Asked by Danny
What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 64000.000N.
ANS= Round to 3 decimal places
Where do I begin?
ANS= Round to 3 decimal places
Where do I begin?
Answers
Answered by
bobpursley
safetyfactor=designload/load
now, you have to convert MPa to Newtons, to do that you have to have the crosssectional area.
tensiondesign=550 MPa*areainm^2
now, you have to convert MPa to Newtons, to do that you have to have the crosssectional area.
tensiondesign=550 MPa*areainm^2
Answered by
Danny
Sorry i missed this part of the question:
The steel hanger in question has a cross sectional area of 7.600cm^2
The steel hanger in question has a cross sectional area of 7.600cm^2
Answered by
bobpursley
change that area to m^2, and solve.
Answered by
Danny
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?
Answered by
MathMate
See your other post.
Try to work with stress for comparison.
Try to work with stress for comparison.
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