Hydraulic Cylinder Calculation

First things are your numbers you are confusing--- PSI (pounds/square inch) and metric 700mm cylinders. While it is convertible stay with one or the other.

Second thing how does 4700 ton total equal 1600 (X2), more like 2350 tons on each cylinder.

700 mm bore at 6000 psi comes out to about 1788.48 tons. better than 1600 tons but still not 2350 tons.

deflection on the shaft has not been considered if you allow for this extra tonnage may be required.

Hi Mike L

OOOh! I shall have to watch this one as I agree with your calc's although there may be some mechanical advantage somewhere that we don't know about.

As the OP's name suggests "it's a MISTRY" I think Toya said something similar too!

The total 4700 tones of load is according to FOS = 1.5, while calculating the FOS = 1 this load becomes 3133 Tones, thats why i choose 1600 tones push force capacity Hydraulic Cylinders, The material i want to bend in this bending machine, has following technical specification : Tensile Strength : 320-440 N/mm^2, Yield Strength : 380 N/mm^2 material designation equivalent as ASTM (American) 1040

I do not know how you determined the value because it is function of the distance between the two bottom rolls and of the radius you want to achieve (minimal value).

According to my computations the mentioned force occurs for a distance of 1450mm and a radius of the bend plate of 3800 mm in the middle.

With the Cs =1.5 and an efficiency of 0.95 at the ram level the diameter is 617mm for the 6000 psi you mentioned.

Is that coherent with your basic values for the computation?

I dare make a remark with respect to the mechanical values you give for the plate material: the elastic limit is higher than the lower ultimate stress. This is usually not the case. Are you sure of the values?

Now to take a 1600 tones output force , the pressure will be around at 6000 PSI. kindly correct me,

I'd expect closer to 2800 PSI, specifics are untenable without knowledge of: hot/cold process? desired single pass arc capabilities?

More information required...

I would be interested to know how you estimate the pressure to be closer to 2800 in stead of 6000.

From what he gave this was not possible without some assumptions and detailed computations. Since your results are so far from mines I am very interested to know why.

I'd be interested in the kind of pump that will deliver 6000 PSI continuous and at suffient volume to be reasonably fast?

The OP doesn't need a pump capable of 6000 PSI. The description is of 6000 PSI force, so the actuator would need only hold against this force created as the plate is driven through the pinch roll machine.

It's the motor doing this driving I'd like to see

Instead of having to choose between speed and force you can have both.

Scroll down to Hi Lo pump circuit on this page:

http://www.hydraulicspneumatics.com/200/eBooks/Article/True/44617/

I may be wrong but for such a huge plate I think the bending has to be progressive so that they will need the pressure generator to go from one step to the other.

I also have doubts that the 800 and 700 mm rolls are stiff enough and think that a control of upper roll position (parallelism to the other rolls) will be necessary. Without this control the risk to get cones instead of cylinders is on my opinion real. This will impose due to pressure a pump/cylinder and instruments for position measuring. If the rolls are not stiff enough then even with this control there could be problems.

The motor for the rolls will not be -as power- very big if all of them are on roller bearings friction not too being so important power requirements will be low but the transmission will be of a very high reduction.

The flows being any way small there are pumps available up to 7000 psi. But coming back to the dimension and pressures you mentioned I made the following computations based on the model in the sketch. As you will see the forces are higher than it appears at first view.

Nice...

The OP's description details two (2) 27" cylinders per/side or two cylinders at each end of the top roll.

I'm assuming because of lack of input this machine is comprised of three or four rolls. If three rolls the orientation will differ from your graphic in that two rolls will clamp the sheet with the top roll moving at about 35° - 38° angle towards a driven static and pinch roll.

Yes I was meaning the trans/motor relationship is interesting always.

Is it this way you think?

The right system I saw for small dimensions and the left for bigger up to 4 m wide (conversion factors in previous text). The right has a disadvantage when forces are high since the friction between the "pinch" rolls has to overcome the horizontal component of F1 and this is no light stuff for a plate as mentioned. This could be the reason for a big motor and heavy transmission. The left version needs to offer ONLY the torque able to overcome the friction of all rolls in their bearings. I presume tis is the reason why I saw it for heavier parts to bend.

As usual I may be wrong since I detain not yet the truth, may some time in the future but not now.

Expect comment.

Some of the large projects I've involvement in were all of the right side configuration, by large I mean 3" thickness x 31' A36 plate. In a three roll configuration F2 would be absent but in a four roll F1, F2 and the lower F would be actuated; this one the pinch roll is necessary.

Actually the rendition on the left we used for lite stuff bars, angles and strips etc..

Hi. My name is Sebastián and I´m working on the design a machine simmilar.

