# 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"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 [ ]: