630.000MPa = 630000kN ??????
Factor of safety = ultimate strength/allowable strength=
= ultimate strength x Area/ Load
Load = ultimate strength x Area/ Factor of safety=
=630•10⁶•π•0.03²/4•5=8.9064•10⁴ N=
=89.064 kN
The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm?
ANS = kN (Round to 3 decimal places)
Safety = 5.000
Rod diameter = 3.000cm = 0.03m
Ultimate Strength
= 630.000MPa = 630000kN
Allowable strength
= Ultimate strength/ Factor of safety
= 630000kN/ 0.03m
= 21000000kN
Load = Allowable strength * Area
Area = pi * d^2/ 4
= 21000000kN * 3.1416 * 3^2/ 4
= 21000000kN * 3.1416 * 9/ 4
= 593762400kN/ 4
ANS = 148440.600kN
Please check. Thank you.
3 answers
Wouldn't you use Allowable Strength and not Ultimate Strength?
Factor of Safety = Ultimate Strength/ Allowable Strength
Allowable Strength = Ultimate Strength/ Factor of Safety
Allowable Strength = Load/ Area=
Load= Allowable Strength * Area=
= Allowable Strength * (pi*d^2/4)
Factor of Safety = Ultimate Strength/ Allowable Strength
Allowable Strength = Ultimate Strength/ Factor of Safety
Allowable Strength = Load/ Area=
Load= Allowable Strength * Area=
= Allowable Strength * (pi*d^2/4)
Factor of safety = Ultimate strength/Allowable strength …(1)
Allowable strength =Load/Area ….(2)
Substitute (2) in (1) =>
Factor of safety =
=Ultimate strength x Area/ Load =>
Load = ultimate strength x Area/ Factor of safety=
=630•10⁶•π•0.03²/4•5=8.9064•10⁴ N=
=89.064 kN
Allowable strength =Load/Area ….(2)
Substitute (2) in (1) =>
Factor of safety =
=Ultimate strength x Area/ Load =>
Load = ultimate strength x Area/ Factor of safety=
=630•10⁶•π•0.03²/4•5=8.9064•10⁴ N=
=89.064 kN
0 answers