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+// Problem no 9.5,Page no.235
+
+clc;clear;
+close;
+
+L=4 //m //Length of column
+W=250 //KN //Safe Load
+FOS=5 //Factor of safety
+//d=0.8*D //Internal diameter is 0.8 times Extarnal Diameter
+sigma_c=550 //MPa //Compressive stress
+alpha=1*1600**-1 //constant
+
+//Calculations
+
+P=W*FOS //N //Crippling Load
+
+//A=%pi*4**-1(D**2-d**2) //m**2 //Area of hollow cyclinder
+//After substituting value of d we get
+
+//A=%pi*0.09*D**2
+
+//I=%pi*64**-1*(D**4-d**4) //m**4 //Mo,ent Of Inertia
+//After substituting value of d we get d we get
+
+//I=0.009225*%pi*D**4
+
+//K=(I*A**-1)**0.5 //Radius of Gyration
+//After substituting value of I and A and further simplifying we get
+//K=0.32*D
+
+//Now using the Relation we get
+//P=sigma_c*A*(1+alpha*(l_e*k)**2)**-1 //Rankines Formula
+//Now Substituting values in above equation we get
+//125*10**4=550*10**6*%pi*0.09*D**2*(1+1*1600**-1*((2*0.32)**2)**-1)**-1
+
+//Further simplifying and rearranging we get
+//D**4-0.008038*D**2-0.0001962397=0
+
+a=1
+b=-0.008038
+c=-0.0001962397
+
+X=b**2-4*a*c
+
+D_1=((-b+X**0.5)*(2*a)**-1)**0.5*10**2
+D_2=((-b-X**0.5)*(2*a)**-1)**0.5
+
+//Thus Diameter cannot be negative, discard value of D_2
+d=0.8*D_1
+
+//Result
+printf("The Minimum Diameter is %.2f",d);printf(" cm")