Posted by Danny on .
What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 79000.000N. The steel hanger in question has a cross sectional area of 7.000cm^2.
ANS = (Round to 3 decimal places)
ultimate strength = 550.000MPa
load = 79000.000N
area = 7.000cm^2 = .07m
Not too sure how to start. A little help would be greatly appreciated. Thanks.

Physics (Stress & Strain) 
Elena,
Allowable strength=Load/Area,
Factor of safety =
=ultimate strength/allowable strength=
= ultimate strength x Area/ Load=
=550•10⁶•7•10⁻⁴/79000=4.873