--- title: "Thermal stress in a stepped bar" subject: "Finite Element Method" chapter: 6 key: "semester_8/finite_element_method:thermal-stress-bar-between-walls" source: https://www.bctengineer.com/q/semester_8/finite_element_method/thermal-stress-bar-between-walls site: BCT Engineering --- # Thermal stress in a stepped bar Finite Element Method · Chapter 6: Applications in Solid Mechanics - Asked in: year not given, 8 marks - Video: [A Stepped bar subjected to thermal load in finite element methods | 1D thermal stress bar Problems](https://www.youtube.com/watch?v=9FuEB7ARQTE) — Mahesh Gadwantikar - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/finite_element_method/thermal-stress-bar-between-walls - Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-6 ## Question A stepped steel bar is held between two rigid walls with no gap, at the assembly temperature. The left portion has area 500 mm$^2$ and length 300 mm; the right portion has area 250 mm$^2$ and length 200 mm. $E = 200$ GPa and $\alpha = 12\times10^{-6}/^\circ$C. The temperature rises by 60 °C. Using two bar elements, find the displacement of the junction, the stress in each portion and the wall reaction. Explain how thermal loading enters the FE equations. ## Short answer Junction displacement 0.0617 mm right, stresses -102.9 MPa (left) and -205.7 MPa (right), wall reaction 51.43 kN. Thermal loads enter via equivalent nodal forces {f_T}=EAαΔT{-1,+1}^T derived from initial strain. ## Answer #### Thermal load in FEM If an element is free to expand, there is no stress. If the expansion is prevented, the initial strain $\varepsilon_0 = \alpha\Delta T$ produces an equivalent nodal force vector $$ \{f_T\} = \int_V [B]^T[D]\{\varepsilon_0\}\,dV $$ For a bar element this gives $\{f_T\} = EA\alpha\Delta T\begin{Bmatrix} -1 \\ +1 \end{Bmatrix}$. Stress is found from the mechanical strain only: $\sigma = E(\varepsilon - \alpha\Delta T)$. #### Element data ```text wall|==[A1=500]==o==[A2=250]==|wall L1=300 2 L2=200 1 3 ``` $$ k_1 = \frac{500(200\times10^3)}{300} = 3.333\times10^5\ \text{N/mm},\qquad k_2 = \frac{250(200\times10^3)}{200} = 2.5\times10^5\ \text{N/mm} $$ $$ E A_1\alpha\Delta T = 200\times10^3(500)(12\times10^{-6})(60) = 72\,000\ \text{N},\qquad EA_2\alpha\Delta T = 36\,000\ \text{N} $$ #### Equations $u_1 = u_3 = 0$; only $u_2$ is unknown. The thermal force at node 2 is $+72\,000$ from element 1 and $-36\,000$ from element 2: $$ (k_1 + k_2)\,u_2 = 72\,000 - 36\,000 = 36\,000 $$ $$ u_2 = \frac{36\,000}{5.833\times10^5} = 0.0617\ \text{mm (to the right)} $$ #### Element forces and stresses $$ \begin{aligned} N_1 &= k_1(u_2 - 0) - EA_1\alpha\Delta T = 20\,571 - 72\,000 = -51\,429\ \text{N} \\ N_2 &= k_2(0 - u_2) - EA_2\alpha\Delta T = -15\,429 - 36\,000 = -51\,429\ \text{N} \end{aligned} $$ The two forces are equal, as equilibrium of the junction requires. $$ \sigma_1 = \frac{-51\,429}{500} = -102.9\ \text{MPa},\qquad \sigma_2 = \frac{-51\,429}{250} = -205.7\ \text{MPa} $$ #### Check Free expansion would be $\alpha\Delta T(L_1 + L_2) = 0.36$ mm. It is cancelled by the compression of the two bars: $P\left(\dfrac{300}{500E} + \dfrac{200}{250E}\right) = \dfrac{P(1.4)}{200\times10^3} = 0.36$ gives $P = 51\,429$ N, which agrees. **Answer:** $u_2 = 0.0617$ mm; $\sigma_1 = -102.9$ MPa and $\sigma_2 = -205.7$ MPa (compressive); wall reaction $= 51.43$ kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does temperature rise enter the FE equations for a bar element? A. As an equivalent nodal force EA alpha dT, acting -1 and +1 on the two nodes B. As a change in the length of the element only C. As a change in element stiffness D. It does not enter, since the bar is free Answer: A. Prevented expansion gives initial strain alpha dT and equivalent nodal force EA alpha dT {-1, +1}. (Key point: Thermal load) 2. What is the equation for the single unknown u2? A. (k1 - k2) u2 = 36,000 B. (k1 + k2) u2 = 72,000 + 36,000 C. (k1 + k2) u2 = 72,000 - 36,000 D. k1 u2 = 72,000 Answer: C. Thermal force at node 2 is +72,000 N from element 1 and -36,000 N from element 2, giving u2 = 36,000/5.833x10^5 = 0.0617 mm. (Key point: Reduced equation) 3. How is the element force found? A. N = k(u_j - u_i) - EA alpha dT, using mechanical strain only B. N = EA alpha dT C. N = k u2 + EA alpha dT D. N = k(u_j - u_i), ignoring thermal part Answer: A. Stress comes from mechanical strain, sigma = E(eps - alpha dT), so the thermal part is subtracted, giving N = -51,429 N. (Key point: Stress calculation) 4. How is the result verified? A. The stresses in both portions are equal B. The displacement u2 equals 0.36 mm C. Free expansion 0.36 mm equals compression P(1.4)/200,000, giving P = 51,429 N D. The wall reaction is zero Answer: C. Free expansion alpha dT (L1 + L2) = 0.36 mm is cancelled by compression, giving the same force P as the FE solution. The forces are equal, but the stresses differ (-102.9 and -205.7 MPa). (Key point: Check) ## Related in Finite Element Method - [Stepped bar under axial loads by FEM](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Astepped-bar-displacement-stress) — Finite Element Method, chapter 3 - [Tapered bar analysis using two elements](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Atapered-bar-two-elements) — Finite Element Method, chapter 5 - [Area coordinates and temperature at a point](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Acst-area-coordinates-interpolation) — Finite Element Method, chapter 5 ## Related in other subjects - [Thermal stress in a restrained bar](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Abar-between-rigid-walls-temperature) — Strength of Materials, chapter 2 - [Elongation of a stepped steel bar](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Astepped-bar-axial-elongation) — Strength of Materials, chapter 2 - [Total elongation of stepped-load steel bar](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-bar-total-elongation-abcd) — Strength of Materials, chapter 3 - [Steel-aluminium bar with temperature rise: stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-aluminium-bar-temperature-supports) — Strength of Materials, chapter 3