Posted by **sushil** on Thursday, July 25, 2013 at 12:48pm.

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.

- solid mechanics -
**Elena**, Thursday, July 25, 2013 at 4:05pm
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

- solid mechanics -
**MathMate**, Thursday, July 25, 2013 at 4:08pm
Unit volume expansion, e

=εx+εy+εz

=(1-2ν)(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

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