Hydraulic cylinder for a scissor lift

[durga20]

I am designing a scissor lift which is needed to lift a load of 1.5 ton, the hydraulic cylinder is at an angle of 12 degree from the base, and the cylinder is pivoted at a distance of 45 mm from the centre hinge of the scissor lift, what is the force acting on the scissor lift.

I would be very grateful to you people.



This is how it looks like, L=750, X=The value where the force required is very less

 

[kennyd]

Welcome to TBN

Try the calculators here:Baum Hydraulics Corp :: Spec Calculator

#2 might have your answer?

 

[J_J]

What Kenny said. The angle as drawn, looks about 15 to 20 degrees.Just put the angle of the cylinder into the calculator. That will be your start push force. As the angle changes, your push force should increase, because you are moving toward the 90 degree push force which is the maximum. Although you cyl can push with a lot of force, you see some of those lifts with legs of 1/4, etc. I would suggest to make the legs 3/8, or 1/2 in thick. and since you will put holes in about the middle for the pivot point, I would make the legs about 2 to 3 in wide, to give enough metal around the hole.

How much weight are you wanting to lift? I have one of the motorcycle lifts with a foot pump, and it is kind of stout. The tag is missing, so I don't know the weight limits.

 

[durga20]

Thanks

The lift is intended to lift a load of 1.5 ton, and I need a procedure of how to design the arms of that lift