--- title: "Tensile test and stress-strain diagram" subject: "Material Science" chapter: 3 key: "semester_3/material_science:tensile-test-engineering-stress-strain" source: https://www.bctengineer.com/q/semester_3/material_science/tensile-test-engineering-stress-strain site: BCT Engineering --- # Tensile test and stress-strain diagram Material Science · Chapter 3: Mechanical Properties and their tests - Asked in: year not given, 8 marks - Video: [Tensile Test](https://www.youtube.com/watch?v=D8U4G5kcpcM) — MaterialsScience2000 - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/material_science/tensile-test-engineering-stress-strain - Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-3 ## Question Describe the tensile test. Draw the engineering stress-strain diagram of a mild steel specimen and explain the salient points on it. ## Short answer A tensile test records load vs. extension of a standard specimen to compute engineering stress (load/original area) and strain (extension/original gauge length). Mild steel’s curve shows a linear elastic region (Hooke’s law), an upper/lower yield point (Lüders bands), strain hardening to the ultimate tensile strength where necking starts, and finally decreasing stress until fracture. ## Answer **Tensile test:** a standard specimen (gauge length $L_0$, area $A_0$) is gripped in a universal testing machine and pulled axially at a slow, constant rate until it fractures. The load $P$ and extension $\Delta L$ are recorded. Then $$ \sigma = \frac{P}{A_0}, \qquad e = \frac{\Delta L}{L_0} $$ give the **engineering stress** and **engineering strain**. The test gives the modulus, yield strength, tensile strength and ductility. #### Stress-strain diagram of mild steel ```text stress | D (UTS) | ,-'-. | B C ,' `. | /`--' `. E | / `x (fracture) | / A |/ +------------------------> strain O ``` #### Salient points - **O-A: proportional limit.** Stress is proportional to strain (Hooke's law), $\sigma = E e$. The slope is Young's modulus (about 200 GPa for steel). - **A-B: elastic limit.** Up to this point the specimen returns to its original length on unloading. - **B: upper yield point; C: lower yield point.** Dislocations break away from their atmospheres of interstitial (carbon, nitrogen) atoms, so the stress drops and the specimen extends at nearly constant load (Lüders bands). Permanent deformation starts here. - **C-D: strain hardening.** Stress rises with strain as the dislocation density increases, and elongation is uniform along the gauge length. - **D: ultimate tensile strength (UTS).** The maximum engineering stress, $P_{max}/A_0$. A neck starts to form here. - **D-E: necking.** Deformation concentrates in the neck and its area falls rapidly, so the engineering stress falls although the true stress still rises. - **E: fracture** (cup-and-cone type for ductile steel). #### Properties obtained Yield strength, UTS, percentage elongation $=\dfrac{L_f - L_0}{L_0}\times100$, percentage reduction in area $=\dfrac{A_0 - A_f}{A_0}\times 100$, resilience and toughness (areas under the curve). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is engineering stress defined? A. P / A_i B. P / A_0 C. Delta L / L_0 D. P / L_0 Answer: B. Engineering stress uses the original area A_0. (Key point: Engineering stress) 2. What causes the upper/lower yield point in mild steel? A. Neck formation B. Strain hardening C. Dislocations breaking away from carbon/nitrogen atmospheres D. Fracture starting Answer: C. Dislocations escape interstitial atmospheres, so stress drops and Luders bands form. (Key point: Yield point cause) 3. What happens at the ultimate tensile strength point? A. Fracture occurs B. Yielding begins C. The elastic limit is reached D. Maximum engineering stress; necking starts Answer: D. UTS = Pmax/A0 and necking begins there. (Key point: UTS) 4. Why does engineering stress fall during necking? A. Deformation concentrates in the neck and engineering stress uses original area while load falls B. Material softens elastically C. Young's modulus drops D. True stress also falls Answer: A. The load drops while A_0 is used; true stress still rises. 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