Help an electrical engineering student with a beam deflection question The theoretical formula for my beam deflection is:
$$  v(x) =
\begin{cases}
-\dfrac{Px}{48EI}(3L^2-4x^2),  & 0\le x \le {L\over2}\\[10pt]
-\dfrac{P(x-L)}{48EI}(L^2-8Lx+4x^2), & {L\over2}\lt x \le L
\end{cases}$$
I need to derive the formulae for the slope $v'$, curvature $v''$ and $v'''$
Is this just a simply case of taking the first, second, and third derivative of the original equation? I know absolutely nothing about structural beams...
 A: I don't know much about beams, either, but ...
Looks like $v(x)$ denotes the vertical deflection of the beam at horizontal location $x$. If so, the slope of the beam is $v'(x) = dv/dx$.
So, yes, you just differentiate $v(x)$ with respect to $x$ to get $v'$.
Differentiate a couple more times to get $v''$ and $v'''$, although I can't see the physical relevance of these quantities, so I don't know why they are interesting in engineering.
A: The beam deflection $v$, the derivative of deflection $v'$ (which is the rotate of the beam), and the second derivative of the deflection $v''$ (which is the curvature of the beam) should be continuous through the span $x$. 
You need to make sure $v'(\frac{L}{2}) = \lim_{x\to\frac{L}{2}}v'(x)$ and $v''(\frac{L}{2}) = \lim_{x\to\frac{L}{2}}v''(x)$. This has already been taken care of since the continuity order of $v$ is $C^2$. So you just need to take the derivatives.
Some additional notes: If you know the second or first derivative of a beam deflection and the problem is to find the delfection, you need the meet the continuity reqirement to elimiate the integration constant.
A: Yes to find slope, bending moment, shear and rate of loading you need just to take derivatives along the span $x$ for the updated/revised new function in the new $x$ domain to the right of discontinuity as you have been given.
When shear force and bending moments are abruptly changing the neutral axis is no more the same function. Functions $(f,g)$ of neutral elastic line shown are different.
Supposing a problem of given shear force is posed to find bent beam neutral axis one arrives at the two functions by successive integrations using Euler-Bernoulli Law $ EI v''= M.$
In your case there is a sudden central shear force introduced. That is why beam line functions are different. They are spliced together with zero, first and second order continuities but with no third order discontinuity as shown in sketch below, $  \dfrac{PL^3}{48 EI} $ is taken unity  representing central deflection of a simply supported beam with central load $P.$ Due to simply supported condition bending moment is zero at either end of the beam.
If this is understood it is interesting to learn further on with Prof. Hetenyi's (MIT) analytic procedure expressing the discontinuous deflection curve in terms of a McLaurin power series with


*

*initial/boundary conditions

*discontinuous forces and rates of loading, that runs like:


$$ v(x) = v_0(x)+ v_0'(x) x-\dfrac{M_0}{EI}\dfrac{x^2}{2!}-\dfrac{V_0}{EI}\dfrac{x^3}{3!}-\dfrac{q_0}{EI}\dfrac{x^4}{4!} $$
 
