"Look at Equation 3-6," Dr. Vance pointed. Leo read aloud:

"Material spec says yield shear strength is 60 MPa," Leo said. "We're below yield. So why did it fail?" "Because you didn't check the angle of twist ," Dr. Vance said. "Turn to Equation 3-15."

This story aligns with problems (e.g., 3-1 to 3-42) where students compute shear stress, angle of twist, and design shaft diameters for power transmission.

Setting: Engineering Lab, Coast Guard Inspection Yard. 2:00 AM.

Leo solved: [ d = \sqrt[3]\frac16T\pi \tau_allow ] [ d = \sqrt[3]\frac16(4000)\pi (24\times10^6) = 0.094 \text m \approx 94 \text mm ]

[ \tau_max = \fracTcJ ]