Download - tensor de almansi
![Page 1: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/1.jpg)
Continuum mechanicsI. Kinematics in cartesian coordinates
Ales Janka
office Math [email protected]
http://perso.unifr.ch/ales.janka/mechanics
September 22, 2010, Universite de Fribourg
Ales Janka I. Kinematics
Kinematics: description of position and deformation
initial configuration x = x iei
x i . . . material coordinates
deformed config. y = y iei
y i . . . spatial coordinates
displacement u = y − x
Two possibilities:
Lagrange description:u(x) = y(x)− x
Euler description:u(y) = y − x(y)
e3
e2e1
x
y
dx
y
dy
x
u
u+dux+dxy+dy
Initial config.
Deformed config.
Ales Janka I. Kinematics
![Page 2: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/2.jpg)
Material vs. spatial coordinates: deformation of a 1D rod
Let y1 = y1(x1, t) = [(x1)2 − 1] · t + x1.
Inversely, x1 = x1(y1, t) = 12t
(√1 + 4t(2t + y1)− 1
)
Ales Janka I. Kinematics
1. Lagrange description
x = x iei and y = y iei .
Choice: y = y(x)
Deformation gradient:
F ij (x) =
∂y i
∂x j= y i
,j
dx = dx iei
dy = dy iei =∂y i
∂x jdx jei
dy = dx j F ij (x) gi = dx j gj
e3
e1
e3
e2e1
x
y
dx
y(x)
dy
y(x+dx)
g2
g3
e2g1
^g
2
^g1
^g3
=
==
Initial config.
Deformed config.
Ales Janka I. Kinematics
![Page 3: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/3.jpg)
1. Lagrange description: how to measure deformation?
The ”edge” dx in initial configuration is deformed to dy
Convenient measure of deformation:
dy2−dx2 = (dx i gi )·(dx j gj)−(dx igi )·(dx jgj) = dx i dx j (gij−gij)
where gij = gi · gj and gij = gi · gj
are metric tensors (matrices)
Green strain tensor:
εij(x) =1
2(gij − gij)
εij(x) =1
2
(F k
i (x) F `j (x) gk` − gij
)
Ales Janka I. Kinematics
1. Lagrange descript.: Green strain tensor in displacement
u(x) = y(x)− x
u = ui gi
dy = dx + du
du =∂ui
∂x jdx j gi
dy =(dx i + ui
,j dx j)
gi
e3
e2e1
x
y
y(x)
y(x+dx)
dy
(x)u
(x)u du
u(x+dx)dx
x
x+dx
Initial config.
Deformed config.
Ales Janka I. Kinematics
![Page 4: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/4.jpg)
1. Lagrange descript.: Green strain tensor in displacement
(dy)2 =
(dx i + ui
,j dx j
)(dxk + uk
,` dx`
)gik
(dy)2−(dx)2 = ui,j dx j dxk gik + uk
,` dx` dx i gik + ui,j uk
,` dx j dx` gik
(dy)2−(dx)2 =(ui ,j + uj ,i + uk,i u
k,j
)dx i dx j
Green strain tensor in displacements:
εij =1
2
(ui ,j + uj ,i + uk,i u
k,j
)Green strain tensor in displacements in cartesian coordinates:
εij =1
2
(∂ui
∂xj+∂uj
∂xi+∂uk
∂xi
∂uk
∂xj
)Ales Janka I. Kinematics
1. Lagrange description: example 1 – rigid body motion
x
u
a
y
e3
e2
e1
Deformed config.
Initial config.
α
y1
y2
y3
=
cosα − sinα 0sinα cosα 0
0 0 1
·x1
x2
x3
+
a1
a2
a3
Ales Janka I. Kinematics
![Page 5: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/5.jpg)
1. Lagrange description: example 1 – rigid body motion
x
u
a
y
e3
e2
e1
Deformed config.
Initial config.
α
u1
u2
u3
=
cosα− 1 − sinα 0sinα cosα− 1 0
0 0 0
·x1
x2
x3
+
a1
a2
a3
ε11 =1
2
(u1,1 + u1,1 +
3∑k=1
uk,1uk,1
)
= cosα− 1 +(cosα− 1)2 + sin2 α
2= 0
Ales Janka I. Kinematics
1. Lagrange description: example 1 – rigid body motion
x
u
a
y
e3
e2
e1
Deformed config.
Initial config.
α
u1
u2
u3
=
cosα− 1 − sinα 0sinα cosα− 1 0
0 0 0
·x1
x2
x3
+
a1
a2
a3
ε12 =1
2
(u1,2 + u2,1 +
3∑k=1
uk,1uk,2
)
=− sinα + sinα
2+− sinα (cosα− 1) + sinα (cosα− 1)
2= 0
Ales Janka I. Kinematics
![Page 6: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/6.jpg)
1. Lagrange description: example 2
−5 −4 −3 −2 −1 0 1 2 3 4 5
−2
−1
0
1
Initial configuration is bent into the deformed configuration
Principal strain of Green strain tensor (Lagrange formulation)?
