SlideShare a Scribd company logo
Equilibrium of a Rigid Body5
Engineering Mechanics:
Statics in SI Units, 12e
Copyright © 2010 Pearson Education South Asia Pte Ltd
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Objectives
• Develop the equations of equilibrium for a rigid body
• Concept of the free-body diagram for a rigid body
• Solve rigid-body equilibrium problems using the
equations of equilibrium
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Conditions for Rigid Equilibrium
2. Free-Body Diagrams
3. Equations of Equilibrium
4. Two and Three-Force Members
5. Free Body Diagrams
6. Equations of Equilibrium
7. Constraints and Statical Determinacy
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.1 Conditions for Rigid-Body Equilibrium
• The equilibrium of a body is expressed as
• Consider summing moments about some other point,
such as point A, we require
  



0
0
OOR
R
MM
FF
   0ORRA MFrM
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.2 Free Body Diagrams
Support Reactions
• If a 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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.2 Free Body Diagrams
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.2 Free Body Diagrams
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.2 Free Body Diagrams
Internal Forces
• External and internal forces can act on a rigid body
• 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.2 Free Body Diagrams
Weight and Center of Gravity
• Each particle has a specified weight
• 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.2 Free Body Diagrams
Procedure for Drawing a FBD
1. 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.2 Free Body Diagrams
3. Identify Each Loading and Give Dimensions
• Indicate dimensions for calculation of forces
• Known forces and couple moments should be properly
labeled with their magnitudes and directions
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 5.1
Draw the free-body diagram of the uniform beam. The
beam has a mass of 100kg.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Free-Body Diagram
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Free-Body Diagram
• 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.3 Equations of Equilibrium
• For equilibrium of a rigid body in 2D,
∑Fx = 0; ∑Fy = 0; ∑MO = 0
• ∑Fx and ∑Fy represent sums of x and y components of
all the forces
• ∑MO represents the sum of the couple moments and
moments of the force components
Please refer to
the Companion
CD for the
animation:
Equilibrium of
a Free Body
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
• For coplanar equilibrium problems,
∑Fx = 0; ∑Fy = 0; ∑MO = 0
• 2 alternative sets of 3 independent equilibrium
equations,
∑Fa = 0; ∑MA = 0; ∑MB = 0
Please refer to
the Companion
CD for the
animation:
Equilibrium of
a Free Body
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.3 Equations of Equilibrium
Procedure for Analysis
Free-Body Diagram
• 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
Please refer to
the Companion
CD for the
animation:
Equilibrium of
a Free Body
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.3 Equations of Equilibrium
Procedure for Analysis
Equations of Equilibrium
• Apply ∑MO = 0 about a point O
• Unknowns moments of are zero about O and a direct
solution the third unknown can be obtained
• Orient the x and y axes along the lines that will provide
the simplest resolution of the forces into their x and y
components
• Negative result scalar is opposite to that was assumed
on the FBD
Please refer to
the Companion
CD for the
animation:
Equilibrium of
a Free Body
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 5.5
Determine the horizontal and vertical components of
reaction for the beam loaded. Neglect the weight of the
beam in the calculations.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Free Body Diagrams
• 600N represented by x and y components
• 200N force acts on the beam at B
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Equations of Equilibrium
NBBNM xxB 424045cos600;0  
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








Copyright © 2010 Pearson Education South Asia Pte Ltd
5.4 Two- and Three-Force Members
Two-Force Members
• When forces are applied at only two points on a
member, the member is called a two-force member
• Only force magnitude must be determined
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.4 Two- and Three-Force Members
Three-Force Members
• When subjected to three forces, the forces are
concurrent or parallel
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 5.13
The 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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Free Body Diagrams
• BD is a two-force member
• Lever ABC is a three-force member
Equations of Equilibrium
Solving,
kNF
kNFA
32.1
07.1


045sin3.60sin;0
040045cos3.60cos;0
3.60
4.0
7.0
tan 1








 



FFF
NFFF
Ay
Ax

Copyright © 2010 Pearson Education South Asia Pte Ltd
5.5 Free-Body Diagrams
Support Reactions
As in the two-dimensional case:
• A force is developed by a support
• A couple moment is developed when rotation of the
attached member is prevented
• The force’s orientation is defined by the coordinate
angles α, β and γ
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.5 Free-Body Diagrams
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.5 Free-Body Diagrams
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 5.14
Several 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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.6 Equations of Equilibrium
Vector Equations of Equilibrium
• For two conditions for equilibrium of a rigid body in
vector form,
∑F = 0 ∑MO = 0
Scalar Equations of Equilibrium
• If all external forces and couple moments are
expressed in Cartesian vector form
∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0
∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.7 Constraints for a Rigid Body
Redundant Constraints
• More support than needed for equilibrium
• Statically indeterminate: more unknown
loadings than equations of equilibrium
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.7 Constraints for a Rigid Body
Improper Constraints
• Instability caused by the improper constraining by the
supports
• When all reactive forces are concurrent at this point,
the body is improperly constrained
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.7 Constraints for a Rigid Body
Procedure for Analysis
Free Body Diagram
• Draw an outlined shape of the body
• Show all the forces and couple moments acting on the
body
• Show all the unknown components having a positive
sense
• Indicate the dimensions of the body necessary for
computing the moments of forces
Copyright © 2010 Pearson Education South Asia Pte Ltd
5.7 Constraints for a Rigid Body
Procedure for Analysis
Equations of Equilibrium
• Apply the six scalar equations of equilibrium or vector
equations
• Any set of non-orthogonal axes may be chosen for this
purpose
Equations 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 5.15
The 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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Free Body Diagrams
• Five unknown reactions acting on the plate
• Each reaction assumed to act in a positive coordinate
direction
Equations of Equilibrium
0981300;0
0;0
0;0



