The displacement field that leads to a state of antiplane shear is (in rectangular Cartesian coordinates)

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

where ${\displaystyle u_{i},~i=1,2,3}$  are the displacements in the ${\displaystyle x_{1},x_{2},x_{3}\,}$  directions.

For an isotropic, linear elastic material, the stress tensor that results from a state of antiplane shear can be expressed as

${\displaystyle {\boldsymbol {\sigma }}\equiv {\begin{bmatrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}&\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}\end{bmatrix}}={\begin{bmatrix}0&0&\mu ~{\cfrac {\partial u_{3}}{\partial x_{1}}}\\0&0&\mu ~{\cfrac {\partial u_{3}}{\partial x_{2}}}\\\mu ~{\cfrac {\partial u_{3}}{\partial x_{1}}}&\mu ~{\cfrac {\partial u_{3}}{\partial x_{2}}}&0\end{bmatrix}}}$ 

where ${\displaystyle \mu \,}$  is the shear modulus of the material.

The conservation of linear momentum in the absence of inertial forces takes the form of the equilibrium equation. For general states of stress there are three equilibrium equations. However, for antiplane shear, with the assumption that body forces in the 1 and 2 directions are 0, these reduce to one equilibrium equation which is expressed as

${\displaystyle \mu ~\nabla ^{2}u_{3}+b_{3}(x_{1},x_{2})=0}$ 

where ${\displaystyle b_{3}}$  is the body force in the ${\displaystyle x_{3}}$  direction and ${\displaystyle \nabla ^{2}u_{3}={\cfrac {\partial ^{2}u_{3}}{\partial x_{1}^{2}}}+{\cfrac {\partial ^{2}u_{3}}{\partial x_{2}^{2}}}}$ . Note that this equation is valid only for infinitesimal strains.

The antiplane shear assumption is used to determine the stresses and displacements due to a screw dislocation.

• Infinitesimal strain theory
• Deformation (mechanics)