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

In summary, the homework statement says that 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.
  • #1
buxZED
12
0

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
 
Physics news on Phys.org
  • #2
Your method looks fine to me, though I got ν=0.27.
 
  • #3
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??
 
  • #4
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:

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

where [itex]A = \pi(D/2)^2[/itex] is the cross-sectional area.
 
  • #5
vela said:
That might look more familiar if you divide both sides by D and rearrange it slightly:

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

where [itex]A = \pi(D/2)^2[/itex] is the cross-sectional area.
why dose using this equation gives me a different answer for passoinse ratio?
 
  • #6
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.
 
  • #7
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?
 

1. What is the Poisson ratio?

The Poisson ratio is a measure of the deformation of a material in response to an applied force. It is defined as the ratio of lateral strain to axial strain, meaning how much a material contracts in the direction perpendicular to the applied force compared to how much it elongates in the direction of the force.

2. How do you calculate the Poisson ratio?

The Poisson ratio can be calculated by dividing the lateral strain by the axial strain. It can also be determined experimentally by measuring the change in thickness and length of a material when a force is applied.

3. What is a typical range for the Poisson ratio?

The Poisson ratio can range from -1 to 0.5, with most materials falling between 0 and 0.5. Materials with a negative Poisson ratio, also known as auxetic materials, exhibit unusual properties such as expanding in the direction perpendicular to an applied force.

4. How does the Poisson ratio affect the behavior of a material?

The Poisson ratio affects the stiffness, strength, and stability of a material. Materials with a higher Poisson ratio are typically more flexible and can withstand larger deformations, while materials with a lower Poisson ratio are stiffer and less prone to buckling under stress.

5. Can the Poisson ratio change under different conditions?

Yes, the Poisson ratio can change depending on the temperature, pressure, and stress applied to a material. For example, some materials may exhibit a negative Poisson ratio at low temperatures and a positive one at high temperatures. The Poisson ratio can also vary depending on the direction of the applied force or the microstructure of the material.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Mechanical Engineering
Replies
5
Views
930
  • Engineering and Comp Sci Homework Help
Replies
3
Views
735
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Other Physics Topics
Replies
2
Views
2K
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
8K
Replies
1
Views
2K
  • Mechanical Engineering
Replies
2
Views
882
  • Introductory Physics Homework Help
Replies
21
Views
6K
Back
Top