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Characteristic strength at 95% confidence

Short answer

Mean of 10 results = 21.60 N/mm², sample standard deviation s = 4.766 N/mm². Characteristic strength f_ck = f_m – 1.65s = 21.60 – 1.65×4.766 = 13.74 N/mm² (≈13.7 N/mm²), assuming normal distribution.

The compressive strength test results of a concrete specimen were found as 16; 17; 19; 21; 22; 25; 26; 27; 28 and 15 N/mm2^2. Calculate the characteristic strength of the test result at 95% confidence level.

Characteristic strength is the value of strength below which not more than 5% of the test results are expected to fall (IS 456, Terminology).

Formula: characteristic strength (95% confidence) fck=fm−1.65 sf_{ck} = f_m - 1.65\,s, with the sample standard deviation s=∑(x−fm)2/(n−1)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, ftarget=fck+1.65 sf_{target} = f_{ck} + 1.65\,s, IS 10262). Assumptions: the results follow the normal distribution, and the sample standard deviation (n−1n-1) is used.

n=10n = 10, ∑x=216\sum x = 216, mean fm=216/10=21.60f_m = 216/10 = 21.60 N/mm2^2.

No.xxx−fmx - f_m(x−fm)2(x-f_m)^2
116-5.6031.36
217-4.6021.16
319-2.606.76
421-0.600.36
522+0.400.16
625+3.4011.56
726+4.4019.36
827+5.4029.16
928+6.4040.96
1015-6.6043.56
Sum204.40
s=204.4010−1=4.766 N/mm2fck=21.60−1.65×4.766=13.74 N/mm2\begin{aligned} s &= \sqrt{\frac{204.40}{10-1}} = 4.766\ \text{N/mm}^2 \\ f_{ck} &= 21.60 - 1.65\times4.766 = 13.74\ \text{N/mm}^2 \end{aligned}

Answer: characteristic strength fck≈f_{ck} \approx 13.7 N/mm2^2.

✓ Answer checked by Claude Sonnet 5.5

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