Wednesday
March 29, 2017

Post a New Question

Posted by on .

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 - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question