NNTBAF
BF
BF
Czzz
yy
xx
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Equations of Equilibrium
• 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Solving,
Az = 790N Bz = -217N TC = 707N
• The negative sign indicates Bz acts downward
• The plate is partially constrained as the supports
cannot prevent it from turning about the z axis if a force
is applied in the x-y plane
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
1. If a support prevents translation of a body, then the
support exerts a ___________ on the body.
A) Couple moment
B) Force
C) Both A and B.
D) None of the above
2. Internal forces are _________ shown on the free body
diagram of a whole body.
A) Always
B) Often
C) Rarely
D) Never
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
3. 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 moment
B) 3 forces and 1 couple moment
C) 3 forces
D) 4 forces
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
4. Internal forces are not shown on a free-body diagram
because the internal forces are_____.
A) Equal to zero
B) Equal and opposite and they do not affect the
calculations
C) Negligibly small
D) Not important
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
5. The three scalar equations  FX =  FY =  MO = 0,
are ____ equations of equilibrium in two dimensions.
A) Incorrect B) The only correct
C) The most commonly used D) Not sufficient
6. A rigid body is subjected to forces.
This body can be considered
as a ______ member.
A) Single-force B) Two-force
C) Three-force D) Six-force
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
7. For this beam, how many support reactions are there
and is the problem statically determinate?
A) (2, Yes) B) (2, No) C) (3, Yes) D) (3, No)
8. The beam AB is loaded as shown: a) how many
support reactions are there on the beam, b) is this
problem statically determinate, and c) is the structure
stable?
A) (4, Yes, No) B) (4, No, Yes)
C) (5, Yes, No) D) (5, No, Yes)
F F F F
FFixed
support
A B
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
9. Which equation of equilibrium allows you to determine
FB right away?
A)  FX = 0 B)  FY = 0
C)  MA = 0
D) Any one of the above.
10. A beam is supported by a pin joint and a roller. How
many support reactions are there and is the structure
stable for all types of loadings?
A) (3, Yes) B) (3, No)
C) (4, Yes) D) (4, No)
AX A B
FBAY
100 lb
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
11. If a support prevents rotation of a body about an
axis, then the support exerts a ________ on the
body about that axis.
A) Couple moment B) Force
C) Both A and B. D) None of the above.
12. When doing a 3-D problem analysis, you have
________ scalar equations of equilibrium.
A) 3 B) 4
C) 5 D) 6
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
13. The rod AB is supported using two cables at B and a
ball-and-socket joint at A. How many unknown
support reactions exist in this problem?
A) 5 force and 1 moment reaction
B) 5 force reactions
C) 3 force and 3 moment reactions
D) 4 force and 2 moment reactions
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
14. If an additional couple moment in the vertical direction is applied
to rod AB at point C, then what will happen to the rod?
A) The rod remains in equilibrium as the cables provide the
necessary support reactions.
B) The rod remains in equilibrium as the ball-and-socket joint will
provide the necessary resistive reactions.
C) The rod becomes unstable as the cables cannot support
compressive forces.
D) The rod becomes unstable since a
moment about AB cannot be restricted.
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
15. A plate is supported by a ball-and-socket joint at A, a
roller joint at B, and a cable at C. How many unknown
support reactions are there in this problem?
A) 4 forces and 2 moments
B) 6 forces
C) 5 forces
D) 4 forces and 1 moment
Copyright © 2010 Pearson Education South Asia Pte Ltd
QUIZ
16. What will be the easiest way to determine the force
reaction BZ ?
A) Scalar equation  FZ = 0
B) Vector equation  MA = 0
C) Scalar equation  MZ = 0
D) Scalar equation  MY = 0

More Related Content

What's hot

Resolution of a force
Resolution of a forceResolution of a force
Resolution of a force
L Roshan Meitei
 
Simple stresses and strain
Simple stresses and strainSimple stresses and strain
Simple stresses and strain
Dhruv Trambadiya
 
engineering mechanics
engineering mechanicsengineering mechanics
engineering mechanics
WaqasAnwar51
 
Statics free body diagram
Statics free body diagramStatics free body diagram
Statics free body diagram
Gourav Kapoor
 
Centroid & Centre of Gravity
Centroid & Centre of GravityCentroid & Centre of Gravity
Centroid & Centre of Gravity
Akash Patel
 
Moment of inertia
Moment of inertia Moment of inertia
Moment of inertia
Ankita Dungarwal
 
Mechanics of materials
Mechanics of materialsMechanics of materials
Mechanics of materials
Self-employed
 
Basics and statics of particles unit i - GE6253 PPT
Basics and statics of particles   unit i - GE6253 PPTBasics and statics of particles   unit i - GE6253 PPT
Basics and statics of particles unit i - GE6253 PPT
THANGA KASI RAJAN S
 
engineering statics :equilibrium
engineering statics :equilibriumengineering statics :equilibrium
engineering statics :equilibrium
musadoto
 
Freebodydigram
FreebodydigramFreebodydigram
Freebodydigram
Gourav Kapoor
 
Resultant of forces
Resultant of forcesResultant of forces
Resultant of forces
HUSSAIN SHAIK
 
Free body diagram
Free body diagramFree body diagram
Free body diagram
Chandresh Suthar
 
INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptx
INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptxINTRODUCTION TO ENGINEERING MECHANICS - SPP.pptx
INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptx
Samirsinh Parmar
 
L01 03
L01 03L01 03
L01 03hitusp
 
Torsion of circular shafts
Torsion of circular shaftsTorsion of circular shafts
Torsion of circular shafts
Yatin Singh
 
Unit 5 rigid body dynamics
Unit 5 rigid body dynamicsUnit 5 rigid body dynamics
Unit 5 rigid body dynamics
THANGA KASI RAJAN S
 
Free body diagrams
Free body diagramsFree body diagrams
Free body diagramsTekZeno
 
