Is there a standard chart in Eurocode for the maximum and minimum bending moment in uniformly loaded beam for indeterminate structures where we can use directly? Many textbooks give different coefficient values for the moment for continuous beam, some wl2/10, some 0.046wl2. I know that hogging and sagging moment are different. Is there an authoritative reference for this?

  • $\begingroup$ Which moment? The maximum positive moment along the span (which isn't at midspan)? The moment at midspan? The negative moment at the central support? And for what load? A uniformly distributed load along the entire beam? $\endgroup$
    – Wasabi
    Commented Jun 20, 2017 at 17:44
  • $\begingroup$ I'll add an analytical answer tomorrow. $\endgroup$
    – Wasabi
    Commented Jun 21, 2017 at 2:36

2 Answers 2


A continuous beam with two spans with uniformly distributed load looks like this:

enter image description here

This is symmetric, and we know by inspection that the rotation of the beam at the central support will be null. So let's make our lives a bit easier and only look at the right half of the beam, considering the central support as fixed. Obviously, this is only possible because the beam and its loading is symmetric.

enter image description here

This beam is still statically indeterminate, however. But let's get rid of the rightmost support for now:

enter image description here

This is an isostatic structure we can trivially solve. However, the deflection at the extremity will obviously be non-null, which isn't correct for our true condition. Let's calculate what that deflection would be, using the fundamental beam equation:

$$\frac{\text{d}^2}{\text{d}x^2}\left(EI\frac{\text{d}^2w}{\text{d}x^2}\right) = q$$ which tells us that the fourth derivative of the deflection is equal to the applied load.

So, integrating the applied load we get the shear force:

$$\begin{gather} Q = \int\limits_0^L q\text{d}x = qx + C_1 \\ Q(0) = C_1 = -qL \\ \therefore Q = qx - qL \end{gather}$$

Integrating the shear force we get the bending moment:

$$\begin{gather} M = \int\limits_0^L (qx - qL)\text{d}x = \frac{qx^2}{2} - qLx + C_2 \\ M(0) = C_2 = \frac{qL^2}{2} \\ \therefore M = \frac{qx^2}{2} - qLx + \frac{qL^2}{2} \end{gather}$$

Integrating the bending moment we get the curvature times the stiffness:

$$\begin{gather} EI\theta = \int\limits_0^L \left(\frac{qx^2}{2} - qLx + \frac{qL^2}{2}\right)\text{d}x = \frac{qx^3}{6} - \frac{qLx^2}{2} + \frac{qL^2x}{2} + C_3 \\ \theta(0) = C_3 = 0 \\ \therefore EI\theta = \frac{qx^3}{6} - \frac{qLx^2}{2} + \frac{qL^2x}{2} \end{gather}$$

Integrating the curvature we get the deflection:

$$\begin{gather} EI\delta = \int\limits_0^L \left(\frac{qx^3}{6} - \frac{qLx^2}{2} + \frac{qL^2x}{2}\right)\text{d}x = \frac{qx^4}{24} - \frac{qLx^3}{6} + \frac{qL^2x^2}{4} + C_4 \\ \delta(0) = C_4 = 0 \\ \therefore EI\delta = \frac{qx^4}{24} - \frac{qLx^3}{6} + \frac{qL^2x^2}{4} \end{gather}$$

Therefore, at the end of the cantilever the deflection is

$$\begin{align} EI\delta(L) &= \frac{qL^4}{24} - \frac{qL^4}{6} + \frac{qL^4}{4} \\ \delta(L) &= \frac{qL^4}{8EI} \end{align}$$

So, in order to regain compatibility with our true condition, we need to bring that deflection down to zero. That's done by the vertical reaction force of the rightmost support. So let's replace that support with a concentrated vertical force and calculate the deflection caused by that force:

$$\begin{gather} Q = \int\limits_0^L 0\text{d}x = C_1 \\ Q(0) = C_1 = -P \\ \therefore Q = -P \\ M = \int\limits_0^L -P\text{d}x = -Px + C_2 \\ M(0) = C_2 = PL \\ \therefore M = -Px + PL \\ \theta = \int\limits_0^L (-Px + PL)\text{d}x = \frac{-Px^2}{2} + PLx + C_3 \\ \theta(0) = C_3 = 0 \\ \therefore EI\theta = \frac{-Px^2}{2} + PLx \\ \delta = \int\limits_0^L \left(\frac{-Px^2}{2} + PLx\right)\text{d}x = \frac{-Px^3}{6} + \frac{PLx^2}{2} + C_4 \\ \delta(0) = C_4 = 0 \\ \therefore EI\delta = \frac{-Px^3}{6} + \frac{PLx^2}{2} \\ \delta(L) = \frac{PL^3}{3EI} \end{gather}$$