Ales Janka I. Kinematics
1. Lagrange description: example 2
−5 −4 −3 −2 −1 0 1 2 3 4 5
−2
−1
0
1
−5 −4 −3 −2 −1 0 1 2 3 4 5
−2
−1
0
1
Wrong - this is Almansi strain (Euler formulation)!Ales Janka I. Kinematics
![Page 7: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/7.jpg)
1. Lagrange description: example 2
−5 −4 −3 −2 −1 0 1 2 3 4 5
−2
−1
0
1
−5 −4 −3 −2 −1 0 1 2 3 4 5
−2
−1
0
1
This is the correct Green strain (Lagrange formulation)Ales Janka I. Kinematics
1. Lagrange description: example 2
−5 −4 −3 −2 −1 0 1 2 3 4 5
−2
−1
0
1
−5 −4 −3 −2 −1 0 1 2 3 4 5
−2
−1
0
1
This is the correct Green strain (Lagrange formulation)Ales Janka I. Kinematics
![Page 8: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/8.jpg)
2. Euler description
x = x iei and y = y iei .
Choice: x = x(y)
Deformation gradient inverse:
F−1 ij =
∂x i
∂y j= x i
,j
dy = dy iei
dx = dx iei =∂x i
∂y jdy j ei
dx = dx j F−1 ij gi = dx j gj
e3
e2e1
x
y
dx
dy
(y)
y+dy
y
e1g1=
g2
e2
=
g1
~
g2
~
g3
~x(y+dy)
(y)x
e3g3 =
Initial config.
Deformed config.
Ales Janka I. Kinematics
2. Euler description: Almansi strain tensor
the deformed ”edge” dy corresponds to the undeformed dx
Difference of their (lengths)2:
dy2−dx2 = (dy igi )·(dy jgj)−(dy i gi )·(dy j gj) = dy i dy j (gij−gij)
where gij = gi · gj and gij = gi · gj
are metric tensors (matrices)
Almansi strain tensor:
Eij(y) =1
2(gij − gij)
Eij(y) =1
2
(gij − F−1 k
i F−1 `j gk`
)
Ales Janka I. Kinematics
![Page 9: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/9.jpg)
2. Euler description: Almansi strain tensor in displacement
u(y) = y − x(y)
u = ui gi
dx = dy − du
du =∂ui
∂y jdy j gi
dx =(dy i − ui
,j dy j)
gi
e3
e2e1
x
y
dy
(y)u
(y)u du
u(y+dy)dx
ydx
(y)
y+dy(y+dy)x
(y)x
Initial config.
Deformed config.
Ales Janka I. Kinematics
2. Euler description: Almansi strain tensor in displacement
(dx)2 =
(dy i − ui
,j dy j
)(dyk − uk
,` dy `
)gik
(dy)2−(dx)2 = ui,j dy j dyk gik + uk
,` dy ` dy i gik − ui,j uk
,` dy j dy ` gik
(dy)2−(dx)2 =(ui ,j + uj ,i − uk,i u
k,j
)dy i dy j
Almansi strain tensor in displacements:
Eij =1
2
(ui ,j + uj ,i − uk,i u
k,j
)Almansi strain tensor in displacements in cartesian coordinates:
Eij =1
2
(∂ui
∂yj+∂uj
∂yi− ∂uk
∂yi
∂uk
∂yj
)Ales Janka I. Kinematics
![Page 10: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/10.jpg)
3. Green and Almansi strain tensors: mutual relation
Green strain tensor:
εij(x) =1
2(gij − gij) =
1
2
(F k
i F `j gk` − gij
)εij =
1
2
(∂ui
∂xj+∂uj
∂xi+∂uk
∂xi
∂uk
∂xj
)Almansi strain tensor
Eij(y) =1
2(gij − gij) =
1
2
(gij − F−1 k
i F−1 `j gk`
)Eij =
1
2
(∂ui
∂yj+∂uj
∂yi− ∂uk
∂yi
∂uk
∂yj
)
Ales Janka I. Kinematics
3. Green and Almansi strain tensors: mutual relation
Relation between F ij = ∂y i
∂x j and F−1 jk = ∂x j
∂yk :
F ij · F−1 j
k =3∑
j=1
∂y i
∂x j
∂x j
∂yk=∂y i
∂yk= δik
by chain rule for the derivatives.
