Question

The three-dimensional stress-strain relationship for an isotropic material is given as \[ \begin{Bmatrix} \sigma_{xx} \sigma_{yy} \sigma_{zz} \tau_{yz} \tau_{xz} \tau_{xy} \end{Bmatrix} = \begin{bmatrix} P & Q & Q & 0 & 0 & 0 \\ Q & P & Q & 0 & 0 & 0 \\ Q & Q & P & 0 & 0 & 0 \\ 0 & 0 & 0 & R & 0 & 0 \\ 0 & 0 & 0 & 0 & R & 0 \\ 0 & 0 & 0 & 0 & 0 & R \end{bmatrix} \begin{Bmatrix} \varepsilon_{xx} \varepsilon_{yy} \varepsilon_{zz} \gamma_{yz} \gamma_{xz} \gamma_{xy} \end{Bmatrix}, \] 

where \(P, Q, R\) are elastic constants; \(\sigma,\tau\) are normal and shear stresses; and \(\varepsilon,\gamma\) are normal and engineering shear strains. Which relation is correct?

• A. \(\displaystyle R=\frac{P-Q}{2}\)
• B. \(\displaystyle R=\frac{Q-P}{2}\)
• C. \(\displaystyle Q=\frac{P-R}{2}\)
• D. \(\displaystyle Q=\frac{R-P}{2}\)

Hint:

For isotropic elasticity with engineering shear, diagonal normal terms map to \(2G+\lambda\) (self) and \(\lambda\) (coupling), while shear terms map directly to \(G\). Set \(R=G\) and back-substitute.

Answer 1

The Correct Option is A

Step 1: Recall isotropic Hooke's law in tensor form.
For small strains, \[ \boldsymbol{\sigma} = \lambda\,\text{tr}(\boldsymbol{\varepsilon})\,\mathbf{I} + 2G\,\boldsymbol{\varepsilon}, \] where \(\lambda\) is Lamé's first parameter and \(G\) is the shear modulus.

Step 2: Match the matrix entries for normal stresses.
From the above, \[ \sigma_{xx}=(2G+\lambda)\varepsilon_{xx}+\lambda\varepsilon_{yy}+\lambda\varepsilon_{zz}. \] Comparing with the given matrix row \(\ \sigma_{xx}=P\varepsilon_{xx}+Q\varepsilon_{yy}+Q\varepsilon_{zz}\), we identify \[ P=2G+\lambda,\qquad Q=\lambda. \] 

Step 3: Match the shear part (engineering shear).
With engineering shear strain \(\gamma_{xy}=2\varepsilon_{xy}\), isotropic Hooke's law gives \(\tau_{xy}=G\,\gamma_{xy}\). Comparing with the given matrix, \(\tau_{xy}=R\,\gamma_{xy}\Rightarrow R=G\). 

Step 4: Eliminate \(\lambda,G\) to relate \(P,Q,R\).
Using \(Q=\lambda\) and \(R=G\) in \(P=2G+\lambda\): \[ P=2R+Q \ \Rightarrow\ R=\frac{P-Q}{2}. \] \(\boxed{R=\dfrac{P-Q}{2}}\)