# Poisson's Ratio

## Homework Statement

A titanium disk (with ##E=107 ~GPa##, Poisson's ratio, ##v=0.34##) precisely ##l_0=8~mm## thick by ##d_0=30~mm## diameter is used as a cover plate in a mechanical loading device. If a ##P=20~kN## load is applied to the disk, calculate the resulting dimensions, ##l_{01}## and ##d_{01}##.

## Homework Equations

##v=- \frac {d\varepsilon_{trans}} {d\varepsilon_{axial}}=- \frac {d\varepsilon_{y}} {d\varepsilon_{x}}=- \frac {d\varepsilon_{z}} {d\varepsilon_{x}}##

##\sigma=\frac F A##

##\sigma=E\cdot \varepsilon##

## The Attempt at a Solution

I had a go working through the problem, but my final depth came out as ##7.999~mm## which seems really wrong. I'm confused which axis is which, or even which axis is being used. Which is trans and which is axial? Or is it ##z## and ##x## we're dealing with here?

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Chestermiller
Mentor

## Homework Statement

A titanium disk (with ##E=107 ~GPa##, Poisson's ratio, ##v=0.34##) precisely ##l_0=8~mm## thick by ##d_0=30~mm## diameter is used as a cover plate in a mechanical loading device. If a ##P=20~kN## load is applied to the disk, calculate the resulting dimensions, ##l_{01}## and ##d_{01}##.

## Homework Equations

##v=- \frac {d\varepsilon_{trans}} {d\varepsilon_{axial}}=- \frac {d\varepsilon_{y}} {d\varepsilon_{x}}=- \frac {d\varepsilon_{z}} {d\varepsilon_{x}}##

##\sigma=\frac F A##

##\sigma=E\cdot \varepsilon##

## The Attempt at a Solution

I had a go working through the problem, but my final depth came out as ##7.999~mm## which seems really wrong. I'm confused which axis is which, or even which axis is being used. Which is trans and which is axial? Or is it ##z## and ##x## we're dealing with here?
Les’s see your work. I get 7.998 mm.

• WhiteWolf98
Les’s see your work. I get 7.998 mm.
##\sigma = \frac F A = \frac {20×10^3} {\pi (15×10^{-3})^2} = 28.3×10^6 ~ Nm^{-2}##

##\sigma = E \cdot \varepsilon##

##\varepsilon = \frac \sigma E = \frac {28.3×10^6} {107×10^9}=2.64×10^{-4} ##

## v=- \frac {\varepsilon_{lat}} {\varepsilon_{long}}=0.34 → \varepsilon_{lat}=-(0.34)(2.64×10^{-4})=-8.99×10^{-5}##

##\varepsilon_{lat}=\frac {\Delta D} {D_0}##

##{\Delta D}=(-8.99×10^{-5})(8~mm)=-7.19×10^{-4}~mm##

##Hence ~D_f = 7.999 ~mm##

Truth be told, I watched someone do a similar problem, though it was with a cylinder for them. I just did the exact same as them, but used my numbers instead. I don't know why these equations are used or how to find the new length. If I was asked to do this again in an exam, I could probably do it, but I want to understand why

Chestermiller
Mentor
The way I interpreted this problem was that the load was compressional, and in the 8 mm thickness direction of the disk. So the strain in the 8 mm dimension was -0.000264, and the strain in the diameter direction was +0.0000899.

• WhiteWolf98
The main thing I don't understand is the formula: ##v=- \frac {\varepsilon_{lat}} {\varepsilon_{long}}##. How does one know which is latitudinal and longitudinal? Apologies if it sounds like a silly question. I don't know how to apply the formula

haruspex
Homework Helper
Gold Member
The main thing I don't understand is the formula: ##v=- \frac {\varepsilon_{lat}} {\varepsilon_{long}}##. How does one know which is latitudinal and longitudinal? Apologies if it sounds like a silly question. I don't know how to apply the formula
Longitudinal means in the direction of the applied stress. "Along the force".

• WhiteWolf98 and Chestermiller
Longitudinal means in the direction of the applied stress. "Along the force".
I see! That clears things up, thank you! And Chestermiller too lol