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Example2
: tofindcompression loadand installation type for acylinderwithdiameter
D
=50
mm, strokeL
=700
mm
andapressure
p
=6
bar
.
Figure 18
DiagramA
In correspondencewith the value
p
=6
bar
,we followahorizontal segment until the inclined “diameter
D
=50
mm
(20
mm
rod)” line, from the
x
axis we note that the value of the compression load is of
F
s
>1000
N
DiagramB
From the above junction, draw a vertical segment that meets the “rod Ø20
mm
” to give a value of
L
=600
mm
(approx.)
Possible installation types
:
Figure 17
Pos. B/1
: this installation type enables the cylinder to develop amaximum stroke length
L
max
=
600
mm *
1,4=
840
mm
Pos. B/2
: this installation type enables the cylinder to develop amaximum stroke length
L
max
=
600
mm *
1,7=
1020
mm
Pos. B/3
: this installation type enables the cylinder to develop amaximum stroke length
L
max
=
600
mm *
2=
1200
mm
Pos. C
: this installation type enables the cylinder to develop amaximum stroke length
L
max
=
600
mm *
2,8=
1680
mm
All connections are possible.
Devices tomodify the cylinder performance
The lever
The strength of a cylinder can bemodifiedwithmechanical
levers
.
Figure 19
Pos. 1
: the lever is a rigid rod with a pivot at the point called the
fulcrum
(fc)
. At the extreme end of this rod, a
Force
F
is applied. The perpendicular distance from the line of action of the Force to the axis of rotation (pivot), is
referred to as the
arm
(b)
.
The product of the Forcemultiplied by the arm generates a component defined as
Moment of a force
.
m
=
F * b
When lifting a load you have two Forces: the power
P
(generated by the cylinder) and resistance
R
(the body’s
weight to be lifted). From the product of each Forcemultiplied by the respective arms you arrive at the “moment of
power” and the “moment of resistance”.
The lever is in equilibriumwhen the twomoments are equal, i.e. when:
bp * P
=
br * R
3
68
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