So now, we can calculate the necessary force to counteract the deflection at the rightmost extremity:

$$\begin{gather} \frac{PL^3}{3EI} = -\frac{qL^4}{8EI} \\ \therefore P = -\frac{3qL}{8} \end{gather}$$

So, looking at the component equations for the bending moment from the uniform load and the reaction force, we can add them together to get the final bending moment equation for the span:

$$\begin{align} M &= \frac{qx^2}{2} - qLx + \frac{qL^2}{2} - Px + PL \\ M &= \frac{qx^2}{2} - qLx + \frac{qL^2}{2} - \left(\frac{-3qL}{8}\right)x + \left(\frac{-3qL}{8}\right)L \\ M &= \frac{qx^2}{2} - \left(qL - \frac{3qL}{8}\right)x - \frac{3qL^2}{8} + \frac{qL^2}{2} \\ M &= q\left(\frac{x^2}{2} - \frac{5Lx}{8} + \frac{L^2}{8}\right) \end{align}$$

The minimum moment is found at the central support and is equal to

$$M(0) = \frac{qL^2}{8}$$

The maximum moment is trickier, since it isn't necessarily at midspan. First we need to calculate the moment's derivative zero to find its position:

$$\begin{align} M' &= q\left(x - \frac{5L}{8}\right) = 0 \\ \therefore x &= \frac{5L}{8} \\ \therefore M_{max} &= M\left(\frac{5L}{8}\right) \\ &= q\left(\frac{\left(\dfrac{5L}{8}\right)^2}{2} - \frac{5L\left(\dfrac{5L}{8}\right)}{8} + \frac{L^2}{8}\right) \\ &= -\frac{9qL^2}{128} \approx -\frac{qL^2}{14.2}\\ \end{align}$$

That approximation is a classic numerical shortcut for the maximum moment on a fixed-and-pinned beam. And obviously all my results assume that $q > 0$ points upwards. So all the loads in the beams at the start of my answer have $q = -1\text{ kN/m}$. If you prefer to think of such loads as positive, just flip the sign in my results.

And then, to confirm our results, here is a beam with $L=5\text{ m}$ (per span) and $q=128\text{ kN/m}$. The equations give us

$$\begin{align} M_{min} &= \frac{qL^2}{8} = -400\text{ kNm}\\ M_{max} &= -\frac{9qL^2}{128} = 225\text{ kNm}\\ \end{align}$$

enter image description here


I would suggest Timoshenko, Strength of Materials is an authoritative reference (used in the UK ...) : This is found on the www, and also, copies can be found in second-hand bookshops but even then they are not cheap...

Link removed to avoid "link rot"...

  • $\begingroup$ So as to avoid link-rot, please expand your answer to include the relevant information from the link provided, at least the most essential points. Feel free to quote directly from the source. $\endgroup$
    – Wasabi
    Commented Jun 20, 2017 at 13:26
  • $\begingroup$ @wasabi I gave an answer to the very last question posed by the O P - "Is there an authoritative reference for this?" Surely answering that question with the information , of arguably one of the most eminent sources, is correct and not, as you say "link rot". $\endgroup$
    – Solar Mike
    Commented Jun 20, 2017 at 13:33
  • $\begingroup$ Link rot is the link breaking because of a change of URL. It has nothing to do with the quality of the information contained in the link. If the link rots, your answer becomes useless for future readers. Which is why it is best practice to quote the relevant parts of the source as well as offer the source itself, if available. Then the link might rot, but the relevant information will still be in your answer. $\endgroup$
    – Wasabi
    Commented Jun 20, 2017 at 13:50
  • 1
    $\begingroup$ Dude, you don't have to remove the link. Just put the actual information from the link on your answer along with the link. $\endgroup$
    – Wasabi
    Commented Jun 20, 2017 at 17:41

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