Three moment equation: derive and apply to fixed beam

Derivation

Consider two adjacent spans AB ($L_1$) and BC ($L_2$) of a continuous beam with supports at the same level and constant $EI$. Let $M_A, M_B, M_C$ be the support moments (sagging positive). On each span the BM diagram is the free (simple beam) diagram plus a trapezoid of support moments. Let $a_1, a_2$ be the areas of the free BM diagrams, $\bar x_1$ the distance of the centroid of $a_1$ from A, and $\bar x_2$ the distance of the centroid of $a_2$ from C.

Slope at B from span AB (moment of the $M/EI$ area about A equals the deviation of A from the tangent at B, which is $\theta_BL_1$ for level supports):

$$\theta_{B1} = \frac{1}{EI}\left[\frac{a_1\bar x_1}{L_1} + \frac{M_AL_1}{6} + \frac{M_BL_1}{3}\right]$$

Slope at B from span BC (taking moments about C):

$$\theta_{B2} = -\frac{1}{EI}\left[\frac{a_2\bar x_2}{L_2} + \frac{M_CL_2}{6} + \frac{M_BL_2}{3}\right]$$

The beam is continuous over B, so $\theta_{B1} = \theta_{B2}$. Multiplying by $6$ gives the three moment equation:

$$M_AL_1 + 2M_B(L_1+L_2) + M_CL_2 = -\frac{6a_1\bar x_1}{L_1} - \frac{6a_2\bar x_2}{L_2}$$

Application: fixed beam AB with a UDL $w$ over the span $L$

A fixed end is treated as a continuous span of zero length beyond the end. Add an imaginary span $A'A$ of length $L_0 = 0$ at A and $BB'$ of length $L_0 = 0$ at B; the end moments $M_A$ and $M_B$ are the unknowns.

For the span AB: $a = \dfrac23\cdot L\cdot\dfrac{wL^2}{8} = \dfrac{wL^3}{12}$, $\bar x = \dfrac L2$, so $\dfrac{6a\bar x}{L} = \dfrac{wL^3}{4}$.

Equation at A (spans $A'A$ and $AB$, the first has zero length):

$$2M_AL + M_BL = -\frac{wL^3}{4}$$

Equation at B (spans $AB$ and $BB'$):

$$M_AL + 2M_BL = -\frac{wL^3}{4}$$

Subtracting gives $M_A = M_B$. Then $3M_AL = -\dfrac{wL^3}{4}$, so

$$M_A = M_B = -\frac{wL^2}{12}$$

Results

• Reactions: $R_A = R_B = \dfrac{wL}{2}$ (symmetry).
• Mid-span moment: $\dfrac{wL^2}{8} - \dfrac{wL^2}{12} = +\dfrac{wL^2}{24}$ (sagging).
• Points of contraflexure at $x = \dfrac{L}{2}\left(1 \pm \dfrac{1}{\sqrt3}\right)$, that is $0.211L$ from each end.

For example, with $w = 20$ kN/m and $L = 6$ m: $M_A = M_B = -60$ kNm, $R_A = R_B = 60$ kN, mid-span $+30$ kNm.