import math
print('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):
Sm=0; ##[lb/in^2] Mean stress
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/math.pi)**(1./3.); ##[in] Diameter of shaft
D1=1.375; ##[in] Next higher available is 1.375 in. so use D1
print'%s %.2f %s '%('\n The required diameter of rotating shaft is ',D1,' in.');
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-5.2 Page No.95\n');
Su=90000.; ##[lb/in^2] Ultimate strength (Appendix 8)
Sy=37000.; ##[lb/in^2] Yield strength (Appendix 8)
Sni=34000.; ##[lb/in^2] Endurance limit (Appendix 8)
SF=1.6; ##[] Safety factor
F=1000.; ##[lb] Load
L=12.; ##[in] Length of cantilever beam
Mmax=F*L; ##[lb*in] Maximum bending moment
Mmin=0.; ##[lb*in] Minimum bending moment
Csize=0.85 ##[] Assuming 0.5<D<2 in
Ctype=1.; ##[] Bending stress
Csurface=1.; ##[] As surface is polished
Malt=(Mmax-Mmin)/2.; ##[lb*in] Alternating bending moment
Mmean=(Mmax+Mmin)/2.; ##[lb*in] Mean bending moment
Sn=Csize*Csurface*Ctype*Sni; ##[lb/in^2] Modified endurance limit
Z=((Mmean/Sy)+(Malt/Sn))*SF; ##[in^3] Section modulus
D=(32.*(Z)/math.pi)**(1./3.); ##[in] Diameter of bar
print'%s %.2f %s '%('\n The required diameter of bar using the soderberg method is ',D,' in.');
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-5.3 Page No.97\n');
Su=90000.; ##[lb/in^2] Ultimate strength (Appendix 8)
Sy=37000.; ##[lb/in^2] Yield strength (Appendix 8)
Sni=34000.; ##[lb/in^2] Endurance limit (Appendix 8)
SF=1.6; ##[] Safety factor
F=1000.; ##[lb] Load
L=12.; ##[in] Length of cantilever beam
Mmax=F*L; ##[lb*in] Maximum bending moment
Mmin=0.; ##[lb*in] Minimum bending moment
Csize=0.85 ##[] Assuming 0.5<D<2 in
Ctype=1.; ##[] Bending stress
Csurface=1.; ##[] As surface is polished
Malt=(Mmax-Mmin)/2.; ##[lb*in] Alternating bending moment
Mmean=(Mmax+Mmin)/2.; ##[lb*in] Mean bending moment
Sn=Csize*Csurface*Ctype*Sni; ##[lb/in^2] Modified endurance limit
Z=((Mmean/Su)+(Malt/Sn))*SF; ##[in^3] Section modulus
D=(32.*(Z)/math.pi)**(1/3.); ##[in] Diameter of bar
print'%s %.2f %s '%('\n The required diameter of bar using the soderberg method is ',D,' in.');
##Note that the modified Goodman results in a less conservative size as would be expected from figure 5.10
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-5.4 Page No.98\n');
Su=95000.; ##[lb/in^2] Ultimate strength
Sy=60000.; ##[lb/in^2] Yield strength
SF=1.5; ##[] Safety factor
Fmax=1000.; ##[lb] Maximum load
Fmin=-6000.; ##[lb] Minimum load
Fmean=(Fmax+Fmin)/2.; ##[lb] Mean load
Fmean=abs(Fmean); ##[lb] Considering absolute value
Falt=(Fmax-Fmin)/2.; ##[lb] Alternating load
Csize=1. ##[] Assuming b<0.5 in
Ctype=0.8 ##[] Axial stress
Csurface=0.86 ##[] Machined surface Figure 5.7b
Sn=Csize*Csurface*Ctype*(0.5*Su); ##[lb/in^2] Modified endurance limit
A=((Fmean/Sy)+(Falt/Sn))*SF; ##[in^2] Area of cross section of rod
b=math.sqrt(A); ##[in] Side of square cross section
print'%s %.2f %s '%('\n The required square size in the center section is ',b,' in.');
import math
print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-5.5 Page No.100\n');
Su=80000.; ##[lb/in^2] Ultimate strength
Sy=71000.; ##[lb/in^2] Yield strength
D=0.6; ##[in] Diameter of shaft
d=0.5; ##[in] Diameter of shaft at notch
r=0.05; ##[in] Radius of notch
Z=math.pi*d**3/16.; ##[in^3] Polar section modulus
Tmax=200.; ##[lb*in] Maximum load
Tmin=0.; ##[lb*in] Minimum load
Smax=Tmax/Z; ##[lb/in^2] Maximum stress
Smin=Tmin/Z; ##[lb/in^2] Minimum stress
Smean=(Smax+Smin)/2.; ##[lb/in^2] Mean stress
Salt=(Smax-Smin)/2.; ##[lb/in^2] Alternating stress
Csize=0.85; ##[] Assume 0.5<D<2 in
Csurface=0.88; ##[] Machined surface Figure 5.7b
Ctype=0.6; ##[] Torsional stress
Sn=Csize*Csurface*Ctype*(0.5*Su); ##[lb/in^2] Modified endurance limit
Kt=1.32; ##[] (D/d)=1.2, (r/d)=0.1 from Appendix 6c
N=1/(Smean/(0.5*Sy)+Kt*Salt/Sn); ##[] Safety factor
print'%s %.2f %s '%('\n The factor of safety for this design is ',N,'');
import math
print('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=1/(Smean/(0.5*Sy)+Kt*Salt/Snn); ##[] Factor of safety
print'%s %.2f %s '%('\n The new factor of safety for this condition is .',N,'');