forces

acting on the ith particle about the arbitrary point O

By the equilibrium equation and distributive law of vector cross product,

ri X (Fi + fi) = ri X Fi + ri X fi = 0



Similar equations can be written for other particles of the body

Adding all these equations vectorially,

∑ri X Fi + ∑ri X fi = 0 Second term = 0 since internal

forces occur in equal but opposite collinear pairs

Resultant moment of each pair of forces about point O is zero

Using notation ∑MO = ∑ri X Fi, ∑MO = 0



Equations of Equilibrium for Rigid Body∑F = 0∑MO = 0

A rigid body will remain in equilibrium provided the sum of all the external forces acting on the body = 0 and sum of moments of the external forces about a point = 0

For proof of the equation of equilibrium, - Assume body in equilibrium



- Force system acting on the body satisfies the equations ∑F = 0 and ∑MO = 0- Suppose additional force F’ is applied to the body

∑F + F’ = 0∑MO + MO’= 0

where MO’is the moment of F’ about O- Since ∑F = 0 and ∑MO = 0, we require F’ = 0 and MO’- Additional force F’ is not required and equations ∑F = 0 and ∑MO = 0 are sufficient

5.2 Free-Body Diagrams5.2 Free-Body Diagrams

FBD is the best method to represent all the known and unknown forces in a system

FBD is a sketch of the outlined shape of the body, which represents it being isolated from its surroundings

Necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when equations of equilibrium are applied


Support Reactions If the support prevents the translation of a

body in a given direction, then a force is developed on the body in that direction

If rotation is prevented, a couple moment is exerted on the body

Consider the three ways a horizontal member, beam is supported at the end- roller, cylinder- pin- fixed support


Support ReactionsRoller or cylinderPrevent the beam from

translating in the vertical direction

Roller can only exerts a force on the beam in the vertical direction


Support ReactionsPin The pin passes through a hold in the

beam and two leaves that are fixed to the ground

Prevents translation of the beam in any direction Φ

The pin exerts a force F on the beam in this direction


Support ReactionsFixed Support This support prevents both

translation and rotation of the beam

A couple and moment must be developed on the beam at its point of connection

Force is usually represented in x and y components


Cable exerts a force on the bracket

Type 1 connections

Rocker support for this bridge girder allows horizontal movements so that the bridge is free to expand and contract due to temperature

Type 5 connections


Concrete Girder rest on the ledge that is assumed to act as a smooth contacting surface

Type 6 connections

Utility building is pin supported at the top of the column

Type 8 connections


Floor beams of this building are welded together and thus form fixed connections

Type 10 connections


External and Internal Forces A rigid body is a composition of particles,

both external and internal forces may act on it

For FBD, internal forces act between particles which are contained within the boundary of the FBD, are not represented

Particles outside this boundary exert external forces on the system and must be shown on FBD

FBD for a system of connected bodies may be used for analysis


Weight and Center of Gravity When a body is subjected to gravity, each

particle has a specified weight For entire body, consider gravitational

forces as a system of parallel forces acting on all particles within the boundary

The system can be represented by a single resultant force, known as weight W of the body

Location of the force application is known as the center of gravity


Weight and Center of GravityCenter of gravity occurs at the

geometric center or centroid for uniform body of homogenous material

For non-homogenous bodies and usual shapes, the center of gravity will be given


Idealized ModelsNeeded to perform a correct force

analysis of any objectCareful selection of supports, material,

behavior and dimensions for trusty results

Complex cases may require developing several different models for analysis


Idealized Models Consider a steel beam used to support

the roof joists of a building For force analysis, reasonable to

assume rigid body since small deflections occur when beam is loaded

Bolted connection at A will allow for slight rotation when load is applied => use Pin


Support at B offers no resistance to horizontal movement => use Roller

Building code requirements used to specify the roof loading (calculations of the joist forces)

Large roof loading forces account for extreme loading cases and for dynamic or vibration effects

Weight is neglected when it is small compared to the load the beam supports


Idealized Models Consider lift boom, supported by

pin at A and hydraulic cylinder at BC (treat as weightless link)

