# Chapter3-Stress and Deformation¶

## Ex1-pg41¶

In :
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.1   Page No-41 \n');

F=20000.;               ##[lb] Load applied to steel bar
L=6.;                   ##[in] Length of steel bar
d=1.;                   ##[in] Diameter of steel bar
A=math.pi*(d**2)/4.;         ##[in^2] Area of cross section of steel bar
E=30.*10**6;             ##[lb/in^2] Modulus of elasticity for AISI 1020 hot-rolled steel
Sy=30000.;              ##[lb/in^2] Yield limit

S=F/A;                 ##[lb/in^2] Stress in bar
print'%s %.2f %s '%('\na. Stress in bar=',S,' lb/in^2.');

delta=F*L/(A*E);       ##[in] Change in length of bar
print'%s %.2f %s '%('\nb. bar shorten by ',delta,' in.');

if Sy>S:
print'%s %.2f %s %.2f %s  '%('\nc. The stress of ',S,' psi is less than Sy of ',Sy,' psi, so it will'and '\n   return to its original size because the yield limit was not exceeded.');
else:

##Note: The deviation of answer from the answer given in the book is due to round off error.(In the book values are rounded while in scilab actual values are taken)

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-3.1   Page No-41

a. Stress in bar= 25464.79  lb/in^2.

b. bar shorten by  0.01  in.

c. The stress of  25464.79  psi is less than Sy of  30000.00
return to its original size because the yield limit was not exceeded.


## Ex2-pg43¶

In :
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.2 Page No.43\n');

b=2.;                      ##[in] Width of beam
h=2.;                      ##[in] Height of beam
I=(b*h**3)/12.;             ##[in^4] Moment of inertia
F=3000.;                   ##[lb] Load applied to beam
L=36.;                     ##[in] Length of beam
c=1.;                      ##[in] Distance of outer most fiber from neutral axis
E=30*10**6;                ##[lb/in^2] Modulus of elasticity
Sy=30000.;                 ##[lb/in^2] Yield strength
Su=55000.;                 ##[lb/in^2] Ultimate strength
SF=2.;                     ##[] Safety factor based on ultimate stress

M=F*L/4.;                  ##[lb*in] Bending moment
S=(M/I)*c;                ##[lb/in^2] Bending stress

##Note-In the book I=1.33 in^4 is used instead of I=1.3333333 in^2

print'%s %.2f %s '%('\na. The maximum stress in beam is ',S,' lb/in^2');

delta=-F*L**3/(48.*E*I);     ##[in] Maximum deflection in this beam

print'%s %.2f %s '%('\nb. The maximum deflection in this beam is ',delta,' in.');

if Sy>S:
print'%s %.2f %s %.2f %s  '%('\nc. Yes, the stress of ',S,' lb/in^2'and ' is less than the yield of Sy=',Sy,' lb/in^2.');
else:
print'%s %.2f %s  %.2f %s '%('\nc. No, the stress of ',S,' lb/in^2'and ' is greater than the yield of Sy=',Sy,' lb/in^2');

Sall=Su/SF;                ##[lb/in^2] Allowable stress

if Sall>S:
print'%s %.2f %s '%('\nd. It is acceptable because allowable stress is greater than the acttual stress of ',S,' lb/in^2.');
else:
print'%s %.2f %s '%('\nd. Design is not acceptable because allowable stress is less than the actual stress of ',S,' lb/in^2.')

##Note: The deviation of answer from the answer given in the book is due to round off error.(In the book values are rounded while in scilab actual values are taken)

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-3.2 Page No.43

a. The maximum stress in beam is  20250.00  lb/in^2

b. The maximum deflection in this beam is  -0.07  in.

c. Yes, the stress of  20250.00  is less than the yield of Sy= 30000.00  lb/in^2.

d. It is acceptable because allowable stress is greater than the acttual stress of  20250.00  lb/in^2.


## Ex3-pg45¶

In :
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.3 Page No.45\n');

Su=80.*10**3;          ##[lb/in^2] Ultimate strength
d=0.5;               ##[in] Diameter of pin
As=math.pi*d**2./4.;        ##[in^2] Area of cross section of pin

Ss=F/(2.*As);         ##[lb/in^2] Shear stress

if 0.5*Su>=Ss:
print('Pin would not fail');
else:
if 0.6*Su>=Ss:
print('Pin would not fail');
else:
print('\n Actual stress is too high and the pin would fail.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-3.3 Page No.45

Actual stress is too high and the pin would fail.


