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Before we start, some basic mathematic theorem.
Divergence
∇⋅v=vi,i
∇⋅T=Tij,iej
Gradient (dim+1)
∇ϕ=ϕ,iei
∇v=vi,jei⊗ej
∇T=Tij,kei⊗ej⊗ek
Curl (dim=)
∇×v=εijkvj,iek
∇×T=εijkTmj,iek⊗em
The product rule of differentiation
∵σ⋅ν=σijνjei
∴∇⋅(σ⋅ν)=(σijνj),i=σij,iνj+σijνj,i
also
(∇⋅σ)⋅ν=σij,iνj
∇ν:σ=νi,jσji
or as Prof. Zohdi suggested
∇ν:σ=νi,jσij
but since σ is symmetric
σij=σji
therefore
∇⋅(σ⋅ν)=(∇⋅σ)⋅ν+∇ν:σ(1)
Divergence Theorem
∫∂RϕndA=∫R∇⋅ϕdV
∫∂Rv⋅ndA=∫R∇⋅vdV(2)
∫∂RTndA=∫R∇⋅TdV
or
∫∂RϕnidA=∫Rϕ,idV
∫∂RvinidA=∫Rvi,idV
∫∂RTijnjdA=∫RTij,jdV
Stokes theorem
∫Cϕdx=∫Sn×∇ϕdA
∫Cv⋅dx=∫Sn⋅∇×vdA
∫CTdx=∫S(∇×T)TndA
or
∫Cϕdxi=∫Sεijknjϕ,kdA
∫Cvidxi=∫Sniεijkvk,jdA
∫CTijdxj=∫SεjpqTiq,pnjdA
1
The SMALL deformation of the body is governed by (strong form):
∇⋅(IE:∇u)+ρg=0
since
σ=IE:∇u
let
f=ρg
thus
∇⋅σ+f=0
so we can write
∫Ω(∇⋅σ+f)⋅νdΩ=0
using (1)
∫Ω(∇⋅(σ⋅ν)−∇ν:σ)dΩ+∫Ωf⋅νdΩ=0
using (2)
∫Ω∇ν:σdΩ=∫Ωf⋅νdΩ+∫∂Ωσ⋅ν⋅ndA
because
t=σ⋅n
therefore
∫Ω∇ν:σdΩ=∫Ωf⋅νdΩ+∫Γtt⋅νdA
so we have the weak form
Find u,u∣Γu=u∗, such that ∀ν,ν∣Γu=0∫Ω∇ν:IE:∇udΩ=∫Ωf⋅νdΩ+∫Γtt⋅νdA
where u∗ is the applied boundary displacement on Γu, t=t∗ on Γt
since the integrals must be finite, we improve the weak form as below
Find u∈H1(Ω),u∣Γu=u∗, such that ∀ν∈H1(Ω),ν∣Γu=0∫Ω∇ν:IE:∇udΩ=∫Ωf⋅νdΩ+∫Γtt⋅νdA
where
H1(Ω):=[H1(Ω)]3
explained as below
similar to the definition of u∈H1(Ω) which states
u∈H1(Ω) if ∣∣u∣∣H1(Ω)2:=∫Ωu,ju,jdΩ+∫ΩuudΩ<∞
the definition of u∈H1(Ω) states as below
u∈H1(Ω) if ∣∣u∣∣H1(Ω)2:=∫Ωui,jui,jdΩ+∫ΩuiuidΩ<∞
2
should u∗ be different from the applied boundary displacement on Γu, weak form can be stated as below
Find u∈H1(Ω), such that ∀ν∈H1(Ω)∫Ω∇ν:IE:∇udΩ=∫Ωf⋅νdΩ+∫Γtt⋅νdA+P⋆∫Γu(u∗−u)⋅νdA(3)
or
Find u∈H1(Ω), such that ∀ν∈H1(Ω)∫Ωνi,jIEijkluk,ldΩ=∫ΩfiνidΩ+∫ΓttiνidA+P⋆∫Γuui∗νi−uiνidA
where the (penalty) parameter P⋆ is a large positive number.
3
To have
{ϕ^i(ζ1,ζ2,ζ3)=1ϕ^i(ζ1,ζ2,ζ3)=0 , ζ1=ζi1,ζ2=ζi2,ζ3=ζi3 , others
for each ϕ^i 8 functions (for each of the 8 points) with 8 unknowns, aka. ζ1mζ2nζ3k where m,n,k=0,1 can be derived, leading to the only solution as following
⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧ϕ^1=81(1−ζ1)(1−ζ2)(1−ζ3)ϕ^2=81(1+ζ1)(1−ζ2)(1−ζ3)ϕ^3=81(1+ζ1)(1+ζ2)(1−ζ3)ϕ^4=81(1−ζ1)(1+ζ2)(1−ζ3)ϕ^5=81(1−ζ1)(1−ζ2)(1+ζ3)ϕ^6=81(1+ζ1)(1−ζ2)(1+ζ3)ϕ^7=81(1+ζ1)(1+ζ2)(1+ζ3)ϕ^8=81(1−ζ1)(1+ζ2)(1+ζ3)
4
IE has the following symmetries
IEijkl=IEklij=IEjikl=IEijlk
allowing us to rewrite (3) in a more compact matrix form
∫Ω([D]{ν})T[IE]([D]{u})dΩ=∫Ω{ν}T{f}dΩ+∫Γt{ν}T{t∗}dA+P⋆∫Γu{ν}T{u∗−u}dA(4)
where
[D]:=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡∂x1∂00∂x2∂0∂x3∂0∂x2∂0∂x1∂∂x3