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+// Problem no 10.2,Page No.250
+
+clc;clear;
+close;
+
+//Taking moment at Pt A we get
+R_B=100*8*4**-1
+
+//Applying the summation of vertical forces  we get
+R_AV=-R_B
+
+//Applying the summation of horizontal forces  we get
+R_H=100
+
+//joint B
+
+//Applying the summation of vertical forces  we get
+F_CB=R_B
+
+//Applying the summation of horizontal forces  we get
+F_AB=0 //As there is no force to balance in horizontal direction
+
+//joint A
+
+//Applying the summation of horizontal forces  we get
+F_AC=R_H*(cos(45*%pi*180**-1))**-1
+
+//Applying the summation of vertical forces  we get
+F_AD=200-F_AC*sin(45*%pi*180**-1)
+
+//joint C
+
+//Applying the summation of vertical forces  we get
+F_EC=200-F_AC*cos(45*%pi*180**-1)
+
+//Applying the summation of horizontal  forces  we get
+F_DC=F_AC*cos(45*%pi*180**-1)
+
+//joint D
+
+//Applying the summation of horizontal  forces  we get
+F_DE=F_DC*(cos(45*%pi*180**-1))**-1
+
+//DF and EF are null members at this joint as each member individually has nothing to balance
+F_DF=0
+F_EF=0
+
+//Result
+printf("The Forces in Each members are as follows:F_AB = %.1f kN",F_AB)
+printf("\n                                          :F_CB = %.1f kN (compressive)",F_CB)
+printf("\n                                          :F_AC %.2f",F_AC);printf(" KN(Tensile)")
+printf("\n                                          :F_AD=%.1f kN (Tensile)",F_AD)
+printf("\n                                          :F_EC=%.1f kN N(Compressive)",F_EC)
+printf("\n                                          :F_DC=%.1f kN N(Compressive)",F_DC)
+printf("\n                                          :F_DE %.2f",F_DE);printf(" KN(Tensile)")
+printf("\n                                          :F_DF = %.f kN",F_DF)
+printf("\n                                          :F_EF = %.f kN",F_EF)