Simplificações de Forças e Binários

Prof. Ettore Baldini-Neto

• Algumas vezes é conveniente reduzir um sistema de forças e momentos binários atuando em um corpo por um sistema equivalente mais simples, que consiste de uma força resultante e de um momento binário resultante atuando em relação a um dado ponto.

• Um sistema é equivalente quando os efeitos externos produzidos por ele sobre o corpo são os mesmos que aqueles causados pelas forças e momentos binários originiais.

• Neste contexto, os efeitos externos referem-se a movimentos de translação/rotação caso o corpo esteja livre para se mover, ou à reação em suportes se o corpo estiver fixo.

Entendendo melhor..

160 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

4.7 Simplification of a Force and CoupleSystem

Sometimes it is convenient to reduce a system of forces and couple momentsacting on a body to a simpler form by replacing it with an equivalent system,consisting of a single resultant force acting at a specific point and a resultantcouple moment.A system is equivalent if the external effects it produces ona body are the same as those caused by the original force and couplemoment system. In this context, the external effects of a system refer to thetranslating and rotating motion of the body if the body is free to move, or itrefers to the reactive forces at the supports if the body is held fixed.

For example, consider holding the stick in Fig. 4–34a, which issubjected to the force F at point A. If we attach a pair of equal butopposite forces F and –F at point B, which is on the line of action of F,Fig. 4–34b, we observe that –F at B and F at A will cancel each other,leaving only F at B, Fig. 4–34c. Force F has now been moved from A to Bwithout modifying its external effects on the stick; i.e., the reaction at thegrip remains the same. This demonstrates the principle of transmissibility,which states that a force acting on a body (stick) is a sliding vector sinceit can be applied at any point along its line of action.

We can also use the above procedure to move a force to a point that is noton the line of action of the force. If F is applied perpendicular to the stick, asin Fig. 4–35a, then we can attach a pair of equal but opposite forces F and –Fto B, Fig. 4–35b. Force F is now applied at B, and the other two forces, F at Aand –F at B, form a couple that produces the couple moment ,Fig. 4–35c. Therefore, the force F can be moved from A to B provided acouple moment M is added to maintain an equivalent system. This couplemoment is determined by taking the moment of F about B. Since M isactually a free vector, it can act at any point on the stick. In both cases thesystems are equivalent which causes a downward force F and clockwisecouple moment M = Fd to be felt at the grip.

M = Fd

F F

F!FAB

(a)

AB

F

(b) (c)

Fig. 4–34

F F

!FA

d

(a)

FF

m " Fd

(b) (c)

Fig. 4–35

160 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

4.7 Simplification of a Force and CoupleSystem

Sometimes it is convenient to reduce a system of forces and couple momentsacting on a body to a simpler form by replacing it with an equivalent system,consisting of a single resultant force acting at a specific point and a resultantcouple moment.A system is equivalent if the external effects it produces ona body are the same as those caused by the original force and couplemoment system. In this context, the external effects of a system refer to thetranslating and rotating motion of the body if the body is free to move, or itrefers to the reactive forces at the supports if the body is held fixed.

For example, consider holding the stick in Fig. 4–34a, which issubjected to the force F at point A. If we attach a pair of equal butopposite forces F and –F at point B, which is on the line of action of F,Fig. 4–34b, we observe that –F at B and F at A will cancel each other,leaving only F at B, Fig. 4–34c. Force F has now been moved from A to Bwithout modifying its external effects on the stick; i.e., the reaction at thegrip remains the same. This demonstrates the principle of transmissibility,which states that a force acting on a body (stick) is a sliding vector sinceit can be applied at any point along its line of action.

We can also use the above procedure to move a force to a point that is noton the line of action of the force. If F is applied perpendicular to the stick, asin Fig. 4–35a, then we can attach a pair of equal but opposite forces F and –Fto B, Fig. 4–35b. Force F is now applied at B, and the other two forces, F at Aand –F at B, form a couple that produces the couple moment ,Fig. 4–35c. Therefore, the force F can be moved from A to B provided acouple moment M is added to maintain an equivalent system. This couplemoment is determined by taking the moment of F about B. Since M isactually a free vector, it can act at any point on the stick. In both cases thesystems are equivalent which causes a downward force F and clockwisecouple moment M = Fd to be felt at the grip.

M = Fd

F F

F!FAB

(a)

AB

F

(b) (c)

Fig. 4–34

F F

!FA

d

(a)

FF

m " Fd

(b) (c)

Fig. 4–35

160 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

4.7 Simplification of a Force and CoupleSystem

Sometimes it is convenient to reduce a system of forces and couple momentsacting on a body to a simpler form by replacing it with an equivalent system,consisting of a single resultant force acting at a specific point and a resultantcouple moment.A system is equivalent if the external effects it produces ona body are the same as those caused by the original force and couplemoment system. In this context, the external effects of a system refer to thetranslating and rotating motion of the body if the body is free to move, or itrefers to the reactive forces at the supports if the body is held fixed.

For example, consider holding the stick in Fig. 4–34a, which issubjected to the force F at point A. If we attach a pair of equal butopposite forces F and –F at point B, which is on the line of action of F,Fig. 4–34b, we observe that –F at B and F at A will cancel each other,leaving only F at B, Fig. 4–34c. Force F has now been moved from A to Bwithout modifying its external effects on the stick; i.e., the reaction at thegrip remains the same. This demonstrates the principle of transmissibility,which states that a force acting on a body (stick) is a sliding vector sinceit can be applied at any point along its line of action.

We can also use the above procedure to move a force to a point that is noton the line of action of the force. If F is applied perpendicular to the stick, asin Fig. 4–35a, then we can attach a pair of equal but opposite forces F and –Fto B, Fig. 4–35b. Force F is now applied at B, and the other two forces, F at Aand –F at B, form a couple that produces the couple moment ,Fig. 4–35c. Therefore, the force F can be moved from A to B provided acouple moment M is added to maintain an equivalent system. This couplemoment is determined by taking the moment of F about B. Since M isactually a free vector, it can act at any point on the stick. In both cases thesystems are equivalent which causes a downward force F and clockwisecouple moment M = Fd to be felt at the grip.

M = Fd

F F

F!FAB

(a)

AB

F

(b) (c)

Fig. 4–34

F F

!FA

d

(a)

FF

m " Fd

(b) (c)

Fig. 4–35

A força F foi movida de A para B sem modificar seus efeitos externos sobre o bastão, ou seja, a reação no ponto de contato é a mesma.

Este fato demonstra o princípio da transmissibilidade que diz que a força F é um vetor deslizante desde que pode ser aplicada em qualquer ponto sob sua linha de ação no bastão.

160 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

4.7 Simplification of a Force and CoupleSystem

Sometimes it is convenient to reduce a system of forces and couple momentsacting on a body to a simpler form by replacing it with an equivalent system,consisting of a single resultant force acting at a specific point and a resultantcouple moment.A system is equivalent if the external effects it produces ona body are the same as those caused by the original force and couplemoment system. In this context, the external effects of a system refer to thetranslating and rotating motion of the body if the body is free to move, or itrefers to the reactive forces at the supports if the body is held fixed.

For example, consider holding the stick in Fig. 4–34a, which issubjected to the force F at point A. If we attach a pair of equal butopposite forces F and –F at point B, which is on the line of action of F,Fig. 4–34b, we observe that –F at B and F at A will cancel each other,leaving only F at B, Fig. 4–34c. Force F has now been moved from A to Bwithout modifying its external effects on the stick; i.e., the reaction at thegrip remains the same. This demonstrates the principle of transmissibility,which states that a force acting on a body (stick) is a sliding vector sinceit can be applied at any point along its line of action.

We can also use the above procedure to move a force to a point that is noton the line of action of the force. If F is applied perpendicular to the stick, asin Fig. 4–35a, then we can attach a pair of equal but opposite forces F and –Fto B, Fig. 4–35b. Force F is now applied at B, and the other two forces, F at Aand –F at B, form a couple that produces the couple moment ,Fig. 4–35c. Therefore, the force F can be moved from A to B provided acouple moment M is added to maintain an equivalent system. This couplemoment is determined by taking the moment of F about B. Since M isactually a free vector, it can act at any point on the stick. In both cases thesystems are equivalent which causes a downward force F and clockwisecouple moment M = Fd to be felt at the grip.

M = Fd

F F

F!FAB

(a)

AB

F

(b) (c)

Fig. 4–34

F F

!FA

d

(a)

FF

m " Fd

(b) (c)

Fig. 4–35

160 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

4.7 Simplification of a Force and CoupleSystem

Sometimes it is convenient to reduce a system of forces and couple momentsacting on a body to a simpler form by replacing it with an equivalent system,consisting of a single resultant force acting at a specific point and a resultantcouple moment.A system is equivalent if the external effects it produces ona body are the same as those caused by the original force and couplemoment system. In this context, the external effects of a system refer to thetranslating and rotating motion of the body if the body is free to move, or itrefers to the reactive forces at the supports if the body is held fixed.

