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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
commit7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch)
treedbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /1205/CH6/EX6.6/S_6_6.sce
parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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initial commit / add all books
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+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")
+
+