Callister Ch11

download Callister Ch11

of 45

  • date post

    14-Apr-2018
  • Category

    Documents

  • view

    286
  • download

    14

Embed Size (px)

Transcript of Callister Ch11

  • 7/29/2019 Callister Ch11

    1/45

    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)

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    2/45

    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)

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    3/45

    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

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    4/45

    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.)

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    5/45

    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).

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    6/45

    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 .

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    7/45

    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)

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    8/45

    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.)

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    9/45

    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.)

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  • 7/29/2019 Callister Ch11

    10/45

    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.

    Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only tostudents enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permittedby Sections 107 or 108 of the 1976 United States Copyright Act without the permi