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A beam of length $L$ is held a $x = 0, L$. A force $F$ in the middle ($x = L/2$) pushes the beam.

The Euler-Bernoulli equation can be written as such:

$EI\frac{\partial^4 y(x)}{\partial x^4} = F\delta(x - L/2)$

Taking $\alpha = \frac{F}{EI}$, integrating to get the deflection $y(x)$, I get:

$y(x) = (\alpha + C_1) \frac{x^3}{6} + C_2 \frac{x^2}{2} + C_3 x + C_4$

Because $y(0)= 0$, then $C_4 = 0$ and $dy(0)/dx = 0$, then $C_3 = 0$

Similarly, $y(L)= 0$ so that

$0 = (\alpha + C_1) \frac{L^3}{6} + C_2 \frac{L^2}{2}$

$C_2 = -(\alpha + C_1)\frac{L}{3}$

Now, $\partial y(L)/\partial x = 0$ so that

$0 = (\alpha + C_1) \frac{L^2}{2} + C_2 L = (\alpha + C_1) \frac{L^2}{2} -(\alpha + C_1)\frac{L^2}{3}$

$C_1 = - \alpha$

$C_2 = 0$

And thus, $y = 0$

What am I doing wrong?

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    $\begingroup$ If its held at both ends, just try to use symmetry with zero rotation at the middle and calculating just one half. $\endgroup$ Commented Oct 29, 2022 at 18:13
  • $\begingroup$ When you integrated, what is the term corresponding to the integral of $F\ \delta(x-L/2)$? Is it $\alpha x^3/6$? That is probably where the mistake is. $\endgroup$
    – AJN
    Commented Oct 30, 2022 at 8:28
  • $\begingroup$ @AJN the integral of the dirac delta is 1 which is then integrated 3 other times (hence $x^{3}/6$. Is this correct? $\endgroup$
    – mooder
    Commented Oct 31, 2022 at 12:15
  • $\begingroup$ Point load divides beam into parts with different deflection functions, so in this case you need one function for the first half and another function for the second half. When you try to put zero deflection and rotation at endpoints into one deflection function, it does not work. $\endgroup$ Commented Oct 31, 2022 at 17:16

1 Answer 1

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The integral of dirac delta is $\int \delta(x) dx = u(x)$ where $u$ is the unit step function; not $1$.

So the integral of shifted dirac delta would be $\int \delta(x-x_0) dx = u(x-x_0)$.

In fact Wkipedia seems to cover exactly this scenario.

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