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Unformatted text preview: ANSWERS TO SELECTED PROBLEMS APPLIED ELASTICTY ADVANCED STRENGTH OF MATERIALS FOURTH EDITION (THIRD EDITION PROBLEM#s IN BRACKETS) Chapter 1 1.1 (1.1) P all = 4.27kN 1.2 (1.2) a) τ x’y’ = -16.08 MPa, b) τ x’y’ = -25 MPa. 1.6 (1.5) F = 67.32 kN/m 3 ??? (1.9) a) σ 1 = 121 MPa, σ 2 = -71 MPa, =96 MPa max τ φ p = -19.3 , φ s = 25.7 b) = 200 MPa, = -50 MPa, = 125 MPa 1 σ 2 τ max τ φ p = 26.58 , φ s = 71.55 1.18 (1.14) P = 3 π pr 2 1.23 (1.16) σ 1 = 37.84 MPa, σ 2 = -4.18 MPa, 1.29 (1.22) σ x = -40 MPa, σ y = 20 MPa; 1.39 (1.29) p all = 493 kPa. 1.40 (1.30) , I MPa ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − 6 16 . 16 99 . 7 16 . 16 392 . 5 66 . 2 99 . 7 66 . 2 342 . 25 1 =26 MPa, I 2 =-349 MPa 2 , I 3 =-6464MPa 3 1.43 (1.33) 2416 . , 1688 . , 9556 . 479 . 44 , 418 . 28 , 016 . 66 1 1 1 3 2 1 = = = − = = = n m l MPa MPa MPa σ σ σ 1.49 (1.38) 2 2 2 , 1 2 2 xy y x y x τ σ σ σ σ σ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ± + = 1.53 (1.42) σ =51.42 MPa, =51....
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assignment.ans - ANSWERS TO SELECTED PROBLEMS APPLIED...

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