Page 406 - Marks Calculation for Machine Design
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P1: Naresh
                          January 4, 2005
                 Brown˙C09
        Brown.cls
                  388
                            U.S. Customary 15:28  APPLICATION TO MACHINES  SI/Metric
                  Step 7. Substitute the mean shear stress (τ m )  Step 7. Substitute the mean shear stress (τ m )
                  from step 5 and the alternating shear stress (τ a )  from step 5 and the alternating shear stress (τ a )
                  from step 6, and the given endurance limit (S e )  from step 6, and the given endurance limit (S e )
                  and ultimate shear stress (S us ) in the Goodman  and ultimate shear stress (S us ) in the Goodman
                  theory given in Eq. (7.34) as      theory given in Eq. (7.34) as
                             τ a  τ m  1                      τ a  τ m  1
                               +    =                           +    =
                             S e  S us  n                     S e  S us  n
                     39.6 kpsi  60.5 kpsi  1          511 MPa  805 MPa  1
                           +        =                       +        =
                     60 kpsi  140 kpsi  n             420 MPa  980 MPa  n
                                      1                                1
                        (0.660) + (0.432) =              (1.217) + (0.821) =
                                      n                                n
                                      1                                1
                                1.092 =                          2.038 =
                                      n                                n
                                 1                                1
                            n =     = 0.92 (unsafe)          n =  2.038  = 0.49 (very unsafe)
                               1.092
                    The fact that the factor-of-safety (n) is  The fact that the factor-of-safety (n) is much
                  less than 1, means the spring must be re-  less than 1, means the spring must be re-
                  designed.                          designed.
                  9.3 FLYWHEELS
                  Flywheels store and release the energy of rotation, called inertial energy. The primary
                  purpose of a flywheel is to regulate the speed of a machine. It does this through the amount
                  of inertia contained in the flywheel, specifically the mass moment of inertia. Flywheels are
                  typically mounted onto one of the axes of the machine, integral with one of the rotating
                  shafts. Therefore, it is the mass moment of inertia about this axis that is the most important
                  design parameter. As stated in the introduction to this chapter, too much inertia in the
                  flywheel design and the system will be sluggish and unresponsive, too little inertia and
                  the system will lose momentum over time. The inertia has to be just right! Determining
                  the right amount of inertia is the main purpose of the disussion that follows.

                  9.3.1 Inertial Energy of a Flywheel
                  Shown in Fig. 9.8 is a solid disk flywheel integral to a rotating shaft supported by appropriate
                  bearings at each end. The applied torque (T ) produces an angular acceleration, denoted (α),
                  which in turn produces an angular velocity, denoted by (ω).

                                   t
                                              Flywheel
                                                                      T

                                                                       a, w


                                         L
                            FIGURE 9.8  Solid disk flywheel on a rotating shaft.
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