6161103 4.9 further reduction of a force and couple system

4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System

� Consider special case where system of forces and moments acting on a rigid body reduces at point O to a resultant force FR = ∑F and MR = ∑ MO, which are perpendicular to one anotherperpendicular to one another

� Further simplify the system by moving FR to another point P either on or off the body


� Location of P, measured from point O, can be determined provided FR and MR known

� P must lie on the axis bb, which is perpendicular to the line of action at FR and the aa axisthe line of action at FR and the aa axis

� Distance d satisfies MRo = Frd or d = MRo/Fr


� With FR located, it will produce the same resultant effects on the body

� If the system of forces are concurrent, coplanar or parallel, it can be reduced to a single resultant force actingparallel, it can be reduced to a single resultant force acting

� For simplified system, in each case, FR and MRo will always be perpendicular to each other


Concurrent Force Systems

� All the forces act at a point for which there is no resultant couple moment, so that point P is no resultant couple moment, so that point P is automatically specified


Coplanar Force Systems � May include couple moments directed perpendicular

to the plane of forces

� Can be reduced to a single resultant force

When each force is moved to any point O in the x-y � When each force is moved to any point O in the x-y plane, it produces a couple moment perpendicular to the plane


Coplanar Force Systems

� For resultant moment,

MRo = ∑M + ∑(r X F)

� Resultant moment is perpendicular to resultant force � Resultant moment is perpendicular to resultant force

� FR can be positioned a distance d from O to create this same moment MRo about O


Parallel Force Systems � Include couple moments that are perpendicular to the

forces

� Can be reduced to a single resultant force� Can be reduced to a single resultant force

� When each force is moved to any point O in the x-y plane, it produces a couple moment components only x and y axes


Parallel Force Systems � For resultant moment,

MRo = ∑MO + ∑(r X F)

� Resultant moment is perpendicular to resultant force � Resultant moment is perpendicular to resultant force

� FR can be positioned a distance d from O to create this same moment MRo about O


� Three parallel forces acting on the stick can be replaced a single resultant force FR acting at a distance d from the grip

� To be equivalent,

F = F + F + FFR = F1 + F2 + F3

� To find distance d,

FRd = F1d1 + F2d2 +F3d3

Procedure for Analysis � Establish the x, y, z axes and locate the resultant

force an arbitrary distance away from the origin of the coordinates

Force Summation


Force Summation� For coplanar force system, resolve each force

into x and y components� If the component is directed along the positive x

or y axis, it represent a positive scalar� If the component is directed along the negative x

or y axis, it represent a negative scalar

Procedure for Analysis Force Summation

� Resultant force = sum of all the forces in the system

Moment Summation


Moment Summation

� Moment of the resultant moment about point O = sum of all the couple moment in the system plus the moments about point O of all the forces in the system

� Moment condition is used to find location of resultant force from point O

Reduction to a Wrench � In general, the force and couple moment system

acting on a body will reduce to a single resultant force and a couple moment at o that are not


force and a couple moment at o that are not perpendicular

� FR will act at an angle θ from MRo

� MRo can be resolved into one perpendicular M┴and the other M║ parallel to line of action of FR

Reduction to a Wrench � M┴ can be eliminated by moving FR to point P

that lies on the axis bb, which is perpendicular to both MRo and FR

� To maintain equivalency of loading, for distance


� To maintain equivalency of loading, for distance from O to P, d = M┴/FR

� When FR is applied at P, moment of FR tends to cause rotation in the same direction as M┴

Reduction to a Wrench � Since M ║ is a free vector, it may be moved to P so that it

is collinear to FR

� Combination of collinear force and couple moment is called a wrench or screw


called a wrench or screw

� Axis of wrench has the same line of action as the force

� Wrench tends to cause a translation and rotation about this axis

Reduction to a Wrench

� A general force and couple moment system acting on a body can be reduced to a wrench

� Axis of the wrench and the point through which


� Axis of the wrench and the point through which this axis passes can always be determined

Example 4.16

The beam AE is subjected to a system of coplanar forces. Determine the magnitude, direction and location on the beam of a resultant force which is


location on the beam of a resultant force which is equivalent to the given system of forces measured from E

Solution

Force Summation

Σ=→+ FF xRx ;


↓=−=

+−=

Σ=→+

→==

+=

NN

NNF

FF

NN

NNF

Ry

yRy

Rx

0.2330.233

20060sin500

;

0.3500.350

10060cos500

o

o

Solution

� For magnitude of resultant force

5.420

)0.233()0.350()()( 2222

=

+=+= RyRxR

N

FFF


� For direction of resultant force

o7.33

0.350

0.233tantan

5.420

11

=

=

=

=

−−

Rx

Ry

F

F

N

θ

Solution

- Moment Summation

� Summation of moments about point E,MM ERE ;Σ=


md

mNmN

mNmNNdN

MM ERE

07.50.233

1.1182

)5.2)(200()5.0)(100(

)0)(60cos500()4)(60sin500()0(350)(0.233

;

==

−−

+=+

Σ=oo

Example 4.17

The jib crane is subjected to three coplanar

forces. Replace this loading by an equivalent

resultant force and specify


resultant force and specify

where the resultant’s line of

action intersects the column

AB and boom BC.

