What is the factor of safety for a steel hanger having an ultimate strength of 70000N per cm^2 and supporting a load of 87500N. The hanger has a cross sectional area of 5cm^2.
Factor of safety =
ultimate stress/ allowable stress
Have I done this correctly?
Stress = force / area
Load is 87500N which is just a force.
You need to divide the force by the area resisting the tensile stress.
Factor of safety is then
ultimate stress / stress under load
Stress = force/ area
= 87500N/ 5cm^2
= 70000/ 17500
ANS = 4
Is that correct?posted by Danny