Normal stress and strain
Normal stress and strainNormal stress and strain
Normal stress and strain
Shivendra Nandan
 
TORSION (MECHANICS OF SOLIDS)
TORSION (MECHANICS OF SOLIDS)TORSION (MECHANICS OF SOLIDS)
TORSION (MECHANICS OF SOLIDS)
Er.Navazhushen Patel
 

What's hot (20)

Resolution of a force
Resolution of a forceResolution of a force
Resolution of a force
 
Simple stresses and strain
Simple stresses and strainSimple stresses and strain
Simple stresses and strain
 
engineering mechanics
engineering mechanicsengineering mechanics
engineering mechanics
 
Statics free body diagram
Statics free body diagramStatics free body diagram
Statics free body diagram
 
Centroid & Centre of Gravity
Centroid & Centre of GravityCentroid & Centre of Gravity
Centroid & Centre of Gravity
 
Moment of inertia
Moment of inertia Moment of inertia
Moment of inertia
 
Mechanics of materials
Mechanics of materialsMechanics of materials
Mechanics of materials
 
Basics and statics of particles unit i - GE6253 PPT
Basics and statics of particles   unit i - GE6253 PPTBasics and statics of particles   unit i - GE6253 PPT
Basics and statics of particles unit i - GE6253 PPT
 
Introduction to Engineering Mechanics
Introduction to Engineering MechanicsIntroduction to Engineering Mechanics
Introduction to Engineering Mechanics
 
engineering statics :equilibrium
engineering statics :equilibriumengineering statics :equilibrium
engineering statics :equilibrium
 
Freebodydigram
FreebodydigramFreebodydigram
Freebodydigram
 
Resultant of forces
Resultant of forcesResultant of forces
Resultant of forces
 
Free body diagram
Free body diagramFree body diagram
Free body diagram
 
INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptx
INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptxINTRODUCTION TO ENGINEERING MECHANICS - SPP.pptx
INTRODUCTION TO ENGINEERING MECHANICS - SPP.pptx
 
L01 03
L01 03L01 03
L01 03
 
Torsion of circular shafts
Torsion of circular shaftsTorsion of circular shafts
Torsion of circular shafts
 
Unit 5 rigid body dynamics
Unit 5 rigid body dynamicsUnit 5 rigid body dynamics
Unit 5 rigid body dynamics
 
Free body diagrams
Free body diagramsFree body diagrams
Free body diagrams
 
Normal stress and strain
Normal stress and strainNormal stress and strain
Normal stress and strain
 
TORSION (MECHANICS OF SOLIDS)
TORSION (MECHANICS OF SOLIDS)TORSION (MECHANICS OF SOLIDS)
TORSION (MECHANICS OF SOLIDS)
 

Viewers also liked

M2 equilibrium of_rigid_bodies
M2 equilibrium of_rigid_bodiesM2 equilibrium of_rigid_bodies
M2 equilibrium of_rigid_bodieskochadaiyaan
 
Dynamics of particles , Enginnering mechanics , murugananthan
Dynamics of particles , Enginnering mechanics , murugananthanDynamics of particles , Enginnering mechanics , murugananthan
Dynamics of particles , Enginnering mechanics , murugananthan
Murugananthan K
 
Engineering Mechanics Presentation correction 2
Engineering Mechanics Presentation correction 2Engineering Mechanics Presentation correction 2
Engineering Mechanics Presentation correction 2Prof. S.Rajendiran
 
Diploma i em u iv centre of gravity & moment of inertia
Diploma i em u   iv centre of gravity & moment of inertiaDiploma i em u   iv centre of gravity & moment of inertia
Diploma i em u iv centre of gravity & moment of inertia
Rai University
 
Problem1 Engineering mechanics
Problem1 Engineering mechanicsProblem1 Engineering mechanics
Problem1 Engineering mechanics
Prof. S.Rajendiran
 
Introduction to design
Introduction to designIntroduction to design
Introduction to design
THANGA KASI RAJAN S
 
Two mark question and answer on Engineering Mechanics
Two mark question and answer on Engineering MechanicsTwo mark question and answer on Engineering Mechanics
Two mark question and answer on Engineering MechanicsProf. S.Rajendiran
 
University problems on Engineering Mechanics solved in differrent way
University problems on Engineering Mechanics solved in differrent wayUniversity problems on Engineering Mechanics solved in differrent way
University problems on Engineering Mechanics solved in differrent wayProf. S.Rajendiran
 
2 d equilibrium-split
2 d equilibrium-split2 d equilibrium-split
2 d equilibrium-split
sharancm2009
 
University problems on Engineering Mechanics solved in differrent way part II
University problems on Engineering Mechanics solved in differrent way part IIUniversity problems on Engineering Mechanics solved in differrent way part II
University problems on Engineering Mechanics solved in differrent way part IIProf. S.Rajendiran
 
Step by step Engineering Mechanics updated
Step by step Engineering Mechanics updatedStep by step Engineering Mechanics updated
Step by step Engineering Mechanics updatedProf. S.Rajendiran
 
Structure analysis assignment 4 centroid-moment of inertia
Structure analysis assignment 4 centroid-moment of inertiaStructure analysis assignment 4 centroid-moment of inertia
Structure analysis assignment 4 centroid-moment of inertia
The University of Lahore
 
Engineering Mechanics made simple
Engineering Mechanics made simpleEngineering Mechanics made simple
Engineering Mechanics made simpleProf. S.Rajendiran
 
Unit 5 Friction
Unit 5 FrictionUnit 5 Friction
Unit 5 Friction
THANGA KASI RAJAN S
 
Chapter4: Statics - Torques and equilibrium
Chapter4: Statics - Torques and equilibriumChapter4: Statics - Torques and equilibrium
Chapter4: Statics - Torques and equilibrium
Said Azar
 

