Asked by dmkp
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.
Answers
Answered by
Elena
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
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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