Questions about derivation of equation

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In summary, the conversation is about the derivation of an equation in linear elasticity. The equation in question is \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). The conversation also includes given relationships for e_{xx}, e_{yy}, e_{zz}, e_{xy}, e_{yz}, and e_{zx}. The participants discuss the process of deriving this equation and the use of cyclic indices. The conversation concludes with speculation on the origins of the equation.
  • #1
roldy
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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].
 
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  • #2
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
I understand that. The problem I'm having is deriving the first equation from the other equations.
 
  • #4
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
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
(just got up :zzz:)
roldy said:
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
Why are you opposed to cycling the indices?
 
  • #8
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
roldy said:
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
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
roldy said:
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).
 

FAQ: Questions about derivation of equation

What is the derivation of an equation?

The derivation of an equation is the process of mathematically deriving or proving the equation using known mathematical principles and formulas.

Why is it important to understand the derivation of an equation?

Understanding the derivation of an equation allows us to see the logical and mathematical steps that were used to arrive at the equation, making it easier to apply and manipulate the equation in different scenarios.

How do you derive an equation?

To derive an equation, you must start with a set of known equations and apply mathematical operations such as addition, subtraction, multiplication, and division, as well as rules such as the distributive property and the power rule, to arrive at the desired equation.

Can an equation have multiple derivations?

Yes, an equation can have multiple derivations, as there may be different ways to arrive at the same equation using different mathematical principles and formulas. However, the end result should be the same.

Is it possible to derive new equations?

Yes, it is possible to derive new equations by combining and manipulating existing equations using mathematical operations and rules. This is often how new mathematical principles and formulas are discovered.

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