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Math Help - Maxima minima calculus help

  1. #1
    Member astartleddeer's Avatar
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    Maxima minima calculus help

    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:

     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 =  30_{10^{6}}Nm^{2}

    F = 100kN

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

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

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

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

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

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

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

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

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

    Now I'm lost on the position?

    Thanks in advance
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  2. #2
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    Re: Maxima minima calculus help

    ... the deflection at a distance x from one end is given by ...
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  3. #3
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    Re: Maxima minima calculus help

    So it's 5 then?
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  4. #4
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    Re: Maxima minima calculus help

    Yes, the mid-point.
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    Re: Maxima minima calculus help

    Surely, knowing the load was in the middle we could have said immediately that the maximum deflection would be when x=5
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  6. #6
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    Re: Maxima minima calculus help

    I don't know how the formula for y is derived, but had the expression in brackets been 3Lx-4x^{3}, for example, the maximum deflection would not have been at the mid-point, (though I would have to admit that from a practical point of view I would have expected it to be).
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