278_Problem CHAPTER 9

# 278_Problem CHAPTER 9 - PROBLEM 9.185 CONTINUED Then 0 a a...

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PROBLEM 9.185 CONTINUED Then 24 111 2 2 2 1 22 4 xy aa I m x y m x y ta ta ρ  =+ = =   2 22 yx I myz ( ) 2 zx z x I mzx I mz x ′′ + From Problem 9.170 () 4 2 1 72 zx I ta =− Then 42 4 11 1 1 1 72 2 3 3 24 zx I ta ta a a ta ρρ + = ( a ) Equation 9.56 ( ) ) 32 2 2 2 xyz x yy zz x x x KI I I I I I I IIII K −+ + + + + −−− ( 222 20 x y z x yz y zx z xy xy yz zx III II I I I −− = Substituting 22 2 34 2 4 513 51 1 3 35 1 1 0 1 224 1 41 2 4 2 4 Kt a K t a K         + + + +                  3 4 3 1 00 0 1 4 44 ta   =  Simplifying and letting 4 a ρζ = yields 5 479 125 0 3 576 1152 = +− = ζζ ζ Solving numerically. .. 4 0.203032 or 0.203 a == W 4 0.698281 or 0.698 a W 4 33 0.765354 or 0.765 a W ( b ) Equations 9.54a and 9.54b ( )( ) ( ) ( ) 0 xx x y y z x
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