088235 2088235 v 0 0 v0 0 the displacement is found

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: + C 3 + C 4 6Ë 4¯ 4¯ 2 Ë 4¯ 4¯ 6 2 Ë Ë Applying the first two boundary conditions, w(0) = 0 fi C 4 = 0 w¢(0) = 0 fi C 3 = 0 3 2 Applying the next two B.C., EIw¢¢¢( L) - kw( L) = 0 fi Fo + C1 + Fo k 27 L3 M o k L2 C1 kL3 C 2 kL2 + =0 6 EI 64 2 EI 16 6 EI 2 EI EIw¢¢( L) - kq w¢( L) = 0 F k 9 L2 M o k L kq C1 L2 kq C 2 L 3Fo L - M o + C1 L + C 2 - o q + + =0 4 2 EI 16 EI 4 EI 2 EI fi- This produces a 2x2 linear system È kL3 1Í 6 EI Í 2 Í L + kq L Í 2 EI Î È ˘ Ê 9kL3 ˆ kL2 M o kL2 ˘ Fo Á1 Í ˙ Á 128 EI ˜ - 32 EI ˜ ˙ Ë ¯ 2 EI ˙ È C1 ˘ = Í ˙ kq L ˙ ÍC 2 ˙ Í Ê 3L kq 9 L2 ˆ k L ˆ˙ Ê Î˚ ˜ + M o Á1 + q ˜ ˙ 1+ + Í Fo Á Á EI ˙ ˚ 32 EI ˜ Ë 4 EI ¯˙ Í Ë4 ¯ Î ˚ (b) Bending moment distribution Lˆ Ê Lˆ 3L ˆ Ê Ê M y = EIw¢¢ = - Fo Á x - ˜ stpÁ x - ˜ - M o stpÁ x ˜ + C1 x + C 2 4¯ Ë 4¯ 4¯ Ë Ë (c) if M o = Fo L Fo 2L FL kq = o 2 k= then, setting a = Fo L2 , We can substitute these values into our 2x2 linear system EI 13a ˘ a 2˘ È È Í 1 - 256 ˙ Í 1 - 12 ˙ È C1 ˘ 4L = Fo Í Í Ê aˆ a ˙ ÍC 2 ˙ 7 17 a ˆ˙ Î˚ Í LÊ + Í LÁ1 + ˜ 1 + ˙ Á ˜˙ 2˙ Í Ë 4¯ Í Ë 4 64 ¯˙ Î ˚ Î ˚ Solving for the two constants of integration, 3Fo 512 + 454a + 21a 2 C1 = 32 48 + 32a + a 2 ( ( ) ) ) C2 = Fo L 2304 - 244a + 29a 2 64 48 + 32a + a 2 ( ( ) Therefore, the displacement (non-dimensional) is EI 1Ê x 1ˆ Lˆ 1Ê x 3ˆ 3L ˆ x3 x2 Ê Ê w( x) = - Á - ˜ stpÁ x - ˜ - Á - ˜ stpÁ x + C2 ˜ + C1 6Ë L 4¯ 4 ¯ 2Ë L 4¯ 4¯ Fo L3 6 Fo L3 2 Fo L3 Ë Ë 3 2 11. (i) A* =  (ii) Ei Ai = 2 + 3 * 4 + 1.5 * 2 = 17 in 2 Eo z* = 0 y o* = (iii) ÊE * I yy =  Á ÁE Ëo ˆ Ê1 ˆ Ê1 ˆ Ê1 ˆ ˜ I yy + z 2 A i = 1 * Á * 1 * 2 3 ˜ + 3Á * 2 * 2 3 ˜ + 1.5Á * 1 * 2 3 ˜ = 5.66 in 4 ˜ Ë 12 ¯ Ë 12 ¯ Ë 12 ¯ ¯i 1 A* ÊEˆ  Á E ˜ (y ) A Á˜ Ë o ¯i oi i = 1 (1 * 1.5 * 2 + 1.5 * -1.5 * 2) = -.088235 17 [ ÊE * I zz =  Á ÁE Ëo ˆ Ê1 ˆ ˜ I zz + y 2 A i = 1 * Á * 2 * 13 + 1.588235 2 * 2 ˜ + ˜ Ë 12 ¯ ¯i [ Ê1 ˆ Ê1 ˆ 3Á * 2 * 2 3 + .088235 2 * 4 ˜ + 1.5Á * 2 * 13 + 1.4112 * 2 ˜ = 15.5343in 4 Ë 12 ¯ Ë 12 ¯ (iv) Axial deflection Eo A*u ¢¢(x ) = 0 Eo A*u ¢(x ) = C1 Eo A*u (x ) = C1 x + C 2 Using the following boundary conditions, u (0) = 0 fi C 2 = 0 EA* u ¢( L) = 100 fi C1 = 100 Therefore, Eo A* u ( x) = 100 x u ( x) = (v) 100 x = 5.882 ¥ 10 -7 x 6 10 ¥ 10 ( ) 17 Eo I z*z v ¢¢¢¢( x) = -10 integrating four times, and applying the following boundary conditions v ¢¢¢( L) = 0 v ¢¢( L) = -100 ⋅ 2.088235 = -208.8235 v ¢(0) = 0 v(0) = 0 The displacement is found to be v( x) = 1 * Eo I zz Ê 5 4 500 3 ˆ x - 25104.41175 x 2 ˜ Á- x + 3 Ë 12 ¯ v( x) = -2.68223 ¥ 10 -4 x 4 + 1.07289 ¥ 10 -6 x 3 - 1.61606 ¥ 10 -4 x 2 (vi) The w displacement is * Eo I yy w¢¢...
View Full Document

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.

Ask a homework question - tutors are online