How to find a value for poisson's ratio

[shortshanks]

Homework Statement

The question is: A rod of 50mm in diameter and 0.8m in length is subjected to an axial load of 150kn. If the rod entends by 1.4mm and there is a decrease in diameter of 0.01mm, determine: The value of poissons ratio?

Homework Equations

v=lateral strain/longitudinal strain

e= change/original

The Attempt at a Solution

My attempt is: e(lat)=0.01/0.050
=0.2
e(long)= change in length/original length
= 0.014/0.8
= 1.75x 10-3

 

[p21bass]

It appears that you didn't convert delta_d from 0.01mm to 0.00001m.

 

[vela]

And you converted 1.4 mm to meters incorrectly as well.

 

[shortshanks]

how about: 1.4mm/800mm = 1.75x10-3 (0.00175) (longitudinal)
-0.1mm/0.50mm = -2x10-3 (0.002) (lateral)

lateral/longitudinal= -0.002/0.00175 = -1.1428

 

[vela]

Is the decrease in diameter 0.1 mm or 0.01 mm?

 

[shortshanks]

oops! 0.01mm.
-0.01mm/0.050mm = -0.02mm (lateral)
1.4mm/800 = 1.75x10-3 (0.00175) (longitudinal)

lateral/longitudinal = -0.02/0.00175 = -11.428

Im not sure if this is right though as -11.428 does not tallie up with any ratio for any material for poissons ratio. What do you think?

 

[shortshanks]

Can anyone let me know if I am heading in the right direction with my answer please?

 

[Mapes]

Check your lateral strain calculation again, particularly the units, and note that Poisson's ratio is defined as the negative of the ratio.

 

[shortshanks]

0.001/0.050 = 0.02 (lateral)
1.4/800 = 1.75x10-3 (0.0075) (longitudinal)

0.02/0.0075 = 2.666
I think I finally have it, can someone confirm please?
If I do then thank you to everyone who helped!

 

[Mapes]

Check your lateral strain calculation again, particularly the units.