md1_project1_spring2014

8/10/2019 MD1_Project1_Spring2014

1/14


2/14

This

report

d oc u m en t s t h e analysis of an off- loading station lo ca te d a t t h e en d of

a paper

rol l ing machine.

his

off- loading station

design

uses a V-linkage to effect ively transfer fi n is h ed p a p er ro ll s f rom t h e m a c h in e _

onveyer

to

th e

forklift

t ruck.

Th e

off- loading station consists of

tw o V-l inks t ha t

are welded to a steel tube

t ha t

I S

xed

to a

shaft

rotated by an air cyl inder

through

a crank

arm.

Th e

purpose

of t h is s tu dy is

to design th e

V-l inks,

rod, an d

shaft

with

th e appropr iate

dimensions

to

satisfy a safety

factor t ha t is

closest

to 1.2.

Th e

weight of

th e p a p er

roll

is

to

be determined. Along with

t he n ot ch

sensit ivi ty, fatigue

stress

factor,

max imum an d min imum

bending moments ,

an d

th e

mean an d alternating

norma l stresses

Of

e

V-l inks.

F or t h e Cylinder

rod,

th e

compressive

force at

starting

posit ion, an d th e critical C 0 l T | p l S S l V 8 .

load

m u s t

determined.

Additional ly,

th e torque

m ean an d alternating

shear

stresses,

an d

polar

m o m e n t

of inert ia m us t

be

Furthermore,

th e

dimensions of

th e

V-links,

rod,

an d

shaft can

manipula ted

to

provide a factor of

safety

losest

to m i n i m um give.

g(0Q_ ./

/

1


3/14


4/14


5/14

rom th e initial problem

statement , the maximum deflect ion on

an

arm

wil l

occur j us t a fte r

it

begins th e

_ _ _ _

ransfer an d

before it completes the transfer .

At

this

t ime

th e

roll is in

contact with both V-link arms, an d -

hen

it ls another posi t ion on

th e arm s th e deflection

t ip w il l

b e supported. With

that said th e

arm

can be

reated a s

a

canti lever

beam. Another

th ing to take into consideration is t h e f ac t that th e m ax force in th e

od

occurs w hen th e transfer starts. Additionally, th e max torque will

o cc u r w h en t h e

l ink

is horizontal an d

e p a p er

is

located

a s

shown

in figure

6 .

- u - m a

|

l

of Paper Rol l:

i t F

-

olving

fo r

th e

weight

of

t he p a p er

roll

1

4

IIn

__ __Qq

E e l w =

2 -

( 0 0 2

- 1 0 * )

- p - 9 - 1 - . . .

[ N 1

here W

is

weight [N], O D is th e outer d iameter

of

th e roll [ m ], lD

is

th e inner diameter of th e roll [m ], p

is

th e

density of

th e

roll [ 3 - ] , an d Ln, is

length

of th e roll [m ]

- - 1 ?

v-]ink

A1-mysis:

Figure

2 : V- linkage F B D

T he m ax b en d in g m o m en t in th e

arm

c a n b e found by ( this can b e seen in

fig.2

an d fig.3)

E q . 2 Mm ,

( 2 + - Z ; )

Where ta

is

the th ickness of th e

V-link

arm

[m ]

an d D m b e

is th e

diameter of th e

steel

tube

[m ]

Solv ing

for

_-.-

Q

l e e

is.

- . r _ -

L

OD

ta

E q s

r .=T+; I

s

e

l

a----0

Determin ing

th e

max

bending moment

in

th e V-link arm

E q _ 3

Mm =

Mam - %

Figure 3

Where Mm, is equivalent t o th e

max

bending

force [m4]

Determining th e m in im um bending m o m en t in th e V-link a rm, M m, [m4]

Eq04 Z0

Th e

m o m ent of inertia is n ow calculated s o that th e normal stresses c an b e found

E wcftaa

q

5 fv =T Table

1:

Neubers Cons t on t for Steels

.

. . .

R9 . . .

S i - I 1 (km

F

lruc )

W h e re

1 ,, I S th e

m o m e n t ofmertia

of

th e V-link [75],

an d wars

th e width

of th e

link

ar m

[m ]

t - ..

