12.524 2005 10 12 LE07b: Some elastic solutions of geologic interest
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Elasticity of Rocks
Assigned Reading:
Gueguen, Yves and V. Palcisukas, Introduction to the Physics of Rocks, Chapter III, p.
53-61; Chapter IV. 73-92. (Essentially a discussion of some of the material we
have covered over the last couple of week. I hope this reading is easier no than it
would have been before we started.)+
Resource reading:
Atkinson BK (1987) Fracture Mechanics of Rock. Academic Press, London UK, pp 534
Chapters 1&11
Johnson AM (1970) Physical Processes in Geology. Freeman Cooper, San Francisco CA,
Chapters 9-11.
Clyne, T. W., and P. J. Withers, An Introduction to Metal Matrix Composites, Cambridge
University Press, 1993.
Hearmon, R. F. S., An Introduction to Applied Anisotropic Elasticity, Oxford University
Press, 1961.
Mavko, G., T. Mukerji, and J. Dvorkin, The Rock Physics Handbook, Cambridge Univ.
Press, 1998.
Watt, J. P., G. F. Davies, and R. J. O’Connell, “The Elastic Properties of Composite
Materials”, Rev. Geophys. Space Phys., 14, 541-563, 1976.
Muskhelishvili, N. I.(Nikolaĭ Ivanovich),1891- Some basic problems of the mathematical
theory of elasticity; fundamental equations, plane theory of elasticity, torsion, and
bending, Groningen, P. Noordhoff, 1963. QA931.M9871 1954
Exact Elastic Treatment of Simple Geometry
Fundamental Equations of Elasticity:
Equilibrium, Compatibility, Constitutive Law
Biharmonic Equation, Airy Stress Function
Particular Solutions
Internally Pressurized Cylinder
Externally Stressed Hole
Externally Stressed Elliptical Hole
Elastic Moduli of Cracked Solids
12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
Basic Equations of Isotropic Elasiticity
Elasticity of isotropic elastic materials
(1 + ν ) σ − ν δ σ
σ ij = λδ ij ε kk + 2µε ij
ε ij =
ij
ij kk
E
E
Equilibrium (Static)
∂σ ij
=0
∂x j
Compatibility
∂ 2ε ij
∂ 2ε jl
∂ 2ε kl
∂ 2ε ik
+
−
−
=0
∂xk ∂xl ∂xi ∂x j ∂x j ∂xl ∂xi ∂xk
Proof:
Strain may be viewed as the differential of the displacement change
vector. Provided that no gaps open up and that no overlaps develop, the
displacements must be continuous.
If the displacement change vector, ∆u = [u1 , u2 , u3 ] , then
∂u
ε11 = 1
∂x1
calculate
and
∂ 2ε11
∂ 3u1
=
∂ 2 x2 ∂ 2 x2 ∂x1
∂ 2ε 22
∂ 2ε12
∂ 2ε 22 ∂ 2ε11
∂ 2ε12
and
.
Notice
that
2
+
=
∂x1∂x2
∂ 2 x1 ∂ 2 x2
∂x1∂x2
∂ 2 x1
In general any problem for the statics of a three
dimensional body loaded externally and with no body
forces is well-posed using these equations as long as
sufficient boundary conditions are given.
