Transcript Chapter 5 Hydraulic Cylinders

Chapter 5
Hydraulic cylinders
rotary (motor)
Actuators
single acting
linear(cylinder)
double acting
‧ single acting
1) small rod
2) rams
‧double acting
‧Telescope-cylinder
‧Limited angle rotary cylinder
§ Design consideration of hydraulic cylinders
1.Output force
FOP=Fh - Fb - Ff ± Fg

Where Fh  Pa  D 2
4
 2
Fb  Pb 
4
( D  d 2 ) (b : back pressure )
2~6 bar
1
Fop )
10
Fg : gravity of piston and rod
Ff : friction force (
2.Cylinder as a pushing unit
D 2
effective area 
4
thus D  1.13
Fop
Pa
3.Cylinder as a pulling unit
 (D2  d 2 )
effective area 
4
4.Calculation of piston rod diameter (d)
a)with respect to material strength

Fop

4
 [ ]
d2
 d  1.13
[ ] 
s
Material strength
n
Factor of safety
Fop
[ ]
b)with respect to velocity ratio of the piston
Q
4Q
v1 

A1 D 2
Q
4Q
v2 

A2  ( D 2  d 2 )
v2
1
 
v1 1  ( d ) 2
D
‧Given 〝D〞, 〝〞 Calculate 〝d〞
5.Check for buckling rupture
Def: slenderness ratio =  k
where  : stroke of piston
(i.e. length of piston rod)
I
d 4
k
A
(for solid rod : I 
diameter of
piston rod
d
, hence k  )
64
4
area moment of inertia
(a) if  d >10 , then buckling rupture of the piston rod
must be considered, when the compression force is
applied to the piston rod.
for stability  Fop 
Fb
n
where Fb :buckling load
n :Factor of safety
Fb = f(mounting situation, slenderness ratio, …)
(b) 〝empirical eqns〞 to determine Fb:
i) 
 m i
k
Fb 
s  A
1
a 
(
)2
k
i
where s : material strength
m : const (for steel m  85)
1
a : const (for steel a 
)
5000
i : const (function of mounting situation)
ii) 
k
 m
i
i 2 EI
Fb 
2
where E: Young’s module (for steel 2.1 106 bar) 4
d
I

I : area moment of inertia(for solid rod 64 )
Ix  I y 
d 4
64
6.Determination of the barrel thickness
From the formula based on thin pipe:
PD
S 
material
2[ ]
strength
where P : max pressure

[ ]  s
D : piston diameter
n
S : barrel thickness
Safety factor
§ End-position cushioning devices
— preventing impact at the end of the stroke
‧Types:1. Cushioning device (inside the cylinder)
2. Shock absorber
(outside the cylinder)
‧Consider Type1(Cushioning device inside the cylinder):
Assume:pressure P during the deceleration is const.
 a
dv
A P

 a0
dt
m
 v0 2  2a0 S  v 2
2a0 S 12
v
  (1  2 )
v0
v0
(const )
where S : stroke ( v0  v )
v0 : initial velocity
v : final velocity
v0 2
if v  0  a0 
2S
 Force required to decelerate the mass m from
v0  0 is:
F  m a0
Ex: Compute the force required to decelerate a mass
m=3.63 kg from v0 =36.576 m/min to 0 within a
distance of 0.0381 m:
<sol>
2
m )
(36.576
v0 2
60
sec
F  m  a0  m
 (3.63kg) 
2S
2  (0.0381m)
 17.649 kN
緩衝裝置
(1) Cushioning device
Check
valve
(2) Shock absorber
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check valve目的在導引
高壓油(當液缸右行伸出)至整個活
塞面積。
5-4 液壓缸之密封圈(seals)
‧常見之密封單元：
(1): 隔離密封圈(主在阻
絕外在之塵污)
(2): 凹槽型密封圈(不需O型
環)
(3): 支持環(防止徑向負荷造
成之卡死)
(4): O型環+PTFE密封圈
‧加工配合之方法一般可 ‧O型環本身一般只當作
參考廠商之型錄
靜態密封之用