7th Edition Chapter 3 Solutions - Mechanics Of Materials

Dr. Vance tossed him a well-worn copy of Mechanics of Materials, 7th Edition . "Open to Chapter 3," she said. "We don't have time for a finite element simulation. We need to do this by hand, using the fundamental torsion formulas."

[ \phi = \fracTLJG ]

Dr. Vance closed the book. "Remember, Leo: Torque isn't just force times distance. It's stress times radius, integrated over area. Chapter 3 is about respecting that integration."

"(T) is torque, (c) is the outer radius, and (J) is the polar moment of inertia. For a solid circle, (J = \frac\pi32 d^4)." Mechanics Of Materials 7th Edition Chapter 3 Solutions

"New shaft diameter: 94 mm," Leo said. The replacement shaft—94 mm solid steel—was installed by 5:30 AM. As the sun rose over the SS Resilient , Leo looked at the Chapter 3 solutions in his textbook. They weren't just answers to odd-numbered problems. They were a map of how materials behave when twisted—elastically at first, then plastically, then fatally.

"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."

"2.4 degrees of twist over 2.5 meters is acceptable," Leo said. "We don't have time for a finite element simulation

"Exactly," said Dr. Vance. "The Resilient was overloaded by cyclic torque. Now go re-design the shaft diameter using Equation 3-9: (J = \pi d^4/32). Solve for (d) using (\tau_allow = 60/2.5 = 24) MPa."

[ \tau_max = \fracTcJ ]

[ \phi = \frac(4000)(2.5)(3.106\times10^-6)(77\times10^9) ] [ \phi = 0.0418 \text radians \approx 2.4 \text degrees ] "Remember, Leo: Torque isn't just force times distance

Where (G) is the shear modulus of elasticity (77 GPa for steel), and (L) is the length of the shaft (2.5 m).

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