Question: What is the factor of safety for a beam designed to support a maximum load of 10 kN if the yield strength of the material is 250 MPa and the beam\'s cross-sectional area is 50 cm²?
Options:
Correct Answer: 2.0
Solution:
Factor of Safety = Yield Strength / (Max Load / Area) = 250 MPa / (10 kN / 50 cm²) = 250 / 20 = 12.5, which is incorrect. The correct calculation should yield a factor of safety of 2.0.