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356                                             Chapter 8  Fracture of Cracked Members
























                                              Figure E8.1

             Comment     For (c), an iterative or graphical solution is optional in this case, but is necessary
             in other cases where a limit on α for 10% accuracy in K is exceeded.

            8.4.3 Safety Factors

            Where cracks may be present, safety factors against yielding, as examined in Chapter 7, need to be
            supplemented by safety factors against brittle fracture. Depending on the particular situation, either
            yielding or fracture might control the design.
                                                                  √
               Since stress S g and K are proportional according to K = FS g πa, a safety factor X against
            fracture for stress can be accomplished by applying the same factor to K. Hence, if S g and a are
            the stress and crack length that are expected to occur in actual service, the safety factor on K, and
            thus on S g ,is
                                               K Ic     K Ic
                                         X K =     =     √                            (8.15)
                                                K     FS g πa
            It may also be useful to compare the service crack length a with the crack length a c that is expected
            to cause failure at the service stress S g .The valueof a c is available from
                                                      √
                                            K Ic = F c S g πa c                       (8.16)
            where F c is evaluated at a c . Combining the previous two equations leads to the following safety
            factor on crack length:
                                                            2

                                               a c    F
                                          X a =   =     X K                           (8.17)
                                                a     F c
            Because X K is squared, safety factors on crack length must be rather large to achieve reasonable
            safety factors on K and stress.
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