# Formula for the energy of elastic deformation

• baw
In summary, the energy (per unit mass) of elastic deformation is derived as follows: - The energy is derived as the sum of the energy due to the individual stresses (i.e. the stress due to the first stress, the stress due to the second stress, and the stress due to the third stress). - The stresses are assumed to be independent of one another.
baw
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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• skrin4.PNG
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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:
baw
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?

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?

## 1. What is the formula for the energy of elastic deformation?

The formula for the energy of elastic deformation is given by U = 1/2kx^2, where U is the elastic potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

## 2. How is the formula derived?

The formula for the energy of elastic deformation is derived from Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. This can be represented mathematically as F = -kx. By integrating this equation with respect to displacement, we can obtain the formula for elastic potential energy.

## 3. What is the significance of the spring constant in the formula?

The spring constant, denoted by k, is a measure of the stiffness of a spring. It determines how much force is required to stretch or compress the spring by a certain distance. A higher spring constant indicates a stiffer spring, meaning it requires more force to produce the same amount of displacement. The value of k is crucial in determining the amount of elastic potential energy stored in a spring.

## 4. Can the formula be applied to materials other than springs?

Yes, the formula for the energy of elastic deformation can be applied to any elastic material, not just springs. This includes materials such as rubber bands, elastic cords, and even the human body. As long as the material exhibits elastic behavior, meaning it can return to its original shape after being stretched or compressed, the formula can be used to calculate the energy stored in it.

## 5. How is the formula used in real-world applications?

The formula for the energy of elastic deformation is used in various real-world applications, including engineering, physics, and sports. It is used to design and analyze structures and machines that involve springs, such as suspension systems, shock absorbers, and trampolines. In sports, the formula is used to calculate the potential energy stored in muscles and tendons, which is essential in activities like running and jumping.

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