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Unformatted text preview: AE 321 – Solution to Practice Problems Chapter 8: Plane Problems 1a. The given Airy stress function, ( ) y x , ! , would provide the stress field for this problem if it satisfies the compatibility equations (equilibrium is identically satisfied), i.e. ( ) , 4 = ! " y x ; ! 4 " ! x 4 + 2 ! 4 " ! x 2 ! y 2 + ! 4 " ! y 4 = (1) and ! 4 " ! x 4 = 120 a 51 xy + 24 a 42 y 2 (2a) ! 4 " ! x 2 ! y 2 = 4 a 22 + 24 a 42 x 2 + 36 a 33 xy (2b) ! 4 " ! y 4 = 120 a 15 xy (2c) Substituting (2) in (1), we obtain ! 4 " ! x 4 + 2 ! 4 " ! x 2 ! y 2 + ! 4 " ! y 4 = 8 a 22 + 120 a 51 + 72 a 33 + 120 a 15 ( ) xy + 48 a 42 x 2 + 24 a 42 y 2 = ! a 22 = and a 42 = (3a) 5 a 51 + 3 a 33 + 5 a 15 = or a 33 = ! 5 3 a 51 + a 15 ( ) (3b) Therefore the given Airy stress function has to satisfy Equations (3) to provide the solution of our problem. Then the stress field is given as follows. ! xx = " 2 # " y 2 = 2 a 02 + 6 a 03 y + 6 a 13 xy + 6 a 33 x 3 y + 20 a 15 xy 3 (4a) ! yy = " 2 # " x 2 = 6 a 30 x + 6 a 31 xy + 20 a 51 x 3 y + 6 a 33 xy 3 (4b) ! xy = " # 2 $ # x # y = " a 11 " 3 a 31 x 2 " 5 a 51 x 4 " 3 a 13 y 2 " 9 a 33 x 2 y 2 " 5 a 15 y 4 (4c) In order to determine the coefficients in Equations (4) we have to apply the boundary conditions (Cauchy’s equations): (i) Surface 2 h y = along L x ! ! n j [ ] = 1 [ ] T ( ) ( ) yy n y xy n x T T ! ! = = = = and ! ! (5) Substituting (4) in (5) ! yy  y = h 2 = = 6 a 30 + 3 ha 31 + 3 4 h 3 a 33 " # $ % & ’ x + 20 ha 51 x 3 (6a) ! xy  y = h 2 = = a 11 + 3 4 h 2 a 13 + 5 16 h 4 a 15 " # $ % & ’ + 3 a 31 + 9 4 h 2 a 33 " # $ % & ’ x 2 + 5 a 51 x 4 (6b) From (6a) and (6b), we obtain 51 = a (7) 6 a 30 + 3 ha 31 + 3 4 h 3 a 33 = (8) a 11 + 3 4 h 2 a 13 + 5 16 h 4 a 15 = or a 11 = ! 1 16 12 h 2 a 13 + 5 h 4 a 15 ( ) (9) 3 a 31 + 9 4 h 2 a 33 = or a 31 = ! 3 4 h 2 a 33 (10) (ii) Surface 2 h y ! = along L x ! ! n j [ ] = ! 1 [ ] T T x ! n ( ) = = ! " xy and T y ! n ( ) = P L x = ! " yy (11) Since the exponents of y in expression (4c) are even, it is easy to see that traction in the xdirection will yield to Equation (6b). On the other hand, traction in the y direction yields (taking into account (7)) ! yy  y = " h 2 = " P L x = 6 a 30 " 6 ha 31 " 3 4 h 3 a 33 # $ % & ’ ( x then ! P L = 6 a 30 ! 3 ha 31 ! 3 4 h 3 a 33 (12) From (7) and (12), we obtain a 30 = ! P 12 L (13) Substituting (13) and (10) in (8) or (12) yields a 33 = ! P 3 h 3 L (14) a 31 = P 4 hL (15) (14) and (7) in (3b) a 15 = P 5 h 3 L (16) Now, let’s substitute the known values of the coefficients into the expressions for the components of the stress tensor ! xx = 2 a 02 + 6 a 03 y + 6 a 13 xy " 2 P h 3 L x 3 y + 4 P h 3 L xy 3 (17a) ! yy = " P 2 L x + 3 P 2 hL xy " 2 P h 3 L xy 3 (17b) !...
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 Spring '09
 Lambros
 Stress, Plane Strain

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