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+//find stress after putting the system in lathe

+clc

+//soluton

+//given

+Ds=18//mm//diameter of steel

+Dc1=24//mm//inner  diameter of copper rod initially

+Dc2=40//mm//outer diametr of copper

+Fs=10//N/mm^2//stress in steel rod

+pi=3.14

+As=(pi*Ds^2)/4//mm^2//area of steel rod

+Ac=(pi*(Dc2^2-Dc1^2))/4//mm^2//area of copper rod

+//since tensile load on steel rod is equal to compressive load on copper rod,therefore

+//Fs*As=Fc*Ac,therefore

+Fc=Fs*As/Ac//stress in copper rod//N/mm^2

+//when copper rod is reduced outside diametr changes to 40-1.5*2=37mm,therefore new area is

+Ac1=(pi*(37^2-24^2))//mm^2

+//cross section of other half remain same//if Ac2 is the area of remainder then Ac2=Ac

+//let Fc1 be stress in reduced section,Fc2 be stress in remainder ,Fs1 stress in rod aftre turning

+//since load on copper tube is equal to load on steel tube, therefore Ac1*Fc1=Ac2*Fc2=As*Fs1

+//from above equations Fc1=0.41*Fs1,Fc2=0.32*Fs1

+//let L be the length of steela nd copper  rod ,since total change in length is equal to change inlength of rduced section before and aftr turning adn change in length of remainder section beofre and aftre turning

+//dl=dl1+dl2

+//(Fs-Fs1)*L/Es=(Fc1-Fc)*L/(2*Ec)+(Fc2-Fc)*(L)/(2*Ec)

+//given Es=2Ec

+//10-Fs1=0.41*Fs1-3.16+0.32*Fs1-3.16

+Fs1=(10+3.16+3.16)/(1+0.41+0.32)

+printf("the stress in the rod is,%f N/mm^2",Fs1)
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