The Modulus of Elasticity (E) provides a relationship between: 

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  1. shear stress and shear strain
  2. shear stress and normal stress
  3. normal strain and shear strain
  4. normal stress and normal strain

Answer (Detailed Solution Below)

Option 4 : normal stress and normal strain
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Concept:

Elastic Modulus (E)

When the body is loaded within its elastic limit, the ratio of normal stress and normal strain is constant. This constant is known as Elastic modulus or Young's Modulus.

\({\rm{E}} = \frac{{{\rm{Normal\:Stress}}}}{{{\rm{Normal\:Strain}}}} = \frac{{\rm{\sigma }}}{\epsilon}\)

Additional Information

Rigidity modulus (G)

When a body is loaded within its elastic limit, the ratio of shear stress and shear strain is constant, this constant is known as the shear modulus.

\({\rm{G}} = \frac{{{\rm{Shear\;stress\;}}}}{{{\rm{Shear\;strain}}}} = \frac{{\rm{\tau }}}{\phi }\)

Bulk modulus (K)

When a body is subjected to three mutually perpendicular like stresses of same intensity then the ratio of direct stress and the volumetric strain of the body is known as bulk modulus

\({\rm{K}} = \frac{{{\rm{Direct\;stress}}}}{{{\rm{Volumetric\;strain}}}} = \frac{{\rm{\sigma }}}{{\frac{{{\rm{\delta V}}}}{{\rm{V}}}}}\)

The relationship between E, K, G, and μ is:

\(E = 2G(1 + μ) \)

\(G ={ E\over 2(1 + μ) }\)

\(E = 3K (1 – 2μ) \)

\({\bf{E}} = \frac{{9{\bf{KG}}}}{{3{\bf{K}}\ + \;{\bf{G}}}} \)

\({\bf{\mu }} = \frac{{3{\bf{K}}\ -\ 2{\bf{G}}}}{{2{\bf{G}}\ +\ 6{\bf{K}}}}\)

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