## Ex1-pg93¶

In [1]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-5.1 Page No.93\n');

SF=2.;                     ##[] Safety factor
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.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-5.1 Page No.93

The required diameter of rotating shaft is  1.38  in.


## Ex2-pg95¶

In [2]:
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

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.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-5.2 Page No.95

The required diameter of bar using the soderberg method is  1.82  in.


## Ex3-pg97¶

In [3]:
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

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

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-5.3 Page No.97

The required diameter of bar using the soderberg method is  1.65  in.


## Ex4-pg98¶

In [4]:
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.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-5.4 Page No.98

The required square size in the center section is  0.47  in.


## Ex5-pg100¶

In [5]:
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,'');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-5.5 Page No.100

The factor of safety for this design is  2.41


## Ex6-pg102¶

In [6]:
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,'');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-5.6 Page No.102

The new factor of safety for this condition is . 2.80