- 13 Aug '16 18:57 / 3 editsI'm having trouble wrapping my head around something that is apparently 'obvious' ( as I cannot find explaination) in the derivation for the Hoop Stress acting in a thin walled pressure vessel. Please see page 6, figure 3.3 of the linked pdf.

http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect03.d/IAST.Lect03.pdf

A force balance is applied in the following manner:

2•F_hoop - F_p = 0

2•σ•t•dx - p•(2•R•dx) = 0

σ = p•R/t

I 'm am looking for clarification on why p•dA = 2•R•dx in this derivation, as opposed to p•dA = p•π•R•dx?

Can anyone lead me to water on this? - 13 Aug '16 19:58 / 1 edit

Sorry, no other "pi's" in the derivation. The forced balance isn't explicitly shown in this particular example, so you wont see it. However, I can assure you it is the underlying model.*Originally posted by twhitehead***Not having gone into it in detail, my first guess is there is also a pi in the other force and they cancelled them out without saying anything.**

You have told us which figure, but not where your equation occurs?

EDIT: it has something to do with 2•R•dx being the projected area of the cylindrical surface, I just don't understand why it is used. - 13 Aug '16 20:24 / 1 editThe diagram here is clearer:

http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/thin-walled-pressure-vessels

The force in a particular direction is proportional to the area of a cross section along the pipe, so pi doesn't take part.

So, a non-circular pipe would experience tension proportional to a cross section. A square pipe would experience the most tension in the corners. - 13 Aug '16 22:24

Because the element of area in question is*Originally posted by joe shmo***I'm having trouble wrapping my head around something that is apparently 'obvious' ( as I cannot find explaination) in the derivation for the Hoop Stress acting in a thin walled pressure vessel. Please see page 6, figure 3.3 of the linked pdf.**

http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect03.d/IAST.Lect03.pdf

A force balance i ...[text shortened]... •R•dx in this derivation, as opposed to p•dA = p•π•R•dx?

Can anyone lead me to water on this?*normal*to the direction of increasing angle, here θ. The element is a rectangle of width dx and length equal to the diameter of the tube. The equation he's using is:

F_i = ∫σ_ij dA_j

Where the integral involves the inner product of a rank 2 tensor and a vector valued area element. We also have that:

F_i = ∫p dA_j

By the symmetry of the problem σ_ij is constant in (r, θ, x), the pressure is assumed to be constant as well. So we have only to work out the areas to get:

A) F_r = p • 2πRdx = σ_rr • 2πRdx => σ_rr = p.

B) F_θ = p • 2Rdx = 2 • σ_θθ • tdx => σ_θθ = pR/t

C) F_x = p • πR²dx = σ_xx • 2πRt => σ_xx = pR/2t.

In A the relevant area is the inner surface area of the tube. In B the relevant areas are shown in diagram 3.3 b. For the stress it is the two rectangles of pipe of area tdx each and for the pressure term the rectangle is the one bounded by the two area elements from the pipe. For C the stress term is the cross-sectional area of the annulus of pipe of thickness t and for the pressure term the cross-sectional area of the pipe. - 13 Aug '16 23:58 / 3 edits

You have to be a little careful with that. In this case there are no off diagonal components of the stress tensor so your wording of: "The force in a particular direction is proportional to the area of a cross section along the pipe, so pi doesn't take part.", is good enough, but had there been off diagonal components then the direction the component of force acts in is not normal to the area element for the off diagonal components of stress. Which should be intuitively obvious as a shear involves planes moving along each other. So your following statement about non-circular pipes has two problems, one is the difficulty that because the problem is no longer symmetric we cannot simply write F = ∫ σdA = σ ∫ dA - in other words the force is no longer proportional to the cross-sectional area. The other problem is that there are now shearing forces. If a square pipe is of thickness t and has sides of interior length L, it makes sense to do the problem in Cartesian coordinates with the origin at the centre of the pipe and the z-direction the one a fluid would flow along, then we get:*Originally posted by twhitehead***The diagram here is clearer:**

http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/thin-walled-pressure-vessels

The force in a particular direction is proportional to the area of a cross section along the pipe, so pi doesn't take part.

So, a non-circular pipe would experience tension proportional to a cross section. A square pipe would experience the most tension in the corners.

