Lamé equations - continued
Before continuing Lamé equations I want to think about the special things of Magic Hose. A Magic Hose (regarded as a thick walled elastic tube) extends or stretches along the axial direction by max 3 times the original length when water pressure is applied, not only when one end is closed but even when one end is open for watering while the water is flowing through the hose. Let's us think about this mechanism.
Extension requires deformation (change of the shape). Deformation is made due to strain which relates with stress. Stress is made due to the force applied.
Applied force - Water pressure
Without new or additional stress the hose shape remains unchanged balanced by the forces (Newton's Law). Stress is added by water pressure. The stress can be analyzed to three mutually perpendicular directions. The direction of the force is not the same as the direction of the stress. Or the directions of the force are not the same as the directions of the stress. Or the directions of the forces are not the same as the directions of the stresses.
1) Hoop (tangential, circumferential) stress is the strongest. This is somehow anti-intuitive but the theoretical calculation shows this.
You can feel this hoop stress by the following simple way.
Fold (bend) a piece of A4 paper round to make a tube and hold it by one hand in the middle of the tube to keep the tube shape. Close the one open end by the palm of the other hand. Put air (breath) strongly from your mouth into the tube from the other unclosed end. Then feel the stresses or forces made by this action through the palm of the holding hand. Stronger than you thought.
2) Radial (lateral, transverse) stress is small as compared with 1) Hoop stress. How small is depends on (varies with) the inner diameter and the outer diameter (or the same thing but the inner and outer radii and the wall thickness) and the inside pressure and the outside pressure. In our case the inside pressure increases and decreases while the outside pressure remains unchanged - usually the atmospheric pressure.
3) Axial (longitudinal) stress
In case of Magic Hose, this is the point. We must be careful whether the axial or longitudinal extension is due to the axial (longitudinal) stress or / and some other stress (es) as the seemingly radial expansion is actually largely due to the hoop stress, not the radial stress.
We will continue to check some other articles on Thick wall cylinder or Lamé equations.
From Mechanics of Materials with Program in C (this is a book title)
8.8 Thick Cylinder
Modified by me with my full responsibility
Starting with
σr + r (dσr / dr) = 2A - σr
"
2σr = - (r dσr / dr) + 2A
(My note: Originally <-2a>. It does not make a big difference as <A>, <a> are some value of constant.)
2 (σr - A) = - r dσr / dr
2 (σr - A) / dσr = - r / dr
Reversing the both sides
dσr / 2 (σr - A) = - dr / r
Rearranging
2 dr / r = - dσr / (σr - A)
Please recall < σr + r (dσr / dr) = 2A - σr or r(dσr / dr) + 2σr = 2A> shown at the last part of the last post <Thick walled cylinder Elastic tube - 1>.
Integrating with dr and dσr
2
dr / r = - dσr / (σr - A) 2
(1 / r ) dr = - (1/ (σr - A) ) dσr 2 ln r = - ln (σr - A) + ln B (<ln B> is a constant)
lin r 2 = ln (B / (σr - A) )
r 2 = B / (σr - A)
r 2 (σr - A) = B
σr - A = B / r 2
σr = A + B / r 2
"
Here we have no < where has 2 r σR gone ? > or < where has 2σr gone ? > problem or do not have to use the math trick shown in <Thick walled cylinder Elastic tube - 1>.
One more
Mathematical model of the Lamé Problem for Simplified Elastic Theory applied to
Controlled-Clearance Pressure Balances
Saragosa Silvano
http://arxiv.org/ftp/arxiv/papers/1007/1007.0813.pdf
2. The Lame Problem
Among the theoretical methods, the Simplified Elastic Theory is an analytical approach to evaluate
the pressure distortion coefficient by the solution of a system of differential equations named the
Lame equations. The results of this analysis lead to the determination of the stresses and distortions
for the case of thick walled cylinder. The method is formally known as The Lame Problem and the
differential formulation of this problem is based on the analysis of the equilibrium of an elementary
volume [5] (fig.1).
