--- title: "Strain rosette: principal strains and stresses" subject: "Mechanics of Solids" chapter: 3 key: "semester_5/mechanics_of_solids:strain-rosette-principal-numerical" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/strain-rosette-principal-numerical site: BCT Engineering --- # Strain rosette: principal strains and stresses Mechanics of Solids · Chapter 3: Deformable Body and Strain Tensor - Asked in: year not given, 8 marks - Video: [Understanding Strain Gauge and Rosette Problems!](https://www.youtube.com/watch?v=78XX1OdPOdc) — Simple Civil - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/strain-rosette-principal-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-3 ## Question A 0-45-90 degree strain gauge rosette is bonded on the surface of a steel machine part (free surface, plane stress). The readings are $\varepsilon_0 = 800\times10^{-6}$, $\varepsilon_{45} = 400\times10^{-6}$ and $\varepsilon_{90} = -200\times10^{-6}$. Find (a) the strains $\varepsilon_x$, $\varepsilon_y$ and $\gamma_{xy}$, (b) the principal strains, maximum shear strain and the principal directions, (c) the principal stresses. Take $E = 200$ GPa, $\nu = 0.3$. ## Short answer From ε_0=800×10⁻⁶, ε_45=400×10⁻⁶, ε_90=-200×10⁻⁶ we get ε_x=800×10⁻⁶, ε_y=-200×10⁻⁶, γ_xy=200×10⁻⁶. Principal strains: ε_1=809.9×10⁻⁶, ε_2=-209.9×10⁻⁶, γ_max=1019.8×10⁻⁶ at θ_p=5.65°. Principal stresses: σ_1=164.2 MPa, σ_2=7.3 MPa. ## Answer #### (a) Strain components Gauge $a$ along x (0$^\circ$), $b$ at 45$^\circ$, $c$ along y (90$^\circ$). The strain in direction $\theta$: $$ \varepsilon_\theta = \varepsilon_x\cos^2\theta + \varepsilon_y\sin^2\theta + \gamma_{xy}\sin\theta\cos\theta $$ $$ \begin{aligned} \varepsilon_x &= \varepsilon_0 = 800\times10^{-6} \\ \varepsilon_y &= \varepsilon_{90} = -200\times10^{-6} \\ \varepsilon_{45} &= \tfrac12(\varepsilon_x+\varepsilon_y) + \tfrac12\gamma_{xy} \\ \gamma_{xy} &= 2\varepsilon_{45} - \varepsilon_x - \varepsilon_y = 800 - 800 + 200 = 200\times10^{-6} \end{aligned} $$ #### (b) Principal strains (units $10^{-6}$) $$ \begin{aligned} \varepsilon_{avg} &= \frac{800 - 200}{2} = 300 \\ R &= \sqrt{\left(\frac{\varepsilon_x-\varepsilon_y}{2}\right)^2 + \left(\frac{\gamma_{xy}}{2}\right)^2} = \sqrt{500^2+100^2} = 509.9 \\ \varepsilon_{1,2} &= \varepsilon_{avg} \pm R = 809.9,\ -209.9 \\ \gamma_{max} &= 2R = 1019.8 \\ \tan 2\theta_p &= \frac{\gamma_{xy}}{\varepsilon_x-\varepsilon_y} = \frac{200}{1000} \Rightarrow \theta_p = 5.65^\circ \end{aligned} $$ $\varepsilon_1$ acts at $5.65^\circ$ anticlockwise from the x-axis; $\varepsilon_2$ at $95.65^\circ$. #### (c) Principal stresses (plane stress, $\sigma_z=0$) $$ \begin{aligned} \sigma_1 &= \frac{E}{1-\nu^2}(\varepsilon_1 + \nu\varepsilon_2) = \frac{200\times10^3}{0.91}(809.9 - 0.3\times209.9)\times10^{-6} = 164.2\ \text{MPa} \\ \sigma_2 &= \frac{E}{1-\nu^2}(\varepsilon_2 + \nu\varepsilon_1) = \frac{200\times10^3}{0.91}(-209.9 + 0.3\times809.9)\times10^{-6} = 7.3\ \text{MPa} \end{aligned} $$ The out-of-plane strain is $\varepsilon_z = -\dfrac{\nu}{1-\nu}(\varepsilon_1+\varepsilon_2) = -257\times10^{-6}$ (not measured by the rosette, but present). **Answer:** $\varepsilon_x = 800\times10^{-6}$, $\varepsilon_y = -200\times10^{-6}$, $\gamma_{xy} = 200\times10^{-6}$; $\varepsilon_1 = 810\times10^{-6}$, $\varepsilon_2 = -210\times10^{-6}$, $\gamma_{max} = 1020\times10^{-6}$ at $5.65^\circ$; $\sigma_1 = 164.2$ MPa, $\sigma_2 = 7.3$ MPa. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 0-45-90 rosette, how is gamma_xy found from the readings? A. gamma_xy = eps_0 + eps_90 B. gamma_xy = eps_45 - eps_0 C. gamma_xy = 2 eps_45 - eps_x - eps_y = 200e-6 D. gamma_xy = eps_45/2 Answer: C. From eps_45 = (eps_x + eps_y)/2 + gamma/2: gamma = 800 - 800 + 200 = 200e-6. (Key point: Rosette shear strain) 2. How are principal strains computed? A. eps_avg +/- R with R = sqrt(((ex - ey)/2)^2 + (gamma/2)^2) giving 809.9 and -209.9 B. ex and ey directly C. eps_avg +/- gamma D. Only eps_45 is used Answer: A. Mohr's circle for strain has centre 300 and radius 509.9 (in 1e-6), with gamma/2 on the shear axis. (Key point: Principal strains) 3. Which formula gives sigma_1 from principal strains in plane stress? A. sigma_1 = E eps_1 B. sigma_1 = E/(1 - nu^2) (eps_1 + nu eps_2) C. sigma_1 = E (eps_1 - nu eps_2) D. sigma_1 = E/(1 + nu)(eps_1) Answer: B. With sigma_z = 0, the Hooke relations invert to sigma_1 = E(eps_1 + nu eps_2)/(1 - nu^2) = 164.2 MPa. (Key point: Plane stress from strain) 4. On a free surface with sigma_z = 0, what is eps_z? A. Zero B. Equal to eps_1 C. -nu/(1-nu) (eps_1 + eps_2), non-zero though not measured D. Equal to gamma_max Answer: C. Plane stress still has Poisson thickness strain: -0.3/0.7 x 600e-6 = -257e-6. (Key point: Out-of-plane strain) ## Related in Mechanics of Solids - [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 - [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 - [Strain energy density and thermal stress](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Astrain-energy-density-thermal-constraint) — Mechanics of Solids, chapter 4 - [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 - [Stress transformation equations and Mohr's circle](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aderive-stress-transformation-mohr) — Mechanics of Solids, chapter 2 ## Related in other subjects - [Normal/shear stress on 25 degree plane, principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-30-25deg-mohr-verify) — Strength of Materials, chapter 4, asked 3× - [Stresses on 20 degree plane; principal and max shear](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-20deg-50-40-80) — Strength of Materials, chapter 4, asked 4× - [Stresses on 40 degree plane; principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-40deg-50-30) — Strength of Materials, chapter 4, asked 2× - [Plane stress: principal, max shear, normal](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-80-60-50-mpa) — Strength of Materials, chapter 4, asked 2×