Stress in x direction:
σ x = 100 N / m m 2 \sigma_x=100\ N/mm^2 σ x = 100 N / m m 2
Stress in y direction:
σ y = 50 N / m m 2 \sigma_y=50\ N/mm^2 σ y = 50 N / m m 2
Shear stress:
τ x y = 60 N / m m 2 \tau_{xy}=60\ N/mm^2 τ x y = 60 N / m m 2
Principal stresses:
σ 1 , 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 \sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{(\frac{\sigma_x-\sigma_y}{2})^2+\tau_{xy}^2} σ 1 , 2 = 2 σ x + σ y ± ( 2 σ x − σ y ) 2 + τ x y 2
σ 1 , 2 = 100 + 50 2 ± ( 100 − 50 2 ) 2 + 6 0 2 \sigma_{1,2}=\frac{100+50}{2}\pm\sqrt{(\frac{100-50}{2})^2+60^2} σ 1 , 2 = 2 100 + 50 ± ( 2 100 − 50 ) 2 + 6 0 2
σ 1 = 140 N / m m 2 \sigma_1=140\ N/mm^2 σ 1 = 140 N / m m 2
σ 2 = 10 N / m m 2 \sigma_2=10\ N/mm^2 σ 2 = 10 N / m m 2
Direction of principal plane:
t a n 2 θ p = − σ x − σ y 2 τ x y tan2\theta_p=-\frac{\sigma_x-\sigma_y}{2\tau_{xy}} t an 2 θ p = − 2 τ x y σ x − σ y
t a n 2 θ p = − 100 − 50 2 ⋅ 60 = − 5 / 12 tan2\theta_p=-\frac{100-50}{2\cdot60}=-5/12 t an 2 θ p = − 2 ⋅ 60 100 − 50 = − 5/12
θ p = − 11.31 ° \theta_p=-11.31\degree θ p = − 11.31°
The greatest shear stress:
τ m a x = ∣ σ 1 − σ 2 ∣ / 2 \tau_{max}=|\sigma_1-\sigma_2|/2 τ ma x = ∣ σ 1 − σ 2 ∣/2
τ m a x = ( 140 − 10 ) / 2 = 65 N / m m 2 \tau_{max}=(140-10)/2=65\ N/mm^2 τ ma x = ( 140 − 10 ) /2 = 65 N / m m 2
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