In one of my physics textbooks there is a chapter on the elasticity of materials which contains pretty basic outline about Young's modulus, stress-strain, elastic potential energy and related stuff. There is only one thing stated in the book which I didn't understand, which is this:

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Consider an elastic beam rigidly supported at both its ends in a horizontal fashion, which is loaded with a weight $W$ at the centre. It's length is $l$, breadth is $b$, depth is $d$ and the Young's modulus is $Y$. Then the beam sags by an amount $\delta$ which is given by: $$\delta={Wl^3 \over 4bd^3Y}$$

The book says that it can be derived easily with basic concepts of elasticity and some calculus. What I tried was:

  1. Try to calculate longitudinal strain by approximating the bent beam as a circular arc.
  2. Integrate the shear stress along the beam and approximate the shear modulus $G\approx {Y \over 3}$
  3. Equate the work $(W\delta)$ done on the beam due to the load to the elastic potential energy.

Despite many efforts, I could not arrive at the result. Can anyone please help me in proving this result?

  • $\begingroup$ Your approach was wrong, even if you used the right formula for $y(x)$. The basic problem is, if we equate those two energies -- $W \delta$ and the energy of bending which resists it -- then we're expecting the total potential energy to be 0. Another beam with 0 potential energy is the horizontal beam with no weight on it. Now suddenly put $W$ on it: it will not go to equilibrium but through it, with some kinetic energy. It's not until friction pulls out that energy that it comes to rest at its equilibrium. So the minimum potential energy is not 0, but strictly less than that. $\endgroup$
    – CR Drost
    Mar 12, 2015 at 19:23
  • $\begingroup$ @ChrisDrost I did suspect that that approach was wrong, thanks for clearing that up! $\endgroup$
    – najayaz
    Mar 12, 2015 at 19:27
  • $\begingroup$ If you want, you can use the calculus of variations to find a steady-state y minimizing potential energy. It is not too hard: take the correct expression $U=\int dx\left(F(x)y(x)-YI[y''(x)]^2\right)$ and replace $y(x)\rightarrow y(x)+\delta y(x)$, looking for $\delta U=\int dx\left(F~\delta y-Y~I~\left[(y''+\delta y'')^2-(y'')^2\right]\right)=0$. Ignoring $(\delta y'')^2$ you integrate by parts twice: $\int dx(F-Y~I~y'''')\delta y=0$ can only be $0$ for all $\delta y$ if $F=Y~I~y''''.$ (But, F(x) does have a δ function in it.) $\endgroup$
    – CR Drost
    Mar 12, 2015 at 19:44

2 Answers 2


At every point along the beam, the curvature has to be such that the externally applied bending moment exactly counters the internal stress. This tells you that the curvature is not constant - it is a function of distance to the side (largest in the middle, zero at the wall). This means that your assumption of "circular section" is wrong.

See for example figure 3.16 in this link and associated derivations.

Simplifying the description found there:

From their equation 3.21, the curvature $\rho$ of a beam is related to the bending moment $M$ by

$$\rho = \frac{EI}{M}\tag1$$

Where $E$ is the Young's modulus and $I$ is the second moment of area. For a rectangular beam (not specified in your question, but that's what I am assuming) we can compute $I$ as

$$I = \frac{bd^3}{12}\tag2$$

(see for example this link)

Now we need an expression for the bending moment as a function of position. For points to the left of the center, bending moment is proportional to $Wx/2$ - half the weight (two supports) times the distance from the support.

Knowing that the radius of curvature is (for small deflections) inversely proportional to the second derivative of the shape, we can write

$$\frac{d^2 y}{dx^2} = -\frac{Wx}{2EI}\tag3$$

Integrating twice, we get

$$y = -\frac{Wx^3}{12EI} + Ax + B\tag4$$

If we set $y=0$ at $x=0$, we get $B=0$. Setting the slope of the curve =0 at $x = \frac{\ell}{2}$, we find

$$-\frac{W(\ell/2)^2}{4EI} + A = 0$$ $$A = \frac{W\ell^2}{16EI} \tag5$$

leading to an expression for the deflection

$$y = \frac{Wx}{12EI}\left(\frac{3\ell^2}{4}-x^2\right) \tag6$$

Substituting $x=\frac{\ell}{2}$ into (6), and using expression (2) for $I$, we obtain the deflection you were looking for.

This expression agrees with equation (7) at this reference.

  • 2
    $\begingroup$ The link is good, but the derivation is a bit out of the scope of the book I found the formula in. Do you think there's a simpler derivation? $\endgroup$
    – najayaz
    Mar 12, 2015 at 12:53
  • $\begingroup$ I hope the updated answer helps $\endgroup$
    – Floris
    Mar 12, 2015 at 15:30

The 4's and cubes possibly come from the general physical theory involving a fourth derivative, $$\alpha \frac {d^4y}{dx^4} + \mu ~ \ddot y = W \delta(x).$$ This expression has a Dirac $\delta$-function to locate the weight at $x=0$ and in steady state $\ddot y = 0$. So $y'''(x)$ is discontinuous at $x=0$ by a total amount $W/\alpha$; then in the space from $x=0$ to $l/2$ it takes on the form of a cubic, $y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$.

Using the x-reflection symmetry we can reduce this to a simple problem on $(0, l/2)$ where we have the boundary conditions: $y'''(0) = 6 a_3 = W/(2\alpha)$, $y'(0) = a_1 = 0$, we also know that $a_0 = -\delta$ and then we can choose $a_2$ to force $y(l/2) = 0$.

Once you've worked out this polynomial I'd try to find out what the other condition is that is required at $x = l/2$ to get the relationship that you are looking at. It may be $y''(l/2) = 0$; Wikipedia in the link above calls this a "simply supported end" and it basically "looks like" your figure. It may instead be $y'(l/2) = 0$ which would be a "clamped-to-be-horizontal" end.

  • 2
    $\begingroup$ Much thanks for the answer, but unfortunately, I didn't get a word of that. $\endgroup$
    – najayaz
    Mar 12, 2015 at 12:49

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