I have introduced the following article in another post
<Thick walled cylinder Elastic tube -1> when I was struggling with
Lame equations as
"
I finally found an explanation and an answer to this question - a tricky application of the product rule of differentiation.
Cal Poly San Luis Obispo Mechanical Engineering
ME328 Introduction to Design
Cylinder Stresses
http://www.calpoly.edu/~jridgely/ME328/misc/cylinder.pdf
"
Most of the articles on Lame equations I have checked use an "edge" approach like below.
Fig.1
(This gives a hint later when we consider the direction of the tangential stress σt).
I did not introduce in <Thick
walled cylinder Elastic tube -1> that <Cal Poly San Luis Obispo
Mechanical Engineering, ME328 Introduction to Design, Cylinder
Stresses> uses a thin shell approach.
The
following is a copy of this article with my notes (questions). I cannot
copy of the drawing here so please see the pdf file above or below.
Fig. 2
ME328 Introduction to Design, Cylinder Stresses (bold parts)
"
This handout shows one way of analyzing the stresses in pressurized cylinders, as
presented in class. The method is fashioned after that presented in Shigley and
Mitchell’s Mechanical Engineering Design, 4th Ed., McGraw-Hill, 1983.
General (Thick-walled) Cylinders
The cylinder is defined with its inside and outside radii as a and b respectively.
The internal pressure is pi and the external pressure is po, and the length of section
being analyzed is l. We imagine the cylinder to be composed of an infinite number
of infinitely thin (thickness dr) shells. We balance the forces on each shell, then
integrate these forces for all radii r. In this way we can see how the stresses vary
from the inside radius to the outside radius.
(My note:
The strategy of this method is <We balance the forces on each shell, then integrate these forces for all radii
r. In this way we can see how the stresses vary from the inside radius to the outside radius.>)
"
The forces on the thin shell acting in the y direction must add up to zero. Each of these forces is calculated as stress times area. This means
2σtldr + 2σrlr - 2l(σr + dσr)(r + dr) = 0 ----- (1)
"
The above figure does not show < l >.
σt: Tangential stress
dr:
Thickness of the thin shell and infinitesimal thickness of the thin
shell as stated above <an infinite number of infinitely thin
(thickness
dr) shells>
l : Length of the cylinder (the above figure does not show this)
r : Radius from the axis to the inside of the thin shell
σr:
Radial stress at the inner surface of the thin shell. The above figure
shows the outward direction. This is supposed to be originated from
pi and σr varies with
<r>.
σr + dσr:
Radial stress at the outer surface of the thin
shell. The above figure shows the inward direction. This can be
considered as originated from
po. But po is supposed to be smaller than pi but not stated so. Just saying <analyzing the stresses in pressurized cylinders>. The magnitude of σr of (σr + dσr) is supposed to be the same as the above σr (outward direction). So this (σr) must be considered as the counter stress against the outward σr. The question is what dσr is. Supposing as above (Pi is > Po), dσr is negative. But the direction of σr at the outer surface is opposite to the direction of σr at the inner surface, the expression (σr + dσr) will be fine.
(σr + dσr) varies with
<r> and
dσr is considered very small but exists and in the inward direction to balance the forces.
But most of the other article on thick walled cylinders the directions of the radial stresses are opposite like shown in Fig. 1
and the figure below.
From: http://www.ewp.rpi.edu/hartford/users/papers/engr/ernesto/poworp/Project/4.%20Supporting_Material/Books/32658_10.pdf
Fig. 3
(the text book (Mechanics of Materials, Chapter 10 <Thick Cylinder> ) of this figure says
"
All stresses are assumed tensile. (10.2. Development of the Lame theory)
But the radial tensile stresses are are considered balanced when no inner and outer extra pressures. In case of the pressurized by the inner pressure Fig 2 will be appropriate. The outward stress at the inner surface of the thin shell and the inward stress at the outer surface of the thin shell.
"
Equation (1) is a big jump. To get this equation we need several thoughts and some earlier step equations.
1) First - What does <The forces on the thin shell acting in the
y direction must add up to zero.> mean ?
It says <the forces on the thin shell>. The above figure shows only two forces as
pi (F/Area) and
po (F/Area). But a stress is regarded as a force too - Force / Area like a pressure. So these forces are supposed to be expressed as the stresses on (the surface of) the thin shell. As I stated above,
(
σr + dσr)
can be regarded as the counter stress against the outward
σr (due to the continuous body nature), which in turn is originated from
pi. When < r > reaches the outer surface of the original cylinder (Radius b) (
σr + dσr) disappears and
σr opposes
po and σr = po < pi.
