Let's break down the problem: ### **Given:** - The element is subjected to a **pure shear stress** of \( \tau_{xy} = 65 \) ksi. - The element is rotated **60° clockwise**. - We are to determine the **equivalent state of stress** on the rotated element. --- ## **Step 1: Identify Initial Stresses** The initial stresses: - \( \sigma_x = 0 \) - \( \sigma_y = 0 \) - \( \tau_{xy} = 65 \) ksi --- ## **Step 2: Stress Transformation Equations** For a plane stress element rotated by angle \( \theta \): \[ \begin{align*} \sigma_x' &= \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta \\ \sigma_y' &= \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2} \cos 2\theta - \tau_{xy} \sin 2\theta \\ \tau_{x'y'} &= -\frac{\sigma_x - \sigma_y}{2} \sin 2\theta + \tau_{xy} \cos 2\theta \end{align*} \] Given \( \sigma_x = \sigma_y = 0 \): \[ \begin{align*} \sigma_x' &= \tau_{xy} \sin 2\theta \\ \sigma_y' &= -\tau_{xy} \sin 2\theta \\ \tau_{x'y'} &= \tau_{xy} \cos 2\theta \end{align*} \] --- ## **Step 3: Substitute Values** - \( \tau_{xy} = 65 \) ksi - \( \theta = 60^\circ \) - \( 2\theta = 120^\circ \) \[ \sin 120^\circ = \sqrt{3}/2 \approx 0.866 \] \[ \cos 120^\circ = -1/2 = -0.5 \] Now substitute: \[ \sigma_x' = 65 \times 0.866 = 56.29 \text{ ksi} \] \[ \sigma_y' = -65 \times 0.866 = -56.29 \text{ ksi} \] \[ \tau_{x'y'} = 65 \times (-0.5) = -32.5 \text{ ksi} \] --- ## **Step 4: Final Answer** ### **The equivalent state of stress is:** \[ \boxed{ \begin{align*} \sigma_x' &= 56.3 \text{ ksi} \\ \sigma_y' &= -56.3 \text{ ksi} \\ \tau_{x'y'} &= -32.5 \text{ ksi} \end{align*} } \] **(Rounded to one decimal place)** --- Let me know if you need a stress element sketch for the rotated case!