Suppose that, under the action of external forces, a material point
p
=
(
X
1
,
X
2
,
X
3
)
{\displaystyle \mathbf {p} =(X_{1},X_{2},X_{3})}
in a body is displaced to a new location
q
=
(
x
1
,
x
2
,
x
3
)
{\displaystyle \mathbf {q} =(x_{1},x_{2},x_{3})}
where
x
1
=
A
X
1
+
κ
X
2
;
x
2
=
A
X
2
+
κ
X
1
;
x
3
=
X
3
{\displaystyle x_{1}=A~X_{1}+\kappa ~X_{2}~;~~x_{2}=A~X_{2}+\kappa ~X_{1}~;~~x_{3}=X_{3}}
and
A
{\displaystyle A}
and
κ
{\displaystyle \kappa }
are constants.
A displacement field is called proper and admissible if the Jacobian (
J
{\displaystyle J}
) is greater than zero. If a displacement field is proper and admissible, then the deformation of the body is continuous.
Indicate the restrictions that must be imposed upon
A
{\displaystyle A}
so that the deformation represented by the above displacement is continuous.
The deformation gradient
(
F
)
{\displaystyle (F)}
is given by
F
i
j
=
∂
x
i
∂
X
j
=
[
A
κ
0
κ
A
0
0
0
1
]
{\displaystyle {\begin{aligned}F_{ij}&={\frac {\partial x_{i}}{\partial X_{j}}}\\&={\begin{bmatrix}A&\kappa &0\\\kappa &A&0\\0&0&1\end{bmatrix}}\end{aligned}}}
Therefore, the requirement is that
J
=
det
(
F
)
>
0
{\displaystyle J={\text{det}}(F)>0}
where
J
=
A
2
−
κ
2
{\displaystyle J=A^{2}-\kappa ^{2}}
The restriction is
|
A
|
>
|
κ
|
{\displaystyle {|A|>|\kappa |}}
Suppose that
A
=
0
{\displaystyle A=0}
. Calculate the components of the infinitesimal strain tensor
ε
{\displaystyle {\boldsymbol {\varepsilon }}}
for the above displacement field.
The displacement is given by
u
=
x
−
X
{\displaystyle \mathbf {u} =\mathbf {x} -\mathbf {X} }
. Therefore,
u
=
[
κ
X
2
−
X
1
κ
X
1
−
X
2
0
]
{\displaystyle \mathbf {u} ={\begin{bmatrix}\kappa ~X_{2}-X_{1}\\\kappa ~X_{1}-X_{2}\\0\end{bmatrix}}}
The infinitesimal strain tensor is given by
ε
=
1
2
(
∇
u
+
∇
u
T
)
{\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}u+{\boldsymbol {\nabla }}u^{T})}
The gradient of
u
{\displaystyle \mathbf {u} }
is given by
∇
u
=
[
−
1
κ
0
κ
−
1
0
0
0
0
]
{\displaystyle {\boldsymbol {\nabla }}u={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}}
Therefore,
ε
=
[
−
1
κ
0
κ
−
1
0
0
0
0
]
{\displaystyle {{\boldsymbol {\varepsilon }}={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}}}
Calculate the components of the infinitesimal rotation tensor
W
{\displaystyle \mathbf {W} }
for the above displacement field and find the rotation vector
ω
{\displaystyle {\boldsymbol {\omega }}}
.
The infinitesimal rotation tensor is given by
W
=
1
2
(
∇
u
−
∇
u
T
)
{\displaystyle \mathbf {W} ={\frac {1}{2}}({\boldsymbol {\nabla }}u-{\boldsymbol {\nabla }}u^{T})}
Therefore,
W
=
[
0
0
0
0
0
0
0
0
0
]
{\displaystyle {\mathbf {W} ={\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}}}
The rotation vector
ω
{\displaystyle {\boldsymbol {\omega }}}
is
ω
=
[
0
0
0
]
{\displaystyle {{\boldsymbol {\omega }}={\begin{bmatrix}0\\0\\0\end{bmatrix}}}}
Do the strains satisfy compatibility ?
