Remember: Every bridge, aircraft wing, and artificial hip joint owes its safety to the principles in Dowling’s book. Mastering those principles, with or without the manual, is your responsibility as an engineer. If you are a professor, consider making selected solutions available to your students. If you are a student, form a study group and share the cost of an official Chegg subscription or student manual. And always—always—double-check your units.
However, anyone who has navigated the complex chapters on stress concentrations, cyclic plasticity, or linear elastic fracture mechanics (LEFM) knows that the end-of-chapter problems are notoriously challenging. This is where the enters the spotlight. This companion guide is more than just an answer key—it is a pedagogical tool that decodes the intricate methodologies required to master the subject.
A large titanium alloy plate contains a center crack of length ( 2a = 20 ) mm. The plate is subjected to a tensile stress of 500 MPa perpendicular to the crack. Given ( K_{IC} = 55 ) MPa√m for the alloy, what is the safety factor against brittle fracture? Assume the finite width correction factor ( Y ) for a center crack in an infinite plate is 1.0 for simplicity.
Using ( K_I = \sigma \sqrt{\pi a} ) with ( a = 10 ) mm (half crack length). The student calculates ( K_I = 500 \sqrt{\pi \times 0.01} = 500 \times 0.177 = 88.5 ) MPa√m. That exceeds ( K_{IC} = 55 ), so the safety factor ( SF = 55/88.5 = 0.62 ). The student concludes the plate will fail, but the calculation is correct but misleading—it actually predicts failure, but is the safety factor defined correctly?
