1 files changed, 99 insertions, 0 deletions
diff --git a/3862/CH3/EX3.7/Ex3_7.sce b/3862/CH3/EX3.7/Ex3_7.sce
new file mode 100644
index 000000000..f3534f490
--- /dev/null
+++ b/3862/CH3/EX3.7/Ex3_7.sce
@@ -0,0 +1,99 @@
+clear
+//
+
+// Since all members are 3 m long, all triangles are equilateral and hence all inclined members are at 60° to horizontal. Joint-by-joint analysis is carried out . Then nature of the force is determined. 
+
+//variable declaration
+
+AB=3.0
+BC=AB
+AC=AB
+BD=BC
+CD=BD
+CE=CD
+DE=CE
+EF=DE
+DF=DE
+EG=DE
+FG=DF
+
+theta=60.0*%pi/180 //angles BAC,BCA,DCE,DEC,FEG,FGE,°
+
+PB=40.0    //Vertical Loading at point B,KN
+PD=30.0    //Vertical Loading at point D,KN
+HF=10.0    //Horizontal Loading at point F,KN
+PG=20.0    //Vertical Loading at point G,KN
+
+//joint G
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FGF=PG/sin(theta)
+
+printf("\n FGF= %0.4f KN (Tension)",FGF)
+
+FGE=FGF*cos(theta)
+
+printf("\n FGE= %0.4f KN (Comp.)",FGE)
+
+//joint F
+
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FFG=FGF
+
+printf("\n FFG= %0.4f KN (Comp.)",FFG)
+
+FFE=FGF
+FFD=FGF*cos(theta)+FFE*cos(theta)-HF
+printf("\n FFD= %0.4f KN (Tension)",FFD)
+
+//Now, without finding reaction we cannot proceed. Hence, consider equilibrium of the entire truss
+//moment about point A
+
+RE=((PB*AC/2)-(HF*EF*sin(theta))+(PD*(AC+CE/2))+(PG*(AC+CE+EG)))/(AC+CE)
+
+VA=PB+PD+PG-RE
+
+HA=HF
+
+//joint A
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FAB=VA/sin(theta)
+
+printf("\n FAB= %0.4f KN (Comp.)",FAB)
+
+FAC=FAB*cos(theta)-HF
+
+printf("\n FAC= %0.4f KN (Tension)",FAC)
+
+
+//joint B
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FBC=(PB-FAB*sin(theta))/sin(theta)
+
+printf("\n FBC= %0.4f KN (Comp.)",FBC)
+
+FBA=FAB
+FBD=-FBC*cos(theta)+FBA*cos(theta)
+
+printf("\n FBD= %0.4f KN (Comp.)",FBD)
+
+//joint C
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FCD=FBC*sin(theta)/sin(theta)
+
+printf("\n FCD= %0.4f KN (Tension)",FCD)
+
+FCE=FCD*cos(theta)+FBC*cos(theta)-FAC
+
+printf("\n FCE= %0.4f KN (Comp.)",FCE)
+
+//joint D
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FDE=(PD+FCD*sin(theta))/sin(theta)
+
+printf("\n FDE= %0.4f KN (Comp.)",FDE)