--- title: "Generate stiffness matrix for portal frame" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:stiffness-matrix-frame-3i-i-5m" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-matrix-frame-3i-i-5m site: BCT Engineering --- # Generate stiffness matrix for portal frame Theory of Structures II · Chapter 3: Displacement method - Asked in: 2070 Asar, 5 marks - Video: [How to Compile Stiffness Matrix](https://www.youtube.com/watch?v=8hddX-cOKfw) — Engineering Edu - 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-matrix-frame-3i-i-5m - 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. [Figure: portal frame, beam 3I of 5 m, columns I of 4 m, fixed bases.] ## Short answer Coordinates are rotations at B and C plus sway. Ignoring axial deformation, the stiffness matrix is EI[[3.4,1.2,-0.375],[1.2,3.4,-0.375],[-0.375,-0.375,0.375]], symmetric with k11=k22 and k13=k23. ## Answer **Assumptions:** portal frame A-B-C-D with fixed bases A and D, columns AB and CD of height 4 m and rigidity $EI$, beam BC of span 5 m and rigidity $3EI$. Axial deformations are ignored. Coordinates (positive directions): **1 = rotation of joint B, 2 = rotation of joint C (both clockwise), 3 = horizontal sway of the beam (to the right)**. #### Member stiffness coefficients | Member | EI | L (m) | $4EI/L$ | $2EI/L$ | $6EI/L^2$ | $12EI/L^3$ | |---|---|---|---|---|---|---| | AB, CD | $EI$ | 4 | 1.0000EI | 0.5000EI | 0.3750EI | 0.1875EI | | BC | $3EI$ | 5 | 2.4000EI | 1.2000EI | - | - | #### Column-by-column generation ($k_{ij}$ = force at $i$ for a unit displacement at $j$, others zero) - **Unit rotation at 1** (B turns clockwise by 1): $k_{11}=\dfrac{4EI}{4}+\dfrac{4(3EI)}{5}=3.4000EI$; $k_{21}=\dfrac{2(3EI)}{5}=1.2000EI$; $k_{31}$: the column AB gives a shear $\dfrac{6EI}{4^2}$ which must be resisted by the sway restraint, $k_{31}=-0.3750EI$ (negative: on its own, a clockwise rotation of B would make the top of column AB move to the right, so the restraint must push the girder to the left). - **Unit rotation at 2:** $k_{22}=\dfrac{4(3EI)}{5}+\dfrac{4EI}{4}=3.4000EI$; $k_{12}=1.2000EI$; $k_{32}=-0.3750EI$. - **Unit sway at 3:** $k_{33}=2\times\dfrac{12EI}{4^3}=0.3750EI$; $k_{13}=k_{23}=-\dfrac{6EI}{4^2}=-0.3750EI$. #### Stiffness matrix $$ [K]=EI\begin{bmatrix}3.4000 & 1.2000 & -0.3750 \\ 1.2000 & 3.4000 & -0.3750 \\ -0.3750 & -0.3750 & 0.3750\end{bmatrix} $$ The matrix is symmetric and, because the frame is symmetrical, $k_{11}=k_{22}$ and $k_{13}=k_{23}$. The equilibrium equation for any loading is $\{P\}=[K]\{d\}$ with $\{d\}=\{\theta_B,\theta_C,\Delta\}^T$ and $P$ the equivalent joint loads (applied loads minus fixed-end effects). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What coordinates are used for the portal frame stiffness matrix? A. Rotation of B, rotation of C, and horizontal sway of the beam B. Vertical deflections of B and C and rotation of A C. Rotations of A and D and sway D. Axial deformations of the columns Answer: A. With fixed bases and axial deformation ignored, the free displacements are the two joint rotations and the sway. (Key point: Coordinates chosen) 2. How is k11 = 3.4EI obtained? A. 4EI/4 from column AB plus 4(3EI)/5 from beam BC B. 12EI/4^3 plus 12(3EI)/5^3 C. 2EI/4 plus 2(3EI)/5 D. 4EI/4 only Answer: A. For unit rotation of B, the column gives 4EI/L = 1.0EI and the beam gives 4(3EI)/5 = 2.4EI, total 3.4EI. (Key point: Diagonal term k11) 3. Why is k33 = 0.375EI? A. Two columns each give 12EI/L^3 = 0.1875EI for unit sway B. One column gives 6EI/L^2 C. The beam gives 12(3EI)/5^3 D. It is 4EI/4 plus 2EI/4 Answer: A. A unit sway of the beam bends both columns, each with stiffness 12EI/4^3 = 0.1875EI, so k33 = 0.375EI. (Key point: Sway stiffness k33) 4. Which property do symmetry of the frame and the Maxwell-Betti law give for [K]? A. The matrix is symmetric, and k11 = k22 and k13 = k23 B. The matrix is diagonal C. All off-diagonal terms are positive D. k13 differs from k31 Answer: A. [K] is symmetric (k_ij = k_ji) and the symmetric frame makes the two joint rotations behave identically. 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