This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions: Homework Set 10 1. For Problem 8.3, from a freebody 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 leftend 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 leftend BC. The right end is free, so the rightend BC is simply EAu  x = L = 0 . 2. For Problem 8.4, from a freebody 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 righthand 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 leftend 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...
View
Full
Document
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.
 Spring '10
 Staff

Click to edit the document details