--- title: "Idealized stress-strain diagram of steel" subject: "Design of Steel and Timber Structure" chapter: 1 key: "semester_6/design_of_steel_and_timber_structure:stress-strain-mild-steel" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/stress-strain-mild-steel site: BCT Engineering --- # Idealized stress-strain diagram of steel Design of Steel and Timber Structure · Chapter 1: Steel Structures and their Analysis and Design - Asked in: 2068 Magh (old course), 6 marks - Video: [| Idealized Stress Strain Curves for Steel | Lecture No. 17](https://www.youtube.com/watch?v=lYPMlZcP35w) — GANESH LAKADE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/stress-strain-mild-steel - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-1 ## Question Show the idealized stress-strain diagram for mild steel and high tensile steel used in the design of steel structures. ## Short answer Mild steel is idealized as elastic-perfectly plastic with a yield plateau at f_y, elastic modulus E=2×10^5 N/mm², and yield strain 0.00125 for f_y=250. High tensile steel has no distinct yield point; its yield is the 0.2% proof stress, idealized as elastic up to that stress then flat, with lower ductility. ## Answer The real stress-strain curve is idealised for design (IS 800:2007). #### Mild steel (E250) ```text stress fu | ___------. | _-/ \ fy |-------+==============+ <- yield plateau | /| | / | | / | E = 2x10^5 | / | |__/____|______________|____ strain 0.00125 ~0.014 ~0.2 ``` - Elastic line up to $f_y$ with slope $E=2\times10^5$ N/mm²; yield strain $\varepsilon_y=f_y/E=0.00125$ for $f_y=250$. - Constant stress $f_y$ (yield plateau) up to roughly 10-15 times $\varepsilon_y$ (strain hardening starts at about 0.014). - Strain hardening up to the ultimate stress $f_u$ ($\approx410$ N/mm²) at a strain of about 0.15-0.2, then necking and fracture. - For design the curve is idealised as elastic-perfectly plastic (bilinear): $\sigma = E\varepsilon$ up to $f_y$, then $\sigma=f_y$. #### High tensile steel (E350-E450, cold-worked) ```text stress fu | ___--- | _-/ 0.2%|------_/ proof / | / | / |___/____|__________ strain 0.002 offset ``` - No definite yield point or plateau; the stress rises smoothly. - The yield stress is taken as the **0.2% proof stress** (found by a line parallel to the elastic part, drawn from 0.002 strain). - Idealised as elastic up to the proof stress and then flat (or rising slightly). Ductility is lower than that of mild steel. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the yield strain of E250 steel with E = 2x10^5 N/mm2? A. 0.0125 B. 0.00125 C. 0.014 D. 0.002 Answer: B. epsilon_y = fy / E = 250 / (2x10^5) = 0.00125. (Key point: Yield strain) 2. What is the idealised stress-strain curve of mild steel for design? A. Elastic-perfectly plastic (bilinear): sigma = E*epsilon up to fy, then constant fy B. Smooth curve with no yield point C. Linear up to fu D. Only the strain-hardening portion Answer: A. For design the curve is bilinear: elastic to fy and then flat at fy. (Key point: Idealised curve) 3. How is yield stress defined for high tensile steel which has no definite yield point? A. As the ultimate stress B. As the stress at 0.2% proof stress, from a line parallel to the elastic part offset by 0.002 strain C. As 0.6 times the ultimate stress D. As the stress at 2% strain Answer: B. The 0.2% proof stress, found by drawing a line parallel to the elastic portion from 0.002 strain, is taken as the yield stress. 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