The first part of the equation (1)
<2σtldr> is supposed to be the total tangential stress experienced the cross section of the walls of the thin shell (two separate and symmetrical flat surfaces) as
2σtldr = σt x 2 x
l x
dr
But
this cross section is imaginary. We
must think of stresses of a continuous body though in this case a very
thin shell.
This reminds me of a commonly found drawing of a thin wall cylinder like below.
Fig. 4
This figure does not show but
P = Pressure (blue arrow)
L = Length of the cylinder
D = Diameter = 2
r
t = thickness of the wall
And the equation is
PLD = 2
σtLt or,
σt
= Pr/t ------- (2)
P x Area (LD) is a force. But this force seems very gross
when we see the stresses shown in the first figure including the
directions. But <2
σtLt> is almost the same meaning as <
2σtldr > although in the latter
dr is an infinitesimal thickness for differentiation and integration purpose. Something seems wrong as the object of the analysis is a thick wall, not a thin wall, though using a thin shell. The gross part is that
σt is the same at every part of the cross sections (two flat surfaces).
By the arrows shown Fig 2 the second part of the equation (1)
<2σrlr> are supposed to be the total or the half of the radial stress experienced at the inside curved surface of the thin shell. But
2σrlr =
σrlD (
D = Diameter)
lD is a flat area and an imaginary flat area corresponding to the above imaginary cross section, not the inside curved surface of the thin shell. Something seems wrong as this article declares <
We imagine the cylinder to be composed of an infinite number of infinitely thin (thickness dr) shells. We balance the forces on each shell, then integrate these forces for all radii r. In this way we can see how the stresses vary from the inside radius to the outside radius.>
The same thing (question) to the third part of the equation (1)
<2l(σr + dσr)(r + dr)>.
2l(r + dr) is a flat area and an imaginary flat area corresponding to the above imaginary cross section, and slightly larger (
2ldr) than
2lr or
lD. Something seems wrong again.
From the first drawing we are thinking of the whole thin shell, not flat areas. Look at Equation (1) again.
2σtldr + 2σrlr - 2l(σr + dσr)(r + dr) = 0 ----- (1)
Let's consider the direction of each stress. According to the above figure (Fig. 2)
σt : downward
σr : upward (mostly)
σr + dσr : downward (mostly)
Then
downward (
2σtldr) + upward (
2σrlr) - downward (
2l(σr + dσr)(r + dr)) = 0
The above figure suggests that this should be
downward (
2σtldr) + downward (
2l(σr + dσr)(
r + dr)) - upward (
2σrlr) = 0
This is a question to solve.
The stresses shown in Fig 2 are "compressive", which suggests the internal pressure (
Pi) is higher than the outer pressure (
Po). This is important as the Lame equations are made on the assumption <all the stresses are tensile (supposed to be a non-pressurized thick wall cylinder) as shown in Fig. 1 and Fig. 3.
Mechanics of Materials, Chapter 10 <Thick Cylinder> says in 10.3. <Thick cylinder - Internal pressure only>
"
N.B. - The internal pressure is considered as a negative radial stress since it will produce a radial compression (i.e. thinning) of the cylinder walls and normal stress convention takes compression as negative.
"
Complicated but this means
upward (
2σrlr) is larger than (
2l(σr + dσr)(
r + dr)) negatively. So,
<upward (
2σrlr) - downward (
2l(σr + dσr)(r + dr))> is still negative.
The tangential stress
2σtldr is always positive (tensile). And then
2σtldr + <upward (
2σrlr) - downward (
2l(σr + dσr)(r + dr))
> = 0 (balance)
The remaining problem is the direction of <
2σtldr>. This can be arbitrary as anywhere in the thin shell there is a pair of opposite direction tensile stresses as shown in Fig. 1.
Then the directions of the radial stresses shown in Fig. 2 is regarded as OK. Complicated.
Another problem of this method is that it does not show the difference of the tangential stress on the surfaces (the both sides) of the imaginary cross section shown in Fig. 2. Apparently
σt differs from at <r> to <
r + dr> while the radial stress is treated as it differs <r> to <
r + dr>. Generally
σt at the inner radius is bigger than
σt at the outer radius. But
dr is very small (thin) so this difference can be neglected. Some question remains, however, as
σt is not so small as the radial stresses.
But we are considering the balance of the three flats (two large and two long). And something seems still wrong.
Next,
If we multiply each part of Equation (1) by <π> and rearrange
2πdrlσt + 2πrlσr - 2π(r + dr)l(σr + dσr) = 0
2πrl is the total inner area of the thin shell.
σr = Force
/ 2πrl or
σr2πrl = Force (outward)
2π(r + dr)l is the total outer area of the thin shell.