50

T h e radius of t he n o tc h r[m], is necessary to solve for notch sensitivity

Eq.6 r =

0. 25

- ta

Using Tabl e 6 -6

from

t h e b oo k ( Ta ble 1), to find in terp o late an d fin d th e

necessary

1/E,WhBl8Sut is equiva lent to

ultimate

tensile S trength [ksi]

s ,,

-s,,

Eq.7 \/E

=

\/E1

+

F 2 _ $ u '; ' (\/52

-

\/E1)

To

fin d

th e

notch

sensitiv ity q ,

t h e n o tc h r ad iu s

wil l be

converted to

inches

in

order

fo r th e Eq.8 to give

viab le answer

E

. 8

=

-1 -

q

1 + _ _

olving fo r static stress concentration,

K t,

to fin d the fat igue stress concentration

(see

fig.4)

q . 9

K , =

/ l , , -

4

55

G O

70

8 0

90

1(1)

1

10

12 0

. ,.- __ ._ _-.,-....

0 . 1 3 0

O

11 8

O .

10 8

O

0 93

0130

0

O70

0 . 0 6 2

O

0 55

0 . 0 4 9


6/14

fatigue

stress concentration,

K i, , 1 , , ,

x,=1+q(K, -1) ll .

. .

.- .

.,

. ? 1 1 . , _ L L e s s

.-

-'

*

- '

2 u ~

:1 - . 1 1

u

ci_q-;

4- ?

/ - 1 '

_ _ _ ;. - 1 : , - ; = _ I : _ = _ i q l uS - s ~ 1 -

-

/- ( , -n

-'

'

-

. ..-5I.--{J

. _ . . _ q

.7 2

E

r

I.-{Ii

J

'

'7.

-

P3

I.

I a

= 5 a

9

Q . Pd

L

5

-

.

~

I

=

. .

., ,

F

.

-b

9

.

u u

.0

8'

m l - 1 -or equations 1 1 an d 12 , determining t h e m a x im u m an d

normal stress, where

c is

th e

neutral axis, [MPa]

K

I

q .

1 1 am =

q .

1 2 Um =

2

\. -~

'-i___:

-T -1 ,5,--.....i-_-.,

.

I

-

~ -;

*4\n>|,____1_Q_--

-if-.I..:.g-_-

-

'

r '

or

equations

13

an d

14the

m ean an d

alternating

nomina l

. _ i

-

~ . . c = = . ; 1 - 1 - a u w v - . - = . . . , _ _ _ _ ;

is

found [MP3] e..;....L...-.l-1e; . .-1lILiL..i-..;.;_.;_: F 5

1

q.13

amnom

=

rid

u u

-'7 ';.?r

G1_|Ktna|I1

and:

L10

r

|zo---~ 9

1 K A('J)

where 7

T ,

_ \

M _ _ _ _

I,

If \ h

-.-.,..._ . .. .- -----

--.

--

IIII

INK)? Z0 -0.

33

MD 0 .932

31 - 0 . 3 0 3 0 4

L3 0 0 .958

8 0

0 .2 7 2 6 0

I20

0 . 99 5 9 0

- 0 . 2 3 8

29

I I0 |.0I6$0

0 . 3 1 5 4 8

L0 5 1.0 2 260 -0.19156

l.0l 0.96i5If9 -015-4|?

2

Figure 4:Geometric

Stress-Concentration

Factor

K t

E q . 1

4

O . _ o'mnx_f-Tmln

tlnom_ 2

Eq'15

Kfm

=

Kf Kf(0'ma.x)nom

< Sy

Now

that

k;

an d

Kr ,

is

found,

actual alternating an d

mean

stress

can

be found us ing equation

16

an d 17

Eq16 am

= Kfrn

' amnom

Eq-17

an = Kf ' aanom

_

___11"5 _

.. 1T

Since there are n o other

nonzero

stress

components

a t t h is po in t on th e to p of t he a rm , th e

Vo n

Mises S t r e s s e s , a , . , an d 0 ; } , are

Eq.18

0 1 , ,

=

om

Eq.19

0;

=

oa

Th e

unmodified

endurance l imit,

5;,

is to be determined in

order

to

solve fo r

the corrected endurance

l imit

Eq.20 S e :

= 0 .5

-

S m

Th e load

correction

f a c t o r , C , , , , , , 1 ,

is

equiva lent to

1 because

th e

V-link

arm acts l ike

a

cantilever beam

an d

is a bending force

Eq.21

C , _ , , , , d = 1

To fin d th e

size

correction factor,

C 5 , - , , , . , ,

equations 2 2 an d 2 3

would

be

needed

A95=

'

Wa

'

ta

E q .

2 3

1 1 , ,

=

Th e

equation

represented in equation24,

is used, because

th e d e q is 8m m an d 2 50 m m

E q . 2 4 cm ,

=

1.189-(d,,,) 7

Th e

following

equation fo r surface correct ion factor C5,,

E q .

