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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.

38.5.2 Deflection Due to Concentrated Load The deflection of a ring segment at a concentrated load can be obtained using the first relation of Eq. (38.2). The complete analytical solution is quite lengthy, and so a result is shown here that can be solved using computer solutions of Simpson s approximation. First, define the three solutions to Eq. (38.28) as T1 = C1Fr M1 = C2Fr R1 = C3F (38.42)

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[(C1 C3) sin + C2 cos ]2 d (cos sin sin cos )2 d [(C3 C1) cos + C2 sin C3]2 d

data. I made one change to the play method to specify the hint when creating the stream, as shown in Listing 36-24.

(38.43)

(38.44)

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(38.45)

Listing 3-24. Modified play Method with Hinting - (void)play { // Create audio stream using callback functions. // Third argument is an optional hint to file type. AudioFileStreamOpen(self, propertyListenerCallback, audioDataCallback, self.fileTypeHint, &myAudioStream); // Create the network connection NSURLRequest *networkRequest = [NSURLRequest requestWithURL:self.url]; networkConnection = [[NSURLConnection alloc] initWithRequest:networkRequest delegate:self]; }

(38.46)

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The results of these four integrations should be substituted into y= EI Fr3 AF + BF + (CF + DF) GK EI (38.47)

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to obtain the deflection due to F and at the location of the force F. It is worth noting that the point of maximum deflection will never be far from the middle of the ring, even though the force F may be exerted near one end. This means that Eq. (38.47) will not give the maximum deflection unless = /2. 38.5.3 Segment with Distributed Load The resultant load acting at the centroid B in Fig. 38.8 is W = wr , and the radius r is given by Eq. (38.6), with substituted for . Thus the shear reaction at the fixed end A is R1 = wr /2. M1 and T1, at the fixed ends, can be determined using Castigliano s method. We use Fig. 38.9 to write equations for moment and torque for any section, such as the one at D. When Eq. (38.6) for r is used, the results are found to be M = T1 sin + M1 cos wr2 sin + wr2(1 cos ) 2 wr2 (1 cos ) + wr 2( sin ) 2 (38.48) (38.49)

To provide a hint, I d set the fileTypeHint property after creating the streaming object but before starting playback, as Listing 3-25 shows.

FIGURE 38.8 Ring segment of span angle subjected to a uniformly distributed load w per unit circumference acting downward. Point B is the centroid of the load. The ends are fixed to resist bending moment and torsional moment.

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FIGURE 38.9 A portion of the ring has been isolated here to determine the moment and torque at any section D at angle from the fixed end at A.

These equations are now employed in the same manner as in Sec. 38.5.1 to obtain a11 a12 a21 a22 T1/wr2 b = 1 M1/wr2 b2 (38.50)

Listing 3-25. Providing a Hint for Audio Streaming - (IBAction)play:(id)sender { streamer = [[SimpleStreamer alloc] initWithURL:url]; streamer.fileTypeHint = kAudioFileMP3Type; [streamer play]; }

It turns out that the aij terms in the array are identical with the same coefficients in Eq. (38.28); they are given by Eqs. (38.29), (38.30), (38.32), and (38.33), respectively. The coefficients bk are bk = Xk + where X1 = EI Yk GK (38.51) (38.52) (38.53)

sin2 + sin cos + 2 sin 2 sin2 sin cos + (1 + cos ) 2

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