Asked by Danny
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?
Factor of safety =
ultimate stress/ allowable stress
= 70000/87500
= 0.8
Have I done this correctly?
Answers
Answered by
MathMate
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
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
Answered by
Danny
Stress = force/ area
= 87500N/ 5cm^2
= 17500
= 70000/ 17500
ANS = 4
Is that correct?
= 87500N/ 5cm^2
= 17500
= 70000/ 17500
ANS = 4
Is that correct?
Answered by
MathMate
Correct!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.