Page 376 - Marks Calculation for Machine Design
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P1: Sanjay
                                      15:14
                          January 4, 2005
        Brown.cls
                 Brown˙C08
                                           APPLICATION TO MACHINES
                  358
                                               H
                                                        t             P
                                                                     ×
                                      P                  Top view
                                                                       P
                              L
                                   t                        L o

                                  Front view             Side view
                              FIGURE 8.17  Vertical fillet welds in bending.

                  where the factor 2 represents that there are two areas over which the normal stress acts, and
                  the single term in brackets represents the moment of inertia of the welds about their own
                  centroidal axes.
                    Note that the shear stress (τ shear ) for the weld configuration in Fig. 8.17 would still be
                  the same as for the weld configuration in Fig. 8.16 and given by Eq. (8.79).
                    Once the shear stress (τ shear ) and normal stress (σ bending ) have been determined, then the
                  principal stress (σ 1 ) and maximum shear stress (τ max ) can be found using the methods of
                  Chap. 5.
                    Consider the following example of a welded joint in bending.


                            U.S. Customary                       SI/Metric
                  Example 4. For the fillet weld and loading  Example 4. For the fillet weld and loading
                  configuration shown in Figs. 8.15 and 8.16,  configuration shown in Figs. 8.15 and 8.16,
                  determine the principal stress (σ 1 ) and maxi-  determine the principal stress (σ 1 ) and maxi-
                  mum shear stress (τ max ), where   mum shear stress (τ max ), where
                     P = 800 lb                        P = 3,600 N
                                          ◦
                    H = 0.265 in (0.375 in × cos 45 )  H = 0.7 cm = 0.007 m (1 cm × cos 45 )
                                                                                ◦
                     L = 2.5 in                        L = 6cm = 0.06 m
                    d o = 0.75 in                      d o = 2cm = 0.02 m
                    L o = 1.5 ft = 18 in               L o = 45 cm = 0.45 m
                  solution                           solution
                  Step 1. Substitute the given information in  Step 1. Substitute the given information in
                  Eq. (8.79) to determine (τ shear ) as  Eq. (8.72) to determine (τ shear ) as
                               P                                P
                       τ shear =                        τ shear =
                             2 (HL)                           2 (HL)
                                  800 lb                          3,600 N
                           =                                =
                             2 (0.265 in)(2.5in)              2 (0.007 m)(0.06 m)
                                                                        2
                                   2
                           = 604 lb/in = 0.6 kpsi           = 4,286,000 N/m = 4.3MPa
                           = τ xy                           = τ xy
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