3.FORCE SYSTEMS - Excellence in Engineering...

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Page 1: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

3.FORCE SYSTEMS

Page 2: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

FORCE SYSTEMS

Page 3: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

θ AForce is

characterized by

1. Point of application (A)

2. Magnitude ( 100 N)

3. Direction

Characteristics of a Force

F

θ-

(from x-axis)

Objective: To bring out the characteristics of a Force as applied in Statics

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Sense of Force

x

y

x

y

Tension Compression

Objective: To bring out the two senses in which a Force can act

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RIGID BODIES AND FLEXIBLE BODIES

Flexible Body

Rigid Body

Steel, Wood, Concrete, Stone are rigid bodies, we neglect their deformation in Statics

Foam is Flexible material, which undergoes large deformations under loading

In Statics we deal with

Rigid bodies alone

Objective: To bring out the difference between a Rigid Body and a Flexible Body

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F

F

Hence Pulling is equal to Pushing, provided the Forces are on the same horizontal line (same line of action)

Transmissibility

This is known as the Principle of Transmissibility

Objective: To explain the concept of Transmissibility

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From the viewpoint of Statics, the Arch which is loaded on the top is equivalent top the arch which is loaded from beneath – this is an application of the principle of Transmissibility

Objective: An example which illustrates the principle of Transmissibility

=

P

P

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Red Forces are Applied Forces

Blue Forces are Reaction Forces

The dotted line the Reference Boundary for the Structural System under consideration

Objective: To describe, System Boundary, Applied Forces and Reaction Forces

Page 9: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

The Green Forces are the Internal Forces

Objective: To describe Internal Forces

Page 10: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

Reference Boundary of the Structural System

External Force

Reaction Forces

Internal Forces

Objective: An example to illustrate: System Boundary, External ,Reaction & Internal Forces

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CONCLUSIONS

1. The System boundaries can be defined arbitrarily

2. What the Applied Forces, the Reaction Forces and the Internal Forcesare, will be clarified accordingly

Objective: To explain the idea that the System Boundary is defined, depending on theportion of the structure one wishes to focus on

Page 12: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

Vector Addition

Characteristics of a Vector

An important characteristic of vectors is that they must be added according to the parallelogram law . This is necessary because vectors have both magnitude and direction .

Using parallelogram law, we may add vectors graphically or by trigonometric relationships.

First we will see the graphical method and then the trigonometric method.

Objective: To explain how Vectors can be added graphically

Page 13: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

200

400

200 400 600 800 1000

R=1300#

o6.22=θ

PARALLELOGRAM LAW OF VECTOR ADDITION

600

800

1000

A box is being pulled up by a Force of 900 N and pulled to the right with a Force of 900 N. We want to know the Resultant.

N

N

H=900 N

H=900 N

V=900 N

V=90

0 N

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500#

900#7515

R

θ

PARALLELOGRAM LAW OF ADDITION OF VECTORS

Objective: To illustrate the parallelogram method of addition of vectors using triangles

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z

TIP –TO – TAIL METHODAnother Method of Vector Addition

O

A

B

R

θ

O

A

B

R

θ

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Determining the Resultant by Analytical Method.

Sometimes it is more convenient to determine the Resultant by using the cosine law as shown in the following example.

lb

FFFR

192)259.0)(140)(100(2140100

cos222

22

21

2

=−−+=

−+= φ

o

o

o

8.44

)704.0(sin704.0

192966.0140

105sinsin

sin105sin

1

2

2

=

=

=

×=

=

=

−θ

θ

θ

RF

FR

lbF 1001 =

lbF 1402 =

o30

o45

o75o105=φ

θ

o30

o30

o45lbF 1001 =

lbF 1402 =Rx

y

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VECTOR ADDITION

°35°45

900 lb600 lb

°35°45

600 lb

900 lb

Parallelogram Method

R= 1413 lb, Angle=50.8

600 lb

900 lb

R R

Tip-to-tail method

R=1413 lb, Angle=50.8

Page 18: 3.FORCE SYSTEMS - Excellence in Engineering …profmathews.weebly.com/.../2/5/7125213/3._force_systems.pdfStructural System External Force Reaction Forces Internal Forces Objective:

Graphical addition of Three or More Vectors

F1F2

F3

x

y

x

y

F2

R12

F3

R123

F1