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CSIR-NET 2020 JUNE

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Q) Assume that the poisons solid is subjected to a longitudinal stress while the transverse stress are zero. Then the young’s modules ( E ) in terms of modules of rigidity (µ)will be___

(Thanks to Chandrasekhar, ANU)

 Solutions:

 From the relation

\mu =\frac{E}{2(1+\nu)}

\mu =\frac{E}{2(1+0.25)}   for poissons solid v =0.25

\mu =\frac{E}{2(1.25)}

\mu =\frac{E}{2.5}

E= 2.5\times\mu

Reference : Fundamental of Geophysics, William Lowrie.

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