Idea Transcript
Plain Strain We will derive the transformation equations that relate the strains in inclined directions to the strain in the reference directions. State of plain strain - the only deformations are those in the xy plane, i.e. it has only three strain components εx, εy and γxy. Plain stress is analogous to plane stress, but under ordinary conditions they do not occur simultaneously, except when σx = -σy and when ν = 0
Strain components εx, εy, and γxy in the xy plane (plane strain).
Comparison of plane stress and plane strain.
Transformation Equations for Plain Strain Assume that the strain εx, εy and γxy associated with the xy plane are known. We need to determine the normal and shear strains (εx1 and γx1y1) associated with the x1y1 axis. εy1 can be obtained from the equation of εx1 by substituting θ + 90 for θ.
For an element of size δx δy In the x direction, the strain εx produces an elongation εx δx. The diagonal increases in length by εx δx cos θ. In the y direction, the strain εy produces an elongation εy δy. The diagonal increases in length by εy δy sin θ.
The shear strain γxy produces a distortion. The upper face moves γxy δy. This deformation results in an increase of the diagonal equal to: γxy δy cos θ
The total increase Δd of the diagonal is the sum of the preceding three expressions, thus:
Δd = εx δx cos θ + εy δy sin θ + γxy δy cos θ The normal strain εx1 in the x1 direction is equal to the increase in length divided by the initial length δs of the diagonal.
εx1 = Δd / ds = εx cos θ δx/δs + εy sin θ δy/δs + γxy cos θ δy/δs Observing that δx/δs = cos θ and δy/δs = sin θ
ε X 1 = ε X cos 2 θ + ε Y sin 2 θ + γ XY sin θ cos θ ε X1
⎛ γ XY = ε X cos θ + ε Y sin θ + ⎜ ⎝ 2 2
2
⎞ ⎟ 2 sin θ cos θ ⎠
Shear Strain γx1y1 associated with x1y1 axes. This strain is equal to the decrease in angle between lines in the material that were initially along the x1 and y1 axes. Oa and Ob were the lines initially along the x1 and y1 axis respectively. The deformation caused by the strains εx, εy and γxy caused the Oa and Ob lines to rotate and angle α and β from the x1 and y1 axis respectively. The shear strain γx1y1 is the decrease in angle between the two lines that originally were at right angles, therefore, γx1y1 = α+β.
γ
The angle α can be found from the deformations produced by the strains εx, εy and γxy . The strains εx and γxy produce a cwrotation, while the strain εy produces a ccw-rotation.
Let us denote the angle of rotation produced by εx , εy and γxy as α1 , α2 and α3 respectively. The angle α1 is equal to the distance εx δx sinθ divided by the length δs of the diagonal: α1 = εx sinθ dx/ds α2 = εy cosθ dy/ds α3 = γxy sinθ dy/ds Observing that dx/ds = cos θ and dy/ds = sin θ. The resulting ccwrotation of the diagonal is α = - α1 + α2 - α3 = - (εx – εy) sinθ cosθ - γxy sin2θ
The rotation of line Ob which initially was at at 90o to the line Oa can be found by substituting θ +90 for θ in the expression for α. Because β is positive when clockwise. Thus β = (εx – εy) sin(θ + 90) cos(θ + 90) + γxy sin2(θ +90) β = - (εx – εy) sinθ cosθ + γxy cos2θ Adding α and β gives the shear strain γx1y1 γx1y1 = α + β = - 2(εx – εy) sinθ cosθ + γxy (cos2θ - sin2θ) To put the equation in a more useful form:
γ X 1Y 1 2
= −ε X sin θ cos θ + ε Y sin θ cos θ +
ε X1 = ε X εY1 = ε X
⎛ γ XY cos θ + ε Y sin θ + ⎜ ⎝ 2 ⎛γ sin 2 θ + ε Y cos 2 θ − ⎜ XY ⎝ 2 2
2
γ XY 2
(cos
2
θ − sin θ )
⎞ ⎟ 2 sin θ cos θ ⎠ ⎞ ⎟ 2 sin θ cos θ ⎠
2
⎡ ⎤ 2 ⎡ cos ε θ ⎢ X1 ⎥ ⎢ ε ⎥ = ⎢ sin 2 θ ⎢ Y1 ⎥ ⎢ ⎢ γ X 1Y 1 ⎥ ⎢⎣− sin θ cosθ ⎢⎣ 2 ⎥⎦ ⎡ ⎤ ⎡ ⎤ ⎢ ε X1 ⎥ ⎢ εX ⎥ ⎢ ε ⎥ = [T ]× ⎢ ε ⎥ ⎢ Y1 ⎥ ⎢ Y ⎥ ⎢ γ X 1Y 1 ⎥ ⎢ γ XY ⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎡ ⎤ ⎡ ⎤ ⎢ εX ⎥ ⎢ ε X1 ⎥ ⎢ ε ⎥ = [T ]−1 × ⎢ ε ⎥ ⎢ Y ⎥ ⎢ Y1 ⎥ ⎢ γ XY ⎥ ⎢ γ X 1Y 1 ⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦
sin θ cos 2 θ sin θ cosθ 2
⎡ cos 2 θ [T ] = ⎢⎢ sin 2 θ ⎢− sin θ cosθ ⎣
⎡ ⎢ ε xx ⎢1 [ε ] = ⎢ γ xy ⎢2 ⎢1 γ ⎢⎣ 2 xz
⎡ ⎤ 2 sin θ cosθ ⎤ ⎢ ε X ⎥ ⎥⎢ − 2 sin θ cosθ ⎥ ε Y ⎥ ⎢ ⎥ 2 2 γ cos θ − sin θ ⎥⎦ ⎢ XY ⎥ ⎢⎣ 2 ⎥⎦
(
)
sin 2 θ cos 2 θ sin θ cosθ
1 γ yx 2
ε yy 1 γ yz 2
2 sin θ cosθ − 2 sin θ cosθ cos 2 θ − sin 2 θ
(
⎤ ⎥ ⎥ ⎥ ⎦
)
1 ⎤ γ zx ⎥ 2 1 ⎥ γ zy ⎥ = Strain _ Tensor 2 ⎥ ε zz ⎥ ⎥⎦
Transformation Equations for Plain Strain Using known trigonometric identities, the transformation equations for plain strain becomes:
(ε X
