hw32solutions

hw32solutions - AAE 221 – Aerospace Structures II –...

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Unformatted text preview: AAE 221 – Aerospace Structures II – Spring 2004 Chapter 3.2 – Analytical Solutions of Static Problems Using Energy Methods – Solution 1a. Starting with the GDE, ( 0 ) 3 1 2 ( 0 )  3 1 2 ¡ ¢ ¥ £ ¤ EIw' ' ' ' ( x ) = − Pδ x − Boundary conditions: 2L 3 w(0 ) = 0 EIw' ' (0) = Kw' (0 ) EIw' ' (L ) = 0 EIw' ' ' (L ) = 0 ¦ ¨ § Integrating, EIw' ' ' ( x ) = − Pstp x − ¦ §  ©  2L + C1 3 ¦ ¨ §  ©  EIw' ' ( x ) = − x − ©   © 2L 2L + C1 x + C 2 Pstp x − 3 3 C x2 2L P + 1 + C2 x + C3 stp x − 2 3 2 ¦ ¦ § ¨ § ©   ©  ¨  C x3 C x2 2L P 2L + 1 + 2 + C3 x + C 4 stp x − 3 6 3 6 2 Applying the boundary conditions, we get The first BC above gives C 4 = 0 . The forth BC above gives 0 = − P + C1 C1 = P . 2L − 2 PL The third BC above gives 0 = − L − P + PL + C 2 . 3 3 − 2 PL The second BC above gives Kw' (0 ) = , but since 3 C C − 2 PL − 2 PLEI − 2 PL2 C3 = . EIw' (0) = C 3 w' (0 ) = 3 , K 3 = = EI EI 3 3K 3α " # % & " # % & EIw( x ) = stp x − 3 2L 1 1 2 PL2 + Px 3 − PLx 2 − x 3 6 3 3α 3 2 ( 0 3 1 ( 0 2L 1x 1x 2x EIw( x ) − 1 x 2 = − + − − stp x − 3 6L3 3 6L 3L 3α L PL When x = L, the right-hand-side of the non-dimensionalized equation is a constant, i.e., PL3 − 14 2 − w(L ) = EI 81 3α Therefore, doubling the length increases the deflection by a factor of 8. 3 45 6 ) 2 ) 9 7 8 1 2 ( 0 ) 3 1 2 $ ' $ −P 2L x− 6 3 In non-dimensional form, ' 3 !        ¦ § EIw( x ) = − x − ©  3 ¦ § 2L EIw' ( x ) = − x − 3 ¨   ¨ ¨  2 1b. Applying Castigliano’s theorems, 2 2 1 L M y (x ) 1 M y (0 ) U= dx + 2 0 EI 2K where M y (0 ) = Kw' (0 ) . We can apply a dummy load PD at the end of the beam and then inspect the free body diagram at various sections of the beam. 2L For <x≤L, 3 M y ( x ) = PD (L − x ) For 0 ≤ x ≤ 2L , 3 Substituting, § ¨  2 D 2 Taking the partial derivative of U with respect to PD yields the following expression. ∂U −1 = (14 PLK + 54PEI − 27 PD LK − 81EIPD )L2 ∂PD 81KEI Finally, ! 2 1 PD L3 1 2 + + PD L − PL 162 EI 2K 3 # $ " % 3 & & ( ' 3 1 1 PD L3 1 L3 (− 3PD + 2 P ) U= + 2 EI 81 − PD + P 81 −P D +P 1 ) 0 )   § ¨ 1 2L −x U= PD (L − x ) − P 2 EI 0 3 Integrating,      2L / 3 1 dx + 2 EI P2 L 1 P (L − x ) dx + PD L − 2K 3 2L / 3 2 L ©  2 ¡¢ ¤ ¥ M y ( x ) = PD ( L − x ) − P ©   2L −x 3 £ ¦ 2 ∂U −1 = (14PLK + 54PEI )L2 PD → 0 ∂P 81KEI D Simplifying the above expression, PL3 − 14 2 − w(L ) = EI 81 3α w(L ) = lim 1c. EIw(L ) − 14 2 = − 81 3α PL3 ¡ ¢ ¥ £ ¤ As α tends to infinity, K tends to infinity. The problem becomes a cantilever beam problem. However, when α is zero, the problem is ill-defined. 