# Chapter4-Combined Stress and Failure Theories¶

## Ex1-pg66¶

In [1]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-4.1   Page 66 ')

D=2.;               ##[in] Dia. of short column
L=15.;              ##[in] Length of column

A=(math.pi*D**2)/4.;      ##[in^2] Area of cross section of column
SA=F/A;             ##[lb/in^2] Axial Stress

Z=(math.pi*D**3)/32.;     ##[in^4] Section modulus for bending
M=F*e;              ##[lb*in] Bending moment
SB=M/Z;             ##[lb/in^2] Bemding stress

S=-SA-SB;           ##S=(+-)SA+(+-)SB Max. stress

##The bending stress and axial stress are added on inner side of column

print'%s %.2f %s '%('\n\n Maximum stress in column is ',S,' lb/in^2.\n')

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-4.1   Page 66

Maximum stress in column is  -28647.89  lb/in^2.



## Ex2-pg67¶

In [2]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-4.2   Page 67 ')

F1=800.;             ##[lb] Vertical force
F2=600.;             ##[lb] Horizontal force
D=0.5;              ##[in] Pin diameter
A=(math.pi*D**2)/4.;     ##[in^2] Area of cross section of pin

F=math.sqrt(F1**2+F2**2);   ##[lb] Resultant force on pin
S=F/A;               ##[lb/in^2] Shear stress in pin

##If forces were not perpendicular, they would be added vectorially.
print'%s %.2f %s '%('\n\n Shear stress in pin is ',S,' lb/in^2.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-4.2   Page 67

Shear stress in pin is  5092.96  lb/in^2.


## Ex3-pg68¶

In [3]:
import math

print('MACHINE DESIGN\n Timothy H. Wentzell, P.E.\n Example 4.3   Page no 68');

P=50.;            ##[hp] Power transmitted
N=300.;           ##[rpm] Speed
D=10.;            ##[in] Effective pitch diameter of sprocket
d=1.;             ##[in] Diameter of shaft from figure 4.3
Z=(math.pi*d**3)/16.;  ##[in^3] Section modulus of shaft
A=(math.pi*d**2)/4.;   ##[in^2] Area of cross section

T=(63000./N)*P;  ##[lb*in] Torque required to transmit power
F=T/(D/2.);       ##[lb] Driving force in chain

Ss=F/A;          ##[lb/in^2] Shear stress in shaft

St=T/Z;          ##[lb/in^2] Torsional stress in shaft

S=Ss+St;         ##[lb/in^2] Resultant stress

##Note-There is mistake in addition of Ss and St.

##This value would be compared to shear stress allowable for shaft material

print'%s %.2f %s '%('\n\n The combined stress in 1 inch diameter shaft is ',S,' lb/in^2.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
Example 4.3   Page no 68

The combined stress in 1 inch diameter shaft is  56149.86  lb/in^2.


## Ex4-pg71¶

In [4]:
import math

print('MACHINE DESIGN\n Timothy H. Wentzell, P.E.\n Example 4.4  Page no 71')

P=20.;             ##[hp] Power transmitted by chain drive
n=500.;            ##[rpm] speed
d=8.;              ##[in] Pitch diameter of sprocket
fos=2.;
D=1.25;           ##[in] Diameter of shaft
L=12.;             ##[in] Distance between two supporting bearings
Z1=math.pi*D**3/16.;    ##[in^3] Section modulus for torsion
Z2=math.pi*D**3/32.;    ##[in^3] Section modulus for bending

T=63000.*P/n;      ##[in*lb] Torque on shaft

F=T/(d/2.);        ##[lb] Force in chain

M=F*L/4.;          ##[in*lb] Bending moment in shaft

Ss=T/Z1;          ##[lb/in^2] Torsional shear stress

Sb=M/Z2;          ##[lb/in^2] Bending normal stress

##Note- In the book Sb=9860 lb/in^2 is used instead of Sb=9856.7075 lb/in^2

S=(Sb/2.)+math.sqrt(Ss**2.+(Sb/2.)**2);  ##[lb/in^2] Combined max. stress

Sy=30000.;         ##[lb/in^2]From APPENDIX 4 Page no-470 for AISI 1020 and Hot-rolled steel
FOS=(Sy/2.)/S;     ##[]Actual factor of safty

if S < Sy/2.: ##Strength is greater than combined stress so design is safe
print'%s %.2f %s '%('\n\n Design is acceptable and Combined stress is ',S,' lb/in^2');
else:
print('\n\n Design is not acceptable');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
Example 4.4  Page no 71

Design is acceptable and Combined stress is  13142.28  lb/in^2