50000 stress s xx nc m ci m ci z yy y yz y i

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Unformatted text preview: Applying the boundary condition of u(o)=0, The constant of integration, C is found to be C= Thus, the axial deflection u(x) is PL 1 EAo e È ˘ Í P ÊLˆ 1˙ u ( x) = Á ˜ Í- 1 + x˙ EAo Ë e ¯ Í 1- e ˙ Í L˙ Î ˚ È ˘ EAo Ê 1 ˆ Í 1˙ (b) u ( x) = Á ˜ Í- 1 + x˙ PL Ë e ¯ Í 1- e ˙ Í L˙ Î ˚ (c) The stress is then s ( x) = P Ao xˆ Ê Á1 - e ˜ L¯ Ë -2 or non-dimensionally, AÊ xˆ s ( x ) = o = Á1 - e ˜ PË L¯ -2 15. The GDE for this problem is Lˆ 3L ˆ Ê Ê EIv ¢¢¢¢( x) = - f o + f o stpÁ x - ˜ - f o Ld Á x ˜ 4¯ 4¯ Ë Ë The boundary conditions are v(0) = v( L) = 0 v ¢(0) = 0 EIv ¢¢( L) = 0 Integrating four times with respect to x 1 1Ê Lˆ Lˆ 1 3L ˆ 3L ˆ 1 1 Ê Ê Ê 3 2 EIv( x) = fo x4 + f o Á x - ˜ stpÁ x - ˜ - f o LÁ x ˜ stpÁ x ˜ + C1 x + C 2 x + C 3 x + C 4 24 24 Ë 4¯ 4¯ 6 4¯ 4¯ 6 2 Ë Ë Ë 4 3 Applying the boundary conditions C 4 = C3 = 0 1249 fo L 2048 - 289 C2 = f o L2 2048 C1 = Thus the non dimensional deflection is 4 4 3 EIv( x) 1 È Ê x ˆ Ê x 1 ˆ Ê x 1 ˆ˘ 1 Ê x 3 ˆ Ê x 3 ˆ 1249 x 289 = - Á ˜ + Á - ˜ stpÁ - ˜˙ - Á - ˜ stpÁ - ˜ + Í 4 24 Í Ë L ¯ Ë L 4 ¯ fo L Ë L 4 ¯˙ 6 Ë L 4 ¯ Ë L 4 ¯ 2048 L 2048 Î ˚ The moment is 2 2 EIv ¢¢( x) 1 È Ê x ˆ Ê x 1 ˆ Ê x 1 ˆ˘ Ê x 3 ˆ Ê x 3 ˆ 1249 x 289 = Í- Á ˜ + Á - ˜ stpÁ - ˜˙ - Á - ˜ stpÁ - ˜ + 4 2 Í Ë L¯ Ë L 4¯ fo L Ë L 4 ¯˙ Ë L 4 ¯ Ë L 4 ¯ 2048 L 2048 Î ˚ 16. Boundary conditions: w(0 ) = 0 w' (0 ) = 0 w' ' (L ) = 0 EIw' ' ' (L ) = kw(L ) Governing differential equation: Lˆ Ê EIw' ' ' ' = - M 0 d' Á x - ˜ 2¯ Ë Lˆ Ê EIw' ' ' = - M 0 dÁ x - ˜ + C1 2¯ Ë Lˆ Ê EIw' ' = - M 0 stpÁ x - ˜ + C1 x + C 2 2¯ Ë Lˆ Ê Lˆ 1 Ê EIw' = - M 0 Á x - ˜ stpÁ x - ˜ + C1 x 2 + C 2 x + C 3 2¯ Ë 2¯ 2 Ë Lˆ Ê Áx - ˜ Lˆ 1 1 2¯ Ê EIw = - M 0 Ë stpÁ x - ˜ + C1 x 3 + C 2 x 2 + C 3 x + C 4 2 2¯ 6 2 Ë fi C1 = kw(L ), C 2 = M 0 - Lkw(L ), C 3 = C 4 = 0 2 Therefore, we have EIw(x ) = - M0 Ê Lˆ Lˆ 1 12 Ê 3 Á x - ˜ stpÁ x - ˜ + kw(L )x + x (M 0 - Lkw(L )) 2Ë 2¯ 2¯ 6 2 Ë 2 We need the expression for w(L ) Ê L2 ˆ 1 1 Á ˜ + kw(L )L3 + L2 (M 0 - Lkw(L )) Á4˜ 6 2 ˯ 2 (3 8)M 0 L EI fi w(L ) = 3 , where a = 3 kL (a + 1 3) kL EIw(L ) = M0 2 Substituting, we get the deflection 2 2 ÈÊ x ˆ 3 Ê x ˆ 2 ˘ EIw(x ) 1Ê x 1ˆ 1 Ê x 1ˆ 1Ê x ˆ = - Á - ˜ stpÁ - ˜ + Á ˜ + ÍÁ ˜ - 3Á ˜ ˙ 2Ë L 2¯ M 0 L2 Ë L 2 ¯ 2 Ë L ¯ 16(a + 1 3) ÍË L ¯ ËL¯ ˙ Î ˚ For bending moment, Lˆ Ê EIw' ' (x ) = - M 0 stpÁ x - ˜ + kw(L )x + M 0 - Lkw(L ) 2¯ Ë EIw' ' (x ) 3 Ê x 1ˆ Êx ˆ fi = - stpÁ - ˜ + 1 + Á - 1˜ M0 8(a + 1 3)Ë L ¯ Ë L 2¯ The...
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This note was uploaded on 01/18/2010 for the course AE 322 taught by Professor Lambros during the Spring '04 term at University of Illinois at Urbana–Champaign.

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