2a. This is a statically indeterminate problem with a relax spring constraint (Rs) as shown in the free body diagrams below. For section b-c, we sum the moments about point c (counterclockwise positive). M c = M 2 + Py − R s y = 0 M 2 = y (R s − P ) For section a-b, we sum the forces (upward positive) and the moments about point aa (counterclockwise positive). F = Fw + 2 Lq 0 − (R s − P ) = 0 Fw = −2 Lq 0 + R s − P ¢ ¡ £ M aa = M 1 − (Rs − P )(2 L − x ) − M 1 = (Rs − P )(2 L − x ) + 2L − x q 0 (2 L − x ) = 0 2 The torsional moment (twist) is M t = L (R s − P ) Applying Castigliano’s theorems, 2L L 2L 2 2 2 1 M 12 1 M2 1 Mt 1 Rs U= dx + dx + dx + 2 0 EI 2 0 EI 2 0 µJ 2K Taking the partial derivative of U with respect to Rs yields ∂U − 2 L4 Kq 0 − 5 L3 KP + 5L3 KRs + Rs EI = ∂Rs EIK Setting the above expression to zero gives 2 L4 Kq 0 + 5 L3 KP Rs = − − 5 L3 K − EI Substituting Rs back into U, we have ¦ ¦ ¦ ¥ ¤ q0 (2L − x )2 2 1 (20q 2 KL5 + 8q 2 EIL2 + 20q 0 PEIL + 25P 2 EI )L3 10 (5L3 K + EI )EI Hence, the deflection at the point where load P is applied is given by ∂U L3 (2q 0 L + 5P ) = w= ∂P 5 L3 K + EI U= 3. Because the wing is symmetric, we can consider only half of the wing. Introducing a dummy load P0 at x = L, the free body diagram at any x-station becomes The distributed load in terms of g is given as L The corresponding moment is ¡ 2 0 L L   x3 x4 x3 x4 M (x ) = P x + 2 P0 f 0 x − + f 02 − 3L 12 L2 3L 12 L2 Therefore, the internal energy is     2 y 2 0 2     §   = P0 x + f 0  ¦ ¦ x3 x4 − 3L 12 L2 or § ¨  = P0 x + f 0 ¦ ¦ © © x 2 (x − γ ) 1 − 2γ + γ 2 dγ ¨   © ©     M y ( x ) = P0 x + x (x − γ ) f (γ )dγ 0 ¡ £ ¤ f (γ ) = f 0 1 − £ ¤ = f0 1 − 12 L − 2 Lγ + γ 2 2 L ¢ ¥ ¢ (L − γ )2 ¥ ( ) 2 1 2 U= M y ( x )dx 2 EI 0 L3 1 110 f 02 L2 + 1197 f 0 LP0 + 3780 P02 2 EI 11340 Taking the partial derivative of U with respect to P0 gives 1 L3 ∂U (120 P0 + 19 f 0 L ) = ∂P0 2 EI 180 = ∂U 19 f 0 L4 = w(L ) = lim P0 → 0 ∂P 360 EI 0 4a. Starting with the GDE, L ( ) µJφ ' ' ( x ) = − mt = Boundary conditions: m0 x L µJφ ' (0 ) = Kφ (0 ) µJφ ' (L ) = 0 m0 x 2 + C1 2L m x3 µJφ ( x ) = 0 + C1 x + C 2 6L Integrating, µJφ ' ( x ) = The boundary condition at x = L gives us the relation C1 = The boundary condition at x = 0 gives us the relation − m0 L KC 2 − m 0 LµJ = C1 = C2 = 2 2K µJ Substituting, ¢ £ ©   ©   − m0 L . 