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Chapter 3Static of rigid bodies
3.0 Introduction
Most bodies in elementary mechanics are assumed to be rigid,i.e., the actual deformations are small and do not affect theconditions of equilibrium or motion of the body.
Forces acting on rigid bodies are divided into two groups:
External forces
Internal forces External forces generally cause translation i.e. linear motion
and/or rotation (motion about a pivot) of the rigid body
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Chapter 3Static of rigid bodies
3.1 Principle of Transmissibility
Principle of transmissibility states that the condition of rest ormotion of a rigid body is unaffected if a force, F acting on a point
A is moved to act at a new point, B provided that the point B lieson the same line of action of that force.
NOTE: F and F are equivalent forces.
Moving the point of application of
the force F to the rear bumper
does not affect the motion or the
other forces acting on the truck.
A
B
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Chapter 3Static of rigid bodies
3.2 Moment of Force scalar formulation
The effect of a force on a rigid body arethe moment of force (also called torque).This momentof a force provides ameasure of the tendency of the force tocause a body to rotate about the point oraxis.
Consider horizontal force Fx, which actsperpendicular to the handle of the wrenchand is located dy from the point O
Fx tends to turn the pipe about the z axis
The larger the force or the distance dy,the greater the turning effect
Torque tendency of
rotation caused by Fxor simple moment (Mo) z
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Chapter 3Static of rigid bodies
For magnitude ofMO,
MO = Fdwhere d = moment arm or perpendiculardistance from the axis at point O to its lineof action of the force
Units for moment is N.m
Direction ofMO is specified by using righthand rule
- Fingers of the right hand are curled tofollow the sense of rotation when forcerotates about point O
- Thumb points along the moment axis togive the direction and sense of themoment vector
- Moment vector is upwards andperpendicular to the shaded plane
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Chapter 3Static of rigid bodies
MO is shown by a vector arrow with acurl to distinguish it from force vector
Moment of a force does not alwayscause rotation
Force F tends to rotate the beamclockwise about A with moment
MA = FdA
Force F tends to rotate the beamcounterclockwise about B withmoment
MB = FdB Hence support at A prevents the
rotation
Resultant moment, MRo = addition ofthe moments of all the forcesalgebraically since all moment forcesare collinear
MRo = Fd
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Chapter 3Static of rigid bodies
3.3 Cross Product (Vector Product)
Concept of the moment of a force about apoint is more easily understood throughapplications of the vector productorcross
product.
Vector product of two vectors P and Q isdefined as the vectorVwhich satisfies the
following conditions: Line of action ofVis perpendicular to
plane containing P and Q.
Magnitude ofVis
Direction ofVis obtained from the right-hand rule.
Vector products:
- are not commutative,
- are distributive,
- are not associative,
QPPQ
2121 QPQPQQP
SQPSQP
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Chapter 3Static of rigid bodies
kQPQP
jQPQPiQPQP
xyyx
zxxzyzzy
Vector products of Cartesian unit vectors,
0
0
0
kkikjjki
ijkjjkji
jikkijii
Vector products in terms of rectangularcoordinates
kQjQiQkPjPiPV zyxzyx
zyx
zyx
QQQ
PPP
kji
A shorthand way to represent this calculation is
by the use of the determinant
Determinant = Sum of Products - Sum of
Products
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Chapter 3Static of rigid bodies
3.4 Moment of Force - vector formulation
Moment of force F about point O can beexpressed using cross product
MO = rX F
where rrepresents position vector from O
to any pointlying on the line of action ofF
For magnitude of cross product,
MO = rFsin
where is the angle measured between
tails ofrand F
x y z
x y z
i j k
A A A
B B B
C=AB
x y z x y
x y z x y
i j k i j
A A A A A
B B B B B
( )( ) ( )( ) ( )( )y z z y z x x z x y y xA B A B i A B A B j A B A B k
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Chapter 3Static of rigid bodies
Treat ras a sliding vector. Since d= r
sin,MO = rFsin = F(rsin) = Fd
Direction and sense ofMO aredetermined by right-hand rule
3.5 Principle of moments
Also known as Varignons Theorem
Moment of a force about a point isequal to the sum of the moments of theforces components about the point
ForF = F1 + F2,MO = rX F1 + rX F2= rX (F1 + F2)
= rX F
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Chapter 3Static of rigid bodies
Resultant moment of forces
about point O can be
determined by vector addition
O x y z
x y z
i j kr r r
F F F
M = r F
MRo = (rx F)
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Chapter 3Static of rigid bodies
3.6 Moment of couple
Two forces F and
F are said toform a couple if they have thesame magnitude, parallel lines ofaction but directed in oppositedirection and separated by aperpendicular distance, d.
