08 Couples Equivalence

7/29/2019 08 Couples Equivalence

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Copyright G Golub & G Milano Couples & Equivalence 1

THE COUPLE MOMENT

A couple is formed by two equal but opposite forces with

different lines of action.

x

y

z

F

-F Consider the moment,Mo,about an arbitrary point, O,

of the two equal & opposite

forces.O

The two forces form a plane.


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x

y

z

F

-Fr1

r2

r1/2

d

ConsiderMoof the couple:

Where eP is in the direction perpendicular to the plane formed by

the two forces by definition of the cross-product.

O

eP

Mo =r1 xF

= (r1 -r2) xF

=r1/2 xF = r1/2sinqFeP

+r2 x (-F)

= d

q

r1/2sin =d

FeP

Mo =r1 xF -r2 x (F)

r2 x (-F)= -r2 x (F)sketch & use basic definition

r1 x F -r2 x (F)= (r1 -r2) x F

linearity of vector algebra

(r1 -r2) x F =r1/2 x F

By definition of addition


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x

y

z

F

-F

r1

r2

r1/2

d

MA =r1 xF +r2 x (-F)

= (r1 -r2) xF

=r1/2 xF = dFeP

MA = dFeP

O

Repeat the process with another arbitrary point A.

A

Take the moment of the

couple about thepoint, A.

The moment of a couple is independent of the point of

application.


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SAMPLE PROBLEMFind the resultant of the twocouples shown:

x

y

z

34

12

10#

10#

60o

12#12#

FindM10

by summing moments

about point(12, 4, 3)

-12i

-4j

(12, 4, 3)

M10 = -12i x 10(.5j - .866k)

The 10# force lies in the yz plane.

10#

60o

5j

-8.66k

M10 = 120(-0.5k - 0.866j)


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x

y

z

34

12

10#

10#

60o

12#12#

Find M12 by summing moments

about point (12, 4, 0)

M12 = -4j x 12i = 48k

-12i

-4j (12, 4, 0)


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x

y

z

34

12

10#

10#

60o

12#12#

M10

= 120(-0.5k - 0.866j)

M12 = 48k

R =M10 +M12 = -12k - 104j

x

y

z 60#48#

-12i

-4j

104#


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By observation:

x

y

z

34

12

10#

10#

60o

12#12#

Consider M

magnitude-.866j-.5k

48k

M10= 10# X 12 = 120#

-sin30kcos30j = - 0.5k - 0.866j

Direction is perpendicular to the plane of themoment, clearly in the y-z plane and the

inverse of the unit vector of the 10# force.

30o


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By observation:

x

y

z

34

12

10#

10#

60o

12#12#

M12 = (4 x 12)k = 48k

This result may be found by

computation or by visualization.

-.87j-.5k

48k

M12= 4 x 12# = 48#

Direction is perpendicular to the plane of the

moment, the x-y plane or the z direction.

Magnitude M12


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TRANSLATION OF A FORCE TO AN

ALTERNATE POINT

Consider a force,F, acting on a body at point, A.

F

A


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At point, B, add and subtract the same force,F, which

changes nothing.

F

-F

F

AB

Obviously,Fis equivalent to, the same asF +F+-F


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Regrouping the forces into a force and couple:

F

-F

F

ABObviously,F is

equivalent toF + dFen

d

F

A


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Redrawing the force and couple:

dFen

F

AB

Obviously,F (acting at A) is equivalent

toF (acting at B) + dFen


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EQUALITY AND EQUIVALENCE

200# 100# 100#

The two systems shown are equivalent. They produce thesame resultant force on the same line of action. They

produce the same acceleration on the truck.

Two force systems are said to be equal if they are identical.

Two force systems are said to be equivalent if they producethe same mechanical effect.


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Two force systems aremechanically equivalentiffthey both

reduce to the same force and couple at an arbitrary point, O.

F1

F2 F3

F4

F

5

O O

The two force systems are equivalent iff:


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EQUIVALENT FORCE SYSTEMS

F1

F2

F3

F4

F5

O

O

r1

r2

r4

r5

r3

There is equivalence iff:

F1 +F2 =F3 +F4 +F5

and

r1xF1 +r2xF2 =

r3xF3 +r4xF4 +r5xF5

The resultant force in each caseis the sameThe resultant moment about an

arbitrary point is the same.


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These two systems are equivalent:

.A

F

F

A

d

-dF

1

2

F(1) =F(2) obvious

MA(1) = 0 =MA(2) = dF -dF = 0

F

In 2, move F a distance, d.

Add a couple of magnitude -dF


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A force may be replaced by a system of a force and a

couple. Obviously, the reverse must also be true.

.A

F

.

F

A

d

-Fd


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To help you, you are given twomagical 1,000# forces that you can place anywhere on the

seesaw and move as you walk.

