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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.

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1 Answer 1

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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.

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