Calculate factor of safely on a steel bar

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Girn261

Homework Statement



Calculate stress, strain, ∆l, factor of safety

Diameter=10cm
Length=10m
Load=600N
E=2256Gpa
Elastic limit=518Mpa

Homework Equations

The Attempt at a Solution


Area=.7854x(.10^2)

Stress=load/area , 600N/.007854m^2 = 76.4Kpa

E = stress/strain , 2256x10^9Pa = 76394.19Pa/strain , strain = 3.386×10^-8

Strain=∆I/L , 3.386×10^-8 = ∆I/10m , ∆I = 3.386x10^-7

Factor of safety = elastic limit/Max working stress
518x10^6Pa/76394.19Pa = 6780.62

Why am I getting some a huge number for factor of safety? Can't figure out what I'm doing wrong
 
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I don't know if you are doing anything wrong, it's just that 600 N is a very small load acting over a relatively large area.
 
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