Answer :

Solution :

σ_x = 3P, σ_y = -2P and τ_xy = √2 P

S_y = 350 MPa, N = 1

Principal stress are calculated

$\frac{1}{2}[({\sigma }_x-{\sigma }_y)\pm \sqrt{({\sigma }_x+{\sigma }_y{)}^2+(2{\tau }_{xy}{)}^2}]$

$\frac{1}{2}[(3P-2P)\pm \sqrt{(3P+2P{)}^2+(2\times \sqrt{2}P{)}^2}]$

$\frac{1}{2}[P\pm \sqrt{33}P]$

σ₁ = 3.375 P, σ₂ = - 2.375 P

According to Maximum distortion energy theory :

(σ₁ - σ₂)² + (σ₂ - σ₃)² + (σ₃ - σ₁)² ≤ $2{\left(\frac{S_y}{N}\right)}^2$

here σ₃ = 0, on simplyfying we get

σ₁² + σ₂² - σ₁σ₂ = $\frac{S_y}{N}$

(3.375 P)² + ( - 2.375 P)² - (3.375 P)( - 2.375 P) = (350/1)²

11.4 P² + 5.64 P² + 8.015 P² = 350²

25.055 (P)² = (350)²

P = 350 / 5 = 70 MPa