Viewers also liked (17)

centroid
centroidcentroid
centroid
 
M2 equilibrium of_rigid_bodies
M2 equilibrium of_rigid_bodiesM2 equilibrium of_rigid_bodies
M2 equilibrium of_rigid_bodies
 
Dynamics of particles , Enginnering mechanics , murugananthan
Dynamics of particles , Enginnering mechanics , murugananthanDynamics of particles , Enginnering mechanics , murugananthan
Dynamics of particles , Enginnering mechanics , murugananthan
 
Engineering Mechanics Presentation correction 2
Engineering Mechanics Presentation correction 2Engineering Mechanics Presentation correction 2
Engineering Mechanics Presentation correction 2
 
Diploma i em u iv centre of gravity & moment of inertia
Diploma i em u   iv centre of gravity & moment of inertiaDiploma i em u   iv centre of gravity & moment of inertia
Diploma i em u iv centre of gravity & moment of inertia
 
Problem1 Engineering mechanics
Problem1 Engineering mechanicsProblem1 Engineering mechanics
Problem1 Engineering mechanics
 
Introduction to design
Introduction to designIntroduction to design
Introduction to design
 
Two mark question and answer on Engineering Mechanics
Two mark question and answer on Engineering MechanicsTwo mark question and answer on Engineering Mechanics
Two mark question and answer on Engineering Mechanics
 
University problems on Engineering Mechanics solved in differrent way
University problems on Engineering Mechanics solved in differrent wayUniversity problems on Engineering Mechanics solved in differrent way
University problems on Engineering Mechanics solved in differrent way
 
2 d equilibrium-split
2 d equilibrium-split2 d equilibrium-split
2 d equilibrium-split
 
University problems on Engineering Mechanics solved in differrent way part II
University problems on Engineering Mechanics solved in differrent way part IIUniversity problems on Engineering Mechanics solved in differrent way part II
University problems on Engineering Mechanics solved in differrent way part II
 
Step by step Engineering Mechanics updated
Step by step Engineering Mechanics updatedStep by step Engineering Mechanics updated
Step by step Engineering Mechanics updated
 
Structure analysis assignment 4 centroid-moment of inertia
Structure analysis assignment 4 centroid-moment of inertiaStructure analysis assignment 4 centroid-moment of inertia
Structure analysis assignment 4 centroid-moment of inertia
 
Engineering Mechanics made simple
Engineering Mechanics made simpleEngineering Mechanics made simple
Engineering Mechanics made simple
 
Unit 5 Friction
Unit 5 FrictionUnit 5 Friction
Unit 5 Friction
 
Chapter4: Statics - Torques and equilibrium
Chapter4: Statics - Torques and equilibriumChapter4: Statics - Torques and equilibrium
Chapter4: Statics - Torques and equilibrium
 
Chapter 4
Chapter 4Chapter 4
Chapter 4
 

Similar to Rigid body equilibrium

Engineering Mechanics: Statics.ppt
Engineering Mechanics: Statics.pptEngineering Mechanics: Statics.ppt
Engineering Mechanics: Statics.ppt
Muhammad Sohail Hasan
 
Engineering Mechanics Chapter 5 Equilibrium of a Rigid Body
Engineering Mechanics  Chapter 5  Equilibrium of a Rigid BodyEngineering Mechanics  Chapter 5  Equilibrium of a Rigid Body
Engineering Mechanics Chapter 5 Equilibrium of a Rigid Body
AhmadHajasad2
 
intro.ppt
intro.pptintro.ppt
intro.ppt
anbh30
 
Ch03
Ch03Ch03
4 intro to fbd chap3.1and3.2
4  intro to fbd chap3.1and3.24  intro to fbd chap3.1and3.2
4 intro to fbd chap3.1and3.2
VigneshN59
 
LECTURE 1 W1 Intro Stress_SHH.ppt
LECTURE 1 W1 Intro Stress_SHH.pptLECTURE 1 W1 Intro Stress_SHH.ppt
LECTURE 1 W1 Intro Stress_SHH.ppt
MOHDMUIZZBINMOHDSHUK
 
Lecture notes on trusses
Lecture notes on trussesLecture notes on trusses
Lecture notes on trusses
PUP Lopez College of Civil Engg.
 
3 equilibrium of rigid bodies fe
3 equilibrium of rigid bodies fe3 equilibrium of rigid bodies fe
3 equilibrium of rigid bodies fe
GOKULANATHAN V
 
9789810682446 slides chapter 11
9789810682446 slides chapter 119789810682446 slides chapter 11
9789810682446 slides chapter 11
Chiranjeeva Rao Seela
 
analysis-of-structure.ppt
analysis-of-structure.pptanalysis-of-structure.ppt
analysis-of-structure.ppt
Anas A
 
System Isolation and the Free-Body Diagram
System Isolation and the Free-Body DiagramSystem Isolation and the Free-Body Diagram
System Isolation and the Free-Body Diagram
Shahzaib Farooq
 
L04 5.pdf (trusses)
L04 5.pdf (trusses)L04 5.pdf (trusses)
L04 5.pdf (trusses)
nur izzaati
 
Structural Analysis- Beam.pdf
Structural Analysis- Beam.pdfStructural Analysis- Beam.pdf
Structural Analysis- Beam.pdf
CtKamariahMdSaat
 
truses and frame
 truses and frame truses and frame
truses and frameUnikl MIMET
 
Analysis of plane truss unit 5
Analysis of plane  truss unit 5Analysis of plane  truss unit 5
Analysis of plane truss unit 5
Parimal Jha
 

Similar to Rigid body equilibrium (20)

Engineering Mechanics: Statics.ppt
Engineering Mechanics: Statics.pptEngineering Mechanics: Statics.ppt
Engineering Mechanics: Statics.ppt
 
