- #1

Mech King

- 73

- 0

Hi guys,

I want to check that I am going about this problem in the right way:

The attached image shows a parallelogram type frame with a hydraulic ram located from corner to corner as shown (E to C). There is a load applied to a platform in the far bottom right hand side of the drawing. This load is -4635N. The frame is pin supported and fixed at the base E.

I have worked out that the force on the ram must be 16141N. I have taken moments about the Pivot D. Since the ram is the only thing stopping the frame from collapsing once the load is applied, then the ram and the applied load are the only factors to take into account to determine the ram force:

Ram force = [(640+450)X (-4635)]/313 = -16141N.

Note that the 313 mm above is the perpendicular distance from the ram centreline to the pivot D.

I have then gone on to use the method of sections to resolve the forces in the beams.

The platform creates a normal reaction “Fr” just below A and also an X and Y force at A.

Fr= (-4635x450)/150 = 13905N

Summing forces in the X and Y direction will give the force components at A, these are:

FAx = -13905N, FAy = 4635N.

It is at this point where I get a bit confused. I could use the method of sections to resolve the forces along member ABC. I presume I am right in assuming that given that members EB, CD and the ram (EC) are all two force members. I am torn between this method or could I simply treat the parallelogram BCDE with the Ram EC as two triangles and resolve the forces and then apply these forces the member ABC.

I have got my self into a bit of a pickle, on what seemed like a straight forward problem. Could somebody please advise me on how to proceed and whether my current approach is correct?

This has been bugging me for days.

Any advice will be a life saver,

Many thanks

Amir

I want to check that I am going about this problem in the right way:

The attached image shows a parallelogram type frame with a hydraulic ram located from corner to corner as shown (E to C). There is a load applied to a platform in the far bottom right hand side of the drawing. This load is -4635N. The frame is pin supported and fixed at the base E.

I have worked out that the force on the ram must be 16141N. I have taken moments about the Pivot D. Since the ram is the only thing stopping the frame from collapsing once the load is applied, then the ram and the applied load are the only factors to take into account to determine the ram force:

Ram force = [(640+450)X (-4635)]/313 = -16141N.

Note that the 313 mm above is the perpendicular distance from the ram centreline to the pivot D.

I have then gone on to use the method of sections to resolve the forces in the beams.

The platform creates a normal reaction “Fr” just below A and also an X and Y force at A.

Fr= (-4635x450)/150 = 13905N

Summing forces in the X and Y direction will give the force components at A, these are:

FAx = -13905N, FAy = 4635N.

It is at this point where I get a bit confused. I could use the method of sections to resolve the forces along member ABC. I presume I am right in assuming that given that members EB, CD and the ram (EC) are all two force members. I am torn between this method or could I simply treat the parallelogram BCDE with the Ram EC as two triangles and resolve the forces and then apply these forces the member ABC.

I have got my self into a bit of a pickle, on what seemed like a straight forward problem. Could somebody please advise me on how to proceed and whether my current approach is correct?

This has been bugging me for days.

Any advice will be a life saver,

Many thanks

Amir