King Fahd University of Petroleum & Minerals
Mechanical Engineering Department
ME 215 : Materials Science for ME
Fall Semester (061)
Homework # 3 - SOLUTION
Mr. B.J. Abdul Aleem
Due: Wedneday: November 22, 2006
6.4 We are asked to compute the maximum length of a cylindrical titanium
alloy specimen that is deformed elastically in tension. For a cylindrical
specimen
2
do 
A o =   
 2 
where do is the original diameter. Combining Equations (6.1), (6.2), and
(6.5) and solving for lo leads to
2
lo =
E  do l
4F
107 x 109 N / m2 ()3.8 x 10 3 m 0.42 x 103 m
2
=
(4)(2000 N)
= 0.25 m = 250 mm (10 in.)
6.7 (a) This portion of the problem calls for a determination of the maximum
load that can be applied without plastic deformation (Fy). Taking the yield
strength to be 345 MPa, and employment of Equation (6.1) leads to

Fy =  y A o = 345 x 10
6
2
N/m
130 x 10 -6 m2 
= 44,850 N (10,000 lbf)
(b) The maximum length to which the sample may be deformed without
plastic deformation is determined from Equations (6.2) and (6.5) as
 

li = l o 1  

E

345 MPa 
= (76 mm) 1 
= 76.25 mm (3.01 in.)
3

 103 x 10 MPa 

6.9 This problem asks that we calculate the elongation ∆l of a specimen of
steel the stress-strain behavior of which is shown in Figure 6.24. First it
becomes necessary to compute the stress when a load of 65,250 N is
applied as
 =
F
=
A
o
F
2
do 
  
 2 
=
65, 250 N
2
8.5 x 10 3 m
 



2


= 1150 MPa (170, 000 psi)
Referring to Figure 6.24, at this stress level we are in the elastic region on
the stress-strain curve, which corresponds to a strain of 0.0054. Now,
utilization of Equation (6.2) yields
 l =  lo = (0.0054)(80 mm) = 0.43 mm (0.017 in.)
6.29 This problem calls for us to make a stress-strain plot for aluminum,
given its tensile load-length data, and then to determine some of its
mechanical characteristics.
(a) The data are plotted below on two plots: the first corresponds to the
entire stress-strain curve, while for the second, the curve extends just
beyond the elastic region of deformation.
(b) The elastic modulus is the slope in the linear elastic region as
E =
  200 MPa  0 MPa
3
=
= 62.5 x 10 MPa = 62.5 GPa

0.0032  0
9.1 x 106 psi
(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It
intersects the stress-strain curve at approximately 285 MPa (41,000 psi ).
(d)
The tensile strength is approximately 370 MPa (53,500 psi),
corresponding to the maximum stress on the complete stress-strain plot.
(e) The ductility, in percent elongation, is just the plastic strain at fracture,
multiplied by one-hundred. The total fracture strain at fracture is 0.165;
subtracting out the elastic strain (which is about 0.005) leaves a plastic
strain of 0.160. Thus, the ductility is about 16%EL.
(f) From Equation (6.14), the modulus of resilience is just
Ur =
2
y
2E
which, using data computed in the problem, yields a value of
2
Ur =
(285 MPa)

(2) 62.5 x 103 MPa
5

= 6.5 x 10 J/m
3
93.8 in. - lbf /in.3 
6.40 We are asked to compute how much elongation a metal specimen will
experience when a true stress of 415 MPa is applied, given the value of n
and that a given true stress produces a specific true strain. Solution of
this problem requires that we utilize Equation (6.19). It is first necessary
to solve for K from the given true stress and strain. Rearrangement of
this equation yields
K =
T
( )n
T
=
345 MPa
(0.02 )0.22
= 816 MPa (118, 000 psi)
Next we must solve for the true strain produced when a true stress of 415
MPa is applied, also using Equation (6.19). Thus
1/n
 
 T =  T 
 K 
1/0 .22
 l 
415 MPa 
i
= 

= 0.0463 = ln 
l 

816 MPa 
 o 
Now, solving for li gives
li = lo e0.0 463 = (500 mm)e 0.0 463 = 523.7 mm (20.948 in.)
And finally, the elongation ∆l is just
l = li  lo = 523.7 mm  500 mm = 23.7 mm (0.948 in.)
6.D1 For this problem the working stress is computed using Equation (6.24)
with N = 2, as
w =
y
2
=
860 MPa
= 430 MPa (62, 500 psi )
2
Since the force is given, the area may be determined from Equation (6.1),
and subsequently the original diameter do may be calculated as
Ao =
d o 2
F
=   
w
 2 
And
do =
4F