The compatibility equations are
ε
11
,
22
+
ε
22
,
11
−
2
ε
12
,
12
=
0
ε
22
,
33
+
ε
33
,
22
−
2
ε
23
,
23
=
0
ε
33
,
11
+
ε
11
,
33
−
2
ε
13
,
13
=
0
(
ε
12
,
3
−
ε
23
,
1
+
ε
31
,
2
)
,
1
−
ε
11
,
23
=
0
(
ε
23
,
1
−
ε
31
,
2
+
ε
12
,
3
)
,
2
−
ε
22
,
31
=
0
(
ε
31
,
2
−
ε
12
,
3
+
ε
23
,
1
)
,
3
−
ε
33
,
12
=
0
{\displaystyle {\begin{aligned}\varepsilon _{11,22}+\varepsilon _{22,11}-2\varepsilon _{12,12}&=0\\\varepsilon _{22,33}+\varepsilon _{33,22}-2\varepsilon _{23,23}&=0\\\varepsilon _{33,11}+\varepsilon _{11,33}-2\varepsilon _{13,13}&=0\\(\varepsilon _{12,3}-\varepsilon _{23,1}+\varepsilon _{31,2})_{,1}-\varepsilon _{11,23}&=0\\(\varepsilon _{23,1}-\varepsilon _{31,2}+\varepsilon _{12,3})_{,2}-\varepsilon _{22,31}&=0\\(\varepsilon _{31,2}-\varepsilon _{12,3}+\varepsilon _{23,1})_{,3}-\varepsilon _{33,12}&=0\end{aligned}}}
All the equations are trivially satisfied because there is no dependence on
X
1
{\displaystyle X_{1}}
,
X
2
{\displaystyle X_{2}}
, and
X
3
{\displaystyle X_{3}}
.
Compatibility is satisfied.
{\displaystyle {\text{Compatibility is satisfied.}}}
Calculate the dilatation and the deviatoric strains from the strain tensor.
The dilatation is given by
e
=
tr
ε
{\displaystyle e={\text{tr}}{\boldsymbol {\varepsilon }}}
Therefore,
e
=
−
2
(Note: Looks like shear only but not really.)
{\displaystyle {e=-2~~~~{\text{(Note: Looks like shear only but not really.)}}}}
The deviatoric strain is given by
ε
d
=
ε
−
tr
ε
3
I
{\displaystyle {\boldsymbol {\varepsilon }}_{d}={\boldsymbol {\varepsilon }}-{\frac {{\text{tr}}~{\boldsymbol {\varepsilon }}}{3}}\mathbf {I} }
Hence,
ε
d
=
[
−
1
3
κ
0
κ
−
1
3
0
0
0
−
2
3
]
{\displaystyle {{\boldsymbol {\varepsilon }}_{d}={\begin{bmatrix}-{\cfrac {1}{3}}&\kappa &0\\\kappa &-{\cfrac {1}{3}}&0\\0&0&-{\cfrac {2}{3}}\\\end{bmatrix}}}}
What is the difference between tensorial shear strain and engineering shear strain (for infinitesimal strains)?
The tensorial shear strains are
ε
12
{\displaystyle \varepsilon _{12}}
,
ε
23
{\displaystyle \varepsilon _{23}}
,
ε
31
{\displaystyle \varepsilon _{31}}
.
The engineering shear strains are
γ
12
{\displaystyle \gamma _{12}}
,
γ
23
{\displaystyle \gamma _{23}}
,
γ
31
{\displaystyle \gamma _{31}}
.
The engineering shear strains are twice the tensorial shear strains.
Briefly describe the process which you would use to calculate the principal stretches and their directions.
Compute the deformation gradient (
F
{\displaystyle \mathbf {F} }
).
Compute the right Cauchy-Green deformation tensor (
C
=
F
T
∙
F
{\displaystyle \mathbf {C} =\mathbf {F} ^{T}\bullet \mathbf {F} }
).
Calculate the eigenvalues and eigenvectors of
C
{\displaystyle \mathbf {C} }
.
The principal stretches are the square roots of the eigenvalues of
C
{\displaystyle \mathbf {C} }
.
The directions of the principal stretches are the eigenvectors of
C
{\displaystyle \mathbf {C} }
.