Chapter 3 Rigid Body

7/28/2019 Chapter 3 Rigid Body

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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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.

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Chapter 3 Rigid Body