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clear all; clc;
disp("Scilab Code Ex 4.16 : ")
//Given:
yield = 250; //MPa
r = 4; //mm
width = 40; //mm
thick = 2; //mm
//a)
r_h = r/(width - (2*r));
w_h = width/(width - (2*r));
K = 1.75;
area = (thick*(width - (2*r))*10^-6);
P_y = (yield*10^6*area)/K;
P_y = P_y/1000;
//b)
P_p = (yield*10^6*area);
P_p = P_p/1000;
//Display:
printf("\n\nThe maximum load P that does not cause the steel to yield = %1.2f kN",P_y);
printf('\nThe maximum load that the bar can support = %1.2f kN',P_p);
//-------------------------------------------------------------------------END----------------------------------------------------------------------
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