HW4solution - AE 321 – Solution to Homework #4 Chapter 3:...

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Unformatted text preview: AE 321 – Solution to Homework #4 Chapter 3: Strain 1. Given the displacement field u x x 2 20 10 4 m, u y 2 yz 10 3 m, u z z 2 xy 10 3 m (a) Before deformation, the distance between 7 , 5 , 2 , , P z y x P and 9 , 8 , 3 , , Q z y x Q is given by m 741 . 3 14 4 9 1 7 9 5 8 2 3 2 2 2 ds . The position of points P and Q after deformation is determined using the following relation u x X (5.1) Using Equation (5.1), the positions after deformations are m 039 . 7 070 . 5 0024 . 2 m 10 5 2 7 10 7 5 2 10 20 2 7 5 2 3 2 3 4 2 z y x P z y x z y x P P e e e X e e e e e e u x X m 057 . 9 144 . 8 0029 . 3 m 10 8 3 9 10 9 8 2 10 20 3 9 8 3 3 2 3 4 2 z y x Q z y x z y x Q Q e e e X e e e e e e u x X The distance between the given points, i.e. P and Q , after deformation is m 8109 . 3 039 . 7 057 . 9 070 . 5 144 . 8 0024 . 2 0029 . 3 2 2 2 dS . Therefore, the change in distance between P and Q is m 0692 . 7417 . 3 8109 . 3 ds dS . (b) Lagrangian strain tensor E ij ij L 1 2 u i , j u j , i u k , i u k , j (5.2) Infinitesimal strain tensor ij 1 2 u i , j u j , i (5.3) The components of the Lagrangian strain are 5 10 5 4 x 10 4 4 x 2 100 y 2 5 10 7 xy 5 10 4 y 1 2 10 3 z 5 10 4 4 z 10 3 4 z 2 x 2 5 10 4 2 y x 1 2 10 3 z 2 10 3 z 10 3 y 2 z 2 The components of the infinitesimal strain are 2 10 4 x 5 10 4 y 2 10 3 z 5 10 4 2 y x 2 10 3 z . Note that this is nothing but the above results for E ij , with all second (and higher) order terms neglected. (c) Rotation tensor ij 1 2 u i , j u j , i (5.4) ij 5 10 4 y 5 10 4 2 y x 5 10 4 y 5 10 4 2 y x (d) The Lagrangian and the infinitesimal strain tensors are each evaluated at 3 , 1 , 2 z y x : E ij 10 6 400.58 1 503 1 6020 2012 503 2012 6020...
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This note was uploaded on 03/07/2011 for the course AE 311 taught by Professor Dutton,j during the Spring '08 term at University of Illinois, Urbana Champaign.

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HW4solution - AE 321 – Solution to Homework #4 Chapter 3:...

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