--- title: "Stiffness matrix: frame with coordinates 1, 2, 3" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:stiffness-frame-coords-123-15knm-10kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-coords-123-15knm-10kn site: BCT Engineering --- # Stiffness matrix: frame with coordinates 1, 2, 3 Theory of Structures II · Chapter 3: Displacement method - Asked in: 2069 Asar, 10 marks - Video: [How To Make a Stiffness Matrix || Most Imp Topic Of Structural Analysis](https://www.youtube.com/watch?v=b63PB97uXJY) — THE CIVIL BANDA - 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-coords-123-15knm-10kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Generate the stiffness matrix for the frame shown in figure below with respect to coordinates 1, 2 and 3 and use it to analyse the frame if the forces 5 kNm and 4 kN are acting at coordinates 1 and 3 respectively in addition to the external loads as shown in figure. Take EI is constant for all members. [Figure: portal frame; left column 6 m (3 m + 3 m) with 10 kN horizontal at mid-height, fixed base; beam of 3 m with 15 kN/m UDL; right column 4 m, fixed base; coordinates 1 (rotation), 2 (horizontal), 3 (rotation) marked at the top joints.] ## Short answer Coordinates are theta_B, sway, theta_C. K=EI[[2,-0.1667,0.6667],[-0.1667,0.2431,-0.375],[0.6667,-0.375,2.3333]]. Solving with equivalent loads gives M_AB=-9.12, M_BA=8.13, M_BC=-3.13, M_CB=14.00, M_CD=-10.00, M_DC=-9.35 kNm. ## Answer **Assumptions (figure partly illegible):** A(0,0), B(0,6), C(3,6), D(3,2). The left column AB (6 m, 3 m + 3 m) carries the 10 kN horizontal load (to the right) at mid-height; the beam BC is 3 m long with 15 kN/m downward; the right column CD is 4 m. A and D are fixed. EI is the same for all members. **Coordinates: 1 = rotation of joint B (clockwise), 2 = horizontal translation (sway) of the beam (to the right), 3 = rotation of joint C (clockwise).** The additional forces at coordinates 1 and 3 are taken as a 5 kNm and a 4 kNm clockwise moment (a 'kN' at coordinate 3 cannot act on a rotation, so the 4 is read as kNm). #### Stiffness matrix $[K]$ (units EI) Terms: $\psi_{AB}=\Delta/6$, $\psi_{CD}=\Delta/4$. - $k_{11}=\dfrac{4EI}{6}+\dfrac{4EI}{3}=2.0000EI$; $k_{13}=k_{31}=\dfrac{2EI}{3}=0.6667EI$; $k_{33}=\dfrac{4EI}{3}+\dfrac{4EI}{4}=2.3333EI$ - $k_{12}=-\dfrac{6EI}{6^2}=-0.1667EI$; $k_{32}=-\dfrac{6EI}{4^2}=-0.3750EI$; $k_{22}=\dfrac{12EI}{6^3}+\dfrac{12EI}{4^3}=0.2431EI$ $$ [K]=EI\begin{bmatrix}2.0000 & -0.1667 & 0.6667 \\ -0.1667 & 0.2431 & -0.3750 \\ 0.6667 & -0.3750 & 2.3333\end{bmatrix} $$ #### Equivalent loads from the member loads (fixed-end effects) Fixed-end moments: BC $\mp\dfrac{15\times3^2}{12}=\mp11.25$ kNm; AB (10 kN at 3 m, horizontal) $\mp\dfrac{PL}{8}=\mp7.5$ kNm (sign for a horizontal load to the right on a column); the equivalent lateral force at the girder level is the fixed-end shear at B, $10/2=5$ kN. $$ \{F_0\}=\begin{Bmatrix}3.750\\5.000\\-11.250\end{Bmatrix}\ (\text{kNm, kN, kNm}),\qquad \{F\}=\{F_0\}+\begin{Bmatrix}5\\0\\4\end{Bmatrix}=\begin{Bmatrix}8.750\\5.000\\-7.250\end{Bmatrix} $$ #### Analysis $[K]\{d\}=\{F\}$ gives $$ EI\theta_B=6.7426,\quad EI\Delta=23.1753,\quad EI\theta_C=-1.3090 $$ #### End moments (member equations, kNm, clockwise +) - $M_{AB}=\frac{2EI}{6}\left(\theta_{B} - (0.5000)\Delta_1\right) - 7.50$ - $M_{BA}=\frac{2EI}{6}\left(2\theta_{B} - (0.5000)\Delta_1\right) + 7.50$ - $M_{BC}=\frac{2EI}{3}\left(2\theta_{B} + \theta_{C}\right) - 11.25$ - $M_{CB}=\frac{2EI}{3}\left(2\theta_{C} + \theta_{B}\right) + 11.25$ - $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$ - $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$ | End | $M$ (kNm) | End | $M$ (kNm) | |---|---|---|---| | AB | -9.12 | BA | 8.13 | | BC | -3.13 | CB | 14.00 | | CD | -10.00 | DC | -9.35 | Checks: - Joint B: (8.13) + (-3.13) = 5.00 kNm (applied clockwise moment 5.00) ✓ - Joint C: (14.00) + (-10.00) = 4.00 kNm (applied clockwise moment 4.00) ✓ #### Bending moment ordinates (kNm) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | -9.12 | -8.13 | 6.38 at 3.00 m | | BC | -3.13 | -14.00 | 8.75 at 1.26 m | | CD | -10.00 | 9.35 | - | Reactions: A: $H=-5.16$, $V=18.88$, $M=9.12$; D: $H=-4.84$, $V=26.12$, $M=9.35$. **Answer: $[K]$ as above; with the loads given, $M_{AB}=-9.12$, $M_{BA}=8.13$, $M_{BC}=-3.13$, $M_{CB}=14.00$, $M_{CD}=-10.00$, $M_{DC}=-9.35$ kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What do coordinates 1, 2 and 3 represent in the analysis? A. Rotation of B, sway of the beam, rotation of C B. Three vertical deflections C. Rotations of A, B and C D. Horizontal reactions at A, D and C Answer: A. Coordinate 1 is rotation of joint B, 2 is horizontal translation of the beam, 3 is rotation of joint C. (Key point: Meaning of coordinates) 2. How is k22 = 0.2431EI obtained? A. 12EI/6^3 + 12EI/4^3 from the two columns B. 6EI/6^2 + 6EI/4^2 C. 4EI/6 + 4EI/4 D. 12EI/3^3 Answer: A. A unit sway bends both columns: 12EI/216 + 12EI/64 = 0.0556EI + 0.1875EI = 0.2431EI. (Key point: Diagonal term k22) 3. How is k11 = 2.0EI obtained? A. 4EI/6 from column AB plus 4EI/3 from beam BC B. 4EI/6 plus 4EI/4 C. 2EI/6 plus 2EI/3 D. 4EI/3 plus 4EI/4 Answer: A. Unit rotation of B stiffens column AB (4EI/6 = 0.667EI) and beam BC (4EI/3 = 1.333EI), total 2.0EI. (Key point: Diagonal term k11) 4. How is the load vector for the analysis built? A. Equivalent loads from member loads (fixed-end effects) plus the additional forces at the coordinates B. Only the additional forces C. Only the member loads D. Reactions at the supports Answer: A. {F} is the sum of the equivalent joint forces from loads on the members and the additional forces applied directly at the coordinates. 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