lect 3 force analysis

8/2/2019 Lect 3 Force Analysis

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Force Systems 2D


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Lecture Overview

- Classification of Vectors

- Vector addition, parallelogram law

- Decomposition of vectors, force components

- Moment of a vector

- Force couples

1.)

2.)


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Statics concerning rigid bodies

- focus on sliding vectors

- principle of transmissibility

(force can be applied on any point along its line of action without

changing its external effect on a rigid body)


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Vector Addition

1. parallelogram law

2. triangle rule

3. analytic method


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Vector Addition graphical method

the parallelogram law resultant force


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2 forces replaced by a single resultant force

A

B

scale


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A

B

scale


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B

A

point of intersection


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B

A


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Line of action of R through point of intersectionof A and B

B

A R

scale


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the triangle rule (from parallelogram law)


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line of action of one force to be moved!

A

B


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for two forces

B

A

RB

A

R

ABBARrrrrr

+=+=


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for three forces

...CABCBAR =++=++=rrrrrrr

B

A

R

C

B

AR

C


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location of line of action of resultant force R?

B

A

R

C

B

A

C

Result of addition:Initial situation:


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location of line of action of resultant force R?

B

A

C

B

A

C

Initial situation:


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lines of action point of intersection

B

A

C

B

A

C

Initial situation:


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B

C

A

B

A

C

Initial situation:


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B

A

C

B

A

C

Initial situation:


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result for 2 forces A and B : A + B = RAB

location of resultant force RAB

B

A

C

RABB

A

C

Initial situation:


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location of RAB and C

C

RABB

A

C

Initial situation:


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lines of action

C

RABB

A

C

Initial situation:


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point of intersection

C

RAB

B

A

C

Initial situation:


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RABC

B

A

C

Initial situation:


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resultant force of the 3 forces A, B and C

CRAB

RABC B

A

C

Initial situation:


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location of line of action of resultant force

CRAB

RABC B

A

C

Initial situation:


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Initial situation:

B

A

C


location of line of action of resultant force

B

A

R

C

RABC


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situation at

particle:

B

A

C


e.g.: equilibrium at particle:

addition of all vectors = 0

B

A

R

C

R

resultant

reaction force


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Recalling equilibrium at particle


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Vector Addition analytic method


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Vector Addition analytic method

trigonometric rules

cosBA2BAR222

+=

R

sin

B

sin=

A

B

RB

magnitude

Inclination or R from A


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Vector Decomposition

1. components along arbitrary axis

a) graphical method

b) analytical method

2. rectangular components

vector addition by rectangular components


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Vector Decomposition graphical method



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vector components along 2 axis of arbitrary incline

given: Force F, 2 axis of arbitrary incline a and b

a

F

b


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a

b



applying parallelogram law

F

V t D iti hi l th d


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a

b



Fa

FFb

V t D iti l ti l th d


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Vector Decomposition analytical method

V t D iti l ti l th d


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a

b

Vector Decomposition analytical method


F

sin

F

sin

F

sin

ba

==

Fa

B

FFb

by trigonometric rules:

Fsin

sinFa

= F

sin

sinFb =and

components in

a and b directions:

)(180 o +=


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R t l C t


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Rectangular Components

F

x

y



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horizontal component of F:

F

x

y

cosFFx =



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horizontal component of F:

vertical component of F:magnitude of F:

F

x

y

cosFFx =

sinFFy =2

y

2

x FFF +=

Fx

Fy


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Vector addition by rectangular components


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Vector addition by rectangular components

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

( ) ( )2212

21R yyxx FFFFR +++==


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example 1


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example 1

given: detail of a symmetric cable support

= 30, P = 10 kN

determine: tensile force (horizontal) and shear force

(vertical) exerted to the connection.

example 1


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example 1

= 30, P = 10 kN

equilibrium condition:

