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SEAL KITS & TECHNICAL DATA
Seal Kits & Technical Data
8
NORTH AMERICAN CYLINDER & ACTUATOR CATALOG
>
Release 8.5
374
Symbols description
Sizing guide: formulas and examples
Sizing guide for Shock Absorbers Series SA
Symbol Unit
Description
m
friction coefficient
a
(
rad)
angle of incline
q
(
rad)
side load angle
w
(
rad/s)
angular velocity
A
(
m)
width
B
(
m)
thickness
C
(/
hr)
impact cycles per hour
D
(
cm)
cylinder diameter
d
(
cm)
piston rod diameter
E
d
(
Nm)
drive energy per cycle
E
k
(
Nm)
kinetic energy per cycle
E
t
(
Nm)
total energy per cycle
E
tc
(
Nm)
total energy per hour
F
(
N)
propelling force
Symbol Unit
Description
Fm
(
N)
maximum shock force
g
(
m/s
2
)
gravity acceleration (9.81 m/s
2
)
h
(
m)
hight
m
(
kg)
mass to be decelerated
Me
(
kg)
effective mass
P
(
bar)
operating pressure
R
(
m)
radius
Rs
(
m)
shock absorber mounting distance
from rotation center
S
(
m)
stroke (shock absorber)
T
(
Nm)
driving torque
t
(
s)
deceleration time
v
(
m/s)
velocity of impact mass
vs
(
m/s)
impact velocity at shock absorber
In order to select the correct dimensions of Shock absorbers the following
parameters are needed:
-
Weight of the impact object m
(
kg)
-
Impact speed
v
(
m/s)
-
Propelling or thrust force
F
(
N)
-
No. of impact cycles per hour C
(/
hr)
Some formulas
5.
Cylinder’s traction force
F =
D
2
·
π
·
P · g/100
4
6.
Cylinder’s thrust force
F =
(
D
2
-
d
2
) ·
π
·
P · g/100
4
7.
Maximum shock force (approx.)
Fm = 1.2 E
t
/
S
8.
Total energy consumption per hour Etc = Et · C
9.
Mass:
Me = 2E
t
/
v
2
Some formulas
1.
Kinetic energy
E
k
= mv
2
/2
2.
Drive energy
E
d
= F · S
3.
Total energy
E
t
=E
k
+E
d
4.
Free fall speed
v = √ (2g*h)
Calculation:
Example 1: Horizontal impact
Application data:
v
= 1.0 m/s
m
= 50 kg
S
= 0.01 m
C
= 1500 cycles/h
The adequate Shock Absorber to use in this case is Mod. SA 2015 according
to the technical data where we find that E
t
(
max)=59 Nm, E
tc
(
max)=38000
Nm/h and Me(max)=120 kg.
E
k
=
mv
2
=
50
.
1
2
= 25 Nm
2 2
E
t
= Ek = 25 Nm
E
tc
= Et
.
C = 25
.
1500
= 37500 Nm/h
Me
=
2
Et
=
2
.
25
= 50 kg
v
2
1
2
Example 2: Horizontal impact with propelling force
Application data:
m
= 40 kg
P
= 6 bar
S
= 0.01 m first hypothesis SA 1210
v
= 1.2 m/s
D
= 50 mm
C
= 780 cycles/h
To facilitate the calculation, the pressure in the empty
cylinder chamber is not considered (safety condition)
Calculation:
The adequate Shock Absorber to use in this case is Mod.SA 2015 according
to the technical data where we find that Et (max)=59 Nm, Etc (max) = 38000
Nm/h and Me(max)=120 kg.
E
k
=
mv
2
=
40
.
1,2
2
= 28,8 Nm
2
2
Consider the shock absorber with the lowest E
t
but superior to 28.8 Nm:
mod. SA 2015 S=0.015 m
E
d
= F
.
S =
D
2 .
π
.
P
.
g/100
.
S =
50
2
.
π
.
6
.
9,81/100
.
0,015
= 17,3 Nm
4
4
E
t
= E
k
+ E
d
= 28,8 + 17,3 = 46,1 Nm
E
tc
= E
t
.
C = 46,1
.
780
= 35958 Nm/h
Me
=
2
E
t
=
2
.
46,1
= 64,0 Kg
v
2
1,2
2