2 5 C,,,,.;

=

A -

( . S , ,, )

Th e

temperature

correction

factor, C r a m p , is equivalent

to

one because

then surrounding environmen t

is

at

room

temperature

Cfgmp=

1

Th e reliability

correction

factor

is

0.897, because

th e

reliability is

at 90 %

EQ-27 Creliab = 0-897

Th e

corrected

endurance l imit, S e ,

is

n ow fo un d b y

Eq.28

S e

= Cioad

'

Csize

' Csurf '

Ctemp ' Creliab

' Se

The fat igue factor

of safety,

N f,

can

n ow be fo un d b y

s

2 9 N

= s 5

q

f

0;'$u;+O:n-5;

T he m ax

deflection at

t ip, d m a x

W

'_'(Marm)z'(Marm3'f-v)

E q . 3 0 6 m a x ' z

] , * 5 - , 1

5


7/14

R o d Analysis:

e

cylinder

rod

is

given

to

be

at fixed-p inned condi tion. Therefore, th e

effective

length,L,, ,

can

be

found

using

th e A I S C

recommended

quation

1 :.

_ q~ . .

- _ - . ~ .

T oF ]

M

Leff = '- ' L;-ad

Determin ing

t h e M om e n t

of inertia, where dm d is

th e diameter of

th e

rod

E q . 3 2 , =

,,

. $ 1 2 2 :

s 4

I-4

Determin ing

th e

Area

of

th e

ro d t o

fin d

gyration

radius

of

th e

rod

'2

E q . 3 3 A r = ,, . L

i = 1

LM

[

. 7 < ~ 1

--

- -

. .

F E

Determin ing th e

Slenderness

ratio, S r, to

then

fin d

the slenderness ratio

at

_ _

;_---

tangent

poin t , ( 5 , ) , - _ , i J-3:

F i gur e 5:

FB D

ofCrank Ar m & Cylinder R od

Sr

:

-ail

(STJD

=

11 ' -

Th e

following

equation determines th e critical buckl ing load

of

th e rod,

Pa ,

where S 3 , ,

is

the comprehensive

yield

strength

.

2

E q . 3 7 P , _ . , = .4 , - (3,,

-

)

Calcu la t ing

th e

torque of

th e

V-link arm,

' I , , , to fin d

th e m a x im u m t orq u e, T m ,

E q . 3 8

T ,

=

N

-32

q . 3 9 T , , , , , , , =

Moment of th e

maximum

compressive force,

M

I - c , , , , , , , , where L , - m , ,, , is th e

length

of th e crank, 6 ,, is th e position

angle

of

th e

crank arm, an d

9 ,.

is th e position angle

of

th e ro d

Eq-4'0 MFcomp =

Lcrank

' 5in(9c _er)

Th e

max

compressive

force, F c o m p ,

c a n then b e found with th e following equation

E q . 4 1 F =

@-

M F c c m 1 p

Determining

th e Safety factor, ca n then be

easily found

by

th e following

equation

E q . - i 2

N ;

=

Om p

6


8/14

Analysis:

? m m l m u m

T O T Q H B . T m . is to

be

determined in order to solve fo r Mean torque,

T m ,

an d Alternating tor que, T ,

Tmin = (Tu ' ' 2

T (Tmnx'*Tmln)

m = 2

Ta

;_

(rmaxgrmln)

ow

that

th e m ea n an d alternating torque

are

fo un d t h e shear

stresses

can b e calculated, where

rm

is th e

mean

shear

stress,

1 , ,

IS

th e

shear

stress, and Dshaf, is

th e diameter

of th e

shaft

. 4 6 = . _ ; ~ _ ' =

rm T m

27

Ta=

Ta

q

P

-

on M is es m ea n a nd alternatin stress can

b e found

usin th e followin e

uations 4 8

an d

4 9, where

0 3 ; , is

th e

mean

vo n

Mises stress,

an d 0 ,,

S 8

8 q

th e alternating vo n Mises

stress

J7

= 1/3 ~

(fm)2

git

= 1/3 - ('['a)2

h e

load correction factor of t h e sh af t c an b e

found

using

equation

21.

h e s iz e correction factor of th e

shaft

can be

found

using equation 24 , except substituting th e d e q with Dshaft.

h e temperature correct ion factor of th e shaft can b e

found

using

equation

2 6.

Th e

reliability

correct ion factor of t h e sh aft c a n be found using equation 2 7.

The surface correct ion factor of t h e sh aft can b e found using equation 25 .

Th e uncorrected endurance l imit of t h e sh aft can b e f ou n d by u s in g e qu a tion

20.

Th e

corrected endurance

l imi t

of t h e sh aft c a n b e fo un d b y

using

equation

28.

Th e fatigue

factor

of safety

of t h e sh aft can

be found

by

using

equation 29.

