Handouts

ME 301-Soylu-H1

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only one joint between 2 bodies.

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REVIEW OF COMPLEX NUMBERS

1) z = r ei

exponential form

where, i = 1

Euler’s identity : e i

= cos i sin (EU)

z = r ( cos + i sin ) = r cos + i r sin

z = x + i y orthogonal form

2)

x y

x

y

Real

Imaginary

r

z

Complex numbers

may be used to

express 2D vectors.

’

A

B

r

Real axis

ME 301-Soylu-H7

AB = r ei

BA = - AB = - r ei

or,

BA = r ei ’

= r ei ( + )

Let

z1 = r1 1ie

z2 = r2 2ie

3)

z1 z2 = ( r1 cos1 + i r1 sin1 ) ( r2 cos2 + i r2 sin2 )

= ( r1 cos1 r2 cos2 ) + i (r1 sin1 r2 sin2 )

4) z1 z2 = r1 r2 ) θθ i( 21 e

5) 2

1

z

z =

2

1

r

r ) θθ i( 21 e

6) i ei

= ei ( / 2 )

ei

= ei( + / 2 )

form to be used in writing

the loop closure equations

eiθ i e

iθ

1 Re

eiθ Unit vector

ME 301-Soylu-H8

z Complex conjugate of z .

7) 1z = r1 1ie

= r1 cos1 – i r1 sin1

8) z1 z2 = 1z 2z

9) d

d e

i = i e

i [

dt

d e

i = ( i e

i ) (

dt

d ) ]

10) dt

dz1 = dt

d( r1 1i

e

)

= ( dt

dr1 ) ( 1ie

) + ( r1 ) ( i 1ie

dt

d 1 )

11) ei

+ e-i

= 2cos

12) ei

- e-i

= 2 i sin

13) ei0

= 1

ei ( /2)

= i

ei

= -1

ei (3 /2)

= -i

Proof by

Euler’s theorem

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CHAPTER 5

FORCE ANALYSIS IN MACHINERY

BASIC CONCEPTS :

Force : Action of one body on another. Force is a vector

quantity which has :

- Magnitude

- Direction

- Line of Action (LA)

- Point of application.

Mechanics of deformable bodies Point of application is

important, since we are interested in deformations and

stresses.

Rigid (i.e., undeformable) body mechanics Point of

application is not important. one can “ slide ” a force

along its LA.

LA

F

( For rigid body mechanics ,

e.g., force analysis )

ijF

Force applied by body i on body j.

jiF

Force applied by body j on body i.

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Newton’s third law ijF

= - jiF

and LA’s are the

same.

Representations of forces in 2D :

Imaginary (y)

Fy

F

Fx

Real (x)

F

= Fx i

+ Fy j

= Fei

= F <

F = 22 )()( yx FF F

x = F cos

= ATAN2( F

x / F , F

y / F ) F

y = F sin

Note that if

F

= F <

then

- F

= -F < = F < +

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Moment of a force about a point :

F

r

r

A

AM

= r

F

= r

F

= …….. = -Fd k

or,

AM

= r

F

= ( r < r ) ( F < F ) = rF sin(F - r ) k

Couple : Two, equal, opposite and non-collinear forces.

A . B .

d

F

r

- F

AM

= BM

= ….. = M

M

= r

F

= r

F

= …. = - Fd k

x

y

r

x

y

d

F

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