Chapter 9:Mechanical Properties of Metals¶
Example 9.1 Page 310¶
In [1]:
#initiation of variable
from math import acos, asin, sqrt, cos, sin
E=110.0*10**3 #Young's modulus of Copper in MPa
sigma=276.0 #Applied stress in MPa
l_o=305.0 #Original length in mm
#calculation
del_l=sigma*l_o/E #Deformation
#result
print" Elongation obtained is %.2f mm \n" %del_l
Example 9.2 Page 312¶
In [3]:
#initiation of variable
from math import acos, asin, sqrt, cos, sin, pi
del_d=-2.5*10**-3 #Deformation in diameter in mm
d_0=10.0 #Initial diameter in mm
v=0.34 #Poisson ratio for brass
E=97*10**3 #Modulus of elasticity in MPa
#calculation
e_x=del_d/d_0 # Strain in x-direction
e_z=-e_x/v
sigma=e_z*E # Stress produced
F=sigma*pi*(d_0**2)/4
#result
print" Strain in x-direction is %.1e" %e_x
print" Strain in z-direction is %.2e" %e_z
print" Applied force is %d N" %F
print " Answer in book is 5600N. It is due to rounding off at intermediate steps"
Example 9.3 Page 316¶
In [6]:
#initiation of variable
from math import acos, asin, sqrt, cos, sin, pi, floor
sigma_2=150.0 # Stress at a point in MPa
sigma_1=0.0 # Stress at a point in MPa
epsilon_2=0.0016 # Strain at a point in MPa
epsilon_1=0.0 # Strain at a point in MPa
d_0=12.8*10**-3 #Initial Diameter in m
sigma=450.0*10**6 #tensile strength in MPa
l0=250.0; #Initial length in mm
e=0.06; #strain
# Part A
#calculation
E=(sigma_2-sigma_1)/(epsilon_2-epsilon_1) #Young's Modulus = stress/strain
#result
print " Modulus of elasticity is %.1f GPa which is very close to its true value 97 GPa" %(E/10**3)
# Part C
A_0=pi*d_0**2/4.0
F=sigma*A_0
dl=e*l0
#result
print" Maximum load sustained is %d N" %(floor(F/10)*10)
#part D
print" Change in length is %d mm" %dl;
Example 9.4 Page 323¶
In [8]:
#initiation of variable
from math import acos, asin, sqrt, cos, sin, pi, floor
di=12.8 #Initial diameter in mm
df=10.7 #Final diameter in mm
sigma=460.0*10**6 #Tensile strength
# Part A
#calculation
RA = ((di**2-df**2)/di**2)*100 #Ductility in terms of Reduction Area
#result
print" Percent reduction in area is %d" %RA
# Part B
A_o=pi*di**2*10**-6/4.0
F=sigma*A_o
Af=pi*df**2/4
sigma_t=F/Af
#result
print" True stress at failure is %d MPa" %sigma_t
print "Answer in book is 660. It is due to founding off at intermediate stage"
Example 9.5 Page 323¶
In [9]:
#initiation of variable
from math import acos, asin, sqrt, cos, sin, pi, floor, log
sigma_t=415.0 #True stress in MPa
et=0.1 #True strain
K=1035.0 # Bulk modulus in MPa
#calculation
n=log(sigma_t/K)/log(et)
#result
print" Strain hardening coefficient is %.2f" %n;
Example 9.6 Page 332¶
In [21]:
#initiation of variable
%pylab inline
import numpy as np
import matplotlib.pyplot as plt
from math import acos, asin, sqrt, cos, sin, pi, floor, log
n=[0.0, 1.0, 2.0, 3.0, 4.0, 5.0]
T_S=[510.0, 520.0, 512.0, 515.0, 522.0, 530]
#First and Last points are arbitrary to plot the required points
#partA
plt.plot(n,T_S,'ro')
plt.xlabel('sample number')
plt.ylabel('Tensile Strength (Mpa)')
plt.title('Tensile Strength vs Sample Number')
plt.show()
#Mean Tensile strength
i=2;
TS_mean=0;
for i in range (1,5):
TS_mean=TS_mean+(T_S[i]/4);
print " Average tensile strength is %d MPa." %TS_mean;
#Standard Deviation
#partB
j=0
std=0
for i in range (1,5):
std=std+((T_S[i]-TS_mean)**2/(4.0-1))
print" Standard deviation is %.1f MPa" %(sqrt(std))
Example 9.7 (Design Problem 9.1) Page 334¶
In [24]:
#initiation of variable
import numpy as np
import matplotlib.pyplot as plt
from math import acos, asin, sqrt, cos, sin, pi, floor, log
f = 220000.0 # Maximum load in N
sigma_y = 310.0 # Minimum field strength in MPa
sigma_t = 565.0 # Tensile strength strength in MPa
N= 5.0 # Factor of safety
#calculation
sigma_w = sigma_y*1.0e6/N
d = 2*sqrt (f/(2*pi*sigma_w))# Calculation of diameter
#result
print" Diameter of each of two rods should be %.1f mm" %(d*1e3)