γ XY + ε Y ) (ε X − ε Y ) cos 2θ + sin 2θ ε X1 = + 2 2 2 (ε X − ε Y ) γ X 1Y 1 γ XY =− sin 2θ + cos 2θ 2 2 2 These equations are counterpart of the equations for plane stress where εx1, εx, γx1y1 and γxy correspond to σx1, σx, τx1y1 and τxy respectively. There are also counterparts for principal stress and Mohr’s circle. εx1 + εy1 = εx+ εy
Principal Strains The angle for the principal strains is The value for the principal strains are
γ XY tan 2θ P = ε X − εY
( ε X + εY ) + ε1 =
⎛ ε X − ε Y ⎞ ⎛ γ XY ⎞ ⎜ ⎟ +⎜ ⎟ 2 ⎝ ⎠ ⎝ 2 ⎠
(ε X + ε Y ) ε2 = −
⎛ ε X − ε Y ⎞ ⎛ γ XY ⎞ ⎜ ⎟ +⎜ ⎟ 2 ⎝ ⎠ ⎝ 2 ⎠
2
2
2
2
2
2
Maximum Shear The maximum shear 2 2 strains in the xy plane γ MAX ⎛ ε X − ε Y ⎞ ⎛ γ XY ⎞ =+ ⎜ ⎟ +⎜ ⎟ or γ MAX = (ε1 − ε 2 ) are associated with 2 ⎝ 2 ⎠ ⎝ 2 ⎠ axes at 45o to the γ MAX (ε1 − ε 2 ) = directions of the 2 2 principal strains:
Mohr’s Circle for Plane Strain
Example An element of material in plane strain undergoes the following strains: εx=340x10-6 εy=110x10-6 γxy=180x10-6 Determine the following: (a) the strains of an element oriented at an angle θ = 30o ; (b) the principal strains and (c) the maximum shear strains. Solution
ε X1 = γ X 1Y 1 2
(ε X
+ ε Y ) (ε X − ε Y ) γ + cos 2θ + XY sin 2θ 2 2 2 (ε − ε Y ) γ =− X sin 2θ + XY cos 2θ 2 2
Then εx1 = 225x10-6 + (115x10-6) cos 60o + (90x10-6) sin 60o = 360x10-6 ½ γx1y1 = - (115x10-6) (sin 60o ) + ( 90x10-6)(cos 60o) = - 55x10-6 Therefore γx1y1 = - 110x10-6 The strain εy1 can be obtained from the equation εx1 + εy1 = εx+ εy εy1 = (340 + 110 -360)10-6 = 90x10-6 (b) Principal Strains The principal strains are readily ε1 = determine from the following equations: ε1 = 370x10-6
ε2 = 80x10-6
ε2 =
(ε X + ε Y ) 2
(ε X + ε Y ) 2
⎛ ε X − εY ⎞
+ ⎜ ⎝
2
2
⎛ γ XY ⎞
2
⎟ ⎟ +⎜ 2 ⎠ ⎠ ⎝
⎛ ε − ε Y ⎞ ⎛ γ XY ⎞ − ⎜ X ⎟ ⎟ +⎜ 2 2 ⎠ ⎠ ⎝ ⎝ 2
2
(c) Maximum Shear Strain The maximum shear strain is calculated from the equation: ½ γmax = SQR[((εx – εy)/2)2 + ( ½ γxy)2]
or
γmax = (ε1 – ε2 )
Then γmax = 290x10-6 The normal strains of this element is εaver = ½ (εx + εy) = 225x10-6
For 3-D problems Which is a symmetrical matrix. As in the case of stresses: ⎡ ⎢ε x − ε ⎢1 ⎢ γ yx ⎢2 ⎢ 1γ ⎢⎣ 2 zx
εy −ε
εx −ε
1 γ xy 2
1 γ yx 2 1 γ zx 2
εy −ε
1 γ xy 2 1 γ zy 2
1 γ zy 2
⎤ ⎥ k 0 ⎥⎡ ⎤ ⎡ ⎤ ⎥ ⎢ l ⎥ = ⎢0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣m⎥⎦ ⎢⎣0⎥⎦ ⎥ εz −ε ⎥⎦ 1 γ xz 2 1 γ yz = 0 2 1 γ xz 2 1 γ yz 2
εz −ε
⎡ ⎢ εx ⎢1 [ε ] = ⎢ γ yx ⎢2 ⎢1 γ ⎢⎣ 2 zx
ε 1 〉ε 2 〉ε 3
1 γ xy 2
εy 1 γ zy 2
1 ⎤ γ xz ⎥ 2 1 ⎥ γ yz ⎥ 2 ⎥ ε z ⎥⎥ ⎦
ε 3 − I1 ⋅ ε 2 + I 2 ⋅ ε − I 3 = 0 I1 = ε x + ε y + ε z ⎛ γ xy ⎞ ⎛ γ xz ⎞ ⎛ γ yz ⎞ ⎟⎟ ⎟⎟ − ⎜ ⎟ − ⎜⎜ I 2 = ε x ⋅ ε y + ε y ⋅ ε z + ε x ⋅ ε z − ⎜⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2
2
2
⎛ γ xy ⎞ ⎛ γ yz ⎞ ⎛ γ xy ⎞ ⎛ γ xz ⎞ ⎛ γ yz ⎞ ⎛ γ xz ⎞ ⎟⎟ ⎟⎟ − ε y ⋅ ⎜ ⎟ − ε z ⋅ ⎜⎜ ⎟⎟ − ε x ⋅ ⎜⎜ ⎟⎟ ⋅ ⎜ ⎟ ⋅ ⎜⎜ I 3 = ε x ⋅ ε y ⋅ ε z + 2 ⋅ ⎜⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2
2
p
Dilatation (Volume strain) Under pressure: the volume will change
ΔV Δ= = εx +εy +εz V
p
p V-ΔV
p
2
Strain Deviator
Mean strain
Δ εx +εy +εz = 3 3
It produces a volume change (not a shape change)
1 1 1 ⎡ ⎤ − Δ ε γ γ xy xz ⎥ ⎢ x 3 2 2 ⎢ 1 ⎥ Strain Deviator Matrix 1 1 [Dε ] = ⎢ γ yx ε y − Δ γ yz ⎥ 3 2 ⎢ 2 ⎥ 1 1 ⎥ ⎢ 1γ γ zy ε z − Δ⎥ 1 ⎡ ⎤ zx ⎢ 0 0 ⎥ 2 3 ⎦ ⎣ 2 ⎢ε 1 − 3 Δ ⎢ ⎥ 1 0 ⎥ [Dε ] = ⎢ 0 ε2 − Δ 3 ⎢ ⎥ 1 ⎥ ⎢ 0 0 ε 3 − Δ⎥ ⎢⎣ 3 ⎦
Application : Strain Gauge and Strain Rosette
Relationships between Stress and Strain
(Hooke’s Law) When strains are small, most of materials are linear elastic. Tensile: σ=Εε Shear:
Poisson’s ratio
Δl z εz = l0 z
Δl x εx = − l0 x Nominal lateral strain (transverse strain)
lateral strain ε x =− Poisson’s ratio: ν = − tensile strain εz
τ=Gγ
Relationships between Stress and Strain An isotropic material has a stress-strain relationships that are independent of the orientation of the coordinate system at a point. A material is said to be homogenous if the material properties are the same at all points in the body For isotropic materials Elastic Stress-Strain Relationships
Uniaxial
Principal Stresses σ 1 = Eε1 σ2 = 0 σ3 = 0
Principal Strains ε1 =
σ1 E
ε 2 = −ν ε 3 = −ν
σ1 E
σ1 E
Uniaxial Stresses
⎧σ x ⎫ ⎪σ ⎪ ⎪ y⎪ ⎪⎪σ z ⎪⎪ [σ ] = ⎨ ⎬ ⎪τ yz ⎪ ⎪τ zx ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭
⎧σ x ⎫ ⎪0⎪ ⎪ ⎪ ⎪⎪ 0 ⎪⎪ [σ ] = ⎨ ⎬ ⎪0⎪ ⎪0⎪ ⎪ ⎪ ⎪⎩ 0 ⎪⎭
⎧εx ⎫ ⎪ε ⎪ ⎪ y⎪ ⎪⎪ ε z ⎪⎪ [ε ] = ⎨ ⎬ ⎪γ yz ⎪ ⎪γ zx ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭
⎧ε x ⎫ ⎪ε ⎪ ⎪ y⎪ ⎪⎪ε z ⎪⎪ [ε ] = ⎨ ⎬ ⎪0⎪ ⎪0⎪ ⎪ ⎪ ⎪⎩ 0 ⎪⎭
εx =
σ σx E
= Eε x
x
ε y = −ν
