King Fahd University of Petroleum & Minerals
Mechanical Engineering Department
ME 205: MATERIALS SCIENCE
Fall Semester 2006-2007 (061)
Quiz # 4
Dated: November 28, 2006
Name: _________________
ID # ____________
Prob. # 1
An alloy subjected to true stress of 350 MPa deforms plastically and corresponding
true strain is 0.34. The yield strength of the alloy is 200 MPa and the tensile strength
is 380 MPa. This alloy has a strength coefficient equal to 485 MPa. How much a
specimen of this alloy will elongate when true stress of 300 MPa is applied if the
original length is 200 mm.
Solution:
K = 485 MPa
t = 350 MPa
εt = 0.34
∆l = ?
t = 300 MPa
lo = 200 mm
Since the applied stress value is more than the Yield strength therefore material will
experience plastic deformation.
In Plastic range:
 t  Knt
(1)
Substitute these values in Eq. (1) and find n
ln  t  ln K ln 350  ln 485
n

 0.3
ln  t
ln .34
Now Equation (1) can be written as
 t  485  t0.3
(2)
Substitute t = 300 MPa in Eq. (2) and determine corresponding εt
 
t   t 
 485 
1 / 0.3
 300 


 485 
1 / 0.3
 0.202
l 
 t  ln  i 
 l0 
Therefore li  l 0 exp(  t )  200  exp( 0.202)  244.77 mm
l  li  l0  244.77  200  44.77 mm
King Fahd University of Petroleum & Minerals
Mechanical Engineering Department
ME 205: MATERIALS SCIENCE
Fall Semester 2006-2007 (061)
Quiz # 4
Dated: November 12, 2006
Name: _________________
ID # ____________
Prob. # 1
An alloy subjected to true stress of 300 MPa deforms plastically and corresponding
true strain is 0.27. The yield strength of the alloy is 200 MPa and the tensile strength
is 380 MPa. This alloy has a strength coefficient equal to 400 MPa. How much a
specimen of this alloy will elongate when true stress of 280 MPa is applied if the
original length is 250 mm.
Solution:
K = 400 MPa
t = 300 MPa
εt = 0.27
∆l = ?
t = 280 MPa
lo = 250 mm
Since the applied stress value is more than the Yield strength therefore material will
experience plastic deformation.
In Plastic range:
 t  Knt
(1)
Substitute these values in Eq. (1) and find n
ln  t  ln K ln 300  ln 400
n

 0.22
ln  t
ln .27
Now Equation (1) can be written as
 t  400  t0.22
(2)
Substitute t = 280 MPa in Eq. (2) and determine corresponding εt
 
 280 
t   t 

 0.198

 400 
 400 
l 
 t  ln  i 
 l0 
Therefore li  l 0 exp(  t )  250  exp( 0.198)  304.74 mm
1 / 0.22
1 / 0.22
l  li  l0  304.74  250  54.74 mm