barcode generator vb.net source code Since M1 and M2 are equal, this equation can be solved to give in Software

Encoder EAN13 in Software Since M1 and M2 are equal, this equation can be solved to give

Since M1 and M2 are equal, this equation can be solved to give
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CURVED BEAMS AND RINGS 38.14
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CLASSICAL STRESS AND DEFORMATION ANALYSIS
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FIGURE 38.6 Section of ring of span angle with distributed load.
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M1 = wr2
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1 cos ( /2) sin 1 cos
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(38.18)
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Example 2. A ring has a uniformly distributed load and is supported by three equally spaced reactions. Find the deflection midway between supports. Solution. If we place a load Q midway between supports and compute the strain energy using half the span, Eq. (38.7) becomes y= U 2r = Q EI
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2r M d + GK Q
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T d Q
(38.19)
Using Eq. (38.18) with = 2 /3 gives the moment at a support due only to w to be M1 = 0.395 wr2. Then, using a procedure quite similar to that used to write Eqs. (38.11) and (38.12), we find the moment and torque due only to the distributed load at any section to be Mw = wr2 1 0.605 cos Tw = wr2 0.605 sin sin 3 + cos 3 3 (38.20)
In a similar manner, the force Q results in additional components of MQ = Qr (0.866 cos sin ) 2 (38.21) Qr TQ = (0.866 sin 1 + cos ) 2
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CURVED BEAMS AND RINGS 38.15
CURVED BEAMS AND RINGS
Then
MQ Q
r (0.866 cos sin ) 2
TQ r = (0.866 sin 1 + cos ) Q 2 And so, placing the fictitious force Q equal to zero, Eq. (38.19) becomes y= wr4 EI +
/3 0 /3 0
1 0.605 cos
sin (0.866 cos sin ) d 3 + cos (0.866 sin 1 + cos ) d 3 3 (38.22)
wr 4 GK
0.605 sin
When this expression is integrated, we find y= w r 4 0.141 0.029 + 2 EI GK (38.23)
38.5 RING SEGMENTS WITH FIXED ENDS
A ring segment with fixed ends has a moment reaction M1, a torque reaction T1, and a shear reaction R1, as shown in Fig. 38.7a. The system is indeterminate, and so all three relations of Eq. (38.2) must be used to determine them, using zero for each corresponding displacement. 38.5.1 Segment with Concentrated Load The moment and torque at any position are found from Fig. 38.7b as M = T1 sin + M1 cos R1r sin + Fr sin ( ) T = T1 cos + M1 sin R1r(1 cos ) + Fr[1 cos ( )] These can be simplified; the result is M = T1 sin + M1 cos R1r sin + Fr cos sin Fr sin cos T = T1 cos + M1 sin R1r(1 cos ) Fr cos cos Fr sin sin + Fr Using Eq. (38.3) and the third relation of Eq. (38.2) gives U r = M1 EI Note that
(38.24)
(38.25)
M r d + M1 GK M = cos M1 T = sin M1
T d = 0 M1
(38.26)
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CURVED BEAMS AND RINGS 38.16
CLASSICAL STRESS AND DEFORMATION ANALYSIS
FIGURE 38.7 (a) Ring segment of span angle loaded by force F. (b) Portion of ring used to compute moment and torque at position .
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CURVED BEAMS AND RINGS 38.17
CURVED BEAMS AND RINGS
Now multiply Eq. (38.26) by EI and divide by r; then substitute. The result can be written in the form
(T1 sin + M1 cos R1r sin ) cos d + Fr +
(cos sin sin cos ) cos d
EI GK
[ T1 cos + M1 sin R1r(1 cos )] sin d Fr
(cos cos + sin sin 1) sin d = 0
(38.27)
Similar equations can be written using the other two relations in Eq. (38.2). When these three relations are integrated, the results can be expressed in the form a11 a21 a31 where a11 = sin2 EI sin2 GK EI ( + sin cos ) GK EI ( + sin cos 2 sin ) GK EI ( sin cos ) GK (38.29) (38.30) (38.31) (38.32) (38.33) EI [2(1 cos ) sin2 ] GK (38.34) (38.35) (38.36) (38.37) a12 a22 a32 a13 a23 a33 T1/Fr R1/F b1 (38.28) b3
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