⎡ 1 ⎢ E ⎢ υ ⎧ε x ⎫ ⎢− ⎪ε ⎪ ⎢ E ⎪ y⎪ ⎢ υ ⎪⎪ε z ⎪⎪ ⎢− E ⎨ ⎬=⎢ ⎪0⎪ ⎢ 0 ⎪0⎪ ⎢ ⎪ ⎪ ⎢ 0 ⎪⎩ 0 ⎪⎭ ⎢ ⎢ ⎢⎣ 0
−
σx E
υ
E 1 E
−
ε z = −ν
υ
E 0
− −
υ E
υ
E 1 E 0
σx E
0
0
0
0
0
0
1 G
0
0
0
0
1 G
0
0
0
0
⎤ 0⎥ ⎥ 0 ⎥ ⎧σ x ⎫ ⎥⎪ 0 ⎪ ⎪ ⎪ 0 ⎥⎪ 0 ⎪ ⎥⎪ ⎪ ⎥⎨ 0 ⎬ 0 ⎥⎪ ⎪ ⎥⎪ 0 ⎪ 0 ⎥ ⎪⎪⎩ 0 ⎪⎪⎭ ⎥ 1⎥ G ⎥⎦
Principal Stresses Biaxial
σ2 σ3 Triaxial
E (ε 1 + νε 2 ) 1+ν 2 E (ε 2 + νε 1 ) = 1+ν 2 = 0
σ1 =
Principal Stresses
ε 1 (1 − ν ) + ν (ε 2 + ε 3 ) σ1 = E 1 − ν − 2ν 2 ε 2 (1 − ν ) + ν (ε 1 + ε 3 ) σ2 = E 1 − ν − 2ν 2 ε 3 (1 − ν ) + ν (ε 1 + ε 2 ) σ3 = E 1 − ν − 2ν 2
Principal Strains σ1 σ2 ε 1 = −ν E
ε2 =
σ2 E
ε 3 = −ν
E
−ν
σ1
σ1
−ν
E
E
σ2 E
Principal Strains
ε1 = ε2 = ε3 =
σ1 E
σ2 E
σ3 E
−ν
σ2
−ν
σ1
−ν
σ1
E
E
E
−ν
σ3
−ν
σ3
−ν
σ2
E
E
E
Triaxial Stresses
⎧σ x ⎫ ⎪σ ⎪ ⎪ y⎪ ⎪⎪σ z ⎪⎪ [σ ] = ⎨ ⎬ ⎪τ yz ⎪ ⎪τ zx ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭ ⎧εx ⎫ ⎪ε ⎪ ⎪ y⎪ ⎪⎪ ε z ⎪⎪ [ε ] = ⎨ ⎬ ⎪γ yz ⎪ ⎪γ zx ⎪ ⎪ ⎪ ⎪⎩γ xy ⎪⎭
⎧σ x ⎫ ⎪σ ⎪ ⎪ y⎪ ⎪⎪σ z ⎪⎪ [σ ] = ⎨ ⎬ ⎪0⎪ ⎪0⎪ ⎪ ⎪ ⎪⎩ 0 ⎪⎭ ⎧ε x ⎫ ⎪ε ⎪ ⎪ y⎪ ⎪⎪ε z ⎪⎪ [ε ] = ⎨ ⎬ ⎪0⎪ ⎪0⎪ ⎪ ⎪ ⎪⎩ 0 ⎪⎭
ν ν 1 σx − σy − σz E E E ν ν 1 εy = − σx + σy − σz E E E ν ν 1 εz = − σx − σ y + σz E E E
εx =
⎡ 1 ⎢ E ⎢ υ ε ⎧ x ⎫ ⎢− ⎪ε ⎪ ⎢ E ⎪ y⎪ ⎢ υ ⎪⎪ε z ⎪⎪ ⎢− E ⎨ ⎬=⎢ ⎪0⎪ ⎢ 0 ⎪0⎪ ⎢ ⎪ ⎪ ⎢ 0 ⎪⎩ 0 ⎪⎭ ⎢ ⎢ ⎢⎣ 0
−
υ
E 1 E
−
υ
E 0
− −
υ E
υ
E 1 E 0
0
0
0
0
0
0
1 G
0
0
0
0
1 G
0
0
0
0
⎤ 0⎥ ⎥ 0 ⎥ ⎧σ x ⎫ ⎥ ⎪σ y ⎪ ⎪ ⎪ ⎥ 0 ⎪σ ⎪ ⎥⎪ z ⎪ ⎥⎨ 0 ⎬ 0 ⎥⎪ ⎪ ⎥⎪ 0 ⎪ 0 ⎥ ⎪⎪⎩ 0 ⎪⎪⎭ ⎥ 1⎥ G ⎥⎦
For an isotropic material, the principal axes for stress and the principal axes for strain coincide. 1 εx −εy =
γ XY tan 2θ ε = ε X − εY
γ xy =
τ xy
τ xy
E
(1 −ν )(σ x − σ y )
G
2τ xy γ XY E G tan 2θ ε = = = ⋅ = tan 2θσ ε X − ε Y 1 (1 −ν )(σ − σ ) 2G (1 − ν ) (σ x − σ y ) x y E
1 (σ x −νσ y ) E 1 ε y = (σ y −νσ x ) E
Plane Stress ε x =
γ xy =
Plane Strain
τ xy
εz = −
ν
γ yz = 0
E
(σ
x
+σ y )
γ zx = 0
G
[
]
σz =
[
]
τ yz = 0
E (1 −ν )ε x +νε y σx = (1 +ν )(1 − 2ν ) E (1 −ν )ε y +νε x σy = (1 +ν )(1 − 2ν ) τ xy = Gγ xy
νE
(1 +ν )(1 − 2ν )
τ xz = 0
(ε
x
+εy)
Tensors and Elasticity Common misconception Æ Cubic materials are isotropic, i.e. they have the same properties in every direction. Many properties are isotropic in cubic crystal, but elasticity, electrostriction and magnetostriction are anisotropic even in cubic crystals. Example turbine blade (single crystal). Ni based with cuboidal Ni3Ti intermetallic. It shows variation in the elastic constant with the directions in the material.
Some polycrystalline materials develop preferred orientations during processing. They will show a degree of anisotropy that is dependent on the degree of preferred orientation or texture.
Tensor: A specific type of matrix representation that can relate the directionality of either a material property (property tensors – conductivity, elasticity) or a condition/state (condition tensors – stress, strain). Tensor of zero‐rank: scalar quantity (density, temperature). Tensor of first‐rank: vector quantity (force, electric field, flux of atoms). Tensor of second‐rank: relates two vector quantities (flux of atoms with concentration gradient).
Tensor third‐rank: relates vector with a second rank tensor (electric field with strain in a piezoelectric material) Tensor Fourth‐rank: relates two second rank tensors (relates strain and stress – Elasticity) The key to understanding property or condition tensors is to recognize that tensors can be specified with reference to some coordinate system which is usually defined in 3‐D space by orthogonal axes that obey a right‐hand rule. Rotation Matrix and Euler Angles: Several schemes can be used to produce a rotation matrix. The three Euler angles are given as three counterclockwise rotations: (a)A rotation about a z‐axis, defined as φ1 (b)A rotation about the new x‐axis, defined as Φ (c)A rotation about the second z‐position, defined as φ2
The rotation matrix a is given by the matrix multiplication of the rotation matrices of each individual rotations: 0 0 ⎤ ⎡ cos φ1 sin φ1 0⎤ ⎡ cos φ2 sin φ2 0⎤ ⎡1 [a ] = aφ 2 ⋅ aΦ ⋅ aφ1 = ⎢⎢− sin φ2 cosφ2 0⎥⎥ × ⎢⎢0 cos Φ sin Φ ⎥⎥ × ⎢⎢− sin φ1 cosφ1 0⎥⎥ ⎢⎣ 0 0 1⎥⎦ 0 1⎥⎦ ⎢⎣0 − sin Φ cos Φ ⎥⎦ ⎢⎣ 0 cos φ2 sin φ1 + cos Φ cos φ1 sin φ2 sin Φ sin φ2 ⎤ ⎡ cos φ2 cos φ1 − cos Φ sin φ1 sin φ2 [a ] = ⎢⎢− sin φ2 cosφ1 − cos Φ sin φ1 cosφ2 − sin φ2 sin φ1 + cos Φ cosφ1 cosφ2 sin Φ cosφ2 ⎥⎥ ⎢⎣ − sin Φ cos φ1 sin Φ sin φ1 cos Φ ⎥⎦
Stereographic Projection Crystallographic directions, plane normals and planes can be all represented in the stereographic projection.
