Stress in cylinder
Stress in cylinder
(OP)
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
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





RE: Stress in cylinder
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)
RE: Stress in cylinder
RE: Stress in cylinder
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
RE: Stress in cylinder
You still can't do a simple F/A to get stress and apply your factor of safety?
Regards,
Cockroach
RE: Stress in cylinder
RE: Stress in cylinder
RE: Stress in cylinder
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
RE: Stress in cylinder
RE: Stress in cylinder
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
RE: Stress in cylinder
RE: Stress in cylinder
RE: Stress in cylinder
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
RE: Stress in cylinder
catch 22 ?