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+//to find 1.)diameter of the rods,2.)extension in each rod in the length of 2.5m

+clc

+//solution

+//given

+P=3.5*10^6//N//load applied

+f1=85//(N/mm^2)// safe stress

+E=210*10^3//(N/mm^2)//young's modulus

+l=2.5*10^3//mm

+pi=3.14

+//1)diameter of rod(d)

+//let d be diameter of rods in mm

+//since both rods carries equal load ,therefore load on single rod is

+P1=P/2//N

+d=sqrt(4*P1/(f1*pi))//using f1=P/A//mm

+printf("the diameter of rods is,%f mm\n",d)

+//2)extension in rod

+//let x be extension in rod

+//E=(P1*l)/(A*x)

+ //P1/A=f1

+x=(f1*l)/E

+printf("the extension of rod is,%f mm",x)
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