2. Benchmark solution#
2.1. Calculation method#
For the problem studied, the displacement \(u\) is radial is therefore of shape \(u=[u\mathrm{,0}\mathrm{,0}]\).
From this we deduce the general shape of the strain tensor in large deformations:
as well as the expression for the stress tensor, which is simply written if we take into account the fact that \(J=\text{det}F=1\) for an incompressible problem:
sigma =-p {I} _ {d} _ {d} +mu {b} ^ {d}
, that is:
- left{begin {array} {c} {
sigma} _ {mathit {rr}}mathrm {=}mathrm {=}mathrm {-} p+mu (frac {2} {3} {(1+utext {“})}} ^ {2}}mathrm {-}mathrm {-}}mathrm {-}}frac {1} {3} {3} {(1+frac {u} {r})} ^ {2}mathrm {-}frac {1} {3})\ {sigma} _ {thetatheta}mathrm {=}mathrm {=}mathrm {-}frac {1} {3} {(1+utext {“})}} ^ {2})} ^ {2})} ^ {2}} +frac {2} {3} {3} {r})} ^ {2} {r})} ^ {2}mama thrm {-}frac {1} {3})\ {sigma} _ {mathit {zz}}mathrm {=}mathrm {=}mathrm {-}frac {1} {3} {{3} {(1+utext {“})}}mathrm {“})}} ^ {2}mathrm {-}frac {1} {3} {(1+frac {u} {r})} ^ {2} +frac {2} {3})\ {sigma} _ {rtheta}mathrm {=} {sigma} _ {mathit {riz}}mathrm {=} {sigma} _ {sigma} _ {theta z} _ {theta z}}mathrm {=} 0end {array} _ {theta z}}mathrm {=} right.
Writing balance equations leads to the verification of a single equation:
sigma {text {“}} _ {mathit {rr}}} +frac {{sigma} _ {mathit {rr}}mathrm {-} {sigma}} _ {sigma} _ {thetatheta}}} {r}mathrm {=} 0
which makes it possible to determine the pressure \(p\) knowing the radial displacement field \(u\):
2.2. Particularization of the solution#
The incompressibility condition is written as \(\text{det}F=1\) with \(F=\left[\begin{array}{ccc}1+u\text{'}& 0& 0\\ 0& 1+\frac{u}{r}& 0\\ 0& 0& 1\end{array}\right]\). The displacement \(u\) therefore verifies the following differential equation:
The imposed load is as follows \(u={U}_{0}\) in \(r=a\).
So the solution on the go is:
- left{begin {array} {c} {c} {u} _ {u} _ {r} =-r+rsqrt {{r} ^ {2} + {U} _ {0} ({U} _ {0}} +mathrm {2a})}
\ {u} _ {theta} = {u} _ {z} =0end {array}right.
The deformation tensor therefore has the following expression:
- left{begin {array} {c} {b} {b} _ {mathrm {rr}}} =frac {{r} ^ {2}} {{r} ^ {2}} + {U} _ {2} + {U} _ {0}}}\
- {b} _ {thetatheta} =frac {{r} ^ {2} + {U} _ {0} ({U} _ {0} +mathrm {2a})} {{r} ^ {2}}\
- {b} _ {mathrm {zz}} =1\
{b} _ {rtheta} = {b} _ {ztheta} = {b} _ {theta z} =0end {array}right.
