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Math Help - Caculus Genius wanted,,

  1. #1
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    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.

    HELP Please!!!!
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  2. #2
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    Quote Originally Posted by gina_x View Post
    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.

    HELP Please!!!!
    Hi

    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)
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  3. #3
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    can you explain more please ?where is the maximum and what is the diffrence?
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  4. #4
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    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}
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