Solution3

Solution3 - AE 321 Solution to Practice Problems Chapter 3:...

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AE 321 – Solution to Practice Problems 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 . 0 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)
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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 0 " 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 [ ] = 0 0 5 " 10 # 4 y 0 0 5 " 10 # 4 2 y + x ( ) # 5 " 10 # 4 y # 5 " 10 # 4 2 y + x ( ) 0 $ % & & & & ( ) ) ) ) (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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! ij [ ] = 10 " 6 400 0 500 0 6000 " 2000 500 " 2000 6000 # $ % % % & ( ( ( Clearly, by comparing the components of the Lagrangian and the infinitesimal strain tensors, we can conclude that E ij [ ] ! " ij [ ] . This means that in this case one can assume small deformation gradients. (e) Compatibility equations 0 , , , , = ! ! + ik jl jl ik ij kl kl ij (5.5) Only 6 independent equations are obtained from (5.5), namely, ( ) 0 0 2 0 0 0 2 2 2 2 = ! + " = # # # ! # # # + # # # y x x x y y xy yy xx $ ( ) 0 0 2 0 0 0 2 2 2 2 = ! + " = # # # ! # # # + # # # z y y y z z yz zz yy ( ) 0 0 2 0 0 0 2 2 2 2 = ! + " = # # # ! # # # + # # # z x z z x x xz xx zz 0 0 0 0 0 0 2 2 2 = ! ! + " = # # # ! # # # ! # # # + # # # z y x x y x z x xx yz xz xy 0 0 0 0 0 0 2 2 2 2 = ! ! + " = # # # ! # # # ! # # # + # # # x z y y z y x y yy zx yx yz 0 0 0 0 0 0 2 2 2 2 = ! ! + " = # # # ! # # # ! # # # + # # # y x z z x z y z zz xy zy zx Therefore, the given displacement field is compatible (which may not be obvious upon immediate inspection of the displacement function given).
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Solution3 - AE 321 Solution to Practice Problems Chapter 3:...

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