Mécanique Continue: Exercices sur les Tenseurs de Déformations

θ
0−θh1−Ri
r0iy0y=y0+θh1−Ri r0ix0avec r0=qx2
0+y2 0
[ ]
[ ]
(x0;y0= 0) (x0= 0; y0) (x0;y0=x0)
(1,2,3)
=
1 3 −2
3 1 −2
−2−2 6 
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= 0.04xo0.02
0.02 −0.02yo!avec xo, yoen cm
(o)
(0,0)
[0,5] ×[0,5]cm2
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u xoyo
u= x−xo
y−yo!= −θ[1 −Ri
ro]yo
θ[1 −Ri
ro]xo!= −θ[1 −Ri(x2
o+y2 o)−1/2]yo
θ[1 −Ri(x2
o+y2 o)−1/2]xo!
u
∇u= ∂ux
∂xo
∂ux
∂yo
∂uy
∂xo
∂uy
∂yo!=
−Riθxoyo
r3
o−θ+Riθ
ro−Riθy2
o
r3
o
θ−Riθ
ro+Riθx2
o
r3
o
Riθxoyo
r3
o
L=I+∇u= ∂ux
∂xo
∂ux
∂yo
∂uy
∂xo
∂uy
∂yo!=
1−Riθxoyo
r3
o−θ+Riθ
ro−Riθy2
o
r3
o
θ−Riθ
ro+Riθx2
o
r3
o1 + Riθxoyo
r3
o
B
B=LTL=
1−Riθxoyo
r3
o
θ−Riθ
ro+Riθx2
o
r3
o
−θ+Riθ
ro
−Riθy2
o
r3
o
1 + Riθxoyo
r3
o! 1−Riθxoyo
r3
o
−θ+Riθ
ro
−Riθy2
o
r3
o
θ−Riθ
ro+Riθx2
o
r3
o
1 + Riθxoyo
r3
o!
B= B11 B12
B21 B22 !
B11 = 1+θ2+R2
iθ2
r6
o
(x4
o+x2 oy2 o)+2R2
iθ2
r4
o
(x2
o)+2Riθ
r3
o
(θx2
o−xoyo)+R2
iθ2
r2
o−2Riθ2
ro
B12 =−2R2
iθ2
r4
o
(xoyo) + Riθ
r3
o
(x2
o−y2 o) + R2
iθ2
r6
o
(x3
oyo+xoy3 o)
B21 =−2R2
iθ2
r4
o
(xoyo) + Riθ
r3
o
(x2
o−y2 o) + R2
iθ2
r6
o
(x3
oyo+xoy3 o)
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B22 = 1+θ2+R2
iθ2
r6
o
(y2
o+x2 oy2 o)−2R2
iθ2
r4
o
(y2
o)+ 2Riθ
r3
o
(θy2
o+xoyo)+ R2
iθ2
r2
o−2Riθ2
ro
=1
2(B−I) = 1 2 B11 −1B12
B21 B22 −1!
lin
lin =1
2(∇u+ (∇u)T) = 
−Riθyoxo
r3
o
Riθ(x2
o−y2 o)
2r3
o
Riθ(x2
o−y2 o)
2r3
o
Riθxoyo
r3
o
me θ2
λ lin λ0=λ/(Riθ
r3
o)
det (lin −λI)=0→detRiθ
r3
o −xoyo−λ0x2
o−y2 o
2
x2
o−y2 o
2xoyo−λ0!= 0
(λ02−x2
oy2 o)−(x2
o−y2 o)2
4= 0
I, II
I=λ1=θRi
2ro
II =λ2=−θRi
2ro
.
linni=λini
i= 1,2λ1, λ2n1, n2
λ1
−Riθyoxo
r3
o
Riθ(x2
o−y2 o)
2r3
o
Riθ(x2
o−y2 o)
2r3
o
Riθxoyo
r3
o
 n1x
n1y!= Riθ
2ron1x
Riθ
2ron1y!
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eI=1
√2ro xo−yo
xo+yo!
λ2
eII =1
√2ro xo+yo
−xo+yo!
(Oxyz)
a)(xo;yo= 0) ⇒ro=xo
eI=xo
√2ro
(1,1) = 1
√2(1,1) eII =xo √2ro
(1,−1) = 1
√2(1,−1)
b)(xo= 0; yo)⇒ro=yo
eI=yo
√2ro
(−1,1) = 1
√2(−1,1) eII =yo √2ro
(1,1) = 1
√2(1,1)
c)(xo;xo=yo)⇒ro=√2xo=√2yo
eI=1
√2ro
(0,√2xo) = (0,1) eII =1
√2ro
(√2xo,0) = (1,0)
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