Let's break down the problem: ### **Given:** - \(\sigma_x = 90\) MPa (tension) - \(\sigma_y = -60\) MPa (compression) - \(\tau_{xy} = 30\) MPa (shear) ### **Required:** 1. Principal stresses (\(\sigma_1, \sigma_2\)) 2. Maximum in-plane shear stress (\(\tau_{\text{max}}\)) --- ## **Step 1: Principal Stresses Formula** \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] --- ## **Step 2: Substitute Values** - \(\sigma_x = 90\) MPa - \(\sigma_y = -60\) MPa - \(\tau_{xy} = 30\) MPa \[ \frac{\sigma_x + \sigma_y}{2} = \frac{90 + (-60)}{2} = \frac{30}{2} = 15 \text{ MPa} \] \[ \frac{\sigma_x - \sigma_y}{2} = \frac{90 - (-60)}{2} = \frac{150}{2} = 75 \text{ MPa} \] \[ \sqrt{ (75)^2 + (30)^2 } = \sqrt{ 5625 + 900 } = \sqrt{6525} \approx 80.78 \text{ MPa} \] --- ## **Step 3: Calculate Principal Stresses** \[ \sigma_1 = 15 + 80.78 = 95.78 \text{ MPa} \] \[ \sigma_2 = 15 - 80.78 = -65.78 \text{ MPa} \] --- ## **Step 4: Maximum In-Plane Shear Stress** \[ \tau_{\text{max}} = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] \[ \tau_{\text{max}} = 80.78 \text{ MPa} \] --- ## **Final Answers:** ### **Principal Stresses:** - \(\boxed{\sigma_1 = 95.78\ \text{MPa}}\) - \(\boxed{\sigma_2 = -65.78\ \text{MPa}}\) ### **Maximum In-Plane Shear Stress:** - \(\boxed{\tau_{\text{max}} = 80.78\ \text{MPa}}\) Let me know if you need the principal planes or a Mohr's circle sketch!

**1. Introduction** Understanding the state of stress at a point within a material is fundamental in the field of solid mechanics, especially in the analysis of plane stress conditions. Plane stress occurs in thin structures where stresses perpendicular to the plane are negligible, simplifying three-dimensional stress states to two dimensions. In such conditions, the internal forces are described by normal stresses (\(\sigma_x, \sigma_y\)) acting perpendicular to the plane and shear stresses (\(\tau_{xy}\)) acting within the plane. The key concepts involved include principal stresses, which are the maximum and minimum normal stresses at a point, and shear stresses, which act tangentially to the planes. The principal stresses are important because they indicate the maximum normal stresses the material experiences, and their orientations reveal the planes where shear stress is zero. The maximum in-plane shear stress is equally significant because it identifies the highest shear force acting within the plane, which influences failure modes such as shear fracture. Understanding how to determine principal stresses and maximum shear stress from given stress components involves using specific formulas derived from the equilibrium and Mohr's circle concepts. These calculations help engineers evaluate the strength, safety, and failure potential of materials under complex loading conditions. **Explanation:** This introduction provides the theoretical foundation necessary to comprehend the problem. Recognizing the significance of principal stresses and shear stresses enables a systematic approach to analyze the stress state, guiding the calculations that follow. It connects core concepts of stress transformation with their practical applications in material failure analysis. --- **2. Presentation of Relevant Formulas and Representation of Given Data** **Relevant Formulas:** - **Principal Stresses (\(\sigma_1, \sigma_2\)):** \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] - **Maximum In-Plane Shear Stress (\(\tau_{max}\)):** \[ \tau_{max} = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] **Representation of Given Data:** | Parameter | Description | Value | Units | |------------|--------------|--------|--------| | \(\sigma_x\) | Normal stress in x-direction | 90 | MPa | | \(\sigma_y\) | Normal stress in y-direction | -60 | MPa | | \(\tau_{xy}\) | Shear stress | 30 | MPa | **Rationale:** The formulas are derived from the equilibrium conditions and Mohr's circle analysis, providing a direct method to compute principal stresses and shear stresses from known stress components. Accurate representation of data ensures precise calculations. **Explanation:** Using these formulas allows transforming the given stress components into principal stresses and maximum shear stress, which are critical in assessing material response under in-plane loading. --- **3. A Detailed Step-by-Step Solution** **Step 1: Calculate the average normal stress** \[ \frac{\sigma_x + \sigma_y}{2} = \frac{90 + (-60)}{2} = \frac{30}{2} = 15 \text{ MPa} \] *This value represents the mean normal stress at the point, serving as the central point for principal stress calculations.* **Step 2: Calculate the difference of normal stresses divided by 2** \[ \frac{\sigma_x - \sigma_y}{2} = \frac{90 - (-60)}{2} = \frac{150}{2} = 75 \text{ MPa} \] *This difference indicates the extent of anisotropy in normal stresses, influencing the principal stress magnitude.* **Step 3: Calculate the radicand for the principal stresses** \[ \sqrt{ (75)^2 + (30)^2 } = \sqrt{ 5625 + 900 } = \sqrt{6525} \approx 80.78 \text{ MPa} \] *This term combines the effects of normal and shear stresses to determine the maximum and minimum normal stresses.* **Step 4: Determine the principal stresses** \[ \sigma_1 = 15 + 80.78 = 95.78 \text{ MPa} \] \[ \sigma_2 = 15 - 80.78 = -65.78 \text{ MPa} \] *These represent the maximum and minimum normal stresses experienced at the point.* **Step 5: Calculate the maximum in-plane shear stress** \[ \tau_{max} = 80.78 \text{ MPa} \] *This is the maximum shear stress within the plane, occurring at an orientation where shear stress reaches this value.* --- **Explanation:** Each step involves straightforward algebraic manipulations based on the stress components. Calculating the average and difference normal stresses sets the stage for applying the principal stress formula. Using the Pythagorean relation captures the combined effects of normal and shear stresses, culminating in the determination of principal stresses and maximum shear stress, which are pivotal for failure analysis and design considerations. --- **Conclusion** The analysis of the given plane stress state yields principal stresses of approximately **95.78 MPa** (tensile) and **-65.78 MPa** (compressive), indicating the maximum and minimum normal stresses at the point. The maximum in-plane shear stress is approximately **80.78 MPa**, representing the highest shear force acting within the plane. These values are obtained using standard transformation formulas derived from Mohr's circle concepts, providing essential insights into the stress conditions that influence material behavior and failure potential under the given loading scenario.