The moment produced by acouple is called a couple moment.
A couple moment is a free vectorthat can be applied at any pointwith the same effect.
( ) ( ) 0net
F = F F
F
d
-FB
A
FdrFM
Fr
Frr
FrFrM
BA
BA
sin
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Chapter 3Static of rigid bodies
Two couples are equivalent if they
produce the same moment For resultant moment of two
couples at point P,
MR = M1 + M2 For more than 2 moments,
MR = (rX F)
The combination of the force and
couple is referred to as a force-
couple system.
The three forces may be replaced by
an equivalent force vector and couple
vector.
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Chapter 3Static of rigid bodies
Equivalent Force-Couple
O AF = FO A OA AM = M r F
A system of forces may be replaced by a collection of
force-couple systems acting a given pointO
The force and
couple vectors may
be combined into aresultant force
vector and a
resultant couple
vector,
FrMFRR
O
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Chapter 3Static of rigid bodies
3.7 Reactions at Supports and Connections
For a rigid body in static equilibrium, the external forces and moments are balanced andwill impart no translational or rotational motion to the body.
The necessary and sufficient condition for the static equilibrium of a body are that theresultant force and couple from all external forces form a system equivalent to zero,
Resolving each force and moment into its rectangular components leads to 6 scalarequations which also express the conditions for static equilibrium,
For two-dimensional problem, the moment about the z-direction must be zero in order topreserve the static equilibrium condition.
00 FrMF O
000
000
zyx
zyx
MMM
FFF
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Reactions equivalent to aforce with known line of
action.
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Reactions equivalent to a
force of unknown direction
and magnitude.
Reactions equivalent to a
force of unknown directionand magnitude and a
couple of unknown
magnitude
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Chapter 3Static of rigid bodies
First step in the static equilibrium analysis of arigid body is identification of all forces acting
on the body with a free-bodydiagram.
Select the extent of the free-body and
detach it from the ground and all other
bodies.
3.8 Free-Body Diagram
Indicate point of application, magnitude,
and direction of external forces, including
the rigid body weight.
Indicate point of application and assumed
direction of unknown applied forces.These usually consist of reactions through
which the ground and other bodies oppose
the possible motion of the rigid body.
Include the dimensions necessary to
compute the moments of the forces.
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Chapter 3Static of rigid bodies
3.9 Equilibrium of a Rigid Body in Two Dimensions
For all forces and moments acting on atwo-dimensional structure,
Ozyxz MMMMF 00
Equations of equilibrium become
000 Ayx MFF
whereA is any point in the plane of
the structure.
The 3 equations can be solved for no
more than 3 unknowns.
The 3 equations can not be augmented
with additional equations, but they can be
replaced
000 BAx MMF
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Chapter 3Static of rigid bodies
3.10 Statically Indeterminate Reactions
More unknowns
than equations
Fewer unknowns
than equations,partially constrained
Equal number unknowns
and equations but
improperly constrained
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Chapter 3Static of rigid bodies
3.11 Equilibrium of a Two-Force Body
Consider a plate subjected to two forces F1and F2
For static equilibrium, the sum of moments
aboutA must be zero. The moment ofF2must
be zero. It follows that the line of action ofF2
must pass throughA.
Similarly, the line of action ofF1 must pass
through B for the sum of moments about B to
be zero.
Requiring that the sum of forces in any
direction be zero leads to the conclusion that
F1 and F2must have equal magnitude but
opposite sense.
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Chapter 3Static of rigid bodies
Consider a rigid body subjected to forces acting
at only 3 points.
3.12 Equilibrium of a Three-Force Body
Assuming that their lines of action intersect, the
moment of F1 and F2about the point of
intersection represented by D is zero. Since the rigid body is in equilibrium, the sum of
the moments ofF1, F2, and F3 about any axis
must be zero. It follows that the moment ofF3about D must be zero as well and that the line of
action ofF3 must pass through D.
The lines of action of the three forces must be
concurrent or parallel.