Graduate Lame parameter mu = shear modulus derivation (rogue factor of 2)

[Twigg]

Hello,

I am trying and failing to derive that the shear modulus $G$ is equal to the Lame parameter $\mu$. I start with the linear, symmetric, isotropic stress-strain formula: $$\sigma = \lambda \mathrm{tr}(\epsilon) \mathrm{I} + 2\mu \epsilon$$ I then substitute a simple (symmetric) shear strain: $$\epsilon = \epsilon_{xy} (\hat{x} \otimes \hat{y} + \hat{y} \otimes \hat{x} )$$ But then I end up with $$\sigma_{xy} = 2\mu \epsilon_{xy}$$ or equivalently $$G = 2\mu$$ What did I goof up?

Thanks!

 

[Frabjous]

μ=G. Your Hooke‘s law equation is correct. The true shear strain equals one-half the engineering shear strain. They always mess me up.
https://en.wikipedia.org/wiki/Infinitesimal_strain_theory#Infinitesimal_strain_tensor
https://en.wikipedia.org/wiki/Hooke's_law#Isotropic_materials
You need the ½ so that you have a tensor.