Assume rigid material with density known

For design loading P, idealized model is used for force analysis

Average dimensions used to specify the location of the loads and supports


Procedure for Drawing a FBD1. Draw Outlined Shape Imagine body to be isolated or cut free

from its constraints Draw outline shape

2. Show All Forces and Couple Moments Identify all external forces and couple

moments that act on the body


Procedure for Drawing a FBD Usually due to

- applied loadings- reactions occurring at the supports or at points of contact with other body- weight of the body

To account for all the effects, trace over the boundary, noting each force and couple moment acting on it

3. Identify Each Loading and Give Dimensions Indicate dimensions for calculation of

forces


Procedure for Drawing a FBD Known forces and couple moments

should be properly labeled with their magnitudes and directions

Letters used to represent the magnitudes and direction angles of unknown forces and couple moments

Establish x, y and coordinate system to identify unknowns


Example 5.1Draw the free-body diagram of the

uniform beam. The beam has a mass of 100kg.


SolutionFree-Body Diagram


Solution Support at A is a fixed wall Three forces acting on the beam at A denoted as

Ax, Ay, Az, drawn in an arbitrary direction Unknown magnitudes of these vectors Assume sense of these vectors For uniform beam,

Weight, W = 100(9.81) = 981N acting through beam’s center of gravity, 3m from A


Example 5.2Draw the free-body diagram of the foot lever. The operator applies a vertical force to the pedal so that the spring is stretched 40mm and the force in the short link at B is 100N.


Solution Lever loosely bolted to frame at A Rod at B pinned at its ends and acts as

a short link For idealized model of the lever,


Solution Free-Body Diagram

Pin support at A exerts components Ax and Ay on the lever, each force with a known line of action but unknown magnitude


Solution Link at B exerts a force 100N acting in the

direction of the link Spring exerts a horizontal force on the lever

Fs = ks = 5N/mm(40mm) = 200N Operator’s shoe exert vertical force F on the

pedal Compute the moments using the dimensions

on the FBD Compute the sense by the equilibrium

equations


Example 5.3Two smooth pipes, each having a mass of 300kg, are supported by the forks of the tractor. Draw the free-body diagrams for each pipe and both pipes together.


SolutionFor idealized models,

Free-Body Diagram of pipe A


Solution For weight of pipe A, W = 300(9.81) = 2943N Assume all contacting surfaces are smooth,

reactive forces T, F, R act in a direction normal to tangent at their surfaces of contact

Free-Body Diagram at pipe B


Solution*Note: R represent the force of A on B, is equal and opposite to R representing the force of B on

A Contact force R is considered an internal force,

not shown on FBD Free-Body Diagram of both pipes


Example 5.4Draw the free-body diagram of the unloaded platform that is suspended off the edge of the oil rig. The platform has a mass of 200kg.


Solution Idealized model considered in

2D because by observation, loading and the dimensions are all symmetrical about a vertical plane passing through the center

Connection at A assumed to be a pin and the cable supports the platform at B


SolutionDirection of the cable and average

dimensions of the platform are listed and center of gravity has been determined

Free-Body Diagram


SolutionPlatform’s weight = 200(9.81) =

1962NForce components Ax and Ay along

with the cable force T represent the reactions that both pins and cables exert on the platform

Half of the cables magnitudes is developed at A and half developed at B


Example 5.5The free-body diagram of each object

is drawn. Carefully study each solution

and identify what each loading represents.


Solution

5.3 Equations of Equilibrium5.3 Equations of Equilibrium

For equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0

∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on the body

∑MO represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body


Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, ∑Fx = 0;

∑Fy = 0; ∑MO = 0 can be used Two alternative sets of three independent

equilibrium equations may also be used∑Fa = 0; ∑MA = 0; ∑MB = 0

When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis


Alternative Sets of Equilibrium EquationsConsider FBD of an arbitrarily shaped

bodyAll the forces on FBD may be

replaced by an equivalent resultant force FR = ∑F acting at point A and a

resultant moment MRA = ∑MA If ∑MA = 0 is satisfied, MRA = 0


Alternative Sets of Equilibrium Equations If FR satisfies ∑Fa = 0, there is no

component along the a axis and its line of axis is perpendicular to the a axis

If ∑MB = 0 where B does not lies

on the line of action of FR, FR = 0

Since ∑F = 0 and ∑MA = 0, the

body is in equilibrium

Alternative Sets of Equilibrium Equations A second set of alternative equations is

