<Dd> σ n = 1 2 (σ x + σ y) + 1 2 (σ x − σ y) cos ⁡ 2 θ + τ x y sin ⁡ 2 θ (\ displaystyle \ sigma _ (\ mathrm (n)) = (\ frac (1) (2)) (\ sigma _ (x) + \ sigma _ (y)) + (\ frac (1) (2)) (\ sigma _ (x) - \ sigma _ (y)) \ cos 2 \ theta + \ tau _ (xy) \ sin 2 \ theta \, \!) </Dd> <Dl> <Dd> τ n = − 1 2 (σ x − σ y) sin ⁡ 2 θ + τ x y cos ⁡ 2 θ (\ displaystyle \ tau _ (\ mathrm (n)) = - (\ frac (1) (2)) (\ sigma _ (x) - \ sigma _ (y)) \ sin 2 \ theta + \ tau _ (xy) \ cos 2 \ theta \, \!) </Dd> </Dl> <Dd> τ n = − 1 2 (σ x − σ y) sin ⁡ 2 θ + τ x y cos ⁡ 2 θ (\ displaystyle \ tau _ (\ mathrm (n)) = - (\ frac (1) (2)) (\ sigma _ (x) - \ sigma _ (y)) \ sin 2 \ theta + \ tau _ (xy) \ cos 2 \ theta \, \!) </Dd> <P> These equations indicate that in a plane stress or plane strain condition, one can determine the stress components at a point on all directions, i.e. as a function of θ (\ displaystyle \ theta \, \!), if one knows the stress components (σ x, σ y, τ x y) (\ displaystyle (\ sigma _ (x), \ sigma _ (y), \ tau _ (xy)) \, \!) on any two perpendicular directions at that point . It is important to remember that we are considering a unit area of the infinitesimal element in the direction parallel to the y (\ displaystyle y \, \!) - z (\ displaystyle z \, \!) plane . </P>

A state of pure shear is one in which the normal stresses are zero. consider