Respuesta :
Answer:
Diameter of the cylinder will be [tex]d=2.998\times 10^4m[/tex]
Explanation:
We have given young's modulus of steel [tex]E=207GPa=207\times 10^9Pa[/tex]
Change in length [tex]\Delta l=0.38mm[/tex]
Length of rod [tex]l=500mm[/tex]
Load F = 11100 KN
Strain is given by [tex]strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}[/tex]
We know that young's modulus [tex]E=\frac{stress}{strain}[/tex]
So [tex]207\times 10^9=\frac{stress}{7.6\times 10^{-4}}[/tex]
[tex]stress=1573.2\times 10^{-5}N/m^2[/tex]
We know that stress [tex]=\frac{force}{artea }[/tex]
So [tex]1573.2\times 10^{-5}=\frac{11100\times 1000}{area}[/tex]
[tex]area=7.055\times 10^{8}m^2[/tex]
So [tex]\frac{\pi }{4}d^2=7.055\times 10^{8}[/tex]
[tex]d=2.998\times 10^4m[/tex]
