HW9 Solutions

# HW9 Solutions - Mechanics of Aircraft structures C.T Sun...

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Mechanics of Aircraft structures C.T. Sun 5.1 Find the flexural shear flow produced by the transverse shear force in the beam with the thin-walled section given by Fig. 5.30. N V z 1000 = Figure 5.30 Thin-walled section with a side cut Solution: (a) Assume that the transverse shear force acts through the shear center, and, thus, no torsional effect exists. Assume that the loss of material at the cut is negligible. Hence the centroid of the cross-sectional area is obviously at the center as shown in Fig. 5.30. The shear flow is obtained by y z s I Q V q = (5.1.1) where is the first moment of area (the area measured along the wall from an free edge to the current position of interest), and is the vertical distance from the centroid of to the y -axis. We have c s A z A zdA Q s = = ∫∫ s A c z s A 4 6 3 3 10 6676 . 6 ] ) 002 . 0 2 . 0 )( 002 . 0 1 . 0 ( ) 002 . 0 2 . 0 )( 002 . 0 1 . 0 [( 12 1 m I y × = + + = (b) Setup the shear flow contour as shown in the figure below. 5.1.1

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Mechanics of Aircraft structures C.T. Sun (1) On m s 1 . 0 ~ 0 : 1 2 1 1 1 001 . 0 ) 2 )( ( s s t s z A zdA Q c s A = = = = ∫∫ s 2 1 5 6 2 1 10 5 . 1 10 6676 . 6 001 . 0 1000 s s I Q V q y z s × = × × = = At , 0 1 = s 0 0 10 5 . 1 2 5 = × × = s q At , (The negative sign means that the actual direction of this shear flow is opposite to contour direction of . 1 . 0 1 = s m N q s / 1500 1 . 0 10 5 . 1 2 5 = × × = 1 s --- ANS (2) On m s 1 . 0 ~ 0 : 2 The first moment Q for this segment must include the entire first moment of the segment covered by contour s 1 . Thus, 2 4 5 2 2 10 2 10 ) 1 . 0 ( 002 . 0 ) 1 . 0 ( 001 . 0 s s Q × + = + = 2 4 6 2 4 5 10 3 1500 10 6676 . 6 ) 10 2 10 ( 1000 s s I Q V q y z s × = × × + × = = Note that the distribution of the shear flow is linear along the contour. At , 0 2 = s m N q s / 1500 = At , m s 1 . 0 2 = m N q s / 4500 1 . 0 10 3 1500 4 = × × = --- ANS 5.1.2
Mechanics of Aircraft structures C.T. Sun (3) On m s 1 . 0 ~ 0 : 3 In a similar manner, the first moments for the previous two segments must be added to that for the additional area along contour s 3 . We have ) 5 ( 10 2 10 3 ) 5 . 0 1 . 0 ( 002 . 0 ) 1 . 0 ( 002 . 0 ) 1 . 0 ( 001 . 0 2 3 3 4 5 3 3 2 2 s s s s Q × + × = + + = 2 3 3 6 2 3 3 4 5 150000 30000 4500 10 6676 . 6 )] 5 ( 10 2 10 3 [ 1000 s s s s I Q V q y z s + = × × + × × = = At , 0 3 = s m N q s / 4500 ) 0 ( 150000 ) 0 ( 30000 4500 2 = + = At , m s 1 . 0 3

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HW9 Solutions - Mechanics of Aircraft structures C.T Sun...

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