This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mechanics of Aircraft structures C.T. Sun 2.3 Consider the displacement field in a body: u = 0.02x + 0.02y – 0.01z cm v = + 0.01y – 0.02z cm w = 0.01x + 0.01z cm Find the locations of the two points (0,0,0) and (5,0,0) after deformation. What is the change of distance between these two points after deformation? Calculate the strain components corresponding to the given displacement field. Use the definition of xx ε to estimate the change of distance between the two points. Compare the two results. Solution: (a) Consider the point (0,0,0), after deformation :  ' , , = + = = = = z y x u x x  ' , , = + = = = = z y x v y y  ' , , = + = = = = z y x w z z So the corresponding location after deformation is (0,0,0) (b) Consider the point (5,0,0), after deformation : 1 . 5 5 02 . 5  ' , , 5 = × + = + = = = = z y x u x x  ' , , 5 = + = = = = z y x v y y 05 . 5 01 .  ' , , 5 − = × − = + = = = = z y x w z z (c) The change of distance between these two points after deformation. (1) before deformation: distance between (0,0,0) and (5,0,0), 5 = before D (2) after deformation: distance between (0,0,0) and ( 5.1 , 0 , 0.05 ), ( ) ( ) ( ) 100245 . 5 05 . 1 . 5 2 2 2 = − − + − + − = after D (3) change of distance D δ : 100245 . 5 100245 . 5 = − = − = before after D D D δ (d) Calculate the strain components 2.3.1 Mechanics of Aircraft structures C.T. Sun 02 . = ∂ ∂ = x u xx ε , 02 . 02 . = + = ∂ ∂ + ∂ ∂ = x v y u xy γ , 01 . = ∂ ∂ = y v yy ε , 02 . 02 . − = + − = ∂ ∂ + ∂ ∂ = y w z v yz γ , 01 . = ∂ ∂ = z w zz ε 02 . 01 . 01 . − = − − = ∂ ∂ + ∂ ∂ = x w z u xz γ (e) The normal strain in the xdirection is 02 , x u xx = ∂ ∂ = ε . The change of distance between the two points can be estimated by => ( ) 1 . ) 5 ( 02 ....
View
Full Document
 Fall '08
 Chen
 Force, Aircraft structures C.T., structures C.T. Sun, Aircraft structures C.T. Sun

Click to edit the document details