Page 317 - Mechanical Behavior of Materials
P. 317

318                               Chapter 7  Yielding and Fracture under Combined Stresses


             where X CM = 1.00 is substituted so that the point of fracture is analyzed. Solving for τ xy yields



                                            τ xy = 134.6MPa                           Ans.

             This value gives σ 1 ,σ 2 =−150 ± τ xy =−15.4, −284.6 MPa. Hence, Eq. 7.65 gives ¯σ NP = 0
             and infinite X NP , so the maximum normal stress component does not control, and the preceding
             solution is valid.
                 (b) Proceeding as before, except for substituting X CM = 2.00, gives τ xy = 129.1MPa
             (Ans.). This value corresponds to σ 1 ,σ 2 =−20.9, −279.1 MPa, so Eq. 7.65 again gives ¯σ NP =
             0, and this solution is also valid.

             Comment     In the solution for (b), the safety factor of X CM = 2.00 is, in effect, applied
             both to the pressure and to τ xy . Due to increased pressure making fracture more difficult, it
             turns out that only a small decrease in τ xy is needed to achieve the safety factor. From an
             engineering viewpoint, if the pressure is considered not to vary, it would be wise to apply
             the desired safety factor of X = 2.00 to only the shear stress that is allowed to vary, so that
             the solution for (b) becomes τ xy = 134.6/2.00 = 67.3 MPa. This would be the same as a load
             factor design approach, with Y p = 1.00 applied to the pressure and Y τ = 2.00 applied to τ xy .
             (See Section 7.6.2.)



            7.9 ADDITIONAL COMMENTS ON FAILURE CRITERIA

            To gain additional perspective on the subject of this chapter, we will engage in some limited further
            discussion on brittle versus ductile behavior and on time-dependent effects.


            7.9.1 Brittle Versus Ductile Behavior
            Engineering materials that are commonly classed as ductile are those for which the static strength
            in engineering applications is generally limited by yielding. Many metals and polymers fit into this
            category. In contrast, the usefulness of materials commonly classed as brittle is generally limited by
            fracture. In a tension test, brittle materials exhibit no well-defined yielding behavior, and they fail
            after only a small elongation, on the order of 5% or less. Examples are gray cast iron and certain
            other cast metals, and also stone, concrete, other ceramics, and glasses.
               However, normally brittle materials may exhibit considerable ductility when tested under
            loading such that the hydrostatic component σ h of the applied stress is highly compressive. Such
            an experiment can be conducted by testing the material in a chamber that is already pressurized,
            as in Fig. 4.27. The surprising result of large plastic deformations in a normally brittle material is
            illustrated by some stress–strain curves for limestone in Fig. 7.22.
               Also, materials normally considered ductile fail with increased ductility if the hydrostatic stress
            is compressive, or reduced ductility if it is tensile. For example, although the initial yielding of
            metals is insensitive to hydrostatic stress, the point of fracture is affected. Data showing this for
            a steel are given in Fig. 7.23, where the true fracture stress and strain are seen to increase with
   312   313   314   315   316   317   318   319   320   321   322