What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 64000.000N.
The steel hanger in question has a cross sectional area of 7.600cm^2
ANS= Round to 3 decimal places
Safety factor = design load/ load
= 550.000MPa = 550000000Pa
= 550000000/ 64000
= 8593.75
tension design = 550MPa * 0.076M^2
= 18.000
Have I messed thing up?Please let me know if I am on the right track. Thanks.
If I were to do the calculations, I would always work with stress, because ultimate stress is constant from problem to problem if the same material is used. It helps to spot errors, if any.
Ultimate strength = 550 mPa.
Stress under load = 64000/(7.6*10^-4)
=84.2 mPa
Now find the factor of safety from
F=ultimate strength / stress
Note: The conversion factor from cm² to m²l is 10^-4, as explained in
http://www.jiskha.com/display.cgi?id=1372041297
You are on the right track, but there are a few mistakes in your calculations.
To determine the factor of safety, you should use the formula:
Factor of safety = Ultimate strength / Design load
Ultimate strength = 550,000 MPa = 550,000,000 Pa (since 1 MPa = 1,000,000 Pa)
Design load = 64,000 N
Now, let's calculate the factor of safety:
Factor of safety = 550,000,000 Pa / 64,000 N
Converting the units, we have:
Factor of safety = (550,000,000 / 1,000,000) Pa / 64,000 N
= 550 Pa / 64,000 N
= 8.594
So, the factor of safety is 8.594 (rounded to 3 decimal places).
To calculate the tension design, you need to multiply the ultimate strength (in Pa) by the cross-sectional area (in m²):
Tension design = Ultimate strength * Cross-sectional area
Here, cross-sectional area = 7.600 cm² = 7.600 × 10⁻⁴ m² (since 1 cm² = 0.0001 m²)
Now, let's calculate the tension design:
Tension design = 550,000,000 Pa * 7.600 × 10⁻⁴ m²
= 418,000 N
So, the tension design is 418,000 N.
Therefore, please make the necessary changes in your calculations.