Let's break this down step by step: --- ## **Given Stress Tensor** \[ \begin{bmatrix} -100 & & -80 \\ & 20 & \\ -80 & & 20 \end{bmatrix} \text{ MPa} \] --- # **(a) Principal Stresses (Eigenvalues):** Principal stresses are found by solving the characteristic equation: \[ \left| \sigma_{ij} - \lambda \delta_{ij} \right| = \] \[ \begin{vmatrix} -100-\lambda & & -80 \\ & 20-\lambda & \\ -80 & & 20-\lambda \end{vmatrix} = \] ### **Step 1: Expand the Determinant** Expand along the second row (since it has two zeros): \[ (20-\lambda) \cdot \begin{vmatrix} -100-\lambda & -80 \\ -80 & 20-\lambda \end{vmatrix} = \] Expand the 2x2 determinant: \[ (20-\lambda)\left[(-100-\lambda)(20-\lambda) - (-80)^2\right] = \] ### **Step 2: Set Up the Equation** \[ (20-\lambda) \left[(-100-\lambda)(20-\lambda) - 640\right] = \] Now expand \((-100-\lambda)(20-\lambda)\): \[ (-100-\lambda)(20-\lambda) = -200 + 100\lambda - 20\lambda + \lambda^2 = -200 + 80\lambda + \lambda^2 \] So, \[ (20-\lambda) \left[ -200 + 80\lambda + \lambda^2 - 640 \right] = \] \[ (20-\lambda) \left[ \lambda^2 + 80\lambda - 840 \right] = \] ### **Step 3: Solve for Eigenvalues** - First root: \( \lambda_1 = 20 \) MPa - The quadratic: \[ \lambda^2 + 80\lambda - 840 = \] Use quadratic formula: \[ \lambda = \frac{-80 \pm \sqrt{80^2 - 4 \cdot 1 \cdot (-840)}}{2} \] \[ \lambda = \frac{-80 \pm \sqrt{640 + 33600}}{2} \] \[ \lambda = \frac{-80 \pm \sqrt{40000}}{2} \] \[ \sqrt{40000} = 200 \] So, \[ \lambda_2 = \frac{-80 + 200}{2} = \frac{120}{2} = 60 \text{ MPa} \] \[ \lambda_3 = \frac{-80 - 200}{2} = \frac{-280}{2} = -140 \text{ MPa} \] ### **Principal Stresses (Final Answer):** \[ \boxed{ \begin{aligned} \sigma_1 &= 60\ \text{MPa} \\ \sigma_2 &= 20\ \text{MPa} \\ \sigma_3 &= -140\ \text{MPa} \end{aligned} } \] --- # **(b) Octahedral Stresses and Maximum Shear Stress** ## **1. Octahedral Normal Stress (\(\sigma_{oct}\))** \[ \sigma_{oct} = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3} \] \[ \sigma_{oct} = \frac{60 + 20 + (-140)}{3} = \frac{-60}{3} = \boxed{-20\ \text{MPa}} \] --- ## **2. Octahedral Shear Stress (\(\tau_{oct}\))** \[ \tau_{oct} = \frac{1}{3} \sqrt{ (\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2 } \] Compute each term: - \((\sigma_1 - \sigma_2)^2 = (60 - 20)^2 = 160\) - \((\sigma_2 - \sigma_3)^2 = (20 - (-140))^2 = (160)^2 = 25600\) - \((\sigma_3 - \sigma_1)^2 = (-140 - 60)^2 = (-200)^2 = 40000\) Sum: \( 160 + 25600 + 40000 = 67200 \) \[ \tau_{oct} = \frac{1}{3} \sqrt{67200} \] \[ \sqrt{67200} \approx 259.23 \] \[ \tau_{oct} \approx \frac{259.23}{3} \approx \boxed{86.41\ \text{MPa}} \] --- ## **3. Maximum Shearing Stress** \[ \tau_{max} = \frac{\sigma_{max} - \sigma_{min}}{2} \] \[ \tau_{max} = \frac{60 - (-140)}{2} = \frac{200}{2} = \boxed{100\ \text{MPa}} \] --- # **Summary Table** | Quantity | Value | |-----------------------------|--------------------| | Principal Stresses (\(\sigma_1, \sigma_2, \sigma_3\)) | 60 MPa, 20 MPa, -140 MPa | | Octahedral Normal Stress | -20 MPa | | Octahedral Shear Stress | 86.41 MPa | | Maximum Shear Stress | 100 MPa | --- If you need further explanation for any step, let me know!