Questions about derivation of equation

  • Thread starter roldy
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  • #1
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Main Question or Discussion Point

I'm studying linear elasticity and I came across an equation that I having problems figuring out the derivation. I want to understand this in case I may need to know it later in the course. The equation is as follows:

[itex]\frac{\partial^2e_{zz}}{\partial x \partial y} = \frac{\partial}{\partial z}\left(\frac{\partial e_{yz}}{\partial x} + \frac{\partial e_{zx}}{\partial y} - \frac{\partial e_{xy}}{\partial z}\right)[/itex]

I'm trying to understand this derivation with the given relationships below.

[itex]e_{xx} = \frac{\partial U_x}{\partial x}[/itex]
[itex]e_{yy} = \frac{\partial U_y}{\partial y}[/itex]
[itex]e_{zz} = \frac{\partial U_z}{\partial z}[/itex]

[itex]e_{xy} = 1/2\left(\frac{\partial U_y}{\partial x} + \frac{\partial U_x}{\partial y}\right)[/itex]
[itex]e_{yz} = 1/2\left(\frac{\partial U_z}{\partial y} + \frac{\partial U_y}{\partial z}\right)[/itex]
[itex]e_{zx} = 1/2\left(\frac{\partial U_x}{\partial z} + \frac{\partial U_z}{\partial x}\right)[/itex]

The only thing that makes sense is that the derivative with respect to x and y was taken on [itex]e_{zz}[/itex].
 

Answers and Replies

  • #2
tiny-tim
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hi roldy! :smile:

if you expand [itex]\left(\frac{\partial e_{yz}}{\partial x} + \frac{\partial e_{zx}}{\partial y} - \frac{\partial e_{xy}}{\partial z}\right)[/itex] by using those definitions,

you'll get 6 terms, of which 4 should cancel :wink:
 
  • #3
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I understand that. The problem I'm having is deriving the first equation from the other equations.
 
  • #4
tiny-tim
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ok, so what did you get when you expanded [itex]\left(\frac{\partial e_{yz}}{\partial x} + \frac{\partial e_{zx}}{\partial y} - \frac{\partial e_{xy}}{\partial z}\right)[/itex] ? :smile:
 
  • #5
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I would get [itex]\frac{\partial^2 e_{zz}}{\partial x \partial y}[/itex] after simplifying. The question that I have is how to get the other relationships (without using cyclic indices) for [itex]\frac{\partial^2 e_{xx}}{\partial y \partial z}[/itex] and [itex]\frac{\partial^2 e_{yy}}{\partial z \partial x}[/itex] just by using the equations below. What is the procedure?

[itex]e_{xy} = 1/2\left(\frac{\partial U_y}{\partial x} + \frac{\partial U_x}{\partial y}\right)[/itex]
[itex]e_{yz} = 1/2\left(\frac{\partial U_z}{\partial y} + \frac{\partial U_y}{\partial z}\right)[/itex]
[itex]e_{zx} = 1/2\left(\frac{\partial U_x}{\partial z} + \frac{\partial U_z}{\partial x}\right)[/itex]
 
  • #6
tiny-tim
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(just got up :zzz:)
I would get [itex]\frac{\partial^2 e_{zz}}{\partial x \partial y}[/itex] after simplifying.
i don't understand :redface:

why is there no U in there? :confused:
 
  • #7
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Why are you opposed to cycling the indices?
 
  • #8
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There is no U in the final equation because I end up with [itex]\frac{\partial^3 U_z}{\partial x \partial y \partial z}[/itex] which when letting [itex]e_{zz} = \frac{\partial U_{zz}}{\partial z}[/itex] the solution becomes [itex]\frac{\partial e_{zz}}{\partial x \partial y}[/itex]

I'm opposed to using cyclic indices because I would like to know how this equation came to be. Sure, I could use cyclic indices for the other equations but if I am asked on a test to derive this relationship without cyclic indices I wouldn't know where to begin.
 
  • #9
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There is no U in the final equation because I end up with [itex]\frac{\partial^3 U_z}{\partial x \partial y \partial z}[/itex] which when letting [itex]e_{zz} = \frac{\partial U_{zz}}{\partial z}[/itex] the solution becomes [itex]\frac{\partial e_{zz}}{\partial x \partial y}[/itex]

I'm opposed to using cyclic indices because I would like to know how this equation came to be. Sure, I could use cyclic indices for the other equations but if I am asked on a test to derive this relationship without cyclic indices I wouldn't know where to begin.
So your real question is "How did anyone ever think of this?" I think you already know the answer. What's your best guess? Incidentally, I've had more than a nodding acquaintance with stress analysis during my career, and I don't ever remember any practical application where this type of relationship was used.
 
  • #10
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You are correct. I know we need this equation for constraining requirements. My best guess is that "they" assumed a relationship with all the shear strains. Then they figured out whether or not a particular shear strain is added or subtracted.
 
  • #11
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You are correct. I know we need this equation for constraining requirements. My best guess is that "they" assumed a relationship with all the shear strains. Then they figured out whether or not a particular shear strain is added or subtracted.
I think that some person was just sitting around "playing" around with the equations, as many of us often do, and they came up with this result (which they found very interesting).
 
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