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12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
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Consider plane strain and plane stress (i.e. 2-D) without body
forces:
Equilibrium
∂σ 11 ∂σ 12
+
=0
∂x1
∂x2
∂σ 21 ∂σ 22
+
=0
∂x1
∂x2
Compatibility
∂ 2ε 22 ∂ 2ε11
∂ 2ε12
+
=2
∂ 2 x1 ∂ 2 x2
∂x1∂x2
Elasticity
ε11 =
(1 + ν ) σ
11
E
(1 + ν )σ 12
ε12 =
E
−
ν
E
(σ 11 + σ 22 ) =
σ 11
E
−
νσ 22
E
Equilibrium, Compatibility and Elasticity require
⎛ ∂2
∂2 ⎞
σ + σ 22 ) = 0 = ∇ 2σ ii
⎜ 2 +
2 ⎟ ( 11
∂x2 ⎠
⎝ ∂x1
Airy Stress Function:
Assume that there exists a function, Φ, the Airy stress function, such that
∂2Φ
= σ 11
∂x2 2
∂2Φ
= σ 22
∂x12
∂2Φ
−
= σ 12
∂x1∂x2
Then from above,
∇4Φ = 0
Biharmonic Equation
12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
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Biharmonic Equation: Method of solution is 1.)Find a
general Airy’s stress function with appropriate
symmetry; 2.) Use differential equations to get stress;
3.) Use boundary conditions to find arbitrary
constants: 4.) Use Hook’s law to get strains; 5.)
Integrate to get displacements
Elasticity in Problems with Cylindrical and Spherical Symmetry
In problems with cylindrical symmetry, r, θ, z:
Strain:
∂ur
1 ∂uθ ur
,
+ ,
εθθ =
∂r
r ∂θ
r
1 ⎛ ∂u
1 ∂ur uθ ⎞
1 ⎛ 1 ∂u z ∂uθ ⎞
,
= ⎜ θ +
− ⎟ , εθ z = ⎜
+
2 ⎝ ∂r r ∂θ
2 ⎝ r ∂θ
∂z ⎟⎠
r ⎠
ε rr =
ε rθ
∂u z
∂z
∂u ⎞
1 ⎛ ∂u
ε rz = ⎜ r + z ⎟
2 ⎝ ∂z
∂r ⎠
ε zz =
The gradient operator and the Laplacian in cylindrical coordinates are
∂
1 ∂
∂
+ eˆθ
+ eˆ z
∇ = eˆ r
∂r
∂z
r ∂θ
∂2
1 ∂ ⎛ ∂ ⎞ 1 ∂2
r
+
+
∇2 =
⎜
⎟
r ∂r ⎝ ∂r ⎠ r 2 ∂θ 2 ∂z 2
Equilibrium:
In problems with cylindrical symmetry, r, θ, z:
∂σ rr 1 ∂σ rθ ∂σ zθ σ rr − σ θθ
+
+
+
=0
r ∂θ
r
∂r
∂z
σ
∂σ rθ 1 ∂σ θθ ∂σ zθ
+
+
+ 2 rθ = 0
r ∂θ
r
∂r
∂z
∂σ zr 1 ∂σ zθ ∂σ zz σ zr
+
+
+
=0
r ∂θ
r
∂r
∂z
Spherical symmetry, r, θ, φ,
Special case that displacements occur in the radial direction only:
Strain:
In problems with spherical symmetry, r, θ, φ, for the special case that
displacements occur in the radial direction only:
∂u
u
ε rr = r ,
εφφ = εθθ = r ,
ε rθ = εθ z = ε rz = 0
r
∂r
12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
Equilibrium:
∂σ rr 1
+ ( 2σ rr − σ θθ − σ φφ ) = 0,
∂r
r
∂σ θθ ∂σ φφ
=
= 0,
∂θ
∂φ
5
σ θθ = σ φφ
Example 1: Stress around a circular pipe (tube)
Assume plane strain and plane stress, that the pipe is internally pressured
with no applied stresses.
New boundary conditions are
σrr=-P and σrθ=0 at r=ro
σrr= σrθ=0 at r=∞
Notice that Φ must be independent of θ.
Now, let
r=et
Φ=emt and substitute into the biharmonic.