For example, consider holding the stick in Fig. 4–34a, which issubjected to the force F at point A. If we attach a pair of equal butopposite forces F and –F at point B, which is on the line of action of F,Fig. 4–34b, we observe that –F at B and F at A will cancel each other,leaving only F at B, Fig. 4–34c. Force F has now been moved from A to Bwithout modifying its external effects on the stick; i.e., the reaction at thegrip remains the same. This demonstrates the principle of transmissibility,which states that a force acting on a body (stick) is a sliding vector sinceit can be applied at any point along its line of action.

We can also use the above procedure to move a force to a point that is noton the line of action of the force. If F is applied perpendicular to the stick, asin Fig. 4–35a, then we can attach a pair of equal but opposite forces F and –Fto B, Fig. 4–35b. Force F is now applied at B, and the other two forces, F at Aand –F at B, form a couple that produces the couple moment ,Fig. 4–35c. Therefore, the force F can be moved from A to B provided acouple moment M is added to maintain an equivalent system. This couplemoment is determined by taking the moment of F about B. Since M isactually a free vector, it can act at any point on the stick. In both cases thesystems are equivalent which causes a downward force F and clockwisecouple moment M = Fd to be felt at the grip.

M = Fd

F F

F!FAB

(a)

AB

F

(b) (c)

Fig. 4–34

F F

!FA

d

(a)

FF

m " Fd

(b) (c)

Fig. 4–35

160 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

4.7 Simplification of a Force and CoupleSystem

Sometimes it is convenient to reduce a system of forces and couple momentsacting on a body to a simpler form by replacing it with an equivalent system,consisting of a single resultant force acting at a specific point and a resultantcouple moment.A system is equivalent if the external effects it produces ona body are the same as those caused by the original force and couplemoment system. In this context, the external effects of a system refer to thetranslating and rotating motion of the body if the body is free to move, or itrefers to the reactive forces at the supports if the body is held fixed.

For example, consider holding the stick in Fig. 4–34a, which issubjected to the force F at point A. If we attach a pair of equal butopposite forces F and –F at point B, which is on the line of action of F,Fig. 4–34b, we observe that –F at B and F at A will cancel each other,leaving only F at B, Fig. 4–34c. Force F has now been moved from A to Bwithout modifying its external effects on the stick; i.e., the reaction at thegrip remains the same. This demonstrates the principle of transmissibility,which states that a force acting on a body (stick) is a sliding vector sinceit can be applied at any point along its line of action.

We can also use the above procedure to move a force to a point that is noton the line of action of the force. If F is applied perpendicular to the stick, asin Fig. 4–35a, then we can attach a pair of equal but opposite forces F and –Fto B, Fig. 4–35b. Force F is now applied at B, and the other two forces, F at Aand –F at B, form a couple that produces the couple moment ,Fig. 4–35c. Therefore, the force F can be moved from A to B provided acouple moment M is added to maintain an equivalent system. This couplemoment is determined by taking the moment of F about B. Since M isactually a free vector, it can act at any point on the stick. In both cases thesystems are equivalent which causes a downward force F and clockwisecouple moment M = Fd to be felt at the grip.

M = Fd

F F

F!FAB

(a)

AB

F

(b) (c)

Fig. 4–34

F F

!FA

d

(a)

FF

m " Fd

(b) (c)

Fig. 4–35

Quando o ponto de apoio não está na linha de ação da força

As forças F em A e -F em B formam um binário que produz o momento M=F.d.

Portanto, a força F pode ser movida de A para B desde que seja adicionado um momento M para manter o sistema equivalente. Este momento é calculado tomando-se a ação de F em relação ao ponto B.

Como M é um vetor livre ele pode atuar em qualquer ponto do bastão.

Sistema de forças e de momentos binários4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 161

4

System of Forces and Couple Moments. Using the abovemethod, a system of several forces and couple moments acting on abody can be reduced to an equivalent single resultant force acting at apoint O and a resultant couple moment. For example, in Fig. 4–36a, O isnot on the line of action of , and so this force can be moved to pointO provided a couple moment is added to the body.Similarly, the couple moment should be added to thebody when we move to point O. Finally, since the couple moment Mis a free vector, it can just be moved to point O. By doing this, we obtainthe equivalent system shown in Fig. 4–36b, which produces the sameexternal effects (support reactions) on the body as that of the force andcouple system shown in Fig. 4–36a. If we sum the forces and couplemoments, we obtain the resultant force and the resultantcouple moment Fig. 4–36c.

Notice that is independent of the location of point O; however,depends upon this location since the moments and aredetermined using the position vectors and Also note that isa free vector and can act at any point on the body, although point O isgenerally chosen as its point of application.

We can generalize the above method of reducing a force and couplesystem to an equivalent resultant force acting at point O and aresultant couple moment by using the following two equations.

(4–17)

The first equation states that the resultant force of the system isequivalent to the sum of all the forces; and the second equation statesthat the resultant couple moment of the system is equivalent to the sumof all the couple moments plus the moments of all the forces about point O. If the force system lies in the x–y plane and any couplemoments are perpendicular to this plane, then the above equationsreduce to the following three scalar equations.

©MO©M

FR = ©F

(MR)O = ©MO + ©M

(MR)O

FR

(MR)Or2.r1

M2M1

(MR)OFR

(MR)O = M + M1 + M2,FR = F1 + F2

F2

M2 = r2 * F2

M1 = r1 * FF1

O

F1

(a)

F2

r2

r1

M

(b)

O(c)

!

O

F1

F2

M

M2 ! r2 " F2

M1 ! r1 " F1

FR

MRO

!

u

Fig. 4–36

(4–18)

Here the resultant force is determined from the vector sum of its twocomponents and (FR)y.(FR)x

(FR)x = ©Fx

(FR)y = ©Fy

(MR)O = ©MO + ©M

4.7

SIM

PLIF

ICAT

ION

OF

AFO

RCE

AN

DC

OU

PLE

SYST

EM1

61

4

Syst

em o

f Fo

rces

and

Cou

ple

Mom

ents

.U

sing

the

abo

ve

met

hod,

a sy

stem

of

seve

ral

forc

es a

nd c

oupl

e m

omen

ts a

ctin

g on

a

body

can

be

redu

ced

to a

n eq

uiva

lent

sin

gle

resu

ltan

t fo

rce

acti

ng a

t a

poin

tOan

d a

resu

ltan

t cou

ple

mom

ent.

For

exam

ple,

in F

ig.4

–36a

,Ois

not

on t

he li

ne o

f ac

tion

of

,and

so

this

for

ce c

an b

e m

oved

to

poin

t

Opr

ovid

ed a

cou

ple

mom

ent

is a

dded

to

the

body

.

Sim

ilarl

y,th

e co

uple

mom

ent

shou

ld b

e ad

ded

to t

he

body

whe

n w

e m

ove

to p

oint

O.F

inal

ly,s

ince

the

coup

le m

omen

t M

is a

free

vec

tor,

it c

an ju

st b

e m

oved

to p

oint

O.B

y do

ing

this

,we

obta

in

the

equi

vale

nt s

yste

m s

how

n in

Fig

.4–3

6b,w

hich

pro

duce

s th

e sa

me

exte

rnal

eff

ects

(su

ppor

t rea

ctio

ns)

on th

e bo

dy a

s th

at o

f the

forc

e an

d

coup

le s

yste

m s

how

n in

Fig

.4–3

6a.I

f w

e su

m t

he f

orce

s an

d co

uple

mom

ents

,we

obta

in th

e re

sult

ant f

orce

an

d th

e re

sult

ant

coup

le m

omen

t Fi

g.4–

36c.

Not

ice

that

is

inde

pend

ent o

f the

loca

tion

of p

oint

O;h

owev

er,

depe

nds

upon

th

is

loca

tion

sinc

e th

e m

omen

ts

and

are

dete

rmin

ed u

sing

the

posi

tion

vect

ors

and

Als

o no

te th

atis

a fr

ee v

ecto

r an

d ca

n ac

t at

any

poi

nton

the

bod

y,al

thou

gh p

oint

Ois

gene

rally

cho

sen

as it

s po

int o

f app

licat

ion.

We

can

gene

raliz

e th

e ab

ove

met

hod

of r

educ

ing

a fo

rce

and

coup

le

syst

em t

o an

equ

ival

ent

resu

ltan

t fo

rce

acti

ng a

t po

int

Oan

d a

resu

ltan

t cou

ple

mom

ent

by u

sing

the

follo

win

g tw

o eq

uati

ons.

