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Characteristic strength of concrete cubes

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.

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.

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) 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=11n = 11, ∑x=263\sum x = 263, mean fm=263/11=23.91f_m = 263/11 = 23.91 N/mm2^2.

No.xxx−fmx - f_m(x−fm)2(x-f_m)^2
126+2.094.37
222-1.913.64
326+2.094.37
427+3.099.55
523-0.910.83
624+0.090.01
722-1.913.64
822-1.913.64
928+4.0916.74
1018-5.9134.92
1125+1.091.19
Sum82.91
s=82.9111−1=2.879 N/mm2fck=23.91−1.65×2.879=19.16 N/mm2\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 fck≈f_{ck} \approx 19.2 N/mm2^2.

✓ Answer checked by Claude Sonnet 5.5

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