Slide #: 1 Chapter 3 Rigid Bodies : Equivalent Systems of Forces Part -2

Slide #: 1

Chapter 3Chapter 3

Rigid Bodies : Rigid Bodies : Equivalent Systems Equivalent Systems of Forcesof ForcesPart -2Part -2

Slide #: 2

CoupleCouple

• Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple.

Slide #: 3

Moment of A CoupleMoment of A Couple

• Moment of the force F,

F

- F

O

A

rA

B

rB

• Moment of the force -F,

• Moment of the couple,

MO= rA x F

MO= rB x (-F)

MO= rA x F + rB x (-F)

MO= (rA - rB) x F

MO= r x F

r

Slide #: 4

Moment of A CoupleMoment of A Couple

• Moment of the couple,

FdrFM

Fr

Frr

FrFrM

BA

BA

sin

• The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vector that can be applied at any point with the same effect.

Where d is the perpendicular distance between the lines of action of F and -F

Slide #: 5

Problem 3.69 on page 116Problem 3.69 on page 116

A piece of plywood in which several holes are being drilled has been secured to a workbench by means of two nails. Knowing that the drill applies a 12 Nm couple of moment on the plywood, determine the magnitude of resulting forces applied to the nails if they are located to

(a)At A and B

(b)At A and C

Slide #: 6


(a)Nails are at A and B

12 Nm = (0.45m) * F

---> F = 26.67 N

-F= 26.67 N

F=26.67 N

Slide #: 7

Equivalent CouplesEquivalent Couples

Two couples will have equal moments if

• 2211 dFdF

• the two couples lie in parallel planes, and

• the two couples have the same sense or the tendency to cause rotation in the same direction.

Slide #: 8

Equivalent CouplesEquivalent Couples

6 m 20 N20 N

M

1 m

120 N

120 N

M

1 m

120 N

120 N

M

Slide #: 9

Non- Equivalent CouplesNon- Equivalent Couples

6 m 20 N20 N

M

1 m

120 N

120 NM

Slide #: 10

Addition of CouplesAddition of Couples

• Consider two intersecting planes P1 and P2 with each containing a couple

222

111

planein

planein

PFrM

PFrM

• Resultants of the vectors also form a couple

21 FFrRrM

• By Varigon’s theorem

21

21

MM

FrFrM

• Sum of two couples is also a couple that is equal to the vector sum of the two couples

Slide #: 11

Couples Can Be Represented by Couples Can Be Represented by VectorsVectors

• A couple can be represented by a vector with magnitude and direction equal to the moment of the couple.

• Couple vectors obey the law of addition of vectors.

• Couple vectors are free vectors, i.e., the point of application is not significant.

• Couple vectors may be resolved into component vectors.

Slide #: 12

Sample Problem 3.6 on page 113Sample Problem 3.6 on page 113

Determine the components of the single couple equivalent to the couples shown.

Slide #: 13


compute the sum of the moments of the four forces about D.

SOLUTION:

ikjkjM

FxrFxrMMMM

D

DDD

lb 20in. 12in. 9lb 30in. 18

)()( 221121

k

jiM

in.lb 180

in.lb240in.lb 540

Slide #: 14


ALTERNATIVE SOLUTION:

• Attach equal and opposite 20 lb forces in the +x direction at A, thereby producing 3 couples for which the moment components are easily computed.

Slide #: 15


• Attach equal and opposite 20 lb forces in the +x direction at A

• The three couples may be represented by three couple vectors,

in.lb 180in. 9lb 20

in.lb240in. 12lb 20

in.lb 540in. 18lb 30

z

y

x

M

M

M

k

jiM

in.lb 180

in.lb240in.lb 540

Slide #: 16

Resolution of a Force Into a Force at Resolution of a Force Into a Force at OO and a and a CoupleCouple

• Force vector F can not be simply moved to O without modifying its action on the body.

