hw21_Solutions - AAE 221 Aerospace Structures II Spring...

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AAE 221 – Aerospace Structures II – Spring 2004 Chapter 2.1 – Beam Bending and Extension Solutions
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Problem 0
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1. (a) The Young’s modulus of the three sections are 7 3 2 7 1 10 1 10 3 ¥ = = ¥ = E E E First, the Young’s modulus of aluminum is chosen to be the reference modulus. The areas of each sub-component are at A at A at A 2 8 4 3 2 1 = = = The modulus weighted area is ˜ ˜ ˆ = * i i A E E A o The distances from the origin of the coordinate axis to the m.w. centroid are found from ˜ ˜ ˆ = ˜ ˜ ˆ = * * * * i i i i i i A y E E y A A z E E z A ) ( ) ( o o o o o o where the quantity o z etc. are the distances from the origin to the centroid of each sub- component. The location of the centroid is found to be symmetry) to (due 20 2 22 248 22 38 21 t a y t t a z = = = + = * * o o The moments and product of inertia can be found by first calculating the modulus weighted moments of inertia about our coordinate system origin, and then applying the parallel axis theorem to shift to the centroidal axis. The m.w. moments of inertia can be found from
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( ) [ ] ( ) [ ] + ˜ ˜ ˆ = + ˜ ˜ ˆ = * * i zz z z i yy y y A y I E E I A z I E E I 2 2 o o o o o o o o where the terms yy I , etc. are the moments of inertia of each sub-component about its own centroidal axis. For a symmetric rectangular piece, 3 12 1 bh I yy = where the base b is parallel to the axis of the moment of inertia under consideration. For this problem, 4 4 3 344020 3 87400 t I t I z z y y = = * * o o o o From this, the m.w. moments of inertia are found by applying the parallel axis theorem. ( ) ( ) * * * * * * * * - = - = A y I I A z I I z z zz y y yy 2 2 o o o o o o Plugging in the appropriate values found previously, these were found to be 4 4 3 80020 363 427240 t I t I zz yy = = * * The product of inertia, 0 = * yz I due to symmetry. (b) For the various sub-components ( ) ( ) ( ) 7 3 2 7 1 10 1 0 10 1 ¥ = D = D ¥ - = D T T T a
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The thermal axial load is [ ] D ˜ ˜ ˆ = i T A T E E E N ) ( a o o The moments due to bending are [ ] [ ] D ˜ ˜ ˆ - = D ˜ ˜ ˆ - = i i i zT i i i yT TA y E E E M TA z E E E M o o o o Plugging in the appropriate values and centroidal distances, 3 7 3 7 2 6 10 0 . 2 10 4 . 1 10 1 t M t M t N zT yT T ¥ = ¥ = ¥ - = Finally, the induced thermal stress may be found from T E D = s Point A is in an aluminum sub-component, thus, 7 10 3 ¥ = E psi. The thermal stress is ( )( ) psi 10000 10 1 10 3 3 7 = ¥ ¥ = - Point B is in the steel sub-component, thus 7 10 1 ¥ = E psi. The thermal stress is ( ) psi 30000 10 1 10 1 3 7 - = ¥ - ¥ = - 2. The area, as a function of x is : 2 2 4 3 1 ) ( ˜ ˜ ˆ ˜ ˆ - = L x a x A since the stress is
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( ) ( ) dx du E x A P x = = s 2 2 4 3 1 ˜ ˜ ˆ ˜ ˆ - = L x a P dx du E 2 2 4 3 1 ˜ ˜ ˆ ˜ ˆ - = L x Ea P dx du Therefore, the displacement u(x) is ( ) - ˜ ˆ - = dx L x Ea P x u 2 2 4 3 1 integrating, u(x) becomes 1 1 2 4 3 1 3 4 ) ( C L x L Ea P x u + ˜ ˆ - ˜ ˆ - - = - We can apply the boundary condition of zero displacement at x=0 0 ) 0 ( = u 0 3 4 ) 0 ( 1 2 = + ˜ ˆ = C L Ea P u 2 1 3 4 Ea PL C - = The displacement u(x) is then ˜ ˜ ˜ ˜ ˆ - + - = L x Ea PL x u 4 3 1 1 1 3 4 ) ( 2
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To plot the displacement of the non-dimensional displacements divide u(x) by 2 Ea PL ˜ ˜ ˜ ˜ ˆ - + - = L x PL Ea x u 4 3 1 1 1 1 3 4 ) ( 2 3. Cross section has width a and height 2a. The area of the cross-section is A = 2a 3 .
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hw21_Solutions - AAE 221 Aerospace Structures II Spring...

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