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+clear all; clc;
+
+
+disp("Scilab Code Ex 1.14 : ")
+
+//Given:
+shear_allow = 55; //MPa
+l_ac = 200; //mm
+l_cd= 75; //mm
+l_de = 50; //mm
+l_ce = l_cd + l_de;
+load_d =15; //kN
+load_e = 25; //kN
+
+//Internal Shear Force:
+//summation Mc = 0
+
+f_ab = ((load_d*l_cd +load_e*(3/5)*l_ce)/l_ac);
+c_x =-load_d + (load_e*(4/5)); //resolving C in x dir
+c_y = load_d + (load_e*(3/5)); //resolving C in y dir
+
+f_c = sqrt(c_x^2 + c_y^2); //kN
+V = f_c/2;
+
+//Required Area
+A = ((V*10^3)/(shear_allow)); //A = V/Allowable shear in mm^2
+d = ((sqrt((4*A)/%pi))) // Area = (%pi\4)d^2 in mm^2
+
+chosen_d = ceil(ceil(d))+1;
+
+//Displaying Results:
+
+
+printf("\n\nThe force at AB           = %.2f kN",f_ab);
+printf("\nThe resultant force at C  = %.2f kN",f_c);
+printf("\nThe area of pin           = %.2f mm^2",A);
+printf("\nThe diameter of pin       = %.2f mm",chosen_d);
+
+//---------------------------------------------------------------END--------------------------------------------------------------------------------------
+
+
+