2.001 - MECHANICS AND MATERIALS I
Lecture #14
Prof. Carol Livermore
Recall from last time:
Normal strains, changes in length
u(x, y, z) = ux (x, y, z)î + uy (x, y, z)ĵ + uz (x, y, z)k̂
∂ux
(x, y, z)
∂x
∂uy
=
(x, y, z)
∂y
xx =
yy
zz =
∂uz
(x, y, z)
∂z
Shear Strain
γxy = θ =
1
δ
L
γxy = θ1 + θ2 =
γxy =
δ1
δ2
+
L
L
∂uy
∂ux
+
∂y
∂x
xy = yx =
γxy
2
∂uy
1 ∂ux
xy =
+
2 ∂y
∂x
∂uz
1 ∂uy
yz =
+
2 ∂z
∂y
∂uz
1 ∂ux
xz =
+
2 ∂z
∂x
xx xy xz [ ] = yx yy yz .
zx zy zz What kind of strain you see (normal vs. shear) and its magnitude depend on
the relative orientation of deformation and coordinates.
EXAMPLE:
xy = 0
2
xy = 0
Relationship between σ and Recall uniaxial loading:
= δ/L
xx = δ/L
σ = E = σxx = Exx
σ0 = Exx
Look at thicker bar:
xx =
δ
; yy = −(something)
L
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In this case:
xx = axial
yy = lateral
lateral = −νaxial , whereν is Poisson’s Ratio (unitless).
Typically ν ≈ 0.3
Range 0 ≤ ν ≤ 0.5
Note: ν = 0.5 ⇒ incompressible
Microstructure view of Poisson’s Ratio
Recall Young’s Modulus
Vertical bonds do not extend
Diagonal bonds do extend
May be able to minimize energy
This leads to Poisson’s ratio, how this bond stretching energy is minimized.
Equations of Linear, Isotropic Elasticity
Linearity (E,ν are not a function of loading)
Superposition (A property of linearity)
σa ⇒ a
σb ⇒ b
ασa + βσb ⇒ αa + βb α and β are scalar constants.
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Isotropic:
Material properties are the same in all orientations.
Examples of anisotropic materials
Wood (against the grain, with the grain)
Single crystals (depends on which crystal direction)
Relationship between σ and (the constitutive equations) is not orientation
dependent.
Elastic:
Deformation is removed when load is released (deformation is fully recoverable)
axial =
σaxial
in uniaxial loading
E
lateral = −νaxial
Consider:
σxx , σyy , σzz =
−
σxy = σyz = σxz = 0
σyy
σzz
σxx
−ν
−ν
E
E
E
σxx
σyy
σzz
yy = −ν
+
−ν
E
E
E
σxx
σyy
σzz
zz = −ν
−ν
+
E
E
E
1
σxx − ν(σyy + σzz )
xx =
E
1
yy =
σyy − ν(σxx + σzz )
E
1
zz =
σzz − ν(σxx + σyy )
E
xx =
Multi-axial stress-strain relationships among normal stresses and strains for linear isotropic elastic materials.
What about shear stresses and strains?
σxy = Gγxy where G is the shear modulus with units of Pa
5
G=
E
2(1 + ν)
For linear, isotropic elastic materials, 2 material constants fully define a material.
1
σxy
G
1
= σxz
G
1
= σyz
G
γxy =
γxz
γyz
1
σxy
2G
1
σxz
xz =
2G
1
σyz
yz =
2G
Equations of linear isotropic elasticity (aka. Constitutive Relationships)
xy =
1
σxx − ν(σyy + σzz )
E
1
=
σyy − ν(σxx + σzz )
E
1
=
σzz − ν(σxx + σyy )
E
xx =
yy
xx
1
σxy
2G
1
σxz
xz =
2G
1
σyz
yz =
2G
EXAMPLE: Block in a frictionless channel
xy =
6
What are all stresses and strains?
No shears due to frictionless
Boundary Conditions:
xx = 0
σxx = ?
yy = Given σyy = ?
zz = ?
σzz = 0
1
=
σxx − ν(σyy + σzz )
E
xx
So:
1
σxx − νσyy
E
1
=
σyy − ν(σxx + σzz )
E
0=
yy
So:
yy =
zz =
1
σyy − νσxx
E
1
σzz − ν(σxx + σyy )
E
So:
zz =
1
−ν(σxx + σyy )
E
Solve:
σxx = νσyy
Plug in:
yy
=
=
1
σyy − ν(νσyy )
E
1
σyy − ν 2 σyy
E
σyy =
yy E
(1 − ν 2 )
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Plug in:
σxx =
νyy E
(1 − ν 2 )
Plug in:
zz =
yy E
ν νyy E
1
+
−ν(σxx + σyy ) −
E
E (1 − ν 2 ) (1 − ν 2 )
ν+1
zz = −ν
(1 + ν)(1 − ν)
zz =
−ν
yy
1−ν
EXAMPLE: Hydrostatic Pressure
Q: What is change in volume?
σxx = σyy = σzz = −p
Initial Volume:
Vi = dxdydz
Final Volume:
(1 + xx )dx(1 + yy )dy(1 + zz )dz
ΔV = Vf − Vi
ΔV = (1 + xx )(1 + yy )(1 + zz )Vi − Vi
For small strains:
2 , 3 ≈ 0
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So:
(1 + xx + yy + zz )Vi − Vi
xx
1
−p − ν(−p − p)
=
E
=
1
σxx − ν(σyy + σzz )
E
−1(1 − 2ν)
p
E
−1(1 − 2ν)
p
yy =
E
−1(1 − 2ν)
p
zz =
E
3(1 − 2ν)
−3(1 − 2ν)
pVi
ΔV 1 −
p Vi − V i =
E
E
xx =
ΔV
−3(1 − 2ν)
=
p
Vi
E
−p
E
=
=k
3(1 − 2ν)
ΔV /Vi
k is the bulk modulus.
Note: If ν = 0.5,
ΔV
Vi
= 0 (Incompressible) and k → ∞
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