(4–1

7)

The

fir

st e

quat

ion

stat

es t

hat

the

resu

ltant

for

ce o

f th

e sy

stem

is

equi

vale

nt t

o th

e su

m o

f al

l th

e fo

rces

;and

the

sec

ond

equa

tion

stat

es

that

the

resu

ltant

cou

ple

mom

ent o

f the

sys

tem

is e

quiv

alen

t to

the

sum

of a

ll th

e co

uple

mom

ents

pl

us th

e m

omen

ts o

f all

the

forc

es

abou

t po

int

O.I

f th

e fo

rce

syst

em li

es in

the

x–y

plan

e an

d an

y co

uple

mom

ents

are

per

pend

icul

ar t

o th

is p

lane

,th

en t

he a

bove

equ

atio

ns

redu

ce to

the

follo

win

g th

ree

scal

ar e

quat

ions

.

©M

O©

MF R=

©F

(MR

) O=

©M

O+

©M

(MR

) OF R

(MR

) Or 2

.r 1

M2

M1

(MR

) OF R

(MR

) O=

M+

M1+

M2,F R

=F 1

+F 2

F 2M

2=

r 2*

F 2M

1=

r 1*

FF 1

O

F 1

(a)

F 2

r 2

r 1

M

(b)

O(c

)

!

O

F 1

F 2

M

M2

!r 2

"F 2

M1

!r 1

"F 1

F R

MR

O

!

u

Fig.

4–36

(4–1

8)

Her

e th

e re

sulta

nt f

orce

is

dete

rmin

ed f

rom

the

vec

tor

sum

of

its t

wo

com

pone

nts

and

(FR

) y.

(FR

) x

(FR

) x=

©F x

(FR

) y=

©F y

(MR

) O=

©M

O+

©M

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 161

4

System of Forces and Couple Moments. Using the abovemethod, a system of several forces and couple moments acting on abody can be reduced to an equivalent single resultant force acting at apoint O and a resultant couple moment. For example, in Fig. 4–36a, O isnot on the line of action of , and so this force can be moved to pointO provided a couple moment is added to the body.Similarly, the couple moment should be added to thebody when we move to point O. Finally, since the couple moment Mis a free vector, it can just be moved to point O. By doing this, we obtainthe equivalent system shown in Fig. 4–36b, which produces the sameexternal effects (support reactions) on the body as that of the force andcouple system shown in Fig. 4–36a. If we sum the forces and couplemoments, we obtain the resultant force and the resultantcouple moment Fig. 4–36c.

Notice that is independent of the location of point O; however,depends upon this location since the moments and aredetermined using the position vectors and Also note that isa free vector and can act at any point on the body, although point O isgenerally chosen as its point of application.

We can generalize the above method of reducing a force and couplesystem to an equivalent resultant force acting at point O and aresultant couple moment by using the following two equations.

(4–17)

The first equation states that the resultant force of the system isequivalent to the sum of all the forces; and the second equation statesthat the resultant couple moment of the system is equivalent to the sumof all the couple moments plus the moments of all the forces about point O. If the force system lies in the x–y plane and any couplemoments are perpendicular to this plane, then the above equationsreduce to the following three scalar equations.

©MO©M

FR = ©F

(MR)O = ©MO + ©M

(MR)O

FR

(MR)Or2.r1

M2M1

(MR)OFR

(MR)O = M + M1 + M2,FR = F1 + F2

F2

M2 = r2 * F2

M1 = r1 * FF1

O

F1

(a)

F2

r2

r1

M

(b)

O(c)

!

O

F1

F2

M

M2 ! r2 " F2

M1 ! r1 " F1

FR

MRO

!

u

Fig. 4–36

(4–18)

Here the resultant force is determined from the vector sum of its twocomponents and (FR)y.(FR)x

(FR)x = ©Fx

(FR)y = ©Fy

(MR)O = ©MO + ©M

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 161

4

System of Forces and Couple Moments. Using the abovemethod, a system of several forces and couple moments acting on abody can be reduced to an equivalent single resultant force acting at apoint O and a resultant couple moment. For example, in Fig. 4–36a, O isnot on the line of action of , and so this force can be moved to pointO provided a couple moment is added to the body.Similarly, the couple moment should be added to thebody when we move to point O. Finally, since the couple moment Mis a free vector, it can just be moved to point O. By doing this, we obtainthe equivalent system shown in Fig. 4–36b, which produces the sameexternal effects (support reactions) on the body as that of the force andcouple system shown in Fig. 4–36a. If we sum the forces and couplemoments, we obtain the resultant force and the resultantcouple moment Fig. 4–36c.

Notice that is independent of the location of point O; however,depends upon this location since the moments and aredetermined using the position vectors and Also note that isa free vector and can act at any point on the body, although point O isgenerally chosen as its point of application.

We can generalize the above method of reducing a force and couplesystem to an equivalent resultant force acting at point O and aresultant couple moment by using the following two equations.

(4–17)

The first equation states that the resultant force of the system isequivalent to the sum of all the forces; and the second equation statesthat the resultant couple moment of the system is equivalent to the sumof all the couple moments plus the moments of all the forces about point O. If the force system lies in the x–y plane and any couplemoments are perpendicular to this plane, then the above equationsreduce to the following three scalar equations.

©MO©M

FR = ©F

(MR)O = ©MO + ©M

(MR)O

FR

(MR)Or2.r1

M2M1

(MR)OFR

(MR)O = M + M1 + M2,FR = F1 + F2

F2

M2 = r2 * F2

M1 = r1 * FF1

O

F1

(a)

F2

r2

r1

M

(b)

O(c)

!

O

F1

F2

M

M2 ! r2 " F2

M1 ! r1 " F1

FR

MRO

!

u

Fig. 4–36

(4–18)

Here the resultant force is determined from the vector sum of its twocomponents and (FR)y.(FR)x

(FR)x = ©Fx

(FR)y = ©Fy

(MR)O = ©MO + ©M

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 161

4

System of Forces and Couple Moments. Using the abovemethod, a system of several forces and couple moments acting on abody can be reduced to an equivalent single resultant force acting at apoint O and a resultant couple moment. For example, in Fig. 4–36a, O isnot on the line of action of , and so this force can be moved to pointO provided a couple moment is added to the body.Similarly, the couple moment should be added to thebody when we move to point O. Finally, since the couple moment Mis a free vector, it can just be moved to point O. By doing this, we obtainthe equivalent system shown in Fig. 4–36b, which produces the sameexternal effects (support reactions) on the body as that of the force andcouple system shown in Fig. 4–36a. If we sum the forces and couplemoments, we obtain the resultant force and the resultantcouple moment Fig. 4–36c.

Notice that is independent of the location of point O; however,depends upon this location since the moments and aredetermined using the position vectors and Also note that isa free vector and can act at any point on the body, although point O isgenerally chosen as its point of application.

We can generalize the above method of reducing a force and couplesystem to an equivalent resultant force acting at point O and aresultant couple moment by using the following two equations.

(4–17)

The first equation states that the resultant force of the system isequivalent to the sum of all the forces; and the second equation statesthat the resultant couple moment of the system is equivalent to the sumof all the couple moments plus the moments of all the forces about point O. If the force system lies in the x–y plane and any couplemoments are perpendicular to this plane, then the above equationsreduce to the following three scalar equations.

©MO©M

FR = ©F

(MR)O = ©MO + ©M

(MR)O

FR

(MR)Or2.r1

M2M1

(MR)OFR

(MR)O = M + M1 + M2,FR = F1 + F2

F2

M2 = r2 * F2

M1 = r1 * FF1

O

F1

(a)

F2

r2

r1

M

(b)

O(c)

!

O

F1

F2

M

M2 ! r2 " F2

M1 ! r1 " F1

FR

MRO

!

u

Fig. 4–36

(4–18)

Here the resultant force is determined from the vector sum of its twocomponents and (FR)y.(FR)x

(FR)x = ©Fx

(FR)y = ©Fy

(MR)O = ©MO + ©M

Se as forças estiverem no plano xy e os momentos perpendiculares a este plano as equações anteriores reduzem-se à

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 161

4

System of Forces and Couple Moments. Using the abovemethod, a system of several forces and couple moments acting on abody can be reduced to an equivalent single resultant force acting at apoint O and a resultant couple moment. For example, in Fig. 4–36a, O isnot on the line of action of , and so this force can be moved to pointO provided a couple moment is added to the body.Similarly, the couple moment should be added to thebody when we move to point O. Finally, since the couple moment Mis a free vector, it can just be moved to point O. By doing this, we obtainthe equivalent system shown in Fig. 4–36b, which produces the sameexternal effects (support reactions) on the body as that of the force andcouple system shown in Fig. 4–36a. If we sum the forces and couplemoments, we obtain the resultant force and the resultantcouple moment Fig. 4–36c.

Notice that is independent of the location of point O; however,depends upon this location since the moments and aredetermined using the position vectors and Also note that isa free vector and can act at any point on the body, although point O isgenerally chosen as its point of application.

