Problem 1: Part 9: Elastic-plastic tangent modulus
edit
Assume that the elastic response of the material is linear, i.e.,
C
=
λ
1
⊗
1
+
2
μ
I
.
{\displaystyle {\boldsymbol {\mathsf {C}}}=\lambda {\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu {\boldsymbol {\mathsf {I}}}~.}
Derive the expression for the elastic-plastic tangent modulus for a von Mises yield
condition with Johnson-Cook flow stress for a linear elastic material using the expressions
that you have derived in the previous parts.
The elastic-plastic tangent modulus is given by
C
ep
=
C
−
(
(
C
:
f
σ
)
⊗
(
C
:
f
σ
)
f
σ
:
C
:
f
σ
−
2
3
f
α
ε
p
:
f
σ
‖
ε
p
‖
−
χ
ρ
C
p
f
T
σ
:
f
σ
)
.
{\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}={\boldsymbol {\mathsf {C}}}-\left({\cfrac {({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}}{f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}-{\sqrt {\cfrac {2}{3}}}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\cfrac {\chi }{\rho ~C_{p}}}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}}}\right)~.}
From the previous parts
f
σ
=
3
2
n
;
f
α
=
−
n
B
α
n
−
1
[
1
−
(
T
−
T
0
T
m
−
T
0
)
]
;
f
T
=
(
1
T
m
−
T
0
)
[
σ
0
+
B
α
n
]
.
{\displaystyle f_{\boldsymbol {\sigma }}={\sqrt {\cfrac {3}{2}}}~\mathbf {n} ~;~~f_{\alpha }=-n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~;~~f_{T}=\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~.}
Therefore,
C
:
f
σ
=
3
2
(
λ
1
⊗
1
+
2
μ
I
)
:
n
=
3
2
(
λ
tr
(
n
)
1
+
2
μ
n
)
=
3
2
2
μ
n
.
{\displaystyle {\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}={\sqrt {\cfrac {3}{2}}}~(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):\mathbf {n} ={\sqrt {\cfrac {3}{2}}}~(\lambda ~{\text{tr}}(\mathbf {n} )~{\boldsymbol {\mathit {1}}}+2~\mu ~\mathbf {n} )={\sqrt {\cfrac {3}{2}}}~2~\mu ~\mathbf {n} ~.}
Some of the results used in the above derivation are shown below.
Recall (from previous homework):
1
⊗
1
:
n
=
δ
i
j
δ
k
l
n
k
l
=
n
k
k
δ
i
j
=
tr
(
n
)
1
{\displaystyle {\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}:\mathbf {n} =\delta _{ij}~\delta _{kl}~n_{kl}=n_{kk}~\delta _{ij}={\text{tr}}(\mathbf {n} )~{\boldsymbol {\mathit {1}}}}
and
I
:
n
=
1
2
(
δ
i
k
δ
j
l
+
δ
i
l
δ
j
k
)
n
k
l
=
1
2
(
δ
j
l
n
i
l
+
δ
j
k
n
k
i
)
=
1
2
(
n
i
j
+
n
j
i
)
=
n
i
j
=
n
{\displaystyle {\boldsymbol {\mathsf {I}}}:\mathbf {n} ={\frac {1}{2}}~(\delta _{ik}~\delta _{jl}+\delta _{il}~\delta _{jk})n_{kl}={\frac {1}{2}}~(\delta _{jl}~n_{il}+\delta _{jk}~n_{ki})={\frac {1}{2}}~(n_{ij}+n_{ji})=n_{ij}=\mathbf {n} }
(we have used the symmetry of the stress tensor above.)
Also,
tr
(
n
)
=
tr
(
s
‖
s
‖
)
=
tr
(
s
)
‖
s
‖
{\displaystyle {\text{tr}}(\mathbf {n} )={\text{tr}}\left({\cfrac {\mathbf {s} }{\lVert \mathbf {s} \rVert _{}}}\right)={\cfrac {{\text{tr}}(\mathbf {s} )}{\lVert \mathbf {s} \rVert _{}}}}
Now,
s
=
s
i
j
=
σ
i
j
−
1
3
σ
k
k
δ
i
j
⟹
tr
(
s
)
=
s
i
i
=
σ
i
i
−
1
3
σ
k
k
δ
i
i
=
σ
i
i
−
1
3
σ
k
k
3
=
σ
i
i
−
σ
k
k
=
0
{\displaystyle \mathbf {s} =s_{ij}=\sigma _{ij}-{\frac {1}{3}}~\sigma _{kk}~\delta _{ij}\qquad \implies \qquad {\text{tr}}(\mathbf {s} )=s_{ii}=\sigma _{ii}-{\frac {1}{3}}~\sigma _{kk}~\delta _{ii}=\sigma _{ii}-{\frac {1}{3}}~\sigma _{kk}~3=\sigma _{ii}-\sigma _{kk}=0}
Therefore,
tr
(
n
)
=
0
.
