<Dl> <Dd> ∑ F x ′ = σ n d A − σ x d A cos 2 ⁡ θ − σ y d A sin 2 ⁡ θ − τ x y d A cos ⁡ θ sin ⁡ θ − τ x y d A sin ⁡ θ cos ⁡ θ = 0 σ n = σ x cos 2 ⁡ θ + σ y sin 2 ⁡ θ + 2 τ x y sin ⁡ θ cos ⁡ θ (\ displaystyle \ (\ begin (aligned) \ sum F_ (x') & = \ sigma _ (\ mathrm (n)) dA - \ sigma _ (x) dA \ cos ^ (2) \ theta - \ sigma _ (y) dA \ sin ^ (2) \ theta - \ tau _ (xy) dA \ cos \ theta \ sin \ theta - \ tau _ (xy) dA \ sin \ theta \ cos \ theta = 0 \ \ \ sigma _ (\ mathrm (n)) & = \ sigma _ (x) \ cos ^ (2) \ theta + \ sigma _ (y) \ sin ^ (2) \ theta + 2 \ tau _ (xy) \ sin \ theta \ cos \ theta \ \ \ end (aligned))) </Dd> </Dl> <Dd> ∑ F x ′ = σ n d A − σ x d A cos 2 ⁡ θ − σ y d A sin 2 ⁡ θ − τ x y d A cos ⁡ θ sin ⁡ θ − τ x y d A sin ⁡ θ cos ⁡ θ = 0 σ n = σ x cos 2 ⁡ θ + σ y sin 2 ⁡ θ + 2 τ x y sin ⁡ θ cos ⁡ θ (\ displaystyle \ (\ begin (aligned) \ sum F_ (x') & = \ sigma _ (\ mathrm (n)) dA - \ sigma _ (x) dA \ cos ^ (2) \ theta - \ sigma _ (y) dA \ sin ^ (2) \ theta - \ tau _ (xy) dA \ cos \ theta \ sin \ theta - \ tau _ (xy) dA \ sin \ theta \ cos \ theta = 0 \ \ \ sigma _ (\ mathrm (n)) & = \ sigma _ (x) \ cos ^ (2) \ theta + \ sigma _ (y) \ sin ^ (2) \ theta + 2 \ tau _ (xy) \ sin \ theta \ cos \ theta \ \ \ end (aligned))) </Dd> <P> However, knowing that </P> <Dl> <Dd> cos 2 ⁡ θ = 1 + cos ⁡ 2 θ 2, sin 2 ⁡ θ = 1 − cos ⁡ 2 θ 2 and sin ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ (\ displaystyle \ cos ^ (2) \ theta = (\ frac (1 + \ cos 2 \ theta) (2)), \ qquad \ sin ^ (2) \ theta = (\ frac (1 - \ cos 2 \ theta) (2)) \ qquad (\ text (and)) \ qquad \ sin 2 \ theta = 2 \ sin \ theta \ cos \ theta) </Dd> </Dl>

A positive shear stress produces a positive shear strain