solid mechanics

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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 -

    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 -

    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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