mechanical properties of materials

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A cylindrical steel bar is required to support a load of 650 N. the required bar must be 0.17 m in length and must not deform by more than 0.015 mm. given that the Young's modulus of steel is 210 GPa, calculate the minumum diameter of the bar.

  • mechanical properties of materials -

    σ=Eε
    σ=F/A=4F/πd²
    ε= ΔL/L
    4F/πd²=ΔL/L
    d=sqrt{4FL/πEΔL} =
    =sqrt{4•650•0.17/π•210•10⁹•0.015•10⁻³}=
    =6.68•10⁻³ m

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