Engineering Mechanics Chapter 5 Equilibrium of a Rigid Body
Engineering Mechanics  Chapter 5  Equilibrium of a Rigid BodyEngineering Mechanics  Chapter 5  Equilibrium of a Rigid Body
Engineering Mechanics Chapter 5 Equilibrium of a Rigid Body
 
intro.ppt
intro.pptintro.ppt
intro.ppt
 
Mech chapter 03
Mech chapter 03Mech chapter 03
Mech chapter 03
 
Chapt 04a beams
Chapt 04a beamsChapt 04a beams
Chapt 04a beams
 
Ch03
Ch03Ch03
Ch03
 
4 intro to fbd chap3.1and3.2
4  intro to fbd chap3.1and3.24  intro to fbd chap3.1and3.2
4 intro to fbd chap3.1and3.2
 
Mm210(1)
Mm210(1)Mm210(1)
Mm210(1)
 
LECTURE 1 W1 Intro Stress_SHH.ppt
LECTURE 1 W1 Intro Stress_SHH.pptLECTURE 1 W1 Intro Stress_SHH.ppt
LECTURE 1 W1 Intro Stress_SHH.ppt
 
Lecture notes on trusses
Lecture notes on trussesLecture notes on trusses
Lecture notes on trusses
 
3 equilibrium of rigid bodies fe
3 equilibrium of rigid bodies fe3 equilibrium of rigid bodies fe
3 equilibrium of rigid bodies fe
 
9789810682446 slides chapter 11
9789810682446 slides chapter 119789810682446 slides chapter 11
9789810682446 slides chapter 11
 
Weebly
WeeblyWeebly
Weebly
 
Equilibrium
EquilibriumEquilibrium
Equilibrium
 
analysis-of-structure.ppt
analysis-of-structure.pptanalysis-of-structure.ppt
analysis-of-structure.ppt
 
System Isolation and the Free-Body Diagram
System Isolation and the Free-Body DiagramSystem Isolation and the Free-Body Diagram
System Isolation and the Free-Body Diagram
 
L04 5.pdf (trusses)
L04 5.pdf (trusses)L04 5.pdf (trusses)
L04 5.pdf (trusses)
 
Structural Analysis- Beam.pdf
Structural Analysis- Beam.pdfStructural Analysis- Beam.pdf
Structural Analysis- Beam.pdf
 
truses and frame
 truses and frame truses and frame
truses and frame
 
Analysis of plane truss unit 5
Analysis of plane  truss unit 5Analysis of plane  truss unit 5
Analysis of plane truss unit 5
 

More from Taral Soliya

Loads on beam
Loads on beamLoads on beam
Loads on beam
Taral Soliya
 
Forces acting on the beam with shear force & bending moment
Forces acting on the beam with shear force & bending momentForces acting on the beam with shear force & bending moment
Forces acting on the beam with shear force & bending moment
Taral Soliya
 
Bending stress
Bending stressBending stress
Bending stress
Taral Soliya
 
basics of Tool steel
basics of Tool steelbasics of Tool steel
basics of Tool steel
Taral Soliya
 
The van der waals gas
The van der waals gasThe van der waals gas
The van der waals gas
Taral Soliya
 
Nondestructive test
Nondestructive testNondestructive test
Nondestructive test
Taral Soliya
 
Non destructive testing basics
Non destructive testing basicsNon destructive testing basics
Non destructive testing basics
Taral Soliya
 
Non destructive testing
Non destructive testingNon destructive testing
Non destructive testing
Taral Soliya
 
Magnetic particle inspection
Magnetic particle inspectionMagnetic particle inspection
Magnetic particle inspection
Taral Soliya
 
Iron iron-carbide phase diagram
Iron iron-carbide phase diagramIron iron-carbide phase diagram
Iron iron-carbide phase diagram
Taral Soliya
 
Introduction to non destructive testing
Introduction to non destructive testingIntroduction to non destructive testing
Introduction to non destructive testing
Taral Soliya
 
Cast iron
Cast ironCast iron
Cast iron
Taral Soliya
 
Alloy steel
Alloy steelAlloy steel
Alloy steel
Taral Soliya
 
Laser & i'ts applications
Laser & i'ts applicationsLaser & i'ts applications
Laser & i'ts applications
Taral Soliya
 
Laser & its applications
Laser & its applicationsLaser & its applications
Laser & its applications
Taral Soliya
 
Laser & it's applications
Laser & it's applicationsLaser & it's applications
Laser & it's applications
Taral Soliya
 
Wind turbines
Wind turbinesWind turbines
Wind turbines
Taral Soliya
 
Wind turbine types
Wind turbine typesWind turbine types
Wind turbine types
Taral Soliya
 
Wind power
Wind powerWind power
Wind power
Taral Soliya
 
Wind energy technology
Wind energy technologyWind energy technology
Wind energy technology
Taral Soliya
 

More from Taral Soliya (20)

Loads on beam
Loads on beamLoads on beam
Loads on beam
 
Forces acting on the beam with shear force & bending moment
Forces acting on the beam with shear force & bending momentForces acting on the beam with shear force & bending moment
Forces acting on the beam with shear force & bending moment
 
Bending stress
Bending stressBending stress
Bending stress
 
basics of Tool steel
basics of Tool steelbasics of Tool steel
basics of Tool steel
 
The van der waals gas
The van der waals gasThe van der waals gas
The van der waals gas
 
Nondestructive test
Nondestructive testNondestructive test
Nondestructive test
 
Non destructive testing basics
Non destructive testing basicsNon destructive testing basics
Non destructive testing basics
 
Non destructive testing
Non destructive testingNon destructive testing
Non destructive testing
 
Magnetic particle inspection
Magnetic particle inspectionMagnetic particle inspection
Magnetic particle inspection
 
Iron iron-carbide phase diagram
Iron iron-carbide phase diagramIron iron-carbide phase diagram
Iron iron-carbide phase diagram
 
