Transcript
Continuum mechanicsI. Kinematics in cartesian coordinates
Ales Janka
office Math 0.107ales.janka@unifr.ch
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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