--- title: "Variational approach for one-dimensional heat conduction" subject: "Finite Element Method" chapter: 5 key: "semester_8/finite_element_method:variational-approach-heat-conduction" source: https://www.bctengineer.com/q/semester_8/finite_element_method/variational-approach-heat-conduction site: BCT Engineering --- # Variational approach for one-dimensional heat conduction Finite Element Method · Chapter 5: Interpolation Functions - Asked in: year not given, 8 marks - Video: [Variational Approach in FEM | Introduction to Variational Methods (Part 3)](https://www.youtube.com/watch?v=gtCTefXSsk8) — Mech Ujala - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/variational-approach-heat-conduction - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-5 ## Question Using the variational approach, derive the element equations for steady one-dimensional heat conduction in a rod of length $L$, cross-section $A$, conductivity $k$, with internal heat generation $Q$ per unit volume. Use a linear two-node element. ## Short answer The variational functional is minimized with linear shape functions, giving element stiffness (kA/L)[[1,-1],[-1,1]] and load (QAL/2){1,1}. Nodal heat flows q1, q2 are added as natural boundary conditions. Convection at an end adds hA to stiffness and hAT_infinity to load. ## Answer #### Governing equation and functional Steady 1D conduction with heat generation: $$ kA\frac{d^2T}{dx^2} + QA = 0 $$ The equivalent functional (analogous to the total potential energy) is $$ \Pi = \int_0^L \left[\frac{kA}{2}\left(\frac{dT}{dx}\right)^2 - QA\,T\right]dx $$ The solution of the differential equation makes $\Pi$ stationary (this can be shown by taking $\delta\Pi = 0$ and integrating by parts, which gives back the governing equation). #### Element approximation For a linear element with nodal temperatures $T_1$, $T_2$: $$ T = N_1T_1 + N_2T_2,\quad N_1 = 1 - \frac{x}{L},\quad N_2 = \frac{x}{L} $$ $$ \frac{dT}{dx} = \frac{1}{L}[-1\ \ 1]\begin{Bmatrix} T_1 \\ T_2 \end{Bmatrix} = [B]\{T\} $$ #### Minimising the functional Substitute into $\Pi$: $$ \Pi = \frac{1}{2}\{T\}^T\left[\int_0^L kA\,[B]^T[B]\,dx\right]\{T\} - \{T\}^T\int_0^L QA\,[N]^T dx $$ Setting $\dfrac{\partial\Pi}{\partial T_i} = 0$ gives $[k]\{T\} = \{f\}$, where $$ [k] = \int_0^L kA\,[B]^T[B]\,dx = \frac{kA}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$ $$ \{f\} = \int_0^L QA\begin{Bmatrix} N_1 \\ N_2 \end{Bmatrix}dx = \frac{QAL}{2}\begin{Bmatrix} 1 \\ 1 \end{Bmatrix} $$ #### Element equation $$ \frac{kA}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} T_1 \\ T_2 \end{Bmatrix} = \frac{QAL}{2}\begin{Bmatrix} 1 \\ 1 \end{Bmatrix} + \begin{Bmatrix} q_1 \\ q_2 \end{Bmatrix} $$ where $q_1$, $q_2$ are nodal heat flows at the ends (natural boundary conditions). The generated heat is shared equally between the two nodes. Convection at an end with coefficient $h$ and ambient temperature $T_\infty$ adds $hA$ to the stiffness term at that node and $hAT_\infty$ to the load. The assembly, boundary conditions ($T$ specified at a node) and solution follow the same steps as for the bar problem. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which functional is stationary for steady 1D conduction with generation Q? A. Pi = integral of [ kA T^2 - QA ] dx B. Pi = integral of [ kA (dT/dx) + QA T ] dx C. Pi = integral of [ (kA/2) T - Q (dT/dx) ] dx D. Pi = integral of [ (kA/2)(dT/dx)^2 - QA T ] dx Answer: D. The functional is analogous to total potential energy: a gradient-squared 'strain energy' term minus a generation 'work' term. Its variation gives kA T'' + QA = 0. (Key point: Functional) 2. For the linear two-node element, what is [B] = dT/dx per unit nodal temperature? A. (L)[-1 1] B. [-1 1] C. (1/L)[1 1] D. (1/L)[-1 1] Answer: D. With N1 = 1 - x/L and N2 = x/L, dT/dx = (T2 - T1)/L. (Key point: Element approximation) 3. What does minimising the functional give for the element conductivity matrix? A. (kA/L)[[1,1],[1,1]] B. (kA L)[[1,-1],[-1,1]] C. (kA/L)[[1,-1],[-1,1]] D. (k/AL)[[2,-1],[-1,2]] Answer: C. [k] = integral of kA [B]^T[B] dx = (kA/L)[[1,-1],[-1,1]]. (Key point: Conductivity matrix) 4. What is the nodal load from uniform heat generation Q? A. QA/L at each node B. QAL at node 1 only C. (QAL/2) at each node D. QAL/3 at node 1 and 2QAL/3 at node 2 Answer: C. Integral of QA N_i dx = QAL/2 for each node, so generated heat is shared equally. 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