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+// Problem no 14.5,Page No.330
+
+clc;clear;
+close;
+H=30 //m //height of Dam
+a=2 //m //top width
+b=5 //m //bottom width
+
+h=29 //m //height of water
+rho_w=9810 //N/m**3 
+rho_mason=22560 //N/m**3
+sigma1=0 //KN/m**3
+sigma2=880 //KN/m**3
+
+//Calculations
+
+//Let L=1 m (length of dam)
+L=1
+
+//weight of dam
+//W=(a+b)*2**-1*L*H*rho_mason*10**-3
+//After substituting values and Further simplifying we get
+//W=338.4*(a+b) //equation1
+
+//Pressure at B=0, Sinc etension at base has just been avoided
+
+//Eccentricity
+e=b*6**-1  //as sigma1=0
+
+//Pressure at C
+//sigma2=W2*b**-1*(1+6*e*b**-1)
+//After substituting values and Further simplifying we get
+//W=440*b
+
+//From equation1,440*b=338*(a+b)
+//b=3.33*a
+
+//the distance of C.Gof dam
+//x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+//After substituting values and Further simplifying we get
+//x_bar=1.187*a
+
+//distance of pt where resultant cuts the base
+//x=P*W2**-1*H*3**-1
+//After substituting values and Further simplifying we get
+//x=27.214*a**-1
+
+//Now x_bar+x=2*3**-1*b
+//After substituting values and Further simplifying we get
+a=(27.17*(2.22-1.187)**-1)**0.5
+b=3.33*a
+
+//Result
+printf("The top width dam is %.2f",a);printf(" m")
+printf("\n The bottom width dam is %.2f",b);printf(" m")
+
+
+//Plotting the Shear Force Diagram
+X1=[0,L,L]
+Y1=[sigma2,sigma1,0]
+Z1=[0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")