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M1/Fr = b2
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a21 = ( sin cos ) + a31 = ( sin cos ) + a12 = ( + sin cos ) + a22 = a11 a32 = sin2 + a13 = a32 a23 = a31
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EI (3 4 sin + sin cos ) a33 = ( sin cos ) GK b1 = sin sin cos cos sin2 + ( ) sin + EI [cos sin2 GK
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sin sin cos + ( )sin + 2 cos 2 cos ]
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(38.38)
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CURVED BEAMS AND RINGS 38.18
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CLASSICAL STRESS AND DEFORMATION ANALYSIS
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b2 = ( ) cos sin + cos sin cos + sin sin2 + EI [( ) cos sin + 2 sin cos sin cos sin sin2 ] GK (38.39)
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b3 = ( ) cos sin + cos sin cos + sin sin2 + EI [( ) cos sin cos sin cos sin sin2 GK (38.40)
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+ 2(sin + + cos sin sin cos )] For tabulation purposes, we indicate these relations in the form aij = Xij + EI Yij GK bk = Xk + EI Yk GK
(38.41)
Programs for solving equations such as Eq. (38.28) are widely available and easy to use. Tables 38.4 and 38.5 list the values of the coefficients for a variety of span and load angles.
TABLE 38.4 Coefficients aij for Various Span Angles
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CURVED BEAMS AND RINGS 38.19
CURVED BEAMS AND RINGS
TABLE 38.5 Coefficients bk for Various Span Angles and Load Angles in Terms of
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)
Then Eq. (38.2) will have four integrals, which are AF = BF = CF =
0 0 0
[(C1 C3) sin + C2 cos ]2 d (cos sin sin cos )2 d [(C3 C1) cos + C2 sin C3]2 d
(38.43)
(38.44)
(38.45)
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CURVED BEAMS AND RINGS 38.20
CLASSICAL STRESS AND DEFORMATION ANALYSIS
DF =
[1 (cos cos + sin sin )]2 d
(38.46)
The results of these four integrations should be substituted into y= EI Fr3 AF + BF + (CF + DF) GK EI (38.47)
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)
T = T1 cos + M1 sin
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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