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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)
+\ No newline at end of file