Moment distribution principle; frame stiffness matrix

Part 1: Principle of moment distribution

The method (Hardy Cross) solves continuous beams and frames by successive approximation instead of simultaneous equations.

1. Lock all joints. Each loaded member has fixed-end moments (FEM); at each joint the sum of the FEMs is the unbalanced moment.
2. Release a joint: it rotates until in equilibrium, i.e. the balancing moment (= unbalanced moment with reversed sign) is distributed to the members in proportion to the distribution factors $DF=k/\sum k$, with $k=4EI/L$ (far end fixed) or $3EI/L$ (far end hinged).
3. Carry over half of each distributed moment to the far end if it is fixed ($COF=\tfrac12$).
4. Repeat for all joints until the carried-over moments are negligible. Final moment = FEM + distributed + carried-over moments.

Example. Beam ABC, AB = 6 m with 10 kN/m, BC = 4 m, EI constant; A fixed, C simple support. $K_{BA}=EI/6$, $K_{BC}=\tfrac34\cdot EI/4$, so $DF_{BA}=0.471$, $DF_{BC}=0.529$. $FEM_{AB}=-30$, $FEM_{BA}=+30$ kNm. Balancing moment at B = $-30$ kNm, distributed as $-30\times0.471=-14.12$ to BA and $-30\times0.529=-15.88$ to BC; carry-over to A = $-7.06$. Final: $M_{AB}=-37.06$, $M_{BA}=15.88$, $M_{BC}=-M_{BA}=-15.88$ kNm, $M_{CB}=0$.

Part 2: Stiffness matrix for $u_1,v_1,\theta_1,\theta_2$

Assumed geometry (as described): member 1 is horizontal, fixed at its left end (node 1) and joined at its right end to joint 2. Member 2 hangs vertically down from joint 2 to node 3, where the translations are restrained and the rotation $\theta_2$ is free (hinge/roller type end). Both members have the same $E,I,A,L$. Coordinates (positive directions): $u_1$ horizontal (right), $v_1$ vertical (up), $\theta_1$ rotation of joint 2 (anticlockwise), $\theta_2$ rotation of the end of member 2 (anticlockwise).

Use $a=AE/L$ and $b=EI/L^3$. For a prismatic member with six end displacements the local stiffness matrix has the terms $a$ (axial), $12b$, $6bL$, $4bL^2$, $2bL^2$ (bending).

Member 1 (horizontal), contribution at joint 2:

$$\begin{bmatrix}a&0&0\\0&12b&-6bL\\0&-6bL&4bL^2\end{bmatrix}\ \text{(rows/columns }u_1,v_1,\theta_1\text{)}$$

(with the fixed left end, the shear at the right end for a unit upward displacement is $12b$ and the end moment term is $\mp6bL$, depending on the sign convention).

Member 2 (vertical), rotated by 90°: the local axial direction is the global $v$ direction and the local transverse direction is the global $u$ direction. Hence the axial term $a$ appears in the $v_1v_1$ position and the bending terms $12b$ appear for $u_1$:

$$\begin{bmatrix}12b&0&6bL&6bL\\0&a&0&0\\6bL&0&4bL^2&2bL^2\\6bL&0&2bL^2&4bL^2\end{bmatrix}\ \text{(rows/columns }u_1,v_1,\theta_1,\theta_2\text{)}$$

Assembled structure stiffness matrix (sum of the two members):

$$[K]=\begin{bmatrix}a+12b&0&6bL&6bL\\0&a+12b&-6bL&0\\6bL&-6bL&8bL^2&2bL^2\\6bL&0&2bL^2&4bL^2\end{bmatrix}$$

with $\{F\}=\{F_{x1},F_{y1},M_1,M_2\}^T$ and $\{d\}=\{u_1,v_1,\theta_1,\theta_2\}^T$, so that $\{F\}=[K]\{d\}$.

Checks: $[K]$ is symmetric; $K_{33}=8bL^2=4bL^2+4bL^2$ (the two members each contribute $4EI/L$ to the rotation of the joint); $K_{11}=a+12b$ (axial stiffness of member 1 plus the transverse stiffness of member 2) and $K_{22}=a+12b$ (axial of member 2 plus transverse of member 1). The signs of the coupling terms $K_{13}$, $K_{23}$ change if the positive directions or the geometry are mirrored.