project3a - Aircraft Structures I project Phase 3 EAS 4200C...

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Aircraft Structures I project Phase 3 EAS 4200C Roberto Barraza 11/14/2010
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UFID 0087-1745 1. Determine the shear flow due to V and T. Shear flow: ' x z VQ q I  (eqn. 1) Finding the Moment of inertia: Moment of inertia about z: 2 z i i I z A (eqn. 2) z I       22 11 171 50 171 2 50 2 223746 12 12 4 in Now, we find ' q for each S section: At 1 S : 2 1 1 1 1 1 1 ( ) 24.5 24.5 0.5 2 c S Q A x S t S S    Replacing 1 Q and z I into equation 1 we get:   2 ' 1 24.5 0.5 223746 x V S S q W do the same process for '' 23 , qq and ' 4 q . The results are as follows:   2 ' 2 24.5 223746 x VS q   2 33 ' 3 4165 24.5 0.5 223746 x V S S q   4 ' 4 4165 24.5 223746 z q  Since x V passes through the shear center, the twist angle is equal to zero:
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___ 1 0 2 q ds t GA  (eqn. 3) At 1 S :   40 1 0 1 0 0 0.043818 49 x q q dS V q   2 S :   170 2 0 2 0 0 170 1.58226 x q q dS q V 3 S :   49 3 0 3 0 0 49 0.955946 x q q dS q V 4 S :   170 4 0 4 0 0 1.58226 170 x q q dS V q Replacing those values into equation 3: 0 0 0 0 0.043818 49 170 1.58226 49 0.955946 170 1.58226 0 x x x x V q q V q V q V   Solving for 0 q and using the value of 5 4.9239 10 x V  pounds from phase two: 0 0.005006 2464.9 x qV     Until now, we have solved for the shear flow due to the shear force V . Now, we have to find the shear flow due to the torque T: ___ 2 root T T q A (eqn. 4) Where: ___ (49)(170) 8330 A 2 in From phase 2 of this project, we found 7 () 2.0801 10 root AVG T   and 7 3.0769 10 root ell T   . Thus:
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() 1248.56 T AVG q  and 1846.88 T ell q Since the sign of the flow due to torque is negative, this means the flow has a counterclockwise direction.
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This note was uploaded on 08/27/2011 for the course AEROSPACE 3115C taught by Professor Bakcer during the Spring '10 term at University of Florida.

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project3a - Aircraft Structures I project Phase 3 EAS 4200C...

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