1 files changed, 34 insertions, 0 deletions
diff --git a/1205/CH6/EX6.6/S_6_6.sce b/1205/CH6/EX6.6/S_6_6.sce
new file mode 100644
index 000000000..8adfcdbd3
--- /dev/null
+++ b/1205/CH6/EX6.6/S_6_6.sce
@@ -0,0 +1,34 @@
+clc;
+P=3;//kN, Horizontal Force applied at A
+AB=1;//m, perpendicular distance between A and B
+BD=1;//m, perpendicular distance between D and B
+CD=1;//m, perpendicular distance between C and D
+FC=1;//m, perpendicular distance between C and F
+EF=2.4;//m, perpendicular distance between E and F
+//Entire frame
+//Applying sum(M_E)=0
+Fy=P*(AB+BD+CD+FC)/EF;//kN, Y component of reaction at F
+
+
+//Applying sum(Fy)=0
+Ey=-Fy;//kN, Y component of  reaction at E
+
+//Free body member ACE
+//Applying sum(Fy)=0, and sum(M_E)=0 we get 2 equation
+A=[-AB/sqrt(AB^2+EF^2),CD/sqrt(CD^2+EF^2);-EF/sqrt(AB^2+EF^2)*(AB+BD+CD+FC),-EF/sqrt(CD^2+EF^2)];// Matrix of coefficients
+B=[Ey;-P*(AB+BD+CD+FC)];// Matrix B
+X=linsolve(A,B);//kN Solution matrix
+F_AB=X(1);//kN, Forec inmember AB
+F_CD=X(2);//kN, Forec inmember CD
+Ex=-P-EF/sqrt(AB^2+EF^2)*F_AB-EF/sqrt(CD^2+EF^2)*F_CD;//kN, X component of force exerted at E
+//Free body : Entire frame
+//Applying sum(F_X)=0
+Fx=-P-Ex;//kN, X component of force exetered at F
+printf("Components of force exerted at F is Fx=%.1f kN and Fy=%.0f kN \n",Fx,Fy);
+printf("Force in member AB is F_AB=%.1f kN \n",F_AB);
+printf("Force in member CD is F_CD=%.1f kN \n",F_CD);
+printf("Components of force exerted at E is Ex=%.1f kN and Ey=%.1f kN \n",Ex,Ey);
+
+printf("Negative signs shows forces are in negative direction\n")
+
+