ASE 365 - HW 9 Solutions

ASE 365 - HW 9 Solutions - Solutions: Homework Set 10 1....

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Unformatted text preview: Solutions: Homework Set 10 1. For Problem 8.3, from a free-body diagram of an element of the rod of length x , which has axial force EA ( x ) u ( x,t ) acting to the left on the left face, and axial force EA ( x + x ) u ( x + x,t ) acting to the right on the right face, and external force f ( x,t ) x acting to the right, Newtons second law gives m ( x ) x u ( x,t ) = A ( x ) x u ( x,t ) = F x = f ( x,t ) x + EA ( x + x ) u ( x + x,t )- EA ( x ) u ( x,t ) Divide by x , take lim x : m ( x ) u ( x,t ) = f ( x,t ) + lim x parenleftbigg EAu | x + x- EAu | x x parenrightbigg = f ( x,t ) + x parenleftBig EA ( x ) u ( x,t ) parenrightBig PDE: m ( x ) 2 u ( x,t ) t 2- x parenleftbigg EA ( x ) u ( x,t ) x parenrightbigg = f ( x,t ) , < x < L For left-end boundary condition, draw FBD of element of length x at the left end of the rod. Spring force f s = ku (0 ,t ) acts to the left, and axial force EA ( x ) u ( x,t ) acts to the right, along with external force f x . Apply Newtons second law and take lim x . Then m x and f x terms vanish, leaving 0 = EAu | x =0- ku | x =0 as the left-end BC. The right end is free, so the right-end BC is simply EAu | x = L = 0 . 2. For Problem 8.4, from a free-body diagram of an element of the shaft of length x , which has internal torque stress resultant GJ ( x ) ( x,t ) acting as a negative torque (by the right-hand rule) on the left face, and internal torque GJ ( x + x ) ( x + x,t ) acting positively on the right face, and external moment m ( x,t ) x acting positively, M = I gives: I ( x ) x ( x,t ) = J ( x ) x ( x,t ) = M x = m ( x,t ) x + GJ ( x + x ) ( x + x,t )- GJ ( x ) ( x,t ) (Here I ( x ) represents the mass moment of inertia of the shaft per unit length, which equals J ( x ), where is the mass density and J ( x ) is the polar area moment of inertia of the cross section.) Divide by x , take lim x : I ( x ) ( x,t ) = m ( x,t ) + lim x parenleftbigg GJ | x + x- GJ | x x parenrightbigg = m ( x,t ) + x parenleftBig GJ ( x ) ( x,t ) parenrightBig 1 PDE: I ( x ) 2 ( x,t ) t 2- x parenleftbigg GJ ( x ) ( x,t ) x parenrightbigg = m ( x,t ) , < x < L For left-end boundary condition, draw FBD of element of length x at the left end of the shaft. Spring torque k 1 (0 ,t ) acts negatively, and internal torque GJ ( x ) ( x,t ) acts positively, along with external moment...
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This note was uploaded on 09/18/2011 for the course ASE 365 taught by Professor Staff during the Spring '10 term at University of Texas at Austin.

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ASE 365 - HW 9 Solutions - Solutions: Homework Set 10 1....

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