2 − 4b. We first apply a dummy load MD at x = L/2, as shown in the diagram below.   m0 L2 − 11 µJ = − 48 2 KL µJ     "  ! At the middle of the beam, the twist angle is m0 L2 1 1 L 11 1 µJ = − − φ 2 22 2 KL µJ 6 8                % # $       ¢ ©   ¥ ¦ ¥ φ (x ) = m 0 L2 1 x µJ 6 L     3 1x 1 µJ − 2L 2 KL ¤¨ ¨ ¡ ¨   § Then we break the shaft into two sections and inspect the free body diagrams. L For < x ≤ L , 2 Since the section is at static equilibrium, the reaction moment Mt(x) must be m ξ2 M t ( x ) = − mt (ξ )dξ = − 0 2L x L L =− x m0 2 (L − x 2 ) 2L L For 0 ≤ x ≤ , 2 m0 ξ 2 M t ( x ) = M D − mt (ξ )dξ = M D − 2L x L ¡ The internal energy is   2 µJ     0 = 2 2 60 M D (2 µJ + KL ) − 5M D Lm 0 (24 µJ + 11KL ) + 2 L3 m 0 (15 LµJ + 8 K ) 240 µJK Taking the partial derivative of U with respect to MD gives ∂U 24 M D (2 µJ + KL ) − Lm0 (24µJ + 11KL ) = ∂M D 48µJK The twist angle at middle is, therefore, ∂U − Lm0 (24µJ + 11KL ) L = lim φ = M D →0 ∂M 2 48µJK D which is equivalent to ¢ ¤ £ § ¥ ¦ ¨ © + ¨ m 2 L2 1 2 M D − M D m0 L + 0 2K 4    U= 1 L/2 2 MD − m2 M D m0 2 1 L − x 2 + 02 L2 − x 2 dx + L 2 µJ 4L ( ) ( )  L =MD − x m0 2 L − x2 2L ( ) M t (0) = M D −  m0 L 2 2 m0 2 L − x 2 dx 2 L / 2 4L L ( ) 5. Consider the free body diagram below. Static equilibrium equations: Fx = P − R H = 0 R H = P  Fy = V + R − V = −R +  0 0 To get the expression for internal energy, we need M(x) and M(y). = R(2 L − x ) + f 0 x 4 − 4 f 0 Lx 3 + 16 f 0 L3 x − 4 L3 (4 f 0 L + 3P ) 12 L2 M ( y ) = Py  M ( x ) = − PL + R(2 L − x ) − 2L (ξ − x ) f (ξ )dξ x  M a = M + PL + xf ( x )dx − 2 RL  2L ¨ 2L ¨ © f ( x )dx 2L f ( x )dx 2L 0 M = − PL + 2 RL − xf ( x )dx 0 ¡ £ ¤ ¡ £ ¤ φ ¢ L m L2 − 11 µJ =0 − 2 µJ 48 2 KL ¥ ¢ ¥ § ¦  Thus, the internal energy is given by ¨ § ¡ ¦ ¦ £ ¤ U= = 1 2 EI 2L M 2 ( x )dx + M 2 ( y )dy 0 L 0 1 L3 7560 R 2 − 252 R(28 f 0 L + 45 P ) + 1840 f 02 L2 + 4536 f 0 LP + 6615 P 2 2 EI 2835 Taking the partial derivative of U with respect to R gives 1 4 L3 ∂U (60R − 28 f 0 L − 45P ) = ∂R 2 EI 45 Setting the above expression to zero gives 7 f L 3P R= 0 + 15 4 Now we can substitute the reaction R into the force and moment balance equations. 2L 13 f 0 L 3P V = − R + f ( x )dx = − 15 4 0 M = − PL + 2 RL − xf ( x )dx = 0 2L 2 PL 2 f 0 L − 2 5 [ Finally, the deflection at the point where load P is applied is simply ∂U 5PL3 2 f 0 L4 w= = − 6 EI 15EI ∂P ¢ ¥ ...
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