Hence:εk` = F i
k · Eij · F j`
Ales Janka I. Kinematics
![Page 11: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/11.jpg)
3. Green and Almansi strain tensors: matrix form
Deformation gradient matrix:
F =
∂y1
∂x1∂y1
∂x2∂y1
∂x3
∂y2
∂x1∂y2
∂x2∂y2
∂x3
∂y3
∂x1∂y3
∂x2∂y3
∂x3
=[F i
j
]and F−1 =
[F−1 i
j
]
Green and Almansi matrix for cartesian coords (gi = ei , gij = δij):
[εij ] =1
2
(FT F− I
)and [Eij ] =
1
2
(I− F−T F−1
)Mutual relations:
[εij ] = FT · [Eij ] · F and [Eij ] = F−T · [εij ] · F−1
Ales Janka I. Kinematics
3. Green and Almansi strain tensors: small deformations
If ui ,j � 1 then uk,i · uk,j is negligible:
εij =1
2
(∂ui
∂x j+∂uj
∂x i+∂uk
∂x i
∂uk
∂x j
)≈ 1
2
(∂ui
∂x j+∂uj
∂x i
)= eij
We can replace Green strain εij by the Cauchy strain eij
Advantage: Cauchy strain eij(u) is linear in u
Eij =1
2
(∂ui
∂y j+∂uj
∂y i− ∂uk
∂y i
∂uk
∂y j
)≈ 1
2
(∂ui
∂y j+∂uj
∂y i
)=
1
2
(∂ui
∂xk
∂xk
∂y j+∂uj
∂yk
∂xk
∂y i
)≈ 1
2
(∂ui
∂x j+∂uj
∂x i
)= eij
because xk = yk − uk and ui ,k · uk,` is negligible.NB: Green and Almansi simplify to the same Cauchy strain!
Ales Janka I. Kinematics
![Page 12: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/12.jpg)
4. Physical meaning of Cauchy strain in cartesian coords
dx dy
yx
e1
e2
e3
xyInitial config. Deformed config.
Special choices of the deformation mode:Let dx = dx1 e1 and dy = dy1 e1. Then:
dy2 − dx2 = (dy1)2 − (dx1)2 = 2 e11 (dx1)2
Hence (for small deformations dx1 + dy1 ≈ 2 dx1):
e11 =(dy1)2 − (dx1)2
2 (dx1)2=
(dy1 − dx1)(dy1 + dx1)
2 (dx1)2≈ dy1
dx1− 1
Meaning of ekk : relative elongation along ek
Ales Janka I. Kinematics
4. Physical meaning of Cauchy strain in cartesian coords
yx
e1
e2
e3
yx
dx2
dx1 dy1
dy2
Initial config. Deformed config.
θ
Special choices of the deformation mode:Let dx = dx1 + dx2 and dy = dy1 + dy2, dxk = dxk ek :
dy2−dx2 = (dy1)2+2 dy1 ·dy2+(dy2)2−(dx1)2−2 dx1 ·dx2−(dx2)2
= 2[e11 (dx1)2 + 2 e12 dx1 dx2 + e22 (dx2)2
]Hence (for small deformations ekk � 1 and θ is small):
e12 =θ
2
dy1
dx1
dy2
dx2≈ θ
2(1 + e11) (1 + e22)
≈ θ
2(1 + e11 + e22 + e11 e22) ≈ θ
2
Meaning of e12: half of shear angle θ in the plane (0, e1, e2)Ales Janka I. Kinematics
![Page 13: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/13.jpg)
4. Physical meaning of Cauchy strain in cartesian coords
yx
e1
e2
e3
yx
dx2
dx1 dy1
dy2
Initial config. Deformed config.
θ
Special choices of the deformation mode:Let dx = dx1 + dx2 and dy = dy1 + dy2, dxk = dxk ek :
dy1 · dy2 = 2 e12 dx1 dx2
= |dy1| · |dy2| cos(π/2− θ) ≈ dy1 dy2 sin θ ≈ dy1 dy2 · θHence (for small deformations ekk � 1 and θ is small):
e12 =θ
2
dy1
dx1
dy2
dx2≈ θ
2(1 + e11) (1 + e22)
≈ θ
2(1 + e11 + e22 + e11 e22) ≈ θ
2
Meaning of e12: half of shear angle θ in the plane (0, e1, e2)Ales Janka I. Kinematics
4. Physical meaning of Cauchy strain in cartesian coords
e1
e2
e3
yxdx3
dy3dy1
dy2dx2dx1
Initial config. Deformed config.
Volume before (dV ) and after (dV ) deformation (small deformations):
dV = dx1 dx2 dx3 dV ≈ dy1 dy2 dy3
Relative change of volume (for small deformations ekk � 1):
dV − dV
dV=
dV
dV− 1 ≈ dy1
dx1
dy2
dx2
dy3
dx3− 1
= (1 + e11)(1 + e22)(1 + e33)− 1 ≈ e11 + e22 + e33
Meaning of trace of eij : relative change of volume
Ales Janka I. Kinematics
![Page 14: tensor de almansi](https://reader031.vdocuments.net/reader031/viewer/2022013114/53fbc3cedab5caa86a8b464a/html5/thumbnails/14.jpg)
5. How to transform areas dS0 → dS?
Nanson’s relation:we know how to transform vectors:
dy i =∂y i
∂x jdx j
we know how to transform volumes:
dV = J·dV0 with J = det
[∂y i
∂x j
]
dx
e2
(x)ye3
e1
x
0dV
dy
dS
dS0
Initial config.
Deformed config.
dV
Idea: complete areas to volumes (for any dx, ie. any dy):
dSi dy i = dV = J · dV0 = J · dS0j dx j = J · dS0j∂x j
∂y idy i
Ales Janka I. Kinematics