1. ## Caculus Genius wanted,,

hi a ll thats a hard one:
a beam of length L is fixed at both ends and loaded uniformly by an amount w per unit length. The deflection, y, at a dsitance,x, from one end is given by:

y= w/2EL (0.5 x^4 - Lx^3 + 0.5 L^2 x^2)

where E and L are constans

i) find the maximum deflection of the beam
ii)The points of contraflexure are those where d2y/dx2=0. Find the distance of any points of contraflexure from the end of the beam.

2. Originally Posted by gina_x
hi a ll thats a hard one:
a beam of length L is fixed at both ends and loaded uniformly by an amount w per unit length. The deflection, y, at a dsitance,x, from one end is given by:

y= w/2EL (0.5 x^4 - Lx^3 + 0.5 L^2 x^2)

where E and L are constans

i) find the maximum deflection of the beam
ii)The points of contraflexure are those where d2y/dx2=0. Find the distance of any points of contraflexure from the end of the beam.

Hi

$y= \frac{w}{2EL} \0.5 x^4 - L x^3 + 0.5 L^2 x^2) = \frac{w}{2EL} \frac{x^2}{2} (x^2 - 2Lx + L^2) = \frac{w}{4EL}\:x^2 {(x-L)}^2" alt="y= \frac{w}{2EL} \0.5 x^4 - L x^3 + 0.5 L^2 x^2) = \frac{w}{2EL} \frac{x^2}{2} (x^2 - 2Lx + L^2) = \frac{w}{4EL}\:x^2 {(x-L)}^2" />

$\frac{dy}{dx} = \frac{w}{4EL}\:\left[2x (x-L)^2 + x^2 \:2(x-L)\right] = \frac{w}{4EL}\:2x(x-L)(2x-L)$

3. can you explain more please ?where is the maximum and what is the diffrence?

4. x being between 0 and L
$x \geq 0$ and $x-L \leq 0$

When $x \leq \frac{L}{2}$, $2x - L \leq 0$ therefore $\frac{dy}{dx} \geq 0$

When $x \geq \frac{L}{2}$, $2x - L \geq 0$ therefore $\frac{dy}{dx} \leq 0$

The maximum deflection of the beam is obtained for $x = \frac{L}{2}$