∑MA = 0; ∑MB = 0; ∑MC = 0 Points A, B and C do not lie on the

same line Consider FBD, if If ∑MA = 0, MRA = 0 ∑MA = 0 is satisfied if line of action of FR

passes through point B ∑MC = 0 where C does not lie on line AB FR = 0 and the body is in equilibrium



Procedure for AnalysisFree-Body Diagram Establish the x, y, z coordinates axes in

any suitable orientation Draw an outlined shape of the body Show all the forces and couple

moments acting on the body Label all the loadings and specify their

directions relative to the x, y axes


Procedure for AnalysisFree-Body DiagramThe 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


Procedure for AnalysisEquations of Equilibrium Apply the moment equation of

equilibrium ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces

The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained


Procedure for AnalysisEquations of Equilibrium When applying the force equilibrium ∑Fx

= 0 and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components

If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD


Example 5.6Determine the horizontal and vertical components of reaction for the beam

loaded. Neglect the weight of the beam in the calculations.


SolutionFBD 600N force is represented by its x and y

components 200N force acts on the beam at B and is

independent of the force components Bx and By, which

represent the effect of the pin on the beam


SolutionEquations of Equilibrium

A direct solution of Ay can be obtained by applying ∑MB = 0 about point B

Forces 200N, Bx and By all create zero moment about B

NB

BN

M

x

x

B

424

045cos600

;0


Solution

NB

BNNNN

F

NA

mAmNmNmN

M

y

y

y

y

y

B

405

020010045sin600319

;0

319

0)7()2.0)(45cos600()5)(45sin600()2(100

;0


SolutionChecking,

NB

mBmN

mNmNmN

M

y

y

A

405

0)7()7)(200(

)5)(100()2.0)(45cos600()2)(45sin600(

;0


Example 5.7The cord supports a force of 500N and wraps

over the frictionless pulley. Determine the tension

in the cord at C and the horizontal and vertical components at pin A.


SolutionFBD of the cord and pulley Principle of action: equal but opposite

reaction observed in the FBD Cord exerts an unknown load

distribution p along part of the pulley’s surface

Pulley exerts an equal but opposite effect on the cord


SolutionFBD of the cord and pulley Easier to combine the FBD of the pulley

and contracting portion of the cord so that the distributed load becomes internal to the system and is eliminated from the analysis



Tension remains constant as cord passes over the pulley (true for any angle at which the cord is directed and for any radius of the pulley

NT

mTmN

M A

500

0)2.0()2.0(500

;0


Solution

NA

NNA

F

NAx

NA

F

y

y

y

x

x

933

030cos500500

;0

250

030sin500

;0


Example 5.8The link is pin-connected at a and rest a smooth support at B. Compute the

horizontal and vertical components of reactions at pin

A


SolutionFBDReaction NB is perpendicular to the

link at BHorizontal and vertical

components of reaction are represented at A



NAx

NA

F

NN

mNmNmN

M

x

x

B

B

A

100

030sin200

;0

200

0)75.0()1(60.90

;0


Solution

NA

NNA

F

y

y

y

233

030cos20060

;0


Example 5.9The box wrench is used to tighten the bolt at A. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt.


SolutionFBDBolt acts as a “fixed support” it exerts

force components Ax and Ay and a torque MA on the wrench at A



NAx

NNA

F

mNM

mNmNM

M

x

x

A

A

A

00.5

060cos3013

552

;0

.6.32

0)7.0)(60sin30()3.0(13

1252

;0


Solution

NmM

mNmNM

MCCW

NA

NNA

F

A

A

y

y

y

y

6.32

07.060sin303.013

1252

;0

0.74

060sin3013

1252

;0


Solution Point A was chosen for summing the

moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are not included in the moment summation

