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+// Problem no 10.1,Page No.249
+
+clc;clear;
+close;
+
+//Consider Equilibrium of joint A
+//As there are no Load applied at A members AC and AB have nothing to Balance 
+//So they are null members
+F_AB=0
+F_AC=0
+
+//Consider Equilibrium of joint B
+
+//Applying the summation of horizontal forces  we get
+F_DB=4*(cos(45*%pi*180**-1))**-1
+
+//Applying the summation of vertical forces  we get
+F_BC=F_DB*sin(45*%pi*180**-1)
+
+//Consider Equilibrium of joint B
+
+//Applying the summation of vertical forces  we get
+F_CE=4*(sin(45*%pi*180**-1))**-1
+
+//Applying the summation of horizontal forces  we get
+F_DC=F_CE*cos(45*%pi*180**-1)
+
+//Result
+printf("The Forces in Each members are as follows:F_AB = %.f kN",F_AB)
+printf("\n                                          :F_AC = %.f kN",F_AC)
+printf("\n                                          :F_DB %.2f",F_DB);printf(" KN(compression)")
+printf("\n                                          :F_BC %.2f",F_BC);printf(" KN(Tension)")
+printf("\n                                          :F_CE %.2f",F_CE);printf(" KN(Tension)")
+printf("\n                                          :F_DC %.2f",F_DC);printf(" KN (compression)" )