=
w

(4)(13,300 N)
 430 x 106 N/ m 2

= 6.3 x 10-3 m = 6.3 mm (0.25 in.)
Chapter 7
7.11 This problem calls for us to determine whether or not a metal single
crystal having a specific orientation and of given critical resolved shear
stress will yield. We are given that  = 60,  = 35, and that the values of
the critical resolved shear stress and applied tensile stress are 6.2 MPa
(900 psi) and 12 MPa (1750 psi), respectively. From Equation (7.1)
 R =  cos  cos  = (12 MPa)(cos 60 )(cos 35) = 4.91 MPa
(717 psi)
Since the resolved shear stress (4.91 MPa) is less that the critical
resolved shear stress (6.2 MPa), the single crystal will not yield.
However, from Equation (7.3), the stress at which yielding occurs
is
y 
 crs s
cos  cos 

cos
6.2 MPa
 15.1MPa (2200 psi)
60cos 35
7.13 This problem asks that we compute the critical resolved shear stress for
silver. In order to do this, we must employ Equation (7.3), but first it is
necessary to solve for the angles  and  from the sketch below.
If the unit cell edge length is a, then
 = tan
-1 a 
  = 45
a 
For the angle , we must examine the triangle OAB. The length of line
OA is just a, whereas, the length of AB is a 2 . Thus,
 = tan
-1 a 2 

 = 54.7 
 a 
And, finally
 crs s =  y (cos  cos )
= (1.1 MPa) cos(54 .7)cos(45) = 0.45 MPa (65.4 psi)
7.15 This problem asks, for a BCC metal, that we compute the applied
stress(s) that are required to cause slip to occur in the [1 1 1] direction on
each of the (110), (011), and (10 1 ) planes. In order to solve this problem
it is necessary to employ Equation (7.3), which means that we need to
solve for the for angles  and  for the three slip systems.
In the sketch below is shown the unit cell and the (110)- [1 1 1] slip
configuration.
The angle  between the [1 1 1] slip direction and the normal to the (110)
plane (i.e., the [110] direction) is 45. Now, in order to determine the
angle between the [1 1 1] direction and the [100] direction (i.e., the
direction of stress application), , we consult the illustration of this same
unit cell as shown below.
 A B 
.
 B C 
For the triangle ABC, the angle  is equal to tan 1 
The length B C
is equal to the unit cell edge length a. From triangle ABD, A B  a 2 ,
a 2 
 = 54.7.
 a 
and, therefore,  
And, solving for the resolved shear
stress from Equation (7.1)
 R   cos cos
 (4.0 MPa)cos(45)cos(54.7)  (4.0 MPa)(0.707)(0.578)  1.63 MPa
The (011)- [1 1 1] slip configuration is represented in the following
sketch.
In this case,  has the same value as previously (i.e., 54.7); however the
value for  is 90. Thus, the resolved shear stress for this configuration is
 R   cos cos

 (4.0 MPa)cos(90)cos(54.7)  (4.0 MPa)(0)(0.578)  0 MPa
And, finally the (10 1 ) - [1 1 1] slip configuration is shown below.
Here, the value of  is 45, which again leads to
 R   cos cos
 (4.0 MPa)cos(45)cos(54.7)  (4.0 MPa)(0.707)(0.578)  1.63 MPa

(b) The most favored slip system(s) is (are) the one(s) that has (have) the
largest  R value. Both (110)- [1 1 1] and (10 1 ) - [1 1 1] slip systems are most
favored since they have the same  R (1.63 MPa), which is larger than the
 R value for (011)- [1 1 1] (viz., 0 MPa).
7.38 (a) Using the data given and Equation (7.7) and taking n = 2, we may
set up two simultaneous equations with do and K as unknowns; thus
3.9 x 10 -2 mm  d2o = (30 min)K
2
6.6 x 10-2 mm  d2o = (90 min)K
2
Solution of these expressions yields a value for do, the original grain
diameter, of
do = 0.01 mm,
Also
K = 4.73 x 10-5 mm2/min
(b) At 150 min, the diameter is computed as
d2
o  Kt
d=
=
2

(0.01 mm)  4.73 x 10
7.D5
5
2

mm / min (150 min) = 0.085 mm
This problem calls for us to explain the procedure by which a
cylindrical rod of steel may be deformed so as to produce a given final
diameter, as well as a specific tensile strength and ductility. First let us
calculate the percent cold work and attendant tensile strength and ductility
if the drawing is carried out without interruption. From Equation (7.6)
%CW =
d o 2
d d 2
      
 2 
 2 
2
d 
  o 
 2 
x 100
2
=
2
15.2 mm 
10 mm 
 
   



 2 
2
2
15.2 mm 
 


2

x 100 = 56%CW
At 56%CW, the steel will have a tensile strength on the order of 920 MPa
(133,000 psi) [Figure 7.17(b)], which is adequate; however, the ductility
will be less than 10%EL [Figure 7.17(c)], which is insufficient.
Instead of performing the drawing in a single operation, let us
initially draw some fraction of the total deformation, then anneal to
recrystallize, and, finally, cold-work the material a second time in order to
achieve the final diameter, tensile strength, and ductility.
Reference to Figure 7.17(b) indicates that 20%CW is necessary to
yield a tensile strength of 840 MPa (122,000 psi). Similarly, a maximum
of 21%CW is possible for 12%EL [Figure 7.17(c)]. The average of these
extremes is 20.5%CW. If the final diameter after the first drawing is d o' ,
then
2
20.5%CW =
 ' 
2
d
10 mm 
  o    

 2 
 2 
 
 ' 2
d
  o 
 2 
 
x 100
And, solving for d o' , yields d o' = 11.2 mm (0.45 in.).