--- title: "Bar stiffness matrix and 2D truss element" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:bar-stiffness-truss-rotation" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/bar-stiffness-truss-rotation site: BCT Engineering --- # Bar stiffness matrix and 2D truss element Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2070 Magh, 10 marks - Video: [TRUSS element (BAR element): Introduction and stiffness matrix 2](https://www.youtube.com/watch?v=CTzRo1gN9DA) — Ram Kinhikar - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/bar-stiffness-truss-rotation - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Formulate the stiffness matrix for a bar element. Rotate the same bar element and formulate the stiffness matrix for a 2D truss element. ## Short answer Local bar stiffness is AE/L times [1 -1; -1 1]. For a 2D truss element rotated by theta, the global stiffness matrix is AE/L times [c^2 cs -c^2 -cs; cs s^2 -cs -s^2; -c^2 -cs c^2 cs; -cs -s^2 cs s^2], with c=cos theta and s=sin theta. ## Answer A bar (rod) element carries only axial force. It has two nodes and one dof per node in its local axis; a 2D truss element is the same bar placed at an angle, with two dofs per node. #### Part 1: Stiffness matrix of the bar element ```text 1 o--------------------o 2 ---> x u1 (A, E, L) u2 ``` Assume a linear displacement field $u(x)=\alpha_1+\alpha_2x$. Applying $u(0)=u_1$ and $u(L)=u_2$: $$ u(x)=N_1u_1+N_2u_2,\qquad N_1=1-\frac{x}{L},\quad N_2=\frac{x}{L} $$ Strain: $$ \varepsilon=\frac{du}{dx}=\begin{bmatrix}-\frac1L&\frac1L\end{bmatrix}\begin{bmatrix}u_1\\u_2\end{bmatrix}=[B]\{u\} $$ Stress $\sigma=E\varepsilon$. The stiffness matrix from the strain energy (or the principle of minimum potential energy): $$ [k]=\int_0^L[B]^TE[B]A\,dx=AE\int_0^L\begin{bmatrix}\frac1{L^2}&-\frac1{L^2}\\-\frac1{L^2}&\frac1{L^2}\end{bmatrix}dx $$ $$ [k]=\frac{AE}{L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix} $$ Direct check: a unit displacement $u_1=1,\ u_2=0$ needs force $AE/L$ at node 1 and $-AE/L$ at node 2 (equilibrium). The matrix is symmetric and singular (rigid-body motion is possible until supports are added). #### Part 2: Rotating the bar - 2D truss element ```text v2 ^ u2 (global) |/ y ^ 2 o | / | / L theta = angle of the bar | / from global x | o 1 |__theta_______> x ``` Let the bar make angle $\theta$ with the global $x$-axis, $c=\cos\theta=\dfrac{x_2-x_1}{L}$, $s=\sin\theta=\dfrac{y_2-y_1}{L}$. Global dofs: $\{d\}=[u_1\ v_1\ u_2\ v_2]^T$; local axial dofs: $\{d'\}=[u_1'\ u_2']^T$. The local axial displacement at a node is the projection of the global displacement on the bar axis: $$ \begin{bmatrix}u_1'\\u_2'\end{bmatrix}=\underbrace{\begin{bmatrix}c&s&0&0\\0&0&c&s\end{bmatrix}}_{[T]}\begin{bmatrix}u_1\\v_1\\u_2\\v_2\end{bmatrix} $$ The strain energy $\tfrac12\{d'\}^T[k']\{d'\}=\tfrac12\{d\}^T[T]^T[k'][T]\{d\}$ must equal $\tfrac12\{d\}^T[k]\{d\}$, so $$ [k]=[T]^T[k'][T] $$ $$ [k]=\frac{AE}{L}\begin{bmatrix}c^2&cs&-c^2&-cs\\cs&s^2&-cs&-s^2\\-c^2&-cs&c^2&cs\\-cs&-s^2&cs&s^2\end{bmatrix} $$ #### Properties - Symmetric; every row sums to zero (equilibrium of a free element). - Singular: three rigid-body modes in the plane (two translations and one rotation). - $\theta=0$: only $u$ terms remain (horizontal bar). $\theta=90^\circ$: only $v$ terms remain (vertical bar). - After finding the nodal displacements, the bar force is $F=\dfrac{AE}{L}\,(u_2'-u_1')=\dfrac{AE}{L}\begin{bmatrix}-c&-s&c&s\end{bmatrix}\{d\}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 1D bar with u(x) = a1 + a2 x, what are the shape functions? A. N1 = x/L, N2 = 1 - x/L B. N1 = 1 - x/L, N2 = x/L C. N1 = 1 - x^2/L^2, N2 = x^2/L^2 D. N1 = 1, N2 = x Answer: B. Applying u(0) = u1 and u(L) = u2 gives N1 = 1 - x/L and N2 = x/L. (Key point: Bar shape functions) 2. What is the bar element stiffness matrix? A. (AE/L) [1 1; 1 1] B. (AL/E) [1 -1; -1 1] C. (AE/L) [1 -1; -1 1] D. (EL/A) [1 -1; -1 1] Answer: C. [B] = [-1/L 1/L], so integrating [B]^T E [B] A over L gives (AE/L)[1 -1; -1 1]. (Key point: Bar stiffness) 3. How is the 2D truss element matrix obtained from the bar matrix? A. [k] = [T][k'] with no transpose B. [k] = [k'] [T]^T C. [k] = [T] [k'] [T]^T D. [k] = [T]^T [k'] [T] with [T] projecting global displacements onto the bar axis Answer: D. Equating strain energy in local and global dofs gives [k] = [T]^T[k'][T], where local u' = c u + s v at each node. (Key point: Transformation) 4. Which statement about the 2D truss element matrix is correct? A. It is symmetric and singular, with three rigid-body modes in the plane B. It is non-symmetric and invertible C. It is singular with only one rigid-body mode D. It is invertible because it has four dofs Answer: A. It is symmetric, rows sum to zero, and it is singular with two translations and one rotation as rigid-body modes. 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