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I am having an embarrassingly hard time with the derivation for the potential energy of an infinitesimal element of an elastic rod of area $A$. The picture shown below is an element of the rod that has been extended by $du$ by the force $F$.

enter image description here

I've tried this derivation several times and have yet to obtain the factor of $\frac{1}{2}$ in

$$dU=\frac{1}{2}AY\big(\frac{du}{dx}\big)^2dx,\tag{1} $$

which is given in my lecture notes. Here is my best attempt so far: From the stress strain relation, $\sigma=Y\epsilon$, where $Y$ is Youngs modulus we obtain:

$$\frac{F}{A}=Y\frac{du}{dx}.\tag{2} $$

I asked a similar question a few days ago, and based on that response I've assumed that the work done in extending the rod element from length $dx$ to $du+dx$ is

$$dW=F\big[ (du+dx)-dx\big]\tag{3} $$

$$=YA\frac{du}{dx}du.\tag{4} $$

And I can make it look more like the answer given in my lecture notes if I divide and multiply by $dx$:

$$dW=YA\big(\frac{du}{dx}\big)^2dx.\tag{5}$$

And I am assuming that $dW=dU$ here ($F=-\frac{dU}{du} $ but forget about the negative sign)

I have tried other approaches but they make even less sense to me. A formula that I think will come in handy is the deflection at section x of a rod:

$$\delta(x)=\frac{F}{AY}x ,\tag{6}$$

and I think an integral will come in to play but I'm not sure what I'm integrating over any more (I'm dealing with an infinitesimal element of the rod). So how do I use the infinitesimal Work to find the infinitesimal change in potential energy here to get that factor of $\frac{1}{2}$ (assuming that it does in fact belong in $dU$)?

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OP is pondering why the factor $\color{Red}{\frac{1}{2}}$ should be in eq. (1). OP seems aware of that it is related to the factor $\color{Red}{\frac{1}{2}}$ in the elastic potential energy $$\Delta U ~=~\color{Red}{\frac{1}{2}}k(\Delta u)^2 \tag{A} $$ of a spring, but precisely how?

Answer:

  1. The $\Delta$ on the left-hand side of eq. (A) has to be properly understood. Imagine that we before the experiment have drawn an $x$-axis on the unstretched string. (Therefore when the string is stretched, the $x$-labels get deformed. In fluid dynamics we would say that we have adapted the Lagrangian (as opposed to the Eulerian) picture.) Let $\Delta x$ refer to a certain interval of the string. Then $\Delta U$ in eq. (A) is the total potential energy in that part of the string, hence the half.

  2. To avoid paradoxes with dividing infinitesimal quantities, we assume that $\Delta x$ is small, but not infinitesimal small. (For this reason, we do not use the notations $\delta x$ or $dx$.)

  3. We next need to relate the microscopic spring constant $k$ to the macroscopic Young modulus $Y$. By comparison of the formula
    $$\frac{|F|}{A}~=~Y\frac{|\Delta u|}{\Delta x},\tag{2}$$ and Hooke's law $$ |F| ~=~ k|\Delta u|,\tag{B}$$ we see that we should identify

$$k~\stackrel{(2)+(B)}{=}~ \frac{YA}{\Delta x} .\tag{C} $$

  1. Finally, combine eqs. (A) & (C) to achieve OP's sought-for formula

$$\text{Elastic potential energy density}~=~ \frac{1}{A}\frac{\Delta U}{\Delta x} ~\stackrel{(A)+(C)}{=}~\color{Red}{\frac{1}{2}} Y \left(\frac{\Delta u}{\Delta x} \right)^2.\tag{1}$$

For further details, see Wikipedia and Refs. 1-2.

References:

  1. H. Goldstein, Classical Mechanics; 2nd ed; Section 12.1.

  2. H. Goldstein, Classical Mechanics; 3rd ed; Section 13.1.

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The force is proportional to the extension:

$$ F = kx $$

where we subsume all the various constants like Young's modulus and area into the constant $k$. We know $dW = Fdx$, so:

$$ dW = k x dx $$

and integrating this gives:

$$ W = \tfrac{1}{2}kx^2 + C $$

If we define the work to be zero when the extension is zero the constant $C$ is zero, and we get the relation between work and extension complete with the factor of $1/2$.

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  • $\begingroup$ what is equal to $dU$ here? $\endgroup$ – Julien Apr 1 '14 at 15:18
  • $\begingroup$ The potential energy is equal to the work done. Since energy is conserved, if you do work on the system that energy must go into potential energy. So $dU$ and $dW$ are effectively the same. $\endgroup$ – John Rennie Apr 1 '14 at 15:20
  • $\begingroup$ Are you saying there is no factor of $\frac{1}{2}$ in front of $dU$ in the situation I described? (I can't reconcile the two any other way) $\endgroup$ – Julien Apr 1 '14 at 15:26
  • $\begingroup$ I suppose my original question was misleading towards the end. $\endgroup$ – Julien Apr 1 '14 at 15:56
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I got some help, I will post it for the few people that ever run into this (please correct me if I abused the notation):

$$dW=-dU=-\int_0^{du}Fd(du)=-\int_0^{du}AY\frac{du}{dx}d(du)$$ $$ dU= \int_0^yAY\frac{y}{dx}dy$$ $$=\frac{1}{2}AY\frac{y^2}{dx}$$ $$=\frac{1}{2}AY\frac{du^2}{dx}$$ $$=\frac{1}{2}AY(\frac{du}{dx})^2dx$$

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