Posted by **Anonymous** on Thursday, July 25, 2013 at 11:29am.

A steel block 200 mm X 20 mm x 20 mm is subjected to a tensile force of 40 kN in the direction of its length. Determine the change in volume, if E is 205 kN/mm2 and Poisson's ratio = 0.3.

- Solid Mechanics -
**Elena**, Thursday, July 25, 2013 at 3:45pm
E=205 kN/mm²= 205•10⁹ N/m²,

μ = 0.3

σ=E•ε(longitudinal)

F/A = E•ε(longitudinal)

ε=ε(longitudinal) = F/A•E=

=40000/(20•10⁻³)²•205•10⁹=4.9•10⁻⁴.

If a=0.2 m, b=c=0.002 m, ε=4.9•10⁻⁴,

V=abc = 0.2•0.002•0.002 =8•10⁻⁷ m³

The deformed volume is

V₁ =(1+ ε)a•(1- μ ε)b•(1- μ ε)c .

Neglecting powers of ε, the deformed volume

V₁ =(1+ ε - 2•μ•ε)V .

The change in volume is

ΔV = ε(1- 2•μ)V =

=4.9•10⁻⁴(1 - 2•0.3)•8•10⁻⁷ =

=1.57•10⁻¹º m³

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