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+// Problem no 10.6,Page No.258
+
+clc;clear;
+close;
+
+//Calculations
+
+//Taking moment of the Forces about the hinge A
+P=1000*2**0.5*1.2*(0.9)**-1
+
+//Let R_AH be the Horizontal component of the reaction at A
+R_AH=P-1000*2**0.5
+R_A=((R_AH)**2+(1000*2**0.5)**2)**0.5
+
+//Resolving the forces vertically we get
+R_AV=1000*2**0.5 //vertical component of the reaction at A
+
+//joint A
+
+//Resolving vertically we get
+F_BA=1000*2**0.5*(sin(30*%pi*180**-1))**-1
+
+//Resolving horizontally we get
+F_AD=2000*2**0.5*3**0.5*2**-1-1000*2**0.5*3**-1 //N
+
+//Joint C
+
+BD=1.2*sin(30*%pi*180**-1)
+BE=0.6*sin(30*%pi*180**-1)
+ED=0.6*cos(30*%pi*180**-1)
+CE=0.9-0.52
+
+theta=atan(BE*CE**-1)*(180*%pi**-1)
+
+F_CB=P*(sin(38.29*%pi*180**-1))**-1
+
+//Resolving vertically
+F_CD=F_CB*cos(theta*%pi*180**-1)
+
+//Joint D
+
+//Resolving horizontally
+F_DB=(F_AD-1000*2**0.5)*(cos(60*%pi*180**-1))**-1
+
+//Result
+printf("The Pull in chain is %.2f",P);printf(" N")
+printf("\n Force in the each members are as follows:F_BA %.2f",F_BA);printf(" KN(compressive)")
+printf("\n                                         :F_AD %.2f",F_AD);printf(" KN(Tensile)")
+printf("\n                                         :F_CB %.2f",F_CB);printf(" KN(compression)")
+printf("\n                                         :F_CD %.2f",F_CD);printf(" KN(Tensile)")
+printf("\n                                         :F_DB %.2f",F_DB);printf(" KN(compressive)")