Introduction to non destructive testing
Introduction to non destructive testingIntroduction to non destructive testing
Introduction to non destructive testing
 
Cast iron
Cast ironCast iron
Cast iron
 
Alloy steel
Alloy steelAlloy steel
Alloy steel
 
Laser & i'ts applications
Laser & i'ts applicationsLaser & i'ts applications
Laser & i'ts applications
 
Laser & its applications
Laser & its applicationsLaser & its applications
Laser & its applications
 
Laser & it's applications
Laser & it's applicationsLaser & it's applications
Laser & it's applications
 
Wind turbines
Wind turbinesWind turbines
Wind turbines
 
Wind turbine types
Wind turbine typesWind turbine types
Wind turbine types
 
Wind power
Wind powerWind power
Wind power
 
Wind energy technology
Wind energy technologyWind energy technology
Wind energy technology
 

Recently uploaded

MCQ Soil mechanics questions (Soil shear strength).pdf
MCQ Soil mechanics questions (Soil shear strength).pdfMCQ Soil mechanics questions (Soil shear strength).pdf
MCQ Soil mechanics questions (Soil shear strength).pdf
Osamah Alsalih
 
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptx
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxCFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptx
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptx
R&R Consult
 
The Benefits and Techniques of Trenchless Pipe Repair.pdf
The Benefits and Techniques of Trenchless Pipe Repair.pdfThe Benefits and Techniques of Trenchless Pipe Repair.pdf
The Benefits and Techniques of Trenchless Pipe Repair.pdf
Pipe Restoration Solutions
 
Nuclear Power Economics and Structuring 2024
Nuclear Power Economics and Structuring 2024Nuclear Power Economics and Structuring 2024
Nuclear Power Economics and Structuring 2024
Massimo Talia
 
一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理
一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理
一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理
bakpo1
 
Railway Signalling Principles Edition 3.pdf
Railway Signalling Principles Edition 3.pdfRailway Signalling Principles Edition 3.pdf
Railway Signalling Principles Edition 3.pdf
TeeVichai
 
DESIGN A COTTON SEED SEPARATION MACHINE.docx
DESIGN A COTTON SEED SEPARATION MACHINE.docxDESIGN A COTTON SEED SEPARATION MACHINE.docx
DESIGN A COTTON SEED SEPARATION MACHINE.docx
FluxPrime1
 
AP LAB PPT.pdf ap lab ppt no title specific
AP LAB PPT.pdf ap lab ppt no title specificAP LAB PPT.pdf ap lab ppt no title specific
AP LAB PPT.pdf ap lab ppt no title specific
BrazilAccount1
 
WATER CRISIS and its solutions-pptx 1234
WATER CRISIS and its solutions-pptx 1234WATER CRISIS and its solutions-pptx 1234
WATER CRISIS and its solutions-pptx 1234
AafreenAbuthahir2
 
Top 10 Oil and Gas Projects in Saudi Arabia 2024.pdf
Top 10 Oil and Gas Projects in Saudi Arabia 2024.pdfTop 10 Oil and Gas Projects in Saudi Arabia 2024.pdf
Top 10 Oil and Gas Projects in Saudi Arabia 2024.pdf
Teleport Manpower Consultant
 
ASME IX(9) 2007 Full Version .pdf
ASME IX(9)  2007 Full Version       .pdfASME IX(9)  2007 Full Version       .pdf
ASME IX(9) 2007 Full Version .pdf
AhmedHussein950959
 
The role of big data in decision making.
The role of big data in decision making.The role of big data in decision making.
The role of big data in decision making.
ankuprajapati0525
 
space technology lecture notes on satellite
space technology lecture notes on satellitespace technology lecture notes on satellite
space technology lecture notes on satellite
ongomchris
 
Governing Equations for Fundamental Aerodynamics_Anderson2010.pdf
Governing Equations for Fundamental Aerodynamics_Anderson2010.pdfGoverning Equations for Fundamental Aerodynamics_Anderson2010.pdf
Governing Equations for Fundamental Aerodynamics_Anderson2010.pdf
WENKENLI1
 
Water Industry Process Automation and Control Monthly - May 2024.pdf
Water Industry Process Automation and Control Monthly - May 2024.pdfWater Industry Process Automation and Control Monthly - May 2024.pdf
Water Industry Process Automation and Control Monthly - May 2024.pdf
Water Industry Process Automation & Control
 
Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&B
Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&BDesign and Analysis of Algorithms-DP,Backtracking,Graphs,B&B
Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&B
Sreedhar Chowdam
 
RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...
RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...
RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...
thanhdowork
 
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...
Dr.Costas Sachpazis
 
H.Seo, ICLR 2024, MLILAB, KAIST AI.pdf
H.Seo,  ICLR 2024, MLILAB,  KAIST AI.pdfH.Seo,  ICLR 2024, MLILAB,  KAIST AI.pdf
H.Seo, ICLR 2024, MLILAB, KAIST AI.pdf
MLILAB
 
一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理
一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理
一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理
zwunae
 

Recently uploaded (20)

MCQ Soil mechanics questions (Soil shear strength).pdf
MCQ Soil mechanics questions (Soil shear strength).pdfMCQ Soil mechanics questions (Soil shear strength).pdf
MCQ Soil mechanics questions (Soil shear strength).pdf
 
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptx
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxCFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptx
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptx
 
The Benefits and Techniques of Trenchless Pipe Repair.pdf
The Benefits and Techniques of Trenchless Pipe Repair.pdfThe Benefits and Techniques of Trenchless Pipe Repair.pdf
The Benefits and Techniques of Trenchless Pipe Repair.pdf
 
Nuclear Power Economics and Structuring 2024
Nuclear Power Economics and Structuring 2024Nuclear Power Economics and Structuring 2024
Nuclear Power Economics and Structuring 2024
 
一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理
一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理
一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理
 