And the constraints apply to:
- left{begin {array} {c}
{sigma} _ {mathit {rr}}mathrm {=}mathrm {=}mathrm {-} p+mu (frac {2} {3}}frac {{r} ^ {2}}} {{r}}}} {{r}}} {r} ^ {2}}} {{r}}}} {{r}}} {{r}}} {{r}}} {{r}}} {{r}}} {{r}}} {{r}}} {r}}} {r} ^ {2}}} {{r}}} {r} ^ {2}}} {{r}}} {r} ^ {2}}} {{r}}} {r} ^ {2}} frac {1} {3}frac {{r} ^ {2} ^ {2} + {U} _ {0} ({U} _ {0} +mathrm {2a})} {{r} ^ {2}}}mathrm {2}}}mathrm {-}frac {1} {3})\ {sigma} _ {thetatheta}mathrm {=}mathrm {=}mathrm {-}frac {1} {3}frac {{r} ^ {2}} ^ {2}}} {{r}}} {{r}}} {{r} ^ {2}}} + {{r}} {2}}} {{r}} {2}} + {{r}} {2}}} {{r}} {2}} + {{r}} {2}} {{r}}} {r} ^ {2}} + {{r}} {2}}} {{r}} {2}} + {{r}} {2}} {{r}} {2}} + {{r}frac {2} {3}frac {{r} ^ {2} ^ {2} + {U} _ {0} ({U} _ {0} +mathrm {2a})} {{r} ^ {2}}}mathrm {2}}}mathrm {-}frac {1} {3})} {{r} ^ {2}}} {{r} ^ {2}}}mathrm {2}}}mathrm {2}}}{sigma} _ {mathit {zz}}mathrm {=}mathrm {=}mathrm {-}frac {1} {3}frac {{r} ^ {2}} ^ {2}} {{2}}} {{r}}} {{r}} {2}} +mathrm {2}}}mathrm {-}frac {1} {3} {3}frac {{r} ^ {2} + {U} _ {0} ({U} _ {0} +mathrm {2a})}} {{r} {2}})} {{r} {3})} {{r})} {{r}})} {{r} ^ {2}}} +frac {2} {3})\ {sigma} _ {rtheta}mathrm {=} {sigma} _ {sigma} _ {ztheta}mathrm {=} {sigma} _ {theta z}mathrm {=}} 0end {array}right.
with \(p\) obtained by integrating (2.1) which is equal to:
where \(C\) is a constant
The following numerical values are finally obtained:
in \(r\mathrm{=}0.1\): |
in \(r\mathrm{=}0.2\): |
\(\begin{array}{c}{u}_{r}={6.10}^{-5}\\ {\mathrm{\sigma }}_{\mathit{rr}}=-59.9955\\ {\mathrm{\sigma }}_{\mathrm{\theta }\mathrm{\theta }}=99.9566\\ {\mathrm{\sigma }}_{\mathit{zz}}=19.9326\\ {E}_{\mathit{rr}}=\mathrm{0,0005994604316761909}\\ {E}_{\mathrm{\theta }\mathrm{\theta }}=-0.0006001799999999502\end{array}\) |
|
The transition through the Cartesian system is made using the following relationships:
- begin {array} {c}
- {sigma} _ {mathit {xx}}}mathrm {=} {sigma} _ {mathit {rr}} {mathrm {cos}} ^ {2}theta + {sigma} _ {sigma} _ {sigma} _ {thetatheta}} {thetathetatheta} thetatheta {theta} thetatheta} _ {r theta}mathrm {sin}thetamathrm {cos}theta\
- {sigma} _ {thetatheta}mathrm {=} {sigma} _ {mathit {rr}} {mathrm {sin}} ^ {2}theta + {sigma} _ {sigma} _ {thetatheta} _ {thetatheta} _ {theta} _ {theta} _ {theta} _ {theta}} _ {theta}theta} _ {theta}theta} _ {theta}theta}mathrm {sin}thetamathrm {cos}theta\
{sigma} _ {mathit {zz}}}mathrm {=} {sigma} _ {mathit {rr}}mathrm {sin}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos}thetamathrm {cos} thrm {-} 2 {sigma} _ {rtheta} ({mathrm {cos}}} ^ {2}thetamathrm {-} {mathrm {sin}}}} ^ {2}theta)end {array}
2.3. Reference quantities and results#
The following are compared to the reference values:
trips \((u,v)\) to points \(A\) and \(F\),
constraints \(({\sigma }_{\mathit{xx}},{\sigma }_{\mathit{yy}},{\sigma }_{\mathit{zz}},{\sigma }_{\mathit{xy}})\) at points \(A\) and \(F\),
the Von Mises and Tresca constraints as well as the eigenvalues of the stress tensor at point \(A\).