Suppose that, under the action of external forces, a material point
in a body is displaced to a new location
where
![{\displaystyle x_{1}=A~X_{1}+\kappa ~X_{2}~;~~x_{2}=A~X_{2}+\kappa ~X_{1}~;~~x_{3}=X_{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ea84fb796287845202ddf81b52be7b767bd6af2)
and
and
are constants.
A displacement field is called proper and admissible if the Jacobian (
) 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
so that the deformation represented by the above displacement is continuous.
The deformation gradient
is given by
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba5bd3dab09433449b97b69a968128096d3526cb)
Therefore, the requirement is that
where
![{\displaystyle J=A^{2}-\kappa ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e68acaad8ea351e64419f2aa68b53a5fcc179af7)
The restriction is
![{\displaystyle {|A|>|\kappa |}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0473524bd6099faa63d0ec6629b9f113ae992301)
Suppose that
. Calculate the components of the infinitesimal strain tensor
for the above displacement field.
The displacement is given by
. Therefore,
![{\displaystyle \mathbf {u} ={\begin{bmatrix}\kappa ~X_{2}-X_{1}\\\kappa ~X_{1}-X_{2}\\0\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5186c64b29403becd32a4181003d1773e354f7ce)
The infinitesimal strain tensor is given by
![{\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}u+{\boldsymbol {\nabla }}u^{T})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/560c5baba3a1f89a4e1ecb00042887568566d75c)
The gradient of
is given by
![{\displaystyle {\boldsymbol {\nabla }}u={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5bd78a4237f99241ea637133617b99c1d2117b9)
Therefore,
![{\displaystyle {{\boldsymbol {\varepsilon }}={\begin{bmatrix}-1&\kappa &0\\\kappa &-1&0\\0&0&0\end{bmatrix}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9138df900d03583d759fd606f6bc1cd5604bba6)
Calculate the components of the infinitesimal rotation tensor
for the above displacement field and find the rotation vector
.
The infinitesimal rotation tensor is given by
![{\displaystyle \mathbf {W} ={\frac {1}{2}}({\boldsymbol {\nabla }}u-{\boldsymbol {\nabla }}u^{T})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a10aa7b45711e56d2b61d3f79fd88a29304d3381)
Therefore,
![{\displaystyle {\mathbf {W} ={\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cf894f9ada6139d744c4ba673a6b1d75de853ac)
The rotation vector
is
![{\displaystyle {{\boldsymbol {\omega }}={\begin{bmatrix}0\\0\\0\end{bmatrix}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9ddf4216ebe72754b50c9f604b72f4e88694d80)
Do the strains satisfy compatibility ?
The compatibility equations are
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/906b30a665e048ac4b0220f87e27baf53c0385ed)
All the equations are trivially satisfied because there is no dependence on
,
, and
.
![{\displaystyle {\text{Compatibility is satisfied.}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8524a9d7b0cb15c28de6bef7feb0091cfaff749e)
Calculate the dilatation and the deviatoric strains from the strain tensor.
The dilatation is given by
![{\displaystyle e={\text{tr}}{\boldsymbol {\varepsilon }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0053da2fc48995634470d04b0b58affd87e43bf1)
Therefore,
![{\displaystyle {e=-2~~~~{\text{(Note: Looks like shear only but not really.)}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/badd77483afd86f3cb42cbd2a0c313c3fd4a5e97)
The deviatoric strain is given by
![{\displaystyle {\boldsymbol {\varepsilon }}_{d}={\boldsymbol {\varepsilon }}-{\frac {{\text{tr}}~{\boldsymbol {\varepsilon }}}{3}}\mathbf {I} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/94d6aaab8c51b5c2eac16ba03649060a153f8124)
Hence,
![{\displaystyle {{\boldsymbol {\varepsilon }}_{d}={\begin{bmatrix}-{\cfrac {1}{3}}&\kappa &0\\\kappa &-{\cfrac {1}{3}}&0\\0&0&-{\cfrac {2}{3}}\\\end{bmatrix}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b12278e0fb512196b6546e3278b2884d6c6a64ee)
What is the difference between tensorial shear strain and engineering shear strain (for infinitesimal strains)?
The tensorial shear strains are
,
,
.
The engineering shear strains are
,
,
.
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 (
).
- Compute the right Cauchy-Green deformation tensor (
).
- Calculate the eigenvalues and eigenvectors of
.
- The principal stretches are the square roots of the eigenvalues of
.
- The directions of the principal stretches are the eigenvectors of
.