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 0.5 mm (0.02 in.) and a tip radius of curvature of 5 × 10–3 mm (2 ×
10–4 in.), when a stress of 1035 MPa (150,000 psi) is applied.

Solution

In order to estimate the theoretical fracture strength of this material it is necessary to calculate sm using
Equation 8.1 given that s0 = 1035 MPa, a = 0.5 mm, and rt = 5 ´ 10-3 mm. Thus,

1/2
⎛ a⎞
σ m = 2σ 0 ⎜ ⎟
⎝ ρt ⎠

1/2
⎛ 0.5 mm ⎞
= (2)(1035 MPa) ⎜ ⎟
⎝ 5 × 10−3 mm ⎠

= 2.07 × 104 MPa = 20.7 GPa (3 × 106 psi)

 8.4 An MgO component must not fail when a tensile stress of 13.5 MPa (1960 psi) is applied. Determine the
maximum allowable surface crack length if the surface energy of MgO is 1.0 J/m2. Data found in Table 12.5 may
prove helpful.

Solution

The maximum allowable surface crack length for MgO may be determined using a rearranged form of
Equation 8.3. Taking 225 GPa as the modulus of elasticity (Table 12.5), and realizing that values of sc (13.5 MPa)
and gs (1.0 J/m2) are given in the problem statement, we solve for a, as follows:

2 Eγ s
a=
π σ c2

(2) ( 225 × 109 N/m 2 ) (1.0 N/m)

=
2
(π) (13.5 × 106 N/m 2 )

= 7.9 × 10−4 m = 0.79 mm (0.031 in.)

8.6 An aircraft component 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 300 MPa (43,500 psi) when the
maximum (or critical) internal crack length is 4.0 mm (0.16 in.). For this same component and alloy, will fracture
occur at a stress level of 260 MPa (38,000 psi) when the maximum internal crack length is 6.0 mm (0.24 in.)? Why or
why not?

Solution

We are asked to determine if an aircraft component will fracture for a given fracture toughness (40 MPa m
), stress level (260 MPa), and maximum internal crack length (6.0 mm), given that fracture occurs for the same
component using the same alloy for another stress level and internal crack length. (Note: Because the cracks are
internal, their lengths are equal to 2a.) It first becomes necessary to solve for the parameter Y, using Equation 8.5, for
the conditions under which fracture occurred (i.e., s = 300 MPa and 2a = 4.0 mm). Therefore,

K Ic 40 MPa m
Y= = = 1.68
σ πa ⎛ 4 × 10−3 m ⎞
(300 MPa) (π) ⎜ ⎟
⎝ 2 ⎠

Now we will solve for the product Y σ πa for the other set of conditions, so as to ascertain whether or not this value
is greater than the KIc for the alloy. Thus,

⎛ 6 × 10−3 m ⎞
Y σ π a = (1.68)(260 MPa) (π ) ⎜ ⎟
⎝ 2 ⎠

= 42.4 MPa m (39 ksi in.)

Therefore, fracture will occur since this value ( 42.4 MPa m ) is greater than the KIc of the material, 40 MPa m .

Cyclic Stresses
The S–N Curve

8.16 A fatigue test was conducted in which the mean stress was 70 MPa (10,000 psi), and the stress amplitude
was 210 MPa (30,000 psi).
(a) Compute the maximum and minimum stress levels.
(b) Compute the stress ratio.
(c) Compute the magnitude of the stress range.

Solution

(a) Given the values of sm (70 MPa) and sa (210 MPa) we are asked to compute smax and smin. From

Equation 8.14

σ max + σ min
σm = = 70 MPa
2
Or,

σ max + σ min = 140 MPa (8.14a)

Furthermore, utilization of Equation 8.16 yields

σ max − σ min
σa = = 210 MPa
2

Or,

σ max − σ min = 420 MPa (8.16a)

Simultaneously solving Equations 8.14a and 8.16a leads to

σ max = 280 MPa (40,000 psi)

σ min = − 140 MPa (−20, 000 psi)

(b) Using Equation 8.17 the stress ratio R is determined as follows:

