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Solution. Holzer s procedure can be carried out in a tabulated form as shown in Table 31.1. Two trials are shown in Table 31.1. The calculation can be carried out for more values of , and the resulting T3 can be plotted versus , as shown in Fig. 31.21. The frequencies at which T3 = 0 are then the natural frequencies of the system. Better approximation can be obtained by employing the method of false position [31.1]; if (+) and ( ) are the frequencies when the torque has corresponding values of T3(+) and T3( ), the natural frequency can be obtained by = ( )T3(+) (+)T3( ) T3(+) T3( )

Create table #List(Id int identity(1,1), Item varchar(255)) -- identify Properties associated with asset insert into #List (Item) select Property + '=' + Value + ' ' + Coalesce(Unit, '') + '; ' from InventoryProperty inner join Property on InventoryProperty.PropertyId = Property.PropertyId where InventoryProperty.InventoryId = @intInventoryId -- call sp that converts records to a single varchar exec prConvertTempTbl @chvProperties OUTPUT -- display result select @chvProperties Properties drop table #List return 0 go

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With a little bit of care most plastics can be recycled, and collection of plastics for recycling is increasing rapidly. Plastic recycling faces one huge problem: plastic types must not be mixed for recycling, yet it is extremely dif cult to distinguish one type from another by sight or touch. It is usually very important that plastics are separated prior to recycling. This is due to the fact that plastic recyclers use different processes for different types of plastics, and not all plastics have the same properties, such as melting points. Thus, it is imperative that they can be separated prior to recycling, either by the consumer, the hauler, the processor, or the recycler. Plastic types are generally easy to distinguish, because they are imprinted with a recycling symbol and a number. The recycling symbol veri es that the material can be recycled, and the number corresponds to the type of plastic used.

The mode shape corresponding to a natural frequency can be obtained by recalculating the values of 1, 2, and 3 in the Holzer table. In the example, the first natural frequency is 1 = 141.4214 rad/s, and the corresponding mode shape is { 1, 2, 3} = {1.0, 0.0, 0.5}

10:

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

Geared Systems. When a shaft transmits torque to another through a gear drive of speed ratio n, it is necessary to reduce the geared torsional system to an equivalent single-shaft system to find its natural frequency.The moments of inertia and the stiffness of the equivalent system are obtained through a consideration of the kinetic and potential energies of the system. Consider the geared torsional system in Fig. 31.22a. The speed of the second shaft is 2 = n 1. Assuming massless gears, we see that the kinetic energy of the system is T= 1 2 1 1 J1 1 + J2 n2 2 2 2 (31.77)

In the field of solid waste minimization and recycling, there are two conventions for quantifying the waste and materials. Quantification is either done by weight or volume. If by weight, numbers are given in pounds (lb) or more commonly tons (2000 lb = 1 ton). If by volume, then the numbers are given in cubic yards (yd 3). This book incorporates both conventions, because a scale is not used for all waste audits. Most observations are made by volume and then, using density figures, the approximate weight is calculated.

The nested stored procedure is not business-oriented. It loops through records in the temporary table (which was created in calling the batch or stored procedure) and assembles them into a single varchar variable:

Thus the equivalent mass moment of inertia of disk 2 referred to shaft 1 is n2J2. If disks 1 and 2 are clamped and a torque is applied to gear 1, rotating it through an angle 1, there will be deformations in both shafts 1 and 2. Gear 2 will rotate through an angle 2 = n 1. The potential energy stored in the two shafts is U= 1 1 k 1 2 + k2 n2 2 1 1 2 2 (31.78)

Hence the equivalent stiffness of shaft 2 referred to shaft 1 is n2k2. The equivalent torsional system is shown in Fig. 31.22b, where the stiffness and inertia of one side of the system are multiplied by the square of the speed ratio to obtain the corresponding equivalent values for Holzer calculations.

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