• Attaching equal and opposite force vectors at O produces no net effect on the body.

• The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system.

Slide #: 17

Resolution of a Force Into a Force at Resolution of a Force Into a Force at OO and a Coupleand a Couple

Any force F acting at a point A of a rigid body can be replaced by a force-couple system at an arbitrary point O. It consists of the force F applied at O and a couple of moment MO equal to the moment about point O of the force F in its original position. The force vector F and the couple vector MO are alwaysperpendicular to each other.

O

Ar

F

O

A

FMO

Slide #: 18


Replace the couple and force shown by an equivalent single force applied to the lever

Slide #: 19


Secondly resolve the given force into a force at O and a couple

kNmM

jxjmimFxBOFxrM

o

o

)60(

400N-)260.0()150.0(

We move the force F to O and add a couple of moment about O

First calculate the moment created by the couple

kNmM

ixjFxrM

o

o

)24(

200N-0.12m

Slide #: 20


mmOC

mmmOC

kOCkNmM

jNxjOCiOCFxCOkNmM

o

o

420)(

210210.060cos)(

)400(60cos)()84(

)400(60sin)(60cos)()84(

0

0

00

Slide #: 21

Problem 3.Problem 3.8080 on page 11 on page 1188

A 135 N vertical force P is applied at A to the bracket shown. The bracket is held by two screws at B and C.

(a)Replace P with an equivalent force-couple system at B

(b)Find the two horizontal forces at B and C that are equivalent to the couple of moment obtained in part a

Slide #: 22


kNmM

jNxjiFxrM

B

ABB

875.16

)135(05.0125.0(

Slide #: 23


Find the two horizontal forces at B and C that are equivalent to the couple of moment obtained in part a

NF

FkNmM B

225

*)075.0(875.16

F= 225 N

-F= 225 N

Slide #: 24

Reduction of a System of Forces to One Reduction of a System of Forces to One Force and One CoupleForce and One Couple

Any system of forces can be reduced to a force-couple system at a given point O. First, each of the forces of the system is replaced by an equivalent force-couple system at O.

Then all of the forces are added to obtain a resultant force R, and all of couples are added to obtain a resultant couple vector MO.

In general, the resultant force R and the couple vector MO

will not be perpendicular to each other.

Slide #: 25

Reduction of a System of Forces to One Reduction of a System of Forces to One Force and One CoupleForce and One Couple

First, each of the forces of the system is replaced by an equivalent force-couple system at O.

Then all of the forces are added to obtain a resultant force R, and all of couples are added to obtain a resultant couple vector MO.

FrMFR RO

Slide #: 26

Sample Problem 3.8 on page Sample Problem 3.8 on page

For the beam, reduce the system of forces shown to (a) an equivalent force-couple system at A, (b) an equivalent force couple system at B, and (c) a single force or resultant.

Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium.

SOLUTION:

a) Compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A.

b) Find an equivalent force-couple system at B based on the force-couple system at A.

c) Determine the point of application for the resultant force such that its moment about A is equal to the resultant couple at A.

Slide #: 27

Sample Problem 3.8 on pageSample Problem 3.8 on page

SOLUTION:

a) Compute the resultant force and the resultant couple at A.

jjjj

FR

N 250N 100N 600N 150

jR

N600

ji

jiji

FrM RA

2508.4

1008.26006.1

kM RA

mN 1880

Slide #: 28

Sample Problem 3.8 on pageSample Problem 3.8 on page

b) Find an equivalent force-couple system at B based on the force-couple system at A.

The force is unchanged by the movement of the force-couple system from A to B.

jR

N 600

The couple at B is equal to the moment about B of the force-couple system found at A.

kk

jik

RrMM ABRA

RB

mN 2880mN 1880

N 600m 8.4mN 1880

kM RB

mN 1000

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Slide #: 1 Chapter 3 Rigid Bodies : Equivalent Systems of Forces Part -2