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.
To calculate the factor of safety, you need to divide the ultimate strength of the steel hanger by the allowable strength. The allowable strength is determined by dividing the load on the hanger by its cross-sectional area.
Given:
Load = 65000.000N
Ultimate strength = 630.000MPa = 630000000Pa
Area = 5.300cm = 0.053m
To calculate the allowable strength:
Allowable strength = Load / Area
Allowable strength = 65000.000N / 0.053m
Allowable strength = 1226415.094Pa
Now, we can calculate the factor of safety:
Factor of safety = Ultimate strength / Allowable strength
Factor of safety = 630000000Pa / 1226415.094Pa
Factor of safety ≈ 514.071
Rounding to 3 decimal places, the factor of safety is approximately 514.071.