1 files changed, 68 insertions, 0 deletions
diff --git a/3862/CH3/EX3.4/Ex3_4.sce b/3862/CH3/EX3.4/Ex3_4.sce
new file mode 100644
index 000000000..6f799932b
--- /dev/null
+++ b/3862/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,68 @@
+clear
+//
+
+//Now, we cannot find a joint with only two unknown forces without finding reactions.
+//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero.
+
+//variable declaration
+
+PB=30.0      //vertical load at point B,KN
+PC=50.0      //vertical load at point C,KN 
+PDv=40.0     //vertical load at point D,KN
+PDh=20.0     //Horizontal load at point D,KN
+PF=30.0      //vertical load at point F,KN
+HA=PDh
+
+RE=(PC*4+PDv*8+PDh*4+PF*4)/(8.0)
+
+VA=PB+PC+PDv+PF-RE
+
+//joint A
+//sum of vertical & sum of horizontal forces is zero.
+
+FAB=VA
+FAF=HA
+
+//joint E
+//sum of vertical & sum of horizontal forces is zero.
+
+FED=RE
+FEF=0
+
+//Joint B: Noting that inclined member is at 45°
+//sum of vertical & sum of horizontal forces is zero.
+
+theta=45.0
+FBF=(VA-PB)/sin(theta*%pi/180)
+
+printf("\n FBF= %0.4f KN (Tension) ",FBF)
+
+FBC=FBF*cos(theta*%pi/180)
+
+printf("\n FBC= %0.4f KN (Comp.) ",FBC)
+
+//Joint C: 
+//sum of vertical & sum of horizontal forces is zero.
+
+
+FCF=PC
+
+printf("\n FCF= %0.4f KN (Comp.) ",FCF)
+
+FCD=FBC
+
+printf("\n FCD= %0.4f KN (Comp.) ",FCD)
+
+//Joint D: Noting that inclined member is at 45°
+//sum of vertical & sum of horizontal forces is zero.
+
+theta=45.0
+FDF=(RE-PDv)/cos(theta*%pi/180)
+
+printf("\n FDF= %0.4f KN (Tensile) ",FDF)
+
+//check
+
+FDF=(FCD+PDh)/cos(theta*%pi/180)
+
+printf("\n FDF= %0.4f KN Checked ",FDF)