--- title: "Stiffness matrix: frame, BMD" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:stiffness-frame-2knm-10kn-20knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-2knm-10kn-20knm site: BCT Engineering --- # Stiffness matrix: frame, BMD Theory of Structures II · Chapter 3: Displacement method - Asked in: 2071 Shrawan, 15 marks - Video: [Analysis of Frames by Stiffness Matrix Method - Problem No 9 ( Analysis of Sway Frame )](https://www.youtube.com/watch?v=evNFPj3Lb-c) — Stan Academy - 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/stiffness-frame-2knm-10kn-20knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Generate the stiffness matrix for the frame given below. Use the stiffness matrix generated to draw bending moment diagram. Take EI as constant for all members. [Figure: frame; beam BC of 5 m with 2 kN/m UDL; 10 kN horizontal at B; 20 kNm moment at C; column AB (A fixed); column CD of 6 m with D fixed; 10 kN horizontal at E, 2 m above D.] ## Short answer The stiffness matrix method with unknowns θB, θC, and sway Δ is used. Solving the assembled equations yields the end moments: M_AB = -23.27 kNm, M_BA = -17.89 kNm, M_BC = 17.89 kNm, M_CB = 28.94 kNm, M_CD = -8.94 kNm, M_DC = -29.90 kNm. ## Answer **Assumptions (figure partly illegible):** A(0,0), B(0,6), C(5,6), D(5,0) with both columns 6 m high and EI constant for all members. BC = 5 m carries 2 kN/m downward; 10 kN acts horizontally (to the right) at B; the 20 kNm moment at C is clockwise; a second 10 kN horizontal force acts to the right on CD at E, 2 m above D. A and D are fixed. #### Coordinates 1 = $\theta_B$, 2 = $\theta_C$ (clockwise +), 3 = $\Delta$ (sway to the right). $\psi_{AB}=\psi_{CD}=\Delta/6$. #### Fixed-end moments (clockwise +) - BC: $\mp\dfrac{2\times5^2}{12}=\mp4.167$ kNm - CD (10 kN at $a=4$ m from C, $b=2$ m from D, horizontal): $FEM_{CD}=4.44$, $FEM_{DC}=-8.89$ kNm ($Pab^2/L^2=10\times4\times4/36$, $Pa^2b/L^2=10\times16\times2/36$, sign for a load to the right on a column running downward) #### Stiffness matrix of the frame Member stiffnesses: $4EI/6=0.6667EI$, $2EI/6=0.3333EI$, $6EI/6^2=0.1667EI$, $12EI/6^3=0.0556EI$; beam $4EI/5=0.8EI$, $2EI/5=0.4EI$. - $K_{11}=0.6667+0.8=1.4667EI$, $K_{12}=0.4EI$, $K_{22}=0.8+0.6667=1.4667EI$ - $K_{13}=K_{23}=-0.1667EI$, $K_{33}=2\times0.0556EI=0.1111EI$ $$ [K]=EI\begin{bmatrix}1.4667 & 0.4000 & -0.1667 \\ 0.4000 & 1.4667 & -0.1667 \\ -0.1667 & -0.1667 & 0.1111\end{bmatrix} $$ #### Load vector The loads at the coordinates are the applied moment, the applied lateral force and the equivalent joint loads from the member loads: $$ \{F\}=\{4.167,\ 11.389,\ 12.593\}^T $$ (row 1: $-FEM_{BA}-FEM_{BC}$; row 2: $-FEM_{CB}-FEM_{CD}$ plus the applied 20 kNm; row 3: the 10 kN at B plus the equivalent lateral force from the 10 kN on CD). $$ EI\begin{bmatrix}1.4667 & 0.4000 & -0.1667 \\ 0.4000 & 1.4667 & -0.1667 \\ -0.1667 & -0.1667 & 0.1111\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}4.167 \\ 11.389 \\ 12.593\end{Bmatrix} $$ $$ EI\theta_{B}=16.1268,\quad EI\theta_{C}=22.8976,\quad EI\Delta_{1}=171.8699 $$ #### Final end moments (kNm, clockwise +) | End | $M$ (kNm) | End | $M$ (kNm) | |---|---|---|---| | AB | -23.27 | BA | -17.89 | | BC | 17.89 | CB | 28.94 | | CD | -8.94 | DC | -29.90 | Checks: - Joint B: (-17.89) + (17.89) = 0.00 kNm (applied clockwise moment 0.00) ✓ - Joint C: (28.94) + (-8.94) = 20.00 kNm (applied clockwise moment 20.00) ✓ #### Bending moment diagram ordinates (sagging/inside +, kNm) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | -23.27 | 17.89 | - | | BC | 17.89 | -28.94 | - | | CD | -8.94 | 29.90 | - | Reactions: A: $H=-6.86$, $V=-4.37$, $M=23.27$; D: $H=-13.14$, $V=14.37$, $M=29.90$. **Answer: $M_{AB}=-23.27$, $M_{BA}=-17.89$, $M_{BC}=17.89$, $M_{CB}=28.94$, $M_{CD}=-8.94$, $M_{DC}=-29.90$ kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. K33 = 0.1111 EI is? A. 2EI/5 B. 12EI/6^3 from each of the two columns C. 6EI/36 D. 4EI/6 Answer: B. Sway stiffness is 12EI/L^3 per column: 2 x 0.0556. (Key point: K33) 2. K13 = -0.1667 EI is? A. -12EI/216 B. -2EI/5 C. -6EI/6^2 D. -4EI/6 Answer: C. Coupling between rotation and sway is 6EI/L^2 from the column. (Key point: K13) 3. Beam BC (5 m) contributes to K11? A. 2EI/5 B. 6EI/25 C. 12EI/125 D. 4EI/5 = 0.8EI Answer: D. Rotation stiffness 4EI/L. (Key point: Beam stiffness) 4. Row 3 of the load vector includes? A. 10 kN at B plus equivalent lateral force from 10 kN on CD B. 20 kNm C. Only 10 kN D. UDL Answer: A. Lateral joint loads combine applied force at B and the equivalent force of the member load. (Key point: Load vector) ## Related in Theory of Structures II - [Stiffness matrix: frame with 100 kN, 15 kNm](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-100kn-20knm-15knm) — Theory of Structures II, chapter 3 - [Stiffness matrix: frame with 50 kNm moment](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-40kn-10knm-50knm) — Theory of Structures II, chapter 3 - [Stiffness matrix: frame with 20 kN/m, 10 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-20knm-10kn-2i) — Theory of Structures II, chapter 3 - [Stiffness matrix: BMD of frame, 60 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-60kn-a-fixed-d) — Theory of Structures II, chapter 3 - [Stiffness matrix: BMD of frame](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-60kn-10knm-2ei) — Theory of Structures II, chapter 3 - [Stiffness matrix: frame with coordinates 1, 2, 3](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-coords-123-15knm-10kn) — Theory of Structures II, chapter 3 ## Related in other subjects - [Frame AFD, SFD, BMD (20 kNm, 15 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-20knm-15kn) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-80kn-30knm) — Strength of Materials, chapter 1 - [AFD, SFD, BMD of frame with 60 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-60knm-20knm-12kn) — Applied Mechanics, chapter 6 - [Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30knm-50knm) — Strength of Materials, chapter 1