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ASE 365 - HW 9 Solutions

# ASE 365 - HW 9 Solutions - Solutions Homework Set 10 1 For...

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Solutions: Homework Set 10 1. For Problem 8.3, from a free-body diagram of an element of the rod of length Δ x , which has axial force EA ( x ) u ( x,t ) acting to the left on the left face, and axial force EA ( x + Δ x ) u ( x + Δ x,t ) acting to the right on the right face, and external force f ( x,t x acting to the right, Newton’s second law gives m ( x x ¨ u ( x,t ) = ρA ( x x ¨ u ( x,t ) = Σ F x = f ( x,t x + EA ( x + Δ x ) u ( x + Δ x,t ) - EA ( x ) u ( x,t ) Divide by Δ x , take lim Δ x 0 : m ( x u ( x,t ) = f ( x,t ) + lim Δ x 0 parenleftbigg EAu | x x - EAu | x Δ 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 ) , 0 < x < L For left-end boundary condition, draw FBD of element of length Δ 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 Δ x . Apply Newton’s second law and take lim Δ x 0 . Then m Δ x and f Δ 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 Δ x , which has internal torque stress resultant GJ ( x ) θ ( x,t ) acting as a negative torque (by the right-hand rule) on the left face, and internal torque GJ ( x x ) θ ( x x,t ) acting positively on the right face, and external moment m ( x,t x acting positively, Σ M = gives: I ( x x ¨ θ ( x,t ) = ρJ ( x x ¨ θ ( x,t ) = Σ M x = m ( x,t x + GJ ( x + Δ x ) θ ( x + Δ x,t ) - GJ ( x ) θ ( x,t ) (Here I ( x ) represents the mass moment of inertia of the shaft per unit length, which equals ρJ ( x ), where ρ is the mass density and J ( x ) is the polar area moment of inertia of the cross section.) Divide by Δ x , take lim Δ x 0 : I ( x ) ¨ θ ( x,t ) = m ( x,t ) + lim Δ x 0 parenleftbigg GJθ | x x - GJθ | x Δ x parenrightbigg = m ( x,t ) + ∂x parenleftBig GJ ( x ) θ ( x,t ) parenrightBig 1

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PDE: I ( x ) 2 θ ( x,t ) ∂t 2 - ∂x parenleftbigg GJ ( x ) ∂θ ( x,t ) ∂x parenrightbigg = m ( x,t ) , 0 < x < L For left-end boundary condition, draw FBD of element of length Δ x at the left end of the shaft. Spring torque k 1 θ (0 ,t ) acts negatively, and internal torque GJ x ) θ x,t ) acts positively, along with external moment m Δ x . Apply Σ M = and take lim Δ x 0 .
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