Fig. 1 The state of stress in the axialsymmetric cylindrical structures according to the simplified elastic theory
By introducing cylindrical coordinates (x, θ , r), the inner radius of the elementary volume is r and
the outer radius is r+dr (dr is infinitesimal). The volume is transversely limited by the surfaces
ADHE, BCGF, by the surfaces ABFE , CDHG on two planes rotated of an infinitesimal angle dθ
and by the surfaces ABCD, EFGH defined by the inner radius r and outer radius r+dr respectively.
We can consider the following nomenclature:
- δθ : circumferential stress;
- δr and δr + (δr /dr) dr : radial stresses applied to the surface ABCD and EFGH respectively;
- r dθ dx and (r + dr ) dθ dx : internal surface ABCD and the external surface EFGH of the
elementary volume respectively;
- dr dx : surface ABEF and CDGH
- δr r δθ dx : force applied to internal surface ABCD of the elementary volume;
- (δr + dδr dr/dr) (r + dr) δθ × dx: force applied to the external surface EFGH of the elementary
volume;
- δθ sin (dθ / 2) = δθ (dθ / 2) : vertical component (parallel to the r direction KK’) of the
circumferential stress;
- δθ (dθ / 2) drdx : vertical component of the force applied to the surfaces ABFE and CDHG.
- E and n Young’s Modulus and Poisson’s ratio respectively.
The formulation of the The Lamé Problem is based on the following hypotheses:
i) axial-symmetry: u (in r direction) is the only displacement component;
ii) radial displacement u only depend on the radius r:
u (r, x, q ) = u (r )
iii) shear stresses on the elementary volume must be zero: τxr = τxθ = τrθ = 0;
iv) on the basis of the axial-symmetry and constant thickness, the radial and circumferential stresses only depend on the radius r:
δr (r, x, q ) = δr (r )
δθ (r, x, q ) = δθ (r )
v) it is assumed that the axial stress δx s constant along the length of the transversal section (independent from the coordinates x, r):
∂ δx /∂ r = ∂ δx / ∂ x = 0
vi) axial symmetrical loads: because nothing varies in the θ direction,
vii) the problem is solved by linear-elastic approach.
Now considering the radial force equilibrium of the element in fig.1:
(δr + dδr dr/dr) (r + dr) δθ dx - δr r δθ dx - 2 δθ dx dr sin (dθ / 2)
= δr r δθ dx + δr dr δθ dx + dδr dr r δθ dx /dr + dδr dr δθ dx /dr - δr r δ θdx - δθ dx dr dθ
= δr dr δθ dx + dδr dr r δθ dx /dr + dδr dr δθ dx /dr - δθ dx dr dθ = 0
Neglecting the product of small quantity (dδr/dr)dr 2 δθ dx) and collecting terms, the above equation reduces to:
(my note
δr dr δθ dx + dδr dr δθ dx /dr - δθ dx dr dθ = 0
δr dr + dδr dr /dr - δθ dr = 0
δr + dδr /dr - δθ = 0 )
dδr / dr + (δr - δθ ) / r = 0 (1)
In order to solve the problem there is one equation with two variables δr, δθ. We can determine the
second equation applying the Strain-Displacements Equations:
-----
My notes:
1) Please keep in mind these hypotheses (assumptions), especially symmetry nature.
2) <Radial displacement u > is not shown in Fig. 1. Please see below.
From: http://www.roymech.co.uk/Useful_Tables/Mechanics/Cylinders.html
is strain. t it the tangential strain while r is radial strain. Now back to <the Lame Problem >
-----
the Strain-Displacements Equations are the same as the above two equations.
ɛr = du / dr(2)
ɛθ = (2πr (r + u) - 2πr) / 2πr = u / r
From Hooke’s generalized law we can obtain the following equations:
εr = 1/E (σr - υ (σx + σθ))(3)
εθ = 1/E (σθ - υ (σr + σx))
(my note: again from : http://www.roymech.co.uk/Useful_Tables/Mechanics/Cylinders.html
Initial Assumptions
The following relationships are assumed for
the strains ε1,ε2,
ε3 associated with the stress σ1,
σ2 and σ3.
υ = Poisson's ratio.
(E = Young's module)
Reference Notes on stress and strain
υ = Poisson's ratio.
