Stress on 25 micron Fiber Under Elongation Load

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SUMMARY

The discussion centers on calculating the stress on a 25-micron fiber subjected to a 25 g elongation load. The initial calculation of stress using the formula F/A resulted in 500 MPa, which was incorrect. The correct stress value is 0.125 MPa, achieved by using the full diameter instead of the radius in the area calculation. This highlights the importance of precise measurements in stress calculations.

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  • Understanding of tensile stress calculations
  • Familiarity with basic physics concepts such as force and area
  • Knowledge of unit conversions, particularly between grams and newtons
  • Experience with geometric calculations for circular cross-sections
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  • Review the principles of tensile stress and its calculations
  • Learn about the significance of diameter versus radius in area calculations
  • Explore unit conversion techniques, especially for force measurements
  • Investigate common errors in stress analysis and how to avoid them
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This discussion is beneficial for materials scientists, mechanical engineers, and students studying mechanics who are looking to deepen their understanding of stress analysis in fibers and materials under load.

benndamann33
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This doesn't seem particularly hard, thought it was simply an F/A situation but apparently not. A fiber with a diameter of 25 micro meters is subjected to elongation load of 25 g along the fiber axis. What is the stress on the fiber? Is the applied stress a shear or a tensiles stress. I said it was a tensil stress and went on to compute F/A. F was .025Kg * 9.8N and A was simply ((25*10^-6)/2)^2*pi. The resultant stress was 500 MPa. The correct answer is .125 MPa. Any suggestions??!
 
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I get the same answer as you. Maybe the text answer is wrong?
 
yeah I'm going with text is wrong, .125 is found by not dividing the radius by 2 and then dividing by a factor of 1000...haha only thing I could find out, but thank you!
 

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