HomeWork3 - Home Work #3 Using the generalized Hooke’s...

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Unformatted text preview: Home Work #3 Using the generalized Hooke’s law, prove that the maximum value of Poisson’s ratio, µ , cannot exceed 0.5. Solution The strain-stress relationship in a three-dimensional stressed body is given by 1 σ x − µ ( σ y + σ z ) E 1 ε y = σ y − µ ( σ x + σ z ) E 1 ε z = σ z − µ ( σ x + σ y ) E εx = Summing these trains, one establishes the relationship between the volume expansion, ε , and the sum of normal stresses: ε = εx + ε y + εz Θ = σx +σ y +σz 1 − 2µ ε= Θ E In the case of a uniform hydrostatic pressure p, one has 1 σx =σy =σz = −p = − Θ 3 Hence, 3 ( 1 − 2µ ) p which represents the relation between unit volume expansion ε and E hydrostatic pressure p. As no material would ever expand under hydrostatic pressure, the maximum value of Poisson’s ratio cannot exceed 0.5. ε =− The quantity E is called the modulus of volume expansion or bulk modulus. 3 ( 1 − 2µ ) 1 ...
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This note was uploaded on 09/24/2011 for the course CIVL 7690 taught by Professor Staff during the Summer '10 term at Auburn University.

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