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7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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ME 1204: Engineering Statics
Dr. Faraz Junejo
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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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
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Outline
1. Concept of Rigid Body
2. Conditions for Rigid Equilibrium
3. Free-Body Diagrams
4. Equations of Equilibrium
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Rigid Body
A rigid body can be considered as a combination of a
large number of particles in which all particles remain at
a fixed distance from one another, both before and after
applying a load.
This model is important as material properties of any
body that is assumed to be rigid will not have to be
considered when studying the effects of forces acting on
the body.
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Rigid Body (contd.)
Staticsdeals primarily with the calculation of external
forces which act on rigid bodies in equilibrium.
Determination of the internal deformations belongs to the
study of the mechanics of deformable bodies, which
normally follows statics in the curriculum i.e. Mechanics
or Strength of materialscourse.
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CONDITIONS FOR RIGID-BODY EQUILIBRIUM
In Chapter 3 we only consideredforces acting on a particle
(concurrent forces). In this caserotation is not a concern, soequilibrium could be satisfied by:
Forces on a particle
We will now consider cases where forces arenot concurrent so we are also concerned thatthe rigid body does not rotate. In order for a
rigid body to be in equilibrium, the net forceas well as the net moment about anyarbitrary point O must be equal to zero.
Forces on a rigidbody
F= 0 (no translation)
and MO
= 0 (no rotation)
F= 0 (no translation)
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Conditions for Rigid-Body Equilibrium
The equilibrium of a body is expressed as
0
0
OOR
R
MM
FF
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Conditions for Rigid-Body Equilibrium
(contd.)
Consider summing moments
about some other point, such
as point A, we require
0ORRA MFrM
0
0
OOR
R
MM
FF
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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.
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Support Reactions (contd.)
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Support Reactions (contd.)
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Support Reactions (contd.)
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Summary: Support Reactions
A reactionis a force with known line of action, or a force
of unknown direction, or a moment.
The line of action of the force or direction of the moment
is directly related to the motion that is prevented.
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Recognizing support unknowns in FBDs
S t R ti i 2D (T bl 5 1) i thi t bl !
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Support Reactions in 2D (Table 5-1)
The 2D reactions shown below are the ones shown in Table 5-1 inthe text. As a general rule:
1) if a support prevents translation of a body in a givendirection, thena force is developed on the body in the oppositedirection.2) if rotation is prevented, a couple moment is exerted on thebody in the opposite direction
memorize this table!
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7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Free Body Diagrams (contd.)
Internal Forces
External and internal forces can act on a rigid body
In summary:
For FBD, internal forces act between adjacent particles which are
contained within the boundary of the FBD, are not represented
Particles outside this boundary exert external forces on the
system
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Internal Forces (contd.)
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Free Body Diagrams (contd.)
Weight and Center of Gravity
Each particle has a specified weight
System can be represented by a single resultant force,known as weight Wof the body
Location of the force application is known as the
center of gravity
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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
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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
Procedure for Drawing a FBD (contd.)
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Example: 1
Draw the free-body diagram of the uniform beam. The
beam has a mass of 100kg.
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Example: 1 (contd.)
Free-Body Diagram
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Free-Body Diagrams Description
Support at A is a fixed wall, the wall exerts three reactions
on the beam. These 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 beams center of gravity, 3m from A
Example: 1 (contd.)
EXAMPLE: 2
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EXAMPLE: 2
Smooth pin
Weightless link (see Table 5-1)
Given:The operator applies a verticalforce to the pedal so that thespring is stretched 1.5 in. and theforce in the short link atB is20 lb.
Draw:A an idealized model and free-body diagram of the foot pedal.
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Free-Body Diagrams Description
Example: 2 (contd.)
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Exercise: 1
Draw a FBD of member ABC, which is supported by a smoothcollar at A, rocker at B, and link CD.
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Equations of Equilibrium
For equilibrium of a rigid body in 2D,
Fx= 0; Fy= 0; MO= 0
Fxand Fyrepresent sums of x and y components of
all the forces
MOrepresents the sum of the couple moments and
moments of the force components
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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
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Equations of Equilibrium
Apply MO= 0 about a point O, that l ies at the
in tersect ion of the lines of act ion of two unknown
forces.
In this way, moments of these unknowns are zero
about O and a direct solution forthe third unknown can
be obtained
Procedure for Analysis (contd.)
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Equations of Equilibrium
When applying the force equilibrium equations, orient
the x and yaxes along the lines that will provide the
simplest resolution of the forces into theirxand ycomponents
Negative result implies scalar for a force or couple
moment magnitude is opposite to that was assumed on
the FBD
Procedure for Analysis (contd.)
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Example: 1
Determine the horizontal and vertical components of reaction
on the beam caused by the pin at B and the rocker at A.
Neglect the weight of the beam in the calculations.
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Free Body Diagram
600N represented by x and y components
200N force acts on the beam at B
Example: 1 (contd.)
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Equations of Equilibrium
NBBNFxxx
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
Example: 1 (contd.)
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Example: 2
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Example: 2 (contd.)
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Example: 2 (contd.)
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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.
Only force magnitude must be determined
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Two- Force Members (contd.)
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EXAMPLE OF TWO-FORCE MEMBERS
In the cases above, members AB can be considered as two-
force members, provided that their weight is neglected.
This fact simplifies the equilibrium analysis of some rigid
bodies since the directions of the resultant forces at Aand B
are thus known (along the line joining points A and B).
Th F M b
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Three-Force Members When subjected to three forces, the forces are concurrent
or parallel
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Example: 1
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.
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Example: 1 (contd.)
Free Body Diagrams BDis a two-force member
LeverABCis a three-force member
Equations of Equilibrium
Solving,
kNF
kNFA
32.1
07.1
045sin3.60sin;0
040045cos3.60cos;0
3.604.0
7.0tan 1
FFF
NFFF
Ay
Ax
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A man raises a 10 kg joist, of
length 4 m, by pulling on a
rope.
Find the tension in the ropeand the reaction atA.
SOLUTION:
Create a free-body diagram of the joist.
Note that the joist is a 3 force body
acted upon by the rope, its weight, and
the reaction atA.
The three forces must be concurrent
for static equilibrium. Therefore, the
reaction Rmust pass through the
intersection of the lines of action of the
weight and rope forces. Determine the
direction of the reaction force R.
Utilize a force triangle to determine
the magnitude of the reaction force R.
Example: 2
Example: 2 (contd )
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Create a free-body diagram of the joist.
Determine the direction of the reaction
force R.
636.1
414.1
313.2tan
m2.313m515.0828.2
m515.020tanm414.1)2045cot(
m414.1
m828.245cosm445cos
2
1
AE
CE
BDBFCE
CDBD
AFAECD
ABAF
6.58
Example: 2 (contd.)
7/27/2019 Chap 5.1 Conditions Equations of Equilibrium of a Rigid Body
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Determine the magnitude of the reactionforce R.
38.6sin
N1.98
110sin4.31sin
RT
N8.147
N9.81
R
T
Example: 2 (contd.)
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Exercises
Engineering Mechanics - Statics, R.C. Hibbeler, 12th
Edition
Q5.11Q5.24
Q5.32Q5.37
Q5.41Q5.45
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QUIZ
1. If a support prevents translation of a body, then thesupport 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 bodydiagram of a whole body.
A) AlwaysB) Often
C) Rarely
D) Never
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QUIZ (contd.)
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
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QUIZ (contd.)
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 thecalculations
C) Negligibly small
D) Not important
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QUIZ (contd.)
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