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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 strainstress relationship in a threedimensional 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.
 Summer '10
 Staff

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