Solution

Force Summation

−

−=

Σ=→+

kNkNF

FF xRx

75.13

5.2

;


↓=−=

−

−=

Σ=→+

←=−=

−

−=

NkN

kNNF

FF

kNkN

kNkNF

Ry

yRy

Rx

60.260.2

6.054

5.2

;

25.325.3

75.153

5.2

Solution

� For magnitude of resultant force

16.4

)60.2()25.3()()( 2222

=

+=+= RyRxR

kN

FFF


� For direction of resultant force

o7.38

25.360.2

tantan

16.4

11

=

=

=

=

−−

Rx

Ry

F

F

kN

θ

Solution

Moment Summation

Method 1

� Summation of moments about point A,


� Summation of moments about point A,

my

mkNmkN

mkNmkn

kNykN

MM ARA

458.0

)6.1(54

50.2)2.2(53

50.2

)6.0(6.0)1(75.1

)0(60.2)(25.3

;

=

−

+

−=

+

Σ=

Solution

� Principle of Transmissibility

xkNmkN

MM ARA

)(60.2)2.2(25.3

;

−

Σ=


mx

mkNmkN

mkNmkn

xkNmkN

177.2

)6.1(5

450.2)2.2(

5

350.2

)6.0(6.0)1(75.1

)(60.2)2.2(25.3

=

−

+

−=

−

Solution

Method 2

� Take moments about point A,

;Σ= MM


49.160.225.3

)6.1(5

450.2)2.2(

5

350.2

)6.0(6.0)1(75.1

)(60.2)(25.3

;

=−

−

+

−=

−

Σ=

xy

mkNmkN

mkNmkn

xkNykN

MM ARA

Solution

� To find points of intersection, let x = 0 then y = 0.458m

� Along BC, set y = 2.2m then x =


� Along BC, set y = 2.2m then x = 2.177m

Example 4.18

The slab is subjected to four parallel forces.

Determine the magnitude and direction of the

resultant force equivalent to the given force


resultant force equivalent to the given force

system and locate its point of application on

the slab.

Solution

Force Summation

−−+−=

Σ=↑+

NNNNF

FF

R

R

500400100600

;


↓=−=

−−+−=

NN

NNNNFR

14001400

500400100600

Solution

Moment Summation

NmNmNNyN

MM xRx

)0(500)10(400)5(100)0(600)(1400

;

−=−

−−+=−

Σ=


mx

x

NNmNmNxN

MM

my

y

yRy

00.3

42001400

)0(500)0(400)6(100)8(600)(1400

;

50.2

35001400

=

=

++−=

Σ=

=

−=−

Solution

� A force of FR = 1400N placed at point P (3.00m, 2.50m) on the slab is equivalent to the parallel force system acting on the slab


system acting on the slab

Example 4.19

Three parallel bolting forces act

on the rim of the circular cover

plate. Determine the magnitude


plate. Determine the magnitude

and direction of a resultant

force equivalent to the given

force system and locate its

point of application, P on the

cover plate.

Solution

Force Summation

FFR ;rrrr

rrΣ=


Nk

kkkF

FF

R

R

}650{

150200300

;

r

rrrr

−=

−−−=

Σ=

Solution

Moment Summation

)150()200()300(

;

−+−+−=

Σ=

kXrkXrkXrFXr

MM

CBAR

ORo

rrrrrrr

rrrrrrrr

rr


Equating i and j components,

85.84160650

85.84240650

85.8485.84160240650650

)150()45cos8.045sin8.0(

)20()8.0()300()8.0()650()(

)150()200()300(

−=−

−=

−−+=−

−+−+

−−+−=−+

−+−+−=

y

x

ijijjyjx

kXji

kXjkXikXjyix

kXrkXrkXrFXr CBAR

rrrrrr

rrr

rrrrrrr

oo

Solution

� Solving, x = 0.239m and y = -0.116m

� Negative value of y indicates that the +ve direction is wrongly assumed


wrongly assumed

� Using right hand rule,

)45cos8.0(150)8.0(200650

;

)45sin8.0(150)8.0(300650

;

mNmNx

MM

mNmNx

MM

xRx

yRy

o

o

−=−

Σ=

−=

Σ=

6161103 4.9 further reduction of a force and couple system

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