Difficulty in proof of relationship between $Y$, $B$, $\sigma$ On internet, I was looking for the proof regarding the following relationship between $Y$ young's modulus,$\sigma$ poisson's ratio, and $B$ bulk modulus: $Y=3B(1-2\sigma)$, and I came across the following proof, which I have difficulty in understanding it.

Consider a cube subjected to three equal stresses S on the three faces
The total strain in one direction or along one face due to the application of
volumetric stress S is given as
$$\text{linear Strain} = S/Y – \sigma S/Y –\sigma S/Y = S/Y( 1-2\sigma)$$
So $\text{linear Strain} =(S/E)( 1-2r)$. Now $\text{Volumetric Strain} = 3 \cdot \text{Linear Strain}$.
So $\text{Volumetric Strain} = 3 (S/Y)( 1-2\sigma)$
Now by definition of bulk modulus of elasticity K. We have $K= \frac{\text{volumetric Stress}} {\text{volumetric Strain}}$.
$K= \frac{S}{{(3(S/Y)( 1-2\sigma)}}$ , So $Y=3K(1-2\sigma)$

I am having main doubt regarding linear strain equation, especially why $–\sigma S/Y$ is being added two times. Rest of the proof is understandable to me. So please help me there. Moreover I want to ask is there, any simpler proof for this equation?
Also, while finding/researching, I can't come up with proof for another similar equation: $Y=2\eta(1+\sigma)$, where $\eta$ is shear modulus. So I want help here too.
 A: Hre is a general set of relations:
Start from the isotropic stress-strain relation
$$
\sigma_{ij} = \lambda \delta_{ij} e_{kk} + 2\mu e_{ij}
$$
in terms of the Lame constants $\lambda$, $\mu$.
By considering particular
deformations, we   can express the more directly
measurable bulk modulus, shear modulus,
Young's modulus and
Poisson's ratio in terms of $\lambda$ and $\mu$.
The bulk modulus bulk modulus $\kappa$  is defined by
$$
dP =-\kappa \frac {dV}{V}, 
$$
where  an infinitesimal isotropic external pressure  $dP$ causes a
change $V\to V+dV$ in the volume of the  material.
This applied pressure
corresponds to a   surface stress of
$\sigma_{ij}=-\delta_{ij}\,dP$.
An isotropic expansion away from the origin displaces points $x_i\to x_i+\eta_i$ in the material so
that
$$
\eta_i =\frac 13 \frac {dV}{V}x_i.
$$
The strains
$$
e_{ij}= \frac 12 (\partial_i\eta_j+\partial_j \eta_i)
$$
are therefore given by
$$
e_{ij} = \frac 13 \delta_{ij} \frac {dV}{V}.
$$
Inserting  this  strain into the stress-strain relation gives
$$
\sigma_{ij}= \delta_{ij}(\lambda +\frac 23 \mu)\frac
{dV}{V}= - \delta_{ij} dP.  
$$
Thus
$$
\kappa= \lambda+\frac 23 \mu.
$$
To define  the  shear modulus, we assume a deformation   $\eta_1 =
\theta x_2$, so  $e_{12}=e_{21}=\theta/2$, with all other
$e_{ij}$ vanishing.
The
applied shear stress is $\sigma_{12}=\sigma_{21}$. The
shear modulus,  is defined to be $\sigma_{12}/\theta$. Inserting the strain components   into
the stress-strain relation gives
$$
\sigma_{12} =\mu\theta,
$$
and so the shear modulus is equal to the Lame constant  $\mu$.
We can therefore write the generalized Hooke's law as
$$
\sigma_{ij} = 2\mu(e_{ij} -{\textstyle{\frac 13}} \delta_{ij} e_{kk})
+\kappa e_{kk} \delta_{ij},
$$
which reveals  that  the shear modulus  is associated with the traceless
part of the strain tensor, and the bulk modulus with the
trace.
Young's modulus $Y$ is measured  by stretching a wire of
initial length
$L$ and square cross section of side $W$ under a
tension $T=\sigma_{33}W^2$.
We   define $Y$ so that
$$
\sigma_{33} = Y\frac {dL}{L}.
$$
At the same time as the wire stretches, its
width changes $W\to W+dW$.  Poisson's ratio $\sigma$ is defined by
$$
\frac{dW}{W}=-\sigma \frac{dL}{L},
$$
so that $\sigma$ is positive if the wire gets thinner as it
gets longer. The displacements are
$$
\eta_3 = z \left(\frac {dL}{L}\right),\nonumber\\
 \eta_{1}=x 
\left(\frac{dW}{W}\right)  = - \sigma x
\left(\frac{dL}{L}\right),\nonumber\\
 \eta_{2} =  y
\left(\frac{dW}{W}\right)= - \sigma y
\left(\frac{dL}{L}\right),
$$
so the
strain components  are
$$
e_{33} = \frac {dL}{L},\quad e_{11}=e_{22} =
\frac{dW}{W}=-\sigma e_{33}.
$$
We therefore have
$$
 \sigma_{33} = (\lambda(1-2\sigma) +2\mu)\left(\frac
{dL}{L}\right),
$$
leading to
$$
Y= \lambda(1-2\sigma)+2\mu.
$$
Now, the side of the wire is a free surface with no forces
acting on it, so
$$
0=\sigma_{22}=\sigma_{11} = (\lambda(1-2\sigma) -2\sigma \mu)\left(\frac
{dL}{L}\right).
$$
This tells us that
$$
\sigma =\frac 12 \frac{\lambda}{\lambda+\mu},
$$
and
$$
Y=
\mu\left(\frac{3\lambda+2\mu}{\lambda+\mu}\right).
$$
The desiired  relations
$$
Y= 3\kappa(1-2\sigma),\nonumber\\
= 2\mu(1+\sigma),
$$
now follow by simple algebra from  those above.