The x-component of force on the walls parallel to x (in other words the one stretching from {-L/2, L/2, z} to {L/2, L/2, z}) is:

F_x = σ_xx ∫dydz = 2σ_xx tdz = p∫dydz = pLdz

=> σ_xx = pL/2t

similarly

σ_yy = pL/2t on the walls parallel to y.

On the walls normal to x:

F_x = σ_xx ∫dydz = σ_xx Ldz = p∫dydz = pLdz

=> σ_xx = p.

and again on the walls normal to y we similarly obtain σ_yy = p

For σ_zz we need the cross-sections of the pipe material and it's interior. The area of the interior is L², the area of the pipe material is 4(L + t)t ~ 4Lt, so we get:

σ_zz = pL/4t

This leaves the shearing terms: σ_xy, σ_yz, and σ_zx. The last two must be zero on symmetry grounds.

The total force on any one of the faces is pLdz. This has to be cancelled by forces at the edges, the force at each edge is pLdz/2. The shearing force being transmitted through an area element of the edge with coordinates (x, L/2, z) away from the centre of the face is xpLdz. The area is tdz so we get:

σ_xy = xpdz/tdz = xp/t

Which is zero at the centre point of a face and maximal at the corners, with the maximum stress max(σ_xy) = pL/2t. - 14 Aug '16 09:15 / 1 edit

Yes. I took the assumption of an idealised thin walled vessel a bit too far. Perhaps an oval shape would be more accurate. My point, however, is that the tension being calculated is not dependant on the exact shape of the vessel - hence pi isn't involved. I believe a square pipe whose side was the same as the diameter of the round pipe should have the same tension as the round pipe at the centre of its sides.*Originally posted by DeepThought***You have to be a little careful with that.** - 14 Aug '16 14:29

It seems as though I'm having trouble reconciling the fundamental difference between "F_r" and "F_θ" in your (and everyone elses) explanation. The way im looking at it (albeit incorrectly) is that "F_r" casues the stress "σ_θθ", yet "F_r" does not directly appear in the force balance for "σ_θθ", this was confusing for me.*Originally posted by DeepThought***Because the element of area in question is***normal*to the direction of increasing angle, here θ. The element is a rectangle of width dx and length equal to the diameter of the tube. The equation he's using is:

F_i = ∫σ_ij dA_j

Where the integral involves the inner product of a rank 2 tensor and a vector valued area element. We also have th ...[text shortened]... e annulus of pipe of thickness t and for the pressure term the cross-sectional area of the pipe.

I guess what you are saying is that the forces internal to the broken out tank section are not of "direct" concern for the removed element. Only those externally applied to the element keeping it in static equilibrium are relevant for the force balance?

So in the "θ" direction we are to treat the pressurized fluid as if it were contained by a plane wall of area (2Rdx) with a pressure "p" acting over its entire surface from the pressurized fluid external to the element (that is, it is not important "directly" what force "F_r" is being applied to the tank wall internally in the element, as it is not contributing to the equilibruim of the element.

Am I near the proper interpretation of this yet? - 14 Aug '16 18:54

Your statement depends on the symmetries present. In the case with circular cross-section we have symmetry along both the axial and circumferential (hoop) directions. If we alter the profile so that we lose circular symmetry then we'd expect the stress component σ_θθ to pick up a dependence on θ. What your statement hangs on is an assumption that the ratio of the length of a line passing through the centre to a point on the exterior surface and the shorter distance to the interior surface doesn't depend on θ. Imagine filing one side of a circular pipe down so it's half the thickness along a line parallel to the axial direction we start increasing the load, intuitively we expect the pipe to fail along that line which demonstrates that the stress must be highest at the weakness. However for a pipe with elliptical cross-section and the two ellipses with the same eccentricity the diagonal components of the stress don't change.*Originally posted by twhitehead***Yes. I took the assumption of an idealised thin walled vessel a bit too far. Perhaps an oval shape would be more accurate. My point, however, is that the tension being calculated is not dependant on the exact shape of the vessel - hence pi isn't involved. I believe a square pipe whose side was the same as the diameter of the round pipe should have the same tension as the round pipe at the centre of its sides.**

In the above calculation for a pipe of square cross-section my L is doing the same job as 2R, they come out the same all along the faces, not just at the centre points. So you are right about that. What varies is the shear stress, which is completely absent in the cylindrical case, it's zero at the centre of faces and a maximum at the corners, as we'd expect.

