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What I know:

In general, the work done by a conservative force $F$ is equal to the negative change in a potential energy.

$$\int_1 ^2F.ds = - (U_2 - U_1) $$

A potential energy is always associated with a conservative force.

When choosing the conservative force as the force of gravity we get

$$\int_1 ^2W.ds = -mg(y_2-y_1)=- (U_2 - U_1) $$

When choosing the conservative force as the spring force we get

$$\int_1 ^2F_s.ds = -(\frac{1}{2}kx_2 ^2 -\frac{1}{2}kx_1 ^2)=- (U_2 - U_1) $$

So I conclude that every conservative force's work should give rise to -ve of the difference between the values of a function at two positions and this function is the potential energy function.


What I have trouble with:

When introducing the concept of strain energy (which is also a type of potential energy) my textbook determines the work done by the external gradually applied load P (on a bar) and says that this energy will be stored in the bar as strain energy. It doesn't start with the defining equation of potential energy which is $\int_1 ^2F.ds = - (U_2 - U_1) $.

So, my question is,

how can I use this equation ($\int_1 ^2F.ds = - (U_2 - U_1) $) to come up with the strain energy of the bar?

What will be the conservative force $F$ in this case, which I will be using in this $\int_1 ^2F.ds = - (U_2 - U_1) $ equation to find the potential energy? (I suppose it is internal forces, but still, how can I use those internal forces to come up with a relation of the form $\int_1 ^2F.ds = - (U_2 - U_1) $?)

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2 Answers 2

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What will be the conservative force $F$ in this case, which I will be using in this $\int_1 ^2F.ds = - (U_2 - U_1) $ equation to find the potential energy?

I know you know that the force is $F=-kx$ (Hooke's Law derivation), where $k$ is the spring constant and $x$ is the displacement:

$$\int_1 ^2F\cdot ds = \int_1 ^2-kx\,dx = -\frac{kx^2}{2}=- (U_2 - U_1), $$

giving the well-known expression of $\frac{kx^2}{2}$ for the strain energy.

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  • $\begingroup$ Chemomechanics, I have trouble with deriving the strain energy of a bar using the relation $\int_1 ^2F.ds = - (U_2 - U_1) $. In case of a spring the conservative force which is the spring force (the force that acts on the object that pulls or pushes the spring) is well defined, but in case of a bar what is that conservative force? There must be some conservative force which when integrated should give $-\Delta U$. $\endgroup$ May 26 at 16:21
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    $\begingroup$ Every solid acts as a spring for small deformations. That's obtained from the link. The force field is electrostatic. $\endgroup$ May 26 at 16:47
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In a bar the force causing the deflection $\delta$ is very similar to a spring potential energy under the force, F and displacement x:

$$F=\frac{2U}{\delta }$$

$$U=(AL*\epsilon\sigma)/2=\frac{\sigma^2AL}{2E}$$

  • A = area of the rod
  • $\sigma$ = stress
  • E = young modulus
  • L= length
  • $\epsilon$ = strain
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