Posted by **Danny** on Sunday, June 23, 2013 at 10:34pm.

What is the factory of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 61000.000N. The steel hanger in question has a cross sectional area of 6.500cm^2.

ANS = (Round to 3 decimal places)

6.500cm = 0.065m

Load = 61000.000N

Ultimate strength

= 630.000MPa = 630000000N

Factor of safety

= Ultimate stress/ allowable stress

= 630000000/61000

ANS = 10327.868

Is this done correctly?

- Physics (Stress & Strain) -
**MathMate**, Monday, June 24, 2013 at 6:14am
When you see a factor of safety over 10, it would be a good idea to check your calculations.

Watch out for units conversions for areas and volumes.

In converting 6.5 cm² to m², you must multiply by the (linear factor)², namely 0.01².

The area is therefore 6.5*10^-4.

Using this area, the actual stress is 93.85 mPa.

You have divided the ultimate stress of 630 mPa by the load of 61000N.

This is why we always write units in our calculations.

The factor of safety should be given by 630 mPa / 93.85 mPa.

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