--- title: "Steps of displacement method for a frame" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:steps-displacement-method-frame" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/steps-displacement-method-frame site: BCT Engineering --- # Steps of displacement method for a frame Theory of Structures II · Chapter 3: Displacement method - Asked in: 2068 Chaitra, 8 marks - Video: [SA49: Matrix Displacement Method: Frame Analysis (Joint Loads)](https://www.youtube.com/watch?v=tWSNg0C4SWQ) — Dr. Structure - 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/steps-displacement-method-frame - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Explain with a simple example the steps to follow in solving a frame using the displacement method. ## Short answer Steps: find kinematic indeterminacy, lock joints, compute fixed-end moments, write slope-deflection equations, solve equilibrium for displacements, back-substitute. Example: symmetrical portal with UDL gives M_BA=27, M_BC=-27, M_CB=27, M_CD=-27 kNm. ## Answer The displacement (stiffness) method treats the unknown joint displacements (rotations and translations) as the unknowns. The steps are: 1. **Find the degree of kinematic indeterminacy** (the number of independent joint rotations and joint translations). Neglect axial deformation of members, so joints of a rigid girder have a common sway. 2. **Lock all joints** (add imaginary restraints on rotation and translation). Every member becomes a fixed-ended member. 3. **Compute the fixed-end moments and forces** for the loads on each member (tables), and the equivalent joint loads $\{P\}$ = applied joint loads minus the fixed-end actions. 4. **Write the member stiffness relations** (slope-deflection equations) for every member in terms of the joint displacements: $$ M_{ij}=\frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi\right)+FEM_{ij} $$ 5. **Write the equilibrium equations**, one for each unknown: $\sum M=0$ at each rotating joint and, for sway, the shear (storey) equation. This gives $[K]\{d\}=\{P\}$. 6. **Solve for the displacements** $\{d\}$. 7. **Back-substitute** into the member equations to get the end moments, then find the shears, the axial forces and the support reactions from the equilibrium of the members, and draw the BMD, SFD and AFD. 8. **Check** the joint equilibrium and the overall equilibrium. #### Example Portal frame A-B-C-D, columns 4 m, beam 6 m, EI constant, fixed bases, UDL 12 kN/m on the beam. *Step 1.* Unknowns: $\theta_B$, $\theta_C$. The loading and the frame are symmetrical, so no sway occurs. *Step 3.* $FEM_{BC}=-\dfrac{12\times6^2}{12}=-36.00$ kNm, $FEM_{CB}=+36$ kNm. *Step 4.* With $\theta_A=\theta_D=0$ and $\psi=0$: - $M_{AB}=\frac{2EI}{4}\left(\theta_{B}\right)$ - $M_{BA}=\frac{2EI}{4}\left(2\theta_{B}\right)$ - $M_{BC}=\frac{2EI}{6}\left(2\theta_{B} + \theta_{C}\right) - 36.00$ - $M_{CB}=\frac{2EI}{6}\left(2\theta_{C} + \theta_{B}\right) + 36.00$ - $M_{CD}=\frac{2EI}{4}\left(2\theta_{C}\right)$ - $M_{DC}=\frac{2EI}{4}\left(\theta_{C}\right)$ *Step 5.* $M_{BA}+M_{BC}=0$ and $M_{CB}+M_{CD}=0$ give $$ EI\begin{bmatrix}1.6667 & 0.3333 \\ 0.3333 & 1.6667\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}36.000 \\ -36.000\end{Bmatrix} $$ *Step 6.* $$ EI\theta_{B}=27.0000,\quad EI\theta_{C}=-27.0000 $$ *Step 7.* End moments: $M_{AB}=13.50$, $M_{BA}=27.00$, $M_{BC}=-27.00$, $M_{CB}=27.00$, $M_{CD}=-27.00$, $M_{DC}=-13.50$ kNm. The mid-span moment of the beam is $\dfrac{wL^2}{8}-27.00=27.00$ kNm. *Step 8.* Joint B: $27.00+(-27.00)=0$ ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is done in the step of locking all the joints? A. Imaginary restraints are added so every member becomes fixed-ended and fixed-end actions can be found B. The supports are removed so the frame can move freely C. Hinges are inserted at every joint D. The loads are removed from the frame Answer: A. Locking joints makes each member fixed-ended so tabulated fixed-end moments and forces apply. (Key point: Locking joints) 2. How is the load vector {P} obtained? A. Applied joint loads plus fixed-end actions B. Applied joint loads minus fixed-end actions C. Fixed-end actions only D. Support reactions only Answer: B. The equivalent joint loads are the applied joint loads minus the fixed-end actions. (Key point: Equivalent joint loads) 3. Which equilibrium equations give [K]{d} = {P} for a frame with sway? A. Sum of moments zero at each rotating joint, plus the shear (storey) equation for sway B. Sum of vertical forces zero at each joint only C. Compatibility of rotations only D. Sum of moments zero at the supports only Answer: A. One equation per unknown: moment equilibrium for each joint rotation and a shear equation for each sway. (Key point: Equilibrium equations) 4. In the symmetrical portal example (UDL 12 kN/m, beam 6 m), why are the unknowns only theta_B and theta_C? A. Because the columns are rigid B. Because the loading and frame are symmetrical so no sway occurs C. Because the bases are hinged D. Because the beam has no stiffness Answer: B. Symmetry means no sway, leaving two joint rotations as unknowns. (Key point: Example result) ## Related in Theory of Structures II - [Stiffness method: frame, flexural deformations only](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-10kn-40knm-flexural) — Theory of Structures II, chapter 3 - [Flexibility, stiffness coefficient; suitable method](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aflex-stiff-coefficient-frame-method) — Theory of Structures II, chapter 1 ## Related in other subjects - [BMD of a simple beam and a frame](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abmd-simple-beam-and-frame-10knm) — Strength of Materials, chapter 1 - [Strain energy method for frame displacement](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Astrain-energy-frame-500kn) — Theory of Structures I, chapter 2 - [Frame: horizontal displacement and rotation at roller](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-horizontal-rotation-roller-2079) — Theory of Structures I, chapter 3 - [Frame: roller displacement and hinge rotation](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-roller-hinge-rotation-2071b) — Theory of Structures I, chapter 3