Fluid Power Journal

Calculate Diameters of Load-Holding Cylinders

Calculate the required diameter and standard diameter needed for the cylinders to hold the load shown below.

See Solutions

Calculate the force to hold the load for 30°.

30°

500 lbs. / 2 = 250 lbs. / cylinder                                      2224 N / 2 = 1112 N / side

250 / sine of 30° = 500 lbs.                                             112 / sine of 30°= 2,224 N

 

Sizing of cylinder:

500 # / 100 psi = 5 sq. in.                                                2224 / .69 = 3223 mm²

Since it is retracting, add the rod area to the required area.

5 + .7854 = 5.7854 sq. in.                                               3223 + 499 = 3722 mm²

5.7854 / .7854 = 7.366                                                    3722 / .7854 = 4803

√ 7.366 = 2.71” actual size                                             √4803 = 69 mm required

3” = standard size                                                           80 mm standard 

 

45° Force = 250 / sine of 45° = 353.5 lbs.                   1,573 N

353.5 /100 psi = 3.54 sq. in.

3.54 + .7854 = 4.33 sq. in

/.7854 = 5.5

√5.5 = 2.3”                                                                      58.42 mm required

Standard size = 2.5”                                                      60 mm standard 

 

60° Force = 250 / sine of 60° = 288 lbs.                    1,284 N

288 / 100 = 2.88 sq. in

2.88 + .7854 = 3.67 sq. in

3.67 / .7854 = 4.67

√4.67 = 2.16” required                                                 54.8 mm required

Standard size = 2.5”                                                     60 mm standard

Notes:

  1. When lifting anything in a sling, from a common hook out to the end of perhaps a beam, the load on the chains will be greater than the actual load if the angle is less than 30°. This is true even when hanging a picture on a wall.
  1. 250# / sine of 90° = 250 lbs. per cylinder. That’s the same as lifting straight up per cylinder.

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