A general solution is
Φ=A+Br2+C•ln(r)+Er2•ln(r)
Check to see if Φ satisfies the biharmonic
∂ 2 ⎡ ∂ 2 Φ ∂ 2 Φ ⎤ 1 ∂ ⎡ ∂ 2 Φ 1 ∂Φ ⎤ ∂ 4 Φ 1 ∂ 3 Φ 1 ∂ 2 Φ 1 ∂Φ
+ 2 ⎥+
+
+
−
+
⎢
⎢
⎥=
∂r 2 ⎣ ∂r 2
∂r ⎦ r ∂r ⎣ ∂r 2 r ∂r ⎦ ∂r 4 r ∂r 3 r 2 ∂r 2 r 3 ∂r
Now
∂Φ C
= + 2 Br + Er • ln(r ) + Er
∂r r
C
∂2Φ
= − 2 + 2 B + E • ln(r ) + 3E
2
r
∂r
3
∂ Φ 2C
= 3 + 2 Br + Er • ln(r ) + Er
r
∂r 3
4
6C 2 E
∂Φ
=− 4 − 2
4
r
r
∂r
so...
12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
E⎤ 1 ⎡ C
⎡ 6C 2E ⎤ 1 ⎡ 2C
⎤
∇4Φ = ⎢− 4 − 2 ⎥ + ⎢ 3 + 2 ⎥ + 2 ⎢− 2 + 2B + E • ln(r) + 3E⎥
r ⎦ r⎣r
r⎦ r ⎣ r
⎣ r
⎦
1 ⎡C
⎤
− 3 ⎢ + 2Br + Er • ln(r) + Er ⎥
r ⎣r
⎦
and ⇒ ∇4Φ = 0
Now solve for the constants in the solution using the boundary conditions
1 ⎛ ∂Φ ⎞
σ rr = ⎜
⎟
r ⎝ ∂r ⎠
1 ∂
C
A + Br 2 + C ln(r ) + Er 2 ln(r ) ) = 2 B + 2 + 2 E ln( r ) + E
=
(
r ∂r
r
2
C
∂ Φ
σ θθ = 2 = 2 B − 2 + E (3 + 2 ln(r ))
r
∂r
As r → ∞ , σ rr , σ θθ =0.
So E, B=0.
At r=r0 σ rr =-P
C
σ rr |r = ro = 2 = − P
ro
σ rr = − P
σ θθ
σ rθ
r2
ro 2
ro 2
= −P 2
r
=0
.Remarks: Stress falls off as r2, depends linearly on P
Hoop stresses are tensile.
If fractures occur they occur along planes of maximum
tensile stress.
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12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
Example 2: Stress and Strain Concentration around a Spherical
Hole loaded by a pressure at infinity:
Strain about a spherical hole.
Suppose that a body loaded by pressure σ contains a spherical hole.
By symmetry we suppose that the displacements must be a function of r
only. Then guess that
b
ur = ar + 2
r
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12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
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Pore Compressibility:
Suppose we consider a spherical hole with an pressure σ applied at
infinity.The material is elastic. From symmetry we suppose that the strain
will be in the radial direction only, and that it will be proportional to the
applied pressure:
dR
1 (1 − ν )
ε rr =
= C ⋅ d σ where C =
R
K 0 2 (1 − 2ν )
Then the change in pore volume is dV p = 4π R 2 dR = 3V p
So the volumetric strain, εV is
εV =
dR
R
3 (1 − ν )
dσ
2 K 0 (1 − 2ν )
and the pore stiffness is
dV p
3 (1 − ν )
1
=
=
2 K 0 (1 − 2ν ) Kφ
Vp dσ
If the Poisson’s ratio is 0.25, then 1/Kφ is 2.25; as the solid material
becomes incompressible the pore compressibility becomes large.
B( ν )
ν
12.524, 2005 10 12 LE07b: Some elastic solutions of geologic interest
Stress concentrations around pores, flaws, and inclusions:
Cylindrical Internal P
hole
“
External P
“
Uniaxial load, σ
Spherical
hole
“
Elliptical
hole
“
Internal P
External P
Uniaxial load, σ
External P
Biaxial Load, S
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