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
= 70000/87500
= 0.8
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
= 17500
= 70000/ 17500
ANS = 4
Is that correct?
Correct!
Yes, you have correctly calculated the factor of safety. However, there is a mistake in the formula you mentioned. The correct formula to calculate the factor of safety is:
Factor of safety = Ultimate strength / Applied load
In this case, the ultimate strength is given as 70000 N/cm² and the applied load is 87500 N. So, the factor of safety will be:
Factor of safety = 70000 N/cm² / 87500 N
= 0.8
Therefore, the factor of safety for the steel hanger is 0.8.