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diff --git a/3886/CH4/EX4.5/Ex4_5.sce b/3886/CH4/EX4.5/Ex4_5.sce
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+//Analyse truss
+//Refer fig. 4.12 (a)
+//All triangles are equilateral 
+//applying equilibrium conditions At
+//Joint G
+FGF=20/sind(60)  //kN  (Tension)
+FGE=FGF*cosd(60)  //kN  (Compression)
+//Joint F
+FFE=FGF  //kN  (Compression)
+FFD=FGF*cosd(60)+FFE*cosd(60)-10  //kN  (Tension)
+//Consider eqiulibrium of entire truss
+RE=(-10*3*sind(60)+40*3*cosd(60)+30*(3+3*cosd(60))+20*9)/6  //kN
+VA=(40+30+20)-58.17  //kN
+HA=10  //kN
+//Joint A
+FAB=31.83/sind(60)  //kN  (Compression)
+FAC=36.75*cosd(60)-10  //kN  (Tension)
+//Joint B
+FBC=(40-FAB*sind(60))/sind(60)  //kN  (Compression)
+FBD=36.75*cosd(60)-9.44*cosd(60)  //kN  (Compression)
+//Joint C
+FCD=FBC  //kN  (Tension)
+FCE=9.44*cosd(60)+9.44*cosd(60)-8.38  //kN  (Compression)
+//Joint D
+FDE=(30+FCD*sind(60))/sind(60)  //kN  (Compression)
+printf("\nRequired Forces are:-\nFGF=%.2f kN  (Tension)\nFGE=%.2f kN  (Compression)\nFFE=%.2f kN  (Compression)\nFFD=%.2f kN  (Tension)\nRE=%.2f kN\nVA=%.2f kN\nHA=%.2f kN\nFAB=%.2f kN  (Compression)\nFAC=%.2f  (Tension)\nFBC=%.2f kN  (Compression)\nFBD=%.2f kN  (Compression)\nFCD=%.2f kN  (Tension)\nFCE=%.2f kN  (Compression)\nFDE=%.2f kN  (Compression)",FGF,FGE,FFE,FFD,RE,VA,HA,FAB,FAC,FBC,FBD,FCD,FCE,FDE)
+
+