hw21_Solutions

# The values for e and i and solve the system of

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Unformatted text preview: ng the boundary conditions, * Eo I z*z v ¢¢¢ + Eo I yz w¢¢¢ = C1 * Eo I z*z v ¢¢ + Eo I yz w¢¢ = C 2 * Eo I z*z v ¢ + Eo I yz w¢ = C 3 * Eo I z*z v + Eo I yz w = C 4 Note that the constants of integration are 0 in each case, thus * Eo I z*z v + Eo I yz w = 0 (1) For the second equation, integrating, * * Eo I yy w¢¢¢ + Eo I yz v ¢¢¢ = f z x + C1 C1 = - f z L * * Eo I yy w¢¢ + Eo I yz v ¢¢ = f z x2 - f z Lx + C 2 2 C2 = f z L2 2 x3 x 2 f z L2 - fzL + x + C3 6 2 2 * * Eo I yy w¢ + Eo I yz v ¢ = f z C3 = 0 x4 x 3 f z L2 x 2 Eo I w + Eo I v = f z - fzL + + C4 24 6 22 C4 = 0 * yy * yz Ê x4 x 3 L2 x 2 ˆ ˜ E I w + E I v = fzÁ - L + Á 24 6 4˜ Ë ¯ * o yy * o yz Now, plug in the values for E and I and solve the system of equations for v and w .00228646v + .0045w = 0 (3.15 ¥ 10 )v + (9.90206 ¥ 10 )w = b 8 8 Ê x4 x 3 L2 x 2 ˆ ˜. Where b = f z Á - L + Á 24 6 4˜ Ë ¯ Solving for v and w, v = -.5315590062 ¥ 10 -8 b w = .2700863123 ¥ 10 -8 b Thus, Ê x4 ˆ v( x) = 2.445 ¥ 10 -5 Á - 3.33 x 3 + 100 x 2 ˜ Á 24 ˜ Ë ¯ 4 Êx ˆ w( x) = -1.2424 ¥ 10 -5 Á - 3.33 x 3 + 100 x 2 ˜ Á 24 ˜ Ë ¯ To plot the shear and bending moment diagrams, note the following: (insert diagram) ÂF z = V - 4600(20 - x) = 0 A ÂM Ê 20 - x ˆ = M + (4600(20 - x ))Á ˜=0 Ë2¯ Thus, the shear V, and bending moments are V = 4600(20 - x) N M = -2300(20 - x) 2 N-m 5. The GDE for this problem is EIw¢¢¢¢ = f z ( x) 3L ˆ 2q o x Ê L ˆˆ Ê Ê where the term fz is f z = qo Ld Á x Á1 - stpÁ x - ˜ ˜ ˜+ Á 4¯ LË 2 ¯˜ Ë Ë ¯ The boundary conditions for this problem are w(0) = w( L) = 0 w¢( L) = w¢¢(0) = 0 Integrating this function while noting that ˆ Ú d (x - x )dx = stp(x - x ) o o ˆ ˆ ˆ ˆ Ú stp(x - x )dx = stp(x - x )Ú dx o o The non-dimensional displacement w is EIw(x ) 9 1 1 Ê x 3ˆ 1 Ê x ˆ Ê x 1ˆ Ê x 1ˆ =stpÁ - ˜ + Á ˜ - stpÁ - ˜ x 5 stpÁ - ˜ + 4 128 Ë L 4 ¯ 60 Ë L ¯ 60 Ë L 2 ¯ 480 Ë L 2 ¯ qo L 5 9 1 3 Ê x 3ˆ 1 Ê x 3ˆ Ê x 1ˆ Ê x 1ˆ stpÁ - ˜ x + stpÁ - ˜ x - stpÁ - ˜ x 2 stpÁ - ˜ x 2 32 Ë L 4 ¯ 64 Ë L 2 ¯ 8 Ë L 4¯ 24 Ë L 2 ¯ 3 x3 77 x 139 Ê x ˆ Ê x 3ˆ x Ê x 1ˆ + stpÁ - ˜ + stpÁ - ˜ + Á˜ 6 Ë L 4 ¯ 24 Ë L 2 ¯ 3840 L 3870 Ë L ¯ 3 6. The GDE for this problem is EIv ¢¢¢¢( x) = f y ( x) = The boundary conditions are a2 rg 2 EIv ¢¢(0) = + Kv ¢(0 ) EIv ¢ ¢ ¢(L ) = + kv(L ) v(0) = 0 EIv ¢¢( L) = 0 Integrating, we have EIv ¢¢¢ = f y x + C1 x2 + C1 x + C 2 2 x3 x2 EIv ¢ = f y + C1 + C 2 x + C3 6 2 x4 x3 x2 EIv = f y + C1 + C2 + C3 x + C 4 24 6 2 EIv ¢¢ = f y From v(0) = 0, C4 = 0 From the other B.C., EIv ¢¢(0) = EIC 2 = KC 3 Ê L2 ˆ EIv ¢¢(L ) = EI Á f y + C1 L + C 2 ˜ = 0 Á ˜ 2 Ë ¯ EIv ¢¢¢(L ) = f y L + C1 = ˆ k Ê x4 x...
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