# hw41solutions - AAE 221 Aerospace Structures II Spring 2004...

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AAE 221 – Aerospace Structures II – Spring 2004 Chapter 4 – Introduction to Buckling – Solution 1. We can neglect y effects and assume 0 = y s . From the beam equation, 0 ' ' = + w EI P w , we get 2 2 4 S EI P cr p = . In addition, 0 = = y x (plane stress in x-y plane). ( ) T c s x D - = a e or ( ) T E c s x D - = Then it follows that Twt E wt P x D = = ' where c s - = ' . Then for buckling, ( ) c s S t wt S wt Ewt S EI T - = = = D 2 2 2 2 3 2 2 2 3 ' 12 4 ' 4 . 2a. Let ( ) x v be the vertical displacement and ( ) x q be the lateral loading. The GDE of a beam under axial and lateral loadings was shown in class to be q Pv dx v d EI dx d = + 2 2 2 2 In our case, ( ) ( ) x kv x q - = . Rearranging the above equation yields 0 2 2 4 4 = + + kv dx v d P dx v d EI Let EI P = and EI k = b . The above equation becomes 0 2 2 4 4 = + + v dx v d dx v d

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2b. The solutions to the GDE will be of the form mx e , where the four values of m can be found by solving the characteristic equation 0 2 4 = + + b a m m . Using the quadratic formula, 2 4 2 2 - ± - = m . We can consider the following three cases. CASE I: 4 2 > or kEI P 4 2 > Since 0 4 2 > - , both values of 2 m are real. But note that < - 4 2 since 0 > . Therefore, in both cases, 2 m is negative. As a result, when we take the square root, we shall get a purely imaginary number. 0 2 4 2 2 < - ± - = m (both cases) ± = - - ± = ± = - + ± = 2 2 4 , 3 1 2 2 , 1 2 4 2 4 m i i m i i m CASE II: 4 2 = or kEI P 4 2 = i i m m ± = ± = - = 2 2 2 , 1 2 These are two double roots, so they do not determine the general solution completely. We have to look for trial solutions of the form mx xe v = or mx e x v 2 = . CASE III: 4 2 < or kEI P 4 2 < Now 4 2 - is imaginary. ( ) 2 2 4 2 4 2 / 1 2 2 2 q i re i m i m = - ± - = - ± - = In this case, 2 2 4 2 , 1 2 2 2 2 , 1 = = - + = r r ˜ ˜ ˆ - + = - p 2 1 2 , 1 4 tan 2 2 1 n , n = 0, 1 and
b 2 4 , 3 = r ˜ ˜ ˆ - - + = - a p q 2 1 4 , 3 4 tan 2 2 1 n , n = 0, 1. We are ready to solve the boundary value problem. The general solution is ( ) x m x m x m x m De Ce Be Ae x v 4 3 2 1 + + + = CASE I: ( ) x D x C x B x A x v 2 2 1 1 sin cos sin cos m + + + = with ( ) ( ) ( ) ( ) 0 ' ' 0 ' ' 0 = = = = L v v L v v ( ) ( ) 0 0 0 0 ' ' 0 0 0 2 2 2 1 = = = + = - = = + = C A C A v C A C A v ( ) L L D B L D L B L v 1 2 2 1 sin sin 0 sin sin 0 - = = + = (*) ( ) ( ) 0 sin 0 sin sin 0 sin sin 0 ' ' 2 2 1 2 2 2 2 2 2 2 1 * 2 2 2 1 2 1 = - = + - = + = L D L D L D L D L B L v If 2 1 , then 0 sin 2 = L D . Hence if D = 0 then we get B = 0. We have obtained the trivial solution ( ) 0 = x v . For non-trivial ( ) x v , 2 2 2 2 2 2 2 2 2 2 4 2 2 4 0 sin ˜ ˆ + ˜ ˆ = - = ˜ ˆ - ˜ ˆ = - - = = = L n L n L n L n L n n L L Substituting into the equation and yields the critical load for buckling.

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hw41solutions - AAE 221 Aerospace Structures II Spring 2004...

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