solid mechanics
posted by sushil on .
A steel block 200mm*20mm*20mm is subjected to a tensile force of 40kn in the direction of its length.Determine the change in volume,if Eis 205kn/mm square and poisson's ratio=0.3.

E=205 kN/mm²= 205•10⁹ N/m².
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,
μ = 0.3
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 
Unit volume expansion, e
=εx+εy+εz
=(12ν)(σx+σy+σz)/E
Using
σx=40*10^3 N
σy=σz=0
E=205 GPa
ν=0.3
Volume change
=0.2*0.02*0.02*e
=0.0008e