We can generalize the above method of reducing a force and couplesystem to an equivalent resultant force acting at point O and aresultant couple moment by using the following two equations.

(4–17)

The first equation states that the resultant force of the system isequivalent to the sum of all the forces; and the second equation statesthat the resultant couple moment of the system is equivalent to the sumof all the couple moments plus the moments of all the forces about point O. If the force system lies in the x–y plane and any couplemoments are perpendicular to this plane, then the above equationsreduce to the following three scalar equations.

©MO©M

FR = ©F

(MR)O = ©MO + ©M

(MR)O

FR

(MR)Or2.r1

M2M1

(MR)OFR

(MR)O = M + M1 + M2,FR = F1 + F2

F2

M2 = r2 * F2

M1 = r1 * FF1

O

F1

(a)

F2

r2

r1

M

(b)

O(c)

!

O

F1

F2

M

M2 ! r2 " F2

M1 ! r1 " F1

FR

MRO

!

u

Fig. 4–36

(4–18)

Here the resultant force is determined from the vector sum of its twocomponents and (FR)y.(FR)x

(FR)x = ©Fx

(FR)y = ©Fy

(MR)O = ©MO + ©M

Exemplo 1 (*): Substitua as forças e os sistemas binários pelos sistemas resultantes equivalentes.EXAMPLE 4.14

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 163

4

0.2 m 0.3 m

4 kN 5 kN

3 kN

O

(a)

543

30!

0.1 m

0.1 m

Fig. 4–37

(c)

(FR)y " 6.50 kN

(MR)O " 2.46 kN #m

(FR)x " 5.598 kN

FR

uO

Replace the force and couple system shown in Fig. 4–37a by anequivalent resultant force and couple moment acting at point O.

Using the Pythagorean theorem, Fig. 4–37c, the magnitude of isFR

Its direction is

Ans.

Moment Summation. The moments of 3 kN and 5 kN aboutpoint O will be determined using their x and y components. Referringto Fig. 4–37b, we have

u = tan-1a (FR)y

(FR)xb = tan-1a 6.50 kN

5.598 kNb = 49.3°

u

This clockwise moment is shown in Fig. 4–37c.

NOTE: Realize that the resultant force and couple moment in Fig. 4–37c will produce the same external effects or reactions at thesupports as those produced by the force system, Fig 4–37a.

a

b Ans.= -2.46 kN # m = 2.46 kN # m- A45 B (5 kN) (0.5 m) - (4 kN)(0.2 m)

(MR)O = (3 kN)sin 30°(0.2 m) - (3 kN)cos 30°(0.1 m) + A35 B (5 kN) (0.1 m) + (MR)O = ©MO;

SOLUTIONForce Summation. The 3 kN and 5 kN forces are resolved into theirx and y components as shown in Fig. 4–37b. We have

= -6.50 kN = 6.50 kNT(FR)y = (3 kN)sin 30° - A45 B (5 kN) - 4 kN+ c(FR)y = ©Fy;

= 5.598 kN:(FR)x = (3 kN)cos 30° + A35 B (5 kN):+ (FR)x = ©Fx;

Ans.FR = 21FR2x2 + 1FR2y2 = 215.598 kN22 + 16.50 kN22 = 8.58 kN

(3 kN)cos 30!

(3 kN)sin 30!

y

x

0.2 m 0.3 m

4 kN

(5 kN)

O

(b)

45

35

(5 kN)

0.1 m

0.1 m

Exemplo 2: A estrutura abaixo está submetida a um momento de binário M e às forças F1 e F2. Substitua-os por um sistema equivalente tendo o ponto O como referência.

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 165

4

The structural member is subjected to a couple moment M andforces and in Fig. 4–39a. Replace this system by an equivalentresultant force and couple moment acting at its base, point O.

SOLUTION (VECTOR ANALYSIS)The three-dimensional aspects of the problem can be simplified byusing a Cartesian vector analysis. Expressing the forces and couplemoment as Cartesian vectors, we have

Force Summation.

Ans.

Moment Summation.

Ans.

The results are shown in Fig. 4–39b.

= 5-166i - 650j + 300k6 N # m= 1-400j + 300k2 + 102 + 1-166.4i - 249.6j2

MRO= 1- 400j + 300k2 + 11k2 * 1- 800k2+ 3 i j k

-0.15 0.1 1-249.6 166.4 0

3MRO= M + rC * F1 + rB * F2

MRO= ©M + ©MO

= 5-250i + 166j - 800k6 NFR = F1 + F2 = -800k - 249.6i + 166.4jFR = ©F;

M = -500 A45 B j + 500 A35 Bk = 5-400j + 300k6 N # m

= 300 N c {-0.15i + 0.1j} m21-0.15 m22 + 10.1 m22 d = 5-249.6i + 166.4j6 N= 1300 N2a rCB

rCBb

F2 = 1300 N2uCB

F1 = 5-800k6 N

F2F1

EXAMPLE 4.16

F1 ! 800 N0.1 m

F2 ! 300 N

0.15 m

rB

1 m

y

C

53

4

M ! 500 N " m

O

x

(a)

z

rC

B

Fig. 4–39

yx

z

MRO

FR

(b)

O

4.7 SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 165

4

The structural member is subjected to a couple moment M andforces and in Fig. 4–39a. Replace this system by an equivalentresultant force and couple moment acting at its base, point O.

SOLUTION (VECTOR ANALYSIS)The three-dimensional aspects of the problem can be simplified byusing a Cartesian vector analysis. Expressing the forces and couplemoment as Cartesian vectors, we have

Force Summation.

Ans.

Moment Summation.

Ans.

The results are shown in Fig. 4–39b.

= 5-166i - 650j + 300k6 N # m= 1-400j + 300k2 + 102 + 1-166.4i - 249.6j2

MRO= 1- 400j + 300k2 + 11k2 * 1- 800k2+ 3 i j k

-0.15 0.1 1-249.6 166.4 0

3MRO= M + rC * F1 + rB * F2

MRO= ©M + ©MO

= 5-250i + 166j - 800k6 NFR = F1 + F2 = -800k - 249.6i + 166.4jFR = ©F;

M = -500 A45 B j + 500 A35 Bk = 5-400j + 300k6 N # m

= 300 N c {-0.15i + 0.1j} m21-0.15 m22 + 10.1 m22 d = 5-249.6i + 166.4j6 N= 1300 N2a rCB

rCBb

F2 = 1300 N2uCB

F1 = 5-800k6 N

F2F1

EXAMPLE 4.16

F1 ! 800 N0.1 m

F2 ! 300 N

0.15 m

rB

1 m

y

C

53

4

M ! 500 N " m

O

x

(a)

z

rC

B

Fig. 4–39

yx

z

MRO

FR

(b)

O

Até aqui somente...

• Um corpo pode estar sujeito a um carregamento que está distribuído sobre uma superfície

• Exemplos: pressão do vento sobre um outdoor, pressão da água sobre um tanque etc.

• Pressão indica a intendisidade da carga.

• No SI. N/m2=Pa

Redução de um carregamento distribuído simples

Carregamento uniforme junto a um eixo.4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 183

4

4.9 Reduction of a Simple DistributedLoading

Sometimes, a body may be subjected to a loading that is distributed overits surface. For example, the pressure of the wind on the face of a sign, thepressure of water within a tank, or the weight of sand on the floor of astorage container, are all distributed loadings.The pressure exerted at eachpoint on the surface indicates the intensity of the loading. It is measuredusing pascals Pa (or ) in SI units or in the U.S. Customarysystem.

Uniform Loading Along a Single Axis. The most commontype of distributed loading encountered in engineering practice isgenerally uniform along a single axis.* For example, consider the beam(or plate) in Fig. 4–48a that has a constant width and is subjected to apressure loading that varies only along the x axis. This loading can bedescribed by the function . It contains only one variablex, and for this reason, we can also represent it as a coplanar distributedload. To do so, we multiply the loading function by the width b m ofthe beam, so that , Fig. 4-48b. Using the methods ofSec. 4.8, we can replace this coplanar parallel force system with asingle equivalent resultant force acting at a specific location on thebeam, Fig. 4–48c.

Magnitude of Resultant Force. From Eq. 4–17the magnitude of is equivalent to the sum of all the forces in thesystem. In this case integration must be used since there is an infinitenumber of parallel forces dF acting on the beam, Fig. 4–48b. Since dF isacting on an element of length dx, and w(x) is a force per unit length,then In other words, the magnitude of dF isdetermined from the colored differential area dA under the loadingcurve. For the entire length L,

(4–19)

Therefore, the magnitude of the resultant force is equal to the total area Aunder the loading diagram, Fig. 4–48c.

FR = LLw1x2 dx = LA

dA = A+ TFR = ©F;

dF = w1x2 dx = dA.

