9.2
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 m
using Equation 9.1 given that 0 = 1200 MPa, a = 0.25 mm, and t = 1.2  10-3 mm.
Thus,
 a 1/ 2
 m = 2  0  
 t 
1/2
 0.25 mm 
= (2)(1200 MPa) 
 3

1.2  10 mm 
= 3.5  10 4 MPa = 35 GPa (5.1  106 psi)
9.7
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane
strain fracture toughness of 40 MPa m (36.4 ksi in.).
It has been determined that fracture results at a
stress of 365 MPa (53,000 psi) when the maximum internal crack length is 2.5 mm (0.10 in.).
For this same
component and alloy,
compute thestress level at which fracture will occur for a critical internal crack length of
4.0 mm (0.16 in.).
Solution
This problem asks us to determine the stress level at which an a wing component on an aircraft will
fracture for a given fracture toughness (40 MPa m ) and maximum internal crack length (4.0 mm), given
that fracture occurs for the same component using the same alloy at one stress level (365 MPa) and another
internal crack length (2.5 mm).
It first becomes necessary to solve for the parameter Y for the conditions under
which fracture occurred using Equation 9.5.
Y =
K Ic
=
 a
Therefore,
40 MPa m
= 1.75
2.5  103 m 
(365 MPa) () 

2


Now we will solve for c using Equation 9.6 as

c =

Y
K Ic
=
a
40 MPa m
= 288 MPa (41, 500 psi)
4  103 m 
(1.75) () 

2


9.21
The fatigue data for a brass alloy are given as follows:
Stress Amplitude (MPa)
Cycles to Failure
310
2 × 105
223
1 × 106
191
3 × 106
168
1 × 107
153
3 × 107
143
1 × 108
134
3 × 108
127
1 × 109
(a) Make an S–N plot (stress amplitude versus logarithm of cycles to failure) using these data.
(b) Determine the fatigue strength at 5  105 cycles.
(c) Determine the fatigue life for 200 MPa.
Solution
(a)
The fatigue data for this alloy are plotted below.
(b)
As indicated by the “A” set of dashed lines on the plot, the fatigue strength at 5  105 cycles [log
(5  105) = 5.7] is about 250 MPa.
(c)
As noted by the “B” set of dashed lines, the fatigue life for 200 MPa is about 2  106 cycles (i.e.,
the log of the lifetime is about 6.3).
10.12
A 40 wt% Ni-60 wt% Cu alloy is slowly cooled from 1400C to 1150C.
(a) At what temperature does the first solid phase form?
(b) What is the composition of this solid phase?
(c) At what temperature does the liquid solidify?
(d) What is the composition of this last remaining liquid phase?
Solution
Shown below is the Cu-Ni phase diagram (Figure 10.3a) and a vertical line constructed at a
composition of 40 wt% Ni-60 wt% Cu.
(a)
Upon cooling from 1400C, the first solid phase forms at the temperature at which a vertical line
at this composition intersects the L-( + L) phase boundary--i.e., about 1280C;
(b)
The composition of this solid phase corresponds to the intersection with the L-( + L) phase
boundary, of a tie line constructed across the  + L phase region at 1320C--i.e., 53 wt% Ni-47 wt% Cu;
(c)
Complete solidification of the alloy occurs at the intersection of this same vertical line at 40 wt%
Ni with the (α + L)–α phase boundary--i.e., at about 1230C;
(d)
The composition of the last liquid phase remaining prior to complete solidification corresponds
to the intersection with the L–(α + L) boundary, of the tie line constructed across the α + L phase region at
1270°C--i.e., CL about 29 wt% Ni-71 wt% Cu.
10.17
A 90 wt% Ag-10 wt% Cu alloy is heated to a temperature within the  + liquid phase region.
If the
composition of the liquid phase is 85 wt% Ag, determine:
(a) The temperature of the alloy
(b) The composition of the  phase
(c) The mass fractions of both phases
Solution
(a)
In order to determine the temperature of a 90 wt% Ag-10 wt% Cu alloy for which  and liquid
phases are present with the liquid phase of composition 85 wt% Ag, we need to construct a tie line across the 
+ L phase region of Figure 10.7 that intersects the liquidus line at 85 wt% Ag;
this is possible at about 850C.
(b) The composition of the  phase at this temperature is determined from the intersection of this same
tie line with solidus line, which corresponds to about 95 wt% Ag.
(c) The mass fractions of the two phases are determined using the lever rule, Equations 10.1 and
10.2 with C0 = 90 wt% Ag, CL = 85 wt% Ag, and C = 95 wt% Ag, as
W =


