Lecture 9 Elasticity - Indian Institute of...
Transcript of Lecture 9 Elasticity - Indian Institute of...
Lecture 9
Elasticity
• Einstein summation convention: repetition of index in a term implies summation over the repeated index
aibi = a1b1 + a2b2 + a3b3
σkk = σ11 + σ22 + σ33
• Kronecker delta
1 if i = j δij = 0 if i ≠ j
• Stress tensor is symmetrical
• Units :
Stress = Force / Area
Newton / m2 = Pascal (Pa) 1 MPa = 106 Pa = 10 bars
Deviatoric Stress:
ti j = si j �13
skkdi j
r∂2Q∂t2 = (l+2µ)—2Q
r ∂2
∂t2
✓∂u∂y
� ∂v∂x
◆= µ—2
✓∂u∂y
� ∂v∂x
◆
r ∂2
∂t2
✓∂v∂z
� ∂w∂y
◆= µ—2
✓∂v∂z
� ∂w∂y
◆
r ∂2
∂t2
✓∂w∂x
� ∂u∂z
◆= µ—2
✓∂w∂x
� ∂u∂z
◆
u = —f+—⇥y
—2f =1
V 2p
∂2f∂t2
—2y =1
V 2s
∂2y∂t2
Vp =
sl+2µ
r=
sK + 4
3µr
1
Principal Stress (σ1, σ2, σ3) SHmax or most compressive horizontal principal axis Differential stress (σ1 - σ3)
World Stress Map
World Stress Map
Strain: Response of a body to stress
x δx
L M
δx+δu x+u
L’ M’
exx = ∂ux / ∂x eyy = ∂uy / ∂y ezz = ∂uz / ∂z exy = eyx = ½ (∂ux / ∂y + ∂uy / ∂x) exz = ezx =½ (∂ux / ∂z + ∂uz / ∂x) eyz = ezy = ½ (∂uy / ∂z + ∂uz / ∂y)
Hooke’s law
Stress is proportional to Strain σij = λΘδij + 2µeij
Θ = exx + eyy + ezz δij è Kronecker delta λ and µ è Lame’s parameters
µ è shear modulus
• Young’s modulus, E = σxx / exx
• Bulk modulus, K = p / Θ = λ + ⅔ µ
• Hydrostatic pressure, p = ⅓ (σxx + σyy + σzz)
• Poisson’s ratio, σ = - ezz / exx
Stress and strain• Stress-strain relation:
– Elastic domain:
• Stress-strain relation is linear
• Hooke’s law applies
– Beyond elastic domain:
• Initial shape not recovered when stress is
removed
• Plastic deformation
• Eventually stress > strength of material =>
failure
– Failure can occur within the elastic domain
= brittle behavior
• Strain as a function of time under stress:
– Elastic = no permanent strain
– Plastic = permanent strain
• Our goal: find the relation between stress
and strain
• In elastic domain - stress and strain linear - Hooke’s law applies
• Beyond elastic domain - initial shape not recovered when stress removed - plastic deformation
- eventually failure
No permanent strain
Permanent strain
Flexure - subduction
• In addition to load of overriding plate:
– Sediments
– Non-elastic response
Fowler: The Solid Earth
Elasticity• Let’s assume:
– A rectangular prism with 3 sides defining (O,x,y,z)
– A uniform tension Nz exerted on 2 sides perpendicular to
(O,z)
• When the prism is stretched along (O,z):
– Change in length is proportional to tension: εz = Δh/h ∝ Nz
– One can show experimentally that:
– E = Young’s modulus
– Units of stress = km/m2
– Small E => more elastic
• If the prism is stretched along (O,z), it must shrink along
(O,x,y) to conserve mass:
– One can show experimentally that contraction:
– ν = Poisson’s ratio (dimensionless)
!
"z
=#h
h=1
EN
z
!