MA must be included

Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench


Solution By Newton’s third law, the wrench

exerts an equal but opposite moment or torque on the bolt

For resultant force on the wrench,

For directional sense,

FA acts in the opposite direction on the bolt

1.8600.5

0.74tan

1.740.7400.5

1

22

N

N

NFA


Solution Checking,

0.8.51.6.32.2.19

0)7.0(0.74.6.32)4.0(1312

52

;0

mNmNmN

mNmNmN

MC

Example 5.10Placement of concrete from the truck is accomplished using the chute. Determine the force that the hydraulic cylinder and the truck frame exert on the chute to hold it

in position. The chute and the wet concrete contained along its length have a uniform weight of 560N/m.

5.3 Equations of Equilibrium

5.3 Equations of Equilibrium


Solution Idealized model of the chute Assume chute is pin connected to the

frame at A and the hydraulic cylinder BC acts as a short link


SolutionFBD Since chute has a length of 4m, total

supported weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, G

The hydraulic cylinder exerts a horizontal force FBC on the chute

Equations of Equilibrium A direct solution of FBC is obtained by the

summation about the pin at A


Solution

NAx

NA

F

NF

mN

mNmF

M

x

x

BC

BC

A

7900

07900

;0

7900

0)0625.0(30sin2240

)2(30cos2240)5.0(

;0


Solution

Checking,

0)0625.0(30sin2240)1(30cos2240

)30cos1(2240)5.0(7900

;0

2240

02240

;0

mNmN

mNmN

M

NA

NA

F

B

y

y

y


Example 5.11The uniform smooth rod is subjected to a force

and couple moment. If the rod is supported at A by

a smooth wall and at B and C either at the top or bottom by rollers, determine the reactions at these supports. Neglect the weight of the rod.


SolutionFBD All the support reactions act normal to the

surface of contact since the contracting surfaces are smooth

Reactions at B and C are acting in the positive y’ direction

Assume only the rollers located on the bottom of the rod are used for support



030cos30cos300

;0

030sin30sin

;0

0)8)(30cos300()6(.4000)2(

;0

''

''

''

yy

y

xyy

x

yy

A

BCN

F

ABC

F

mNmCmNmB

M


Solution Note that the line of action of the force

component passes through point A and this force is not included in the moment equation

Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD

kNNC

kNNB

y

y

35.14.1346

10.1000

'

'


Solution Top roller at B serves as the support

rather than the bottom one

NA

ANN

x

x

173

030sin0.100030sin4.1346


Example 5.12The uniform truck ramp has a weight of 1600N ( ≈ 160kg ) and is pinned to the

body of the truck at each end and held in

position by two side cables. Determine the tension in the cables.


Solution Idealized model of the ramp Center of gravity located at the midpoint since

the ramp is approximately uniform

FBD of the Ramp



By the principle of transmissibility, locate T at C

md

md

NT

N

TmT

M A

0154.120sin

2

10sin

5985

0)30cos5.1(1600

)30cos2(20sin)30sin2(20cos

;0


SolutionSince there are two cables

supporting the ramp,

T’ = T/2 = 2992.5N

5.4 Two- and Three-Force Members


Simplify some equilibrium problems by recognizing embers that are subjected top only 2 or 3 forces

Two-Force Members When a member is subject to

no couple moments and forces are applied at only two points on a member, the member is called a two-force member



Two-Force MembersExample Forces at A and B are summed to

obtain their respective resultants FA and FB

These two forces will maintain translational and force equilibrium provided FA is of equal magnitude and opposite direction to FB

Line of action of both forces is known and passes through A and B



Two-Force Members Hence, only the force

magnitude must be determined or stated

Other examples of the two-force members held in equilibrium are shown in the figures to the right



Three-Force Members If a member is subjected to only three forces, it

is necessary that the forces be either concurrent or parallel for the member to be in equilibrium

To show the concurrency requirement, consider a body with any two of the three forces acting on it, to have line of actions that intersect at point O



Three-Force Members To satisfy moment equilibrium about O, the

third force must also pass through O, which then makes the force concurrent

If two of the three forces parallel, the point of currency O, is considered at “infinity”

Third force must parallel to the other two forces to insect at this “point”



