--- title: "Characteristic strength of concrete cubes" subject: "Concrete Technology and Masonry Structure" chapter: 5 key: "semester_5/concrete_technology_and_masonry_structure:characteristic-strength-cubes-26" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/characteristic-strength-cubes-26 site: BCT Engineering --- # Characteristic strength of concrete cubes Concrete Technology and Masonry Structure · Chapter 5: Testing of concrete and quality control - Asked in: 2067 Ashadh (old course), 4+1 marks - Video: [CHARACTERISTIC STRENGTH OF CONCRETE](https://www.youtube.com/watch?v=9VSj9PjHMAk) — AIM CAMPUS - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/characteristic-strength-cubes-26 - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-5 ## Question The compressive strength of concrete cubes as obtained from a laboratory test was as 26, 22, 26, 27, 23, 24, 22, 22, 28, 18, 25. What will be its characteristic strength? State necessary assumptions. ## Short answer For 11 cubes, mean strength = 23.91 N/mm², standard deviation s = 2.879 N/mm². Thus f_ck = 23.91 – 1.65×2.879 = 19.16 N/mm² ≈ 19.2 N/mm², assuming normal distribution. ## Answer Characteristic strength is the strength below which not more than 5% of the test results are expected to fall (IS 456, Terminology). **Formula:** characteristic strength (95% confidence) $f_{ck} = f_m - 1.65\,s$, with the sample standard deviation $s = \sqrt{\sum(x-f_m)^2/(n-1)}$. The factor 1.65 comes from the normal distribution (the same factor is used in mix design, $f_{target} = f_{ck} + 1.65\,s$, IS 10262). Assumptions: the results follow the normal distribution, and the sample standard deviation ($n-1$) is used. $n = 11$, $\sum x = 263$, mean $f_m = 263/11 = 23.91$ N/mm$^2$. | No. | $x$ | $x - f_m$ | $(x-f_m)^2$ | |---|---|---|---| | 1 | 26 | +2.09 | 4.37 | | 2 | 22 | -1.91 | 3.64 | | 3 | 26 | +2.09 | 4.37 | | 4 | 27 | +3.09 | 9.55 | | 5 | 23 | -0.91 | 0.83 | | 6 | 24 | +0.09 | 0.01 | | 7 | 22 | -1.91 | 3.64 | | 8 | 22 | -1.91 | 3.64 | | 9 | 28 | +4.09 | 16.74 | | 10 | 18 | -5.91 | 34.92 | | 11 | 25 | +1.09 | 1.19 | | | | Sum | 82.91 | $$ \begin{aligned} s &= \sqrt{\frac{82.91}{11-1}} = 2.879\ \text{N/mm}^2 \\ f_{ck} &= 23.91 - 1.65\times2.879 = 19.16\ \text{N/mm}^2 \end{aligned} $$ **Answer: characteristic strength $f_{ck} \approx$ 19.2 N/mm$^2$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What assumptions are used? A. Uniform distribution B. Normal distribution and sample standard deviation with n-1 C. Population standard deviation only D. Results are all equal Answer: B. Normal distribution and n-1 deviation are used. (Key point: Assumptions) 2. What is the mean of the 11 results? A. 26.0 N/mm2 B. 22.0 N/mm2 C. 23.91 N/mm2 D. 24.5 N/mm2 Answer: C. 263/11 = 23.91. (Key point: Mean) 3. What divisor is used under the root for s here? A. 11 (n) B. 12 C. 9 D. 10 (n-1) Answer: D. With n = 11, n - 1 = 10 gives s = 2.879. (Key point: Std dev) 4. What is fck? A. 19.2 N/mm2 B. 23.9 N/mm2 C. 28.7 N/mm2 D. 16.5 N/mm2 Answer: A. 23.91 - 1.65 x 2.879 = 19.16. 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