Explanation Given information: The normal strain in Rosette 1 is ε 1 = − 60 × 10 − 6 in ./in . . The Rosette 1 makes an angle with the x -axis is θ 1 = 0 ° . The normal strain in Rosette 2 is ε 2 = + 240 × 10 − 6 in ./in . . The Rosette 2 makes an angle with the x -axis is θ 2 = 45 ° . The normal strain in Rosette 3 is ε 3 = + 200 × 10 − 6 in ./in . . The Rosette 3 makes an angle with the x -axis is θ 3 = 90 ° . The modulus of elasticity of the material is E = 29 × 10 6 psi . The Poisson’s ratio is ν = 0.30 . Calculation: Refer to the Rosette arrangement; The normal strain in x -axis is ε x = ε 1 = − 60 × 10 − 6 in ./in . . The normal strain in y -axis is ε y = ε 3 = 200 × 10 − 6 in ./in . . Find the shearing strain in xy -plane ( γ x y ) using the relation. γ x y = 2 ε 2 − ε 1 − ε 3 Substitute + 240 × 10 − 6 in ./in . for ε 2 , − 60 × 10 − 6 in ./in . for ε 1 , and 200 × 10 − 6 in ./in . for ε 3 . γ x y = 2 ( + 240 × 10 − 6 ) − ( − 60 × 10 − 6 ) − ( 200 × 10 − 6 ) = 340 × 10 − 6 in ./in . Find the normal stress in x -axis ( σ x ) using the relation. σ x = E 1 − ν 2 ( ε x + ν ε y ) Substitute 29 × 10 6 psi for E , 0.30 for ν , − 60 × 10 − 6 in ./in . for ε x , and 200 × 10 − 6 in ./in . for ε y . σ x = 29 × 10 6 1 − 0.30 2 ( − 60 × 10 − 6 + 0.30 × 200 × 10 − 6 ) = 0 Find the normal stress in y -axis ( σ y ) using the relation. σ y = E 1 − ν 2 ( ε y + ν ε x ) Substitute 29 × 10 6 psi for E , 0.30 for ν , 200 × 10 − 6 in ./in . for ε y , and − 60 × 10 − 6 in ./in . for ε x . σ y = 29 × 10 6 1 − 0

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ISBN: 8220100257063

Author: BEER

Publisher: YUZU

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Chapter 7.9, Problem 150P

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Find the magnitudes of the axial force P and the horizontal force Qx.

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Chapter 7.9, Problem 149P

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Find the magnitudes of the axial force P and the horizontal force Q x .

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