Chap5 Hydraulic Cylind

7/23/2019 Chap5 Hydraulic Cylind

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Chapter 5

Hydrauliccylinders



rotary (motor)ctuators single acting

linear(cylinder)

double acting

single acting

1) small rod 2) rams

double acting

Telescope-cylinder

Limited angle rotary cylinder!esign consideration o" hydraulic cylinders

#$utput "orce

%$&'%h - %b - %" %g



here

2*+ bar

#Cylinder as a pushing unit

#Cylinder as a pulling unit

2

4 Dah

π

⋅=&%

):()(4

22 pressurebacb&% d Dbb −⋅= π

rodandpistono" gra.ity%

%"orce"riction% op"

:

)

10

1(:

g

≈

4

2 Dπ =areae/ecti.e

a

op

D &

%

thus 13.1=

4

)( 22 d D −= π

areae/ecti.e



#Calculation o" piston rod diameter (d)a)ith respect to material strength

b)ith respect to .elocity ratio o" the piston

i.en 〝 ! 〞 3 〝φ〞⇒

Calculate 〝 d 〞

nd

sop σ σ σ

π σ =≤= ][][

4

2

%4aterial strength

%actor o" sa"ety

][13.1

σ

opd

%≥⇒

21

1

4

D

Q

A

Q

v π ==

)(

422

2

2d D

Q

A

Qv

−

==

π

21

2

)(1

1

Dd v

v

−

==φ



#Chec "or bucling rupture!e" slenderness ratio '

here stroe o" piston

(i#e# length o" piston rod)

a) i" 617 3 then bucling rupture o" the pistonrod must be considered3 hen the compression"orce is applied to the piston rod#

here %b bucling load

n %actor o" sa"ety

%b ' "(mounting situation3 slenderness ratio3 8)

-

)4,64:(

4 d

k

d

I A

I

k === hencerodsolid"or

π

diametero" pistonrod

area moment o"inertia

d

nb

op%%stability"or ≤⇒



b) 〝 empirical e9ns

〞 to determine %bi)

here σs material strength

m const ("or steel m ≈ :5)

a const ("or steel a ≈ )

const ("unction o" mounting situation)

ii)

here ; <oung=s module ("or steel ≈2#1 ×17+ bar)

> area moment o" inertia("or solid rod)

im ⋅≤-

2)(1k i

a

A s

b

+

⋅=

σ %

5777

1

i

im ⋅>-

2

2

EI ib

π =%

64

4d I

π =

64

4

d I I y xπ

==



#!etermination o" the barrel thicness%rom the "ormula based on thin pipe

here & ma? pressure

! piston diameter

@ barrel thicness

;nd-position cushioning de.ices

pre.enting impact at the end o" the stroe

Types1# Cushioning de.ice (inside the cylinder)

2# @hoc absorber (outside the cylinder)

Consider Type1(Cushioning de.ice inside the cylinder)Assumepressure & during the deceleration is const

][2 σ

D P S

⋅≥

n

sσ σ =][

materialstrength

@a"ety "actor

)(0 constam

P A

dt

dva =

⋅−==⇒

20

2

0 2 vS av =−

21

20

0

0

)2

1(v

S a

v

v−=∴



here @ stroe ( ) initial .elocity

Bnal .elocity

∴ %orce re9uired to decelerate the mass m "rom

is

? Compute the "orce re9uired to decelerate amass

m',#+, g "rom ',+#5+ mDmin to 7 ithina

distance o" 7#7,:1 m

Esol6

vv →0

0v

v

S

vav

20

20

0 =⇒=i"

00 →v

0am ⋅=%

0v

kN m

m

kg

S

vmam

649.17)0381.0(2

)sec60

576.36()63.3(

2

220

0

=⋅

⋅==⋅=%











緩衝裝置

說明：

(1)液缸右側緩衝裝置在左側也有

(只

是圖中未示出 ) 。

(2) 液缸左側之 chec .al.e 目的在導引高! ( "液缸右#$出 ) %&'()*+。

1) Cushioninge.ice

hec

al.e

2) @hoc absorber



5-0 液缸之,-.(seals)

01之,-23：

4567 9:,-. 4;在<

=>在之?@ 6

4A67 BCD,-. 4EF

GDH6

4I67 JKH 4LMNOP

QRS之TU 6

4V67 GDH WXYZ[ ,-.



‧$ DH\]^_只"` ab,-之c

defg之hi^_jklmn之Do

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Chap5 Hydraulic Cylind