2 31
Locating a pole in a stereographic projection: Find the angle of the pole with the three axes:
⎡1 0 0 ⋅ 2 3 1⎤ o cos -1 ⎢ ⎥ = 57.6 ⎣ 1 14 ⎦ ⎡ 0 1 0 ⋅ 2 3 1⎤ o cos -1 ⎢ ⎥ = 36.7 ⎣ 1 14 ⎦ ⎡ 0 0 1 ⋅ 2 3 1⎤ o cos ⎢ ⎥ = 74.5 ⎣ 1 14 ⎦ -1
Example The relationship between orientation and applied stress is important in describing the mechanical performance of many crystalline metals and composites. The relationship between applied stress and crystal direction is essential in interpreting the microscopic deformation mechanisms operating in deforming crystals. ⎡2 0 0 ⎤ Consider a property‐tensor or a condition tensor T [T ] = ⎢0 3 0 ⎥ ⎢ ⎥ in the original {x y z} axes given by:
⎢⎣0 0 − 1⎥⎦
Find the tensor T’ , for the rotation shown below from the initial [100], [010], [001] axes of a cubic crystal
φ1 = 45
o
Φ = 54.7 o
φ2 = 0 o
cos φ2 sin φ1 + cos Φ cos φ1 sin φ2 sin Φ sin φ2 ⎤ ⎡ cos φ2 cos φ1 − cos Φ sin φ1 sin φ2 [a ] = ⎢⎢− sin φ2 cosφ1 − cos Φ sin φ1 cosφ2 − sin φ2 sin φ1 + cos Φ cosφ1 cosφ2 sin Φ cosφ2 ⎥⎥ ⎢⎣ − sin Φ cos φ1 sin Φ sin φ1 cos Φ ⎥⎦ cos 0 sin 45 + cos 54.7 cos 45 sin 0 sin 54.7 sin 0 ⎤ ⎡ cos 0 cos 45 − cos 54.7 sin 45 sin 0 [a ] = ⎢⎢− sin 0 cos 45 − cos 54.7 sin 45 cos 0 − sin 0 sin 45 + cos 54.7 cos 45 cos 0 sin 54.7 cos 0⎥⎥ ⎢⎣ sin 54.7 sin 45 cos 54.7 ⎥⎦ − sin 54.7 cos 45 cos 45 sin 45 0 ⎤ ⎡ 1 ⎡ 1 ⎤ ⎢ ⎥ 0 [a ] = ⎢− cos 54.7 sin 45 + cos 54.7 cos 45 sin 54.7 ⎥ ⎢ 2 ⎥ 2 ⎢ ⎥ ⎢⎣ sin 54.7 sin 45 − sin 54.7 cos 45 cos 54.7 ⎥⎦ 1 1 2
[a ] = ⎢− ⎢ 6 ⎢ 1 ⎢ ⎣ 3
T = [a ]× T × [a ]
T
'
⎡ 1 ⎢ 2 ⎢ 1 [T ' ] = ⎢− ⎢ 6 ⎢ 1 ⎢ ⎣ 3
1 2 1 6 1 − 3
⎤ ⎡ 0 ⎥ ⎢ 2 0 0 ⎡ ⎤ ⎥ ⎢ 2 ⎥ ⎢ × ⎢0 3 0 ⎥⎥ × ⎢ ⎢ 6⎥ 1 ⎥ ⎢⎣0 0 − 1⎥⎦ ⎢ ⎥ ⎢ 3⎦ ⎣
1 2 1 2 0
−
1 6 1 6 2 6
6 1 − 3
⎥ 6⎥ 1 ⎥ ⎥ 3⎦
1 ⎤ 3 ⎥ ⎡ 2.5 0.289 − 0.408⎤ ⎥ 1 ⎥ ⎢ − = ⎢ 0.289 0.167 − 1.65 ⎥⎥ 3⎥ 1 ⎥ ⎢⎣− 0.408 − 1.65 1.33 ⎥⎦ ⎥ 3 ⎦
Isotropic Materials
εx =
σx E
ε y = −ν ε z = −ν
σy
−ν
σx E
σx E
E +
σy
−ν
1 γ yz = τ yz G 1 γ zx = τ zx G 1 γ xy = τ xy G
−ν
E
σz E
−ν
σy E
+
σz E
σz E
ε x = S11σ x + S12σ y + S13σ z + S14τ yz + S15τ zx + S16τ xy
ε y = S 21σ x + S 22σ y + S 23σ z + S 24τ yz + S 25τ zx + S 26τ xy
ε z = S31σ x + S32σ y + S33σ z + S34τ yz + S35τ zx + S36τ xy γ yz = S 41σ x + S 42σ y + S 43σ z + S 44τ yz + S 45τ zx + S 46τ xy
γ zx = S51σ x + S52σ y + S53σ z + S54τ yz + S55τ zx + S56τ xy γ xy = S61σ x + S62σ y + S63σ z + S64τ yz + S65τ zx + S66τ xy ⎧ ε x ⎫ ⎡ S11 ⎪ε ⎪ ⎢ S 21 y ⎪ ⎪ ⎢ ε = S σ ⎪ ⎪ ⎪ ε z ⎪ ⎢ S31 ⎨ ⎬=⎢ ⎪γ yz ⎪ ⎢ S 41 ⎪γ zx ⎪ ⎢ S51 S is the compliance matrix ⎪ ⎪ ⎢ ⎪⎩γ xy ⎪⎭ ⎢⎣ S61
[ ] [ ][ ]
S12 S 22 S32
S13 S 23 S33
S14 S 24 S34
S15 S 25 S35
S 42 S52 S62
S 43 S53 S63
S 44 S54 S64
S 45 S55 S65
S16 ⎤ ⎧σ x ⎫ ⎪σ ⎪ ⎥ S 26 ⎥ ⎪ y ⎪ S36 ⎥ ⎪⎪σ z ⎪⎪ ⎥⎨ ⎬ S 46 ⎥ ⎪τ yz ⎪ S56 ⎥ ⎪τ zx ⎪ ⎥⎪ ⎪ S66 ⎥⎦ ⎪⎩τ xy ⎪⎭
Isotropic Materials An isotropic material has stress-strain relationships that are independent of the orientation of the coordinate system at a point. The isotropic material requires only two independent material constants, namely the Elastic Modulus and the Poisson’s Ratio. ν ν ⎤ ⎡ 1 0 0 0 − − ⎥ ⎢ E E E ⎥ ⎢ ν 1 ν ⎧ ε x ⎫ ⎢− 0 0 0 ⎥ ⎧σ x ⎫ − ⎪σ ⎪ ⎪ε ⎪ ⎢ E E E ⎥ y⎪ ⎪ ⎪ y⎪ ⎢ ν 1 ν ⎥ 0 0 0 − − ⎪⎪ ε z ⎪⎪ ⎢ ⎥ ⎪⎪σ z ⎪⎪ E E E ⎨ ⎬=⎢ ⎥ ⎨τ ⎬ 1 γ ⎪ yz ⎪ ⎢ 0 0 0 0 0 ⎥ ⎪ yz ⎪ G ⎪γ zx ⎪ ⎢ ⎥ ⎪τ zx ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎢ 0 ⎥ 0 0 0 0 τ xy ⎪⎭ ⎪ ⎪⎩γ xy ⎪⎭ ⎢ ⎩ ⎥ G ⎢ 1⎥ 0 0 0 0 0 ⎢⎣ G ⎥⎦
⎡ E (1 − ν ) ⎧σ x ⎫ ⎢⎢ (1 + ν )(1 − 2ν ) ⎪σ ⎪ ⎢ Eν ⎪ y ⎪ ⎢ (1 + ν )(1 − 2ν ) ⎪⎪σ z ⎪⎪ ⎢ Eν ⎨ ⎬=⎢ ⎪τ yz ⎪ ⎢ (1 + ν )(1 − 2ν ) ⎪τ zx ⎪ ⎢ 0 ⎪ ⎪ ⎢ ⎪⎩τ xy ⎪⎭ 0 ⎢ 0 ⎢⎣
Eν (1 + ν )(1 − 2ν ) E (1 − ν ) (1 + ν )(1 − 2ν ) Eν (1 + ν )(1 − 2ν ) 0 0 0
[σ ] = [C ][ε ] C is the elastic or stiffness matrix
Eν 0 (1 + ν )(1 − 2ν ) Eν 0 (1 + ν )(1 − 2ν ) E (1 − ν ) 0 (1 + ν )(1 − 2ν ) 0 G 0 0
0 0
⎤ 0 0⎥ ε ⎥ ⎧⎪ x ⎫ ⎪ ⎥ ε 0 0 ⎪ y⎪ ⎥⎪ ⎪ ⎥⎪ ε z ⎪ 0 0 ⎥ ⎨γ ⎬ yz ⎥⎪ ⎪ 0 0 ⎥ ⎪γ zx ⎪ ⎪ ⎪ ⎥ G 0 ⎪⎩γ xy ⎪⎭ ⎥ 0 G ⎥⎦
The isotropic material requires only two independent material constants, namely the Elastic Modulus and the Poisson’s Ratio.