FR

1FR = ©F2,FR

w(x) = p(x)b N/m

p = p(x) N/m2

lb/ft2N/m2

*The more general case of a nonuniform surface loading acting on a body is consideredin Sec. 9.5.

p

L

p ! p(x)

x

(a)

C

x

FR

b

Fig. 4–48

x

w

O

Lx

dx

dF ! dAw ! w(x)

(b)

x

w

O

C A

Lx

FR

(c)

dF = p(x)bdx

= w(x)dx

F =

Z

Lw(x)dx =

Z

Ada = Area

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 183

4

4.9 Reduction of a Simple DistributedLoading

Sometimes, a body may be subjected to a loading that is distributed overits surface. For example, the pressure of the wind on the face of a sign, thepressure of water within a tank, or the weight of sand on the floor of astorage container, are all distributed loadings.The pressure exerted at eachpoint on the surface indicates the intensity of the loading. It is measuredusing pascals Pa (or ) in SI units or in the U.S. Customarysystem.

Uniform Loading Along a Single Axis. The most commontype of distributed loading encountered in engineering practice isgenerally uniform along a single axis.* For example, consider the beam(or plate) in Fig. 4–48a that has a constant width and is subjected to apressure loading that varies only along the x axis. This loading can bedescribed by the function . It contains only one variablex, and for this reason, we can also represent it as a coplanar distributedload. To do so, we multiply the loading function by the width b m ofthe beam, so that , Fig. 4-48b. Using the methods ofSec. 4.8, we can replace this coplanar parallel force system with asingle equivalent resultant force acting at a specific location on thebeam, Fig. 4–48c.

Magnitude of Resultant Force. From Eq. 4–17the magnitude of is equivalent to the sum of all the forces in thesystem. In this case integration must be used since there is an infinitenumber of parallel forces dF acting on the beam, Fig. 4–48b. Since dF isacting on an element of length dx, and w(x) is a force per unit length,then In other words, the magnitude of dF isdetermined from the colored differential area dA under the loadingcurve. For the entire length L,

(4–19)

Therefore, the magnitude of the resultant force is equal to the total area Aunder the loading diagram, Fig. 4–48c.

FR = LLw1x2 dx = LA

dA = A+ TFR = ©F;

dF = w1x2 dx = dA.

FR

1FR = ©F2,FR

w(x) = p(x)b N/m

p = p(x) N/m2

lb/ft2N/m2

*The more general case of a nonuniform surface loading acting on a body is consideredin Sec. 9.5.

p

L

p ! p(x)

x

(a)

C

x

FR

b

Fig. 4–48

x

w

O

Lx

dx

dF ! dAw ! w(x)

(b)

x

w

O

C A

Lx

FR

(c)

A intensidade da força resultante é igual à area total A sob o diagrama de carregamento.

Posição da força resultante

• Como dF produz um momento de força dM=xdF=xw(x)dx em relação ao ponto O, para o comprimento inteiro podemos escrever.

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 183

4

4.9 Reduction of a Simple DistributedLoading

Sometimes, a body may be subjected to a loading that is distributed overits surface. For example, the pressure of the wind on the face of a sign, thepressure of water within a tank, or the weight of sand on the floor of astorage container, are all distributed loadings.The pressure exerted at eachpoint on the surface indicates the intensity of the loading. It is measuredusing pascals Pa (or ) in SI units or in the U.S. Customarysystem.

Uniform Loading Along a Single Axis. The most commontype of distributed loading encountered in engineering practice isgenerally uniform along a single axis.* For example, consider the beam(or plate) in Fig. 4–48a that has a constant width and is subjected to apressure loading that varies only along the x axis. This loading can bedescribed by the function . It contains only one variablex, and for this reason, we can also represent it as a coplanar distributedload. To do so, we multiply the loading function by the width b m ofthe beam, so that , Fig. 4-48b. Using the methods ofSec. 4.8, we can replace this coplanar parallel force system with asingle equivalent resultant force acting at a specific location on thebeam, Fig. 4–48c.

Magnitude of Resultant Force. From Eq. 4–17the magnitude of is equivalent to the sum of all the forces in thesystem. In this case integration must be used since there is an infinitenumber of parallel forces dF acting on the beam, Fig. 4–48b. Since dF isacting on an element of length dx, and w(x) is a force per unit length,then In other words, the magnitude of dF isdetermined from the colored differential area dA under the loadingcurve. For the entire length L,

(4–19)

Therefore, the magnitude of the resultant force is equal to the total area Aunder the loading diagram, Fig. 4–48c.

FR = LLw1x2 dx = LA

dA = A+ TFR = ©F;

dF = w1x2 dx = dA.

FR

1FR = ©F2,FR

w(x) = p(x)b N/m

p = p(x) N/m2

lb/ft2N/m2

*The more general case of a nonuniform surface loading acting on a body is consideredin Sec. 9.5.

p

L

p ! p(x)

x

(a)

C

x

FR

b

Fig. 4–48

x

w

O

Lx

dx

dF ! dAw ! w(x)

(b)

x

w

O

C A

Lx

FR

(c)

�x̄F = �Z

Lxw(x)dx

x̄ =

RL xw(x)dxRL w(x)dx

x̄ =

RA xdaRA da

x̄ =

RL xw(x)dxRL w(x)dx

Esta coordenada localiza o centro geométrico ou centróide da área sob o carregamento distribuído.

Técnicas de integração para determinar a posição do centróide encontram-se no Capítulo 9.

Uma vez que a posição é determinada, a força resultante passa necessariamente por este ponto na superfície da viga. A força resultante neste caso possui uma intensidade igual ao volume sob a curva de carregamento p=p(x) e uma linha de ação que passa pelo centro geométrico (centróide) deste volume.

Exemplo 1: Determine a intensidade da força resultante equivalente que age sobre o eixo da figura abaixo.

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 185

4

EXAMPLE 4.21

Determine the magnitude and location of the equivalent resultantforce acting on the shaft in Fig. 4–49a.

w ! (60 x2)N/m

(a)

dA ! w dx

2 mx dx

Ox

240 N/m

w

Fig. 4–49

(b)

Ox

w

C

x ! 1.5 m

FR ! 160 N

SOLUTIONSince is given, this problem will be solved by integration.

The differential element has an area ApplyingEq. 4–19,

Ans.

The location of measured from O, Fig. 4–49b, is determined fromEq. 4–20.

Ans.

NOTE: These results can be checked by using the table on the insideback cover, where it is shown that for an exparabolic area of length a,height b, and shape shown in Fig. 4–49a, we have

= 3412 m2 = 1.5 mA = ab

3=

2 m1240 N>m23

= 160 N and x = 34

a

= 1.5 m

x = LAx dA

LAdA

= L2 m

0x160x22 dx

160 N=

60¢x4

4≤ `

0

2 m

160 N=

60¢24

4- 04

4≤

160 N

FRx

= 160 N

FR = LAdA = L

2 m

060x2 dx = 60¢x3

3≤ `

0

2 m

= 60¢23

3- 03

3≤+ T FR = ©F;

dA = w dx = 60x2 dx.

w = w1x2

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 185

4

EXAMPLE 4.21

Determine the magnitude and location of the equivalent resultantforce acting on the shaft in Fig. 4–49a.

w ! (60 x2)N/m

(a)

dA ! w dx

2 mx dx

Ox

240 N/m

w

Fig. 4–49

(b)

Ox

w

C

x ! 1.5 m

FR ! 160 N

SOLUTIONSince is given, this problem will be solved by integration.

The differential element has an area ApplyingEq. 4–19,

Ans.

The location of measured from O, Fig. 4–49b, is determined fromEq. 4–20.

Ans.

NOTE: These results can be checked by using the table on the insideback cover, where it is shown that for an exparabolic area of length a,height b, and shape shown in Fig. 4–49a, we have

= 3412 m2 = 1.5 mA = ab

3=

2 m1240 N>m23

= 160 N and x = 34

a

= 1.5 m

x = LAx dA

LAdA

= L2 m

0x160x22 dx

160 N=

60¢x4

4≤ `

0

2 m

160 N=

60¢24

4- 04

4≤

160 N

FRx

= 160 N

FR = LAdA = L

2 m

060x2 dx = 60¢x3

3≤ `

0

2 m

= 60¢23

3- 03

3≤+ T FR = ©F;

dA = w dx = 60x2 dx.

w = w1x2

Exemplo 2: Um carregamento distribuído de p=800x Pa atua sobre a superfície superior da viga mostrada abaixo. Determine a intensidade e a posição da força resultante equivalente.

