# Step-by Solution ##Given Data** - **Steel rod allowable shear stress \(\taus,allow = 15\, \text{ksi}\) - **Brass rod BC allowable shear stress:** \(\tau_{b,allow} = 8\, \text{ksi}\) *(not directly used for AB)* - **Applied torque at A:** \(T = 10\, \text{kip} \cdot \text{in} = 10,000\, \text{lb} \cdot \text{in}\) - **Find:** Required diameter of rod **AB** so that the shearing stress in steel does not exceed 15 ksi. --- ## **Step 1: Shear Stress Formula for Solid Shaft** Shear stress in a solid circular shaft due to torque: \[ \tau = \frac{T \cdot r}{J} \] Where: - \(T\) = applied torque - \(r\) = outer radius of shaft - \(J\) = polar moment of inertia, for solid shaft: \(J = \frac{\pi}{2} r^4\) Maximum shear stress occurs at the outer surface (\(r = \frac{d}{2}\)), so: \[ \tau_{max} = \frac{T \cdot \left(\frac{d}{2}\right)}{\frac{\pi}{2} \left(\frac{d}{2}\right)^4} = \frac{T \cdot d/2}{\frac{\pi}{2} \cdot \frac{d^4}{16}} \] Simplifying denominator: \[ \frac{\pi}{2} \cdot \frac{d^4}{16} = \frac{\pi d^4}{32} \] So, \[ \tau_{max} = \frac{T \cdot d/2}{\pi d^4 / 32} = \frac{T \cdot d}{2} \cdot \frac{32}{\pi d^4} = \frac{16 T}{\pi d^3} \] --- ## **Step 2: Set Allowable Stress and Solve for \(d\)** Set \(\tau_{max} = \tau_{s,allow}\): \[ \tau_{s,allow} = \frac{16 T}{\pi d^3} \] Solve for \(d\): \[ d^3 = \frac{16 T}{\pi \tau_{s,allow}} \] \[ d = \sqrt[3]{\frac{16 T}{\pi \tau_{s,allow}}} \] --- ## **Step 3: Substitute Values** - \(T = 10,000\, \text{lb} \cdot \text{in}\) - \(\tau_{s,allow} = 15\, \text{ksi} = 15,000\, \text{psi}\) Plug in: \[ d = \sqrt[3]{\frac{16 \times 10,000}{\pi \times 15,000}} \] --- ## **Step 4: Calculate Numerically** Calculate denominator: - \(\pi \times 15,000 \approx 3.1416 \times 15,000 \approx 47,124\) - Numerator: \(16 \times 10,000 = 160,000\) Now, \[ d = \sqrt[3]{\frac{160,000}{47,124}} = \sqrt[3]{3.395} \] Now, take cube root: - \(\sqrt[3]{3.395} \approx 1.5\) --- ## **Final Answer** ### **Required diameter of rod AB:** \[ \boxed{1.50\ \text{in.}} \] --- ## **Summary Table** | Quantity | Value | |-----------------------|---------------------------| | Torque, \(T\) | \(10,000\, \text{lb.in}\) | | Allowable Stress | \(15,000\, \text{psi}\) | | Required Diameter, \(d\) | \(\mathbf{1.50\, \text{in.}}\) | --- **Note:** Only the steel rod AB's allowable stress is used to size AB, since that's where the torque is applied.