Hello,

I'm stuck on the following question. I feel like I've computed the correct results but because my units have cancelled I'm a bit lost. If you refer below (I think) I have computed the correct result for a maximum deflection but I don't know what answer is the position for a maximum deflection. I think it's the first derivative but I'm not sure.

Anyway, here's the question:

If a beam of length L is fixed at the ends and loaded in the centre of the beam by a point load of F Newtons the deflection at a distance x from one end is given by:

$\displaystyle y = \frac{F}{48EI} (3L^{2}x - 4x^{3})$

Where E = Young's Modulus and I = Second Moment of Area of the beam. Find the position and the value of the maximum deflection of the beam if:

L = 10m

EI = $\displaystyle 30_{10^{6}}Nm^{2} $

F = 100kN

$\displaystyle y = \left(\frac{100kN}{(48)(30_{10^{6}}Nm^{2})}\right) ((3)(10m)^{2}x - 4x^{3})$

$\displaystyle y = \left(\frac{1}{14400}\right) (300x - 4x^{3}) $

$\displaystyle y = \frac{1}{48} (x) - \frac{1}{3600} (x^{3}) $

$\displaystyle \frac{dy}{dx} = \frac{1}{48} - \frac{1}{1200} (x^{2}) = 0\ for\ a\ maximum\ or\ minimum\ value $

$\displaystyle x = \sqrt{\frac{1200}{48}} = \pm5$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -600x $

$\displaystyle \frac{d^{2}y}{dx^{2}} = - (600)(5) = -3000\ for\ a\ maximum\ value $

Substituting x= (+5) into the initial equation for a maximum deflection:

$\displaystyle y = \left(\frac{1}{14400}\right) ((300)(5) - (4)(5^{3})) = \frac{5}{72} $

Now I'm lost on the position?

Thanks in advance