186 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

A distributed loading of Pa acts over the top surface ofthe beam shown in Fig. 4–50a. Determine the magnitude and locationof the equivalent resultant force.

p = (800x)

EXAMPLE 4.22

(a)

p

7200 Pa

x

9 m

0.2 m

y

p = 800x Pa

x

Fig. 4–50

w ! 160x N/m

(b)

9 m

x

w 1440 N/m

x

C

FR ! 6.48 kN

3 mx ! 6 m

(c)

SOLUTIONSince the loading intensity is uniform along the width of the beam(the y axis), the loading can be viewed in two dimensions as shown inFig. 4–50b. Here

At note that Although we may again applyEqs. 4–19 and 4–20 as in the previous example, it is simpler to use thetable on the inside back cover.

The magnitude of the resultant force is equivalent to the area of thetriangle.

Ans.

The line of action of passes through the centroid C of this triangle.Hence,

Ans.

The results are shown in Fig. 4–50c.

NOTE: We may also view the resultant as acting through thecentroid of the volume of the loading diagram in Fig. 4–50a.Hence intersects the x–y plane at the point (6 m, 0). Furthermore,the magnitude of is equal to the volume under the loadingdiagram; i.e.,

Ans.FR = V = 1217200 N>m2219 m210.2 m2 = 6.48 kN

FR

FR

p = p1x2FR

x = 9 m - 1319 m2 = 6 m

FR

FR = 1219 m211440 N>m2 = 6480 N = 6.48 kN

w = 1440 N>m.x = 9 m,

= 1160x2 N>mw = 1800x N>m2210.2 m2

186 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

A distributed loading of Pa acts over the top surface ofthe beam shown in Fig. 4–50a. Determine the magnitude and locationof the equivalent resultant force.

p = (800x)

EXAMPLE 4.22

(a)

p

7200 Pa

x

9 m

0.2 m

y

p = 800x Pa

x

Fig. 4–50

w ! 160x N/m

(b)

9 m

x

w 1440 N/m

x

C

FR ! 6.48 kN

3 mx ! 6 m

(c)

SOLUTIONSince the loading intensity is uniform along the width of the beam(the y axis), the loading can be viewed in two dimensions as shown inFig. 4–50b. Here

At note that Although we may again applyEqs. 4–19 and 4–20 as in the previous example, it is simpler to use thetable on the inside back cover.

The magnitude of the resultant force is equivalent to the area of thetriangle.

Ans.

The line of action of passes through the centroid C of this triangle.Hence,

Ans.

The results are shown in Fig. 4–50c.

NOTE: We may also view the resultant as acting through thecentroid of the volume of the loading diagram in Fig. 4–50a.Hence intersects the x–y plane at the point (6 m, 0). Furthermore,the magnitude of is equal to the volume under the loadingdiagram; i.e.,

Ans.FR = V = 1217200 N>m2219 m210.2 m2 = 6.48 kN

FR

FR

p = p1x2FR

x = 9 m - 1319 m2 = 6 m

FR

FR = 1219 m211440 N>m2 = 6480 N = 6.48 kN

w = 1440 N>m.x = 9 m,

= 1160x2 N>mw = 1800x N>m2210.2 m2

186 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

A distributed loading of Pa acts over the top surface ofthe beam shown in Fig. 4–50a. Determine the magnitude and locationof the equivalent resultant force.

p = (800x)

EXAMPLE 4.22

(a)

p

7200 Pa

x

9 m

0.2 m

y

p = 800x Pa

x

Fig. 4–50

w ! 160x N/m

(b)

9 m

x

w 1440 N/m

x

C

FR ! 6.48 kN

3 mx ! 6 m

(c)

SOLUTIONSince the loading intensity is uniform along the width of the beam(the y axis), the loading can be viewed in two dimensions as shown inFig. 4–50b. Here

At note that Although we may again applyEqs. 4–19 and 4–20 as in the previous example, it is simpler to use thetable on the inside back cover.

The magnitude of the resultant force is equivalent to the area of thetriangle.

Ans.

The line of action of passes through the centroid C of this triangle.Hence,

Ans.

The results are shown in Fig. 4–50c.

NOTE: We may also view the resultant as acting through thecentroid of the volume of the loading diagram in Fig. 4–50a.Hence intersects the x–y plane at the point (6 m, 0). Furthermore,the magnitude of is equal to the volume under the loadingdiagram; i.e.,

Ans.FR = V = 1217200 N>m2219 m210.2 m2 = 6.48 kN

FR

FR

p = p1x2FR

x = 9 m - 1319 m2 = 6 m

FR

FR = 1219 m211440 N>m2 = 6480 N = 6.48 kN

w = 1440 N>m.x = 9 m,

= 1160x2 N>mw = 1800x N>m2210.2 m2

Exemplo 3: O material granular exerce um carregamento distribuído sobre a viga como mostra a figura abaixo. Determine a intensidade e a posição da resultante equivalente dessa carga.

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 187

4

EXAMPLE 4.23

The granular material exerts the distributed loading on the beam asshown in Fig. 4–51a. Determine the magnitude and location of theequivalent resultant of this load.

SOLUTIONThe area of the loading diagram is a trapezoid, and therefore thesolution can be obtained directly from the area and centroid formulasfor a trapezoid listed on the inside back cover. Since these formulasare not easily remembered, instead we will solve this problem byusing “composite” areas. Here we will divide the trapezoidal loadinginto a rectangular and triangular loading as shown in Fig. 4–51b. Themagnitude of the force represented by each of these loadings is equalto its associated area,

The lines of action of these parallel forces act through the centroid oftheir associated areas and therefore intersect the beam at

The two parallel forces and can be reduced to a single resultantThe magnitude of is

Ans.

We can find the location of with reference to point A, Fig. 4–51band 4–51c. We require

c

Ans.

NOTE: The trapezoidal area in Fig. 4–51a can also be divided intotwo triangular areas as shown in Fig. 4–51d. In this case

and

NOTE: Using these results, show that again and x = 4 ft.FR = 675 lb

x4 = 9 ft - 1319 ft2 = 6 ft

x3 = 1319 ft2 = 3 ft

F4 = 1219 ft2150 lb>ft2 = 225 lb

F3 = 1219 ft21100 lb>ft2 = 450 lb

x = 4 ft

x16752 = 312252 + 4.514502+ MRA= ©MA;

FR

FR = 225 + 450 = 675 lb+ TFR = ©F;

FRFR.F2F1

x2 = 1219 ft2 = 4.5 ft

x1 = 1319 ft2 = 3 ft

F2 = 19 ft2150 lb>ft2 = 450 lb

F1 = 1219 ft2150 lb>ft2 = 225 lb

9 ft

BA

(b)

50 lb/ft

50 lb/ft

F1 F2

x1x2

BA

(c)

FR

x

F3 F4

50 lb/ft

x3

9 ftx4

(d)

100 lb/ftA

100 lb/ft

50 lb/ft

9 ft

B

A

(a)

Fig. 4–51

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 187

4

EXAMPLE 4.23

The granular material exerts the distributed loading on the beam asshown in Fig. 4–51a. Determine the magnitude and location of theequivalent resultant of this load.

SOLUTIONThe area of the loading diagram is a trapezoid, and therefore thesolution can be obtained directly from the area and centroid formulasfor a trapezoid listed on the inside back cover. Since these formulasare not easily remembered, instead we will solve this problem byusing “composite” areas. Here we will divide the trapezoidal loadinginto a rectangular and triangular loading as shown in Fig. 4–51b. Themagnitude of the force represented by each of these loadings is equalto its associated area,

The lines of action of these parallel forces act through the centroid oftheir associated areas and therefore intersect the beam at

The two parallel forces and can be reduced to a single resultantThe magnitude of is

Ans.

We can find the location of with reference to point A, Fig. 4–51band 4–51c. We require

c

Ans.

NOTE: The trapezoidal area in Fig. 4–51a can also be divided intotwo triangular areas as shown in Fig. 4–51d. In this case

and

NOTE: Using these results, show that again and x = 4 ft.FR = 675 lb

x4 = 9 ft - 1319 ft2 = 6 ft

x3 = 1319 ft2 = 3 ft

F4 = 1219 ft2150 lb>ft2 = 225 lb

F3 = 1219 ft21100 lb>ft2 = 450 lb

x = 4 ft

x16752 = 312252 + 4.514502+ MRA= ©MA;

FR

FR = 225 + 450 = 675 lb+ TFR = ©F;

FRFR.F2F1

x2 = 1219 ft2 = 4.5 ft

x1 = 1319 ft2 = 3 ft

F2 = 19 ft2150 lb>ft2 = 450 lb

F1 = 1219 ft2150 lb>ft2 = 225 lb

9 ft

BA

(b)

50 lb/ft

50 lb/ft

F1 F2

x1x2

BA

(c)

FR

x

F3 F4

50 lb/ft

x3

9 ftx4

(d)

100 lb/ftA

100 lb/ft

50 lb/ft

9 ft

B

A

(a)

Fig. 4–51

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 187

4

EXAMPLE 4.23

The granular material exerts the distributed loading on the beam asshown in Fig. 4–51a. Determine the magnitude and location of theequivalent resultant of this load.

