Calculating Poisson ratio is a way to measure a material's response to stress.

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The discussion revolves around calculating the Poisson ratio for a cylindrical metal specimen under tensile stress. The original poster attempts to compute the ratio using the formula V = (change in D/D original)/(change in L/L original), substituting stress and elastic modulus values. Discrepancies arise as they arrive at a Poisson ratio of 0.21, while others suggest the correct answer should be closer to 0.27 or 0.33. Participants highlight potential errors in calculations or rounding, emphasizing the importance of accuracy in intermediate steps. The conversation concludes with a request for clarification on the calculations to resolve the confusion.
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Homework Statement


question

A ciclindrical metal specimen 10 mm in diameter is stressed elastically in tension. A force of 15000 N produces a reduction in diameter of 0.007 mm. Compute Poisson ratio if its elastic modulus is 100 GPa



Homework Equations



E=stress/strain




The Attempt at a Solution



my attempt

D original = 0.01m
change in D = 0.000007m
F=15000N
E= 100*10^9

V = passonse ratio = (change in D/D original)/(change in L/ L original)

we can replace (change in L/ L original) by (stress/E)

so

V=(change in D/D original)/(stress/E)

(change in D/D original) = (0.000007/0.01)
(stress/E) = (F/Area)/E
A= pi(0.01/2)^2

that gives me an answer of V = 0.21



but the tutorial sheet says the answer is 0.33

can you explain where i have gone wrong
 
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Your method looks fine to me, though I got ν=0.27.
 
vela said:
Your method looks fine to me, though I got ν=0.27.



elongation D = (4*F*V)/(pi*Doriginal*E)

is this a valid eqation??
 
buxZED said:
elongation D = (4*F*V)/(pi*Doriginal*E)

is this a valid eqation??
That might look more familiar if you divide both sides by D and rearrange it slightly:

\frac{\Delta D}{D} = \frac{4 F \nu}{\pi D^2 E} = \frac{F}{\pi(D/2)^2 E}\nu = \frac{F}{AE}\nu

where A = \pi(D/2)^2 is the cross-sectional area.
 
vela said:
That might look more familiar if you divide both sides by D and rearrange it slightly:

\frac{\Delta D}{D} = \frac{4 F \nu}{\pi D^2 E} = \frac{F}{\pi(D/2)^2 E}\nu = \frac{F}{AE}\nu

where A = \pi(D/2)^2 is the cross-sectional area.
why dose using this equation gives me a different answer for passoinse ratio?
 
It shouldn't. I suspect you're just making other errors while calculating or rounding off too much on intermediate steps. After all, I did the exact same calculations you did in your original post and got 0.27 while you got 0.21.
 
vela said:
It shouldn't. I suspect you're just making other errors while calculating or rounding off too much on intermediate steps. After all, I did the exact same calculations you did in your original post and got 0.27 while you got 0.21.

i don't get it an I am confused
can u do a quick run true and tell me they are same?
 
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