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.

Answer 1

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