Let's solve the problem **step by step**: --- ## **Given Data** - \( \sigma_{xx} = 2,950 \) lbf/ft² - \( \sigma_{yy} = 4,900 \) lbf/ft² - \( \sigma_n(AA) = 3,450 \) lbf/ft² (normal stress on AA) - Angle \( \theta = 30^\circ \) with respect to the x-axis We are asked to find the **shear stress** \( \sigma_{xy} \). --- ## **Step 1: Stress Transformation Equations** For a plane at angle \( \theta \): \[ \sigma_n = \sigma_{xx} \cos^2\theta + \sigma_{yy} \sin^2\theta + 2\sigma_{xy} \sin\theta \cos\theta \] where: - \( \sigma_n \): normal stress on the plane \( AA \) - \( \sigma_{xy} \): shear stress (unknown) --- ## **Step 2: Plug in the Known Values** - \( \sigma_n = 3,450 \) lbf/ft² - \( \sigma_{xx} = 2,950 \) lbf/ft² - \( \sigma_{yy} = 4,900 \) lbf/ft² - \( \theta = 30^\circ \) - \( \cos 30^\circ = .866 \) - \( \sin 30^\circ = .5 \) Plug into the equation: \[ 3,450 = 2,950(\cos^2 30^\circ) + 4,900(\sin^2 30^\circ) + 2\sigma_{xy}(\sin 30^\circ \cos 30^\circ) \] --- ## **Step 3: Calculate Trig Functions** - \( \cos^2 30^\circ = (.866)^2 = .75 \) - \( \sin^2 30^\circ = (.5)^2 = .25 \) - \( \sin 30^\circ \cos 30^\circ = .5 \times .866 = .433 \) --- ## **Step 4: Substitute Values** \[ 3,450 = 2,950(.75) + 4,900(.25) + 2\sigma_{xy}(.433) \] \[ 3,450 = 2,212.5 + 1,225 + .866\sigma_{xy} \] \[ 3,450 = 3,437.5 + .866\sigma_{xy} \] --- ## **Step 5: Solve for \( \sigma_{xy} \)** \[ 3,450 - 3,437.5 = .866\sigma_{xy} \] \[ 12.5 = .866\sigma_{xy} \] \[ \sigma_{xy} = \frac{12.5}{.866} \] \[ \sigma_{xy} \approx 14.4 \text{ lbf/ft}^2 \] --- ## **Final Answer** \[ \boxed{14.4 \ \text{lbf/ft}^2} \] --- **The shear stress \( \sigma_{xy} \) is \( \boxed{14.4 \ \text{lbf/ft}^2} \).**