1. **Problem statement:** Given principal stresses $\sigma_{p1} = 7.88$ ksi and $\sigma_{p2} = 30.49$ ksi, and $b=2.15$ with $\theta_p = \frac{1}{2} \tan^{-1}(b) = 32.5^\circ$, use Mohr's circle to find: (a) stresses on plane $a$ at $45^\circ$ rotation, (b) stresses on horizontal and vertical planes, (c) absolute maximum shear stress. 2. **Mohr's circle basics:** - Principal stresses are normal stresses with zero shear. - Mohr's circle plots normal stress $\sigma$ on x-axis and shear stress $\tau$ on y-axis. - Center $C = \frac{\sigma_{p1} + \sigma_{p2}}{2}$. - Radius $R = \frac{|\sigma_{p1} - \sigma_{p2}|}{2}$. - For a plane rotated by angle $\theta$ (physical space), the angle on Mohr's circle is $2\theta$. 3. **Calculate center and radius:** $$ C = \frac{7.88 + 30.49}{2} = \frac{38.37}{2} = 19.185\,\text{ksi} $$ $$ R = \frac{|7.88 - 30.49|}{2} = \frac{22.61}{2} = 11.305\,\text{ksi} $$ 4. **(a) Stresses on plane $a$ at $45^\circ$:** Angle on Mohr's circle: $2\times 45^\circ = 90^\circ$. Normal stress: $$ \sigma_a = C + R \cos 90^\circ = 19.185 + 11.305 \times 0 = 19.185\,\text{ksi} $$ Shear stress: $$ \tau_a = R \sin 90^\circ = 11.305 \times 1 = 11.305\,\text{ksi} $$ 5. **(b) Stresses on horizontal ($0^\circ$) and vertical ($90^\circ$) planes:** - For $\theta=0^\circ$, angle on Mohr's circle is $0^\circ$: $$ \sigma_H = C + R \cos 0^\circ = 19.185 + 11.305 = 30.49\,\text{ksi} $$ $$ \tau_H = R \sin 0^\circ = 0 $$ - For $\theta=90^\circ$, angle on Mohr's circle is $180^\circ$: $$ \sigma_V = C + R \cos 180^\circ = 19.185 - 11.305 = 7.88\,\text{ksi} $$ $$ \tau_V = R \sin 180^\circ = 0 $$ 6. **(c) Absolute maximum shear stress:** Maximum shear stress equals radius $R$: $$ \tau_{max} = R = 11.305\,\text{ksi} $$ **Final answers:** - (a) $\sigma_a = 19.185$ ksi, $\tau_a = 11.305$ ksi - (b) Horizontal plane: $\sigma_H = 30.49$ ksi, $\tau_H = 0$; Vertical plane: $\sigma_V = 7.88$ ksi, $\tau_V = 0$ - (c) $\tau_{max} = 11.305$ ksi