Exercise4.3.1 ...3.40) intoeq.(4.12)toobtain: Z › ‡ ‚†kk† (v) ii +2G†ij† (v) ij · dV...
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Transcript of Exercise4.3.1 ...3.40) intoeq.(4.12)toobtain: Z › ‡ ‚†kk† (v) ii +2G†ij† (v) ij · dV...
Exercise 4.3.1When the material is isotropic then we can substitute eq. (3.40)into eq. (4.12) to obtain:
∫
Ω
(λεkkε
(v)ii + 2Gεijε
(v)ij
)dV =
∫
Ω
Fivi dV +∫
∂Ω
Tivi dS+∫
Ω
E
1− 2ναT∆ε
(v)ii dV. (1)
Dene the dierential operator matrix [D] and the material stiness matrix [E]as follows:
[D] :=
∂
∂x0 0
0∂
∂y0
0 0∂
∂z∂
∂y
∂
∂x0
0∂
∂z
∂
∂y∂
∂z0
∂
∂x
[E] :=
λ + 2G λ λ 0 0 0λ λ + 2G λ 0 0 0λ λ λ + 2G 0 0 00 0 0 G 0 00 0 0 0 G 00 0 0 0 0 G
Furthermore, denote u ≡ u := ux uy uzT and v ≡ v := vx vy vzT .Show that eq. (1) can be written in the following form:
∫
Ω
([D]v)T [E][D]u dV =∫
Ω
vT F dV +∫
∂Ω
vT T dS
+∫
Ω
∂vx
∂x
∂vy
∂y
∂vz
∂z
111
EαT∆
1− 2νdV (2)
where F := Fx Fy FzT is the body force vector and T := Tx Ty TzT isthe traction vector.Working with the left hand side rst, we have:
[D]u =
∂
∂x0 0
0∂
∂y0
0 0∂
∂z∂
∂y
∂
∂x0
0∂
∂z
∂
∂y∂
∂z0
∂
∂x
ux
uy
uz
=
ε11
ε22
ε33
2ε12
2ε23
2ε31
1
Similarly
[D]v =
ε(v)11
ε(v)22
ε(v)33
2ε(v)12
2ε(v)23
2ε(v)31
[E][D]u =
λ + 2G λ λ 0 0 0λ λ + 2G λ 0 0 0λ λ λ + 2G 0 0 00 0 0 G 0 00 0 0 0 G 00 0 0 0 0 G
ε11
ε22
ε33
2ε12
2ε23
2ε31
Then:
[E][D]u =
(λ + 2G)ε11 + λε22 + λε33
λε11 + (λ + 2G)ε22 + λε33
λε11 + λε22 + (λ + 2G)ε33
2Gε12
2Gε23
2Gε31
and nally
([D]v)T [E][D]u =
ε(v)11 ε
(v)22 ε
(v)33 2ε
(v)12 2ε
(v)23 2ε
(v)31
(λ + 2G)ε11 + λε22 + λε33
λε11 + (λ + 2G)ε22 + λε33
λε11 + λε22 + (λ + 2G)ε33
2Gε12
2Gε23
2Gε31
([D]v)T [E][D]u = (λ + 2G)ε11ε(v)11 + λε22ε
(v)11 + λε33ε
(v)11 +
λε11ε(v)22 + (λ + 2G)ε22ε
(v)22 + λε33ε
(v)22 +
λε11ε(v)33 + λε22ε
(v)33 + (λ + 2G)ε33ε
(v)33 +
4Gε12ε(v)12 + 4Gε31ε
(v)12 + 4Gε31ε
(v)31
= λεkkε(v)ii + 2Gεijε
(v)ij
The rst two terms on the right hand side are shown by simply recalling thedenition of the inner vector product in index notation, aT b = aibi = biai.Therefore
vT F = Fivi,
2
andvT T = Tivi.
For the third term on the right hand side we just need to show that
ε(v)ii = ε
(v)11 + ε
(v)22 + ε
(v)33 =
∂ux
∂x+
∂uy
∂y+
∂uz
∂z=
∂vx
∂x
∂vy
∂y
∂vz
∂z
111
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