Railway Signalling Principles Edition 3.pdf
Railway Signalling Principles Edition 3.pdfRailway Signalling Principles Edition 3.pdf
Railway Signalling Principles Edition 3.pdf
 
DESIGN A COTTON SEED SEPARATION MACHINE.docx
DESIGN A COTTON SEED SEPARATION MACHINE.docxDESIGN A COTTON SEED SEPARATION MACHINE.docx
DESIGN A COTTON SEED SEPARATION MACHINE.docx
 
AP LAB PPT.pdf ap lab ppt no title specific
AP LAB PPT.pdf ap lab ppt no title specificAP LAB PPT.pdf ap lab ppt no title specific
AP LAB PPT.pdf ap lab ppt no title specific
 
WATER CRISIS and its solutions-pptx 1234
WATER CRISIS and its solutions-pptx 1234WATER CRISIS and its solutions-pptx 1234
WATER CRISIS and its solutions-pptx 1234
 
Top 10 Oil and Gas Projects in Saudi Arabia 2024.pdf
Top 10 Oil and Gas Projects in Saudi Arabia 2024.pdfTop 10 Oil and Gas Projects in Saudi Arabia 2024.pdf
Top 10 Oil and Gas Projects in Saudi Arabia 2024.pdf
 
ASME IX(9) 2007 Full Version .pdf
ASME IX(9)  2007 Full Version       .pdfASME IX(9)  2007 Full Version       .pdf
ASME IX(9) 2007 Full Version .pdf
 
The role of big data in decision making.
The role of big data in decision making.The role of big data in decision making.
The role of big data in decision making.
 
space technology lecture notes on satellite
space technology lecture notes on satellitespace technology lecture notes on satellite
space technology lecture notes on satellite
 
Governing Equations for Fundamental Aerodynamics_Anderson2010.pdf
Governing Equations for Fundamental Aerodynamics_Anderson2010.pdfGoverning Equations for Fundamental Aerodynamics_Anderson2010.pdf
Governing Equations for Fundamental Aerodynamics_Anderson2010.pdf
 
Water Industry Process Automation and Control Monthly - May 2024.pdf
Water Industry Process Automation and Control Monthly - May 2024.pdfWater Industry Process Automation and Control Monthly - May 2024.pdf
Water Industry Process Automation and Control Monthly - May 2024.pdf
 
Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&B
Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&BDesign and Analysis of Algorithms-DP,Backtracking,Graphs,B&B
Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&B
 
RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...
RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...
RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...
 
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...
 
H.Seo, ICLR 2024, MLILAB, KAIST AI.pdf
H.Seo,  ICLR 2024, MLILAB,  KAIST AI.pdfH.Seo,  ICLR 2024, MLILAB,  KAIST AI.pdf
H.Seo, ICLR 2024, MLILAB, KAIST AI.pdf
 
一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理
一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理
一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理
 