(E = Young's module)
Reference Notes on stress and strain
(my note: We may review this Hooke’s generalized law some time later as this is important)
And from the first equation of (3) we can obtain:
σr = Eɛr + υ (σx + σθ) (4)
Substituting eq. (4) in the second equation of (3) we get:
ɛθ = (1/E) [σθ - υEɛr - υ2σx - υ2σθ - υσx] = (1/E) [σθ (1-υ2) - υσx (1+υ) - υEɛr]
Eɛθ = σθ (1-υ2) - υσx (1+υ) - υEɛr = > Eɛθ + υσx (1+υ) + υEɛr = σθ (1-υ2)
σθ = ((1-υ2)/1) [Eɛθ + υσx (1+υ) + υEɛr]
and collecting terms:
σθ = (E / (1-υ2))(ɛθ + υɛr) + υσx / (1-υ)
Substituting eq. (5) in eq. (4):
σr = Eɛr + υσx + (υE / (1-υ2))(ɛθ + υɛr) +υ2σx / (1-υ) =
(Eɛr(1-υ2) + υE(ɛθ + υɛr)) / (1-υ2) + ((υσx / (1-υ) + υ2σx)) / (1-υ ) =
(Eɛr - Eɛrυ2+ υEɛθ + Eɛrυ2) / (1-υ2) + (υσx - υ2σx + υ2σx) / (1-υ) (5)
With opportune simplifications and collecting terms:
σr = (E / (1-υ2)) (ɛr + υɛθ ) + (υσx) / (1-υ) (6)
Substituting the equations (2) in (5) and (6), the following stress-displacements equations can be
obtained:
σθ = (E / (1-υ2)) [(u / r ) + υ(du /dr)] + (υσx) / (1-υ)
(7)σr = (E / (1-υ2)) [(du /dr) + υ (u / r )] + (υσx) / (1-υ)
Substituting eqs. (7) in eq. (1) and with opportune simplifications, we can obtain:
(E / (1-υ2)) (d /dr) [(du /dr) + υ (u / r )] + (E / (1-υ2))(1/r) [(du /dr) + υ (u / r ) - (u / r) - υ (du /dr)
] = 0
(d2u /dr2) + (υ/r)(du /dr) - υ/(u / r2) + (1/r)(du /dr) + υu /r2 ) - u /r2 - (υ/r) (du /dr)= 0
(my note: Please note that a tricky <Quotient rule> is used here.
(d / dr) υ (u / r ) = υ[(d /dr) (u / r )] =
υ[((du/dr)r - u1)) / r2] = rυ(du/dr)/ r2 - υu/ r2 = (υ/r)(du/dr) - υu/ r2 )
(d2u /dr2) + (υ/r)(du /dr) - υ/(u / r2) + (1/r)(du /dr) + υu /r2 - u /r2 - (υ/r) (du /dr)
= 0
And with opportune simplifications:
(d2u /dr2) + (1/r)(du /dr) - u /r2 = 0
Which can be expressed as"
(d/dr) [(du /dr) + (u / r)] = 0 (8)
Furthermore eq. (8) can be written as:
(d/dr) [(du /dr) + (u / r)] = (d/dr) [(1/r) (r du /dr + u) ] = (d/dr) [(1/r) (r du /dr + (dr/dr) u)] = 0
And finally a more compact form of eq. (8) is obtained:
(d/dr) [(1/r) (d/dr (ur)] = 0 (9)
(my note: <Product rule> is used here. d/dr (ur) = r (du /dr) + u (dr/dr))
Integrating this equation:
(1/r) (d/dr (ur)] = C0
(my note: once more integrating)
ur = C0 (r2 / 2) + C2
u = C0 (r / 2) + C2 /r
By assuming: C0/2 = C1, we can obtain the radial displacements
u = C1 r + C2 /r defined for r ≠ 0 (10)
Substituting eq. (10 ) in stress-displacement equations (7) we can determine the circumferential and
the radial stresses at any point through the thickness of the wall.
In order to determine the integration constants C1 and C2, opportune boundary conditions are
necessary.
"
Lame was a mathematician. Without some omissions of very small values, differentiations and integrations (including some tricky operations) and relatively many assumptions we cannot reach the Lame Equations. Lame Equations are necessary to solve problems at least in text books.
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