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+clear
+//
+
+//Each load is 10 kN and all triangles are equilateral with sides 4 m.
+
+//variable declaration
+
+PB=10.0
+PD=PB
+PF=PD
+AB=4.0
+BC=AB
+AC=BC
+BD=BC
+CD=BC
+DE=CD
+CE=CD
+DF=DE
+EF=DE
+EG=DE
+FG=EF
+//Take section (A)–(A), which cuts the members FH, GH and GI and separates the truss into two parts.
+AG=AC+CE+EG
+BG=CE+EG+AC/2
+DG=EG+CE/2
+FG1=EG/2
+RA=PB*7/2
+RO=RA
+theta=60.0*%pi/180
+//moment at point G
+FFH=(RA*AG-PB*BG-PD*DG-PF*FG1)/(FG*sin(theta))
+printf("\n FFH= %0.4f KN (Comp.)",FFH)
+
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FGH=(RA-PB-PD-PF)/(sin(theta))
+printf("\n FGH= %0.4f KN (Comp.)",FGH)
+
+FGI=FFH+FGH*cos(theta)
+printf("\n FGI= %0.4f KN (Tensile)",FGI)