{\displaystyle {\text{tr}}(\mathbf {n} )=0~.}
Hence,
(
C
:
f
σ
)
⊗
(
C
:
f
σ
)
=
3
2
4
μ
2
n
⊗
n
=
6
μ
2
n
⊗
n
{\displaystyle ({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}={\cfrac {3}{2}}~4~\mu ^{2}~\mathbf {n} \otimes \mathbf {n} =6~\mu ^{2}~\mathbf {n} \otimes \mathbf {n} }
and
f
σ
:
C
:
f
σ
=
3
2
2
μ
n
:
n
=
3
μ
(
s
s
:
s
)
:
(
s
s
:
s
)
=
3
μ
s
:
s
s
:
s
=
3
μ
.
{\displaystyle f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}={\cfrac {3}{2}}~2~\mu ~\mathbf {n} :\mathbf {n} =3~\mu ~\left({\cfrac {\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}\right):\left({\cfrac {\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}\right)=3~\mu ~{\cfrac {\mathbf {s} :\mathbf {s} }{\mathbf {s} :\mathbf {s} }}=3~\mu ~.}
Plugging in expression for
C
ep
{\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}}
C
ep
=
λ
1
⊗
1
+
2
μ
I
−
(
6
μ
2
n
⊗
n
3
μ
−
n
B
α
n
−
1
[
1
−
(
T
−
T
0
T
m
−
T
0
)
]
ε
p
:
n
‖
ε
p
‖
−
3
2
χ
ρ
C
p
(
1
T
m
−
T
0
)
[
σ
0
+
B
α
n
]
σ
:
n
)
.
{\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}-\left({\cfrac {6~\mu ^{2}~\mathbf {n} \otimes \mathbf {n} }{3~\mu -n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:\mathbf {n} }{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi }{\rho ~C_{p}}}~\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~{\boldsymbol {\sigma }}:\mathbf {n} }}\right)~.}
Now,
σ
:
n
=
(
s
+
1
3
tr
(
σ
)
1
)
:
s
s
:
s
=
s
:
s
s
:
s
+
1
3
tr
(
σ
)
1
:
s
s
:
s
=
s
:
s
+
1
3
tr
(
σ
)
tr
(
s
)
s
:
s
=
s
:
s
=
‖
s
‖
{\displaystyle {\boldsymbol {\sigma }}:\mathbf {n} ={\cfrac {(\mathbf {s} +{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\mathit {1}}}):\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}={\cfrac {\mathbf {s} :\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}+{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\cfrac {{\boldsymbol {\mathit {1}}}:\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}={\sqrt {\mathbf {s} :\mathbf {s} }}+{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\cfrac {{\text{tr}}(\mathbf {s} )}{\sqrt {\mathbf {s} :\mathbf {s} }}}={\sqrt {\mathbf {s} :\mathbf {s} }}=\lVert \mathbf {s} \rVert _{}}
Therefore, the elastic-plastic tangent modulus can be written as
C
ep
=
λ
1
⊗
1
+
2
μ
[
I
−
3
μ
n
⊗
n
3
μ
−
n
B
α
n
−
1
[
1
−
(
T
−
T
0
T
m
−
T
0
)
]
ε
p
:
n
‖
ε
p
‖
−
3
2
χ
ρ
C
p
(
1
T
m
−
T
0
)
[
σ
0
+
B
α
n
]
‖
s
‖
]
.
{\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\mathsf {I}}}-{\cfrac {3~\mu ~\mathbf {n} \otimes \mathbf {n} }{3~\mu -n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:\mathbf {n} }{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi }{\rho ~C_{p}}}~\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~\lVert \mathbf {s} \rVert _{}}}\right]~.}
![{\displaystyle f_{\boldsymbol {\sigma }}={\sqrt {\cfrac {3}{2}}}~\mathbf {n} ~;~~f_{\alpha }=-n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~;~~f_{T}=\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c6e83866fb96e00a47ae5c61fc8a3aa31079954)
![{\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}-\left({\cfrac {6~\mu ^{2}~\mathbf {n} \otimes \mathbf {n} }{3~\mu -n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:\mathbf {n} }{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi }{\rho ~C_{p}}}~\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~{\boldsymbol {\sigma }}:\mathbf {n} }}\right)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/650e5b1e6ca8a1c73bfc312ad605fdf6d6cdbf4e)
![{\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\mathsf {I}}}-{\cfrac {3~\mu ~\mathbf {n} \otimes \mathbf {n} }{3~\mu -n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:\mathbf {n} }{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi }{\rho ~C_{p}}}~\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~\lVert \mathbf {s} \rVert _{}}}\right]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36825c83cfa0dbfec6e0e597190c6ae427b058ad)