What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 79000.000N. The steel hanger in question has a cross sectional area of 7.000cm^2.
ANS = (Round to 3 decimal places)
ultimate strength = 550.000MPa
load = 79000.000N
area = 7.000cm^2 = .07m
Not too sure how to start. A little help would be greatly appreciated. Thanks.
To find the factor of safety, we need to compare the ultimate strength of the steel hanger to the actual load it is supporting. The factor of safety is a measure of how much stronger the material is compared to the actual load it is subjected to.
First, let's convert the cross-sectional area from cm^2 to square meters. Since 1 cm^2 is equal to 0.0001 square meters, the cross-sectional area is 0.07 square meters (0.0001 x 7.000).
Now, we can calculate the stress on the hanger by dividing the load by the cross-sectional area:
Stress = Load / Area
Stress = 79000.000 N / 0.07 m^2
To find the factor of safety, divide the ultimate strength by the stress:
Factor of Safety = Ultimate Strength / Stress
Factor of Safety = 550.000 MPa / (79000.000 N / 0.07 m^2)
Now, we need to convert MPa to N/m^2 (Pascal) to have the same units for both the ultimate strength and stress. Since 1 MPa is equal to 1,000,000 N/m^2, the ultimate strength is 550,000,000 N/m^2 (550.000 x 1,000,000).
Factor of Safety = 550,000,000 N/m^2 / (79000.000 N / 0.07 m^2)
After performing the calculation, the factor of safety comes out to approximately 985.003 (rounding to 3 decimal places).