Elasticity required when studying plasticity Youngung Jeong
Constitutive description on elasticity
Elastic constitutive law:πΌπ = π (elastic stiffness πΌ = 200 [GPa] )
Elastic constitutive law:πΌ!"#$π#$ = π!"
Apply it to FORTRAN, (or Python or Excel)
-Exercise 1. [3x1] = [3x3] [3x1]
-Exercise 2. [nx1] = [nxn] [nx1]- (hint): use π΄ ! = π΅ !" πΆ "
Kronecker delta may appear in formula
π = π#$πΏ#$ =1#
1$
π#$πΏ#$ =1$
π%$πΏ%$ +1$
π&$πΏ&$ +1$
π'$πΏ'$
= π%%πΏ%% + π&&πΏ&& + π''πΏ'' =1#
π## =1!
π!! =1$
π$$
= π%% + π&& + π''
Linear isotropic elasticity
Elastic constitutive law (Hookeβs law):πΌ!"#$π#$ = π!" (ππππππ ππππ π‘ππππ‘π¦)
πΌ!"#$ = ππΏ!"πΏ#$ + π πΏ!#πΏ"$ + πΏ!$πΏ"# (ππ ππ‘πππππ ππππ π‘ππππ‘π¦; π‘π€π ππππ π‘πππ‘π π, π)
Replacing πΌ!"#$ to the Hookeβs lawπ!" = πΌ!"#$π#$ = ππΏ!"πΏ#$π#$ + π πΏ!#πΏ"$ + πΏ!$πΏ"# π#$= ππΏ!"π## + π πΏ!#πΏ"$π#$ + πΏ!$πΏ"#π#$ = ππΏ!"π## + π πΏ!#π#" + πΏ!$π"$= ππΏ!"π## + π π!" + π"! = ππΏ!"π## + 2ππ!"
Examplesβ’ In order for a material (with π = 115.384 GPa, π =76.923 GPa) to exhibit below elastic strain, what stress should be given?
π =0.002 0 00 β0.0006 00 0 β0.0006
β’ Hint: use βπEF = ππΏEFπGG + 2ππEFβ
Linear isotropic elasticity (Young, Poisson)
Elastic constitutive law (Hookeβs law):
π!" =1πΈ π!" β π π##πΏ!" β π!"
β’ If you apply below stress to a material (with πΈ = 200 GPa, π = 0.3), in what strain tensor will the material exhibit?
π =200 0 00 0 00 0 0
, the unit of stress is MPa
Notice that a material with πΈ = 200 GPa, π = 0.3 behaves equivalently with a material with π = 115.384 GPa, π = 76.923 GPa
Symmetries; why only two parameters?
β’ π!" = π"! gives πΌ!"#$ = πΌ"!#$ thus, the required number of elastic constants reduces from 3x3x3x3 to 6x3x3.
β’ Similarly, π!" = π"! gives πΌ!"#$ = πΌ!"$# so that we have the required of number of constants 6x6=36
The required number of constants can be further reduced. Consider the elastic energy:π = β« π!"ππ!"
π!" =ππππ!"
= πΌ!"#$π#$
If we apply partial derivative once again, we have(!)
(*"#(*$%= (
(*"#πΌ!"#$π#$ since πΌ is βconstantβ, we have
π&πππ+,ππ!"
= πΌ!"#$ππ#$ππ+,
= πΌ!"#$πΏ#+πΏ$, = πΌ!"+,
Symmetries; why only two parameters?
β’ π = β« π!"ππ!"
β’ π!" =#$#%!"
