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.
Part (a)
edit
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.
Solution
edit
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 |}}
Part (b)
edit
Suppose that
A
=
0
{\displaystyle A=0}
. Calculate the components of the infinitesimal strain tensor
ε
{\displaystyle {\boldsymbol {\varepsilon }}}
for the above displacement field.
Solution
edit
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}}}}
Part (c)
edit
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 }}}
.
Solution
edit
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}}}}
Part (d)
edit
Do the strains satisfy compatibility ?
Solution
edit
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.}}}
Part (e)
edit
Calculate the dilatation and the deviatoric strains from the strain tensor.
Solution
edit
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}}}}
Part (f)
edit
What is the difference between tensorial shear strain and engineering shear strain (for infinitesimal strains)?
Solution
edit
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.
Part (g)
edit
Briefly describe the process which you would use to calculate the principal stretches and their directions.
Solution
edit
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} }
.