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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+w = 550;// lb. per foot run
+f = 1000;// lb/in^2
+l = 20;// feet
+d_limit = 15;// inches
+E = 1.5*10^6;// lb/in^2
+//central ddeflection
+delta = (1/2);// inches
+d = (5/24)*(f/E)*20*12/(1/(2*20*12));// inches
+M = w*l*l*12/8;// lb-inches
+b = M/(f*(1/6)*d^2);// inches
+printf('A section with d = %d inches, b = %d inches will do.',round(d),round(b));
+f1 = (1/(2*20*12))*(d_limit/(l*12))*E/(5/24);// lb/in^2
+b = M/(f1*(1/6)*d_limit^2);// inches
+printf('\n If the deapth of section is limited to %d inches, then \n f = %.1f lb/in^2\n b = %.1f inches',d_limit,f1,b);
+
+//tha answer is correct only, but it is approximated in the text book.