σ min −140 MPa

R= = = − 0.50
σ max 280 MPa

(c) The magnitude of the stress range sr is determined using Equation 8.15 as

σ r = σ max − σ min = 280 MPa − ( − 140 MPa)

 = 420 MPa (60,000 psi)

8.19 A cylindrical 2014-T6 aluminum alloy bar is subjected to compression-tension stress cycling along its axis;
results of these tests are shown in Figure 8.20. If the bar diameter is 12.0 mm, calculate the maximum allowable load

amplitude (in N) to ensure that fatigue failure will not occur at 107 cycles. Assume a factor of safety of 3.0, data in
Figure 8.20 were taken for reversed axial tension-compression tests, and that S is stress amplitude.

Solution

From Figure 8.20, the fatigue strength at 107 cycles for this aluminum alloy is 170 MPa. For a cylindrical
specimen having an original diameter of d0, the stress may be computed using Equation 6.1, which is equal to

F F
σ= =
A0 2
⎛d ⎞
π⎜ 0⎟
⎝ 2⎠

4F
=
π d02

When we divide s by the factor of safety, the above equation takes the form

σ 4F
=
N πd2
0

Solving the above equation for F leads to

σπ d02
F=
4N

Now taking s to be the fatigue strength (i.e., 170 MPa = 170 ´ 106 N/m2) and incorporating values for d0 and N

provided in the problem statement [i.e., 12.0 mm (12.0 ´ 10-3 m) and 3.0, respectively], we calculate the maximum
load as follows:

(170 × 106 N/m 2 )(π )(12.0 × 10−3 m)2

F=
(4)(3.0)

6,400 N

8.20 A cylindrical rod of diameter 6.7 mm fabricated from a 70Cu-30Zn brass alloy is subjected to rotating-bending
load cycling; test results (as S-N behavior) are shown in Figure 8.20. If the maximum and minimum loads are +120
N and –120 N, respectively, determine its fatigue life. Assume that the separation between loadbearing points is 67.5
mm.

Solution

In order to solve this problem we compute the maximum stress using Equation 8.18, and then determine the
fatigue life from the curve in Figure 8.20 for the brass material. In Equation 8.18, F is the maximum applied load,
which for this problem is +120 N. Values for L and d0 are provided in the problem statement—viz. 67.5 mm (67.5 ´

10-3 m) and 6.7 mm (6.7 ´ 10-3 m), respectively. Therefore the maximum stress s is equal to

16FL
σ=
π d03

(16)(120 N)(67.5 × 10−3 m)

=
(π )(6.7 × 10−3 m)3

= 137 × 106 N/m 2 = 137 MPa

From Figure 8.20 and the curve for brass, the logarithm of the fatigue life (log Nf) at 137 MPa is about 6.5, which

means that the fatigue life is equal to

N f = 106.5 cycles = 3 × 106 cycles

Stress and Temperature Effects

8.32 A specimen 975 mm (38.4 in.) long of an S-590 alloy (Figure 8.32) is to be exposed to a tensile stress
of 300 MPa (43,500 psi) at 730°C (1350°F). Determine its elongation after 4.0 h. Assume that the total of both
instantaneous and primary creep elongations is 2.5 mm (0.10 in.).

Solution

From the 730°C line in Figure 8.32, the steady state creep rate εs is about 1.0 ´ 10-2 h-1 at 300 MPa. The
steady state creep strain, es, therefore, is just the product of εs and time as

ε s = εs × (time)

= (1.0 × 10−2 h −1 )(4.0 h) = 0.040

Strain and elongation are related as in Equation 6.2—viz.

Δl
ε=
l0

Now, using the steady-state elongation, l0, provided in the problem statement (975 mm) and the value of es, determined
above (0.040), the steady-state elongation, Dls, is determined as follows:

Δls = l0ε s

= (975 mm)(0.040) = 39.0 mm (1.54 in.)

Finally, the total elongation is just the sum of this Dls and the total of both instantaneous and primary creep elongations

[i.e., 2.5 mm (0.10 in.)]. Therefore, the total elongation is 39.0 mm + 2.5 mm = 41.5 mm (1.64 in.).