Page 303 - Analysis and Design of Machine Elements
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281
                           Steps         Computation                                   Results  Shafts  Units

                           (3) Bending   Produce bending moment diagrams to determine the  M = 165 999  Nmm
                                                                                        H
                           moment        distribution of bending moments in the shaft.  M  = 174 830  Nmm
                           diagram and   Moment in horizontal plane                     V1         Nmm
                           torque diagram  M = 165 999 N mm                            M V2  = 8576  Nmm
                           of the shaft   H                                            M = 241 083  Nmm
                                                                                        1
                                         Moment in vertical plane
                                                                                       M = 166 220  Nmm
                                                                                        2
                                         M V1  = 174 830 N mm, M V2  = 8576 N mm       T = 756 000
                                         Total moment M
                                             √
                                                            2
                                                    2
                                         M =   165999 + 174830 = 241083 N •mm
                                          1
                                             √
                                                          2
                                                    2
                                         M =   165999 + 8576 = 166220N •mm
                                          2
                                         Torque T
                                         T = 756 000 N mm
                                         The bending moment diagram and torque diagram of the
                                         shaft is shown in Figure 10.3.
                           5. Strength analysis
                           (1) Check the  Check the strength at dangerous cross section I that bears     < [   ]
                                                                                        ca  −1
                           shaft strength  the greatest bending moment and torque.
                           by the        Since the shaft rotates in one direction, the torsional shear
                           combination of  stress is regarded as a repeated stress, select    = 0.6.
                           torque and    Calculate the stress at the dangerous cross section according
                           bending       to the equivalent moment as
                           moment            √
                                                 2
                                               M +(  T) 2  √    2            2
                                                 1         241083 +(0.6 × 756000)
                                            =           =
                                          ca
                                                 W               0.1 × 70 3
                                         = 14.98MPa
                                         Since    < [   ], the shaft is safe.
                                              ca
                                                   −1
                           (2) Check     Fatigue strength analysis is performed at locations where  S > [S]
                                                                                        ca
                           fatigue strength  diameter changes, the torque and bending moment are large
                           of the shaft  and stress concentration occurs. Therefore, we check fatigue
                                         strength at the right side of the cross section H.
                                         Bending moment M and bending stress are:
                                               113.6 − 25
                                         M = M         = 188028MPa
                                              1  113.6
                                            M    188028
                                            =  =        = 5.48MPa
                                            W    0.1 × 70 3
                                         Torque T and torsional shear stress are:
                                         T = 756 000 N mm
                                            M     756000
                                            =  =        = 11.0MPa
                                            W    0.2 × 70 3
                                              T
                                         Axial force F and tensile stress are:
                                         F = 1002 N
                                          a
                                             F     1002
                                            =   =       = 0.26MPa
                                                d 2      × 70 2
                                            4     4
                                         From Table 10.1, we have    = 600 MPa,    −1  = 240 MPa,
                                                            b
                                            −1  = 140 MPa.
                                                                                                (continued)
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