SOLUTIONThe area of the loading diagram is a trapezoid, and therefore thesolution can be obtained directly from the area and centroid formulasfor a trapezoid listed on the inside back cover. Since these formulasare not easily remembered, instead we will solve this problem byusing “composite” areas. Here we will divide the trapezoidal loadinginto a rectangular and triangular loading as shown in Fig. 4–51b. Themagnitude of the force represented by each of these loadings is equalto its associated area,

The lines of action of these parallel forces act through the centroid oftheir associated areas and therefore intersect the beam at

The two parallel forces and can be reduced to a single resultantThe magnitude of is

Ans.

We can find the location of with reference to point A, Fig. 4–51band 4–51c. We require

c

Ans.

NOTE: The trapezoidal area in Fig. 4–51a can also be divided intotwo triangular areas as shown in Fig. 4–51d. In this case

and

NOTE: Using these results, show that again and x = 4 ft.FR = 675 lb

x4 = 9 ft - 1319 ft2 = 6 ft

x3 = 1319 ft2 = 3 ft

F4 = 1219 ft2150 lb>ft2 = 225 lb

F3 = 1219 ft21100 lb>ft2 = 450 lb

x = 4 ft

x16752 = 312252 + 4.514502+ MRA= ©MA;

FR

FR = 225 + 450 = 675 lb+ TFR = ©F;

FRFR.F2F1

x2 = 1219 ft2 = 4.5 ft

x1 = 1319 ft2 = 3 ft

F2 = 19 ft2150 lb>ft2 = 450 lb

F1 = 1219 ft2150 lb>ft2 = 225 lb

9 ft

BA

(b)

50 lb/ft

50 lb/ft

F1 F2

x1x2

BA

(c)

FR

x

F3 F4

50 lb/ft

x3

9 ftx4

(d)

100 lb/ftA

100 lb/ft

50 lb/ft

9 ft

B

A

(a)

Fig. 4–51

4.9 REDUCTION OF A SIMPLE DISTRIBUTED LOADING 187

4

EXAMPLE 4.23

The granular material exerts the distributed loading on the beam asshown in Fig. 4–51a. Determine the magnitude and location of theequivalent resultant of this load.

SOLUTIONThe area of the loading diagram is a trapezoid, and therefore thesolution can be obtained directly from the area and centroid formulasfor a trapezoid listed on the inside back cover. Since these formulasare not easily remembered, instead we will solve this problem byusing “composite” areas. Here we will divide the trapezoidal loadinginto a rectangular and triangular loading as shown in Fig. 4–51b. Themagnitude of the force represented by each of these loadings is equalto its associated area,

The lines of action of these parallel forces act through the centroid oftheir associated areas and therefore intersect the beam at

The two parallel forces and can be reduced to a single resultantThe magnitude of is

Ans.

We can find the location of with reference to point A, Fig. 4–51band 4–51c. We require

c

Ans.

NOTE: The trapezoidal area in Fig. 4–51a can also be divided intotwo triangular areas as shown in Fig. 4–51d. In this case

and

NOTE: Using these results, show that again and x = 4 ft.FR = 675 lb

x4 = 9 ft - 1319 ft2 = 6 ft

x3 = 1319 ft2 = 3 ft

F4 = 1219 ft2150 lb>ft2 = 225 lb

F3 = 1219 ft21100 lb>ft2 = 450 lb

x = 4 ft

x16752 = 312252 + 4.514502+ MRA= ©MA;

FR

FR = 225 + 450 = 675 lb+ TFR = ©F;

FRFR.F2F1

x2 = 1219 ft2 = 4.5 ft

x1 = 1319 ft2 = 3 ft

F2 = 19 ft2150 lb>ft2 = 450 lb

F1 = 1219 ft2150 lb>ft2 = 225 lb

9 ft

BA

(b)

50 lb/ft

50 lb/ft

F1 F2

x1x2

BA

(c)

FR

x

F3 F4

50 lb/ft

x3

9 ftx4

(d)

100 lb/ftA

100 lb/ft

50 lb/ft

9 ft

B

A

(a)

Fig. 4–51

Simplificações Adicionais para um sistema de forças e binários.

Um sistema de forças pode ser simplificado a um sistema equivalente resultante desde que as linhas de ação de FR e MR0 sejam perpendiculares entre si. Por conta disto, somente sistemas de forças que são concorrentes, paralelas e coplanares podem ser simplificados.

170 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

4.8 Further Simplification of a Force andCouple System

In the preceding section, we developed a way to reduce a force and couplemoment system acting on a rigid body into an equivalent resultant force

acting at a specific point O and a resultant couple moment .Theforce system can be further reduced to an equivalent single resultant forceprovided the lines of action of and are perpendicular to eachother. Because of this condition, only concurrent, coplanar, and parallelforce systems can be further simplified.

Concurrent Force System. Since a concurrent force system isone in which the lines of action of all the forces intersect at a commonpoint O, Fig. 4–40a, then the force system produces no moment aboutthis point. As a result, the equivalent system can be represented by asingle resultant force acting at O, Fig. 4–40b.FR = ©F

(MR)OFR

(MR)OFR

Coplanar Force System. In the case of a coplanar force system,the lines of action of all the forces lie in the same plane, Fig. 4–41a, andso the resultant force of this system also lies in this plane.Furthermore, the moment of each of the forces about any point O isdirected perpendicular to this plane. Thus, the resultant moment

and resultant force will be mutually perpendicular,Fig. 4–41b. The resultant moment can be replaced by moving theresultant force a perpendicular or moment arm distance d awayfrom point O such that produces the same moment aboutpoint O, Fig. 4–41c. This distance d can be determined from the scalarequation .(MR)O = FRd = ©MO or d = (MR)O>FR

(MR)OFR

FR

FR(MR)O

FR = ©F

F2

FR

F2

F4 F3

O O

(a) (b)

!

Fig. 4–40

Sistema de forças concorrentes:

Como forças concorrentes são aquelas cujas linhas de ação interceptam-se em um ponto comum O, existe um torque em relação a este ponto.

Sistema de forças coplanares

Neste caso, as linhas de ação das forças estão todas no mesmo plano, assim como a força resultante. Além disso os torques produzidos por cada uma das forças ao redor de um ponto O, serão perpendiculares ao plano, assim como o torque resultante. Neste caso, as linhas de ação de FR e MR serão mutuamente perpendiculares. Podemos desclocar FR da distância d (d = M0R/FR) que é perpendicular a O.

4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 171

4

Parallel Force System. The parallel force system shown in Fig. 4–42aconsists of forces that are all parallel to the z axis. Thus, the resultantforce at point O must also be parallel to this axis, Fig. 4–42b.The moment produced by each force lies in the plane of the plate, and sothe resultant couple moment, , will also lie in this plane, along themoment axis a since and are mutually perpendicular. As aresult, the force system can be further reduced to an equivalent singleresultant force , acting through point P located on the perpendicular baxis, Fig. 4–42c. The distance d along this axis from point O requires

.(MR)O = FRd = ©MO or d = ©MO>FR

FR

(MR)OFR

(MR)O

FR = ©F

z

F1 F2

F3O

(a)

z

a O

bb

(b)

FR ! "F

FR ! "F

z

O

d

(c)

a

P

(MR)O

! !

Fig. 4–42

(a) (b) (c)

O

(MR)O

FR

OFR

O d

F3

F4 F1

F2

! !

Fig. 4–41

Sistema de forças paralelas

Neste caso, as linhas de ação das forças são paralelas ao eixo z, assim como a força resultante. Além disso os torques produzidos por cada uma das forças estarão no plano da placa junto ao eixo a, assim como o torque resultante. Neste caso, as linhas de ação de FR e MR serão mutuamente perpendiculares. Pode-se deslocar FR até o ponto P situado em d. (d=MR/FR)

4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 171

4

Parallel Force System. The parallel force system shown in Fig. 4–42aconsists of forces that are all parallel to the z axis. Thus, the resultantforce at point O must also be parallel to this axis, Fig. 4–42b.The moment produced by each force lies in the plane of the plate, and sothe resultant couple moment, , will also lie in this plane, along themoment axis a since and are mutually perpendicular. As aresult, the force system can be further reduced to an equivalent singleresultant force , acting through point P located on the perpendicular baxis, Fig. 4–42c. The distance d along this axis from point O requires

.(MR)O = FRd = ©MO or d = ©MO>FR

FR

(MR)OFR

(MR)O

FR = ©F

z

F1 F2

F3O

(a)

z

a O

bb

(b)

FR ! "F

FR ! "F

z

O

d

(c)

a

P

(MR)O

! !

Fig. 4–42

(a) (b) (c)

O

(MR)O

FR

OFR

O d

F3

F4 F1

F2

! !

