# Chapter 6 Mechanical Properties : part one¶

## Example 6_1 pgno:207¶

In :
from math import pi
F=1000#in lb
Ao=(pi/4)*(0.505)**2#in**2
rho=F/Ao
delta_I=0.001#in
I_o=2#in
e=delta_I/I_o
print"The value in psi is=",rho
print"The value of epselon",e

The value in psi is= 4992.60678261
The value of epselon 0.0005


## Example 6_2 pgno:207¶

In :
# Initialisation of Variables
F=45000;#Force applied on an aluminum rod in lb
e=25000;#the maximum allowable stress on the rod in psi
l2=150;#the minimum length of the rod in in
e1=0.0025;#The strain appiled on rod
sigma=16670;#Stress applied on rod in psi
L=0.25;#The maximum allowable elastic deformation in in
from math import sqrt,pi
#CALCULATIONS
Ao1=F/e;#The required crosssectional area of the rod
d=sqrt((Ao1*4)/pi);#Diameter of rod in in
l1=e1*L;#The maximum length of the rod in in
e2=L/e1;#The minimum strain allowed on rod
Ao2=F/sigma;#The minimum cross-sectional area in in^2
print "The required crosssectional area of the rod in in^2:",Ao1
print "Diameter of rod in in:",d
print "The maximum length of the rod in in:",l1
print "The minimum strain allowed on rod:",e2
print "The minimum cross-sectional area in in^2:",Ao2

The required crosssectional area of the rod in in^2: 1
Diameter of rod in in: 1.1283791671
The maximum length of the rod in in: 0.000625
The minimum strain allowed on rod: 100.0
The minimum cross-sectional area in in^2: 2


## Example 6_3 pgno:213¶

In :
# Initialisation of Variables
sigma1=35000;#Stress applied of aluminum alloy in psi from table 6-1
e1=0.0035;##Strain applied of aluminum alloy from table 6-1
sigma2=30000;#Stress applied of aluminum alloy  in psi
Lo=50;#initial length of  aluminum alloy
#CALCULATIONS
E=sigma1/e1;#Modulus of elasticity  of aluminum alloy
e2=sigma2/E;#Strain applied of aluminum alloy
L=Lo+(e2*Lo);#The length after deformation of bar in in
print "Modulus of elasticity  of aluminum alloy from table 6-1:",E
print "The length after deformation of bar in in",L
print "Strain applied of aluminum alloy:",e2

Modulus of elasticity  of aluminum alloy from table 6-1: 10000000.0
The length after deformation of bar in in 50.15
Strain applied of aluminum alloy: 0.003


## Example 6_4 pgno:214¶

In :
# Initialisation of Variables
Lf=2.195;#Final length after failure
d1=0.505;#Diameter of alluminum alloy in in
d2=0.398;#Final diameter of alluminum alloy in in
Lo=2;#Initial length of alluminum alloy
from math import pi
#CALCULATIONS
A0=(pi/4)*d1**2;#Area of original of alluminum alloy
Af=(pi/4)*d2**2;#Area of final of alluminum   alloy
E=((Lf-Lo)/Lo)*100;#Percentage of Elongation
R=((A0-Af)/A0)*100;#Percentage of Reduction in area
print "Percentage of Elongation:",E
print "Percentage of Reduction in area:",round(R,1)
print"The final length is less than 2.205 in because, after fracture, the elastic strain is recovered."

Percentage of Elongation: 9.75
Percentage of Reduction in area: 37.9
The final length is less than 2.205 in because, after fracture, the elastic strain is recovered.


## Example 6_5 pgno:217¶

In :
# Initialisation of Variables
F=8000.;#.......#Load applied for the aluminum alloy in lb
F2=7600.;#......#Load applied for the aluminum alloy in lb at fracture
dt1=0.505;#.......#diameter of for the aluminum alloy in in
Lot=2.;#.............#Initial length of alluminum alloy
Ff=7600.;#.........#Load applied for the aluminum alloy after fracture in lb
df=0.398;#.......#The diameter at maximum load after fracture
Lf=0.205;#.......#Final length at fracture
from math import pi,log
#CALCULATIONS
Es=F/((pi/4)*dt1**2);#.....#Engineering stress in psiAt the tensile or maximum load
Ts=F/((pi/4)*dt2**2);#.....#True stress in psi At the tensile or maximum load
Ee=(Lt-Lot)/Lot;#........#Engineering strain At the tensile or maximum load
Te=log(Lt/Lot);#........#True strain At the tensile or maximum load
Es2=F2/((pi/4)*dt1**2);#......##Engineering stress At fracture:
Ts2=F2/((pi/4)*df**2);#......#True stress At fracture:
Ee2=Lf/Lot;#..........#Engineering strain At fracture:
Te2=log(((pi/4)*dt1**2)/((pi/4)*df**2));#.......#True strain At fracture:
print "Engineering stress in psiAt the tensile or maximum load",Es
print "True stress in psi At the tensile or maximum load",Ts
print "Engineering strain At the tensile or maximum load",Ee
print "True strain At the tensile or maximum load",Te
print "Engineering stress At fracture:",Es2
print "True stress At fracture",Ts2
print "Engineering strain At fracture:",Ee2
print "True strain At fracture:",round(Te2,3)

Engineering stress in psiAt the tensile or maximum load 39940.8542609
True stress in psi At the tensile or maximum load 41237.025201
Engineering strain At the tensile or maximum load 0.06
True strain At the tensile or maximum load 0.058268908124
Engineering stress At fracture: 37943.8115478
True stress At fracture 61088.2335041
Engineering strain At fracture: 0.1025
True strain At fracture: 0.476


## Example 6_6 pgno:221¶

In :
# Initialisation of Variables
Fs=45000;#.......#The flexural strength of a composite  material in psi
Fm=18*10**6;#........#The flexural modulus of composite material in psi
w=0.5;#.......#wide of sample in in
h=0.375;#......#Height of sample in in
l=5;#..........#Length of sample in in
#CALCULATIONS
F=Fs*2*w*h**2/(3*l);#......#The force required to fracture the material in lb
delta=(l**3)*F/(Fm*4*w*h**3);#.......#The deflection of  the sample at fracture
print "The force required to fracture the material in  lb:",F
print "The deflection of  the sample at fracture in in",round(delta,4)

The force required to fracture the material in  lb: 421.875
The deflection of  the sample at fracture in in 0.0278