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7.5: Beam deflections from applied bending moments

  • Page ID
    7825
  • he second derivative (curvature) of the neutral axis line (dotted line in diagram)

    \[\kappa=\frac{d^{2} y}{d x^{2}}\]

    The approximation involved in equating beam curvature to the curvature of the neutral axis of a beam.

    It follows that

    \[M=\kappa E I=E I \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}\]

    Since the moment at the section concerned can also be written, for a cantilever beam, as M = F (L - x)

    it follows that

    \[E I \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=F(L-x)\]

    This second order differential equation can be integrated (twice), with appropriate boundary conditions, to find the deflection of the beam at different points along its length. For a cantilever beam, this operation is shown below.

    \[E I \frac{\mathrm{d} y}{\mathrm{d} x}=F L x-\frac{F x^{2}}{2}+C_{1}\]

    \(\text { at } x=0, \frac{\mathrm{d} y}{\mathrm{d} x}=0, \text { thus } C_{1}=0\)

    \[E I y=\frac{F L x^{2}}{2}-\frac{F x^{3}}{6}+C_{2}\]

    \(\text { at } x=0, y=0, \text { thus } C_{2}=0\)

    which can be rearranged to give

    \[y=\frac{F x^{2}}{6 E I}(3 L-x)\]

    For example, at the loaded end ( x = L ), this gives

    \[\delta=\frac{F L^{3}}{3 E I}\]

    Symmetrical 3-point bending

    Symmetrical 3-point bend loading

    The bending moment is given by

    \[M=\frac{-F x}{2}\]

    It follows that

    \[E I \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=\frac{-F x}{2}\]

    and the integration procedures lead to

    \[E I \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{F x^{2}}{4}+C_{1}\]

    \(\text { at } x=\frac{L}{2}, \quad \frac{d y}{d x}=0 \text { thus } C_{1}=\frac{F L^{2}}{16}\)

    \[E I y=\frac{-F x^{3}}{12}-\frac{F L^{2} x}{16}+C_{2}\]

    \(\text { at } x=0, y=0 \text { thus } C_{2}=0\)

    so the equation for the deflection is

    \[y=\frac{F x}{48 E I}\left(3 L^{2}-4 x^{2}\right)\]

    and deflection of the centre of the beam is given by

    \[\delta=\frac{F L^{3}}{48 E I}\]

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