GENERAL CATALOGUE API

Cylinders Technical data

Motion force F

The available motion force to the piston rod is: F = Ft - R

Where R represents a force of reaction that comprises numerous factors: friction, form and type of seals, operating pressure, counter-pressure at discharge. The value of R is not easy to quantify as its component factors are not only numerous, but also variable. A cautious estimate for usual applications could be 30% Ft. As shown by the graph illustrated below, which indicates the progress of the pressure values of delivery and discharge during the uniform movement of a cylinder, the delivery value Pm and the discharge value Ps remain constant during the stroke of the cylinder, if we exclude the brief transitory periods: of acceleration following the switching of the distributing valve and cushion at the end of the stroke. The cylinder is thus prevalently subject to a motion force F proportional to Pm and the pushing surface, and to a counter-pressure force Fs proportional to the pressure Ps and the section upon which it acts, both of which are constant. The load reaction Fc must be added to these two forces. In other words, the cylinder, in dynamic equilibrium, will - like all engines in this state - find itself under the action of contrasting forces that balance each other. It will move at a constant speed under the action of a constant force. Ft - Fs - Fa = Fc Where Fs is the counter-pressure force and Fa is a force that bears in mind the friction and reduction of the operating power, to which Ft is linked, which does not reach the static network pressure, as can be seen in the graph. During the transitory acceleration period, the force Fs is very low, as the air is being discharged. As the speed of the piston increases, the air being discharged is compressed and the force Fs increases until the state of equilibrium is reached. For example, we wish to find the cylinder capable of overcoming the load value Fc = 1200 [ N ] The theoretical force Ft must be at least 30% greater. Let’s assume that: Ft = 1600 [ N ] This gives the following result: F = √ 40Ft/ p p      F = √ 40x1600/3,14x6      @ 58 [ mm ] The closest standardized bores turn out to be: 50 mm and 63 mm It is advisable to choose the bore F = 63 mm, also because it enables a reserve of power to be obtained. The uniform movement of the cylinder can be obtained by regulating the air at the discharge. In order to obtain high values, on the other hand, it is necessary to make an appropriate increase in the discharge space in order to obtain accelerated movements, as the equilibrating force of counter-pressure is no longer present.

1 - CYLINDERS

Peak load

In the case of long strokes, the load that can be applied to the piston rod is reduced due to the decrease in resistance at peak load. The lifespan of a cylinder depends largely on its mechanical application. Installation must be performed in such a way as to avoid, or at least minimize,

bending moments and radial loads on the piston rod (the most onerous kind of anchorage is the hinge type). If only axial loads need be applied, the piston rod will be subjected to the peak load during pushing.

As the acceptable peak load is proportional to the diameter of the piston rod d (through the elastic modulus and the inertia moment) and inversely proportional to twice the stroke (length of free inflexion), in the case in which it does not allow the application of the required force, it is necessary to increase the diameter of the piston rod, passing to a suitably larger bore. The choice of the standardized bore that best satisfies the requirements of the application in question is not just linked to the satisfaction of the force to be provided, but also to that of other conditions. These include the need to always have a power reserve (by choosing a larger size) and that of not causing excessive stress to the cushionings.

Air consumption Nl/min

Air consumption air is a working value; it has a significant influence on costs. It is possible to calculate the average air consumption using the following formula: Q = p F 2 /4x 60 c/t x (p+p0) / p0 x 10 -3 x 10 -3 [ nl/min ] Where: Q = air consumption [ nl/min ]

F = bore [ mm ] c = stroke [ mm ]

t = time taken to perform the stroke [ s] p = atmospheric operating pressure [ bar ] p0 = atmospheric pressure: 1 bar For example, we want to calculate the consumption of the following cylinder: d = 50 mm; c = 300 mm; t = 0,45 s; p = 6 bar Q = 3,14 x 25 x 10 2 /4 x (60 x 3 w 10 2 /0,45) x 7 10 -3 x 10 -3 = 550 [ nl/min ]

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