HW12 Solutions

# HW12 Solutions - Mechanics of Aircraft structures C.T Sun...

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Mechanics of Aircraft structures C.T. Sun 7.1 The truss structure consists of two bars connected by a pin-joint (which allows free rotation of the bars). The other ends of the bars are hinged as shown in Fig. 7.26. A weight W is hung at the joint. Find the maximum weight the truss can sustain before buckling occurs. Figure 7.26 Two-bar truss Solution: From equilibrium, axial forces of the bars can be easily determined as θ sin 12 W N = (Tension) sin cos cos 12 13 W N N = = (Compression) Only the compressed bar 13 may suffer buckling when the weight W increases. Since bar 13 is connected with pin at both ends, its buckling load is 2 2 L EI P cr π = When the axial force reaches the critical load , buckling occurs. That is 13 N cr P cr P L EI W N = = = 2 2 max max , 13 sin cos => tan 2 2 max L EI W = --- ANS 7.1.1

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Mechanics of Aircraft structures C.T. Sun 7.2 A bar is built-in at the left end and supported at the tight end by a linear spring with spring constant α . Find the equation for buckling loads. Hint : The boundary conditions are 0 = w and 0 / = dx dw at the left end; and and 0 = M w V = at the right end. Figure 7.27 Bar with a built-in end and an elastically supported end Solution: The equilibrium equation in terms of deflection is 0 2 2 2 4 4 = + dx w d k dx w d (7.2.1) where EI P k = The general solution is 4 3 2 1 cos sin C x C kx C kx C w + + + = Its first, second and third derivatives, respectively, are 3 2 1 sin cos C kx k C kx k C dx dw + = kx k C kx k C dx w d cos sin 2 2 2 1 2 2 = kx k C kx k C dx w d sin cos 3 2 3 1 3 3 + = Boundary conditions: At the left end, , 0 = x 0 = w => 0 4 2 = + C C (7.2.2a) 0 / = dx dw => 0 3 1 = + C k C (7.2.2b) At the right end, L x = , 0 2 2 = = dx w d EI M => (7.2.2c) 0 ) cos sin ( 2 1 2 = + kL C kL C k w dx dw P dx w d EI V = = 3 3 => EI w dx dw k dx w d = + 2 3 3 7.2.1
Mechanics of Aircraft structures C.T. Sun => ) cos sin ( sin cos sin cos 4 3 2 1 2 3 3 2 3 3 2 3 1 C L C kL C kL C EI k C kL k C kL Ck kL k C kL k C + + + = + + + α => ) cos sin ( 4 3 2 1 2 3 C L C kL C kL C EI k C + + + = (7.2.2d) From (7.2.2c), kL C C tan 1 2 = From (7.2.2b),

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## This note was uploaded on 01/18/2012 for the course AAE 352 taught by Professor Chen during the Fall '08 term at Purdue.

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HW12 Solutions - Mechanics of Aircraft structures C.T Sun...

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