--- title: "Strain energy density and thermal stress" subject: "Mechanics of Solids" chapter: 4 key: "semester_5/mechanics_of_solids:strain-energy-density-thermal-constraint" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/strain-energy-density-thermal-constraint site: BCT Engineering --- # Strain energy density and thermal stress Mechanics of Solids · Chapter 4: General Hooke’s Law - Asked in: year not given, 8 marks - Video: [MECHANICS OF SOLIDS MODULE 4 PART 3 STRAIN ENERGY METHOD](https://www.youtube.com/watch?v=DHfilAnGwuU) — IgnitEd - Vishnu Shaji Kanneth - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/strain-energy-density-thermal-constraint - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-4 ## Question (a) At a point in an isotropic steel body the principal stresses are 100 MPa, 50 MPa and -30 MPa. Calculate the strain energy per unit volume and split it into volumetric (dilatation) and distortion parts. Take $E = 200$ GPa, $\nu = 0.3$. (b) A steel cube is completely prevented from expanding in all three directions and is heated by $50^\circ$C. Using the equations of thermo-elasticity, find the stress developed in the cube. Take $\alpha = 12\times10^{-6}$/$^\circ$C. ## Short answer (a) U_0 = 0.03275 N·mm/mm³, volumetric U_v = 0.0048, distortion U_d = 0.02795 N·mm/mm³. (b) Fully restrained cube heated by 50°C develops hydrostatic compressive stress σ = –EαΔT/(1-2ν) = –300 MPa. ## Answer #### (a) Strain energy density $$ U_0 = \frac{1}{2E}\left[\sigma_1^2+\sigma_2^2+\sigma_3^2 - 2\nu(\sigma_1\sigma_2+\sigma_2\sigma_3+\sigma_3\sigma_1)\right] $$ Terms: $\sigma_1^2+\sigma_2^2+\sigma_3^2 = 10000+2500+900 = 13400$; $\sigma_1\sigma_2+\sigma_2\sigma_3+\sigma_3\sigma_1 = 5000 - 1500 - 3000 = 500$. $$ U_0 = \frac{13400 - 2(0.3)(500)}{2\times200\times10^3} = \frac{13100}{4\times10^5} = 0.03275\ \text{N\,mm/mm}^3 $$ **Volumetric part** (change of volume only): $$ U_v = \frac{1-2\nu}{6E}(\sigma_1+\sigma_2+\sigma_3)^2 = \frac{0.4}{1.2\times10^6}(120)^2 = 0.0048\ \text{N\,mm/mm}^3 $$ **Distortion part**: $$ U_d = \frac{1+\nu}{6E}\left[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2\right] = \frac{1.3}{1.2\times10^6}(25800) = 0.02795\ \text{N\,mm/mm}^3 $$ Check: $U_v + U_d = 0.0048+0.02795 = 0.03275 = U_0$. About 85 percent of the energy is distortion energy (used in the von Mises theory). #### (b) Fully constrained heated cube Thermo-elastic equations for isotropic material: $$ \varepsilon_x = \frac{1}{E}\left[\sigma_x - \nu(\sigma_y+\sigma_z)\right] + \alpha\Delta T $$ Total strain is zero in all directions and the stress is the same in all three by symmetry, $\sigma_x=\sigma_y=\sigma_z=\sigma$: $$ \begin{aligned} 0 &= \frac{\sigma(1-2\nu)}{E} + \alpha\Delta T \\ \sigma &= -\frac{E\alpha\Delta T}{1-2\nu} = -\frac{200\times10^3 \times 12\times10^{-6}\times 50}{1-0.6} = -300\ \text{MPa} \end{aligned} $$ For comparison, a bar restrained in one direction only would have $E\alpha\Delta T = 120$ MPa; the triaxial restraint makes the stress 2.5 times larger. The stress is hydrostatic, so it causes no distortion and the cube does not yield by the shear-based theories. **Answer:** (a) $U_0 = 0.03275$ N mm/mm$^3$ (= 32.75 kJ/m$^3$), $U_v = 0.0048$, $U_d = 0.02795$ N mm/mm$^3$; (b) $\sigma = -300$ MPa (compressive, equal in all three directions). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For principal stresses 100, 50, -30 MPa, what are sum of squares and sum of cross products? A. 13400 and 500 B. 13400 and 120 C. 10000 and 500 D. 14400 and 500 Answer: A. 10000 + 2500 + 900 = 13400; 5000 - 1500 - 3000 = 500; these give U_0 = 0.03275 N mm/mm^3. (Key point: Strain energy density) 2. How is the volumetric part computed? A. U_v = (1 + nu)/(6E) x sum of squared differences B. U_v = (1 - 2nu)/(6E) (s1 + s2 + s3)^2 = 0.0048 C. U_v = U_0/2 D. U_v = (s1 + s2 + s3)/E Answer: B. Volume change depends only on the sum of principal stresses: 0.4/(1.2e6) x 120^2 = 0.0048. (Key point: Volumetric energy) 3. What does the distortion part represent and its value? A. Change of volume, 0.0048 B. Change of volume, 0.02795 C. Change of shape at constant volume, 0.02795 N mm/mm^3 D. Change of shape, 0.0048 Answer: C. U_d = U_0 - U_v = 0.02795, about 85 percent of the total, and is the basis of von Mises theory. (Key point: Distortion energy) 4. Why is the stress -300 MPa, 2.5 times E alpha dT = 120 MPa? A. Because E doubles B. Because triaxial restraint gives sigma = -E alpha dT/(1 - 2nu) C. Because the cube yields D. Because alpha triples Answer: B. Setting total strain to zero in all directions gives sigma(1 - 2nu)/E + alpha dT = 0, so sigma = -120/0.4 = -300 MPa. (Key point: Fully restrained heated cube) ## Related in Mechanics of Solids - [Principal stresses from a 3D stress tensor](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Athree-d-stress-invariants-numerical) — Mechanics of Solids, chapter 2 - [Plane stress: principal stresses and oblique plane](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aplane-stress-principal-numerical) — Mechanics of Solids, chapter 2 - [Strain rosette: principal strains and stresses](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Astrain-rosette-principal-numerical) — Mechanics of Solids, chapter 3 - [Contact stress between two steel cylinders](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Asteel-cylinders-hertz-numerical) — Mechanics of Solids, chapter 11 - [Thick cylinder stresses and failure criteria](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Athick-cylinder-stresses-failure-numerical) — Mechanics of Solids, chapter 8 - [Generalised Hooke's law for isotropic materials](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Ageneralised-hooke-law-isotropic) — Mechanics of Solids, chapter 4 ## Related in other subjects - [Max shear for 150 MPa stress limit; principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Adirect-stresses-100t-60c-shear-limit) — Strength of Materials, chapter 4 - [Principal stresses: 50 T, 20 C, 30 shear](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-element-50t-20c-30shear) — Strength of Materials, chapter 4 - [Principal stresses and planes (90, 120, 50 MPa)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-stresses-90-50-120) — Strength of Materials, chapter 4 - [Progressive wave energy per unit volume](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aprogressive-wave-energy-per-volume) — Engineering Physics, chapter 2