Building Materials, Third Edition

QQH f2w—

—

(iii) fav =fck + 
†



1.65

1.65 –

n

= 25 +



1.65

1.65 – 3.56

31

= 25 + 4.82

= 29. 82

Current mean,x =

30 29. 32 24.2

31



= 29.15 N/mm^2

< 29.82 N/mm^2

Hence concrete cannot be accepted straightaway.

Criteria 2

Now, fav =fck +^ †





3

1.65 –

n

= 25 +







3

1 .65 – 3.5 6

31

= 25 + 3.96

= 28. 96 N/mm^2

Thereforex is greater than fav (29.15 > 28.96), hence concrete cannot be rejected also. This is

the case where the engineer–in–charge should be consulted and his decision taken as final.

Example 11.2:First ten results, in N/mm^2 , of a M 15 concrete at a site are 22.3, 20.1, 21.4, 18.7,
17.5, 19.6, 14.7, 15.9, 16.7, 20.5. The test result corresponding to the 11th sample is 22.1 N/mm^2.

Is this concrete acceptable? Assume the quality control to be good.

Solution: xcorresponding to first ten results is 18.74 N/mm^2

As the number of test samples are less than 30, the value of 
 has to be assumed as per Table

11.7,
= 3.5.
Applying provisions of acceptance criteria:

Criteria 1

(i) fck– 1.35 
 = 15 – 1.35 × 3.5 = 10.28 N/mm^2

(ii) 0.8 fck = 0.8 × 15 = 12 N/mm^2

The greater of (i) and (ii) is 12 M/mm^2. The test result of 11th sample is 22.1 N/mm^2 ,

which is greater than 12 N/mm^2 , and hence the concrete is acceptable.

(iii) fav =fck +^ †





1.65

1.65

n