--- title: "Idealized stress-strain, characteristic strength" subject: "Design of RCC Structure" chapter: 3 key: "semester_7/design_of_rcc_structure:idealized-stress-strain-characteristic-strength" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/idealized-stress-strain-characteristic-strength site: BCT Engineering --- # Idealized stress-strain, characteristic strength Design of RCC Structure · Chapter 3: Limit State Method of design - Asked in: 2073 Shrawan, 2+2+2 marks - Video: [Idealized Stress Strain Curve of Concrete | Why 0.67 times fck | IS 456-2000](https://www.youtube.com/watch?v=v21ED4zZn6w) — All About Structural Analysis and Design - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/idealized-stress-strain-characteristic-strength - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-3 ## Question What do you understand by idealized stress-strain diagram of concrete and steel bar? Draw idealized stress-strain diagrams. Define characteristic strength of concrete and steel. ## Short answer Concrete idealized curve: parabola to 0.002 strain then horizontal to 0.0035, design stress 0.446fck. Steel (HYSD) idealized curve: linear to 0.8fy then curve to fy with 0.002 plastic strain, design stress 0.87fy. Characteristic strength: fck = 5% fractile 28-day cube strength for concrete; fy = 5% fractile yield/proof stress for steel. ## Answer #### Idealised stress-strain diagram The actual stress-strain curves of concrete and steel are irregular. For design, they are replaced by simplified (idealised) curves given in IS 456:2000. **Concrete** (Fig. 21): a parabola from zero to a strain of 0.002, then a horizontal line up to 0.0035; the peak design stress is $0.67f_{ck}/1.5=0.446f_{ck}$. **Steel:** - HYSD bars (Fig. 23B): linear up to $0.8f_y$ (slope $E_s=2\times10^5$ N/mm$^2$), then a curve up to $f_y$ with a plastic strain of 0.002; the design curve is scaled by $1/1.15$ to peak at $0.87f_y$. - Mild steel (Fig. 23A): linear up to $f_y$, then horizontal (elastic-perfectly plastic); the design curve is capped at $0.87f_y$. ```text Concrete (IS 456 Fig. 21) Steel (Fe415, Fig. 23B) stress stress 0.67fck | _______________ fy | ______ actual | / | 0.87fy| __/ ______ design (design |/ | |_/ / 0.446fck) | |/ / +------------------> +----------------> strain 0.002 0.0035 Es=2x10^5 0.87fy/Es+0.002 ``` #### Characteristic strength - **Concrete:** the compressive strength $f_{ck}$ of 150 mm cubes at 28 days, below which not more than 5% of test results are expected to fall. M20 means $f_{ck}=20$ N/mm$^2$. - **Steel:** the yield stress (or 0.2% proof stress) $f_y$, below which not more than 5% of test results are expected to fall. Fe415 means $f_y=415$ N/mm$^2$. Both are obtained from $f_k=f_m-1.65s$, where $f_m$ is the mean strength and $s$ the standard deviation. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are idealised stress-strain curves used? A. Actual curves are irregular, so simplified curves are used for design B. Actual curves do not exist C. To increase the strength of materials D. Because concrete is perfectly elastic Answer: A. The real curves are irregular, so IS 456 replaces them with simple curves for calculation. (Key point: Why idealise) 2. How is the characteristic strength fck defined? A. Average cube strength at 28 days B. Strength of 150 mm cubes at 28 days below which not more than 5% of results fall C. Strength at 7 days D. Strength below which 50% of tests fall Answer: B. Characteristic strength is a 5% fractile value, lower than the mean. (Key point: Characteristic strength of concrete) 3. What does the IS 456 design curve for steel show? A. Linear to 0.8 fy, then curved to fy, scaled by 1/1.15 to peak at 0.87 fy B. Linear up to 0.87 fy with no curve C. Parabola then plateau at 0.446 fy D. Linear to fy with no scaling Answer: A. The curve is linear to 0.8 fy, then non-linear with 0.002 plastic strain, and the design curve is divided by 1.15. (Key point: Steel idealised curve) 4. Which expression relates characteristic strength to the mean? A. fk = fm + 1.65 s B. fk = fm - 1.65 s C. fk = fm / 1.65 D. fk = fm - s Answer: B. With 1.65 standard deviations below the mean, only 5% of results fall under fk. 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