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                                                                                             Load and Stress Analysis  119
                                               Furthermore, it can be shown that the stress distribution is given by


                                                                                  My
                                                                           σ =                                 (3–64)
                                                                               Ae(r n − y)
                                               where M is positive in the direction shown in Fig. 3–34. The stress distribution given
                                               by Eq. (3–64) is hyperbolic and not linear as is the case for straight beams. The critical
                                               stresses occur at the inner and outer surfaces where y = c i and y = c o, respectively,
                                               and are

                                                                          Mc i           Mc o
                                                                     σ i =         σ o =−                      (3–65)
                                                                          Aer i          Aer o
                                               These equations are valid for pure bending. In the usual and more general case, such as
                                               a crane hook, the U frame of a press, or the frame of a C clamp, the bending moment is
                                               due to a force acting at a distance from the cross section under consideration. Thus, the
                                               cross section transmits a bending moment and an axial force. The axial force is located
                                               at the centroidal axis of the section and the bending moment is then computed at this
                                               location. The tensile or compressive stress due to the axial force, from Eq. (3–22), is then
                                               added to the bending stresses given by Eqs. (3–64) and (3–65) to obtain the resultant
                                               stresses acting on the section.





                            EXAMPLE 3–15       Plot the distribution of stresses across section  A–A of the crane hook shown in
                                               Fig. 3–35a. The cross section is rectangular, with b = 0.75 in and h = 4 in, and the load
                                               is F = 5000 lbf.

                                    Solution   Since A = bh, we have dA = bdr and, from Eq. (3–63),

                                                                          A        bh       h
                                                                                         =                        (1)
                                                                          dA      r o  b     r o
                                                                   r n =      =
                                                                                      dr   ln
                                                                           r        r        r i
                                                                                 r i
                                                                                                          2
                                               From Fig. 3–35b, we see that r i = 2 in, r o = 6 in, r c = 4 in, and A = 3 in . Thus, from
                                               Eq. (1),
                                                                           h       4
                                                                    r n =       =     = 3.641 in
                                                                        ln(r o /r i )  ln  6
                                                                                    2
                                               and the eccentricity is e = r c − r n = 4 − 3.641 = 0.359 in. The moment M is positive
                                               and is M = Fr c = 5000(4) = 20 000 lbf · in. Adding the axial component of stress to
                                               Eq. (3–64) gives

                                                               F       My       5000   (20 000)(3.641 − r)
                                                           σ =    +           =      +                            (2)
                                                                A   Ae(r n − y)  3         3(0.359)r
                                               Substituting values of  r from 2 to 6 in results in the stress distribution shown in
                                               Fig. 3–35c. The stresses at the inner and outer radii are found to be 16.9 and −5.63 kpsi,
                                               respectively, as shown.
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