Anisotropic Materials Up to this point we have limited the study of the properties of materials to isotropic materials. For the most general linearly elastic anisotropic materials, a particular component of stress is assumed to depend of all six components of strain.
σ x = C11ε x + C12ε y + C13ε z + C14γ yz + C15γ zx + C16γ xy Where Cij are constants if the material is homogeneous
⎧σ x ⎫ ⎡C11 ⎪σ ⎪ ⎢ ⎪ y ⎪ ⎢C21 ⎪⎪σ z ⎪⎪ ⎢C31 ⎨ ⎬=⎢ ⎪τ yz ⎪ ⎢C41 ⎪τ zx ⎪ ⎢C51 ⎪ ⎪ ⎢ ⎪⎩τ xy ⎪⎭ ⎢⎣C61
C12 C22 C32 C42 C52 C62
C13 C14 C23 C24 C33 C34 C43 C44 C53 C54 C63 C64
C15 C25 C35 C45 C55 C65
C16 ⎤ ⎧ ε x ⎫ ⎪ε ⎪ ⎥ C26 ⎥ ⎪ y ⎪ C36 ⎥ ⎪⎪ ε z ⎪⎪ ⎥⎨ ⎬ C46 ⎥ ⎪γ yz ⎪ C56 ⎥ ⎪γ zx ⎪ ⎥⎪ ⎪ C66 ⎥⎦ ⎪⎩γ xy ⎪⎭
Taking energy considerations the coefficients of this matrix are symmetric. Hence, instead of 36 independent constant, we have 21 independent constants
⎧σ x ⎫ ⎡C11 ⎪σ ⎪ ⎢ ⎪ y ⎪ ⎢C12 ⎪⎪σ z ⎪⎪ ⎢C13 ⎨ ⎬=⎢ ⎪τ yz ⎪ ⎢C14 ⎪τ zx ⎪ ⎢C15 ⎪ ⎪ ⎢ ⎪⎩τ xy ⎪⎭ ⎢⎣C16
C12
C13
C14
C15
C22 C23 C24 C25
C23 C33 C34 C35
C24 C34 C44 C45
C25 C35 C45 C55
C26
C36
C46
C56
C is referred to as the elastic matrix or stiffness matrix.
C16 ⎤ ⎧ ε x ⎫ ⎪ε ⎪ ⎥ C26 ⎥ ⎪ y ⎪ C36 ⎥ ⎪⎪ ε z ⎪⎪ ⎥⎨ ⎬ C46 ⎥ ⎪γ yz ⎪ C56 ⎥ ⎪γ zx ⎪ ⎥⎪ ⎪ C66 ⎥⎦ ⎪⎩γ xy ⎪⎭
⎡C11 ⎢C ⎢ 12 ⎢C [C ] = ⎢ 13 ⎢C14 ⎢C15 ⎢ ⎢⎣C16
C12 C22
C13 C14 C23 C24
C15 C25
C23 C24 C25 C26
C33 C34 C35 C36
C35 C45 C55 C56
C34 C44 C45 C46
C16 ⎤ C26 ⎥⎥ C36 ⎥ ⎥ C46 ⎥ C56 ⎥ ⎥ C66 ⎥⎦
Hence, we can also write
[ε ] = [S ][σ ] ⎧ ε x ⎫ ⎡ S11 ⎪ε ⎪ ⎢ ⎪ y ⎪ ⎢ S12 ⎪⎪ ε z ⎪⎪ ⎢ S13 ⎨ ⎬=⎢ ⎪γ yz ⎪ ⎢ S14 ⎪γ zx ⎪ ⎢ S15 ⎪ ⎪ ⎢ ⎪⎩γ xy ⎪⎭ ⎢⎣ S16
S12 S 22 S 23 S 24
S13 S 23 S33 S 34
S14 S 24 S 34 S 44
S15 S 25 S35 S 45
S 25 S 26
S 35 S36
S 45 S 46
S55 S56
S16 ⎤ ⎧σ x ⎫ ⎪σ ⎪ ⎥ S 26 ⎥ ⎪ y ⎪ S36 ⎥ ⎪⎪σ z ⎪⎪ ⎥⎨ ⎬ S 46 ⎥ ⎪τ yz ⎪ S56 ⎥ ⎪τ zx ⎪ ⎥⎪ ⎪ S 66 ⎥⎦ ⎪⎩τ xy ⎪⎭
The matrix S is referred to as the compliance matrix and the elements of S are the compliances.
21 elastic constants are required to describe the most general anisotropic material (fully anisotropic). This is in contrast to an isotropic material for which there are only two independent elastic constants (typically the Young Modulus and the Poisson’s ratio).
E (1 − ν ) Eν σx = εx + (ε y + ε z ) (1 + ν )(1 − 2ν ) (1 + ν )(1 − 2ν ) E (1 − ν )ε x + ν (ε y + ε z ) σx = (1 + ν )(1 − 2ν )
[
]
Many materials of practical interest contain certain material symmetries with respect to their elastic properties (elastic symmetries). Other type of symmetries are possible optical, electrical and thermal properties. Let us determine the structure of the elastic matrix for a material with a single plane of elastic symmetry. Crystals whose crystalline structure is monoclinic as examples of materials possessing a single plane of elastic symmetry. Example Iron aluminide, gypsum, talc, ice, selenium Materials with one plane of symmetry are referred to as Monoclinic materials.
Crystal Systems Crystallographers have shown that only seven different types of unit cells are necessary to create all point lattice Cubic a= b = c ; α = β = γ = 90 Tetragonal a= b ≠ c ; α = β = γ = 90 Rhombohedral a= b = c ; α = β = γ ≠ 90 Hexagonal a= b ≠ c ; α = β = 90, γ =120 Orthorhombic a≠ b ≠ c ; α = β = γ = 90 Monoclinic a≠ b ≠ c ; α = γ = 90 ≠ β Triclinic a≠ b ≠ c ; α ≠ γ ≠ β ≠ 90
Monoclinic Materials Let us assume that the z-plane is the plane of elastic symmetry. For such a material the elastic coefficients in the stress-strain law must remain unchanged when subjected to a transformation that represents a reflection in the symmetry plane. For monoclinic materials (due to one plane of elastic symmetry) the number of independent elastic constants is reduced from 21 to 13.
⎧σ x ⎫ ⎡C11 C12 ⎪σ ⎪ ⎢ ⎪ y ⎪ ⎢C12 C22 ⎪⎪σ z ⎪⎪ ⎢C13 C23 ⎨ ⎬=⎢ 0 ⎪τ yz ⎪ ⎢ 0 ⎪τ zx ⎪ ⎢ 0 0 ⎪ ⎪ ⎢ ⎪⎩τ xy ⎪⎭ ⎢⎣C16 C26
C13 C23 C33
0 0 0
0 0 0
0 0 C36
C44 C45 0
C45 C55 0
C16 ⎤ ⎧ ε x ⎫ ⎪ε ⎪ ⎥ C26 ⎥ ⎪ y ⎪ C36 ⎥ ⎪⎪ ε z ⎪⎪ ⎥⎨ ⎬ 0 ⎥ ⎪γ yz ⎪ 0 ⎥ ⎪γ zx ⎪ ⎥⎪ ⎪ C66 ⎥⎦ ⎪⎩γ xy ⎪⎭
KEY TO NOTATION
TRICLINIC (21)
MONOCLINIC (13)
ORTHORHOMBIC (9)
(7)
TETRAGONAL
CUBIC (3)
(6)
HEXAGONAL (5)
(7)
ISOTROPIC (2)
TRIGONAL
(6)
Orthotropic Materials Let us consider a material with a second plane of elastic symmetry. The y-plane and the z-plane are the planes of elastic symmetry and are perpendicular to each other. Again, for such a material the elastic coefficients in the stress-strain law must remain unchanged when subjected to a transformation that represents a reflection in the symmetry plane. For orthotropic materials (due to the two planes of elastic symmetry) the number of independent elastic constants is reduced from 21 to 9.