(
σr + dσr) = Force /
2π(r + dr)l or
(σr + dσr)2π(r + dr)l = Force (inward)
The remaining is 2πdrlσt.
Can we think of any Force ?
By calculation
σt = Force /
2πdrl or
σt2πdrl = Force (which direction ?)
What is
<2πdrl> ? dr is not a radius but the thickness of the thin shell and it is an infinitesimal thickness. dr is treated as < dr
---> 0> (taking the limit). This differs from the thin walled
cylinder theory (calculation) where thickness does not take a limit for
differentiation and integration but is simply used to calculate the two
stresses - Tangent (Hoop) stress and Axial (Longitudinal) stress
relating the pressure (internal), radius (fixed) and thickness (fixed).
Wiki 's <Cylinder Stress> (25th July, 2015)
"
where
- P is the internal pressure
- r is the mean radius of the cylinder.
is the hoop stress.
When the vessel has closed ends the internal pressure acts on them to
develop a force along the axis of the cylinder. This is known as the
axial stress and is usually less than the hoop stress.
Though this may be approximated to
"
-
σθ = σt = Tangential or Hoop stress. P is the internal pressure, not the radial stress. But this suggests that if <t ---> 0> σθ becomes extremely high as P and <
r> have a certain value.
Back to
<π> issue,
Without <π>, this becomes <2drl>, which can be regarded like <2Lt > of <2σtLt > in the 2nd drawing. But now it has <π>. So what does <π> mean here ?
Assuming
2πdrlσt + 2πrlσr - 2π(r + dr)l(σr + dσr) = 0
Divided by
2πl
drσt + rσr - (r + dr)(σr + dσr) = 0
then
drσt + rσr - (rσr + rdσr + drσr + drdσr) = 0
drσt - rdσr - drσr - drdσr = 0
as
drdσr i
s very small and can be deleted. Then
drσt - rdσr - drσr = 0
Divided by
dr
σt - rdσr / dr - σr = 0
Rearranging
σt - σr - r (dσr / dr) = 0
This equation is seen in this article <ME328 Introduction to Design Cylinder Stresses>.
"
We multiply the last terms out, throwing away the d
σrdr term because it’s the
product of two very small quantities. We also cancel out 2
l and simplify:
dσr
σt - σr - r ----- = 0
dr
"
So we must go back to the above question and solve this to show that the above assuming equation
Assuming
2πdrlσt + 2πrlσr - 2π(r + dr)l(σr + dσr) = 0
is viable.
Without <π>, this becomes <2drl>, which can be regarded like PLD (or PL2r) in Fig 4. But it has <π>. So what does <π> mean here ?
This
approach (a thin shell approach) is for thick walled cylinders although
using a thin shell but uses it as an infinitesimal very thin shell for
differentiation and integration not a real thin shell with a fixed
thickness usually denoted <t>. And this thin shell exists inside
the cylinder wall in which the radius ranges from the inner radius (a)
to the outer radius (b) for integration.
Where does <π> come from ?
<π>
is neglected when making Equation (1). Although the figure shows the
stresses applied on the wall of the thin shell - both inside and
outside, when making the equation to balance the forces (add up to
zero), the two flat surface areas are used instead of the walled surface
areas.
2σtldr + 2σrlr - 2l(σr + dσr)(r + dr) = 0 ----- (1)
2rl instead of πrl.
2(r + dr)l instead of π(r + dr)l.
2rl (or DL) is the flat area D x L.
2(r + dr)l (or (D + 2dr)L) is the flat area of (D + 2dr) x L and a little bigger than 2rl (or DL). The bigger portion is 2drL, which is the cross section areas (drL each side) to calculate the tangential stress (σt). And dσr is the additional radial stress against the internal stress (σr). If no
dσr, the internal stress (σr) and the outer stress (σr) shall be balanced out because the magnitude is the same and opposite direction each other. To make three different stresses (σr, σr + dσr and σt) balanced we assume
2σtldr + 2σrlr - 2l(σr + dσr)(r + dr) = 0 ----- (1)
or multiplied by <π>
2πσtldr + 2πσrlr- 2πl(σr + dσr)(r + dr) = 0
----- (1)'
σt is supposed to be large or become very large as compared with σr and σr + dσr, but dr is a very small value (even limited to zero) so 2σtldr or 2πdrlσt will not become very large, in a way being canceled out.
We are interested in Equation (1)'
2πσtldr + 2πσrlr- 2πl(σr + dσr)(r + dr) = 0
2πlr and 2πl(r + dr) are the whole inner surface area and the whole outer surface area of the thin shell and the stress σr and the stress (σr + dσr) are applied to them respectively. These surfaces are not flat and not imaginary either.
