Physics
posted by Danny on .
What is the factor of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 65000.000N. The steel hanger in question has a cross sectional area of 5.300cm.
ANS = (Round to 3 decimal places)
Load = 65000.000N
ultimate strength = 630.000MPa = 600000000Pa
Area = 5.300cm = 0.053m
Factor of safety
= ultimate strength/ Allowable strength
= ultimate strength * Area/ Load
= 630000000 * 0.053/ 65000.000
= 33390000/ 65000.000
= 513.692
ANS = 5.136
Please check. Thank you.

cross sectional area has to be measured in cm² or m² =>
A= 5.300cm² = 5.3•10⁻⁴ m² = 0.00053 m²
Factor of safety = ultimate strength/allowable strength=
= ultimate strength x Area/ Load =
=630•10⁶•5.3•10⁻⁴/65000 =5.130