Bucket link AB on the back hoe is a typical example of a two-force member since it is pin connected at its end provided its weight is neglected, no other force acts on this member

The hydraulic cylinder is pin connected at its ends, being a two-force member



The boom ABD is subjected to the weight of the suspended motor at D, the forces of the hydraulic cylinder at B, and the force of the pin at A. If the boom’s weight is neglected, it is a three-force member

The dump bed of the truck operates by extending the hydraulic cylinder AB. If the weight of AB is neglected, it is a two-force member since it is pin-connected at its end points



Example 5.13The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members are negligible, determine the force of the pin on the lever at A.



Solution FBD Short link BD is a two-force

member, so the resultant forces at pins D and B must be equal, opposite and collinear

Magnitude of the force is unknown but line of action known as it passes through B and D

Lever ABC is a three-force member



Solution FBD For moment equilibrium,

three non-parallel forces acting on it must be concurrent at O

Force F on the lever at B is equal but opposite to the force F acting at B on the link

Distance CO must be 0.5m since lines of action of F and the 400N force are known



Solution Equations of

Equilibrium

Solving,

kNF

kNF

FF

F

NFF

F

A

A

y

A

x

32.1

07.1

045sin3.60sin

;0

040045cos3.60cos

;0

3.604.0

7.0tan 1

5.5 Equilibrium in Three Dimensions (FBD)


Consider types of reaction that can occur at the supports

Support Reactions Important to recognize the symbols used

to represent each of these supports and to clearly understand how the forces and couple moments are developed by each support

As in 2D, a force that is developed by a support that restricts the translation of the attached member



Support Reactions A couple moment is developed when

rotation of the attached member is prevented

Example The ball and socket joint prevents any

translation of connecting member; therefore, a force must act on the member at the point of connection

This force have unknown magnitude Fx, Fy and Fz

Magnitude of the force is given by222zyx FFFF



Support Reactions The force’s orientation is defined by the

coordinate angles α, β and γ Since connecting member is allow to

rotate freely about any axis, no couple moment is resisted by a ball and socket joint

Single bearing supports, single pin and single hinge are shown to support both force and couple moment components



Support Reactions However, if these supports are used

with other bearings, pins or hinges to hold a rigid body in equilibrium and the supports are properly aligned when connected to the body, the force reactions at these supports alone may be adequate for supporting the body

Couple moments become redundant and not shown on the FBD



Ball and socket joint provides a connection for the housing of an earth grader to its frame

Journal bearing supports the end of the shaft



Thrust bearing is used to support the drive shaft on the machine

Pin is used to support the end of the strut used on a tractor



Example 5.14Several examples of objects along with their associated free-body diagrams are shown. In all cases, the x, y and z axes are established and the unknown reaction components are indicated in the positive sense. The weight of the objects is neglected.



Solution


Vector Equations of Equilibrium For two conditions for equilibrium of a rigid

body in vector form, ∑F = 0∑MO = 0

where ∑F is the vector sum of all the external forces acting on the body and ∑MO is the sum of the couple moments and the moments of all the forces about any point O located either on or off the body


Scalar Equations of Equilibrium If all the applied external forces and

couple moments are expressed in Cartesian vector form

∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0

∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0i, j and k components are

independent from one another


Scalar Equations of Equilibrium∑Fx = 0, ∑Fy = 0, ∑Fz = 0 shows that

the sum of the external force components acting in the x, y and z directions must be zero

∑Mx = 0, ∑My = 0, ∑Mz = 0 shows that the sum of the moment components about the x, y and z axes to be zero

5.7 Constraints for a Rigid Body


To ensure the equilibrium of a rigid body, it is necessary to satisfy the equations equilibrium and have the body properly held or constrained by its supports

Redundant Constraints More support than needed for equilibrium Statically indeterminate: more unknown

loadings on the body than equations of equilibrium available for their solution



Redundant ConstraintsExample For the 2D and 3D problems, both are

statically indeterminate because of additional supports reactions

In 2D, there are 5 unknowns but 3 equilibrium equations can be drawn



Redundant ConstraintsExample In 3D, there are 8 unknowns but 6 equilibrium

equations can be drawn Additional equations

involving the physical properties of the body are needed to solve indeterminate problems



