Stress on 25 micron Fiber Under Elongation Load

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The discussion focuses on calculating the stress on a 25-micron fiber under a 25 g elongation load. The initial calculation suggested a tensile stress of 500 MPa, but the correct answer is stated to be 0.125 MPa. Participants debate the method of calculating the area, with one suggesting that the text answer may be incorrect due to a misunderstanding of the radius. The consensus leans towards the idea that the calculation should not involve dividing the radius by 2 and requires a factor adjustment. The conversation highlights the importance of accurate area calculations in stress analysis.
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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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