Fig. 4–41

Exemplo 3: Substitua o sistema de forças e momentos de binário que agem sobre a viga da figura abaixo por uma força resultante equivalente, e encontre sua linha de ação que intercepta a viga, medida a partir do ponto O.

174 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

Replace the force and couple moment system acting on the beam inFig. 4–44a by an equivalent resultant force, and find where its line ofaction intersects the beam, measured from point O.

EXAMPLE 4.17

SOLUTION

Force Summation. Summing the force components,

From Fig. 4–44b, the magnitude of is

Ans.

The angle is

Ans.

Moment Summation. We must equate the moment of aboutpoint O in Fig. 4–44b to the sum of the moments of the force andcouple moment system about point O in Fig. 4–44a. Since the line ofaction of acts through point O, only produces a momentabout this point. Thus,

a

Ans.d = 2.25 m

- [8 kN A35 B ] (0.5 m) + [8 kN A45 B ](4.5 m)

2.40 kN(d) = -(4 kN)(1.5 m) - 15 kN #m+(MR)O = ©MO;

(FR)y(FR)x

FR

u = tan-1a2.40 kN4.80 kN

b = 26.6°

u

FR = 214.80 kN22 + 12.40 kN22 = 5.37 kN

FR

(FR)y = -4 kN + 8 kN A45 B = 2.40 kNc+ c(FR)y = ©Fy;

(FR)x = 8 kN A35 B = 4.80 kN::+ (FR)x = ©Fx;

(a)

O

4 kN

15 kN!m

8 kN

345

1.5 m 1.5 m 1.5 m 1.5 m

0.5 m

y

x

Fig. 4–44

(b)

d

O

FR

(FR)x " 4.80 kN(FR)y " 2.40 kNu

174 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

Replace the force and couple moment system acting on the beam inFig. 4–44a by an equivalent resultant force, and find where its line ofaction intersects the beam, measured from point O.

EXAMPLE 4.17

SOLUTION

Force Summation. Summing the force components,

From Fig. 4–44b, the magnitude of is

Ans.

The angle is

Ans.

Moment Summation. We must equate the moment of aboutpoint O in Fig. 4–44b to the sum of the moments of the force andcouple moment system about point O in Fig. 4–44a. Since the line ofaction of acts through point O, only produces a momentabout this point. Thus,

a

Ans.d = 2.25 m

- [8 kN A35 B ] (0.5 m) + [8 kN A45 B ](4.5 m)

2.40 kN(d) = -(4 kN)(1.5 m) - 15 kN #m+(MR)O = ©MO;

(FR)y(FR)x

FR

u = tan-1a2.40 kN4.80 kN

b = 26.6°

u

FR = 214.80 kN22 + 12.40 kN22 = 5.37 kN

FR

(FR)y = -4 kN + 8 kN A45 B = 2.40 kNc+ c(FR)y = ©Fy;

(FR)x = 8 kN A35 B = 4.80 kN::+ (FR)x = ©Fx;

(a)

O

4 kN

15 kN!m

8 kN

345

1.5 m 1.5 m 1.5 m 1.5 m

0.5 m

y

x

Fig. 4–44

(b)

d

O

FR

(FR)x " 4.80 kN(FR)y " 2.40 kNu

Exemplo 4: O guincho mostrado na figura está sujeito a três forças coplanares. Substitua esse carregamento por uma força resultante equivalente e diga onde a linha de ação intercepta a coluna AB e a lança BC.

4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 175

4

EXAMPLE 4.18

The jib crane shown in Fig. 4–45a is subjected to three coplanar forces.Replace this loading by an equivalent resultant force and specifywhere the resultant’s line of action intersects the column AB andboom BC.

SOLUTIONForce Summation. Resolving the 250-lb force into x and y componentsand summing the force components yields

As shown by the vector addition in Fig. 4–45b,

Ans.

Ans.

Moment Summation. Moments will be summed about point A.Assuming the line of action of intersects AB at a distance y from A,Fig. 4–45b, we have

a

Ans.

By the principle of transmissibility, can be placed at a distance xwhere it intersects BC, Fig. 4–45b. In this case we have

a

Ans.x = 10.9 ft

+ 250 lb A35 B111 ft2 - 250 lb A45 B18 ft2= 175 lb 15 ft2 - 60 lb 13 ft2325 lb 111 ft2 - 260 lb 1x2+MRA= ©MA;

FR

y = 2.29 ft

= 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb A35 B111 ft2 - 250 lb A45 B18 ft2325 lb 1y2 + 260 lb 102+MRA= ©MA;

FR

u = tan-1a260 lb325 lb

b = 38.7° u

FR = 2(325 lb)2 + (260 lb)2 = 416 lb

FRy= -250 lb A45 B - 60 lb = -260 lb = 260 lbT+ cFRy

= ©Fy;

FRx= -250 lb A35 B - 175 lb = -325 lb = 325 lb;:+ FRx

= ©Fx;

6 ft

y

x

5 ft

175 lb60 lb

(a)

250 lb

5 43

3 ft 5 ft 3 ftB

C

A

Fig. 4–45

y

(b)

x

x

FR

FR

y

C

A

260 lb

325 lb

260 lb

325 lbB

u

4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 175

4

EXAMPLE 4.18

The jib crane shown in Fig. 4–45a is subjected to three coplanar forces.Replace this loading by an equivalent resultant force and specifywhere the resultant’s line of action intersects the column AB andboom BC.

SOLUTIONForce Summation. Resolving the 250-lb force into x and y componentsand summing the force components yields

As shown by the vector addition in Fig. 4–45b,

Ans.

Ans.

Moment Summation. Moments will be summed about point A.Assuming the line of action of intersects AB at a distance y from A,Fig. 4–45b, we have

a

Ans.

By the principle of transmissibility, can be placed at a distance xwhere it intersects BC, Fig. 4–45b. In this case we have

a

Ans.x = 10.9 ft

+ 250 lb A35 B111 ft2 - 250 lb A45 B18 ft2= 175 lb 15 ft2 - 60 lb 13 ft2325 lb 111 ft2 - 260 lb 1x2+MRA= ©MA;

FR

y = 2.29 ft

= 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb A35 B111 ft2 - 250 lb A45 B18 ft2325 lb 1y2 + 260 lb 102+MRA= ©MA;

FR

u = tan-1a260 lb325 lb

b = 38.7° u

FR = 2(325 lb)2 + (260 lb)2 = 416 lb

FRy= -250 lb A45 B - 60 lb = -260 lb = 260 lbT+ cFRy

= ©Fy;

FRx= -250 lb A35 B - 175 lb = -325 lb = 325 lb;:+ FRx

= ©Fx;

6 ft

y

x

5 ft

175 lb60 lb

(a)

250 lb

5 43

3 ft 5 ft 3 ftB

C

A

Fig. 4–45

y

(b)

x

x

FR

FR

y

C

A

260 lb

325 lb

260 lb

325 lbB

u

4.8 FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM 175

4

EXAMPLE 4.18

The jib crane shown in Fig. 4–45a is subjected to three coplanar forces.Replace this loading by an equivalent resultant force and specifywhere the resultant’s line of action intersects the column AB andboom BC.

SOLUTIONForce Summation. Resolving the 250-lb force into x and y componentsand summing the force components yields

As shown by the vector addition in Fig. 4–45b,

Ans.

Ans.

Moment Summation. Moments will be summed about point A.Assuming the line of action of intersects AB at a distance y from A,Fig. 4–45b, we have

a

Ans.

By the principle of transmissibility, can be placed at a distance xwhere it intersects BC, Fig. 4–45b. In this case we have

a

Ans.x = 10.9 ft

+ 250 lb A35 B111 ft2 - 250 lb A45 B18 ft2= 175 lb 15 ft2 - 60 lb 13 ft2325 lb 111 ft2 - 260 lb 1x2+MRA= ©MA;

FR

y = 2.29 ft

= 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb A35 B111 ft2 - 250 lb A45 B18 ft2325 lb 1y2 + 260 lb 102+MRA= ©MA;

FR

u = tan-1a260 lb325 lb

b = 38.7° u

FR = 2(325 lb)2 + (260 lb)2 = 416 lb

FRy= -250 lb A45 B - 60 lb = -260 lb = 260 lbT+ cFRy

= ©Fy;

FRx= -250 lb A35 B - 175 lb = -325 lb = 325 lb;:+ FRx

= ©Fx;

6 ft

y

x

5 ft

175 lb60 lb

(a)

250 lb

5 43

3 ft 5 ft 3 ftB

C

A

Fig. 4–45

y

(b)

x

x

FR

FR

y

C

A

260 lb

325 lb

260 lb

325 lbB

u

Atenção: Estudar os exemplos 4.19 (pág.128) e 4.20 (pág. 129) livro do Hibbeler 12a edição.