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 for a steel hanger, you need to compare the ultimate strength of the material to the load it is supporting.

First, let's convert the cross-sectional area from centimeters to meters:
Area = 0.053 cm² = 0.053 x 10^-4 m² = 5.3 x 10^-6 m²

Now we can calculate the factor of safety using the formula:
Factor of safety = ultimate strength / allowable strength

To find the allowable strength, we need to determine the maximum load the hanger can support without failure. We can use the formula:
Allowable strength = Load / Area

Substituting the given values:
Allowable strength = 65000 N / 5.3 x 10^-6 m² = 1.22641509434 x 10^10 Pa

Now we can calculate the factor of safety:
Factor of safety = 630000000 Pa / (1.22641509434 x 10^10 Pa) ≈ 0.0513847696

Rounding this to three decimal places, the factor of safety is approximately 0.051. Therefore, the provided answer of 5.136 is incorrect.