1. The required force to break a 4 x 8 in. cylinder is 44% of that required to break a

6 x 12 in. cylinder of the same strength solely based on a ratio of the two circular cross-sectional areas.

2. A 4 x 8 in. cylinder weighs about 9 lb compared to a 6 x 12 in. cylinder, which weighs about 30 lb, almost three times as much.

3. compressive strength was the only factor significant in affecting the ratio of 4 x 8 in. cylinder strength to 6 x 12 in. cylinder strength. It is recommended that 4 x 8 in. cylinders may be implemented for quality control and assurance specimen testing if the design strength of concrete is greater than 5,000 psi and the capacity of the testing machine will not allow the testing of 6 x 12 in. cylinders based on the design strength.

Following literature may be of help for you to understand size effect

Dear Hammad

There is this fantastic research thesis that I think answers your question and goes deep into the topic:

http://www.eng.auburn.edu/files/file806.pdf

I hope that it will be of use to you.

The difference in value of tensile strength corresponding to different sizes of cylinder sample is known as statistical size effect, which is explained in Section 2.2.2 in the following paper:

Article Influence of material spatial variability on the shear stren...

Fig. 1 in the paper shows the relationship between fct/fct,ref and A/Aref, where fct,ref is the reference tensile strength corresponding to the crack area Aref. From this figure you can find the conversion factor (1.07) between fct,6x12 and fct,4x8.

fct,4x8 = 1.07*fct,6x12.

For determining the tensile strength it is suggested to use more samples to reduce the statistic uncertainty, e.g. 6 samples. If you have the scatter of your experimental results, you can calculate fct,6x12 based on fct,4x8 with high accuracy.

Thanks to all for helping me in finding the right answer.