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clc;
clear;
mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-5.1 Page No.93\n');
SF=2; //[] Safety factor
F=500; //[lb] Load
L=40; //[in] Length of shaft
Su=95000; //[lb/in^2] Ultimate strength (Appendix 4)
Sy=60000; //[lb/in^2] Yield strength (Appendix 4)
Mmax=F*L/4; //[lb*in] Maximum bending moment
Mmin=-F*L/4; //[lb/in^2] Minimum bending moment
Csurface=1; //[] As surface is polished
Csize=0.85; //[] Assuming 0.5<D<2
Ctype=1; //[] Bending stress
Sn=Csize*Csurface*Ctype*(0.5*Su); //[lb/in^2] Endurance limit
if Mmax==abs(Mmin) then
Sm=0; //[lb/in^2] Mean stress
end
Sa=Sn/SF; //[lb/in^2] As (1/SF)=(Sm/Sy)+(Sa/Sn) from soderberg equation
Sa_Z=(Mmax-Mmin)/2; //[lb*in^2] Product of altenating stress and section modulus
Z=Sa_Z/Sa; //[in^4] Section modulus
D=(32*Z/%pi)^(1/3); //[in] Diameter of shaft
D1=1.375; //[in] Next higher available is 1.375 in. so use D1
mprintf('\n The required diameter of rotating shaft is %f in.', D1);
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