Moment Distribution Method

Use small-deflection, linear-elastic, prismatic Euler–Bernoulli members with constant $EI$ unless stated otherwise. Suppress joint translation in an ordinary non-sway distribution. The examples use clockwise-positive member-end moments at both ends as algebraic moments acting on the member; support moments and plotted sagging-positive bending moments must be converted consistently. At a pin, the member-end bending moment equals zero; a roller's resistance to vertical motion does not make it rotationally fixed.

For member $AB$, $M_{AB}$ is the end action at $A$ of member $AB$; $M_{BA}$ is at $B$ of the same member. A rigid joint without applied external couple satisfies $\sum M_{\mathrm{member\ ends}}=0$. With applied joint couple $M_J^{\mathrm{ext}}$ under the same sign convention, use $\sum M_{\mathrm{ends}}+M_J^{\mathrm{ext}}=0$. End supports restrained against rotation are not released in the distribution table.

$w$ is load intensity (kN/m), $P$ is force (kN), $L$ is span length (m), $E$ is elastic modulus (kN/m$^2$), $I$ is second moment of area (m$^4$) and $EI$ is flexural rigidity (kN·m$^2$). Thus $wL^2$, $PL$ and $EI/L$ times angle produce kN·m. For a fixed–pin UDL span with the right end released, the initial right-end moment is cancelled and half its correcting moment transfers to the left: $M_{AB}^{F,\mathrm{fixed-pin}}=-wL^2/8$ and $M_{BA}=0$. Do not mix the original fixed–fixed $M_{BA}^{F}$ with a modified $3EI/L$ far-pin stiffness.

For Figure 1 at $B$ with constant $EI$: $K_{BA}=4EI/5=0.8EI$ (far $A$ fixed), $K_{BC}=3EI/6=0.5EI$ (far $C$ pinned). Therefore $DF_{BA}=0.8/1.3=0.61538$ and $DF_{BC}=0.5/1.3=0.38462$. The factors sum to 1 at the released joint B. At $A$ the rotation is imposed as zero (no distribution); at $C$ the member end has zero moment after release. It is misleading to declare that every roller has a universal DF of 1: DFs belong to an assembly of rotationally participating members at a joint.

For Figure 1 use the far-pin-adjusted span $BC$ FEM: $M_{BC}^{F}=-12(6^2)/8=-54.00$ kN·m, $M_{CB}=0$; for $AB$, $M_{BA}^{F}=+10(5^2)/12=+20.8333$ kN·m and $M_{AB}^{F}=-20.8333$ kN·m. Joint $B$ has $U=20.8333-54=-33.1667$ kN·m. The balancing correction is $+33.1667$ kN·m.

Check: $M_{BA}+M_{BC}=41.2436-41.2436=0$ (joint equilibrium); $M_{CB}=0$ (released end); the only carry-over is $0.5(20.4103)=10.2051$ to fixed $A$.

A useful classroom stopping criterion is $\max_J|U_J|\leq \varepsilon$ (for example, an explicitly chosen 0.10 kN·m) or a relative criterion against a relevant moment scale. Because carry-over perturbs previously balanced joints, stopping immediately after the last joint is usually premature. In a typical positive-stiffness linear elastic system, repeated equilibrium corrections converge; convergence can be slow for unfavorable relative stiffnesses, and an unstable structure or invalid stiffness model is not repaired by extra iterations.

When the frame can move horizontally, the member slope-deflection equation also contains chord rotation $\psi=\Delta/L$. For a prismatic column with both ends rotationally fixed and a prescribed relative lateral translation $\Delta$, the sway-only end-moment magnitude is $6EI\Delta/L^2$ and total column lateral shear magnitude $12EI\Delta/L^3$. These magnitudes and restraints do not apply unchanged to a frame with free joint rotations.

A rigorous traditional approach superposes (a) actual loads with a fictitious lateral restraint and (b) an unloaded trial-sway case, each balanced rotationally by distribution. If the signed fictitious holding force in (a) is $R$ and the signed lateral force response to the trial translation in (b) is $R'$, combine using a scale that cancels the fictitious holding force. Define the sign of both quantities explicitly: for example, with (b)'s restoring force reported opposite to (a)'s holding force, the correction multiplier is $R/R'$; otherwise it is $-R/R'$. Always confirm final horizontal equilibrium after superposition and reject $R'=0$.

• Hardy Cross, “Analysis of Continuous Frames By Distributing Fixed-End Moments,” ASCE Transactions 96 (1932), DOI 10.1061/TACEAT.0004333.
• R. C. Hibbeler, Structural Analysis, 11th ed., Pearson, 2024, displacement-method/moment-distribution chapters.
• University of Illinois Civil and Environmental Engineering, Hardy Cross historical background.