Introduction to the theory of poro-elasticity

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Top view of the Ekofisk reservoir
Part II : Compaction and subsidence
Subsidence below the platform
Part II : Compaction and subsidence
The Finite-Element method
Part II : Compaction and subsidence
The Finite-Element method : 2D
• Start from the TVW : ˆ ∆T d · u dS =
∂ΩT Ω
ˆ ∆σ : (u )dV .
ˆ u is a kinematically admissible displacement field. I • Material response : ∆σ = C : − δ´pf where distribution of ∆pf is known. • Introduce classical discretization and interpolation for displacement (same for virtual and real) {u h }2 = [N]2×2nel {U} with N the shape functions and {U} the nodal displacement unknowns (nel nodes)
Complement on the theory of poro-elasticity
Who is Maurice Biot ?
References
Part I : The main results of lecture 1
Drained conditions
• Initial stress σ 0 • Stress changes for drained conditions ∆σ = (3K JI + 2GIK) : − αδ∆pf , Compressibility modulus K drained conditions Fluid mass has to change. α is Biot’s constant = 1 −
Part II : Compaction and subsidence
The Finite-Element method : 2D
• Strain operator : { h }4 = [B]4×2nel {U} with t { h } = (
11 , 22 , 33 , 2 12 )
• Material response : {∆σ} = [C } − {δ}´pf I]{ • The finite-element set of equations (single element)
(Lewis et al., 2003)
Part II : Compaction and subsidence
The Finite-Element method
• First mechanical problem : initial equilibrium, σ 0 , ρb, T d , conditions often unknown. • Second mechanical problem : the stress changes ∆σ in response to due to ∆pf and ∆T d ; no changes in body forces. • Superposition principle : solve pb 2 without knowing pb 1 (small perturbation assumption). • The pressure drop is assumed to be known over the whole domain.
Part II : Compaction and subsidence
Flamant’s problem is our pb no 1
ur uθ
= − =
2F1 r 1−ν cos θ log − Pθ sin θ , (5) πE d πE 2νF1 2P r (1 − ν)P sin θ + sin θ log + (sin θ − θ cos θ) , πE πE d πE
K Ks
(1)
with Ks skeleton compressibility
Part I : The main results of lecture 1
Undrained conditions
• Initial stress σ 0 • The stress change for undrained conditions No time for fluid migration ∆σ = (3Ku JI + 2GIK) : Ku is the compressibility modulus, undrained (Ku > K ) The pressure change : ∆pf = −tr(∆σ) B is Skempton’s coefficient • G remains identical for the two descriptions • Simple extension of Hooke’s law with 2 new parameters (Ku and Ks ). B 3 with B = Ku − K , αKu (3) (2)
F1
y
r u θ
θ
ur
x
The constant d corresponds to the coordinate of a point on the vertical axis where the displacement is set arbitrarily to zero.
Part II : Compaction and subsidence
f (x )δ(x − x )dx = f (x)
· u 2 is −u 2 · e2 at the point of application of
The vertical displacement is thus minus the integral over the reservoir on the rhs of (6). 3 • Move the load wrt the reservoir to get the vertical displacement profile : the subsidence.
Position of the field in the North Sea (Lewis et al., 2003)
Part II : Compaction and subsidence
Reservoir, compaction et subsidence
Part II : Compaction and subsidence
Strain :
12
Part II : Compaction and subsidence
BAD NEWS ! !
• The competition is protesting to the local government that your company is changing the stress state at 50 km from the reservoir you are producing, arguing that the official building will be soon damaged by large cracks. • Your computer with the FE program is stolen, you can only use the one inherited from your grand-mother. • The library was burnt down. • The only good news : a copy of Timoshenko and Goodier, The theory of linear elasticity (1934) has been saved.
∂ΩT
T d · u 2 dS + 1 · u 1 dS +

b1 · u 2 + α∆p1 tr( 2 )dV = b2 · u 1 + α∆p2 tr( 1 )dV . (4)
Td 2 ∂ΩT

Part II : Compaction and subsidence
Example of Ekofisk
FE results, plane-strain
Displacement : u2
Part II : and subsidence
FE results, plane-strain
Strain :
11
Part II : Compaction and subsidence
FE results, plane-strain
Part II : Compaction and subsidence
Application of reciprocity theorem
F1
M
y
Problem no 1
M
x
Problem no 2
l
D
w
∆ p f2
2L
The two problems considered for the subsidence analysis. Problem 1 : vertical point force on the top surface, ∆pf = 0 everywhere. Problem 2 : ∆pf = 1 in the reservoir, 0 outside.