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MA = TA =
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(38.10)
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Pir(1 cos i)
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where n = number of forces and reactions together. The proof uses Castigliano s theorem and may be found in Ref. [38.3].
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FIGURE 38.3 Ring loaded by a series of concentrated forces.
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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.
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CURVED BEAMS AND RINGS 38.11
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Example 1. Find the shear force, bending moment, and torsional moment at the location of R3 for the ring shown in Fig. 38.4. Solution. Using the principles of statics, we first find the reactions to be R1 = R2 = R3 = Choosing point A at R3, the reduced loads are P0 = P2 = P3 = 0 R3 = 0 360 P1 = 30 F = 0.0833F 360 2 F 3
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120 120 2 R1 = F = 0.2222F 360 360 3
210 F = 0.5833F 360 240 240 2 R2 = F = 0.4444F 360 360 3
P4 =
Then, using Eqs. (38.10), we find VA = 0. Next, MA =
Pi r sin i
= Fr (0 + 0.0833 sin 30 0.2222 sin 120 + 0.5833 sin 210 0.4444 sin 240 ) = 0.0576Fr In a similar manner, we find TA = 0.997Fr.
FIGURE 38.4 Ring loaded by the two forces F and supported by reactions R1, R2, and R3. The crosses indicate that the forces act downward; the heavy dots at the reactions R indicate an upward direction.
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.
CURVED BEAMS AND RINGS 38.12
CLASSICAL STRESS AND DEFORMATION ANALYSIS
The task of finding the deflection at any point on a ring with a loading like that of Fig. 38.3 is indeed difficult.The problem can be set up using Eq. (38.2), but the resulting integrals will be lengthy. The chances of making an error in signs or in terms during any of the simplification processes are very great. If a computer or even a programmable calculator is available, the integration can be performed using a numerical procedure such as Simpson s rule. Most of the user s manuals for programmable calculators contain such programs in the master library. When this approach is taken, the two terms behind each integral should not be multiplied out or simplified; reserve these tasks for the computer.
38.4.1 A Ring with Symmetrical Loads A ring having three equally spaced loads, all equal in magnitude, with three equally spaced supports located midway between each pair of loads, has reactions at each support of R = F/2, M = 0.289Fr, and T = 0 by Biezeno s theorem.To find the moment and torque at any location from a reaction, we construct the diagram shown in Fig. 38.5. Then the moment and torque at A are M = M1 cos R1r sin = Fr (0.289 cos 0.5 sin ) T = M1 sin R1r (1 cos ) = Fr (0.289 sin 0.5 + 0.5 cos ) (38.11)
(38.12)
FIGURE 38.5 The positive directions of the moment and torque axes are arbitrary. Note that R1 = F/2 and M1 = 0.289Fr.
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.
CURVED BEAMS AND RINGS 38.13
CURVED BEAMS AND RINGS
Neglecting direct shear, the strain energy stored in the ring between any two supports is, from Table 38.2, U=2
/3 0
M 2 r d +2 2EI
/3 0
T 2r d 2GK
(38.13)
Castigliano s theorem states that the deflection at the load F is y= U 2r = F EI
M 2r d + F GK
T d F
(38.14)
From Eqs. (38.11) and (38.12), we find M = r(0.289 cos 0.5 sin ) F T = r(0.289 sin 0.5 + 0.5 cos ) F When these are substituted into Eq. (38.14), we get y= Fr3 A B + 2 EI GK (38.15)
which is the same as Eq. (38.5). The constants are A=4 B=4
/3 0 /3 0
(0.289 cos 0.5 sin )2 d (38.16) (0.289 sin 0.5 + 0.5 cos ) d
These equations can be integrated directly or by a computer using Simpson s rule. If your integration is rusty, use the computer.The results are A = 0.1208 and B = 0.0134. 38.4.2 Distributed Loading The ring segment in Fig. 38.6 is subjected to a distributed load w per unit circumference and is supported by the vertical reactions R1 and R2 and the moment reactions M1 and M2. The zero-torque reactions mean that the ring is free to turn at A and B. The resultant of the distributed load is W = wr ; it acts at the centroid: r= 2r sin ( /2) (38.17)
By symmetry, the force reactions are R1 = R2 = W/2 = wr /2. Summing moments about an axis through BO gives M(BO) = M2 + Wr sin wr 2 M1 cos ( ) sin = 0 2 2
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