Posted by Danny on .
What is the factory of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 61000.000N. The steel hanger in question has a cross sectional area of 6.500cm^2.
ANS = (Round to 3 decimal places)
6.500cm = 0.065m
Load = 61000.000N
Ultimate strength
= 630.000MPa = 630000000N
Factor of safety
= Ultimate stress/ allowable stress
= 630000000/61000
ANS = 10327.868
Is this done correctly?

Physics (Stress & Strain) 
MathMate,
When you see a factor of safety over 10, it would be a good idea to check your calculations.
Watch out for units conversions for areas and volumes.
In converting 6.5 cm² to m², you must multiply by the (linear factor)², namely 0.01².
The area is therefore 6.5*10^4.
Using this area, the actual stress is 93.85 mPa.
You have divided the ultimate stress of 630 mPa by the load of 61000N.
This is why we always write units in our calculations.
The factor of safety should be given by 630 mPa / 93.85 mPa.