Euler-Bernouilli for a beam with force in the middle

Question

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?

Answer 1

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.