6.13 In Section 2.6 it was noted that the net bonding energy EN between two isolated positive and negative
ions is a function of interionic distance r as follows:
A B
+
r rn
EN = !
where A, B, and n are constants for the particular ion pair. Equation 6.25 is also valid for the bonding
energy between adjacent ions in solid materials. The modulus of elasticity E is proportional to the slope of
the interionic force–separation curve at the equilibrium interionic separation; that is,
" dF %
E !$ '
# dr &r
o
Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for
the two-ion system) using the following procedure:
1. Establish a relationship for the force F as a function of r, realizing that
F=
dEN
dr
2. Now take the derivative dF/dr.
3. Develop an expression for r0, the equilibrium separation. Since r0 corresponds to the value of r
at the minimum of the EN-versus-r curve (Figure 2.8b), take the derivative dEN/dr, set it equal to zero, and
solve for r, which corresponds to r0.
4. Finally, substitute this expression for r0 into the relationship obtained by taking dF/dr.
Solution
This problem asks that we derive an expression for the dependence of the modulus of elasticity, E,
on the parameters A, B, and n in Equation 6.25. It is first necessary to take dEN/dr in order to obtain an
expression for the force F; this is accomplished as follows:
"B%
" A%
d$ '
d $! '
dE N
# r&
# n&
F =
=
+ r
dr
dr
dr
=
A
r2
!
nB
r (n +1)
The second step is to set this dE /dr expression equal to zero and then solve for r (= r0). The algebra for
N
this procedure is carried out in Problem 2.14, with the result that
(6.25)
! A $1/(1 ' n)
r0 = # &
" nB %
Next it becomes necessary to take the derivative of the force (dF/dr), which is accomplished as follows:
! A$
!
nB $
d# &
d #'
&
dF
"r2 %
" r (n +1) %
=
+
dr
dr
dr
=!
2A
r3
+
(n)(n + 1)B
r (n + 2)
Now, substitution of the above expression for r0 into this equation yields
! dF $
2A
(n)(n + 1) B
+
# & ='
3/(1'
n)
" dr %r
! A$
! A $(n + 2) /(1' n)
0
# &
# &
" nB %
" nB %
which is the expression to which the modulus of elasticity is proportional
6.23 A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile
load of 27,500 N. It must not experience either plastic deformation or a diameter reduction of more than
7.5 × 10-3 mm. Of the materials listed as follows, which are possible candidates? Justify your choice(s).
Modulus of Elasticity
(GPa)
Yield Strength
(MPa)
Poisson’s Ratio
Aluminum alloy
70
200
0.33
Brass alloy
101
300
0.34
Steel alloy
207
400
0.30
Titanium alloy
107
650
0.34
Material
Solution
This problem asks that we assess the four alloys relative to the two criteria presented. The first
criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is
applied; this means that the stress corresponding to this load not exceed the yield strength of the material.
Upon computing the stress
! =
F
=
A0
F
#d &
"% 0 (
$2'
2
=
27,500 N
# 10 ) 10*3 m &2
"%
(
2
$
'
= 350 ) 10 6 N/m2 = 350 MPa
Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.
Relative to the second criterion (i.e., that Δd be less than 7.5 × 10-3 mm), it is necessary to
calculate the change in diameter ∆d for these three alloys. From Equation 6.8
$d
d
E $d
! = "
= " 0 ="
%
#z
% d0
E
#x
Now, solving for ∆d from this expression,
!d = "
# $ d0
E
For the steel alloy
!d = "
(0.30)(350 MPa)(10 mm)
= " 5.1 # 10 -3 mm
207 # 10 3 MPa
Therefore, the steel is a candidate.
For the Ti alloy
!d = "
(0.34)(350 MPa)(10 mm)
= " 11.1 # 10 -3 mm
107 # 10 3 MPa
Hence, the titanium alloy is not a candidate.
8.2 Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the
propagation of an elliptically shaped surface crack of length 0.25 mm (0.01 in.) and having a tip radius of
curvature of 1.2 × 10-3 mm (4.7 × 10-5 in.) when a stress of 1200 MPa (174,000 psi) is applied.
Solution
In order to estimate the theoretical fracture strength of this material it is necessary to calculate σm
using Equation 8.1 given that σ0 = 1200 MPa, a = 0.25 mm, and ρt = 1.2 × 10-3 mm. Thus,
# a &1/ 2
! m = 2! 0 % (
$ "t '
# 0.25 mm &1/2
= (2)(1200 MPa)%
( = 3.5 ! 10 4 MPa = 35 GPa ( 5.1 ! 10 6 psi)
$ 1.2 ! 10"3 mm '
8.6 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture
toughness of 35 MPa m (31.9 ksi in. ). It has been determined that fracture results at a stress of 250
MPa (36,250 psi) when the maximum (or critical) internal crack length is 2.0 mm (0.08 in.). For this same
component and alloy, will fracture occur at a stress level of 325 MPa (47,125 psi) when the maximum
internal crack length is 1.0 mm (0.04 in.)? Why or why not?
