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+clear
+//
+
+//To determine reactions, consider equilibrium equations
+
+//variable declaration
+//all Vertical loading are in KN
+PL1=200.0
+PL2=200.0
+PL3=150.0
+PL4=100.0
+PL5=100.0
+
+//length in m
+UL1=6.0
+UL2=8.0
+UL3=9.0
+UL4=UL2
+UL5=UL1
+
+L1=6.0
+L2=6.0
+L3=6.0
+L4=6.0
+L5=6.0
+L6=6.0
+
+//moment at point LO
+
+R2=(PL1*L1+PL2*(L1+L2)+PL3*(L1+L2+L3)+PL4*(L1+L2+L3+L4)+PL5*(L1+L2+L3+L4+L5))/(L1+L2+L3+L4+L5+L6)
+
+R1=PL1+PL2+PL3+PL4+PL5-R2
+
+//Take the section (1)–(1) and consider the right hand side part.
+
+U3U4=sqrt((1**2)+(UL1**2))
+theta1=asin(1/U3U4)
+
+L3U4=sqrt((UL1**2)+(UL2**2))
+theta2=asin(6/L3U4)
+
+//moment at U4
+
+FL3L4=(R2*(L5+L6)-PL4*L4)/UL4
+
+printf("\n FL3L4= %0.1f KN (Tension)",FL3L4)
+
+//moment at L3
+FU4U3=(-PL4*L4-PL5*(L4+L5)+R2*(L4+L5+L6))/(cos(theta1)*UL3)
+printf("\n FU4U3= %0.1f KN (Comp.)",FU4U3)
+
+//sum of horizontal forces
+FL4L3=FL3L4
+FU4L3=(-FL4L3+FU4U3*cos(theta1))/sin(theta2)
+printf("\n FU4L3= %0.1f KN (Tension)",FU4L3)