"x = "y = #$
ENz x
y
z
O
-Nz
Nz
Elasticity• Let’s assume:
– A rectangular prism with 3 sides defining (O,x,y,z)
– A uniform tension Nz exerted on 2 sides perpendicular to
(O,z)
• When the prism is stretched along (O,z):
– Change in length is proportional to tension: εz = Δh/h ∝ Nz
– One can show experimentally that:
– E = Young’s modulus
– Units of stress = km/m2
– Small E => more elastic
• If the prism is stretched along (O,z), it must shrink along
(O,x,y) to conserve mass:
– One can show experimentally that contraction:
– ν = Poisson’s ratio (dimensionless)
!
"z
=#h
h=1
EN
z
!
"x = "y = #$
ENz x
y
z
O
-Nz
Nz
Elasticity• Let’s assume:
– A rectangular prism with 3 sides defining (O,x,y,z)
– A uniform tension Nz exerted on 2 sides perpendicular to
(O,z)
• When the prism is stretched along (O,z):
– Change in length is proportional to tension: εz = Δh/h ∝ Nz
– One can show experimentally that:
– E = Young’s modulus
– Units of stress = km/m2
– Small E => more elastic
• If the prism is stretched along (O,z), it must shrink along
(O,x,y) to conserve mass:
– One can show experimentally that contraction:
– ν = Poisson’s ratio (dimensionless)
!
"z
=#h
h=1
EN
z
!
"x = "y = #$
ENz x
y
z
O
-Nz
Nz
Young’s modulus, E = σz / εz
Elasticity• Let’s assume:
– A rectangular prism with 3 sides defining (O,x,y,z)
– A uniform tension Nz exerted on 2 sides perpendicular to
(O,z)
• When the prism is stretched along (O,z):
– Change in length is proportional to tension: εz = Δh/h ∝ Nz
– One can show experimentally that:
– E = Young’s modulus
– Units of stress = km/m2
– Small E => more elastic
• If the prism is stretched along (O,z), it must shrink along
(O,x,y) to conserve mass:
– One can show experimentally that contraction:
– ν = Poisson’s ratio (dimensionless)
!
"z
=#h
h=1
EN
z
!
"x = "y = #$
ENz x
y
z
O
-Nz
Nz
Poisson’s ratio, Perfectly incompressible material,
Poisson’s ratio
• Poisson's ratio = ratio oftransverse to longitudinalnormal strain underuniaxial stress, in thedirection of stretchingforce, with:
– Tensile deformation positive
– Compressive deformationnegative
• All common materialsbecome narrower in crosssection when they arestretched => Poisson’s ratiopositive.
!
"z
=#h
h=1
EN
z
!
"x = "y = #$
ENz
!
"# = $%x
%z Poisson’s ratio
• If Vs = 0, ν = 0.5:
– Either a fluid (shear waves
do not propagate through
fluids)
– Or material that maintains
constant volume regardless
of stress = incompressible.
• Vs ~ 0 is characteristic of
a gas reservoir.
!
" =(Vp
2# 2Vs
2)
2(Vp
2#Vs
2)
A negative Poisson's ratio change is associated with the top of a gas zone,
Poisson’s ratio
Examples:
– Perfectly incompressible material => ν = 0. 5 (no volume change)
– If ν > 0.5 => volume increase under compression = dilatant.
– If ν < 0 => become thicker when stretched = auxetic materials (somepolymer foams)
– Rubber: ν = 0.5, Cork: ν = 0 (why is cork used to close glassbottles?)
– Earth’s interior ν = 0.24-0.27
– Granite: ν = 0.2-0.3
– Carbonate rocks ν ~ 0.3
– Sandstones ν ~ 0.2
– Shale ν > 0.3
– Coal ν ~ 0.4.
E. Calais notes
• Young’s modulus, E = σxx/εxx
• Shear modulus, μ = ½ σxy / εxy
• Poisson’s ratio, ν = εxx /εzz
• Lame’s parameter, λ
• Bulk modulus, K = p / Θ where hydrostatic pressure, p = ⅓ (σxx + σyy + σzz)
There are 5 elastic constants:
K = E / 3(1-2ν)
μ = E / 2(1+ν)
λ = Eν / (1+ν)(1-2ν)