Solution
We are asked to determine if an aircraft component will fracture for a given fracture toughness (35
MPa m ), stress level (325 MPa), and maximum internal crack length (1.0 mm), given that fracture occurs
for the same component using the same alloy for another stress level and internal crack length. It first
becomes necessary to solve for the parameter Y, using Equation 8.5, for the conditions under which fracture
occurred (i.e., σ = 250 MPa and 2a = 2.0 mm). Therefore,
Y =
K Ic
=
! "a
35 MPa m
= 2.50
% 2 # 10$3 m (
(250 MPa) (") '
*
2
&
)
Now we will solve for the product Y ! "a for the other set of conditions, so as to ascertain whether or
not this value is greater than the KIc for the alloy. Thus,
% 1 # 10$3 m (
Y ! " a = (2.50)(325 MPa) (") '
*
2
&
)
= 32.2 MPa m
(29.5 ksi in.)
Therefore, fracture will not occur since this value (32.3 MPa m ) is less than the KIc of the material,
35 MPa m .
8.9 Calculate the maximum internal crack length allowable for a 7075-T651 aluminum alloy (Table 8.1)
component that is loaded to a stress one half of its yield strength. Assume that the value of Y is 1.35.
Solution
This problem asks us to calculate the maximum internal crack length allowable for the 7075-T651
aluminum alloy in Table 8.1 given that it is loaded to a stress level equal to one-half of its yield strength.
For this alloy, K Ic = 24 MPa
m
(22 ksi
in. ) ; also, σ = σy/2 = (495 MPa)/2 = 248 MPa (36,000 psi).
Now solving for 2ac using Equation 8.7 yields
2
2
2 # K Ic &
2 ) 24 MPa m ,
2ac = %
. = 0.0033 m = 3.3 mm (0.13 in.)
( = +
! $ Y" '
! * (1.35)(248 MPa) -
8.12 Following is tabulated data that were gathered from a series of Charpy impact tests on a ductile cast
iron.
Temperature (°C)
Impact Energy (J)
–25
124
–50
123
–75
115
–85
100
–100
73
–110
52
–125
26
–150
9
–175
6
(a) Plot the data as impact energy versus temperature.
(b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the
average of the maximum and minimum impact energies.
(c) Determine a ductile-to-brittle transition temperature as that temperature at which the impact
energy is 80 J.
Solution
(a) The plot of impact energy versus temperature is shown below.
(b) The average of the maximum and minimum impact energies from the data is
Average =
124 J + 6 J
= 65 J
2
As indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature
according to this criterion is about –105°C.
(c) Also, as noted on the plot by the other set of dashed lines, the ductile-to-brittle transition
temperature for an impact energy of 80 J is about –95°C.
8.16 An 8.0 mm (0.31 in.) diameter cylindrical rod fabricated from a red brass alloy (Figure 8.34) is
subjected to reversed tension-compression load cycling along its axis.
If the maximum tensile and
compressive loads are +7500 N (1700 lbf) and -7500 N (-1700 lbf), respectively, determine its fatigue life.
Assume that the stress plotted in Figure 8.34 is stress amplitude.
Solution
We are asked to determine the fatigue life for a cylindrical red brass rod given its diameter (8.0
mm) and the maximum tensile and compressive loads (+7500 N and -7500 N, respectively). The first thing
that is necessary is to calculate values of σmax and σmin using Equation 6.1. Thus
! max =
=
7500 N
$ 8.0 " 10#3 m '2
(!) &
)
2
%
(
Fmax
=
A0
!7500 N
$ 8.0 # 10!3 m '2
(") &
)
2
%
(
# d &2
"% 0 (
$2'
= 150 " 10 6 N/m2 = 150 MPa (22, 500 psi)
! min =
=
Fmax
Fmin
# d &2
"% 0 (
$2'
= ! 150 # 10 6 N/m2 = ! 150 MPa (!22,500 psi)
Now it becomes necessary to compute the stress amplitude using Equation 8.16 as
!a =
! max " ! min 150 MPa " ("150 MPa)
=
= 150 MPa (22,500 psi)
2
2
From Figure 8.34, f for the red brass, the number of cycles to failure at this stress amplitude is about 1 ×
105 cycles.