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

enter image description here

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

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.


1 Answer 1


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

Take a look at the last two parts of your loading equation.

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

Would look something like this:

enter image description here

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.



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

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

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

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.


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