T h e

internal

to rques

c an be

found us ing equations 50 an d

51 ,

t hese torques ar e needed to

solve

fo r maximum angle of

twist

of

th e

shaf t \H Mr

Eq-50 T1

= Tmax

E q . 5 1 T , = r, , , , , , , -

5%

h e polar m o m ent

of

inertia, /,

can

be fo un d b y t he fol lowing equation

_ _(Dshaf t )

E q . 5 2

1-

1 1 - 3 2

Determin ing

th e

max

angular tw ist of th e shaft

,

L

'

(T 0.25 T 0.D5)

twwlimqq: W

Shaft

11.6 + 2

K

7

Figure 6:

FB D M a x t orq u e


9/14

G i v e n

D a t a :

O D =

0 .9

m (outer diameter of roll)

D

= 0.22 m ( inner diameter of roll)

u

=

3.23

m

( length of

roll)

,,,,;, =

L , - o n

( length

of

shaft)

= 984% (density of ro ll )

u b , = 0.16 m (diameter of

tube)

, =

1

m

( length

of

V- lin k arm)

n .

o .: 3 _ = , > ' * a

I - J .

fl

ii

= 0. 5 m ( le n gt h of cyl inder rod)

a n k

= 0 .3

m ( le n gt h

of

crank)

= 4 5

(start ing

angle

position of

crank

arm)

= 8

(start ing

angle

position

of cyl inder rod)

ia b = 9 0 % ( p erc en t of

reliability)

2

25

MATLAB

Printed

Results:

1)

2 )

3 )

4 )

The

weight

of the paper roll is 18650.1979

N

V-link

Design:

Thickness is 0 .050 m an d Width is .050 m

Th e notch

sensitivity is

0.9102,/0

/

The fatigue stress

concentration factor is 1.38

:mi///

The

m ax

bending moment is

3683.4141

N*m.and the n bending moment

is 0.000000

N *m

The mean normal stress is

122.12 MP a

and the a

ternating

normal stress is

122.l2259

HP a

C1oad=1.0000

v/%

Csize=0.B306

Csurf=0.B183

Ctemp=l.0OO0 /

Cre1iab=0.8970

q ._ I

The corrected

endurance

lmt is 191. a t E QLrk_

_

The

fatigue safety

factor

is 1.1994

iQ:

ii

T

T

bk

The deflection

at the

tip

is -0.0060

m

Cylinder R od Design:

The

The

The

0

/

diameter

is

0.018 m,the

effective

length

is,0/400

m , and the slenderness ratio 18 88.

compressive

force at

the

starting position/ 49067\394 N

critica compressive load is 63890.4l94-N

The safety factor against

buckling is

1.3021

Shaft Design:

Th e

diameter

is 0 . 0 8 1 m an d th e polar

m o m e n t

of inertia is 4.226s-03.-are

The torque is 8858.B44QvNm

The

m e a n

shear stress

is 0.000000

and the alternating shear stress

is 84.897 HPa'

Cload=1. oooo /

Csize=0.776

Z

Csurf=O.8183

Ctemp=1.0000

h

Creliab=0.897O

The corrected

endurance

lmt is 1.78 e N/m2

The

fatigue

safety factor

is

1.2149

The maximum angle

of twist

is 2.4006

d

rees

8

C U I II )

ID


10/14

dm

Th e

loading

station design

is possible

under

t he g iven parameters .

Ou t of

th e three

components , it can

be

t ha t

th e

V-linkage

would require

th e most materia l.

Th e carbon steel material is th e per fec t materia l to

used,

because it is a strong

material t ha t

should be

able

to

perform

t h e t as k of

transferring

t he p a p er rolls. Th e

of

th e

V- l inks are

0 .05 meters in thickness an d width , p rov id ing

a

factor of safety of

1.1994

an d a -

meter deflection a t th e t ip . T o get a safety factor of 1.3021, th e cyl inder rod only needed a diameter of

meters.

Th e

calculated comp ressive

force

at

th e

starting

posit ion

is

49067

N

ewtons, whereas

th e

critical

loa d was 63890 Newtons.

Therefore, even with

th e

small d imens ions , the cyl inder rod is

able

hold a

ot buckle

during th e

process

of

t rans fer ring tha t s h aft t o th e forklift truck. Additional ly, th e shaft h ad

a

safety

of 1.2149

with

a diameter

of

only

0 .08 1

meters.

With these

dimensions

of

th e shaft, th e

max imum

angle

of

ended up

being

2 .4

degrees and

th e polar

m o m e n t of inertia

ended up being

4 . 2 3 >

md1_project1_spring2014

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