hw21_Solutions

As a approaches infinity the problem is equivalent to

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Unformatted text preview: 3 x2 Áy + C1 + C2 + C3 x + C 4 ˜ Á 24 ˜ EI Ë 6 2 ¯ Solving this system of three equations for the constants of integration, È 0 Í Í L Í EI L3 Í+ 6 Îk the constants of integration are: EI 1 L2 2 È ˘ ˘ Í ˙ 0 - K ˙ È C1 ˘ Í ˙ 2 L ˙ 0 ˙ ÍC 2 ˙ = Í - f y ˙Í ˙ Í ˙ 2 L ˙ ÍC 3 ˙ Í EIf y L f y L4 ˙ Î˚ ˚ Í ˙ 24 ˚ Îk 2 3 1 Lf y 24 KEI + 12 L kEI = 5 KL k C1 = 8 3KEI + 3L2 kEI + kL3 k 2 3 1 KL f y EI 12 EI + L k 8 3KEI + 3L2 kEI + kL3 k ( ) C2 = ( ) 2 3 1 L f y EI 12 EI + L k C3 = 8 3KEI + 3L2 kEI + kL3 k ( ) Alternately, using the non dimensional parameters a and b characterizing the stiffness of the rotational and linear springs respectively, a= KL EI kL2 b= EI C1 = C2 = C3 = f y 24a + 12 b + 5ab 8 3a + 3b + ab 12a + ab 8 3a + 3b + ab fy 12 + b 8 3a + 3b + ab fy Thus, 2 EIv( x) 1 Êxˆ 1 24a + 12 b + 5ab Ê x ˆ 1 12a + ab Ê x ˆ 1 12 + b Êxˆ =- Á ˜ + Á ˜Á ˜Á˜ 2 4 24 Ë L ¯ 48 3a + 3b + ab Ë L ¯ 16 3a + 3b + ab Ë L ¯ 8 3a + 3b + ab Ë L ¯ a rgL 4 3 2 The first plot was done with (a,b)=(5,10), the second with (a,b)= (•,0 ) As a approached 0, the problem becomes ill-defined. As a approaches infinity the problem is equivalent to a rigidly supported cantilever beam for which the deflection of the beam as the end is - f y L4 8 EI and is the solution’s limit. Since the cross-section is known, the moment of inertia can be found. A strip of material parallel to the base has the area dA = ydz = (a - z )dz By definition, a 2 a I y = Ú z dA = Ú z 0 0 2 (a - z ) a4 dz = 12 Using the parallel axis theorem the moment of inertia I = I - Ad 2 = The final answer is therefore Ea 2 v( x) 1 Êxˆ 1 24a + 12 b + 5ab =- Á ˜ + 4 24 Ë L ¯ 48 3a + 3b + ab 18 rgL 4 a4 a2 12 2 a4 Êaˆ Á˜= 36 Ë3¯ 2 1 12a + ab Êxˆ Á ˜Ë L ¯ 16 3a + 3b + ab 3 12 + b Êxˆ 1 Á ˜Ë L ¯ 8 3a + 3b + ab 2 Êxˆ Á˜ ËL¯ For b =0, Ea 2 v(a ) 11 =- 4 8 2a 18 rgL 7. The boundary conditions for this problem are w(0) = w( L) = 0 EIw¢¢(0) = EIw¢¢( L) = 0 The governing differential equation is then EIw¢¢¢¢( x) = f ( x) = q ( x) - kw( x) 8. The GDE for this problem is EAu ¢¢( x) = - f x (x ) where the distributed force term is 3L ˘ È Ê 2x ˆ Ê L ˆ f x ( x) = po Ld Í x - ˙ + po Á1 - ˜ stpÁ - x ˜ 4˚ L ¯ Ë2 Î Ë ¯ This may be rewritten as po - 2 po 2p Ê Lˆ Ê Lˆ 3L ˆ Ê x + o Á x - ˜ stpÁ x - ˜ + po Ld Á x ˜ L LË 2¯ Ë 2¯ 4¯ Ë Therefore, the GDE becomes EAu ¢¢( x) 2 2Ê Lˆ Ê Lˆ 3L ˆ Ê = -1 + x - Á x - ˜ stpÁ x - ˜ - Ld Á x ˜ po L LË 2¯ Ë 2¯ 4¯ Ë The boundary conditions are u (0) = 0 ~ EAu ' ( L) = N ( L) Integrating, EAu ¢( x) x2 1Ê Lˆ Lˆ 3L ˆ Ê Ê = -x + x - Á x - ˜ stpÁ x - ˜ - LstpÁ x ˜ + C1 po L LË 2¯ 2¯ 4¯ Ë Ë EAu ( x) x2 x3 1Ê Lˆ Lˆ 3L ˆ Ê 3L ˆ Ê Ê =+ xÁ x - ˜ stpÁ x - ˜ - LÁ x ˜ stpÁ x ˜ + C1 x + C 2 po 2 3L 3L Ë 2¯ 2¯ 4¯ Ë 4¯ Ë Ë 3 2 From the boundary condition u(0) = 0, C2 = 0 Using the second B.C., ˆ Ê L2 ˆ k Ê L2 L2 1 Ê L3 ˆ ÊLˆ ÁÁ ˜ - Lp o + C1 po = Á ˜ - LÁ ˜ + C1 L ˜ po + Á4˜ Á8˜ ˜ EA Á...
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