A plane stress element in a structural member under loading has
$$σ_x = 4P, σ_y$$​ = 2P and $$τ_{xy}$$​ = $$\sqrt{3}$$​ P , where P > 0 . The yield strength of the material is 300 MPa. If the member is designed using the maximum shear stress theory, then the value of P at which yielding starts is

Correct option is B

​$$\textbf{Given:} \\\sigma_x = 4P \\\sigma_y = 2P \\\tau_{xy} = \sqrt{3}P \\\text{Yield strength } (\sigma_y): 300\, \text{MPa} \\\textbf{Design theory:} \text{ Maximum shear stress theory (Tresca)} \\\\{\textbf{Compute Principal Stresses}} \\\text{For plane stress, the principal stresses } (\sigma_1, \sigma_2) \text{ are:} \\\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \\\\\text{Substitute the given stresses:} \\\sigma_{1,2} = \frac{4P + 2P}{2} \pm \sqrt{ \left( \frac{4P - 2P}{2} \right)^2 + (\sqrt{3}P)^2 } \\\\\text{Simplify:} \\\sigma_{1,2} = 3P \pm \sqrt{P^2 + 3P^2} = 3P \pm 2P \\\\\text{Thus:} \\\sigma_1 = 5P, \quad \sigma_2 = P \\(\sigma_3 = 0 \text{ for plane stress}) \\\\{\textbf{Apply Maximum Shear Stress Theory}} \\\text{The Tresca criterion states yielding begins when:} \\\tau_{\max} = \frac{\sigma_1 - \sigma_2}{2} \geq \frac{\sigma_y}{2} \\\\\text{Substitute } \sigma_1 = 5P \text{ and } \sigma_2 = P: \\\frac{5P - P}{2} = 2P \geq \frac{300}{2} = 150 \\\\\text{Solve for } P: \\2P \geq 150 \Rightarrow \boxed{P \geq 75\, \text{MPa}}$$​​