The theoretical formula for my beam deflection is:

$ v(x) = \begin{cases} -\frac{Px}{48EI}(3L^2-4x^2), & 0\le x \le {L\over2}\\ -\frac{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...

  • $\begingroup$ This is going back a while for me. Basically the deflection is a parabola and the formula changes once you are over halfway past the beam because the beam is not curving up, not down. So yes, I believe you are just taking the first, second, and third derivatives. E is elastic modulus of the beam material and I is the moment of inertia. L is your beam length and x is your position on the beam you are solving for. There are no other 'variables' that you would need to consider when taking the derivatives (so you don't have to take it with respect to multiple variables. Just x $\endgroup$ – Prevost Sep 24 '16 at 1:46
  • $\begingroup$ YES. And also note that t the internal moment is $$M(x) = E I \frac{{\rm d}^2}{{\rm d}x^2} v(x)$$. If you did it right then the moment is zero at pin joints. $\endgroup$ – John Alexiou Sep 25 '16 at 15:24


The first derivative of the deflection is equal to the tangent of the deflection, which for small deflections can be approximated as equal to the angle of rotation of the beam at each point.

The second derivative (times $EI$) is the bending moment along the beam.

The third derivative (times $EI$) is the shear force along the beam.

The fourth derivative (times $EI$) is the distributed load along the beam.

| improve this answer | |
  • $\begingroup$ It makes sense now. I guess if your background is mechanical engineering it's a fairly straight forward question... $\endgroup$ – stuart Sep 25 '16 at 23:21

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