## Ex4-pg46¶

In :
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.4 Page No.46\n');

hp=10.;             ##[hp] Power transmitted
rpm=1750.;          ##[rpm] Turning speed
d=0.5;             ##[in] Diameter of shaft
L=12.;              ##[in] Length of shaft
G=11.5*10**6        ##[lb/in^2] shear modulus of elasticity
Su=62000.;          ##[lb/in^2]

T=63000.*hp/rpm;    ##[in*lb] Torque transmitted
Z=math.pi*d**3/16.;      ##[in^3] Polar section modulus
Ss=T/Z;            ##[lb/in^2] Torsional shear stress

##Note- In the book Z=0.025 in^3 is used instead of Z=0.0245437 in^3

print'%s %.2f %s '%('\na. Stress in the shaft is ',Ss,' lb/in^2.')

J=math.pi*d**4/32.;      ##[in^4] Polar moment of inertia

##Note- In the book J=0.0061 in^4 is used instead of J=0.0061359 in^4

print'%s %.2f %s '%('\nb. The angular deflection of shaft would be ',theta,' radians');

SF=3;               ##[] Safety factor based on ultimate strength

Zd=T/(0.5*Su/SF);   ##[in^3] Polar section modulus required for SF=3
Dd=(16.*Zd/math.pi)**(1/3.); ##[in] Diameter of shaft required Z=%pi*d^3/16

print'%s %.2f %s '%('\nc. Diameter of shaft required is ',Dd,' in.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-3.4 Page No.46

a. Stress in the shaft is  14667.72  lb/in^2.

b. The angular deflection of shaft would be  0.06  radians

c. Diameter of shaft required is  0.56  in.


## Ex5-pg53¶

In :
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.5 Page No.53\n');

I=math.pi*d**4/64.;              ##[in^4] Moment of inertia
A=math.pi*d**2/4.;               ##[in^2] Area of cross section
E=30*10**6;                 ##[lb/in^2] Modulus of elasticity

print'%s %.2f %s '%('\n The radius of gyration ',r,' in.');

K=1;                       ##[] End support condition factor

Le=K*L;                    ##[in] Effective length

print'%s %.2f %s '%('\n Effective length is ',Le,' in');

SR=Le/r;                   ##[] Slenderness ratio

print'%s %.2f %s '%('\n Slenderness ratio is ',SR,'.')

Sy=42000.;                 ##[lb/in^2] Yield strength

Cc=math.sqrt(2*math.pi**2*E/Sy);    ##[] Column constant

print'%s %.2f %s '%('The column constant is',Cc,'');

if SR>Cc:
print('\n Slenderness ratio is greater than column constant, so use the euler formula')

I=math.pi*d**4/64.;             ##[in^4] Moment of inertia

print'%s %.2f %s '%('\n The moment of inertia is ',I,' in^4');

Pc=math.pi**2*E*I/Le**2;        ##[lb] Critical force

##Note- In the book I=0.0075 in^4 is used instead of I=0.0074901 in^4

print'%s %.2f %s '%('\n The critical force is ',Pc,' lb.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-3.5 Page No.53

The radius of gyration  0.16  in.

Effective length is  30.00  in

Slenderness ratio is  192.00 .
The column constant is 118.74

Slenderness ratio is greater than column constant, so use the euler formula

The moment of inertia is  0.01  in^4

The critical force is  2464.16  lb.


## Ex6-pg55¶

In :
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-3.6 Page No.55\n');

L=60.;                    ##[in] Length of column
Sy=36000.;                ##[lb/in^2] Yield strength
SF=2.;                    ##[]Safty factor
E=30*10**6;               ##[lb/in^2] Modulus of elasticity

A=2.26;                  ##[in^2] Area of cross section (Appendix 5.4)
I=0.764;                 ##[in^4] Moment of inertia (Appendix 5.4)

K=0.65;                  ##[] End support condition factor from Figure 3.8
Le=K*L;                  ##[in] Effective length

print'%s %.2f %s '%('\n The effective length is ',Le,' in.');

SR=Le/r;                 ##[] Slenderness ratio

print'%s %.2f %s '%('\n The slenderness ratio is ',SR,'');

Cc=math.sqrt(2*math.pi**2*E/Sy);   ##[] Column constant

print'%s %.2f %s '%('\n The column constant is ',Cc,'');

if Cc>SR :
print('\n The column constant is greater than slenderness ratio, so use the Johnson formula.');

F=(A*Sy/SF)*(1.-Sy*SR**2/(4.*math.pi**2*E));

print'%s %.2f %s '%('\n The acceptable load for a safty factor of 2 is ',F,' lb.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-3.6 Page No.55

The effective length is  39.00  in.

The slenderness ratio is  67.08

The column constant is  128.25

The column constant is greater than slenderness ratio, so use the Johnson formula.

The acceptable load for a safty factor of 2 is  35116.52  lb.