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Characteristic strength from test results

Short answer

Mean of 10 results = 28.50 N/mm², standard deviation s = 2.799 N/mm². Characteristic strength at 95% confidence is f_ck = 28.50 – 1.65×2.799 = 23.88 N/mm² (≈23.9 N/mm²).

Define characteristic strength of concrete. The test results of a compressive strength test are given as follows: 30, 28, 25, 27, 23, 29, 31, 30, 30, 32 (MPa). What will be the characteristic strength of the concrete? Make necessary assumptions.

Definition: the characteristic strength of concrete is the value of the compressive strength of 150 mm cubes at 28 days below which not more than 5% of the test results are expected to fall (IS 456, Terminology). With a normal distribution, fck=fm−1.65 sf_{ck} = f_m - 1.65\,s.

Calculation

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=285\sum x = 285, mean fm=285/10=28.50f_m = 285/10 = 28.50 N/mm2^2.

No.xxx−fmx - f_m(x−fm)2(x-f_m)^2
130+1.502.25
228-0.500.25
325-3.5012.25
427-1.502.25
523-5.5030.25
629+0.500.25
731+2.506.25
830+1.502.25
930+1.502.25
1032+3.5012.25
Sum70.50
s=70.5010−1=2.799 N/mm2fck=28.50−1.65×2.799=23.88 N/mm2\begin{aligned} s &= \sqrt{\frac{70.50}{10-1}} = 2.799\ \text{N/mm}^2 \\ f_{ck} &= 28.50 - 1.65\times2.799 = 23.88\ \text{N/mm}^2 \end{aligned}

Answer: characteristic strength fck≈f_{ck} \approx 23.9 N/mm2^2.

✓ Answer checked by Claude Sonnet 5.5

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