Problem_05_05

Problem_05_05 - Problem 5.5 Given: By means of Eq. 5.6 and...

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Problem 5.5 Given: By means of Eq. 5.6 and 5.7 and the formula for shear stress in a beam of rectangular cross section given in Problem 5.4, show that the correction coefficient k is 1.20. Solution: UV U 0 d = Eq. 5.6 U 0 1 2E σ xx 2 σ yy 2 + σ zz 2 + ν E σ xx σ yy σ yy σ zz + σ xx σ zz + () 1 2G σ xy 2 σ zy 2 + σ xz 2 + + = Eq. 5.7 From Problem 5.4 σ zy τ max 1 4y 2 h 2 = 3V y 2A 1 2 h 2 = h is the hieght of the beam P Q z y V V M M dz x y dA b h U s V σ zy 2 d = z A 9V y 2 8A 2 G 1 2 h 2 2 d d = Let dA b dy = where b is the thickness of the beam.
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