246_Problem CHAPTER 9

246_Problem CHAPTER 9 - PROBLEM 9.170 CONTINUED Now...

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PROBLEM 9.170 CONTINUED Now uv u v I Im u v ′′ =+ Therefore 24 11 1 2 2 2 1 22 4 xy aa I m x y m x y ta ta ρ ′ ′  ′′ ′′ = =   () 222 1 22 yz a I my z myz ta a ρρ =″ + = = ( ) 2 zx z x I m z x = ″″+ + ″″ From Sample Problem 9.6 4 2 area 1 72 I a  =−  Then 4 area 1 72 I tI t a == Then 2 1 a Ia a 42 11 1 2 1 72 2 2 3 3 ta ta a a +− + 4 5 8 ta By observation 1 3 AB = ijk λ Now, Equation 9.46 2 AB x x y y z z xy xy y z zx z x II I I I I I λ λλ =++− 2 4 19 1 5 1 3 1 12 3 4 33 3 ta + 1 1 3 3   −−     51 1 2 8 −− − or 4 5 12 AB I ta = W 0 0
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