# Step-by-Step Solution Let's solve the problem in a structured manner. --- ## **Given Data** - Stresses at a critical section: - \( \sigma_x = -80 \) MPa (compressive) - \( \sigma_y = 90 \) MPa (tensile) - \( \tau_{xy} = 60 \) MPa (shear) - Ultimate strength of material (\( S_u \)): **350 MPa** - Poisson's ratio (\( \nu \)): **.20** - Material: **brittle** - **Ask:** Use **2 different yield criteria** to determine the factor of safety (FoS). --- ## **Step 1: Choose Yield (Failure) Criteria** For **brittle materials**, the two most common failure criteria are: 1. **Maximum Normal Stress (Rankine) Theory** 2. **Mohr’s Theory (Coulomb-Mohr)** ### Why these? - **Maximum Normal Stress Theory**: Suitable for brittle materials that fail due to normal stress. - **Mohr’s Theory**: Considers both tensile and compressive strengths and is more accurate for brittle materials. --- ## **Step 2: Find Principal Stresses** Given: - \( \sigma_x = -80 \) MPa - \( \sigma_y = 90 \) MPa - \( \tau_{xy} = 60 \) MPa Principal stresses (\( \sigma_1, \sigma_2 \)): \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2} \] Plug in values: - \( \frac{\sigma_x + \sigma_y}{2} = \frac{-80 + 90}{2} = 5 \) MPa - \( \frac{\sigma_x - \sigma_y}{2} = \frac{-80 - 90}{2} = -85 \) MPa So, \[ \sigma_{1,2} = 5 \pm \sqrt{(-85)^2 + (60)^2} \] \[ = 5 \pm \sqrt{7225 + 360} \] \[ = 5 \pm \sqrt{10825} \] \[ = 5 \pm 104.03 \] - \( \sigma_1 = 5 + 104.03 = 109.03 \) MPa - \( \sigma_2 = 5 - 104.03 = -99.03 \) MPa --- ## **Step 3: Maximum Normal Stress Theory (Rankine)** - Failure occurs if **maximum principal stress** exceeds ultimate strength. - Factor of Safety (FoS): \[ \text{FoS}_{\text{Rankine}} = \frac{S_u}{|\text{max principal stress}|} = \frac{350}{109.03} = 3.21 \] --- ## **Step 4: Mohr’s Theory (Coulomb-Mohr Criterion)** For brittle materials, failure occurs when the Mohr's circle touches the failure envelope. \[ \text{FoS}_{\text{Mohr}} = \frac{S_u}{\text{max}(|\sigma_1|, |\sigma_2|)} \] But, for materials with different strengths in tension and compression, this is a more accurate approach. However, since only ultimate strength is given, we take the maximum absolute value of principal stresses. \[ \text{FoS}_{\text{Mohr}} = \frac{350}{\text{max}(109.03, 99.03)} = \frac{350}{109.03} = 3.21 \] *(If compressive strength was different, we'd use the relevant value)* --- ## **Step 5: Summary Table** | Criterion | Max Principal Stress (MPa) | Ultimate Strength (MPa) | Factor of Safety | |----------------------|---------------------------|-------------------------|------------------| | Rankine | 109.03 | 350 | 3.21 | | Mohr (Coulomb-Mohr) | 109.03 | 350 | 3.21 | --- ## **Step 6: Brief Discussion** - **Rankine criterion** is suitable for brittle materials as they fail due to high normal stresses. - **Mohr's criterion** is more general and considers both tensile and compressive stresses, important if the material's ultimate strengths differ in tension and compression. - Here, both criteria give the same FoS as only one ultimate strength is provided. --- ## **Final Answers** ### **(a) Factor of Safety by Rankine (Maximum Normal Stress) Theory:** \[ \boxed{3.21} \] ### **(b) Factor of Safety by Mohr (Coulomb-Mohr) Theory:** \[ \boxed{3.21} \] ### **(c) Why these criteria?** - Both are most applicable for **brittle materials**. - They address failure under maximum principal stress and combined stress states, matching how brittle materials actually fail. --- **If you have compressive and tensile ultimate strengths separately, use those in Mohr's criterion!**