Elasticity
14.0 Relation Between $Y$, $B$, $\eta$ and $\sigma$
14.0 Relation Between $Y$, $B$, $\eta$ and $\sigma$
1. $$B = \frac{Y}{{3\left( {1 - 2\sigma } \right)}}$$
Proof: Consider a cube of unit length. Let F be the longitudinal extensive force on each face of the cube and is acting outwards as shown in the Fig. E. 28.
Initial volume of cube = $1uni{t^3}$
We know, $$\begin{equation} \begin{aligned} Y = \frac{{Stress}}{{Strain}} \\ Strain = \frac{{Stress}}{Y} \\\end{aligned} \end{equation} $$$$\frac{{\Delta l}}{l} = \frac{{\frac{F}{A}}}{Y}$$$$\Delta l = \frac{F}{{AY}}\left( {l = 1unit,{\text{as cube is of unit length}}} \right)$$
Therefore,
Extension of along $x$ axis is $${\frac{F}{{AY}}}$$
Compression along $x$ axis due to extension of $y$ axis is $$ \frac{{\sigma F}}{{AY}}$$
Compression along $x$ axis due to extension of $z$ axis is $$ \frac{{\sigma F}}{{AY}}$$
Therefore, Total change in length along $x$ axis is
$$\begin{equation} \begin{aligned} \Delta {l_x} = \frac{F}{{AY}} - \frac{{2\sigma F}}{{AY}} \\ \Delta {l_x} = \frac{F}{{AY}}\left( {1 - 2\sigma } \right) \\ {\left( {{L_f}} \right)_x} = 1 + \frac{F}{{AY}}\left( {1 - 2\sigma } \right) \\\end{aligned} \end{equation} $$
Similarly, $$\begin{equation} \begin{aligned} {\left( {{L_f}} \right)_y} = 1 + \frac{F}{{AY}}\left( {1 - 2\sigma } \right) \\ {\left( {{L_f}} \right)_z} = 1 + \frac{F}{{AY}}\left( {1 - 2\sigma } \right) \\\end{aligned} \end{equation} $$
Final volume of the cube is $${V_f} = {\left[ {1 + \frac{F}{{AY}}\left( {1 - 2\sigma } \right)} \right]^3}$$
Since $Y>>F$ therefore, $${V_f} = 1 + 3\frac{F}{{AY}}\left( {1 - 2\sigma } \right)$$
As, $$\begin{equation} \begin{aligned} \Delta V = {V_f} - {V_i} \\ \Delta V = 1 + 3\frac{F}{{AY}}\left( {1 - 2\sigma } \right) - 1\left( {{V_i} = 1,{\text{as cube is a unit cube}}} \right) \\ \Delta V = \frac{{3P}}{Y}\left( {1 - 2\sigma } \right)\left( {AsP = \frac{F}{A}} \right) \\\end{aligned} \end{equation} $$
As we know that, $$\begin{equation} \begin{aligned} B = \frac{P}{{\frac{{\Delta V}}{{{V_i}}}}} \\ B = \frac{P}{{\frac{{\frac{{3P}}{Y}\left( {1 - 2\sigma } \right)}}{1}}} \\ B = \frac{Y}{{3\left( {1 - 2\sigma } \right)}} \\\end{aligned} \end{equation} $$
2. $$\eta = \frac{Y}{{2\left( {1 + \sigma } \right)}}$$
Proof: Consider a cube of unit length. Let the cube be subjected to a tensile (longitudinal) force $F$ on faces perpendicular to the $x$ axis and an equal compressive force $F$ is applied on faces perpendicular to the $y$ axis.
Therefore,
Extension along $x$ axis is $$\frac{F}{{AY}}$$
Extension along $x$ axis due to compression along $y$ axis is $$\frac{{\sigma F}}{{AY}}$$
Total change in length along $x$ axis is $$\frac{F}{{AY}}\left( {1 + \sigma } \right)$$
Tensile strain here is the ratio of extension along $x$ axis to the original length of the cube.
$$\begin{equation} \begin{aligned} {\text{Tensile Strain}} = \frac{{{\text{Extension along x axis}}}}{{{\text{Original length of the cube}}}} = \frac{{\frac{F}{{AY}}\left( {1 + \sigma } \right)}}{1} \\ {\text{Tensile Strain}} = \frac{F}{{AY}}\left( {1 + \sigma } \right) \\\end{aligned} \end{equation} $$
Compression along $y$ axis is $$\frac{F}{{AY}}$$
Compression along $y$ axis due to extension along $x$ axis is $$\frac{{\sigma F}}{{AY}}$$
Total change in length along $y$ axis is $$\frac{F}{{AY}}\left( {1 + \sigma } \right)$$
Compressive strain here is the ratio of compression along $y$ axis to the original length of the cube.
$$\begin{equation} \begin{aligned} {\text{Compressive Strain}} = \frac{{{\text{Compression along y axis}}}}{{{\text{Original length of the cube}}}} = \frac{{\frac{F}{{AY}}\left( {1 + \sigma } \right)}}{1} \\ {\text{Compressive Strain}} = \frac{F}{{AY}}\left( {1 + \sigma } \right) \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} {\text{Shear Strain}}\left( \theta \right) = {\text{Tensile strain + Compressive Strain}} \\ \theta = \frac{F}{{AY}}\left( {1 + \sigma } \right) + \frac{F}{{AY}}\left( {1 + \sigma } \right) \\ \theta = \frac{{2F}}{{AY}}\left( {1 + \sigma } \right) \\\end{aligned} \end{equation} $$
Now as we know,
$$\begin{equation} \begin{aligned} {\text{Modulus of rigidity}}\left( \eta \right) = \frac{{{\text{Shearing Stress}}}}{{{\text{Shearing Strain}}}} \\ \eta = \frac{{\frac{F}{A}}}{{\frac{{2F\left( {1 + \sigma } \right)}}{{AY}}}} \\ \eta = \frac{Y}{{2\left( {1 + \sigma } \right)}} \\\end{aligned} \end{equation} $$
