# How to deal with inverse distributed load ramp starts and ends somewhere along the beam?

I know how to solve it by moment line methode but i want to use singularity function here and i'm very confused. My strategy consist of writing the correspond singularity function for each element but i have absolutely no clue how can i write the right singularity function for the inverse ramp distributed load at the right side?

Are you familiar with the Dirac delta function, Heaviside step function and the Ramp function?

You could use Laplace Transforms together with those three handy functions:

Dirac Delta function

$$\delta(x)= \left\{\begin{array}{l}+\infty\qquad&x=0\\0&x\neq0\end{array}\right.$$ The Dirac Delta function is equal to infinity, at $x=0$, and $0$ everywhere else. Respectively $\delta(x-a)=+\infty$ if $x=a$.

With the Euler-Bernoulli beam theory, where $$EI\frac{d^4w}{dx^4}=q(x)$$ a static interpretation of the Dirac function is, that a point load $Q$ at point a represents a distributed load with infinite density, thus can be written as $q(x)=Q\cdot \delta(a-x)$

Heaviside step function

The step function is the integral of the Dirac Delta function $$H(x)=\int_{-\infty}^x\delta(s)ds$$ or $$H(x)= \left\{\begin{array}{l}1\qquad&x\geq0\\0&x<0\end{array}\right.$$ Thus, a distributen load, with magniuted $q$, between $x=a$ and $x=b$ can be written as $q(x)=q\left[ H(x-a)-H(x-b) \right]$

Ramp function

The ramp function again is the integral of the Heaviside function. $$R(x)= \left\{\begin{array}{l}x\qquad&x\geq0\\0&x<0\end{array}\right.$$ Thus, a distributied load , starting at $x=a$, with a slope $m$ can be written as $q(x)=m\cdot R(x-a)$

The following distributed load, for example:

This load can be written as: $$q(x)=\frac{q}{b-a}\left[R(x-a)-R(x-b)\right]-\frac{q}{d-c}\left[R(x-c)-R(x-d)\right]$$ please see attached file for the derivation

Further explanation of the Dirac Delta function

Take, for example the point force $Q$. That force can be approximated by a distributed load $Q=q\cdot dx$, therefore $q=\frac{Q}{dx}.$ Now, as a point load acts on an interval $dx\to0$, we can say $\lim_{dx\to0} q=+\infty$

This is also represented, if you look at a shear force diagram. We know, that $V'(X)=q(x)$, and we know that a point load $Q$ (external force or support) at point $a$ causes a "jump" in the shear force diagram, according to its magnitude, this can be represented by the Heaviside function. $V=Q\cdot H(x-a)$, now $$V'(x)=q(x)=(Q\cdot H(x-a))'=Q\cdot H'(x-a)$$ As explained above, the derivative of the Heaviside function is the Dirac delta function, therefore $$q(x)=Q\cdot \delta(x-a)$$

• Thanks for your elegant answer. Could you please explain more about the static interpretation of Dirac function ? the rest is clear. Jan 23 '18 at 21:16
• I added some further interpretation/explanation to the Dirac Delta function at the end of my answer, I hope my explanation is comprehensible. Jan 24 '18 at 11:28