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Does anyone know how to derive the deflection formula $\delta = \dfrac{-0.0068wL^4}{EI}$ for a continuous beam spans over 4 supports?

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  • $\begingroup$ Have you tried anything yourself? Without some more information on your part, no one can guess where you are having trouble. $\endgroup$
    – hazzey
    Sep 8 '16 at 19:52
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I assume you mean a beam with three spans of equal length $L$ under the effect of a uniform distributed load, like the one displayed below. Also, I assume you're asking about the maximum displacement on the lateral (non-central) spans.

enter image description here

This is a statically indeterminate structure, so this'll take some work.

For starters, we can notice this structure is symmetric, and can therefore be simplified to the following (the rectangle represents a rotational restraint, to represent that there's no rotation at the central midspan):

enter image description here

This, however, is still statically indeterminate. So now let's pretend that rotational restraint doesn't exist, which turns this into an isostatic simply-supported beam with cantilever. Actually, let's get rid of the cantilever as well, and replace it with a concentrated force and moment at the support, equal to $M=-\dfrac{q\left(\frac{L}{2}\right)^2}{2}$.

enter image description here

So now let's calculate the displacement at the middle of the simply-supported span.

First we need to calculate the reactions (still considering the cantilever): $$\begin{align} \sum V &= V_1 + V_2 + \frac{3}{2}qL = 0 \\ \sum M_1 &= LV_2 + \frac{9}{8}qL^2 = 0 \\ \therefore V_2 &= -\frac{9}{8}qL \\ V_1 &= -\frac{3}{8}qL \end{align}$$

Now that we have that, we can start integrating our way up to the deflection. $$\begin{align} Q &= \int\limits_0^L q\text{d}x \\ &= qx + C \\ Q(0) &= C = V_1 = -\frac{3}{8}qL \\ M &= -\int\limits_0^L Q\text{d}x \\ &= -(\frac{q}{2}x^2 - \frac{3}{8}qLx + C) \\ M(0) &= C = 0 \\ EI\theta &= \int\limits_0^L M\text{d}x \\ &= -\frac{q}{6}x^3 + \frac{3}{16}qLx^2 + C_1 \\ EI\delta &= \int\limits_0^L \theta\text{d}x \\ &= -\frac{q}{24}x^4 + \frac{1}{16}qLx^3 + C_1x + C_2 \\ \delta(0) &= C_2 = 0 \\ \delta(L) &= -\frac{q}{24}L^4 + \frac{1}{16}qL^4 + C_1L = 0 \\ \therefore C_1 &= -\frac{q}{48}L^3 \\ \therefore \delta &= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{16} -\frac{L^3x}{48}\right) \end{align}$$


This, however, is not our deflection equation, since we still need to deal with that rotation restriction at the end of our cantilever, which we'd ignored so far. To do that, we need to get the rotation at the free end of the cantilever. So:

$$\begin{align} M_c &= q\left(\dfrac{x^2}{2} - \dfrac{Lx}{2} + \dfrac{L^2}{8}\right) \\ EI\theta_c &= \int\limits_0^L M_c\text{d}x \\ &= q\left(\dfrac{x^3}{6} - \dfrac{Lx^2}{4} + \dfrac{L^2x}{8} + C\right) \\ EI\theta_c(0) &= C = EI\theta(L) \\ &= qL^3\left(-\frac{1}{6} + \frac{3}{16} - \frac{1}{48}\right) = 0 \\ EI\theta_c\left(\frac{L}{2}\right) &= q\left(\dfrac{\left(\frac{L}{2}\right)^3}{6} - \dfrac{L\left(\frac{L}{2}\right)^2}{4} + \dfrac{L^2\left(\frac{L}{2}\right)}{8}\right) \\ &= \dfrac{qL^3}{48} \end{align}$$

So, if there were no rotational restriction, the cantilever'd have a rotation of $\dfrac{qL^3}{48}$ at its free end. The rotational restriction must therefore apply a concentrated bending moment sufficient to cancel out that rotation. Now, if a concentrated bending moment $M'$ is applied at the end of the cantilever, it will generate a uniform diagram along the cantilever and a linear diagram along the simply-supported span.

So now we need to redo everything all over again.

$$\begin{align} \sum V &= V_1 + V_2 = 0 \\ \sum M_1 &= LV_2 + M' = 0 \\ \therefore V_2 &= -\frac{M'}{L} \\ V_1 &= \frac{M'}{L} \\ Q &= \int\limits_0^L 0\text{d}x \\ &= C \\ Q(0) &= C = V_1 = \frac{M'}{L} \\ M &= -\int\limits_0^L Q\text{d}x \\ &= \frac{M'}{L}x + C \\ M(0) &= C = 0 \\ EI\theta &= \int\limits_0^L M\text{d}x \\ &= \frac{M'}{2L}x^2 + C_1 \\ EI\delta &= \int\limits_0^L \theta\text{d}x \\ &= \frac{M'}{6L}x^3 + C_1x + C_2 \\ \delta(0) &= C_2 = 0 \\ \delta(L) &= \frac{M'}{6L}L^3 + C_1L = 0 \\ \therefore C_1 &= -\frac{M'L}{6} \\ \therefore \delta &= \frac{M'}{EI}\left(\frac{x^3}{6L} -\frac{Lx}{6}\right) \end{align}$$

And now for the cantilever:

$$\begin{align} EI\theta_c &= \int\limits_0^L M_c\text{d}x \\ &= M'x + C \\ EI\theta_c(0) &= C = EI\theta(L) \\ &= \frac{M'L}{3} \\ EI\theta_c\left(\frac{L}{2}\right) &= \frac{5M'L}{6} \end{align}$$

Now we can find the requisite applied bending moment at the end of the cantilever.

$$\begin{align} \frac{5M'L}{6} &= \dfrac{qL^3}{48} \\ \therefore M' &= \dfrac{qL^2}{40} \end{align}$$


Almost there!

Now all that's left to do is sum up the deflection equations for the simply-supported span. We end up with

$$\begin{align} \delta &= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{16} -\frac{L^3x}{48}\right) + \frac{M'}{EI}\left(\frac{x^3}{6L} -\frac{Lx}{6}\right) \\ &= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{16} -\frac{L^3x}{48}\right) + \frac{q}{EI}\left(\frac{Lx^3}{240} -\frac{L^3x}{240}\right) \\ &= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{15} -\frac{L^3x}{40}\right) \\ \end{align}$$

The maximum value, however, is not at the midpoint. So now we have to find the minimum via the derivative:

$$\delta' = \frac{q}{EI}\left(-\frac{x^3}{6} + \frac{Lx^2}{5} -\frac{L^3}{40}\right) = 0$$

You then use Wolfram Alpha to find that the minimum occurs at $x\approx0.44604 L$. Throw that into the full equation for the deflection and you get

$$\delta = -\dfrac{0.00688421327975355744qL^4}{EI}$$

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  • $\begingroup$ Nice effort. Good explanation. +1 $\endgroup$
    – xCodeZone
    Sep 9 '16 at 6:09

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