Visualization of Equivalence Process

Consider that you are standing on a seesaw at its middle.

You weigh 200#.

you

The support at the middle will destruct if

it detects a weight greater/less than 200#. The seesaw is

built over a pond filled with hungry alligators. To reach

safety, you must walk to the end of the seesaw. If the seesaw

tilts, you are alligator food.


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Visualization of Equivalence Process

you


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Consider that you are standing on a seesaw at its middle. You weigh200#. The support at the middle will destruct if it detects a weightgreater/less than 200#. The seesaw is built over a pond filled with hungryalligators. To reach safety, you must walk to the end of the seesaw. If the

seesaw tilts, you are alligator food. To help you, you are given twomagical 1,000# forces that you can place anywhere on the seesaw andmove as you walk.

you

What do you do?


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You 200#

Place the 1,000# forces at the left end of the seesaw one acting

up and one acting down. The total load at the pin is still 200#

acting downward.

1,000#

1,000#

As you slowly slide to the right, you drag the upward 1,000#

force with you at 1/5 the speed.

A l l lid h i h d h d 1 000#


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You200#

1,000#

1,000#

As you slowly slide to the right, you drag the upward 1,000#

force with you at 1/5 the speed.

When you have moved 1 to the right, the upward 1,000# force

moves 0.2 to the right.

Wh h d 1 h i h h d 1 000# f


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You200#

1,000#

1,000#

When you have moved 1 to the right, the upward 1,000# force

moves 0.2 to the right.

0.2

You200#

1

At this point, view equilibrium:

SFx = +1,0001,000200 + 200 reaction unchanged

SMpin = - 200 x 1 + 1,000 x 0.2 =0

C i i h i h hil i h l 1 000#


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1,000#

Continue moving to the tight while moving the lower 1,000#

force 1/5 the distance at the same time.

You200#

The two 1,000# forces comprise a couple that counters the

tendency to rotate the seesaw while you slowly move to leave

to safety.

SAMPLE PROBLEM


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SAMPLE PROBLEMReplace the force shown by aforce couple system at C.

The forklift is carrying a

3000#load as shown.

Find an equivalent force

couple system at the axle

of the front wheel.

This allows us to evaluate

the force on the front

wheel as well as the

tendency of the weight to

cause the truck to rotate

about the front wheel.

x

y 3000#

5

C


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1st - translate the force to

the axle.

x

y3000#

5

3000#

3000#

Add and subtract 3,000#

forces at C.

The two forces shown

as dashed lines form a

couple


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2nd - the couple at the

axle equals the momentof the original force

about the axle or the

couple of the two dashed

line forces.

Maxle = 5i x -3,000j= -15,000#k

x

y3000#

5Maxle

3000#


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Maxle = -15,000k#

x

y3000#

5Maxle

The force couple system at the axle are equivalent to the

force of the three boxes shown.

REDUCTION OF A SERIES OF FORCES TO A


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REDUCTION OF A SERIES OF FORCES TO A

SINGLE FORCE AND COUPLE

F1

F2

F3

r1

r2

r3

O

The force system shown

can be reduced to a

single force and couple

about an arbitrary

point.R = SFi i=1..3

Mo = rix Fi i=1..3

SFi =F1 +F2 +F3 i=1..3

To clarify the notation used:


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O

R = SFi

i=1..3

Mo = rix Fi i=1..3

R

Mo

In general,R &Mo are

not orthogonal.


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O

R

Mo

The two systems are equivalent. They produce the same

resultant force and couple about an arbitrary point, O.

F1

F3

r1

r2

r3

O

F2

SPECIAL CASES CONCURRENT FORCES


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SPECIAL CASES -- CONCURRENT FORCES

F1

F2

F3

O

R

A set of concurrent forcesalways reduces to a single force at

the point of concurrency.

Summing moments about the point of concurrency always

equals zero.

Mo = rix Fi = 0 sinceri

is always zero.

hence, R = SFi


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COPLANAR FORCES A set of coplanar forces - allforces lie in a plane,

always reduces to either

a single force or

a single moment.

F


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F1

F2

F3

r1

r2

r3

O Given a set of coplanar forces, it

is equivalent to the following:

O

R

Mo

Summing forces and moments

about an arbitrary point, O, mustalways yield a resultant,R in the

plane, plus anMo perpendicular

to the plane.

