-1
$\begingroup$

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) $?)

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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 $-\Delta U$. $\endgroup$
    – Harshit Rajput
    May 26 at 16:21
  • 1
    $\begingroup$ Every solid acts as a spring for small deformations. That's obtained from the link. The force field is electrostatic. $\endgroup$
    – Chemomechanics
    May 26 at 16:47
1
$\begingroup$

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
Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.