Improper Constraints Instability of the body caused by the

improper constraining by the supports In 3D, improper constraining occur when

the support reactions all intersect a common axis

In 2D, this axis is perpendicular to the plane of the forces and appear as a point

When all reactive forces are concurrent at this point, the body is improperly constrained

Improper ConstraintsExample From FBD, summation of moments about the x

axis will not be equal to zero, thus rotation occur

In both cases, impossible to solve completely for the unknowns



Improper Constraints Instability of the body also can be

caused by the parallel reactive forcesExampleSummation of

forces along the x axis will not be equal to zero





Improper Constraints Instability of the body also can be caused

when a body have fewer reactive forces than the equations of equilibrium that must be satisfied

The body become partially constrainedExample If O is a point not located on line AB,

loading condition and equations of equilibrium are not satisfied



Improper Constraints Proper constraining requires

- lines of action of the reactive forces do not insect points on a common axis- the reactive forces must not be all parallel to one another

When the minimum number of reactive forces is needed to properly constrain the body, the problem is statically determinate and equations of equilibrium can be used for solving



Procedure for AnalysisFree Body Diagram Draw an outlined shape of the body Show all the forces and couple

moments acting on the body Establish the x, y, z axes at a

convenient point and orient the axes so that they are parallel to as many external forces and moments as possible

Label all the loadings and specify their directions relative to the x, y and z axes



Procedure for AnalysisFree Body Diagram In general, show all the unknown

components having a positive sense along the x, y and z axes if the sense cannot be determined

Indicate the dimensions of the body necessary for computing the moments of forces



Procedure for AnalysisEquations of Equilibrium If the x, y, z force and moment

components seem easy to determine, then apply the six scalar equations of equilibrium,; otherwise, use the vector equations

It is not necessary that the set of axes chosen for force summation coincide with the set of axes chosen for moment summation

Any set of nonorthogonal axes may be chosen for this purpose



Procedure for AnalysisEquations of Equilibrium Choose the direction of an axis for

moment summation such that it insects the lines of action of as many unknown forces as possible

In this way, the moments of forces passing through points on this axis and forces which are parallel to the axis will then be zero

If the solution yields a negative scalar, the sense is opposite to that was assumed



Example 5.15The homogenous plate has a mass of 100kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a ball and socket joint at N, and a cord at C, determine the components of reactions at the supports.



SolutionFBDFive unknown reactions acting on the

plateEach reaction assumed to act in a

positive coordinate direction




Moment of a force about an axis is equal to the product of the force magnitude and the perpendicular distance from line of action of the force to the axis

Sense of moment determined from right-hand rule

0981300;0

0;0

0;0

NNTBAF

BF

BF

Czzz

yy

xx



Solution

Components of force at B can be eliminated if x’, y’ and z’ axes are used

0)3(.200)5.1(981)5.1(300

;0

0)2()2(300)1(981;0

0.200)3()3()5.1(981)5.1(300

;0

0)2()1(981)2(;0

'

'

mTmNmNmN

M

mAmNmNM

mNmAmBmNmN

M

mBmNmTM

C

y

zx

zz

y

ZCx



SolutionSolving,

Az = 790N Bz = -217N TC = 707N

The negative sign indicates Bz acts downward The plate is partially constrained since the

supports cannot prevent it from turning about the z axis if a force is applied in the x-y plane

Example 5.16The windlass is supported by a thrust bearing at A and a smooth journal bearing at B, which are properly aligned on the shaft. Determine the magnitude of the vertical

force P that must be applied to the handle to maintain equilibrium of the 100kg bucket. Also, calculate the reactions at the bearings.





SolutionFBDSince the bearings at A and B are

aligned correctly, only force reactions occur at these supports




0

0)8.0(

;0

3.424

0)4.0)(6.377()8.0()5.0(981

;0

6.377

0)30cos3.0()1.0(981

;0

y

y

z

z

z

y

x

A

mA

M

NA

mNmAmN

M

NP

mPmN

M




NB

B

F

B

B

F

A

F

z

z

z

y

y

y

x

x

934

06.3779813.424

;0

0

00

;0

0

;0

Example 5.17Determine the tension in cables BC and

BD and the reactions at the ball and socket

joint A for the mast.





