AEROSP
5.a

5.a - Ae 315 Homework 5 Solutions 2008/2/8 We have studied...

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Ae 315 Homework 5 Solutions - 2008/2/8 We have studied all the elements of a theory of elasticity. Here we put everything together with the simplest problems: We know the displacements and we want to confirm the other equations. Preliminaries These are just some items that set-up or describe the operating environment of this program. In[1]:= \$Version Out[1]= 6.0 for Mac OS X x86 H 32 - bit L H April 20, 2007 L In[2]:= DateList @D Out[2]= 8 2008,2,7,14,54,56.002228 < In[3]:= Off @ General:: spell D ; Off @ General:: spell1 D ; Off @ Solve:: svars D ; Off @ ParametricPlot:: ppcom D ; Off @ ParametricPlot3D:: ppcom D In[8]:= Needs @ "Units`" D Needs @ "PhysicalConstants`" D Needs @ "BarCharts`" D ;Needs @ "Histograms`" D ;Needs @ "PieCharts`" D Needs @ "VectorAnalysis`" D Common functions for all problems Here you are given a dispalcement field and you need to interperet what it means. ü reference region. This is the undeformed configuration. Here I have to assume some numerical values for some of the parameters so that the plotting can be accomplished. Here the radius "a" is 1.0, the length "c" is 5. I do not plot the inside radius. Also, note that I make use of polar coordinates to parameterize the "surface". In[12]:= aValue = 1.0; cValue = 5.0; surface = 8 aValueCos @ q D ,aValueSin @ q D ,z < ; Note that I plot the "z" direction first so my cylinder comes out nearly horizontal rather than vertical. I do this with the "Rotate- Right" command. NOTE: I am only plotting 1/2 of the section where x 3 >= 0. The equations that are provided only work on this sub-region.

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In[15]:= ParametricPlot3D @ RotateRight @ surface D , 8 q , -p , p < , 8 z,0,cValue <D Out[15]= ü displacement First, lets define the displacement field: (Note I define the displacements here in such a manner so I can automate certain opera- tions.) Here I'm going to try andd make things a bit more clear in how the displacements are defined. In[16]:= u @ 1 [email protected] x1 _ ,x2 _ ,x3 _ < , 8 a_ , b_ , g_ , d_ < , 8 m _ , n_ ,a _ ,c _ <D : = 1 2 a I x3 2 - n x2 2 + n x1 2 M - n b x1 - g x3x2 + d n 2 H c - x3 L I x1 2 - x2 2 M - x3 3 6 + cx3 2 2 ; u @ 2 [email protected] x1 _ ,x2 _ ,x3 _ < , 8 a_ , b_ , g_ , d_ < , 8 m _ , n_ ,a _ ,c _ <D : = n a x1x2 - n b x2 + g x3x1 + d n H c - x3 L x1x2; u @ 3 [email protected] x1 _ ,x2 _ ,x3 _ < , 8 a_ , b_ , g_ , d_ < , 8 m _ , n_ ,a _ ,c _ <D : = -a x1x3 + b x3 - d - 3 4 + n 2 a 2 x1 + 1 4 I x1 3 - 3 x1x2 2 M + x1x2 2 + x3x1 c - x3 2 ; Here I have separated out each symbol so its easier to determine what the equation is doing. Here "x" is a vector of the coordi- nates and "p" is a column matrix representing the parameters " g ", " d " and " n ", "c", and "a" respectively. To check this out: In[19]:= u @ 1 [email protected] x 1 ,x 2 ,x 3 < , 8 a , b , g , d < , 8 m, n ,a,c <D Out[19]= -b n x 1 - g x 2 x 3 + 1 2 a I n x 1 2 - n x 2 2 + x 3 2 M + d 1 2 n I x 1 2 - x 2 2 M H c - x 3 L + cx 3 2 2 - x 3 3 6 In[20]:= u @ 2 [email protected] x 1 ,x 2 ,x 3 < , 8 a , b , g , d < , 8 m, n ,a,c <D Out[20]= -b n x 2 + a n x 1 x 2 + d n x 1 x 2 H c - x 3 L + g x 1 x 3 In[21]:= u @ 3 [email protected] x 1 ,x 2 ,x 3 < , 8 a , b , g , d < , 8 m, n ,a,c <D Out[21]= b x 3 - a x 1 x 3 - d a 2 - 3 4 - n 2 x 1 + x 1 x 2 2 + 1 4 I x 1 3 - 3x 1 x 2 2 M + x 1 c - x 3 2 x 3 In[22]:= uVector @ x _ ,p _ ,g _ D : = Table @ u @ i [email protected] x,p,g D , 8 i,1,3 <D Thus the new position is: 2 5.a.nb
In[23]:= ClearAll @ y D ; y @ x _ ,p _ ,g _ D : = x + uVector @ x,p,g D ü strain Here I define a strain function that can be used to interperet the above displacement field: In[25]:= ClearAll @ e D ; e @ i _ ,j _ [email protected] x _ ,p _ ,g _ D : = 1 2 I x Q j U u @ i [email protected] x,p,g D + x Q i U u @ j [email protected] x,p,g DM ; This is where indicial notation really comes in handy.

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