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+// Problem no 10.5,Page No.256
+
+clc;clear;
+close;
+
+BC=6 //m
+
+//Calculations
+
+AB=2*BC*(3**0.5)**-1
+
+//Taking moment about B we get
+R_A=-(-2000*3-1000*6)*(12*(3**0.5)**-1)**-1 //reaction at the roller support A
+
+//The resultant of all the three Loads is 4000 N acting at right angle to BC at D
+
+//Resolving it vertically we have
+V=4000*sin(60*%pi*180**-1)
+
+//Resolving it horizontal we have
+H=4000*cos(60*%pi*180**-1)
+
+//Applying the summation of vertical forces  we get
+R_B_v=V-R_A
+
+//Applying the summation of horizontal  forces  we get
+R_B_h=H
+R_B=((R_B_v)**2+(R_B_h)**2)**0.5
+
+tan_theta=R_B_v*R_B_h**-1
+
+//Joint B
+
+//Applying the summation of vertical forces  we get
+F_BD=1000*(3**0.5)*2
+
+//Applying the summation of horizontal  forces  we get
+F_BE=R_B_h+F_BD*cos(30*%pi*180**-1)
+
+//Joint D
+F_DE=2000 //N
+F_CD=F_BD
+
+//Consider equilibrium of truss to the Left of section 2-2
+F_CE=R_A*AB*(sin(30*%pi*180**-1)*6)**-1
+
+//Joint A
+
+//Applying the summation of vertical forces  we get
+F_AC=R_A*(sin(60*%pi*180**-1))**-1
+
+//Applying the summation of horizontal  forces  we get
+F_AE=F_AC*cos(60*%pi*180**-1)
+
+//Result
+printf("Forces in Each members are as follows:F_BD %.2f",F_BD);printf(" KN(compression)")
+printf("\n                                      :F_BE %.2f",F_BE);printf(" KN(Tension)")
+printf("\n                                      :F_DE %.2f",F_DE);printf(" KN(compression)")
+printf("\n                                      :F_CD %.2f",F_CD);printf(" KN(compression)")
+printf("\n                                      :F_CE %.2f",F_CE);printf(" KN(Tension)")
+printf("\n                                      :F_AC %.2f",F_AC);printf(" KN(compression)")
+printf("\n                                      :F_AE %.2f",F_AE);printf(" KN(Tension)")