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Antiplane shear (or antiplane strain) is the state of strain that is obtained when the displacement field is of the form

${\displaystyle u_{1}=u_{2}=0,~~u_{3}=u_{3}(x_{1},x_{2})}$

There is only an out of plane displacement.

Since the strains are given by

${\displaystyle \varepsilon _{ij}={\cfrac {1}{2}}(u_{i,j}+u_{j,i})={\cfrac {1}{2}}\left({\cfrac {\partial u_{i}}{\partial x_{j}}}+{\cfrac {\partial u_{j}}{\partial x_{i}}}\right)}$

we have

${\displaystyle {\begin{aligned}\varepsilon _{11}&={\cfrac {\partial u_{1}}{\partial x_{1}}}=0\\\varepsilon _{22}&={\cfrac {\partial u_{2}}{\partial x_{2}}}=0\\\varepsilon _{33}&={\cfrac {\partial u_{3}}{\partial x_{3}}}=0\\\varepsilon _{23}&={\cfrac {1}{2}}\left({\cfrac {\partial u_{2}}{\partial x_{3}}}+{\cfrac {\partial u_{3}}{\partial x_{2}}}\right)={\cfrac {1}{2}}~u_{3,2}\\\varepsilon _{31}&={\cfrac {1}{2}}\left({\cfrac {\partial u_{3}}{\partial x_{1}}}+{\cfrac {\partial u_{1}}{\partial x_{3}}}\right)={\cfrac {1}{2}}~u_{3,1}\\\varepsilon _{12}&={\cfrac {1}{2}}\left({\cfrac {\partial u_{1}}{\partial x_{2}}}+{\cfrac {\partial u_{2}}{\partial x_{1}}}\right)=0\end{aligned}}}$

Therefore, for antiplane shear, the only nonzero strains are the out-of-plane shear strains

${\displaystyle \varepsilon _{23}={\cfrac {1}{2}}{\cfrac {\partial u_{3}}{\partial x_{2}}}~;~~\varepsilon _{13}={\cfrac {1}{2}}{\cfrac {\partial u_{3}}{\partial x_{1}}}}$