12.005 Lecture Notes 15
Elasticity
So far:
Stress → angle of repose vs accretionary wedge
Strain → reaction to stress → but how?
Constitutive relations
τ ij = τ ij (ε kl ) ; ε ij = ε ij (τ kl )
For example,
Elasticity
Isotropic
Anisotropic
Viscous flow
Isotropic
Anisotropic
Power law creep
Viscoelasticity
Trade offs:
simplicity
↔
realism
constant
variable
isotropic
anisotropic
elastic, viscous
viscoelastic
history
history dependent
independent
Tensors
Most physical quantities that are important in continuum mechanics like temperature,
force, and stress can be represented by a tensor. Temperature can be specified by stating
a single numerical value called a scalar and is called a zeroth-order tensor. A force,
however, must be specified by stating both a magnitude and direction. It is an example of
a first-order tensor. Specifying a stress is even more complicated and requires stating a
magnitude and two directions—the direction of a force vector and the direction of the
normal vector to the plane on which the force acts. Stresses are represented by secondorder tensors.
Tensors are quantities independent of coordinate system.
Coordinate System
x3'
x3
P
Direction Cosines
x2'
x2
x1
x1'
Figure 15.1
Figure by MIT OCW.
x1
x2
x3
x1'
α11
α12
α13
x2'
α21
α22
α23
x3'
α31
α32
α33
α ij = cos φij
where φij is the angle of primed to original.
xi ' = α ij x j
xi = α ji x j '
α ij =
∂xi ' ∂x j
=
∂x j ∂xi '
Tensors:
a. 0th order (scalar) – quantity dependent only on position
b. 1st order (31 components)
Ai ' = α ij Aj
c. 2nd order (32 = 9 components) Aij ' = α isα jk Ask
d. 3rd order (33 = 27 components) Aijk ' = α isα jtα kp Astp
e. 4th order (34 = 81 components) Aijkl ' = α isα jtα kpα lq Astpq
Linear Elastic Solid
τ ij = cijkl ekl
cijkl is elastic modulus tensor
cijkl = constants ( ≠ history, ≠ displacement, ≠ time)
34 = 81 components
τ ij = τ ji , eij = e ji Æ cuts to 36
Strain Energy
1
∂U
U = τ ij eij ⇒ τ ij =
2
∂eij
Write in terms of powers of eij
U = α + βij eij + γ ijkl eij ekl
0 (to avoid spontaneous expansion, contraction)
∂U
= ( γ ijkl + γ klij ) ekl
∂eij
↓
cijkl = cklij ⇒ 21 individual components.
21 components – data fitting – ugh!
Lots of available information, but lots of hard work.
Reality – 21 components (triclinic)
↓
Simplicity – derive in homework
Isotropic
cijkl = λδ ijδ kl + µ (δ ik δ jl + δ ilδ jk )
λ and µ are Lame parameters.
τ ij = λ ekk δ ij + 2µ eij
Can also express
eij = Sijklτ kl
↑
Compliance tensor
Aside
e11 = S1132τ 32
↑
triclinic, trigonal
3
2
1
Figure 15.2
Figure by MIT OCW.
Sheer leads to lengthening.
Isotropic – can relate cijkl and Sijkl directly
τ ii = λ ekkδ ii + 2µ eii = ( 3λ + 2µ ) eii
2 µ eij = τ ij −
λδ ij
τ kk
2 µ + 3λ
Conventional moduli:
1. Hydrostatic comp.
Figure 15.3
Figure by MIT OCW.
τ ij = − pδij
τ ii = − pδii = 3λ ekk + 2µeii
= −3 p = (3λ + 2µ )eii
−
VP
p
=−
≡K
∆V
eii
where K = λ + 2 / 3µ is bulk modulus.
2. Uniaxial stress
Figure by MIT OCW.
τ 11 = T
other τ ij = 0
2µe1 = T −
λ
2µ + 3λ
T
T
≡ E (sometimes Y )
e1
where E =
µ (2µ + 3λ )
is Young’s modulus
µ+λ
Hook’s law:
T = Ee
e22 e33
=
≡ −ν
e11 e11
2µe22 = −
This is called Poisson’s ratio.
λ
2µ + 3λ
τ 11 ⇒ ν =
λ
2(µ + λ )
θ = e11 + e22 + e33 = e11 (1− 2ν )
fluid: µ → 0 ⇒ ν →
1
2
most material: ν = 0.2 − 0.3
1
ν=
⇒ λ =µ
4
It is Poisson solid.
steel: ν ; 0.3 − 0.33
seismically measured
λ + 2µ
vp =
ρ
vs =
µ
ρ
compare v p , vs → ν → discriminate rock types
3. Simple shear
τ 12 = τ 21 = τ
τ 12 = 2µe12 = 2Ge12
where G is shear modulus.
Note: Among λ, µ, K, ν , E, G only two are independent.
Useful forms:
τ 11 = (λ + 2µ )e11 + λ e22 + λ e33
τ 22 = λ e11 + (λ + 2µ )e22 + λ e33
τ 33 = λ e11 + λ e22 + (λ + 2µ )e33
or
e11 =
τ 11
E
e22 = −
e33 = −
−
ντ 22
ντ 11
E
ντ 11
E
E
−
ντ 33
E
+
τ 22
−
ντ 22
E
E
−
ντ 33
+
E
τ 33
E