Topic: Strength of Materials
A plane stress element in a structural member has under loading $sigma_x = 4P$, $sigma_y = 2P$ and $tau_{xy} = sqrt{3}P$, 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
- A. 60 MPa
- B. 75 MPa
- C. 100 MPa
- D. 120 MPa
Answer: B. 75 MPa
Explanation: Principal stresses are calculated as $sigma_{1,2} = (sigma_x + sigma_y)/2 \pm sqrt{((sigma_x - sigma_y)/2)^2 + tau_{xy}^2}$. Substituting the values: $(4P + 2P)/2 \pm sqrt{((4P - 2P)/2)^2 + (sqrt{3}P)^2} = 3P \pm sqrt{P² + 3P^2} = 3P \pm 2P$. Thus, $sigma_1 = 5P$ and $sigma_2 = P$. Maximum shear stress $tau_{max} = (sigma_1 - sigma_2)/2 = (5P - P)/2 = 2P$. According to the Maximum Shear Stress Theory (Tresca), yielding starts when $tau_{max} = S_y / 2$. Therefore, $2P = 300 / 2 = 150$, giving $P = 75$ MPa.