SolutionFBDFive unknown force

magnitudes




0)9

6707.0()

9

61000()

9

3707.0(

0

;0

9

6

9

6

9

3

707.0707.0

}1000{

kTTAjTAiTTA

TTFF

F

kTjTiTr

rTT

kTiTT

kAjAiAF

NjF

DCzDyDCx

DCA

DDDBD

BDDD

CCC

zyxA




0)96

96

93

707.0707.01000(6

0)(

;0

096

707.0;0

096

1000;0

093

707.0;0

kTjTiTkTiTjXk

TTFXr

M

TTAF

TAF

TTAF

DDDCC

DCB

A

DCzz

Dyy

DCxx



Solution

Solving,

NA

NA

NA

NT

NT

TTM

TM

jTTiT

z

y

x

D

C

DCy

Dx

DCD

1500

0

0

1500

707

0224.4;0

060004;0

0)224.4()60004(

Example 5.18Rod AB is subjected to the 200N force. Determine the reactions at the ball and socket joint A and the tension in cables BD and BE.





SolutionFBD




0)200()()(

0

;0

}200{

kAjTAiAA

FTTFF

F

NkF

iTT

iTT

kAjAiAF

zDyEx

DEEA

DD

EE

zyxA




Since rC = 1/2rB,0)()221()200()115.0(

0)(

;0

0200;0

0;0

0;0

jTiTXkjikXkji

TTXrFXr

M

AF

TAF

TAF

DE

DEBC

A

zz

Dyy

Exx



Solution

Solving,

NA

NA

NA

NT

NT

TTM

TM

TM

kTTjTiT

z

y

x

E

D

EDz

Ey

Dx

EDED

200

100

50

50

100

02;0

01002;0

02002;0

0)2()1002()2002(

Example 5.19The bent rod is supported at A by a journal bearing, at D by a ball and socket joint, and at B by means of cable BC. Using only one equilibrium equation, obtain a direct solution for the tension in cable BC. The bearing

at A is capable of exerting force components only in the z and y directions since it is properly

aligned.





SolutionFBDSix unknown Three force

components caused by ball and socket joint

Two caused by bearingOne caused by cable



SolutionEquations of Equilibrium Direction of the axis is defined by the

unit vector

0)(

707.0707.0

21

21

rXFuM

ji

jirr

u

DA

DA

DA




NT

kTiTji

kXj

kTjTiTXjji

XWrXTru

B

BB

BBB

EBB

572857.0

5.490

0]286.0)5.4908577.0).[(707.0707.0(

0)]981()5.0(

)7.0

6.0

7.0

3.0

7.0

2.0()1).[(707.0707.0(

0)(

Chapter SummaryChapter Summary

Free-Body Diagram Draw a FBD, an outline of the body

which shows all the forces and couple moments that act on the body

A support will exert a force on the body in a particular direction if it prevents translation of the body in that direction

A support will exert a couple moment on a body is it prevents rotation


Free-Body DiagramAngles are used to resolved forces

and dimensions used to take moments of the forces

Both must be shown on the FBD


Two Dimensions∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be

applied when solving 2D problemsFor most direct solution, try

summing the forces along an axis that will eliminate as many unknown forces as possible


Three Dimensions Use Cartesian vector analysis when

applying equations of equilibrium Express each known and unknown force

and couple moment shown on the FBD as a Cartesian vector

Set the force summation equal to zero Take moments about point O that lies on

the line of action of as many unknown components as possible


Three Dimensions From point O, direct position vectors to

each force, then use the cross product to determine the moment of each force

The six scalar equations of equilibrium are established by setting i, j and k components of these force and moment sums equal to zero

Chapter ReviewChapter Review

Engineering Mechanics: Statics Chapter 5: Equilibrium of a Rigid Body.

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Transcript of Engineering Mechanics: Statics Chapter 5: Equilibrium of a Rigid Body.