Nonlinear finite elements/Homework 7/Hints: Difference between revisions

From testwiki
Jump to navigation Jump to search
imported>MaintenanceBot
m Subpage navbar
 
(No difference)

Latest revision as of 01:13, 26 July 2017



Hints 1: Index notation

Index notation:

σij=2μεij+λεkkδij.

If j=i

σii=2μεii+λεkkδii=2μεkk+3λεkk=(2μ+3λ)εkk
σkk=(2μ+3λ)εkk

Dummy indices are replaceable.

Hint 2: Index notation

Index notation:

σij=2μεij+λεkkδij.

Multiply by δij:

σijδij=2μεijδij+λεkkδijδijσjj=2μεii+λεkkδiiσkk=2μεkk+3λεkkσkk=(2μ+3λ)εkk

Multiplication by δij leads to replacement of one index.

Aijδkl=?Aijδjl=?

Hint 3: Index notation

Index notation:

σ=σij𝐞i𝐞jε=εij𝐞i𝐞j

From the definition of dyadic product, we can show

(𝐚𝐛):(𝐮𝐯)=(𝐚𝐮)(𝐛𝐯)

Contraction gives:

σ:ε=(σij𝐞i𝐞j):(εkl𝐞k𝐞l)=σijεkl(𝐞i𝐞j):(𝐞k𝐞l)=σijεkl(𝐞i𝐞k)(𝐞j𝐞l)=σijεklδikδjl=σijεij

Hint 4: Tensor product

Index notation:

ε=εij𝐞i𝐞j𝖢=Cijkl𝐞i𝐞j𝐞k𝐞l

Definition of dyadics products:

(𝐚𝐛)𝐱=(𝐛𝐱)𝐚(𝐚𝐛𝐜)𝐱=(𝐜𝐱)(𝐚𝐛)(𝐚𝐛𝐜𝐝)𝐱=(𝐝𝐱)(𝐚𝐛𝐜)

We can show that

(𝐚𝐛𝐜𝐝):(𝐮𝐯)=((𝐚𝐛𝐜𝐝)𝐯)𝐮

Contraction gives:

𝖢:ε=(Cijkl𝐞i𝐞j𝐞k𝐞l):(εmn𝐞m𝐞n)=Cijklεmn(𝐞i𝐞j𝐞k𝐞l):(𝐞m𝐞n)=Cijklεmn((𝐞i𝐞i𝐞k𝐞l)𝐞n)𝐞m=Cijklεmn(𝐞l𝐞n)(𝐞i𝐞j𝐞k)𝐞m=Cijklεmnδln(𝐞k𝐞m)(𝐞i𝐞j)=Cijklεmnδlnδkm𝐞i𝐞j=Cijklεkl

Hint 5 : Tensor product

Tensor Product of two tensors:

𝑨=Aij𝐞i𝐞j𝑩=Bkl𝐞k𝐞l

Tensor product:

𝑨𝑩=(Aij𝐞i𝐞j)(Bkl𝐞k𝐞l)=AijBkl𝐞i𝐞j𝐞k𝐞l

Hint 6: Vector transformations

Change of basis: Vector transformation rule

vi'=Lijvj

Lij are the direction cosines.

L11=𝐞1'𝐞1;L12=𝐞1'𝐞2;L13=𝐞1'𝐞3L21=𝐞2'𝐞1;L22=𝐞2'𝐞2;L23=𝐞2'𝐞3L31=𝐞3'𝐞1;L32=𝐞3'𝐞2;L33=𝐞3'𝐞3

In matrix form

𝐯'=𝐋𝐯;𝐯=𝐋T𝐯';𝐋𝐋T=𝐈𝐋T=𝐋1

Other common form: Vector transformation rule

vi'=Qjivj
Q11=𝐞1𝐞1';Q12=𝐞1𝐞2';Q13=𝐞1𝐞3'Q21=𝐞2𝐞1';Q22=𝐞2𝐞2';Q23=𝐞2𝐞3'Q31=𝐞3𝐞1';Q32=𝐞3𝐞2';Q33=𝐞3𝐞3'

In matrix form

𝐯'=𝐐T𝐯;𝐯=𝐐𝐯';𝐐𝐐T=𝐈𝐐T=𝐐1

Hint 7: Tensor transformations

Change of basis: Tensor transformation rule

Tij'=LipLjqTpq

where Lij are the direction cosines.

In matrix form,

𝐓'=𝐋𝐓𝐋T

Other common form

Tij'=QpiQqjTpq

In matrix form,

𝐓'=𝐐T𝐓𝐐

Template:Subpage navbar