chap4 force system resultant

7/28/2019 CHAP4 Force System Resultant

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CHAPTER 4Force System Resultant


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1.Moment

A measure of the tendency of the force to cause a body torotate about the point or axis.

Torque (T)

Bending moment (M)

M MPT

4.1 Moment of a Force - - - Scalar Formulation


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2. Vector quantity

do

Lime of action (sliding vector)

(1) Magnitude ( N-m or lb-ft)

Mo = Fd

d = moment arm or perpendicular distance from point Oto the line of action of force.

(2) DirectionRight-Hard rule

A. Sense of rotation ( Force rotates about Pt.O)

Curled fingers

B. Direction and sense of moment

Thumb


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3.Resultant Moment of Coplanar Force System

ROM

nF

3F

1F2

F

do1do2 do3

don

1. Definition

(1) magnitude of

4.2 Cross Product

(2)Direction of

perpendicular to the plane containing A & B


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2. Law of operation

(1)

(2)

(3)

3. Cartesian Vector Formulation

(1) Cross product of Cartesian unit vectors.

jikikj

kkjiji

kkjjii

;

90sin

0

j

i


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(2) Cross product of vector A & B in Cartesian vector form

k)BAB(A

jBABAi)B-AB(A

jkBAikBAjkBA

jiBAkjBAjiBA

kBjBiBkAjAiABA

kBjBiBB

kAjAiAA

xyyx

xzxzyzzy

yzyzzy

xyzxyx

zyxzyx

zyx

zyx

)(

)()(

kBB

AAj

BB

AAi

BB

AA

BBB

AAAkji

BA

kAjAiAA

yx

yx

zx

zx

zy

zy

zyx

zyx

zyx


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4.3 Moment of a ForceVector Formulation

F

d

o

(1) Magnitude

Mo=|Mo|=| r x F | =| r|| F | sin=F r sin

=F d

(2) Direction

Curl the right-hand fingers from r toward F (r cross F ) and

the thumb is perpendicular to the plane containing r and F.

1. Moment of a force F about pt. O

Mo=

r

xF

where r = A position vector from pt. O to any pt. on

the line of action of force F .


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4.4 Principle of moments

Varignons theorem

The moment of a force about a point is equal to the sum of the

moment of the forces components about the point .

r

o

F1

F2

Mo=r x FF = F1+F2

Mo= r x (F1+F2)= r x F1+ r x F2

= MO1+MO2

F


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1. ObjectiveFind the component of this moment along a specified axis passes

through the point about which the moment of a force is computed.2. Scalar analysis (See textbook)

3. Vector analysis

4.5 Moment of a force about a specified Axis

Momentaxis

b

b

a

a

A

MaMo= r x F

F

O

Axis of projection

Point O on axis aa


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(1) Moment of a force F about point 0

Mo= rF

Here, we assume that bb axis is the moment axis ofMo

(2) Component of Mo onto aa axis

Ma = Maua

Ma=Mocos

=Moua=( rF ) ua=trip scalar product

HereMa=magnitude ofMaua= unit vector define the direction of aa axis


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4.Method of Finding Moment about a specific axis

(1) Find the moment of the force about point O

Mo= r x F

(2) Resolving the moment along the specific axis

Ma = Maua= (Moua) ua

=[ua ( r x F )]ua

zyx

zyx

azayax

FFF

rrr

UUU

FrUa

)(M a

kFjFiFF

r

kUjUiUU

zyx

azayaxa

krjrir zyx

If

then


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d

4.6 moment of a couple

3. Vector Formulation

F

r

d

F

M= r x F

|M|=M=|r x F |=r F sin

=F d

1. Definition ( couple)

Two parallel forces have the same magnitude, opposite

distances, and are separated by a perpendicular distance d.

2. Scalar Formulation

(1) Magnitude M=Fd

(2) Direction & sense (Right-hand rule)

Thumb indicates the direction

Curled fingers indicates the sense of rotation


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(1) The couple moment is equivalent to the sum of the momentof both couple forces about any arbitrary point 0 in space.

Br A

F

-F

rB r

Ao

Mo= rAx( -F )+ rB x F

=(-rA+rB) x F

=r x F= M

(2) Couple moment is a free vector which can act at any point in

space.

oo

B

A

-FF rMo=Mo= r x F=M

Remark:


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4. Equivalent Couples

The forces of equal couples lie either in the same plane or in planesparallel to one another.

5. Resultant couple moment

Apply couple moment at any point p on a body and add

them vectorially.

AB

M1

M2

M1

M2

MR=M= r x F

A

B

F -F

-FF

plane A // plane BF-F

d

d


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4.7 Equivalent system

1. Equivalent system

When the force and couple moment system produce the same

external effects of translation and rotation of the body as theirresultant , these two sets of loadings are said to be equivalent.

2. Principle of transmissibility

The external effects on a rigid body remain unchanged,when a

force, acting a given point on the body, is applied to another point

lying on line of action of the force.

PF

P

F A

line of action

Same external effect

Internal effect ?

Internal stresses are different.


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3. Point O is on the line of action of the force

A equivalentF

-F o

A

o

A

Sliding vector

4. Point O is not on the line of action of the force

o

A

Original system

line ofaction

ro

-F

F

FCouple moment

oAP

M=r x FMc= r x F

Force on Point A

=Force on point O + couple moment on any point p.

Original system

o

F

Fequivalent

F

A

F


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Example:

A A

F F

o o

AA

FF

o o X P

d

Mo= F d

M= F d (Free vector)

Point O is on the line of action of the force

Point O is not on the line of action of the force


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4.8 Resultant of a force & couple system

1. Objective

Simplify a system of force and couple moments to

their resultants to study the external effects on thebody.

2. Procedures for Analysis

(1)Force summations

FR=F1+F2++F

(2)Moment summations

MR0= MC+r1o*F1+r2o*F2= MC+ M0

MC:Couple moment in the system

Mo:Couple moment about pt.O of the force in the system.


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4.9 Further Reduction of a force & couple system

1. Simplification to a single Resultant Force

(1)Condition

FR MR0 orFR*MR0 = 0

(2)Force system

A. Concurrent Force system

B.Coplanar Force System

F2 F1 FR

Equivalent

P =

System

Fn no couple moment

y F1,F2,F3 on xy plane

F3 M1&M2:z direction MR0=MC+ r * F P MR0

F2 x => => d=

FR

F1 FR=F


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z z FR z

F1 r2 F2 MR0 FR= F

r1 y = y = y

M1 p o

r3 F3 x x MR0

x M2 d = --------------

|FR|d=|MR0| FR

C. Parallel Force System

1. F1 // F2//// Fn

2. MR0 perpendicular to FR, MR0=M+r*F

2. Reduction to a wrench

(1) Condition: FR MR0

MR0=M +M//

M = moment component FR

M// = moment component // FR


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(2) Wrench (or Screw)

An equivalent system reduces a simple resultant

force FRand couple moment MR0 at pt.0 to a

collinear force FRand couple moment M// at pt.

o

o

o

a

aa

a

a

a

b

b

b

b b

b

FRMRo

FRFR

pp

M//

M//

chap4 force system resultant

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