= πΌ!"&'π&'
β’ If we apply partial derivative once again, we haveβ’ ##$
#%$%#%!"= #
#%$%πΌ!"&'π&' since πΌ is βconstantβ, we have
β’ !!"!#"#!#$%
= πΌ$%&'!#&'!#"#
= πΌ$%&'πΏ&(πΏ') = πΌ$%()
β’ We could do the 2nd order derivative in a different way (say, instead of !!"!#"#!#$%
we could have done !!"!#$%!#"#
= !!#$%
!"!#"#
= !!#$%
!"!#"#
= πΌ()$%β’ The two cases (regardless of the order of derivative) should give equivalent result
so thatβ’ πΌ$%() = πΌ()$%
β’ This summarizes our finding on the symmetries in elastic tensor:
Reduction to Voigt notationβ’ π!" = πΌ&%%%π%% + πΌ&%%&π%& + πΌ&%%'π%' + πΌ&%&%π&% + πΌ&%&&π&& + πΌ&%&'π&' +πΌ&%'%π'% + πΌ&%'&π'& + πΌ&%&'π''
β’ π!" =
πΌ&%%%πΌ&%%&πΌ&%%'πΌ&%&%πΌ&%&&πΌ&%&'πΌ&%'%πΌ&%'&πΌ&%''
π%%π%&π%'π&%π&&π&'π'%π'&π''
=
πΌ&%%%2πΌ&%%&2πΌ&%%'βπΌ&%&&2πΌ&%&'ββ
πΌ&%''
π%%π%&π%'βπ&&π&'ββπ''
=
πΌ&%%%πΌ&%&&πΌ&%''2πΌ&%&'2πΌ&%%'2πΌ&%%&
π%%π&&π''π&'π%'π%&
β’
ππ =
πΌ&%%%πΌ&%&&πΌ&%''πΌ&%&'πΌ&%%'πΌ&%%&
π%%π&&π''πΎ&'πΎ%'πΎ%&
with πΎ%& = 2π%& and so forthπ() =
πΌ(),)πΌ(),(πΌ(),+πΌ(),,πΌ(),-πΌ(),.
π)π(π+πΎ,πΎ-πΎ.
with 1,1 β 1 , 2,2 β 2 , 3,3 β (3)2,3 β 4 , 1,3 β 5 , 1,2 β (6)
Reduction to Voigt notation
π&% =
πΌ&%,%πΌ&%,&πΌ&%,'πΌ&%,0πΌ&%,1πΌ&%,2
π%π&π'πΎ0πΎ1πΎ2
with 1,1 β 1 , 2,2 β 2 , 3,3 β (3)2,3 β 4 , 1,3 β 5 , 1,2 β (6)
π&% =
πΌ&%,%πΌ&%,&πΌ&%,'πΌ&%,0πΌ&%,1πΌ&%,2
π%π&π'π0π1π2
with 1,2 β (2,1) β (6)
π2 =
πΌ2,%πΌ2,&πΌ2,'πΌ2,0πΌ2,1πΌ2,2
π%π&π'π0π1π2
with 1,2 β (2,1) β (6)
Reduction to Voigt notationπ!" = πΌ!"#$π#$
π! = πΌ!"π"
π%%π&&π''π&'π%'π%&
=
πΌ%%%%πΌ&&%%πΌ''%%πΌ&'%%πΌ%'%%πΌ%&%%
πΌ%%&&πΌ&&&&πΌ''&&πΌ&'&&πΌ%'&&πΌ%&&&
πΌ%%''πΌ&&''πΌ''''πΌ&'''πΌ%'''πΌ%&''
πΌ%%&'πΌ&&&'πΌ''&'πΌ&'&'πΌ%'&'πΌ%&&'
πΌ%%%'πΌ&&%'πΌ''%'πΌ&'%'πΌ%'%'πΌ%&%'
πΌ%%%&πΌ&&%&πΌ''%&πΌ&'%&πΌ%'%&πΌ%&%&
π%%π&&π''2π&'2π%'2π%&
π%π&π'π0π1π2
=
πΌ%%πΌ&%πΌ'%πΌ0%πΌ1%πΌ2%
πΌ%&πΌ&&πΌ'&πΌ0&πΌ1&πΌ2&
πΌ%'πΌ&'πΌ''πΌ0'πΌ1'πΌ2'
πΌ%0πΌ&0πΌ'0πΌ00πΌ10πΌ20
πΌ%1πΌ&1πΌ'1πΌ01πΌ11πΌ21
πΌ%2πΌ&2πΌ'2πΌ02πΌ12πΌ22
π%π&π'π0π1π2
How many constants are required?
π%%π&&π''π&'π%'π%&
=
πΌ%%%%πΌ&&%%πΌ''%%πΌ&'%%πΌ%'%%πΌ%&%%
πΌ%%&&πΌ&&&&πΌ''&&πΌ&'&&πΌ%'&&πΌ%&&&
πΌ%%''πΌ&&''πΌ''''πΌ&'''πΌ%'''πΌ%&''
πΌ%%&'πΌ&&&'πΌ''&'πΌ&'&'πΌ%'&'πΌ%&&'
πΌ%%%'πΌ&&%'πΌ''%'πΌ&'%'πΌ%'%'πΌ%&%'
πΌ%%%&πΌ&&%&πΌ''%&πΌ&'%&πΌ%'%&πΌ%&%&
π%%π&&π''2π&'2π%'2π%&
How many constants do we need?
π!!π""π##
=πΌ!!!! πΌ!!""
