Calculate factor of safely on a steel bar

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SUMMARY

The discussion focuses on calculating the factor of safety for a steel bar subjected to a load of 600N. The calculations involve determining stress, strain, and elongation (∆l) using the given diameter of 10cm, length of 10m, Young's modulus (E) of 2256 GPa, and an elastic limit of 518 MPa. The calculated stress is 76.4 kPa, strain is 3.386×10^-8, and the factor of safety is determined to be 6780.62, which is attributed to the relatively small load acting over a large area.

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Engineering students, materials scientists, and professionals involved in structural analysis and design will benefit from this discussion, particularly those focused on stress analysis and safety factors in engineering applications.

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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