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+// Problem no 14.2,Page No.327
+
+clc;clear;
+close;
+D=2 //m //External Diameter
+d=1.5 //m //Internal Diameter
+P=1600 //N/m**2 //N/m**2 //Wind Pressure
+W=19200 //N/m**2 //Weight of masonry
+
+//Calculations
+
+//Let H be max height of dam
+
+//W2=%pi*4**-1*(D**2-d**2)*H*W //weight of chimney
+//W2=26400*H
+
+//Eccentricrty
+x=(D**2+d**2)*(8*D)**-1
+
+//P2=H*D*P //Lateral thrust of wind on chimney
+//P2=3200*H
+
+//Now by using the relation we get P*W**-1=x*(H*2**-1)**-1
+//After substituting values and Further simplifying we get
+H=0.39*2*26400*3200**-1
+
+//result
+printf("The Height of Dam is %.2f",H);printf(" m")