Page 103 - Analysis and Design of Machine Elements
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Detachable Joints and Fastening Methods
                          Figure E3.2. The coefficient of friction between the bearing bottom surface and the floor  81
                          is 0.16. The allowable bearing stress between the bearing and floor is [   ] = 125 MPa.
                                                                                      p
                          Select proper bolts for the connection.



                                                          F

                           150                           45˚




                                                                     75
                                              120
                                             160
                                             200





                          Figure E3.2 A sliding bearing bolted to a smooth floor by a group of two bolts.



                          Solution:


                           Steps            Computation                        Results      Units

                          1. Structural design  Propose the number of bolts and pattern.
                                            According to the structure, two identical
                                            bolts are used and a tightening torque is also
                                            applied.
                          2. Force analysis  (1) Determine the forces on the bolted joint.
                                            Resolve load F into vertical and horizontal
                                            components, which are the axial and shear
                                            force on the joints, respectively.
                                            Axial force
                                                      ∘            ∘           F Σv  = 3536  N
                                            F Σv  = F sin 45 = 5000 × sin 45 = 3536N
                                            Shear force                        F  = 3536    N
                                                      ∘
                                                                    ∘
                                            F Σh  = F cos 45 = 5000 × cos 45 = 3536N  Σh
                                            Overturning moment to be resisted by the
                                            bolted joints                                   N⋅mm
                                                                               M = 530 400
                                            M = F Σh  × 150 = 530400N • mm
                                            (2) Determine the force on each individual  F = 1768  N
                                                                                a
                                            bolt under the axial force
                                                F   3536
                                            F =  Σv  =  N = 1768N
                                             a
                                                z     2
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