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HW10 Solutions

# HW10 Solutions - Mechanics of Aircraft structures C.T Sun...

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Mechanics of Aircraft structures C.T. Sun 5.5 Find the shear flow for the three-stringer section shown in Fig. 5.33 for and . Given shear modulus N V z 5000 = 0 = y V GPa G 27 = , find the twist angle per unit length. Also determine the shear center. Is the shear flow statically determinate? Figure 5.33 Single-cell closed section Solution: (a) Assume that the thin sheets are ineffective in bending. Let stringer 2 be the reference point for the location of the centroid of this four-stringer section. We have the horizontal and vertical distances of the centroid from stringer 2 as m cm A y A y i i i i i c 2667 . 0 6667 . 26 ) 10 ( 3 ) 80 )( 10 ( = = = = m cm A z A z i i i i i c 1 . 0 10 ) 10 ( 3 ) 10 )( 10 ( ) 20 )( 10 ( = = + = = Now we set up the y and z axes with the origin at the centroid. The moments of inertia are 4 5 2 4 2 10 2 ) 1 . 0 )( 10 10 ( 2 m z A I i i i y × = × = = 4 4 2 4 2 4 2 10 2667 . 4 ) 2667 . 0 8 . 0 )( 10 10 ( ) 2667 . 0 )( 10 10 ( 2 m y A I i i i z × = × + × = = (b) Shear flows Since this cross-section is symmetric with respect to y axis, the shear center is located on the y axis. Hence only the y position of the shear center needs to be found. We first 5.5.1

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Mechanics of Aircraft structures C.T. Sun make a fictitious cut between stringers 1 and 3 and consider the shear flow as the superposition of two shear flow systems as shown in the figure below. + (1) First, calculate the shear flows by assuming a cut in the wall between stringers 1 and 3. Then , and the shear on the cut section is calculated by using 0 ' 31 = q y i z i I Q V q = ' (5.5.1) where is the first moment of stringer area , and is the vertical distance from the centroid of to the y -axis. We obtain = = i k k k i z A Q 1 k A k z k A m N I Q V q y z / 25000 10 2 ) 1 . 0 10 10 ( 5000 ' 5 4 1 12 = × × × = = 0 10 2 ) 1 . 0 10 10 1 . 0 10 10 ( 5000 ' 5 4 4 2 23 = × × × × × = = y z I Q V q (2) Adding the shear flow q 0 from the second part, we have the total shear flow as 0 0 12 12 25000 ' q q q q + = + = 0 0 23 23 ' q q q q = + = 0 0 31 31 ' q q q q = + = The resulting moment of the total shear flow must be equal to the moment produced by . Taking moment about stringer 1, we have z V 0 1914 . 0 4 . 785 2 ) 2 . 0 )( 8 . 0 ( 2 ) 25000 )( ) 2 . 0 ( 8 ( 2 2 2 0 0 0 0 2 23 123 12 12 = + = + + = + = × q q q q A q A V z π => m N q / 4103 0 = Note: 12 A is the area enclosed by the curved sheet and the line connecting stringers 1 and 2; 123 A is the area enclosed by the lines connecting stringers 1, 2, and 3. Also note that the shear flow passes stringer 1 and, thus, does not produce any moment. 5.5.2
Mechanics of Aircraft structures C.T. Sun The final shear flows are m N q q / 20897 25000 0 12 = + = m N q q / 4103 0 23 = = m N q q / 4103 0 31 = = --- ANS (c) Twist angle per unit length The equation of the twist angle per unit length is = ds t q A G 2 1 θ (5.5.2) We can obtain the twist angle per unit length as m m rad ds t q A G deg/ 10 7 . 5 / 10 97 . 9 10 17 . 5 54 . 51 ) 001 . 0 )( 2 / 8 . 0 2 . 0 2 . 0 8 )( 10 27 ( 2 ) 806 . 0 ( 4103 2 ) 2 2 . 0 ( 20896 2 1 4 6 6 2 9 × = × = × = × + × × × + × = = π --- ANS (d) Shear center To determine the horizontal location of the shear center, we assume that the shear force acts through the shear center which is assumed to be located at a horizontal distance to the right of stringer 1. We now rewrite the moment equation as z V e e q q q q A q A e

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HW10 Solutions - Mechanics of Aircraft structures C.T Sun...

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