CHAPTER 4 Force System Resultant

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# CHAPTER 4 Force System Resultant - PowerPoint PPT Presentation

CHAPTER 4 Force System Resultant. P. T. M. M. 4.1 Moment of a Force - - - Scalar Formulation. 1.Moment A measure of the tendency of the force to cause a body to rotate about the point or axis. â€¢ Torque (T) æ‰­åŠ› â€¢ Bending moment (M) å½Žæ›²åŠ›çŸ©.

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CHAPTER 4

Force System Resultant

P

T

M

M

4.1 Moment of a Force - - - Scalar Formulation

1.Moment

A measure of the tendency of the force to cause a body to

rotate about the point or axis.

• Torque (T) 扭力

• Bending moment (M) 彎曲力矩

2. Vector quantity

d

o

Lime of action (sliding vector)

(1) Magnitude ( N-m or lb-ft) Mo = Fd d = moment arm or perpendicular distance from point O to the line of action of force.

(2) Direction Right-Hard rule A. Sense of rotation ( Force rotates about Pt.O) Curled fingers B. Direction and sense of moment Thumb

3.Resultant Moment of Coplanar Force System

don

do1

do3

do2

4.2 Cross Product

1. Definition

(1) magnitude of

(2)Direction of

perpendicular to the plane containing A & B

j

i

2. Law of operation

(1)

(2)

(3)

3. Cartesian Vector Formulation

(1) Cross product of Cartesian unit vectors.

4.3 Moment of a Force – Vector Formulation

d

F

o

1. Moment of a force F about pt. O

Mo= r x F

where r = A position vector from pt. O to any pt. on the line of action of force F .

(1) Magnitude

Mo=|Mo|=| r x F | =| r|| F | sinθ=F r sinθ

=F d

(2) Direction

Curl the right-hand fingers from r toward F (r cross F ) and the thumb is perpendicular to the plane containing r and F.

4.4 Principle of moments

F1

r

F2

o

Mo=r x F

F = F1+F2

Mo= r x (F1+F2)

= r x F1+ r x F2

= MO1+MO2

Varignon’s theorem

The moment of a force about a point is equal to the sum of the moment of the force’s components about the point .

4.5 Moment of a force about a specified Axis

a

b

Ma

Ө

Mo= r x F

b’

A

Moment axis

a’

1. Objective

Find the component of this moment along a specified axis passes through the point about which the moment of a force is computed.

2. Scalar analysis (See textbook)

3. Vector analysis

Point O on axis aa’

O

Axis of projection

Moment of a force F about point 0

• Mo= r × F
• Here, we assume that bb’ axis is the moment axis of Mo
• (2) Component of Mo onto aa´ axis
• Ma = Ma ua
• Ma=Mo cosθ
• =Mo●ua=( r × F ) ●ua
• =trip scalar product
• Here
• Ma=magnitude of Ma
• ua= unit vector define the direction of aa´ axis

If

then

4.Method of Finding Moment about a specific axis

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

Mo= r x F

(2) Resolving the moment along the specific axis

Ma = Maua

= (Mo•ua) ua

=[ua •( r x F )]ua

4.6 moment of a couple

d

d

θ

F

F

r

1. Definition ( couple) 偶力矩

Two parallel forces have the same magnitude, opposite distances, and are separated by a perpendicular distance d.

2. Scalar Formulation

(1) Magnitude M=Fd

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

•Thumb indicates the direction

•Curled fingers indicates the sense of rotation

3. Vector Formulation

M= r x F

|M|=M=|r x F |=r F sinθ

=F d

-F

r

A

rB

F

rA

o

B

B

F

-F

r

o’

A

Mo=Mo’= r x F=M

o

Remark:

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

Mo= rAx( -F )+ rB x F

=(-rA+rB) x F

=r x F= M

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

4. Equivalent Couples

The forces of equal couples lie either in the same plane or in planes

parallel to one another.

F

-F

-F

F

A

plane A // plane B

F

d

-F

d

B

M2

M1

A

B

M2

M1

5. Resultant couple moment

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

MR=ΣM=Σ r x F

4.7 Equivalent system

A

F

P

P

F

1. Equivalent system

When the force and couple moment system produce the same “external” effects of translation and rotation of the body as their resultant , these two sets of loadings are said to be equivalent.

2. Principle of transmissibility

The external effects on a rigid body remain unchanged,when a force, acting a given point on the body, is applied to another point lying on line of action of the force.

line of action

Same external effect

Internal effect ?

Internal stresses are different.

F

F

F

-F

-F

o

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

A

A

A

equivalent

equivalent

o

o

o

Original system

Sliding vector

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

F

Couple moment

F

Mc= r x F

M=r x F

A

A

F

r

P

A

F

o

line of action

o

Original system

Force on Point A

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

o

o

A

A

F

F

F

F

o

o

X P

A

A

d

M= F d (Free vector)

Mo= F d

Example:

Point O is on the line of action of the force

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

4.8 Resultant of a force & couple system

1. Objective

Simplify a system of force and couple moments to

their resultants to study the external effects on the

body.

2. Procedures for Analysis

(1)Force summations

FR=F1+F2+……+ΣF

(2)Moment summations

MR0= ΣMC+r1o*F1+r2o*F2= ΣMC+ ΣM0

MC:Couple moment in the system

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

y F1,F2,F3 on xy plane

F3 M1&M2:z direction MR0=ΣMC+Σr * F P MR0

F2 x => => d=

FR

F1 FR=ΣF

F2 F1 FR

Equivalent

P =

System

Fn no couple moment

4.9 Further Reduction of a force & couple system

1. Simplification to a single Resultant Force

(1)Condition

FRMR0 or FR*MR0 = 0

(2)Force system

A. Concurrent Force system

B.Coplanar Force System

F1 r2 F2MR0FR= ΣF

r1 y = y = y

M1 p o

r3F3 x x MR0

x M2d = --------------

|FR|d=|MR0|FR

C. Parallel Force System

1. F1 // F2 //……// Fn

2. MR0 perpendicular to FR, MR0=ΣM+Σr*F

2. Reduction to a wrench

(1) Condition: FRMR0

MR0=M +M//

M = moment component FR

M// = moment component // FR

(2) Wrench (or Screw)

An equivalent system reduces a simple resultant

force FR and couple moment MR0 at pt.0 to a

collinear force FR and couple moment M// at pt.

FR

MRo

a

b

o

b

FR

FR

a

a

M//

a

b

M//

o

b

p

o

p

b

b

a

a