⎧σ x ⎫ ⎡C11 C12 ⎪σ ⎪ ⎢ ⎪ y ⎪ ⎢C12 C22 ⎪⎪σ z ⎪⎪ ⎢C13 C23 ⎨ ⎬=⎢ 0 ⎪τ yz ⎪ ⎢ 0 ⎪τ zx ⎪ ⎢ 0 0 ⎪ ⎪ ⎢ 0 ⎪⎩τ xy ⎪⎭ ⎢⎣ 0
C13 C23 C33
0 0 0
0 0 0
0 0 0
C44 0 0
0 C55 0
⎤⎧ ε x ⎫ ⎥⎪ ε ⎪ ⎥⎪ y ⎪ ⎥ ⎪⎪ ε z ⎪⎪ ⎥⎨ ⎬ 0 ⎥ ⎪γ yz ⎪ 0 ⎥ ⎪γ zx ⎪ ⎥⎪ ⎪ C66 ⎥⎦ ⎪⎩γ xy ⎪⎭ 0 0 0
Materials possessing two perpendicular planes of elastic symmetry must also possess a third mutually perpendicular plane of elastic symmetry. Materials having three mutually perpendicular planes of elastic symmetry are referred to as orthotropic (orthogonally anisotropic) materials. Long Fiber Composite
Transversely Isotropic Materials Materials that are isotropic in a plane. Transversely isotropic materials require five independent material constants. 0 0 ⎤⎧ ε ⎫ ⎧σ x ⎫ ⎡C11 C12 C13 0 x ⎢ ⎥ ⎪σ ⎪ C12 C11 C13 0 ⎪ε ⎪ 0 0 ⎥⎪ y ⎪ ⎪ y⎪ ⎢ ⎥⎪ ε ⎪ 0 0 ⎪⎪σ z ⎪⎪ ⎢C13 C13 C33 0 ⎪ z⎪ ⎢ ⎥ = ⎨ ⎬ ⎨ ⎬ 0 0 0 0 0 C 44 ⎥ ⎪γ yz ⎪ ⎪τ yz ⎪ ⎢ ⎥⎪ ⎪ 0 0 0 C44 0 ⎪τ zx ⎪ ⎢ 0 γ zx ⎢ ⎥ ( C11 − C12 ) ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪⎩γ xy ⎪⎭ 0 0 0 0 0 ⎪⎩τ xy ⎪⎭ ⎣ 2 ⎦
Isotropic Materials The isotropic material requires only two independent material constants, namely the Elastic Modulus and the Poisson’s Ratio. 0 0 0 ⎡C11 C12 C12 ⎤ ⎥⎧ ε x ⎫ ⎧σ x ⎫ ⎢C C C 0 0 0 11 12 ⎥⎪ ⎪ ⎪σ ⎪ ⎢ 12 ⎥⎪ ε y ⎪ 0 0 0 ⎪ y ⎪ ⎢C12 C12 C11 ⎥ ⎪⎪ ε ⎪⎪ ⎪⎪σ z ⎪⎪ ⎢ ( C11 − C12 ) z 0 0 0 0 0 ⎢ ⎥ ⎨ ⎬= ⎨ ⎬ 2 τ ⎥ ⎪γ yz ⎪ ⎪ yz ⎪ ⎢ (C11 − C12 ) ⎥ ⎪γ ⎪ ⎪τ zx ⎪ ⎢ 0 0 0 0 0 zx ⎢ ⎥ 2 ⎪ ⎪ ⎪ ⎪ ⎪⎩τ xy ⎪⎭ ⎢ ( C11 − C12 ) ⎥ ⎪⎩γ xy ⎪⎭ 0 0 0 0 ⎢0 ⎥ 2 ⎣ ⎦ C11 =
E (1 − ν ) (1 + ν )(1 − 2ν )
C12 =
Eν (1 + ν )(1 − 2ν )
(C11 − C12 ) = 2
E =G 2(1 + ν )
Engineering Material Constants for Orthotropic Materials The quantities appearing in the coefficient matrix can be written in terms of well understood engineering constants such as the Young Modulus and the Poisson’s ratio. σ x = E xε x For the x, y and z coordinate axes we can write: Where the Young Modulus in the x-, y- and zdirections are not necessarily equal. Any extension in the x-axis is accompanied by a contraction in the y- and z- axis. However, this quantities are not necessarily equal in orthotropic materials. Where νxy is the contraction in the y-direction due to the stress in the x-direction
σ y = E yε y σ z = Ezε z
ε y = −ν xyε x ε z = −ν xzε x
[ε ] = [S ][σ ]
If all three stresses are applied simultaneously, then: ν yx ν zx 1 εx = σx − σ y − σz Ex
Ey
⎧ ε x ⎫ ⎡ S11 ⎪ε ⎪ ⎢ ⎪ y ⎪ ⎢ S12 ⎪⎪ ε z ⎪⎪ ⎢ S13 ⎨ ⎬=⎢ ⎪γ yz ⎪ ⎢ 0 ⎪γ zx ⎪ ⎢ 0 ⎪ ⎪ ⎢ ⎪⎩γ xy ⎪⎭ ⎢⎣ 0
Ez
ν xy
ν zy 1 εy = − σx + σy − σz Ex Ey Ez ν yz
ν xz
S12 S 22 S 23
S13 S 23 S33
0 0 0
0 0 0
0 0 0
0 0 0
S 44 0 0
0 S55 0
⎤ ⎧σ x ⎫ ⎥ ⎪σ ⎪ ⎥⎪ y ⎪ ⎥ ⎪⎪σ z ⎪⎪ ⎥⎨ ⎬ 0 ⎥ ⎪τ yz ⎪ 0 ⎥ ⎪τ zx ⎪ ⎥⎪ ⎪ S 66 ⎥⎦ ⎪⎩τ xy ⎪⎭ 0 0 0
1 εy = − σx − σy + σz Ex Ey Ez
S11 =
1 Ex
S12 = −
ν yx Ey
=−
ν xy Ex
Comparing with the compliance matrix for orthotropic materials: 1 1 S 22 = S33 = Ey Ez
S13 = −
ν zx Ez
=−
ν xz Ex
S 23 = −
ν zy Ez
=−
ν yz Ey
Where νxy is the contraction in the y-direction due to the stress in the x-direction
Whereas with 1 γ xy = τ xy isotropic materials Gxy the relationship 1 between shear S 44 = G yz stress and shear strain is the same in ⎡ 1 any coordinate ⎢ ⎢ Ex planes, for ⎢ ν xy orthotropic ⎧ ε x ⎫ ⎢− E x materials these ⎪ε ⎪ ⎢ relationships are not ⎪ y ⎪ ⎢ − ν xz ⎪⎪ ε z ⎪⎪ ⎢ E x the same. ⎨ ⎬=⎢ ⎪γ yz ⎪ ⎢ 0 ⎪γ zx ⎪ ⎢ ⎪ ⎪ ⎢ ⎪⎩γ xy ⎪⎭ ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢⎣
γ yz
1 τ yz = G yz
1 γ zx = τ zx Gzx
1 S55 = Gzx −
ν yx
Ey 1 Ey
−
ν yz Ey
− −
ν zx Ez
ν zy
Ez 1 Ez
1 S 66 = Gxy 0
0
0
0
0
0 0
0
0
1 G yz
0
0
0
1 Gzx
0
0
0
0
⎤ 0 ⎥ ⎥ ⎥ 0 ⎥ ⎧σ ⎫ x ⎥ ⎪σ ⎪ ⎥⎪ y ⎪ 0 ⎥ ⎪σ ⎪ ⎪ z⎪ ⎥⎨ ⎬ τ yz ⎪ ⎥ ⎪ 0 ⎥ ⎪τ zx ⎪ ⎥⎪ ⎪ 0 ⎥ ⎪⎩τ xy ⎪⎭ ⎥ 1 ⎥ ⎥ Gxy ⎥⎦
⎡ 1 − ν yzν zy ⎢ E EΔ y z ⎧σ x ⎫ ⎢ ⎪σ ⎪ ⎢ν xy + ν xzν zy ⎪ y⎪ ⎢ E E Δ x z ⎪⎪σ z ⎪⎪ ⎢ ⎨ ⎬ = ⎢ν xz + ν xyν yz ⎪τ yz ⎪ ⎢ Ex E y Δ ⎪τ zx ⎪ ⎢ 0 ⎪ ⎪ ⎢ ⎪⎩τ xy ⎪⎭ ⎢ 0 ⎢ 0 ⎢⎣
Δ=
ν yx + ν zxν yz
ν zx + ν yxν zy
E y Ez Δ 1 − ν zxν xz Ex Ez Δ ν yz + ν xzν yx Ex E y Δ 0 0
E y Ez Δ ν zy + ν zxν xy Ex Ez Δ 1 − ν xyν yx Ex E y Δ 0 0
0
0
0
0
0
0
0
0
G yz 0
0 Gzx
0
0
⎤ 0 ⎥ ⎥⎧ ε x ⎫ ⎥⎪ ⎪ 0 ⎥⎪ ε y ⎪ ⎥ ⎪⎪ ε z ⎪⎪ 0 ⎥ ⎨γ yz ⎬ ⎥⎪ ⎪ ⎥ 0 ⎥ ⎪⎪γ zx ⎪⎪ 0 ⎥ ⎪⎩γ xy ⎪⎭ ⎥ Gxy ⎥⎦
