As per IS 2386 (1963), the aggregate crushing value is calculated by using a single sized aggregate, which is:
- Apassing 40 mm and retained on 20 mm sieve
- Bpassing 12.5 mm and retained on 10 mm sieveCorrect
- Cpassing 4.75 mm and retained on 2.36 mm sieve
- Dpassing 20 mm and retained on 10 mm sieve
Solution
Concept
The standard aggregate crushing value test uses aggregate passing the 12.5 mm IS sieve and retained on the 10 mm IS sieve.
Formula
ACV (%) = (W2 / W1) × 100, where W1 = weight of surface-dry sample, W2 = weight of fraction passing 2.36 mm sieve
Step-by-step solution
1. Recall IS 2386 (Part IV) - 1963 for mechanical properties of aggregate. 2. Standard crushing value test: aggregate passing 12.5 mm and retained on 10 mm IS sieve. 3. Fill the cylinder in 3 layers, 25 tamps each, and apply 40 t load gradually in 10 minutes. 4. Sieve the crushed material on the 2.36 mm sieve and find % fines. 5. So the correct size is 12.5 mm to 10 mm. Answer (B).
Explanation
IS 2386 Part IV gives the crushing value test on single-sized aggregate passing 12.5 mm and retained on 10 mm sieve. A 40 tonne load is applied, and the fines passing 2.36 mm are expressed as a percentage of the sample weight. A lower value means a stronger aggregate.
Why the other options are wrong
A: 40-20 mm is too large for the standard mould. C: 4.75-2.36 mm is fine aggregate, not used for this test. D: 20-10 mm is not the specified single size.
Shortcut
Crushing and impact value tests both use 12.5 mm to 10 mm aggregate.
Common mistake
Confusing the test size (12.5-10 mm) with the 2.36 mm sieve used to collect the crushed fines.
Correct answer: B) passing 12.5 mm and retained on 10 mm sieve