WL =
C0  CL
90  85
=
= 0.50
C  CL
95  85
C  C0
C  CL
=
95  90
= 0.50
95  85
10.32
For a lead-tin alloy of composition 78 wt% Sn-22 wt% Pb and at 180C do the following:
(a) Determine the mass fraction of α phase.
(b) Determine the mass fraction of  phase.
(c) Determine the mass fraction of primary  microconstituent.
(d) Determine the mass fraction of eutectic microconstituent.
(e) Determine the mass fraction of eutectic .
Solution
(a & b) This portion of the problem asks that we determine the mass fractions of  and  phases for an
78 wt% Sn-22 wt% Pb alloy (at 180C).
In order to do this it is necessary to employ the lever rule using a tie
line that extends entirely across the  +  phase field.
= 97.8 wt% Sn, and Ceutectic = 61.9 wt% Sn.
From Figure 10.7 and at 180C, C = 18.3 wt% Sn, C
Therefore, the two lever-rule expressions are as follows:
W =
(a)
W =
(b)
C   C0
C   C
=
97 .8  78
= 0.249
97 .8  18 .3
C 0  C
78  18.3
=
= 0.751
C   C 97.8  18.3
(c & d) Now it is necessary to determine the mass fractions of primary  and eutectic
microconstituents for this same alloy.
This requires us to utilize the lever rule and a tie line that extends from
the maximum solubility of Pb in the  phase at 180C (i.e., 97.8 wt% Sn) to the eutectic composition (61.9 wt%
Sn).
Thus
(c) And, finally, we are asked to compute the mass fraction of eutectic , We. This quantity is simply
the difference between the mass fractions of total  and primary  as
We = W – W' = 0.751 – 0.448 = 0.303
10.53
Consider 1.0 kg of austenite containing 1.15 wt% C and cooled to below 727C (1341F).
(a) What is the proeutectoid phase?
(b) How many kilograms each of total ferrite and cementite form?
(c) How many kilograms each of pearlite and the proeutectoid phase form?
(d) Schematically sketch and label the resulting microstructure.
Solution
(a)
The proeutectoid phase will be Fe3C since 1.15 wt% C is greater than the eutectoid composition
(0.76 wt% C).
(b)
form.
For this portion of the problem, we are asked to determine how much total ferrite and cementite
Application of the appropriate lever rule (of the form of Equation 10.12) expression yields
W =
CFe3C  C 0
CFe3C  C
6.70  1.15
= 0.83
6.70  0.022
=
which, when multiplied by the total mass of the alloy (1.0 kg), gives 0.83 kg of total ferrite.

Similarly, for total cementite,
WFe3C =
C0  C
1.15  0.022
=
= 0.17
CFe3C  C
6.70  0.022
And the mass of total cementite that forms is (0.17)(1.0 kg) = 0.17 kg.

(c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form.
Applying Equation 10.22, in which C 1' = 1.15 wt% C

Wp =
which corresponds to a mass of 0.93 kg.

6.70  C 1'
6.70  0.76
=
6.70  1.15
= 0.93
6.70  0.76
Likewise, from Equation 10.23
WFe3C' =
C1'  0.76
5.94
=
1.15  0.76
= 0.07
5.94
which is equivalent to 0.07 kg of the total 1.0 kg mass.

(d) Schematically, the microstructure would appear as:
10.59
Consider 2.9 kg of a 99.5 wt% Fe–0.5 wt% C alloy that is cooled to a temperature just below the
eutectoid.
(a) How many kilograms of proeutectoid ferrite form?
(b) How many kilograms of eutectoid ferrite form?
(c) How many kilograms of cementite form?
Solution
In this problem we are asked to consider 2.9 kg of a 99.5 wt% Fe-0.5 wt% C alloy that is cooled to a
temperature below the eutectoid.
(a)
Equation 10.21 must be used in computing the amount of proeutectoid ferrite that forms.
W ’ =
Thus,
0.76  C 0’ 0.76  0.50
=
= 0.351
0.74
0.74
Or, (0.351)(2.9 kg) = 1.02 kg of proeutectoid ferrite forms.
(b) In order to determine the amount of eutectoid ferrite, it first becomes necessary to compute the
amount of total ferrite using the lever rule applied entirely across the  + Fe C phase field (of the form of
3
Equation 10.12), as
W =
C Fe3C  C 0’
C Fe3C  C
which corresponds to (0.928)(2.9 kg) = 2.692 kg.
=
6.70  0.50
= 0.928
6.70  0.022
Now, the amount of eutectoid ferrite is just the difference
between total and proeutectoid ferrites, or
2.692 kg – 1.02 kg = 1.67 kg
(c) With regard to the amount of cementite that forms, again application of the lever rule across the
entirety of the  + Fe3C phase field, leads to
WFe3C =
C 0’  C
0.50  0.022
=
= 0.072
C Fe3C  C 6.70  0.022
which amounts to (0.072)(2.9 kg) = 0.208 kg cementite in the alloy.