Pset3 Solution - 1. 2 2 2003 . 4 in d A = = π , psi A F a...

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Unformatted text preview: 1. 2 2 2003 . 4 in d A = = π , psi A F a a 3 10 02 . 36 × = = σ psi E a a 6 10 96 . 21 × = = ε σ , 296 . = − = a L ε ε ν 2. lbf in M ⋅ = × = 500 10 50 . psi wt M I My 3556 75 . 5 . 1 500 6 6 2 2 = × × = = = σ strain E − = = µ σ ε 7 . 119 1 1 2 νε ε − = Since the two gages are in adjacent arms, ε ν ε ε 4 ) 1 ( | | 4 | | 2 1 F F e e i o + = − = ∆ 67 . 1 ) 1 ( 4 = + ∆ = ε ν i o e e F 3. (a) For the rectangular rosette, we have y c xy y x o x b x a ε ε γ ε ε θ ε ε ε ε = + + = = = = 2 2 ) 45 ( ' therefore 60 2 680 320 = − − = = = − = = y x b xy c y a x ε ε ε γ ε ε ε ε in µ-strain (b) Solving ) ( 1 ) ( 1 x y y y x x E E νσ σ ε νσ σ ε − = − = we get psi E psi E psi E xy xy x y y y x x 232 ) 1 ( 2 6411 ) ( 1 1340 ) ( 1 2 2 = + = = + − = − = + − = γ ν τ νε ε ν σ νε ε ν σ (c) psi psi xy y x y x 1347 6418 3882 2536 ) 2 ( 2 2 1 2 2 2 , 1 − = = ± = + − ± + = σ σ τ σ σ σ σ σ 4. Based on the rectangular rosette reading, we can find 4....
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This note was uploaded on 01/12/2011 for the course MAE 3800 taught by Professor Feng during the Spring '10 term at Missouri (Mizzou).

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Pset3 Solution - 1. 2 2 2003 . 4 in d A = = π , psi A F a...

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