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clc;
clear;
mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-5.6 Page No.102\n');
//From Example Problem 5.5
Sy=71000; //[lb/in^2] Yield strength
Smax=8148.7331 ; //[lb/in^2] Maximum stress
Smin=0; //[lb/in^2] Minimum stress
Smean=(Smax+Smin)/2; //[lb/in^2] Mean stress
Salt=(Smax-Smin)/2; //[lb/in^2] Alternating stress
Sn=18000; //[lb/in^2] Modified endurance strength
Kt=1.32 //[] Stress concentration factor
Nd=100000; //[cycles] Desired life
Snn=Sn*(10^6/Nd)^0.09; //[lb/in^2]
N=inv(Smean/(0.5*Sy)+Kt*Salt/Snn); //[] Factor of safety
mprintf('\n The new factor of safety for this condition is %f.',N);
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