Whether single-acting or double-acting, the theoretical thrust force in a pneumatic cylinder depends on the cylinder diameter, friction force, sealing elements, and air pressure.
Force Calculation in a Pneumatic Cylinder
The thrust force of cylinders differs between the forward (extend) and backward (retract) strokes. On the forward stroke, the full piston area (A) is effective, while on the retract stroke there’s a loss equal to the piston rod’s cross-sectional area. To increase the force a cylinder can apply, at least one of the pressure or area values must be increased.
P = F / A
F = P . A (Basic formula)
If we also factor in cylinder efficiency
F = P . A . η
F = Piston force (N)
P = Gauge pressure
A = Effective piston area
d1 = Piston diameter
d2 = Piston rod diameter
η = cylinder efficiency (90%)
Forward (Extend) Force:
F= P . A . η
F = P . (∏ . d12/ 4) . η
Return (Retract) Force:
F= P . A . η
F = P . (∏ . d12– d22 / 4) . η
NOTE: In a single-acting cylinder, spring force must also be factored into the calculation. When the spring force in a single-acting cylinder is taken into account, it’s calculated as follows:
F = P . A – Ff
Ff = spring force (3-20% of the cylinder force (F))
Sample Question
In a pneumatic circuit, the working pressure is 8 bar. Given that the double-acting cylinder to be used has a piston diameter of d1=50mm and a piston rod diameter of d2=20mm, calculate the forces the piston can apply on the forward and return strokes. (Piston efficiency will be taken as 90%.)
Solution
Since 1 bar = 0.1 N/mm², to get the result in Newtons, the “p” value in the basic formula is divided by 10.
F= P . A . η
Forward (Extend) Force:
F = P/10 . A . η
F= 0.8 . (∏ . d12/ 4) . η
F= 0.8. (3.14 . 502/4) . 0.90
F= 1413 N
Return (Retract) Force:
F = P/10 . A . η
F= 0.8 . (∏ . d12-d22/ 4) . η
F= 0.8. (3.14 . 502 – 202 /4) . 0.90
F= 1187 N
Force Formula
Pneumatic cylinder force is calculated using F = P × A, where P is pressure and A is the piston’s surface area.
Effect of Diameter
As piston diameter increases, since the area grows with the square of the diameter, the force produced increases disproportionately.
Friction Loss
In real-world applications, around 5-10% of the calculated theoretical force is lost to seal and friction losses.
Related Questions
On the forward stroke, the piston’s entire surface is exposed to pressure. On the return stroke, since the area occupied by the piston rod is no longer available, the effective surface shrinks, making the return force lower than the forward force.
According to the formula, force depends on the product of pressure and piston area. This means either the air pressure supplied to the system must be increased, or the cylinder’s piston diameter must be made larger.
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