8.17 A 12.5 mm (0.50 in.) diameter cylindrical rod fabricated from a 2014-T6 alloy (Figure 8.34) is
subjected to a repeated tension-compression load cycling along its axis. Compute the maximum and
minimum loads that will be applied to yield a fatigue life of 1.0 ! 107 cycles. Assume that the stress plotted
on the vertical axis is stress amplitude, and data were taken for a mean stress of 50 MPa (7250 psi).
Solution
This problem asks that we compute the maximum and minimum loads to which a 12.5 mm (0.50
in.) diameter 2014-T6 aluminum alloy specimen may be subjected in order to yield a fatigue life of 1.0 !
107 cycles; Figure 8.34 is to be used assuming that data were taken for a mean stress of 50 MPa (7250 psi).
Upon consultation of Figure 8.34, a fatigue life of 1.0 ! 107 cycles corresponds to a stress amplitude of 160
MPa (23,200 psi). Or, from Equation 8.16
! max " ! min = 2! a = (2)(160 MPa) = 320 MPa (46, 400 psi)
Since "m = 50 MPa, then from Equation 8.14
! max + ! min = 2! m = (2)(50 MPa) = 100 MPa (14, 500 psi)
Simultaneous solution of these two expressions for "max and "min yields
"max = +210 MPa (+30,400 psi)
"min = –110 MPa (–16,000 psi)
Now, inasmuch as ! =
Fmax =
Fmin =
" d %2
F
(Equation 6.1), and A0 = ! $ 0 ' then
#2&
A0
! max " d 20 (210 # 10 6 N / m2 ) (") (12.5 # 10$3 m)
=
4
4
! min " d 20 (#110 $ 10 6 N / m2 ) (") (12.5 $ 10#3 m)
=
4
4
2
= 25,800 N (6000 lb f )
2
= # 13,500 N (#3140 lb f )
8.24 Briefly explain the difference between fatigue striations and beachmarks both in terms of (a) size and
(b) origin.
Solution
(a)
With regard to size, beachmarks are normally of macroscopic dimensions and may be
observed with the naked eye; fatigue striations are of microscopic size and it is necessary to observe them
using electron microscopy.
(b) With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue
striation is corresponds to the advance of a fatigue crack during a single load cycle.
8.25 List four measures that may be taken to increase the resistance to fatigue of a metal alloy.
Solution
Four measures that may be taken to increase the fatigue resistance of a metal alloy are:
(1) Polish the surface to remove stress amplification sites.
(2) Reduce the number of internal defects (pores, etc.) by means of altering processing and
fabrication techniques.
(3) Modify the design to eliminate notches and sudden contour changes.
(4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot
peening.
12.43 A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum
possible radius of the specimen without fracture, given that the applied load is 425 N (95.5 lbf), the flexural
strength is 105 MPa (15,000 psi), and the separation between load points is 50 mm (2.0 in.).
Solution
We are asked to calculate the maximum radius of a circular specimen of MgO that is loaded using
three-point bending. Solving for R from Equation 12.7b
# F L &1/3
f (
R =%
%$ ! fs " ('
which, when substituting the parameters stipulated in the problem statement, yields
$ (425 N) (50 ! 10"3 m) '1/3
R =&
)
% (105 ! 10 6 N / m2 ) (#) (
= 4.0 ! 10
-3
m = 4.0 mm (0.16 in.)
12.45 (a) A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide
having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5 mm (0.10 in.)
and the support point separation distance is 30 mm (1.2 in.), predict whether or not you would expect the
specimen to fracture when a load of 620 N (140 lbf) is applied. Justify your prediction.
(b) Would you be 100% certain of the prediction in part (a)? Why or why not?
Solution
(a) This portion of the problem asks that we determine whether or not a cylindrical specimen of
aluminum oxide having a flexural strength of 390 MPa (56,600 psi) and a radius of 2.5 mm will fracture
when subjected to a load of 620 N in a three-point bending test; the support point separation is given as 30
mm. Using Equation 12.7b we will calculate the value of "; if this value is greater than "fs (390 MPa),
then fracture is expected to occur. Employment of Equation 12.7b yields
! =
=
FL
"R 3
(620 N) (30 ! 10"3 m)
= 379 ! 10 6 N/m2 = 379 MPa (53,500 psi)
(#) (2.5 ! 10"3 m) 3
Since this value is less than the given value of "fs (390 MPa), then fracture is not predicted.
(b) The certainty of this prediction is not 100% because there is always some variability in the
flexural strength for ceramic materials, and since this value of " (379 MPa) is relatively close to "fs (390
MPa) then there is some chance that fracture will occur.
12.D1 Gallium arsenide (GaAs) and gallium phosphide (GaP) both have the zinc blende crystal structure
and are soluble in one another at all concentrations. Determine the concentration in weight percent of GaP
that must be added to GaAs to yield a unit cell edge length of 0.5570 nm. The densities of GaAs and GaP
are 5.316 and 4.130 g/cm3, respectively.
Solution
This problem asks that we determine the concentration (in weight percent) of GaP that must be
added to GaAs to yield a unit cell edge length of 0.5570 nm. The densities of GaAs and GaP were given in
the problem as 5.307 and 4.130 g/cm3, respectively. To begin, it is necessary to employ Equation (13.1),
and solve for the unit cell volume, VC, for the GaP-GaAs alloy as
VC =
n' Aave
!ave N A
where Aave and #ave are the atomic weight and density, respectively, of the InAs-GaAs alloy. Inasmuch as
both of these materials have the zinc blende crystal structure, which has cubic symmetry, VC is just the
cube of the unit cell length, a. That is
VC = a3 = (0.5570 nm)3
= (5.570 ! 10"8 cm)3 = 1.728 ! 10"22 cm3
It is now necessary to construct expressions for Aave and #ave in terms of the concentration of indium
arsenide, CInAs using Equations 4.11a and 4.10a. For Aave we have
Aave =
=
100
CGaP
+
AGaP
GaP
)
AGaAs
100
C
GaP
100.69 g/mol
whereas for #ave
! C
(100
+
(100
! C
GaP
)
144.64 g/mol
! ave =
=
100
CGaP
+
!GaP
" CGaP )
!GaAs
(100
100
CGaP
+
4.130 g/cm3
! CGaP )
5.316 g/cm3
(100
Within the zinc blende unit cell there are four formula units, and thus, the value of n' in Equation 12.1 is 4;
hence, this expression may be written in terms of the concentration of GaP in weight percent as follows:
VC = 1.728 ! 10-22 cm3
=
n' Aave
!ave N A
"
%
$
'
100
'
(4 formula units/unit cell) $
CGaP
(
100 ! CGaP ) '
$
+
$# 100.69 g/mol
144.64 g/mol '&
=
"
%
$
'
100
$
' (6.022 ( 10 23 formula units/mol)
CGaP
(
100 ! CGaP ) '
$
$# 4.130 g/cm3 + 5.316 g/cm3 '&
And solving this expression for CGaP leads to CGaP = 33.7 wt%.