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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clear all; clc;
+
+disp("Scilab Code Ex 1.15 : ")
+
+//Given:
+P= 20; //kN
+d_hole = 40; //mm
+normal_allow = 60; //MPa
+shear_allow = 35; //MPa
+
+
+//Diameter of Rod:
+area1 = (P*10^3)/(normal_allow*10^6); //Area in m^2
+d = ((sqrt((4*area1)/%pi))*1000); // Area = (%pi\4)d^2
+
+
+//Thickness of disc:
+V = P;
+area2 = (V*10^3)/(shear_allow*10^6); //Area in m^2
+thickness = (area2*10^6)/(d_hole*%pi);// A = pi*d*t
+
+
+printf("\n\nThe cross sectional area of disc = %.8f m^2",area1);
+printf("\nThe diameter of rode = %.2f mm",d);
+printf("\nThe thickness of disc = %.2f mm",thickness);
+
+//------------------------------------------------------------------------END------------------------------------------------------------------------------------
+