Since allri andFi lie in the plane,

then allrixFi are perpendicular

to the plane, by definition.

equivalent to


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F1

F2

F3

r1

r2

r3

O

O

R

Mo

O

R

d

equivalent to

Equivalent to

Shifting the force a distance, equal to

Mo/R, gives an equivalent force

system of a single force. Rd = Mo

PARALLEL FORCES IN SPACE A set of parallel


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PARALLEL FORCES IN SPACE A set of parallel

forces in space always reduce to a single force or a single

couple

=> x

y

z

R

Mo

R=F1 +F2+ F3

Mo =r1xF1 +r2xF2

+r3xF3

x

y

z

F1

F2

F3r1

r2

r3


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Translate R to a new location, A, such that r xR =Mo

x

y

z

R

M

o

x

y

z

r

R

r x R = Mo=>

The two systems are seen to be equivalent,

R =R

Mo =Mo =rxR =Mo

SAMPLE PROBLEM The tugs shown each apply four


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SAMPLE PROBLEM Thetugs shown, each apply fourkilo-Newton forces to the liner as shown. The captain

understands the effect of a force and a couple at the mainmast,

point, O. Find the equivalent force system at point, O.

The forces of the tugs are shown on the following diagram.

O

O


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45o

3

4

60o

4kN 4kN 4kN

4kN

.O 100m50m 30m20m x

y

O


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45o

3 4

60o

4kN 4kN 4kN

4kN

. 100m50m 30m20m x

y

O

Find an equivalent force

system atOR,

Mo25m

25m=>

O

R

Mo


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.707*4

.707*4.866*4

. 5*4

. 6*4

. 8*4

The components of the forces are shown inred.

45o

3

4

60o

4kN 4kN 4kN

4kN

. 100m50m 30m20m x

y

R

Mo 25m25m=>

O

R . 8*4


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R = 4(.707)(i +j) + 4j + 4(-0.5i +0.866j) - 4(0.6i +0.8j)

= 4i(0.707 - 0.5 -0.6) + 4j(0.707 + 1 +0.866 - 0.8)= -1.572i + 7.092j kN

Mo = 4(.707)(20+25)k + 4(100)k - 4(.5)25k +4(.866)200 -

4(.6)25k - 4(.8)70k = 886k k-Nm

45o

3

4

60o

4kN 4kN 4kN

4kN

.O

100m50m30m

20m x

y

Mo

25m

25m

.707*4

.707*4 .866*4

. 5*4

. 6*4

Going counter-clockwise around the ship:

TWO FORCE MEMBERS


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TWO FORCE MEMBERS

Consider the large crane shown lifting a 2 tongirder. The boom extension at the top is hinged atone end and has a pulley at the other end.


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hinge

pulley

Consider the free bodydiagram of the upper pieceof the boom.


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The upper boom is subject to forces from themain

boom, thestandoff pieceand thecablethat passesover thepulleyat the end.

Drawing a free body diagram of the upper boom .

boom

standoff pulley


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Fboom

Fbrace T1cable force

T2

1

2

Summing moments about point 1, T1 and T2will causemoments

0)( 211/21 TTxrM

cable force

r2/1

thatmust total to zero for equilibrium.


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Fboom

Fbrace T1

T2

1

2

0)(x 211/21 TTrM

r2/1 is not null. T1 + T2 is not null in general.

r2/1

ThereforeT1 +T2 must have a line of action thatpasses through point 1 & thus causes no moment.


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Fboom

Fbrace T1

T2

1

2

Similarly, Fbrace+ Fboommust sum to a vector with line of actionfrom point 1 to point 2 for the boom to be in equilibrium, since, themoment about point 2 produces the same result on Fbrace& Fboomas summing moments about point 1.

T1 +T2

Replaced by:


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Fboom

Fbrace T1

T2

1

2

T1 +T2

Fbrace+ Fboom

In a two force member,

loads applied at only two points on the body,

the resultant of the forces applied at each pointmust equal a vector with line of action connectingthe two points.


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Fboom

Fbrace T1

T2

1

2

T1 +T2

Fbrace+ Fboom

T1 +T2=Fbrace+ Fboom

Forces are equal opposite and collinear

This is a very important principle which is the basis for

trusses, a very important structural element.

THREE FORCE BODIES in a Plane


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THREE FORCE BODIES in a Plane

Consider the planar body shown with three appliedforces.

FR

S


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Consider the intersection ofFandRat point A.

FR

S

A


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Summing moments about A, FandRproduce nomoment.

F

R

S

A

Thus, the moment causedbySmust also be zero forequilibrium.

The exception to thisrule occurs when thethree forces are parallel.

Therefore, Smust passthrough A. The threeforces must be concurrent


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FR

S

A

Emphasizing this point:

Assume thatSdoesnt pass

through A as shown.

d

S obviously causes a moment

about A which equals dxS which

is greater than 0.

This means the body is not in

equilibrium.

Therefore, d=0, orS must pass

through A.

SF ilib i h F S R


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F

R

For equilibrium, whereF, S, R

are parallel,

F+R+S= 0

& the moments about anypoint must be zero.

e.g. assume that S=20#, F=10# R=10#

It is clear that F+R+S= 0

If the space betweenS& Fis 1, then the spacebetweenS& Rmust also be 1 for equilibrium.

Sum moments about a point onF. 20#(1)-10#(2)=0

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