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TABLE 37.1 Properties of Beams (Continued)
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37.9 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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TABLE 37.1 Properties of Beams (Continued)
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37.10 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.
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TABLE 37.1 Properties of Beams (Continued)
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37.11 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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TABLE 37.1 Properties of Beams (Continued)
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DEFLECTION
TABLE 37.1 Properties of Beams (Continued)
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DEFLECTION 37.14
CLASSICAL STRESS AND DEFORMATION ANALYSIS
FIGURE 37.2 Frame loaded by three forces.
member has been analyzed and the partial derivatives obtained. The following example demonstrates the technique. Example 1. Find the downward deflection of point D of the frame shown in Fig. 37.3. Solution. A force analysis of the system gives an upward reaction at E of RE = 225 + 3F2. The reaction at A is downward and is RA = 75 2F2. The strain energy for member CE is UCE =
2 RA 2AE
The partial deflection is taken with respect to F2 because the deflection at D in the direction of F2 is desired. Thus UCE 2RA RA = 2AE F2 F2 Also, RA = 2 F2 Thus Eq. (2) becomes UCE (75 2F2 )(30) 37 500 = ( 2) = 0.2E E F2 (3) (2)
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DEFLECTION 37.15
DEFLECTION
FIGURE 37.3 Frame loaded by two forces. Dimensions in inches: ACE = 0.20 in2 ; IAD = 0.18 in4 ; E = 30 106 psi.
Note that we were able to substitute the value of F2 in Eq. (3) because the partial derivative had been taken. The strain energy stored in member ABCD will have to be computed in three parts because of the change in direction of the bending moment diagram at points B and C. For part AB, the moment is MAB = RAx = (75 2F2)x The strain energy is
UAB =
2 M AB dx 2EI
Taking the partial derivative with respect to F2 as before gives UAB = F2
2MAB MAB dx 2EI F2
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DEFLECTION 37.16
CLASSICAL STRESS AND DEFORMATION ANALYSIS
But MAB = 2x F2 Therefore, Eq. (5) may be written UAB 1 = F2 EI =
x(75 2F2 )( 2x) dx
1 0.18E
250x2 dx =
100 000 E
where the value of F2 again has been substituted after taking the partial derivative. For section BC, we have MBC = RAx F1(x 6) = 1800 225x 2F2x MBC = 2x F2 UBC = F2 = =
2MBC MBC dx 2EI F2
1 EI
(1800 225x 2F2 x)( 2x) dx
1 0.18E
( 3600x + 850x2 ) dx =
145 926 E
Finally, section CD yields MCD = (24 x)F2 UCD = F2 = = = Then yD = = UCE UAB UBC UCD + + + F2 F2 F2 F2 1 (37 500 + 100 000 + 145 926 + 758 519) 30(10)6 (when rounded)
24 8
MCD = (24 x) F2
2MCD MCD dx 2EI F2
1 EI
F2(24 x)2 dx
8 24
1 0.18E
(57 600 4800x + 100x2 ) dx
758 519 E
= 0.0347 in
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DEFLECTION 37.17
DEFLECTION
37.4.1 Redundant Members A frame consisting of one or more redundant members is statically indeterminate because the use of statics is not sufficient to determine all the reactions. In this case, Castigliano s theorem can be used first to determine these reactions and second to determine the desired deflection. Let R1, R2 , and R3 be a set of three indeterminate reactions. The deflection at the supports must be zero, and so Castigliano s theorem can be written three times. Thus U =0 R1 U =0 R2 U =0 R3 (37.10)
and so the number of equations to be solved is the same as the number of indeterminate reactions. In setting up Eqs. (37.10), do not substitute the numerical value of the particular force corresponding to the desired deflection. This force symbol must appear in the reaction equations because the partial derivatives must be taken with respect to this force when the deflection is found. The method is illustrated by the following example. Example 2. Find the downward deflection at point D of the frame shown in Fig. 37.4. Solution. Choose RB as the statically indeterminate reaction. A static force analysis then gives the remaining reactions as RA = RC = 0.625(F RB) (1)
FIGURE 37.4 Frame loaded by a single force. Dimensions in millimeters: AAD = ACD = 2 cm2, ABD = 1.2 cm2, E = 207 GPa, F = 20 kN.
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