Reviewing the calculations shown here I think that an error exists in this expression:

Frtheor=(M/L)*(1+0.5*(t+d)/R)

I think it should be:

Frtheor=2*M/L*(1+0.5*(t+d)/R)

Becose "e" is (L/2)/(1+0.5*(t+d)/R).

Sorry to reopen an old issue, but did not want to miss this as it can create confusion.

Let me know if I'm wrong. Thank you very much, very good forum.

Greetings!

May i know how to get the Dc = 685.8mm?

Thanks a lot!

As you noticed my work was not considerd as "good answer" so please ask for explanations at other sources

I am confuse, many confusion, now my confusion about top roll diameter also, Re-calculating everything from the beginning.

How many rolls total, how many rolls driven and dia of each roll, please?

Bottom Two roll have positive gear box drive, Approximately 3 m/min

This is technical data sheet from " J. IRIZAR COMPANY LTD." CANADA, i have downloaded from the net, this can also help.

Is it the one you mean?

It looks very similar to the concept I assumed. It seems that both principles are used. On my opinion this concept requires less power for the rolls rotation.

Could you please mail your sketch with dimensions since as you have seen without dimensions it is quite difficult to check if your computations are or are not right.

Images were loading before, but now its unavailable, i don't know why?

Dimension Top Roll : Ø 800mm Bottom Roll : Ø 700mm Width : 7000mm Bottom Roll Supporting rolls :Ø 300mm Bottom Roll Center : 900mm (Adjustable 1000mm Max. To 700mm Min.) Rolling Speed : 2.5 to 3.5 m/min

how can someone manufacture any types of machine just to see its picture?? its tuff to copy a machine design, because there is no dimension, no details in the picture, i am not copying from that J. Irizar Bending machine picture, but i can say, my concept is also same as that machine which is in the picture, In that Irizar, bottom two rolls having one stand (Housing), but i am designing two different Stand(Housing) for each bottom rolls, this stands are movable with help of Hydraulic Cylinder.

I would like to share my this project to all of you, whatever visitors saying, i really don't mind that, this is i am posting, a foundation frame and Side wall base frame (non drive side). if someone need, CAD Drawing with dimension, feel free to ask.

DEAR KAUSHAL,

WE HAVE ONE SMALL FABRICATION UNIT, WE DONT EFFORD COSTLY READY MACHINE FROM ANY SUPPLIER, SO WE WANT TO DEVELOPE MACHINE AT OUR FACTORY ONLY.PLATE WIDTH IS 2000 MM, PLATE THICKNESS 16 MM,MATERIAL IS MS, PREPUNCHING REQUIRED.IF YOU HAVE DESIGN AND SKETCH AND WILLING TO SHARE YOUR KNOWLEDGE ,PLS SEND ON MY MAIL ID dsp1010@hotmail.com (3 roll plate bending machine, shell making machine)

kindly send the picture & cad drawing of drive side also

I would appreciate CAD drawings in AutoCAD .dwg Format.

Thanks on advance

Hi. My name is Sebastián and I´m working on the design a machine simmilar.

Reviewing the calculations shown here I think that an error exists in this expression:

Frtheor=(M/L)*(1+0.5*(t+d)/R)

I think it should be:

Frtheor=2*M/L*(1+0.5*(t+d)/R)

Becose "e" is (L/2)/(1+0.5*(t+d)/R).

Sorry to reopen an old issue, but did not want to miss this as it can create confusion.

Let me know if I'm wrong. Thank you very much, very good forum.

Greetings!

Hi Sebastian, you rae right there is an error unfortunately I cannot explain whare it comes from!

Thank you for looking in such a detailed way to the threads when you think that it took 2 years to have somebody to notice it, it says a lot about the interest of participants!

Congratulations.

Now you are partially right:

M= F1*e → F1= M/e with e= R*sinφ and sin φ= 0.5*(L/R)/a with a= 1+0.5*(t+d)/R

F1= M/(R*0.5*(L/R)/a) = 2*a*M/L

F= 2*F1*cos φ =4*M/L*(a^2-0.25*(L/R)^2)^0.5

The other component F2= F1*sinφ leads to a central section compression and thus modifies slightly the stress distribution and moment value.

If you have any other question related to your project do not hesitate to ask I shall be dekighted to answer as far as I can. Better you use the "private channel".

Hi nickname. Thank you very much for check this so quickly. Your work has been useful for me to check mine. In my case I have made the hypothesis of elastic-perfectly plastic material and considered resistant plastic module:
Wp = (B * t ^ 2) / 4
The internal moment is in my case:
M = σy * (B * t ^ 2) / 4
The difference is negligible.
An important difference happen with the force, so I began to investigate arriving at:

F1=(M/L)*(2+(t+d)/R)

Thanks again for your predisposition. Keep in touch.
Regards