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Tutorial Solution Sheet 2

# Tutorial Solution Sheet 2 - Imperial College of Science...

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Sheet 2 Solution Page 1 Imperial College of Science, Technology & Medicine Department of Aeronautics Second Year Structural Mechanics and Dynamics II Tutorial Sheet 2 Solutions Question 1 General shear flow distribution = s 0 y x s 0 x y ds x t I S ds y t I S q But dA ds t = , the element of cross-section of the tube. For an integration over the complete tube cross-section, S , this becomes = S y x S x y dA x I S dA y I S q The definition of the centroidal axes is that 0 dA x dA y S S = = Hence the shear flow must be zero at the end of the cross-section. Question 2 The section is symmetric so that only the x D is required since the shear centre must lie on the line of symmetry.

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Sheet 2 Solution Page 2 1 2 3 45 6 0 D 1 x = 1 - 2 () 2 s 0 s 0 12 x s s 64 ds s 32 2 ds y t D s 32 y + = = = = ∫∫ 1280 D 2 x = 2 - 3 s 96 1280 ds 48 2 1280 D 48 y s 0 23 x + = = = 5888 D 3 x = 3 – 4 2 s 0 34 x s s 96 5888 ds s 48 2 5888 D s 48 y + = + = + = Everything else follows from symmetry. 8192 D 3582 D 576 D c 34 x c 23 x c 12 x = =
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Tutorial Solution Sheet 2 - Imperial College of Science...

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