Let's solve the problem step by step as per your instructions. --- ### **Step 1: Conceptual Introduction** When a stress element is rotated, the normal and shear stresses on its faces change with the orientation. To find the new stresses at a rotated angle, we use transformation equations for plane stress. This process allows engineers to determine the stress state at any orientation, which is crucial for failure analysis and design. **Explanation:** The original stress element has known normal and shear stresses. By rotating the element (in this case, 10° counterclockwise), the stresses on the faces change due to the orientation shift. Using transformation equations, we can mathematically calculate the new values of normal (σ) and shear (τ) stresses on the rotated axes. --- ### **Step 2: Formulae** The formulas for plane stress transformation after a rotation by angle θ are: \[ \begin{align*} \sigma_{x'} &= \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) + \tau_{xy} \sin(2\theta) \\ \sigma_{y'} &= \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) - \tau_{xy} \sin(2\theta) \\ \tau_{x'y'} &= -\frac{\sigma_x - \sigma_y}{2} \sin(2\theta) + \tau_{xy} \cos(2\theta) \end{align*} \] **Explanation:** These equations use the original normal stresses (\(\sigma_x, \sigma_y\)) and shear stress (\(\tau_{xy}\)), and the angle of rotation (θ, in radians). The trigonometric terms account for the effect of rotation on the stress components, giving the new stresses in the rotated orientation. --- ### **Step 3: Step-by-Step Calculation for 1st Question (\(\sigma_{x'}\))** Given: - \(\sigma_x = 8\) ksi (horizontal, rightward) - \(\sigma_y = 6\) ksi (vertical, upward) - \(\tau_{xy} = 13.5\) ksi (counterclockwise positive per sign convention) - θ = 10° (convert to radians for calculation: θ = 10 × π/180 = 0.1745 rad) Now, calculate: \[ \begin{align*} \sigma_{x'} &= \frac{8 + 6}{2} + \frac{8 - 6}{2} \cos(2 \times 10^\circ) + 13.5 \sin(2 \times 10^\circ) \\ &= 7 + 1 \cdot \cos(20^\circ) + 13.5 \cdot \sin(20^\circ) \\ \end{align*} \] Calculate values: - \(\cos(20^\circ) = 0.9397\) - \(\sin(20^\circ) = 0.3420\) \[ \sigma_{x'} = 7 + 0.9397 + 13.5 \times 0.3420 \] \[ = 7 + 0.9397 + 4.617 \] \[ = 12.56 \ \text{ksi} \] **Explanation:** We substituted the known values into the first stress transformation equation, evaluated the trigonometric functions, and performed the arithmetic step by step to arrive at the new normal stress on the x'-face. --- ### **Step 4: Step-by-Step Calculation for 2nd Question (\(\sigma_{y'}\))** \[ \begin{align*} \sigma_{y'} &= \frac{8 + 6}{2} - \frac{8 - 6}{2} \cos(20^\circ) - 13.5 \sin(20^\circ) \\ &= 7 - 0.9397 - 4.617 \\ &= 7 - 0.9397 - 4.617 \\ &= 1.44 \ \text{ksi} \end{align*} \] **Explanation:** Similarly, we used the second transformation formula, substituting the same values, but notice the signs change as per the formula. The careful arithmetic yields the new normal stress on the y'-face. --- ### **Step 5: Step-by-Step Calculation for 3rd Question (\(\tau_{x'y'}\))** \[ \begin{align*} \tau_{x'y'} &= -\frac{8 - 6}{2} \sin(20^\circ) + 13.5 \cos(20^\circ) \\ &= -1 \cdot 0.3420 + 13.5 \cdot 0.9397 \\ &= -0.3420 + 12.686 \\ &= 12.34 \ \text{ksi} \end{align*} \] **Explanation:** The third equation gives the new shear stress on the rotated element. The calculation is direct, using the sine and cosine of 20°, and combining with the original shear stress as per the formula. --- ### **Final Answer** - \(\sigma_{x'} = 12.56\) ksi (tensile) - \(\sigma_{y'} = 1.44\) ksi (tensile) - \(\tau_{x'y'} = 12.34\) ksi (CCW on positive x'-face) **These are the normal and shear stresses after a 10° counterclockwise rotation of the element.**

Let's carefully re-verify the calculations step by step. --- ### **Step 1: Restate given data** - Original stresses: - \(\sigma_x = 8\, \text{ksi}\) - \(\sigma_y = 6\, \text{ksi}\) - \(\tau_{xy} = 13.5\, \text{ksi}\) - Rotation angle: - \(\theta = 10^\circ\) --- ### **Step 2: Convert angle to radians** \[ 2\theta = 2 \times 10^\circ = 20^\circ \] - \(\sin(20^\circ) \approx 0.3420\) - \(\cos(20^\circ) \approx 0.9397\) --- ### **Step 3: Apply the transformation formulas** \[ \begin{cases} \sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) + \tau_{xy} \sin(2\theta) \\ \sigma_{y'} = \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) - \tau_{xy} \sin(2\theta) \\ \tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2} \sin(2\theta) + \tau_{xy} \cos(2\theta) \end{cases} \] --- ### **Step 4: Calculate \(\sigma_{x'}\)** \[ \sigma_{x'} = \frac{8 + 6}{2} + \frac{8 - 6}{2} \times 0.9397 + 13.5 \times 0.3420 \] \[ = 7 + 1 \times 0.9397 + 13.5 \times 0.3420 \] \[ = 7 + 0.9397 + 4.617 \] \[ = 12.5567\, \text{ksi} \] **Rounded:** **12.56 ksi** --- ### **Step 5: Calculate \(\sigma_{y'}\)** \[ \sigma_{y'} = 7 - 0.9397 - 4.617 \] \[ = 7 - 0.9397 - 4.617 \] \[ = 7 - 5.5567 \] \[ = 1.4433\, \text{ksi} \] **Rounded:** **1.44 ksi** --- ### **Step 6: Calculate \(\tau_{x'y'}\)** \[ \tau_{x'y'} = -1 \times 0.3420 + 13.5 \times 0.9397 \] \[ = -0.3420 + 12.685 \] \[ = 12.343\, \text{ksi} \] **Rounded:** **12.34 ksi** --- ### **Final verified answers:** | Stress Component | Value (ksi) | Interpretation | |---------------------|--------------|----------------| | \(\sigma_{x'}\) | **12.56** | Tensile | | \(\sigma_{y'}\) | **1.44** | Tensile | | \(\tau_{x'y'}\) | **12.34** | CCW shear on positive x'-face | --- ### **Summary:** The initial calculations were correct, and these verified values match the detailed step-by-step calculations. The key is careful arithmetic and proper use of the transformation equations. **Answer:** **Normal stresses:** \[ \boxed{ \begin{aligned} &\sigma_{x'} = 12.56\, \text{ksi} \\ &\sigma_{y'} = 1.44\, \text{ksi} \end{aligned} } \] **Shearing stress:** \[ \boxed{ \tau_{x'y'} = 12.34\, \text{ksi} } \] These are the stresses after rotating the element through 10° counterclockwise.