Homework Help: mechanical properties of materials

Posted by dmkp on Monday, January 28, 2013 at 12:34pm.

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 - Elena, Monday, January 28, 2013 at 12:49pm
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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