Slope Deflection and Moment Distribution

Displacement methods take joint rotations and translations as the unknowns. The slope-deflection method writes them as equations; moment distribution (Hardy Cross) solves the same problem by iteration, without solving equations.

Slope-deflection equations

For member AB with length L, stiffness EI, end rotations θ_A, θ_B and relative sway Δ between its ends:

M_AB = M_FAB + (2EI/L)(2θ_A + θ_B − 3Δ/L)

M_BA = M_FBA + (2EI/L)(2θ_B + θ_A − 3Δ/L)

M_F are the fixed-end moments (clockwise positive). Write joint equilibrium (sum of member-end moments at each joint = applied moment) and, for sway frames, a shear equation, then solve for θ and Δ.

Stiffness and carry-over

A moment applied at the near end produces, at a fixed far end, half that moment with the same sign (carry-over).

Moment distribution

1. Lock all joints and compute fixed-end moments.
2. Distribution factor of each member at a joint: DF = k/Σk (relative stiffness). DFs at a joint add to 1; DF = 0 at a fixed support and 1 at a pinned end.
3. Release each joint: distribute the unbalanced moment, with opposite sign, in proportion to DFs.
4. Carry over half of each distributed moment to the far (fixed) ends.
5. Repeat until the moments are small; sum each column.

Sway frames (unsymmetric loads or geometry): do a non-sway analysis, then an analysis with an arbitrary sway, and combine them so that the horizontal shear equation is satisfied.

Worked example

A two-span beam ABC: A fixed, B continuous, C fixed; AB = 6 m, BC = 4 m, same EI. Find the distribution factors at B.

• k_BA = 4EI/6 = 0.667EI, k_BC = 4EI/4 = EI.
• DF_BA = 0.667/1.667 = 0.4; DF_BC = 0.6.

Practice