hw21_Solutions

A since s x p du e a x dx e du p 1 2 dx ao x

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Unformatted text preview: ¢( x) = 0 and contains the following boundary conditions w¢¢¢( L) = 0 w¢¢( L) = -100 w¢(0) = 0 w(0) = 0 integrating and applying the boundary conditions results in: w( x) = -8.8235 ¥ 10 -7 x 2 (b) The stress is found from the equation E = Eo ÏNc Ê M cI* - M cI* Ô y yz Á z yy Ì * - yÁ * * *2 ÔA Ë I yy I zz - I yz Ó ˆ Ê M cI* - M cI* z yz ˜ - z Á y zz ˜ Á I* I* - I* 2 yz ¯ Ë yy zz ˆ¸ ˜Ô ˜˝ ¯Ô ˛ s xx () () * * Simplifying, with M z = Eo I zz v ¢¢ and M y = Eo I yy w¢¢ s xx = Ê - 5 x 2 + 1000 x - 50208.8235 ˆ Ê - 100 ˆ¸ E ÏNc ˜ - zÁ - yÁ ˜˝ Ì Á ˜ Eo Ó A* 15.5343 Ë ¯ Ë 5.8 ¯˛ With the coordinates of a, b, and c being (50, 2.088235, 1), (50, -1.911765, 1), and (0, 2.088235, 1), plugging (x, y, z) into the stress equation produces a: s xx = 1732 psi b: s xx = -2311 psi c: s xx = 6773 psi 12. (a) The governing differential equation is of the form EI yy w¢¢¢¢ = f ( x) However, since both distributed forces are acting downward f ( x) = - f o - kw( x) Thus the GDE may be written as EI yy w¢¢¢¢ + kw = - f o with the following boundary conditions w(0) = w¢(0) = w( L) = w¢( L) = 0 (b) This is a non-homogenous differential equation, and the solution may be broken into a homogenous solution and a particular solution, i.e., w( x) = wh ( x) + w p ( x) By inspection, w p ( x) = fo k The homogenous solution may be found by solving the characteristic equation associated with the homogenous equation. wh ( x) = e bx (A1 cos bx + A2 sin bx )+ e - bx (A3 cos bx + A4 sin bx ) where b = k 4 EI Using Mathematica to solve for the four unknown coefficients by applying the boundary conditions, f o Ï - 1 + e bL cos bL + e bL sin bL ¸ Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ f Ï1 - e bL cos bL + e bL sin bL ¸ A2 = o Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ A1 = A3 = e bL A4 = e bL f o Ï e bL - cos bL + sin bL ¸ Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ f o Ï e bL - cos bL - sin bL ¸ Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ To plot, w(x), make the assumptions fo =L k bL = 1 13. The GDE for this problem is EIw¢¢¢¢( x) = -q By integrating this DE successively four times EIw = 1 1 1 qx 4 + C1 x 3 + C 2 x 2 + C 3 x + C 4 24 6 2 The following boundary conditions are applied w(0) = w¢(0) = w( L) = EIw¢¢( L) = 0 From the first two boundary conditions C3 = C 4 = 0 From the last two boundary conditions, a system of equations can be set up È1 3 Í6 L ÍL Î C1 = È1 ˘ 1 2 ˘ È C ˘ Í qL4 ˙ L 1 2 ˙ 2 ˙Í ˙ = Í 1 1 ˙ ÎC 2 ˚ Í qL2 ˙ ˚ Î2 ˚ 5 1 qL C 2 = - qL2 8 8 Thus, the deflection is w( x) = q - 3L2 x 2 + 5 Lx 2 - 2 x 4 48 EI [ Non dimensionalizing, 2 2 4 EI 1 È Êxˆ Êxˆ Êxˆ ˘ w( x) 4 = Í- 3Á ˜ + 5Á ˜ - 2Á ˜ ˙ 48 Í Ë L ¯ qL ËL¯ ËL¯ ˙ Î ˚ The moment is 2 EI 1 È Êxˆ Êxˆ ˘ w( x)¢¢ 4 = Í- 1 + 5Á ˜ - 4Á ˜ ˙ 8Í qL ËL¯ ËL¯ ˙ Î ˚ 14. (a) Since s ( x) = P du =E A( x) dx E du P 1 = 2 dx Ao Ê xˆ Á1 - e ˜ L¯ Ë PÊ xˆ u (x ) = Ú Á1 - e L ˜ dx EAo Ë ¯ -2 integrating this expression gives u ( x) = - P EAo Ê L ˆ -1 1 Á ˜( ) + C Ëe...
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