Px and Py are the rectangular

components of P! fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

Px

PyR= -P

reaction force support

position of bolt

example 1


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e a p e

= 30, P = 10 kN

Px = cosP = cos3010kN = 8.66kN

Py = sinP = sin3010kN = 5.00kN


Px

PyR= -P

example 2


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p


example 2


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p


T

T = 400 N

R = ?R

example 2


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p

T

T = 400 N

R = ?R

600.0N0.5)400N(1cos60TTR ox =+=+=

346.4N23400Nsin60TR 0y ===

692.8NRRR 2y2

x =+=

x

y


example 2


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T

T

R = ?R

600.0N0.5)400N(1cos60TTR ox =+=+=

346.4N23400Nsin60TR 0y ===

692.8NRRR 2y2

x =+=

60x

y


example 2


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T

T

R = ?R

600.0N0.5)400N(1cos60TTR ox =+=+=

346.4N23400Nsin60TR 0y ===

692.8NRRR 2y2

x =+=


60x

y

example 2


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T

T

R = ?R

600.0N0.5)400N(1cos60TTR ox =+=+=

346.4N23400Nsin60TR 0y ===

692.8NRRR 2y2

x =+=


60x

y

example 2


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T

T

R = ?R

600.0N0.5)400N(1cos60TTR ox =+=+=

346.4N23400Nsin60TR 0y ===

692.8NRRR 2y2

x =+=


60x

y

example 2


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T

T

R = ?R

600.0N0.5)400N(1cos60TTR ox =+=+=

346.4N23400Nsin60TR 0y ===

692.8NRRR 2y2

x =+=


60x

y

Lecture Overview


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- Classification of Vectors

- Vector addition, parallelogram law

- Decomposition of vectors, force components

- Moment of a vector

- Force couples

1.)

2.)

Moment


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Moment

1. moment of a force about a point

- scalar product

- cross product

2. force couples

Moment


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moment of a force


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force acting in a plane

F

moment of a force


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line of action and support

F

moment of a force


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line of action ofreaction force

hinge support: no equilibrium - rotation!

F

Fdistance

moment of a force


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axis of rotation

F

d

moment of a force


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Moment of Force F M = Fd [kNm]

d = moment armperpendicular to line of action!

F

d

moment of a force


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sign convention right hand rule (+)

F

M (+)

moment of a force, scalar product


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Moment of Force F M = Fd [kNm]

d = moment arm

F

d

examples - moment of a force


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F

d

M = Fd

left side right sideEquilibrium!


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F

d

M = +Fd M = -Fd

F

d

+ M M -

Equilibrium!


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d d


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d d


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F

d

2F

d

left side right side


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F

d

2F

d

M = +Fd M = -2Fd

Rotation!


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d d/2

left side right sideEquilibrium!


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F

d

2F

M = +Fd M = -2Fd/2

d/2


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1 m

5 m

F ??


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1 m

5 m

500 N

F = ??


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600 N

1 m 3 m

???

Moment


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- scalar product

- cross product

2. force couples

Moment - Cross Product (vector product)


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Moment of a vector V about any point 0

ris a position vector from reference point 0 to any point on

the line of action of vectorV.

VrM0r

r

r

=

zyx

zyx0

VVV

rrr

kji

M

rrr

r

=

0

r

F

sinrF =r

r

rsin

Definition of Cross Product (vector product)


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Determination of resulting vector by three by three matrix with

unit vectors i, j and k.

( ) ( ) ( )kba-bajba-baiba-baBA

xyyxzxxzyzzy

rrrrr

++=

zyx

zyx

bbb

aaa

kjirrr

for 2D system

Moment Varignons Theorem


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Moment of a force F about any point equals the sum of

the moments of the components of the force about the

same point.

by cross product:


Moment Varignons Theorem


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Moment of a force F about any point equals the sum of

the moments of the components of the force about the

same point.

by rectangular moment arms:


example 3

Calculate the magnitude of the moment about the base point 0 in


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Calculate the magnitude of the moment about the base point 0 indifferent ways.

example 3

Calculate the magnitude of the moment about the base point 0 indiff t


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1.) product of force by moment arm

example 3



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example 3



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example 3

Calculate the magnitude of the moment about the base point 0 indifferent ways


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2.) replace force by rectangular

components

x

y

example 3



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Ca cu a e e ag ude o e o e abou e base po 0different ways.

x

y

example 3



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g pdifferent ways.

x

y

example 3



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g pdifferent ways.

x

y

example 3



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g pdifferent ways.

3.) by cross product

x

y

example 3



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different ways.

x

y

example 3



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different ways.

( ) ( ) ( )kba-bajba-baiba-baBA xyyxzxxzyzzyrrrrr

++=

0sin40600-cos40600

042

kji

oo

rrr

= 0= 0

x

y

Moment

f f


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- scalar product

- cross product

2. force couples

Moment

2 f l


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2. force couples


moment of a couple

moment produced by 2 equal and opposite


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moment produced by 2 equal and opposite

forces is known as a couple.

Fd

F

moment of a couple

moment of couple about any point of reference 0.


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F

F

0

d

a

moment of a couple



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o e t o coup e about a y po t o e e e ce 0

F

F

0

d

a

R = F F = 0

M = F(a+d) - FaM = Fd


moment of a couple



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p y p

F

F

0

d

a

R = F F = 0

M = F(a+d) - FaM = Fd


r

cross product: M = r x F

moment of a couple


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Force-couple systems

resolution of a force into a force and a couple equal external effect


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equal external effect.

example 4


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example 4

M = -Fy


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y = -M/F = - 0.375 kNm / 5 kN

y = -75 mm

example 4

M = -Fy


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y = -M/F = - 0.375 kNm / 5 kN

y = -75 mm 75 mm

initial

=


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lect 3 force analysis

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