Recently I had need to prove the beam deflection equation for a simple cantilever beam with a point load at the end. The equation would be in reference to the Euler-Bernoulli theory taught in most basic engineering undergraduate courses. The only caveat is to not use "I" at any point in the derivation, but rather use basic concepts of force, moment, and some conceptual ingenuity to build the equation. The derivation would be specifically to find the angle ($\theta$) that is the result of the intersection of lines projecting from the two beam faces post-deflection. Another starting assumption would be a rectangular beam face.

My problem is that every time I derive the equation, I come out 2x too large. My initial equation is as follows:

$$2\int_0^{h/2} \frac{(r_i + h/2 + y)\theta - l_o}{l_o} (E) (y) (z) dy = Wl_o$$

This simplifies to:

$$\frac{(2)(E) (z)(\theta)}{l_o}\int_0^{h/2} y^2 dy = Wl_o$$

Ultimately leading to:

$$\frac{(E) (z)(h^3)(\theta)}{12l_o}= Wl_o$$

$$ \theta = \frac{12W{l_o}^2}{Ezh^3}$$

As you can see, I come up with an answer that is twice as large as (what I believe to be) the correct answer, and I can't seem to reconcile where I lost a 2.

My thought process behind the initial equation:

  • The initial fraction represents the strain
  • E represents the Elastic Modulus (which needs to be multiplied against the strain to get a force/area (pressure point))
  • For each pressure point you need to multiply by a distance to get the moment (I chose the center axis)
  • To account for the depth of the beam, multiply by $z$
  • Integration from $0$ to $h/2$ is necessary to sum the moment from center axis to top of beam
  • The $2$ represents the bottom half of the face of the beam that is equal in moment to the top half (so double the top half)
  • The $W*l_o$ represents the total moment being accounted for

With all that said, I've racked my brain and can't seem to find the discrepancy. Hopefully it's not a stupid mistake.. Thanks!

Note: This is a X-Post from physics SE.enter image description here

  • $\begingroup$ Its so different from what I am used to seeing. A graphic that labels your terms might help us see your approach better. Its interesting that your off by only a factor of 2.... $\endgroup$
    – ShadowMan
    Apr 5, 2019 at 21:21
  • $\begingroup$ Simple beam deflection posts are on here more than once - have you read them all for pointers? $\endgroup$
    – Solar Mike
    Apr 6, 2019 at 5:54
  • $\begingroup$ Have a look at : engineering.stackexchange.com/a/26586/10902 $\endgroup$
    – Solar Mike
    Apr 6, 2019 at 6:17
  • $\begingroup$ Can't say I've read every article on SE, but I have searched extensively the last week or so and haven't found anything useful. Would you be able to help me in posting a graphic to help everyone understand the situation? I'm unfamiliar in how to do so. $\endgroup$ Apr 7, 2019 at 0:36
  • $\begingroup$ Would you please edit your integrals to add the appropriate differential term ($dy$, maybe, with the limits being $0$ and $h/2$, where $h$ is the beam height?). They don't really have any meaning as is, and I want to make sure I understand your derivation. $\endgroup$ Apr 10, 2019 at 20:12


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