--- title: "Moment distribution principle; frame stiffness matrix" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:principle-moment-dist-stiffness-matrix-u1v1" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/principle-moment-dist-stiffness-matrix-u1v1 site: BCT Engineering --- # Moment distribution principle; frame stiffness matrix Theory of Structures II · Chapter 3: Displacement method - Asked in: 2070 Chaitra, 5+10 marks - Video: [Moment Distribution (MDM) | Part 1 | Concept n Basics](https://www.youtube.com/watch?v=yqO92DRY2Xg) — TheCivilGuy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/principle-moment-dist-stiffness-matrix-u1v1 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Describe with an example the principle of moment distribution. For the frame shown in figure below generate the stiffness matrix that operates on displacements $u_1$, $v_1$, $\theta_1$ and $\theta_2$. Both members are slender and have the same E, I, A and L. Express matrix coefficients in terms of L, $a = AE/L$ and $b = EI/L^3$. [Figure: two members meeting at joint 2: horizontal member from fixed end 1 and vertical member down to fixed/hinged end 3; joint 2 has degrees of freedom $U_1, F_{x1}$ (horizontal), $V_1, F_{y1}$ (vertical), $\theta_1, M_1$ (rotation at 2) and $\theta_2, M_2$ (rotation at 3).] ## Short answer Moment distribution: lock joints, distribute unbalanced moments by DF = k/Σk (k = 4EI/L or 3EI/L), carry over half to fixed far ends. The stiffness matrix for the frame is [K] = [[a+12b, 0, 6bL, 6bL], [0, a+12b, -6bL, 0], [6bL, -6bL, 8bL², 2bL²], [6bL, 0, 2bL², 4bL²]]. ## Answer #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the correct sequence in the moment distribution method? A. Lock joints and find FEMs, release a joint and distribute the balancing moment by DF, carry over half, repeat B. Release all joints, apply loads, solve simultaneous equations C. Find reactions, then draw the SFD, then distribute D. Distribute first, then compute FEMs Answer: A. Joints are locked to get FEMs, then released one by one: balancing moment distributed by DF, half carried over, repeated until negligible. (Key point: Steps of moment distribution) 2. What is the balancing moment at a released joint? A. Equal to the unbalanced moment, same sign B. Equal to the unbalanced moment with reversed sign, distributed in proportion to DF C. Half the unbalanced moment D. The sum of all stiffnesses Answer: B. The joint rotates until in equilibrium, which means applying a moment equal and opposite to the unbalanced moment, shared according to DF. (Key point: Balancing moment) 3. Why is the coefficient K33 = 8bL^2 in the assembled matrix? A. Each of the two members contributes 4EI/L = 4bL^2 to the rotation of the joint B. The vertical member contributes 8EI/L alone C. It is 2bL^2 added to 6bL^2 D. It is a+12b doubled Answer: A. Both members meet at the joint, each giving 4EI/L (= 4bL^2) for a unit rotation, so the total is 8bL^2. (Key point: Diagonal term K33) 4. What is K11 (force at u1 for unit u1) made of? A. Axial stiffness of the horizontal member only (a) B. Axial stiffness a of member 1 plus transverse stiffness 12b of member 2, giving a + 12b C. 12b from both members D. 4bL^2 plus 6bL Answer: B. A horizontal displacement stretches the horizontal member (a) and bends the vertical member (12b), so K11 = a + 12b. 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