But the problem is the first item <2πσtldr>. We can consider this as ldr
is a surface and a flat surface comprising l (length) x dr (thickness) on one side of the imaginary cross section after cutting by half along the axis. And there is the same long rectangular surface the other side, so 2 x. But the two surfaces are imaginary as when the shell is actually cut σt will change. As 2πlr and 2πl(r + dr) are the real surface we may have to think <2πldr> more realistic and at least corresponding 2πlr and 2πl(r + dr).
Assumed <2πσtldr> is corresponding to <2πσrlr> and <2πl(σr + dσr)(r + dr)> in Equation (1)'. But we cannot imaging what <2πldr> is.
Now a new idea.
<ldr> shows an flat surface area in calculation. <l> is relatively long while <dr> is very short.
We do not have to cut the shell but imagine this area. This shell is a
cylinder so it has a symmetry nature. We extend this area 2π times (about 6.28 times, but 2π is an infinite number) . Then we can get <2πldr>. But where do we extend the area ldr ?
We can get this by a simple calculation.
2πl(r + dr) - 2πlr = 2πldr
2πl(r + dr) is the curved outer surface of the shell.
2πlr is the curved inner surface of the shell.
So
2πldr is the difference of the above two surface areas but not a whole surface area as dr is not a radius.
σt is not a radial (direction) stress but Tangent stress. <2πldr> must be perpendicular to σt.
But again this shell has a symmetry nature and a continuous body nature. We do not select or think of a particular one σt at a particular place as it becomes common, general one applicable. σt should be considered to exist everywhere at <ldr> in the shell and the same magnitude.
<2πσrlr> and < - 2πl(σr + dσr)(r + dr)> are whole forces and in the radial opposite direction to be balanced but thy are not fully balanced. To be fully balanced we must add
2πσtldr. But we must consider the direction.
We cannot fully solve the
<π> issue yet but are approaching an conclusion.
See the following figure.
Fig. 5
As we explained above the direction σt may be opposite. But when we think of the whole circle (not a half circle) and consider symmetry of a circle, the direction is not so important.
We are now considering what 2πσtldr means or is. 2πldr can be considered as 2πdrl or 2πdr x l. 2πdr is a very small circle with the radius dr but can be thought out like the one shown in this figure. The magnitude of σt is large, much larger than
σr (and (σr + dσr) as well) but tangential. Equation (1)' (but not Equation (1) ) seems to balance the two total opposing radial stresses on the cylindrical shell and one total tangential stress (σt x 2πdrl) while Equation (1) seems to balance the two total opposite direction radial stresses on the flat long rectangular areas and the same direction two tangential stresses (2σt), one σt on each of the very narrow rectangular (dr wide) area made at the each side. Equation (1) is imaginary while Equation (1)' is more real. Now the point.
We must think of σt tangential along the 2πdr circle. Here we recall
"
N.B. -
The internal pressure is considered as a negative radial stress
since it will produce a radial compression (i.e. thinning) of the
cylinder walls and normal stress convention tales compression as
negative.
"
Then when 2πdr circle is expanding (thinning) the tangential stress is still tensile. So there is σt in the opposite direction as shown in Fig. 5. Opposite but the magnitude is the same due to symmetry. Now we treat σt as a vector - having a magnitude and direction. There are two vectors with the same magnitude (due to symmetry) and almost opposite direction, +σt and - σt . If they are completely opposite,
+σt - σt = 0
But look at Fig. 1 again. The direction of the two σt's. They are not completely opposite (a bit exaggerated in Fig. 1). But even taking a limit ( δθ----> 0) they will not become completely opposite, which means <δθ = 0>. The same consideration can be applied to our case in Fig. 5. These two σt's make a small radial direction (in this case outward) stress (this seems very important). But the direction depends on which pressure is higher, the inner or the outer. And this small radial stress made by the two tangential stresses contributes to make Equation (1)' more meaningful.
------
The above idea is not bad but unrealistic due to the location of this small radius circle. We can move this circle to the thin shell. But the thin shell width is dr not 2dr. So only a half is able to be on the the shell. And we have this dr width on both side.
But <2πσtldr> can be considered or originally being considered ad <2 x πσtldr> because <πσtldr> at both side. <πσtldr> is <2π x dr/2 x σtl>. Now we have a circle of radius <dr/2> at each side. These two are separated but the contribution to the balancing (forces or stresses) remains unchanged as <2πσtldr>. The above argument can apply.
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This post may have some changes or make some additions later.
sptt
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is the engineering normal strain, 
is the original length of the fiber and 
is the final length of the fiber.
