Overall dimensions and technical information are provided solely for informative purposes and may be modified without notice. A|9 Allowed axial load (combined bending and compressing load) This is the maximum load that can be applied axially on the rod tip. Above this value the rod might bend under compression. This value depends on a number of factors such as load size, rod diameter, the distance at which the load is applied (bending and compressing length L) and the conditions under which the load is applied (cylinder mountings). Among the possible conditions, the following three are the most common. The maximum axial load can be calculated in two ways: In an empirical way ( see equations) or by checking the following diagram which shows the worst possible conditions (case 1 & 2) For all other possible mountings alternatives the axial load will surely be higher. With the third equation or using the diagram it is possible to calculate the bending and compression distance. Cylinder ø80 mm Rod diameter ø20 mm Stroke 600 mm Mounting CASE 2 intermediate trunnion: L0=290 mm Carico 2000 N L (distance)= 29+60=89 cm 3 7 4 2 Fk= (p x 2,1 x 10 x 2 ) : (64 x 89 x 5) = 4104 N (Above the 2000 N applied) E = rod material coefficient of elasticity (N/cm2) (steel=2,1x107 N/cm2) d = rod diameter (cm) L = bending and compression distance (cm) C = safety factor (da 2,5 a 5) The same result can be obtained using the below diagram : following the bending and compression distance line relative to 900mm up to the intersection with the 20mm Ø line we obtain 4000N. Also this second example can be resolved using the below diagram: following the bending and compression distance line relative to 900mm up to the intersection with the 4000N maximum load we obtain Ø20 mm. CASE 1 Fixed cylinder and free load stroke stroke stroke stroke Lo Lo Lo Lo L L L L Fk Front or rear flange Foot bracket Fk Fk stroke stroke stroke stroke Lo Lo Lo Lo L L L L Complete trunnion and rod fork Front clavis and rod fork Intermediate trunnion and rod fork Fk Fk Fk CASE 2 Cylinder fixed with pin and loosen load Example: Axial load verification Example: Rod diameter sizing Considering the same conditions as in the above case we need to determinate the rod diameter suitable to withstand a 4000N load d = (4000 x 64 x 892 x 5) / (p3 x 2,1 x 107) = 2 cm The diameter to choose is the next one up : Ø25 mm APPENDIX A Appendix Dimensioning Solutions for pneumatic automation General Catalogue
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