Calculating Dimensions of Titanium Disk Under Load

[WhiteWolf98]

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?

 

[Chestermiller]

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]

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]

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]

Longitudinal means in the direction of the applied stress. "Along the force".

I see! That clears things up, thank you! And Chestermiller too lol