1 − ν xyν yx − ν yzν zy − ν zxν xz − 2ν xyν yzν zx Ex E y Ez
In 2-D ⎡ 1 ⎢ E ⎧εx ⎫ ⎢ x ⎪ ⎪ ⎢ ν xy ⎨ ε y ⎬ = ⎢− ⎪γ ⎪ ⎢ E x ⎩ xy ⎭ ⎢ ⎢ 0 ⎣
−
ν yx
Ey 1 Ey 0
⎤ 0 ⎥ ⎥ ⎧σ x ⎫ ⎥⎪ ⎪ 0 ⎥ ⎨σ y ⎬ ⎥ ⎪τ xy ⎪ 1 ⎥⎩ ⎭ Gxy ⎥⎦
⎡ Ex ⎢ 1 −ν xyν yx ⎧σ x ⎫ ⎢ ⎪ ⎪ ⎢ ν xy E y ⎨σ y ⎬ = ⎢ ⎪τ ⎪ ⎢1 −ν xyν yx ⎩ xy ⎭ ⎢ 0 ⎢ ⎣
⎧σ x ⎫ ⎡C11 C12 ⎪ ⎪ ⎢ ⎨σ y ⎬ = ⎢C12 C22 ⎪τ ⎪ ⎢ 0 0 ⎩ xy ⎭ ⎣
ν yx E x 1 −ν xyν yx Ey 1 −ν xyν yx 0
⎤ 0 ⎥ ⎥⎧ ε x ⎫ ⎥⎪ ⎪ 0 ⎥⎨ ε y ⎬ ⎥ ⎪γ xy ⎪ Gxy ⎥ ⎩ ⎭ ⎥ ⎦
0 ⎤⎧ ε x ⎫ ⎪ ⎪ ⎥ 0 ⎥⎨ ε y ⎬ C33 ⎥⎦ ⎪⎩γ xy ⎪⎭
Neumann’s Principle This is the most important concept in crystal physics. It states; ……………... the symmetry of any physical property of a crystal must include the symmetry elements of the point group of the crystal. This means that measurements made in symmetry‐related directions will give the same property coefficients. Example: NaCl belongs to the m3m group . The [100] and [010] directions are equivalent. Since these directions are physically the same, it should be expected that measurements of permittivity, elasticity or any other physical property will be the same in these two directions.
7 Crystal Systems cubic
monoclinic
orthorhombic tetragonal
triclinic
trigonal
hexagonal
Crystal System
External Minimum Symmetry
Unit Cell Properties
Triclinic Monoclinic Orthorhombic Tetragonal Trigonal Hexagonal Cubic
None One 2‐fold axis, || to b (b unique) Three perpendicular 2‐folds One 4‐fold axis, parallel c One 3‐fold axis One 6‐fold axis Four 3‐folds along space diagonal
a, b, c, al, be, ga, a, b, c, 90, be, 90 a, b, c, 90, 90, 90 a, a, c, 90, 90, 90 a, a, c, 90, 90, 120 a, a, c, 90, 90, 120 a, a, ,a, 90, 90, 90
Anisotropy Factor Cubic Symmetry
For cubic crystals, there are four three‐fold symmetry S11 = S 22 = S33 axes (along the body diagonals) such that: S12 = S 23 = S31 There is a reduction of the nine constants for orthotropic symmetry to three. An anisotropic factor A, can be defined for cubic crystals Using the direction cosines l, m, n for a particular direction, one can determine the elastic properties of a cubic single crystal in a particular direction by the relationship:
S 44 = S55 = S66
2( S11 − S12 ) A= S 44
1 1 ⎤2 2 ⎡ ' = S11 = S11 − 2 ⎢ S11 − S12 − S 44 ⎥l m + m 2 n 2 + l 2 n 2 Ehkl 2 ⎦ ⎣
Isotropy When the anisotropy factor is equal to one, there are just two independent components, e.g. C11 and C12. In this instance, the rigidity or shear modulus G is given by: 1 And λ is given by:
1 G = C44 = (C11 − C12 ) = 2 S 44
λ = C12
These two constants are known as the Lame constants and are used to describe all the elastic constants of isotropic materials 1 Poisson’s ratio can de expressed in terms υ = − S12 = − C12 = S11 C11 + C12 ⎛ G⎞ 2⎜1 + ⎟ of Lame constants: ⎝
The compressibility (β) or bulk modulus (K) relate hydrostatic or mean stress to volume 2G ⎞ ⎛ strain G⎜ 3 + ⎟ E=
λ ⎠ ⎝ ⎛ G⎞ ⎜1 + ⎟ λ⎠ ⎝
K=
1
β
=
σ mean Δ
λ⎠
2G =λ+ 3
Example An orthotropic material has the following properties Ex=7,500ksi, Ey= 2,500ksi, Gxy = 1,250ksi and νxy= 0.25. Determine the principal stresses and strains at a point on a free surface where the following strains were measured: εx=-400μ ; εy=600μ ; γxy=-500μ . Consider plane stress conditions Solution:
⎡ Ex ⎢1 ν ν − xy yx ⎧σ x ⎫ ⎢ ⎪ ⎪ ⎢ ν xy E y ⎨σ y ⎬ = ⎢ ⎪τ ⎪ ⎢1 −ν xyν yx ⎩ xy ⎭ ⎢ 0 ⎢ ⎣
S12 = −
ν yx E x 1 −ν xyν yx Ey 1 −ν xyν yx 0
ν yx Ey
=−
ν xy Ex
⎤ 0 ⎥ ⎥⎧ ε x ⎫ ⎥⎪ ⎪ 0 ⎥⎨ ε y ⎬ ⎥ ⎪γ xy ⎪ Gxy ⎥ ⎩ ⎭ ⎥ ⎦
ν yx Ey
=
ν xy Ex
ν yx
ν xy
0.25 × 2500 = Ey = = 0.083 Ex 7500
⎧σ x ⎫ ⎡ 7660 638.3 0 ⎤ ⎧− 400 x10 −6 ⎫ ⎪ ⎪ ⎪ ⎢ ⎥ −6 ⎪ 0 ⎥ ⎨ 600 x10 ⎬ ⎨σ y ⎬ = ⎢638.3 2553.2 ⎪− 500 x10 −6 ⎪ ⎪τ ⎪ ⎢ 0 ⎥ 0 1250 ⎦⎩ ⎩ xy ⎭ ⎣ ⎭
⎧σ x ⎫ ⎧− 2681 psi ⎫ ⎪ ⎪ ⎪ ⎪ ⎨σ y ⎬ = ⎨1276.6 psi ⎬ ⎪τ ⎪ ⎪ − 625 psi ⎪ ⎭ ⎩ xy ⎭ ⎩
σ 1 = 1372.9 psi σ 2 = −2777.4 psi τ Max = 2075.1 psi
⎡ 1 ⎢ E ⎧ε x ⎫ ⎢ x ⎪ ⎪ ⎢ ν xy ⎨ ε y ⎬ = ⎢− ⎪γ ⎪ ⎢ E x ⎩ xy ⎭ ⎢ ⎢ 0 ⎣
−
ν yx
Ey 1 Ey 0
⎤ 0 ⎥ ⎥ ⎧σ x ⎫ ⎥⎪ ⎪ 0 ⎥ ⎨σ y ⎬ ⎥ ⎪τ xy ⎪ 1 ⎥⎩ ⎭ Gxy ⎥⎦
ε 1 = 659μ ε 2 = −459 μ γ Max = 1118μ
γ XY − 500 = tan 2θ ε = = 0 .5 ε X − ε Y − 400 − 600 2τ xy 2 ⋅ ( −625 ) tan 2θσ = = = 0.316 (σ x − σ y ) − 2681 − 1276 .6 Different angles to obtain the principal stresses and the principal strains.