Rigid body equilibrium

  • 1. Equilibrium of a Rigid Body5 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson Education South Asia Pte Ltd
  • 2. Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Objectives • Develop the equations of equilibrium for a rigid body • Concept of the free-body diagram for a rigid body • Solve rigid-body equilibrium problems using the equations of equilibrium
  • 3. Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter Outline 1. Conditions for Rigid Equilibrium 2. Free-Body Diagrams 3. Equations of Equilibrium 4. Two and Three-Force Members 5. Free Body Diagrams 6. Equations of Equilibrium 7. Constraints and Statical Determinacy
  • 4. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.1 Conditions for Rigid-Body Equilibrium • The equilibrium of a body is expressed as • Consider summing moments about some other point, such as point A, we require       0 0 OOR R MM FF    0ORRA MFrM
  • 5. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Support Reactions • If a 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.
  • 6. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams
  • 7. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams
  • 8. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Internal Forces • External and internal forces can act on a rigid body • 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
  • 9. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Weight and Center of Gravity • Each particle has a specified weight • 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
  • 10. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams Procedure for Drawing a FBD 1. 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
  • 11. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.2 Free Body Diagrams 3. Identify Each Loading and Give Dimensions • Indicate dimensions for calculation of forces • Known forces and couple moments should be properly labeled with their magnitudes and directions
  • 12. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg.
  • 13. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free-Body Diagram
  • 14. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free-Body Diagram • 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
  • 15. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.3 Equations of Equilibrium • For equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 • ∑Fx and ∑Fy represent sums of x and y components of all the forces • ∑MO represents the sum of the couple moments and moments of the force components Please refer to the Companion CD for the animation: Equilibrium of a Free Body
  • 16. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.3 Equations of Equilibrium Alternative Sets of Equilibrium Equations • For coplanar equilibrium problems, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 • 2 alternative sets of 3 independent equilibrium equations, ∑Fa = 0; ∑MA = 0; ∑MB = 0 Please refer to the Companion CD for the animation: Equilibrium of a Free Body
  • 17. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.3 Equations of Equilibrium Procedure for Analysis Free-Body Diagram • 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 Please refer to the Companion CD for the animation: Equilibrium of a Free Body
  • 18. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.3 Equations of Equilibrium Procedure for Analysis Equations of Equilibrium • Apply ∑MO = 0 about a point O • Unknowns moments of are zero about O and a direct solution the third unknown can be obtained • Orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components • Negative result scalar is opposite to that was assumed on the FBD Please refer to the Companion CD for the animation: Equilibrium of a Free Body
  • 19. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.5 Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations.
  • 20. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free Body Diagrams • 600N represented by x and y components • 200N force acts on the beam at B
  • 21. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Equations of Equilibrium NBBNM xxB 424045cos600;0   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        
  • 22. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.4 Two- and Three-Force Members Two-Force Members • When forces are applied at only two points on a member, the member is called a two-force member • Only force magnitude must be determined
  • 23. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.4 Two- and Three-Force Members Three-Force Members • When subjected to three forces, the forces are concurrent or parallel
  • 24. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.13 The 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.
  • 25. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free Body Diagrams • BD is a two-force member • Lever ABC is a three-force member Equations of Equilibrium Solving, kNF kNFA 32.1 07.1   045sin3.60sin;0 040045cos3.60cos;0 3.60 4.0 7.0 tan 1              FFF NFFF Ay Ax 
  • 26. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.5 Free-Body Diagrams Support Reactions As in the two-dimensional case: • A force is developed by a support • A couple moment is developed when rotation of the attached member is prevented • The force’s orientation is defined by the coordinate angles α, β and γ
  • 27. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.5 Free-Body Diagrams
  • 28. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.5 Free-Body Diagrams
  • 29. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.14 Several 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.
  • 30. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution
  • 31. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.6 Equations of Equilibrium Vector Equations of Equilibrium • For two conditions for equilibrium of a rigid body in vector form, ∑F = 0 ∑MO = 0 Scalar Equations of Equilibrium • If all external forces and couple moments are expressed in Cartesian vector form ∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0 ∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0
  • 32. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.7 Constraints for a Rigid Body Redundant Constraints • More support than needed for equilibrium • Statically indeterminate: more unknown loadings than equations of equilibrium
  • 33. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.7 Constraints for a Rigid Body Improper Constraints • Instability caused by the improper constraining by the supports • When all reactive forces are concurrent at this point, the body is improperly constrained
  • 34. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.7 Constraints for a Rigid Body Procedure for Analysis Free Body Diagram • Draw an outlined shape of the body • Show all the forces and couple moments acting on the body • Show all the unknown components having a positive sense • Indicate the dimensions of the body necessary for computing the moments of forces
  • 35. Copyright © 2010 Pearson Education South Asia Pte Ltd 5.7 Constraints for a Rigid Body Procedure for Analysis Equations of Equilibrium • Apply the six scalar equations of equilibrium or vector equations • Any set of non-orthogonal axes may be chosen for this purpose Equations 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
  • 36. Copyright © 2010 Pearson Education South Asia Pte Ltd Example 5.15 The 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.
  • 37. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Free Body Diagrams • Five unknown reactions acting on the plate • Each reaction assumed to act in a positive coordinate direction Equations of Equilibrium 0981300;0 0;0 0;0    NNTBAF BF BF Czzz yy xx
  • 38. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Equations of Equilibrium • 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
  • 39. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Solving, Az = 790N Bz = -217N TC = 707N • The negative sign indicates Bz acts downward • The plate is partially constrained as the supports cannot prevent it from turning about the z axis if a force is applied in the x-y plane
  • 40. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 1. If a support prevents translation of a body, then the support exerts a ___________ on the body. A) Couple moment B) Force C) Both A and B. D) None of the above 2. Internal forces are _________ shown on the free body diagram of a whole body. A) Always B) Often C) Rarely D) Never
  • 41. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 3. 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 moment B) 3 forces and 1 couple moment C) 3 forces D) 4 forces
  • 42. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 4. Internal forces are not shown on a free-body diagram because the internal forces are_____. A) Equal to zero B) Equal and opposite and they do not affect the calculations C) Negligibly small D) Not important
  • 43. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 5. The three scalar equations  FX =  FY =  MO = 0, are ____ equations of equilibrium in two dimensions. A) Incorrect B) The only correct C) The most commonly used D) Not sufficient 6. A rigid body is subjected to forces. This body can be considered as a ______ member. A) Single-force B) Two-force C) Three-force D) Six-force
  • 44. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 7. For this beam, how many support reactions are there and is the problem statically determinate? A) (2, Yes) B) (2, No) C) (3, Yes) D) (3, No) 8. The beam AB is loaded as shown: a) how many support reactions are there on the beam, b) is this problem statically determinate, and c) is the structure stable? A) (4, Yes, No) B) (4, No, Yes) C) (5, Yes, No) D) (5, No, Yes) F F F F FFixed support A B
  • 45. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 9. Which equation of equilibrium allows you to determine FB right away? A)  FX = 0 B)  FY = 0 C)  MA = 0 D) Any one of the above. 10. A beam is supported by a pin joint and a roller. How many support reactions are there and is the structure stable for all types of loadings? A) (3, Yes) B) (3, No) C) (4, Yes) D) (4, No) AX A B FBAY 100 lb
  • 46. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 11. If a support prevents rotation of a body about an axis, then the support exerts a ________ on the body about that axis. A) Couple moment B) Force C) Both A and B. D) None of the above. 12. When doing a 3-D problem analysis, you have ________ scalar equations of equilibrium. A) 3 B) 4 C) 5 D) 6
  • 47. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 13. The rod AB is supported using two cables at B and a ball-and-socket joint at A. How many unknown support reactions exist in this problem? A) 5 force and 1 moment reaction B) 5 force reactions C) 3 force and 3 moment reactions D) 4 force and 2 moment reactions
  • 48. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 14. If an additional couple moment in the vertical direction is applied to rod AB at point C, then what will happen to the rod? A) The rod remains in equilibrium as the cables provide the necessary support reactions. B) The rod remains in equilibrium as the ball-and-socket joint will provide the necessary resistive reactions. C) The rod becomes unstable as the cables cannot support compressive forces. D) The rod becomes unstable since a moment about AB cannot be restricted.
  • 49. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 15. A plate is supported by a ball-and-socket joint at A, a roller joint at B, and a cable at C. How many unknown support reactions are there in this problem? A) 4 forces and 2 moments B) 6 forces C) 5 forces D) 4 forces and 1 moment
  • 50. Copyright © 2010 Pearson Education South Asia Pte Ltd QUIZ 16. What will be the easiest way to determine the force reaction BZ ? A) Scalar equation  FZ = 0 B) Vector equation  MA = 0 C) Scalar equation  MZ = 0 D) Scalar equation  MY = 0