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+// Problem 11.11.6,Page no.278
+
+clc;clear;
+close;
+
+//d=Diameter of rod
+P=500 //KN
+e=0.75 //cm //eccentricity
+
+//calculation
+
+//A=%pi*d**2*4**-1 //cm**2 //Area of rod
+//sigma_d=P*A**-1 //KN/cm**2 //stress due to direct load
+
+//After substituting value and simplifying we get,
+//sigma_d=2000*(%pi*d**2)**-1 //KN/cm**2 
+
+M=P*e //Kn*cm //Moment
+
+//Z=%pi*d**3*32**-1 //cm**3 //section modulus
+//sigma_b=M*Z**-1 //KN/cm**2 //Stress due to moment
+
+//After substituting value and simplifying we get,
+//sigma_b=12000*(%pi*d**3)**-1 //KN/cm**2
+
+//Max stress 
+//sigma=sigma_d+sigma_b 
+
+//After substituting value and simplifying we get,
+//2000*(%pi*d**2)**-1+12000*(%pi*d**3)**-1=12.5
+
+//After simplifying we get,
+//d**3-53.05*d-318.3=0
+
+//From Synthetic Division we get d**2+4.73*d-42.918
+a=1
+b=-4.73
+c=-42.918
+
+X=b**2-(4*a*c)
+
+d_1=(-b+X**0.5)*(2*a)**-1
+d_2=(-b-X**0.5)*(2*a)**-1
+
+//Result
+printf("The minimum diameter of the rod is %.2f",d_1);printf(" cm")