Formula for the energy of elastic deformation

  • #1
baw
9
0
In every book I checked, the energy (per unit mass) of elastic deformation is derived as follows:

## \int \sigma_1 d \epsilon_1 = \frac{\sigma_1 \epsilon_1}{2} ##
and then, authors (e.g. Timoshenko & Goodier) sum up such terms and substitute ##\epsilon ## from generalised Hooke's law i.e.
## \epsilon_1=\frac{1}{E} (\sigma_1 -\nu \sigma_2 -\nu \sigma_3) ##
## \epsilon_2=\frac{1}{E} (\sigma_2 -\nu \sigma_1 -\nu \sigma_3) ##
## \epsilon_3=\frac{1}{E} (\sigma_3 -\nu \sigma_2 -\nu \sigma_1) ##
obtaining:
##V=\frac{1}{2E} (\sigma_1^2 +\sigma_2^2+\sigma_3^2 )-\frac{\nu}{E}(\sigma_1 \sigma_2+\sigma_2 \sigma_3 + \sigma_1 \sigma_3) ##

but... is it correct to substitute generalised Hooke's law after the integration? The formula is obtained as if simple ##\sigma = E \epsilon ## was used. As in the attached figure, it looks like they assume that ##\sigma_x ## has no term independent on ##\epsilon_x ##, despite that Hooke's law can be transformed to:

## \sigma_1=\frac{(\nu -1)E}{(\nu +1)(2 \nu-1)} \epsilon_1 - \frac{\nu E}{(\nu+1)(2\nu -1)}(\epsilon_2+\epsilon_3) ##
##\sigma_2=(...) ##
##\sigma_3=(...) ##

where this term is present. Shouldn't we integrate the above formula? Could someone please, explain me why it is correct?
 

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Answers and Replies

  • #2
Frabjous
Gold Member
985
1,060
It’s energy per unit volume.

Conservation of energy implies that elastic energy is independent of order applied and depends only on final state. Otherwise one could find an order that creates/destroys energy.

The first method takes advantage of this and applies the stresses/strains indendently and then adds them together. There are subtleties to this that I cannot do justice to.

The second method applies everything at once and then integrates.
 
Last edited:
  • #3
baw
9
0
Lets say we applied ##\sigma_1## at first and got ##\epsilon_1## as well as some ##\epsilon_2## and ##\epsilon_3##. The (specific) work done is ##\frac{\sigma_1^2}{2E}##. If we now apply ##\sigma_2## we already have some initial strain, so the plot ##\sigma_2(\epsilon_2)## moves downward by ##\frac{\nu}{E}\sigma_1##. If we now integrate it, we get ##\frac{\sigma_2^2}{2E}-\frac{\nu}{E}\sigma_1 \sigma_2##. Then, ##\sigma_3(\epsilon_3)## is shifted by ##\frac{\nu}{E}(\sigma_1+\sigma_2)## and suma summarum, after the integration we get the right formula. I got it, thanks!

Btw. it means that I just made some mistke in the second method and that's why I didn't got the same answer, doesn't it?
 
  • #4
22,330
5,205
In your starting equations, evaluate the differentials of the strains in terms of the differentials in the three stresses. Then, multiply each differential of strain by its corresponding stress, and add up the resulting 3 equations. What do you get?
 

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