πΌ""""πΌ!!##πΌ"""#πΌ####
π!!π""π##
If the coordinate system happens to give strain and stress all principal values:
exampleβ’ Fe(1-0.025)-Al(0.025) alloyμνμ±κ³μλλ€μκ³Όκ°μ΄μ£Όμ΄μ§λ€.
β’ πΌ)) = 270.71, πΌ%& = 128.03, πΌ00 = 108.77
β’ Fe-Al alloyλ Body-centered cubic κ²°μ ꡬ쑰λ₯Όκ°μ§κ³ , κ²°μ λμΉμ±μμν΄λ€μκ³Όκ°μνμ±κ±°λμνλ€.
β’ λΏλ§μλλΌ, cubic κ²°μ ꡬ쑰μλμΉμ±μΌλ‘μΈν΄ πΌ)) = πΌ(( = πΌ'', πΌ00 =πΌ11 = πΌ22, πΌ%& = πΌ%' = πΌ&'
π%π&π'π0π1π2
=
πΌ%%πΌ&%πΌ'%000
πΌ%&πΌ&&πΌ'&000
πΌ%'πΌ&'πΌ''000
000πΌ0000
0000πΌ110
00000πΌ22
π%π&π'π0π1π2
Example
β’ Fe(1-0.025)-Al(0.025) alloyμλ¨κ²°μ μλ€μκ³Όκ°μνμ±λ³νλ₯ μ΄λνλκΈ°μν΄νμνμλ ₯μνλ?
0.000100
000
000
Convert from Cartesian to Voigt
β’ πΌff(ghijk) = πΌffff
lmnkopimq , πΌrs(ghijk) = πΌrrss
lmnkopimq , πΌtf(ghijk) = πΌrsff
lmnkopimq
β’ Material anisotropyβ’ Symmetry can be represented by an orthogonal second order tensor, β’ πΈ = πEFπiβπu, such that πΈvf = πΈw
β’ The invariance of the stiffness tensor under these transformations (due to symmetry) is:β’ πΌ(xyz) = πΈ β πΈ β πΌ {|} β πΈw β πΈw due to symmetry the resulting
tensor should be equivalent with the original one: πΌ(xyz) β‘ πΌ {|}
http://web.mit.edu/16.20/homepage/3_Constitutive/Constitutive_files/module_3_with_solutions.pdf
Convert from Cartesian to Voigt
β’ πΌff(ghijk) = πΌffff
lmnkopimq , πΌrs(ghijk) = πΌrrss
lmnkopimq , πΌtf(ghijk) = πΌrsff
lmnkopimq
β’ Material anisotropyβ’ Symmetry can be represented by an orthogonal second order tensor, β’ πΈ = πEFπiβπu, such that πΈvf = πΈw
β’ The invariance of the stiffness tensor under these transformations (due to symmetry) is:
β’ πΌ(xyz) = πΈ β πΈ β πΌ {|} β πΈw β πΈw due to symmetry the resulting tensor should be equivalent with the original one: πΌ(xyz) β‘ πΌ {|}
http://web.mit.edu/16.20/homepage/3_Constitutive/Constitutive_files/module_3_with_solutions.pdf
β’ For the case of Monoclinic:
β’ πΈ = π&'π(βπ) =1 0 00 1 00 0 β1
β’ Letβs take a look at the invariance due to symmetry
β’ πΌ(+,-) = πΈ β πΈ β πΌ /01 β πΈ2 β πΈ2 due to symmetry the resulting tensor should be equivalent with the original one: πΌ(+,-) β‘ πΌ /01
β’ In its matrix form:β’ πΌ#$%&
'() = π#*π$'π%+π&,πΌ*'+,+&-
β’ Ex: πΌ"".+#/0 = πΌ"""" = π"*π"'π"+π",πΌ*'+,
+&-
β’ If you look at the matrix form of symmetry operator Q in the above, only diagonal components are non-zero. Therefore, π&' = 0 if π β π.
β’ πΌ(()*&+, = πΌ(((( = π((π((π((π((πΌ((((
*-. = πΌ((((*-.
β’ Therefore, πΌ(()*&+, = πΌ((((
β’ Ex: πΌ"1.+#/0 = πΌ""!2 = π"*π"'π!+π2,πΌ*'+,
+&- = π""π""π!!π22πΌ""!2+&- = 1Γ1Γ1Γ β1 ΓπΌ""!2
+&- = βπΌ""!2+&-
β’ Therefore, in order to satisfy πΌ((/0 = βπΌ((/0, πΌ((/0 should be zero.
monoclinic (one symmetry plane)