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                 224    Mechanical Engineering Design
                  Figure 5–10                         2

                  Octahedral surfaces.
                                              av





                                              oct

                                                                    1
                                              3



                                          equal to the hydrostatic stress σ av. There are eight surfaces symmetric to the principal
                                          directions that contain this stress. This forms an octahedron as shown in Fig. 5–10. The
                                          shear stresses on these surfaces are equal and are called the octahedral shear stresses
                                          (Fig. 5–10 has only one of the octahedral surfaces labeled). Through coordinate trans-
                                          formations the octahedral shear stress is given by 5


                                                              1        2          2          2 1/2
                                                        τ oct =  (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )  (5–16)
                                                              3
                                          Under the name of the octahedral-shear-stress theory, failure is assumed to occur when-
                                          ever the octahedral shear stress for any stress state equals or exceeds the octahedral
                                          shear stress for the simple tension-test specimen at failure.
                                              As before, on the basis of the tensile test results, yield occurs when σ 1 = S y and
                                          σ 2 = σ 3 = 0. From Eq. (5–16) the octahedral shear stress under this condition is
                                                                             √
                                                                               2
                                                                       τ oct =  S y                        (5–17)
                                                                              3
                                          When, for the general stress case, Eq. (5–16) is equal or greater than Eq. (5–17), yield
                                          is predicted. This reduces to
                                                                  2          2         2   1/2
                                                          (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
                                                                                            ≥ S y          (5–18)
                                                                         2
                                          which is identical to Eq. (5–10), verifying that the maximum-octahedral-shear-stress
                                          theory is equivalent to the distortion-energy theory.
                                              The model for the MSS theory ignores the contribution of the normal stresses on
                                          the 45° surfaces of the tensile specimen. However, these stresses are P/2A, and not the
                                          hydrostatic stresses which are  P/3A. Herein lies the difference between the MSS and
                                          DE theories.
                                              The mathematical manipulation involved in describing the DE theory might tend
                                          to obscure the real value and usefulness of the result. The equations given allow the
                                          most complicated stress situation to be represented by a single quantity, the von Mises
                                          stress, which then can be compared against the yield strength of the material through
                                          Eq. (5–11). This equation can be expressed as a design equation by

                                                                              S y
                                                                                                           (5–19)
                                                                         σ =
                                                                              n
                                          5 For a derivation, see Arthur P. Boresi, op. cit., pp. 36–37.
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