# Chapter 6: Mechanical Properties of Metal¶

### Example 6.1 Page No 140¶

In [3]:
E=110*10**3  #Young's modulus of Copper in MPa
sigma=276.0    #Applied stress in MPa
lo=305.0       #Original length in mm

dl=sigma*lo/E

print"Elongation obtained is  ",round(dl,2),"mm"

Elongation obtained is   0.77 mm


### Example 6.2 Page No 142¶

In [7]:
del_d=-2.5*10**-3  #Deformation in dia  in mm
d0=10.0             #Initial dia  in mm
v=0.34           #Poisson ratio for brass

ex=del_d/d0
ez=-ex/v
E=97*10**3        #Modulus of elasticity in MPa
sigma=ez*E
F=sigma*math.pi*(d0**2)/4.0

print"Applied force is ",round(F,0),"N"

Applied force is  5602.0 N


### Example 6.3 Page No 146¶

In [12]:
si2=150           # in MPa
si1=0
e2=0.0016
e1=0
d0=12.8*10**-3    #Initial Diameter in m

E=(si2-si1)/(e2-e1)

A0=math.pi*d0**2/4.0
sig=450*10**6     #tensile strength in MPa
F=sig*A0
l0=250     #Initial lengt in mm
e=0.06     #strain
dl=e*l0

print"Modulus of elasticity is ",round(E/10**3,1),"GPa"
print"From the graph the Yield strength is",l0,"MPa"
print"Maximum load sustained is ",round(F,0),"N/n"
print"Change in length is ",dl,"mm"

Modulus of elasticity is  93.8 GPa
From the graph the Yield strength is 250 MPa
Maximum load sustained is  57906.0 N/n
Change in length is  15.0 mm


### Example 6.4 Page No 153¶

In [14]:
di=12.8     #Initial dia in mm
df=10.7     #Final dia  in mm

import math
RA = ((di**2-df**2)/di**2)*100
Ao=math.pi*di**2*10**-6/4.0
sig=460*10**6    #Tensile strength

F=sig*Ao

Af=math.pi*df**2/4.0
sig_t=F/Af

print"percent reduction in area is ",round(RA,0),"%"
print"True stress is ",round(sig_t,1),"MPa"

percent reduction in area is  30.0 %
True stress is  658.3 MPa


### Example 6.5 Page No 153¶

In [5]:
sig_t=415       #True stress in MPa
et=0.1          #True strain
K=1035.0         # In MPa

n=(math.log(sig_t)-math.log(K))/math.log(et)

print"Strain - hardening coefficient is ",round(n,2)

Strain - hardening coefficient is  0.4


### Example 6.6 Page No 162¶

In [22]:
n=4.0        #No of points
T1=520
T2=512
T3=515
T4=522

Tav=(T1+T2+T3+T4)/n
s=(((T1-Tav)**2+(T2-Tav)**2+(T3-Tav)**2+(T4-Tav)**2)/(n-1))**(0.5)

print"The average Tensile strength is",round(Tav,0),"MPa"
print"The standard deviation  is",round(s,1),"MPa"

The average Tensile strength is 517.0 MPa
The standard deviation  is 4.6 MPa


### Design Example 6.1 ,Page No 164¶

In [18]:
sig_y=310.0      #Minimum yield strength in MPa
N=5.0            # Conservative factor of safety

F=220000/2.0    #Two rods must support half of the total force
sig_w=sig_y/N
d=2*math.sqrt(F/(math.pi*sig_w))

print"Diameter of each of the two rods is ",round(d,1),"mm"

Diameter of each of the two rods is  47.5 mm

In [ ]: