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DEFLECTION 37.18
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CLASSICAL STRESS AND DEFORMATION ANALYSIS
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The frame consists only of tension members, so the strain energy in each member is UAD = UDC = Using Eq. (37.10), we now write 0= 2RA AD RA RB BD RB U = + AADE RB ABDE RB RB (3)
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2 RA AD 2AADE
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2 RB BD 2ABDE
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Equation (1) gives RA/ RB = 0.625. Also, RB/ RB = 1. Substituting numerical values in Eq. (3), except for F, gives 2(0.625)(F RB)(500)( 0.625) RB(400)(1) + =0 2(207) 1.2(207) (4)
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Solving gives RB = 0.369F. Therefore, from Eq. (1), RA = RC = 0.394F. This completes the solution of the case of the redundant member. The next problem is to find the deflection at D. Using Eq. (2), again we write yD = U 2RA AD RA RB BD RB = + AADE F ABDE F F (5)
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For use in this equation, we note that RA/ F = 0.394 and RB/ F = 0.369. Having taken the derivatives, we can now substitute the numerical value of F. Thus Eq. (5) becomes
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In general, when using metric quantities, prefixed units are chosen so as to produce number strings of not more than four members. Thus some preferred units in SI are MPa (N/mm2) for stress, GPa for modulus of elasticity, mm for length, and, say, cm4 for second moment of area. People are sometimes confused when they encounter an equation containing a number of mixed units. Suppose we wish to solve a deflection equation of the form
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64F 3 3 d4E
where F = 1.30 kN, = 300 mm, d = 2.5 cm, and E = 207 GPa. Form the equation into two parts, the first containing the numbers and the second containing the prefixes. This converts everything to base units, including the result. Thus, y= 64(1.30)(300)3 (kilo)(milli)3 3 (2.5) 4 (207) (centi)4 (giga) 103(10 3)3 = 29.48 10 4 m (10 2)4(109)
Now compute the numerical value of the first part and substitute the prefix values in the second. This gives y = (29.48 103) = 2.948 mm Note that we multiplied the result by 103 mm/m to get the answer in millimeters. When this approach is used with Eq. (5), it is found that the result must be multiplied by (10) 2 to get y in millimeters.
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DEFLECTION 37.19
DEFLECTION
yD =
2[0.394(20)](500)(0.394) [0.369(20)](400)(0.369) + 10 2 2(207) 1.2(207)
= 0.119 mm If care is taken to refrain from substituting numerical values for reactions or forces until after partial derivatives are taken, Castigliano s theorem is applicable to statically indeterminate frames containing redundant members.
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DEFLECTION
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Source: STANDARD HANDBOOK OF MACHINE DESIGN
CURVED BEAMS AND RINGS
Joseph E. Shigley
Professor Emeritus The University of Michigan Ann Arbor, Michigan
38.1 BENDING IN THE PLANE OF CURVATURE / 38.2 38.2 CASTIGLIANO S THEOREM / 38.2 38.3 RING SEGMENTS WITH ONE SUPPORT / 38.3 38.4 RINGS WITH SIMPLE SUPPORTS / 38.10 38.5 RING SEGMENTS WITH FIXED ENDS / 38.15 REFERENCES / 38.22
NOTATION
A B C E e F G I K M P Q R r r T U V W w X Area, or a constant Constant Constant Modulus of elasticity Eccentricity Force Modulus of rigidity Second moment of area (Table A.1) Shape constant (Table 36.1), or second polar moment of area Bending moment Reduced load Fictitious force Force reaction Ring radius Centroidal ring radius Torsional moment Strain energy Shear force Resultant of a distributed load Unit distributed load Constant
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