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DoITPoMS Teaching & Learning Packages Introduction to Mechanical Properties of Materials Relation between Young's modulus and shear modulus
Introduction to Mechanical Properties of Materials AimsBefore you startIntroductionTensile test: force and extensionStress, strain and Poisson's ratioElastic deformation: Hooke's law and stiffnessViscoelasticityPlastic deformation: strength and ductilitySlip: resolved shear stress and Schmid factor, Taylor factorHardness and work hardeningCreepFracture: toughnessDuctile-brittle transition temperatureSummaryQuestionsGoing further
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Relation between Young's modulus and shear modulus

stress and strain on a cube

Consider a stress state above acting on a cube of length \( a \). Assume the shear strain \( \gamma \) is small such that the angle between the diagonal and length is approximately the same, i.e. 45°, before and after shearing.

Strain on the diagonal is

\[ \frac{{{\rm{\Delta }}d}}{{\sqrt 2 a}} = \frac{{\gamma a\;{\rm{cos}}45^\circ }}{{\sqrt 2 a}} = \frac{\gamma }{2} \]

From the definition of shear modulus,

\[ G = \frac{\tau }{\gamma } \]

Strain on the diagonal therefore is \( \frac{\tau }{{2G}} \).

Now consider a set of stress state equivalent to the above case.

stress state

By the principle of superposition of two stress pairs, strain on the diagonal is

\[ \frac{{{\sigma _1}}}{E} - \nu \frac{{{\sigma _2}}}{E} = \frac{\tau }{E} - \nu \frac{{ - \tau }}{E} = \frac{\tau }{E}\left( {1 + \nu } \right) \]

where \( E \) is Young’s modulus and \( \nu \) is Poisson’s ratio.

Equating the diagonal strain in two cases gives the relation between Young’s modulus E and shear modulus G:

\[ \frac{\tau }{{2G}} = \frac{\tau }{E}\left( {1 + \nu } \right) \]

\[ E = 2G\left( {1 + \nu } \right) \]

We obtained the relation between Young’s modulus \( E \) and shear modulus \( G \).

We can also relate the bulk modulus \( K \) to Young’s modulus \( E \). Consider a hydrostatic stress state acting on an isotropic cube shown below.

isotropic cube diagram

By the principle of superposition, the linear strain in each orthogonal direction is

\[ {\varepsilon_x} = {\varepsilon_y} = {\varepsilon_z} = \frac{{{\sigma _h}}}{E} - \nu \frac{{{\sigma _h}}}{E} - \nu \frac{{{\sigma _h}}}{E} = \frac{{{\sigma _h}}}{E}\left( {1 - 2\nu } \right) \]

The volumetric strain \( \Delta \) is

\[ \Delta = { \varepsilon_x} + {\varepsilon_y} + {\varepsilon_z}= 3\frac{{{\sigma _h}}}{E}\left( {1 - 2\nu } \right) \]

The proof of the first line can be found in here.

From the definition of bulk modulus,

\[ K = \frac{{{\sigma _h}}}{{\rm{\Delta }}} \]

And equating the volumetric strain \( \Delta \), gives the relation between Young’s modulus E and bulk modulus K:

\[ \frac{{{\sigma _h}}}{K} = 3\frac{{{\sigma _h}}}{E}\left( {1 - 2\nu } \right) \]

\[ E = 3K\left( {1 - 2\nu } \right) \]

Equating the two expressions for Young’s modulus, \( E \) yields a relationship between \( G \) and \( K \),

\[ 2G\left( {1 + \nu } \right) = 3K\left( {1 - 2\nu } \right) \]

Interestingly, for an isotropic material, if two of the four elastic constants, \( E \), \( G \), \( K \), \( \nu\) are known, the others can be calculated.

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