Derive three moment theorem

Statement

For two adjacent spans $L_1$ and $L_2$ of a continuous beam between three supports A, B and C, the support moments are related by the three moment equation (Clapeyron's theorem):

$$M_A L_1 + 2M_B (L_1 + L_2) + M_C L_2 = -\left(\frac{6 a_1 \bar{x}_1}{L_1} + \frac{6 a_2 \bar{x}_2}{L_2}\right)$$

where $a_1, a_2$ are the areas of the free (simply supported) BM diagrams on spans 1 and 2, $\bar{x}_1$ is the distance of the centroid of $a_1$ from A, and $\bar{x}_2$ is the distance of the centroid of $a_2$ from C. If the supports settle (downward $\delta_A, \delta_B, \delta_C$), the term $+6EI\left[\dfrac{\delta_B-\delta_A}{L_1} + \dfrac{\delta_B-\delta_C}{L_2}\right]$ is added to the right side (EI constant).

Derivation (supports at the same level, EI constant)

Let the support moments be $M_A, M_B, M_C$ (sagging positive, so hogging moments are negative). The BM diagram on each span is the sum of the free BM diagram and a trapezoid of support moments.

Step 1: Slope at B in span AB. The deviation of A from the tangent at B equals the moment about A of the $M/EI$ diagram between A and B. The diagram is the free BM diagram (area $a_1$, centroid $\bar{x}_1$ from A) plus two triangles from $M_A$ and $M_B$. With the supports at the same level, this deviation equals $\theta_{B1}L_1$:

$$\theta_{B1}L_1 = \frac{1}{EI}\left[a_1\bar{x}_1 + \frac{M_A L_1}{2}\cdot\frac{L_1}{3} + \frac{M_B L_1}{2}\cdot\frac{2L_1}{3}\right]$$ $$\theta_{B1} = \frac{1}{EI}\left[\frac{a_1\bar{x}_1}{L_1} + \frac{M_A L_1}{6} + \frac{M_B L_1}{3}\right]$$

Step 2: Slope at B in span BC. In the same way, taking moments about C (with $\bar{x}_2$ measured from C), the deviation of C from the tangent at B is $-\theta_{B2}L_2$:

$$\theta_{B2} = -\frac{1}{EI}\left[\frac{a_2\bar{x}_2}{L_2} + \frac{M_C L_2}{6} + \frac{M_B L_2}{3}\right]$$

Step 3: Continuity. The beam is continuous over B, so the slope on the left of B equals the slope on the right: $\theta_{B1} = \theta_{B2}$.

$$\frac{a_1\bar{x}_1}{L_1} + \frac{M_A L_1}{6} + \frac{M_B L_1}{3} = -\frac{a_2\bar{x}_2}{L_2} - \frac{M_C L_2}{6} - \frac{M_B L_2}{3}$$

Multiply by 6 and rearrange:

$$M_A L_1 + 2M_B (L_1 + L_2) + M_C L_2 = -\frac{6a_1\bar{x}_1}{L_1} - \frac{6a_2\bar{x}_2}{L_2}$$

Special results

• UDL $w$ on a span $L$: $\dfrac{6a\bar{x}}{L} = \dfrac{wL^3}{4}$.
• Point load $W$ at distance $a$ from the left support of a span $L$, with $b = L - a$: $\dfrac{6a\bar{x}}{L} = \dfrac{Wab(L+a)}{L}$ when $\bar{x}$ is measured from the left support, and $\dfrac{Wab(L+b)}{L}$ when measured from the right support.

Physical meaning

The equation expresses continuity of slope at the middle support B: the rotation at B caused by the unknown support moments must balance the rotation caused by the loads on the two spans acting as simple beams, so the tangent to the elastic curve is the same on both sides of B. For a beam with $n$ supports it gives $n-2$ equations; the remaining equations come from the end conditions (a fixed end is handled by adding an imaginary span of zero length, that is, an infinitely stiff span, beyond it; a free overhang gives a known support moment). Solving the equations gives the support moments, and the reactions and BM follow from statics.