For an oval shape the pressure is trying to turn our oval cross-section into a circular one, so there'll still be shear components. The area of an ellipse is πab where a and b are the radii of the major and minor axes respectively. Suppose R1 and R2 are the inner major and minor radii and λR1 and λR2 the outer ones and let λ = 1 + s and choose s=t/R1 where t is the thickness along the major axis. The area enclosed by the pipe is A = πR1R2, and the area bounded by the outer surface of the pipe is πλR1λR2 = λ²A, so the area bounded by the inner and outer ellipses is πR1R2(λ² - 1) = λR1R2(s² + 2s). We can neglect the term in s² to obtain:

σ_zz = pπR1R2 / 2sπR1R2 = p / 2s

recalling our definition of s = t/R1, we get

σ_zz = pa/2t

Which is the same as for the circular case, up to a correction term of order s².

The formula for the distance to the edge of an ellipse from the centre point is:

r(θ; a, b) = ab/√[(b cos( θ ))² + (a sin( θ ))²]

Assuming the circumferential stress doesn't vary in the z and r directions we we can try to calculate it. Using the formula for an ellipse:

r(θ; a, b) = ab/√[(b cos( θ ))² + (a sin( θ ))²]

we get a handy expression for r(θ; λR1, λR2):

r(θ; λR1, λR2) = λ²R1R2/√[( λR2 cos( θ ))² + (λR2 sin( θ ))²] = λ r(θ, R1, R2)

so we can write:

σ_θθ( θ ) = p r(θ; R1, R2) / [r(θ; λR1, λR2) - r(θ; R1, R2)] = p / ([r(θ; λR1, λR2)/r(θ; R1, R2) - 1] = p/(λ - 1) = p/s = pa/t

which is identical to the result for the circular case.

By axial symmetry we have that:

σ_zr = σ_rz = σ_θz = σ_zθ = 0

Which leaves σ_θr = σ_rθ and the calculation for this is painful. - 14 Aug '16 19:04

I believe it can all be explained without shear components at all. The oval trying to turn into a circle can be fully explained by the fact that the tension is highest where the cross section is widest, thus is the wall is reasonably flexible it will get 'pulled' into a circle. Hard to fully explain without a diagram. Maybe tomorrow when I have more time.*Originally posted by DeepThought***For an oval shape the pressure is trying to turn our oval cross-section into a circular one, so there'll still be shear components.** - 14 Aug '16 23:19

To do his calculation he's chosen a bisecting plane, see figure 3.3b, the approach is to imagine cutting the object in half and working out what forces would be needed to stop it from moving, but leaving the pressure magically still acting. The circumferential stress times the area of the two exposed faces of container (of combined area 2tdz) is equal to the force pushing the hemicylinder in the*Originally posted by joe shmo***It seems as though I'm having trouble reconciling the fundamental difference between "F_r" and "F_θ" in your (and everyone elses) explanation. The way im looking at it (albeit incorrectly) is that "F_r" casues the stress "σ_θθ", yet "F_r" does not directly appear in the force balance for "σ_θθ", this was confusing for me.**

I guess what you are sayin ...[text shortened]... tributing to the equilibruim of the element.

Am I near the proper interpretation of this yet?*direction normal to the bisecting plane*. This is the key point, the gas exerts a force normal to the container's inner surface at each point on the interior, but, in general, only a component of this force is normal to the bisecting plane. When you get this point you'll understand his calculation. The force on a point on the surface is p dA where dA is a vector valued element of area. To work out the force along the axis normal to the plane of bisection we need ∫pn∙dA = p∫n∙dA where n is the vector normal to the bisecting plane. The integral of this is just equal to the area of the bisecting plane. This means that the force in the direction normal to the bisecting plane on the hemicylinder is equal to the pressure times the area of the bisecting plane.

