# Shear forces along a beam using singularity functions

$\newcommand{\a}[1]{\langle#1\rangle}$ As you can see, the left end of the beam is fixed, and a distributed load is applied to the beam. I calculated the shear forces by conventional means, but when I calculate it with singularity function I get weird results. Here are my calculations:

$$q(x)=-3000\a{x-0}^{-2} + 3000\a{x-0}^{-1} - \frac{2000}{3}\a{x-0}^1 + \frac{2000}{3}\a{x-0}^0$$

The first term comes from reaction torque and the second term comes from reaction normal force. The shear force is just the integral of the $q(x)$.

$$v(x) = 3000\a{x-0}^{-1} - 3000\a{x-0}^{0} + \frac{2000}{6}\a{x-0}^2 - \frac{2000}{3}\a{x-0}^1$$

As you can see the results are false. I'll appreciate any help.

$$\newcommand{\a}[1]{\langle#1\rangle}$$ Loading

$$\frac{2000}{3}\a{x-0}^0 - \frac{2000}{3}\a{x-0}^1$$

Would look something like this:

If we were to superimpose the two graphs they would not look like the triangular loading you have in your picture. If we wanted that loading, we would use:

$$-2000\a{x-0}^0 + \frac{2000}{3}\a{x-0}^1$$

This would make sure that we had -2000kN/m at x=0m, and 0kN/m at x=3m.

Integration

From

$$\int \langle x-a\rangle ^{n}dx={\begin{cases}\langle x-a\rangle ^{{n+1}},&n\leq 0\\{\frac {\langle x-a\rangle ^{{n+1}}}{n+1}},&n\geq 0\end{cases}}$$

$$q(x)=-3000\a{x-0}^{-2} + 3000\a{x-0}^{-1} - 2000\a{x-0}^0 + \frac{2000}{3}\a{x-0}^1$$

$$v(x) = - 3000\a{x-0}^{-1} + 3000\a{x-0}^{0} - 2000\a{x-0}^1 + \frac{1000}{3}\a{x-0}^2$$

Since this is continuous over the entire length, we can simplify to

$$v(x) = 3000 - 2000x + \frac{1000}{3}x^2$$

Which is exactly what we would expect, since there is 3000kN at x=0m, and 0kN at x=3m.

And we can also plot in wolfram alpha.