Stress in cylinder

[petermecheng]

Hello all,
I am new here so please be kind.

I have a problem that i need to solve.

I have a cylinder ID:140mm OD:180mm, so wall thickness is 20mm.

One end is closed (thickness of 20mm) with a 38mm diameter hole in the centre for a thread and the other end is open.

I want to jack the end with the thread hole, for example I will say a force of 100N.

The yield stress of carbon steel 1090 is 250 MPa and the UTS 841 MPa.

I will want a factor of safety of 2.

My question is, how do i calculate the stress the cylinder will be under and whether it will fall below the allowable stress stated above. What equations / assumptions should be used?

Many Thanks

 

[rb1957]

geez man ! don't ask us to be kind ! ... that's like swimming in the shark tank with an open wound ...

so the cyclinder is upside down (closed end at the top) and you're applying a force of 100N at the centre, yes?
sort of like a jack stand ... it not like the cycliner is filled with fluid and under pressure, yes?

for the body of the cyclinder stress = P/A, P = SF*Papp, yes?
but what is the allowable ? Fcy ?? how long is the cyclinder ? ('cause it could buckle)

for the endcap you could slove from first principles (look up Timoshenko "PLates and Shells") or you could look up Roark "Formulas for Stress and Strain" ... he gives a very complicated expression for the moment in the plate ... assuming simply supported edges ...
r = 19mm, a = 80mm, r/a = 0.25, Km = 0.25 (ish), w = 100/(pi*38) = 1 (ish), M = Km*w*a = 0.25*1*80 = 20, stress = 6M/t^2 = 6*20/(20^2) = 0.3MPa
(but then you wouldn't've expected much with only 100N applied)

 

[JStephen]

Refer to Roark's book. Quickest way would be to assume the plate is simply supported outside and loaded inside. If you want to get more involved, you can use the expressions for both plates and cylinders to find the moment at the junction which gives equal rotation to both.

 

[Cockroach]

You can't use "canned recipes" out of engineering books or technical manuals unless they represent the case PeterMechEng is pursuing. Obviously this is more than just a simple pressure vessel application. You have a condition of asymmetric loading, the 100 N load adds a longitudinal stress to the wall, over and above the reactions of the end caps.

Fortunately this is a very well documented problem. You can solve it using the Von Mises-Hencky Theorem and modify the longitudinal stress by adding a normal load to the wall. It's just that there is a tremendous amount of mathematics to the work. I'll give you the equation you need, but not the proof behind the mathematics.


S = sqrt(3 pi^2 P^2 D^4 + 16 F^2) / [pi (D^2 - d^2)

S is the stress, P is internal pressure, D is outside vessel diameter, d is internal pressure vessel diameter and F is the applied load normal to the wall. If F is defined in three dimensional space, then you must use the normal component experienced by the wall and the latter two vectors contribute to shear in their respective directions, i.e. orthogonal to the wall in the radial and hoop planes.

This stress is a principle stress, so any shear has been rotated into the principle plane using Mohr's Sphere. Of consequence is that you can apply the FOS = SY/S directly to the equation thus simplifying the arithmetic once you have your material yeild stress (SY) and computed stress (S) given the pressure vessel geometry and boundary value input load (F).

Obviously thin wall pressure vessel, which is the reduced case to the above theory for F=0 and wall thickness less than ten times the inner diameter, does not apply. Again, the normal load F adds to longitudinal stress which is well beyond the traditional assumptions made in derivation of typical equations.

Regards,
Cockroach

 

[Cockroach]

I re-read your problem and noticed there was no pressure. This is not a pressure vessel problem, just a very simple statics problem. You have a vertical load on a cylinder threaded at one end and open at the other.

You still can't do a simple F/A to get stress and apply your factor of safety?

Regards,
Cockroach

 

[rb1957]

i guess the OP needs to clarify whether the cyclinder is pressurised or not ...

 

[hokie66]

Gee, you guys make a simple question complicated. How could it be pressurised with one end open and a hole in the other end? If the cylinder is short, P/A is about 10 MPa under his 100 kN load. No problem. If it is long, then buckling has to be checked.

 

[Cockroach]

That's the problem Hokie66. The guy is asking for something out of a high school physics textbook and can't do the mathematics. Something wrong here.

As mentioned, it could be as simple as force over wall area. But RB1957 elaborated on "fluid filled" or "pressurized", because the original post was vague. Maybe his oversight, maybe not. That kind of questions the "open ended" argument you refer to.

Finally, the original post is about OD, ID and material properties. Nothing on cylinder length. So what's with the buckling? So of important in the Euler calculation.

Regards,
Cockroach

 

[hokie66]

I'll retreat to the safety of the Structural forums.

 

[Cockroach]

I don't think you need to do that. What we're saying is

Stress = force / area
S = 4 F / pi (D^2 - d^2)
D = 180 mm, d= 140 mm, f = 100 N
S = 400 / 40212 N/mm^2
S = 9.95 kPa

SY = 250 MPa so you have an astronomical FOS. This is because your input force is extremely low, only one hundred newtons, not kilo newtons as mentioned above. So I'm going by the information in your post.

Simple high school physics. Hope it is not your homework problem.

Regards,
Cockroach

 

[hokie66]

My mistake in misreading the load as kN rather than N. We building folks always express loads in kN, and the stress under 100 kN is only 10 MPa, so the FS is still a lot if the member is not too long, which we don't know.

 

[chicopee]

I think that the thread will fail first before the cylinder during the jacking operation assuming the jack is threaded and inserted into the threaded end. That is if I am reading the OP correctly.

 

[Cockroach]

Yeah, I'm getting the drift there are several issues with the original post. But no matter. Force is linear with stress, the metric system simplifies the arithmetic. So you have 9.95 MPa stress in a cylinder of 250 MPa yield. Your factor of safety is just over 25, pretty strong.

Without information on the thread, you can't comment on the state of stress there. I suspect the inner thread relief will have much higher stress because of the thinner wall. That's the best I can do on that.

Hope this helps, clarity in the original post would of been less "stressful" I guess.

Regards,
Cockroach

 

[rb1957]

the problem with clarity is that if you don't know what's important you don't know how much detail to put in (and you'll probably miss the important stuff anyways). but if you know what's important, then you probably know the answer ...

catch 22 ?