I think there are a few ways of getting the result, one might try using energy considerations to find the result (calculate the work done on the container by the gas and this equals the elastic energy stored in the material). An alternative approach is to use a thin walled model in which the casing is treated as a boundary between two fluids, which allows us to use the Young-Laplace equation (see [1] and [2]). The Young-Laplace equation is:

∆p = -γ ∇∙n = γ (1/R1 + 1/R2)

where R1 and R2 are the principle radii of curvature and ∆p is the pressure drop across the boundary, the last term is the divergence of the unit vector normal to the surface. From now I'll follow the author of the Lecture Notes and replace it with p. For a cylinder R1 = R and 1/R2 = 0, for a sphere R1 = R2 = R. γ is the surface tension for the boundary between two fluids and in the approximation we are using is equal to t σ_θθ, which gives us:

p = t σ_θθ (1/R1 +1/ R2)

and so, using the values of R1 and R2 for cylinder and sphere we have:

σ_θθ = pR/t for a cylinder.

σ_θθ = σ_φφ = pR/2t for a sphere.

[1] https://en.wikipedia.org/wiki/Cylinder_stress

[2] https://en.wikipedia.org/wiki/Young%E2%80%93Laplace_equation - 15 Aug '16 00:45

But it's not, see my above calculation, for σ_θθ for a container whose profile is the area enclosed by two concentric identically oriented ellipses with the same eccentricity. The result is pR1/t where t is the thickness of the container along the major axis and does not depend on θ. There simply has to be a shearing force to decrease the eccentricity of the profile of the pipe, which is what will tend to happen if the pressure is increased enough.*Originally posted by twhitehead***I believe it can all be explained without shear components at all. The oval trying to turn into a circle can be fully explained by the fact that the tension is highest where the cross section is widest, thus is the wall is reasonably flexible it will get 'pulled' into a circle. Hard to fully explain without a diagram. Maybe tomorrow when I have more time.**

Something to note, since the stresses are trying to turn the ellipse into a circle, the points where the shearing stress is zero must be away from both the major and minor axes. The reason for this is that the curvature must change at each point on the ellipse until it's a circle. The curvature is higher than a circle of radius R1 at the two points on the major axis and smaller than a circle of radius R2 at the two points on the minor axis. There are 4 points where the ellipse has with the same curvature as a sphere of radius equal to the distance between those points and the centre of the ellipse. The shearing stress will be zero at those points.

To calculate the shearing stress I'd need to get the distance along the circumference as a function of θ which involves doing an elliptic integral. This really isn't trivial. So I can't really present a calculation to demonstrate this. - 15 Aug '16 07:09

I am not saying there are not sheering forces, only that there is no need to consider them. Imagine that the surface is a plastic bag or balloon ie only provides tension but no stiffness. Then the only force we worry about is the force tangent to the surface at any point. The system will automatically end up with the tension uniform everywhere, which is a circle.*Originally posted by DeepThought***But it's not, ......** - 16 Aug '16 01:26

Ok... so each differential element of area has its own normal vector associated with it. The dot product of each elements normal vector is taken with the normal vector of the bisecting plane. This is why some explanations I have saw referred to it as the*Originally posted by DeepThought***To do his calculation he's chosen a bisecting plane, see figure 3.3b, the approach is to imagine cutting the object in half and working out what forces would be needed to stop it from moving, but leaving the pressure magically still acting. The circumferential stress times the area of the two exposed faces of container (of combined area 2tdz) is equal t ...[text shortened]... pedia.org/wiki/Cylinder_stress**

[2] https://en.wikipedia.org/wiki/Young%E2%80%93Laplace_equation*projection*( via**n**•d**A_n**) of the hemicylindrical surface on the bisecting plane.

So, the method of integrating the pressure distribution over the surface and applying the resultant*normally*to the center of presure is only applicable if the surface is planar and (or) the distributions "field lines" (I don't know what these are specifically referred to as in physics, maybe that is a completely improper term) are all parallel to each other( which in the cylindrical/spherical application of the pressure vessel they are not).

I hate to ask, and i'm sure you'll correct me if I'm not,... but do I now seem to be rounding the bases, or did I botch something up in the statement above?

Thanks - 16 Aug '16 09:49 / 1 editSee this diagram for reference:

http://whereitsat.co.za/pipe.png

Assume higher pressure inside the pipe than outside.

Given a pipe, uniform along its length with cross section A and B, the tension in the wall normal to the plane shown in blue is proportional to the area of the plane. The shape of A and B are irrelevant. So if the wall of the pipe is normal to the plane at the contact points (yes my diagram is terrible and doesn't show that properly), then the tension in the wall at that point will be proportional to the area of the plane.