Stretch Forming Force Calculation

[ZotariJohn]

Hi all,

I have started working on a problem at work.

We have installed a machine for 'stretch forming'. The machine clamps a long sheet of aluminium at both ends, while a convex shape steel die is raised in the centre. The die is raised by a hydraulic system, causing the metallic sheet to stretch over it, hence forming the shape of the die. I am trying to calculate the force required to stretch the sheet, without returning to its original shape(springback), and without necking(rupture). Please see the attached image for a visual representation of the tool.

Now for what I have been using for my method:

1) The sheet being stretch formed is Aluminium 2024 T3 temper.
The Ultimate Tensile Strength(UTS) of the material is 470Mpa
The Yield Strength (YS) is 325Mpa
The sheet is 98.4"(2.5m) wide and 0.120"(0.003048m) thick.
ie) Cross Secional Area is 11.808In² (0.00762m²)

2) Formula:
The UTS=Force(N)/Area(m²)
thus Force(N)=UTS*Area(m²)

3) The above equation would lead to the force required to rupture the material but I just want the force to stretch it, so I need to be between the YS and UTS

4) from points 3&4 I now have Force(N)=((UTS+YS)/2)*Area(m²)
This is the average of YSand UTS, multiplied by the area

5) this is fine for a scenario of tensile testing(see image 'i'), but my questions are:

Q1) Is this the same force for the case I present (image '1' & '2')?
Q2) The die is lubricated prior to forming, so do I need to consider frictional force?
Q3) As the sheet is fixed on both ends, does the force that I apply with the ram replicate a force acting tangential to the sheet edges?


Any help would be greatly appreciated,

Thanks in advance,

John

 

[berkshire]

You are taking the sheet to first yield only FTY.

There is a very fine line between first yield and rupture.
It is unusual to stretch form in the T3 temper unless the degree of forming is very small. The material tends to rupture very easily
It is more common to heat and quench, then form in the AQ or W temper.
B.E.

The good engineer does not need to memorize every formula; he just needs to know where he can find them when he needs them. Old professor

 

[Taz99]

Q) Is this the same force for the case I present (image '1' & '2')?
A) No - For a stretch forming operation the vertical load from your hydraulic ram will be reacted by tensile membrane loads within the plane of the Al sheet (think along the lines of a drop of water supported by a stretched sheet of cling film). This is a non-linear large deflection problem.
I don't think you have a great chance of accurately predicting this using a hand calc. Roarks Formula's for Stress and Strain was the first place I thought to look, in Chapter 11 closest I could find was formulas for large deflections of rectangular plates under uniform pressure held at all four sides - not really close to your case.
A non-linear FEA has a much better chance of getting you an estimate of your ram load.

 

[berkshire]

compositeguru
Depending on the age of the machine you have, most often the ram is controled by distance rather than pressure.
With stretching of 2024-T3 at 8 to 10 percent max elongation.
Do you have any movement capability on your jaws? are they Fixed , or power retracting?
B.E.

The good engineer does not need to memorize every formula; he just needs to know where he can find them when he needs them. Old professor

 

[ZotariJohn]

Firstly thank you for your comments,

berkshire - I will ask if it must be in the T3 temper or if any other treatment is acceptable. Also, the machine is quite old, the jaws are not power retracting. They are fixed rotational. The CNC controller does require a distance, but I want to make sure that the force does not exceed the capability of the machine (450 UK tonnes)

Taz99 - I feared that might be the case, It will probably have to model the scenario and run a FEA sim tomorrow.

Cheers,
John

 

[berkshire]

compositeguru (Aerospace),
For getting a first cut at your loads, look at a rope sling on a load suspended by a cable at each end to a center point.
the shallower the angle, the greater the load on the cable, ( In this case your sheet.) so you can see, that as your ram first touches your sheet, you can be applying as much as 11 times the tensile load on the sheet, as the pressure you are putting on the ram.
This ratio will of course fall off rapidly as the ram stretches and wraps the sheet around the form block.
But based on the square inches you showed in your first post and Alcoa's published yield strength of 50,000lbs for 2024 T3, I get 250 tons required.
B.E.

The good engineer does not need to memorize every formula; he just needs to know where he can find them when he needs them. Old professor

 

[mfgenggear]

compositeguru

see this link

http://web.mit.edu/2.810/ts_temp/sheetmetal.ppt

it correlates to your formula
I have some data which I have to dig up
I will post it later

Mfgenggear

 

[ZotariJohn]

berkshire - I ran a simulation over the weekend, and as described in your last post, the initial force is high and reduces in a non linear fashion as the displacement of the ram increases. This is due to the change in angle(as you mentioned) and also the sheet is decreasing in cross sectional area, so requires less force. The edges of the die create high stress points which initially caused a rupture, so widening the jaws has fixed the problem.

mfgenggear - I have taken a look at the link and as you say, the formula given is similar, but if it were accurate for the initial force, it would soon fall outside the plastic region of the material due to change in cross sectional area, and rupture. If you have some data it would be good to compare to my sim.


Many thanks,
John

 

[berkshire]

compositeguru (Aerospace)
It sounds as though you now have a handle on the basics of the machine.
Whilst there is very little spring back on stretch formed parts, it is still there, usually about 1%. On small parts it can be discounted. On bigger parts like fuselage frames, it can be enough to require additional hand straightening.
Good luck with your new toy.
B.E.

The good engineer does not need to memorize every formula; he just needs to know where he can find them when he needs them. Old professor

 

[mfgenggear]

compositeguru

I did not find the data that I had stored.
all I can contribute is this

the percent elongation has to be a minimum in order not to rupture.
or it has to be formed in stages. (IE Anneal then form, if it's heat treat able.)

there is very little spring back because of the initial stretching.

contact the manufacture of the machine (if it is possible) ask them to assist.
there will be a fee, but it will be well to have the experts on hand.
otherwise you will be pulling your hair out by the end of the working day.

To my knowledge the formula F=(Ys + Uts/ 2 * A) is correct as long as the percent of elongation
is with in the requirement. then it will not rupture.
that is very little I can contribute, as I have forgotten it.
again hire experts to get you started, it will make life easier.

Good luck

Mfgenggear