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CHAPTER11

FAILURE

PROBLEM SOLUTIONS

Principles of Fracture Mechanics

11.1 This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an

internal crack. Equation 11.1 is employed to solve this problem, as

m = 20a

t

1/ 2

= (2)(140 MPa)

3.8 x 102 mm

2

1.9 x 104 mm

1/2

= 2800 MPa (400,000 psi)

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11.2 In order to estimate the theoretical fracture strength of this material it is necessary to calculate m

using Equation 11.1 given that 0 = 1035 MPa, a = 0.5 mm, andt= 5 x 10-3 mm. Thus,

m

= 20

a

t

1/ 2

= (2)(1035 MPa)0.5 mm

5 x 103 mm

1/2

= 2.07 x 104 MPa = 207 GPa (3 x 106 psi)

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11.3 We may determine the critical stress required for the propagation of an internal crack in aluminum

oxide using Equation 11.3; taking the value of 393 GPa (Table 12.3) as the modulus of elasticity, we get

c =2Es a

1/ 2

=(2)(393 x 109 N /m2) (0.90 N /m)

()0.4 x 103 m

2

1/ 2

= 33.6 x106 N/m2 = 33.6 MPa

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11.4 The maximum allowable surface crack length for MgO may be determined using Equation 11.3; taking

225 GPa as the modulus of elasticity (Table 12.3), and solving fora, leads to

a =2Es

c2

=(2)(225 x 109 N / m2) (1.0 N /m)

()(13.5 x 106 N / m2

)

2

= 7.9 x 10-4

m = 0.79 mm (0.031 in.)

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11.5 This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when

exposed to a stress of 1030 MPa, given the values ofKIc, Y, and the largest value ofa in the material. This requires

that we solve forc from Equation 11.6. Thus

c =

K

IcY a =

54.8 MPa m

(1.0) ()(0.5 x 103 m)= 1380 MPa (199,500 psi)

Therefore, fracture will not occur because this specimen will tolerate a stress of 1380 MPa (199,500 psi) before

fracture, which is greater than the applied stress of 1030 MPa (150,000 psi).

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11.6 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. It first becomes necessary

to solve for the parameter Y, using Equation 11.5, for the conditions under which fracture occurred (i.e., = 300

MPa anda = 4.0 mm). Therefore,

Y =K

Ic

a=

40 MPa m

(300 MPa ) ()4 x 103 m

2

= 1.68

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,

Y a = (1.68)(260 MPa) ()6 x 103 m

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 .

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11.7 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 (26 MPa m ) and maximum internal crack length (6.0 mm), given that

fracture occurs for the same component using the same alloy at one stress level (112 MPa) and another internal

crack length (8.6 mm). It first becomes necessary to solve for the parameterY for the conditions under which

fracture occurred using Equation 11.5. Therefore,

Y =KIc

a=

26 MPa m

(112 MPa) ()8.6 x 103 m

2

= 2.0

Now we will solve forc

using Equation 11.6 as

c =KIc

Y a

=26 MPa m

(2.0) () 6 x 103 m2

= 134 MPa (19,300 psi)

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11.8 For this problem, we are given values ofKIc (82.4 MPa m ) , (345MPa), andY(1.0) for a largeplate and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is

to solve forac

using Equation 11.7; therefore

ac =1

KIc

Y

2

= 1

82.4 MPa m

(1.0)(345 MPa)

2

= 0.0182 m = 18.2 mm (0.72 in.)

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11.9 This problem asks us to calculate the maximum internal crack length allowable for the Ti-6Al-4V

titanium alloy in Table 11.1 given that it is loaded to a stress level equal to one-half of its yield strength. For this

alloy, KIc = 55 MPa m (50 ksi in. ) ; also, = y/2 = (910 MPa)/2 = 455 MPa (66,000 psi). Now solving for

2ac

using Equation 11.7 yields

2ac =2

KIc

Y

2

=2

55 MPa m

(1.5)(455 MPa)

2

= 0.0041 m = 4.1 mm (0.16 in.)

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11.10 This problem asks that we determine whether or not a critical flaw in a wide plate is subject to

detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m ) , the design

stress (y/2 in which y = 860 MPa), andY= 1.0. We first need to compute the value ofac using Equation 11.7;

thus

ac =1

KIc

Y

2

=1

98.9 MPa m

(1.0)860 MPa

2

2

= 0.0168 m = 16.8 mm (0.66 in.)

Therefore, the critical flaw is subject to detection since this value of ac

(16.8 mm) is greater than the 3.0 mm

resolution limit.

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