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+//Finding required forces
+//Refer fig. 4.16 (a)
+//Symmetry gives
+RE=(15+30+30+30+15)/2  //kN
+RA=RE  //kN
+FAE=30  //kN
+//After construction as shown in ref. fig
+//Taking moment about C
+FAE=(60*5-15*5-30*2.5)/5  //kN  (Tension)
+//assumptions are made as shown in fig. 4.16 (b)
+//Apply equilibrium conditions and solving equations
+FFC=15/0.366  //kN  (Tension)
+FBC=(0.866*40.98+15)/0.707  //kN  (Compression)
+//Lets analyse Joint B
+//Applying equilibrium conditions
+FBF=30*cosd(45)  //kN  (Compression)
+FAB=71.41+21.21  //kN  (Compression)
+//Lets analyse Joint A
+//Applying equilibrium conditions
+FAF=(92.62*sind(45)-45)/sind(30)  //kN  (Tension)
+printf("\nThe required forces are:-\nFAB=%.2d kN (Compression)\nFBC=%.2d kN (Compression)\nFBF=%.2d kN (Compression)\nFAF=%.2d kN (Tension)\nFFC=%.2d kN (Tension)\nFAE=%.2d kN (Tension)",FAB,FBC,FBF,FAF,FFC,FAE)