Example Suppose we start with a state of strain (in μ strain)
Consider an orthotropic material where :
⎡300 50 20 ⎤ ⎢ 50 200 30 ⎥ μ − strain ⎥ ⎢ ⎢⎣ 20 30 100⎥⎦
0 ⎤ ⎡103 55 25 0 0 ⎥ ⎢ 55 50 40 0 0 0 ⎥ ⎢ ⎢ 25 40 75 0 0 0 ⎥ ⎥GPa ⎢ 0 0 45 0 0 ⎥ ⎢ 0 ⎢ 0 0 0 0 10 0 ⎥ ⎥ ⎢ 0 0 0 0 27.6⎥⎦ ⎢⎣ 0
We need to change the strain tensor for a strain vector
⎡300 ⎤ ⎢200⎥ ⎢ ⎥ ⎡300 50 20 ⎤ ⎢ 50 200 30 ⎥ μ − strain = ⎢100 ⎥ × 10− 6 ⎢ ⎥ ⎢ ⎥ 60 ⎢ ⎥ ⎢⎣ 20 30 100⎥⎦ ⎢ 40 ⎥ ⎢ ⎥ ⎣⎢100 ⎦⎥
0 ⎤ ⎡300 ⎤ ⎡103 55 25 0 0 ⎢200⎥ ⎥ ⎢ 55 50 40 0 0 0 ⎢ ⎥ ⎥ ⎢ ⎢100 ⎥ ⎢ 25 40 75 0 0 0 ⎥ −6 × × 10 [σ ] = ⎢ GPa ⎢ ⎥ ⎥ 0 0 45 0 0 ⎥ ⎢ 60 ⎥ ⎢ 0 ⎢ 40 ⎥ ⎢ 0 0 0 0 10 0 ⎥ ⎢ ⎥ ⎥ ⎢ 0 0 0 0 27.6⎥⎦ ⎢⎣100 ⎥⎦ ⎢⎣ 0
⎡44.400⎤ ⎢30.500 ⎥ ⎢ ⎥ ⎡44.4 2.76 0.4 ⎤ ⎢23.000⎥ ⎢2.76 30.5 2.70⎥ MPa [σ ] = ⎢ = MPa ⎥ ⎢ ⎥ 2 . 700 ⎢ ⎥ ⎢⎣ 0.4 2.70 23.0⎥⎦ ⎢ 0.400 ⎥ ⎢ ⎥ ⎢⎣ 2.760 ⎥⎦ Eigen‐values
0 0 ⎤ ⎡22.119 ⎢ 0 ⎥ MPa 30 . 8154 0 ⎢ ⎥ ⎢⎣ 0 0 44.9656⎥⎦
Eigen‐vector (cosines from x‐y‐z angles to the principal axes)
⎡ 0.0217 − 0.1980 − 0.9800⎤ ⎢− 0.3130 0.9296 − 0.1948⎥ ⎥ ⎢ ⎢⎣ 0.9495 0.3110 − 0.0418⎥⎦
Example The orthotropic elastic constants for bovine (cow) femoral (leg) bone has been reported from measurements using ultrasound. The values vary on the basis of the position around the bone and along its length. The elastic constants can be determined using piezoelectric crystals to propagate and measure the speed of sound in the material. Two types of elastic constants can be determined. Propagation of dilatational waves can be used to measure longitudinal stiffness (e.g. C11) and propagation shear waves can be used to measure the shear moduli (e.g. C44)
C11 = ρVdil2
2 C44 = ρVtrans
ρ = density Vdil = wave _ speed _ of _ dilatational _ waves Vtrans = wave _ speed _ of _ transverse _ waves
The approximately reported stiffness values are: 0⎤ ⎡ 14 6.3 4.8 0 0 Find the Young Modulus ⎢6.3 18.4 7 0 0 ⎥ 0 ⎢ ⎥ along the bone length (z‐ ⎢4.8 7 25 0 0 0⎥ ⎢ ⎥ MPa direction)? And along the 0 0 7 0 0⎥ radial direction (x and y ⎢0 ⎢0 directions). 0 0 0 6.3 0 ⎥ ⎢ ⎥ 0 0 0 0 5.3⎥⎦ ⎢⎣ 0 Find the Poisson’s ratios? Convert the stiffness matrix into a compliance matrix. 0 0 0 ⎤ ⎡ 0.086 − 0.026 − 0.009 ⎥ ⎢− 0.026 0 . 07 0 . 014 0 0 0 − ⎥ ⎢ ⎢− 0.009 − 0.014 0.046 0 0 0 ⎥ −1 MPa ⎥ ⎢ 0 0 0 0 . 14 0 0 ⎥ ⎢ ⎢ 0 0 0 0 0.16 0 ⎥ ⎥ ⎢ 0 0 0 0 0.19⎥⎦ ⎢⎣ 0
υ21 = −0.026 ⇒ υ21 = 0.373 E2
E3 =
1 1 = = 21.7GPa S33 0.046
E1 =
1 1 = = 11.6GPa S11 0.086
-
υ12 = −0.026 ⇒ υ12 = 0.3016 E1
E2 =
1 1 = = 14.28GPa S 22 0.07
-
υ13 = −0.009 ⇒ υ13 = 0.1044 E1
-
(100)(uvw)
Cosα = =l Example 2 2 2 1 u +v +w Determine the modulus of elasticity for iron (010)(uvw) β Cos = =m single crystals in the , and 2 2 2 1 u +v +w directions.
10−3 GPa −1 S11 S12 S 44 Fe 8.0 − 2.8 8.60
Cosγ =
1 8.6 ⎞ ⎛ = 8.0 − 2⎜ 8.0 − (− 2.8) − ⎟(0) = 8.0 E100 2 ⎠ ⎝ E100 = 125GPa 1 8.6 ⎞ 1 ⎛ = 8.0 − 2⎜ 8.0 − (− 2.8) − ⎟( + 0 + 0) = 4.75 2 ⎠ 4 E110 ⎝
(001)(uvw)
1 u +v +w Directions l m 100 1 0 1 1 110 2 2 1 1 111 3 3 2
2
2
=n n 0 0 1 3
E110 = 210GPa 1 8.6 ⎞ 1 1 1 ⎛ = 8.0 − 2⎜ 8.0 − (− 2.8) − ⎟( + + ) = 3.7 E111 2 ⎠ 9 9 9 ⎝ E111 = 270GPa