Page 259 - Civil Engineering Project Management, Fourth Edition

Table 19.2
Standard mixes as Table 5 of BS 5328-2:1997 also described as
Standardized Prescribed Concrete in BS 85000:2002
Maximum size of aggregate
Standard Characteristic Constituents Site concreting and reinforcement 239
mix compressive (Cement and 40 mm 20 mm
strength at Total aggregate)
28 days (kg) Workability Workability
2
(N/mm ) Medium High Medium High
ST1 7.5 Cement 180 200 210 230
Aggregate 2010 1950 1940 1880
ST2 10 Cement 210 230 240 260
Aggregate 1980 1920 1920 1860
ST3 15 Cement 240 260 270 300
Aggregate 1950 1900 1890 1820
ST4 20 Cement 280 300 300 330
Aggregate 1920 1860 1860 1800
ST5 25 Cement 320 340 340 370
Aggregate 1890 1830 1830 1770
Workability – slump (mm) 50–100 80–170 25–75 65–135

mix proportions for size of aggregate, and workability. Table 19.2 reproduces
Table 5 of BS 5328-2. The lower grades of characteristic strength 7.5, 10.0 and
2
15N/mm are intended for use in mass concrete filling to strip footings,
blinding concrete, and similar.
Designated mixes are for mixes to meet special requirements, such as for
sulphate resisting concrete, etc. and also for ready-mix concrete for which the
supplier is required to hold a current product conformity certificate, to BS EN
150 9001.
The characteristic strength of a mix is defined as – ‘that value of strength
below which 5per cent of…strength measurements…are expected to fall’. On
a statistical basis, cube strength test results on a given mix are found to follow a
‘normal distribution’ that is 50 per cent of test results are above the mean X, and
50 per cent below. If only 5 per cent of results are to fall below a required value
P, then the mean strength X must obviously be higher than P. From the charac-
teristics of a normal distribution curve, the value of X has to be 1.64S higher
than P, that is, X (P 1.64S) to achieve not more than 5 per cent results below
1
P, where S is the standard deviation of the test results obtained. If only 2.5 per
cent of results are to fall below P, then X must be (P 1.96S).

1 The standard deviation, S, is defined by

2
2
(X x ) (X x ) L (X x ) 2
n
S 1 2
n 1
where there are n test results x 1 , x 2 , …,x n and X is their mean value.