If aggregates completely pass through a sieve of size 75 mm and are retained on a sieve of size 60 mm, the aggregates will be known as elongated aggregate if its length is not less than
Correct Answer: D. 121.5 mm
📚 Detailed Explanation: Elongation Threshold Calculation
The elongation criterion from IS 2386 (Part I) states that an aggregate particle is classified as elongated if its longest dimension exceeds 1.8 times (= 9/5 times) the mean sieve size of the size fraction it belongs to.
Step-by-step calculation:
1. Upper sieve = 75 mm, Lower sieve = 60 mm
2. Mean dimension = (75 + 60) ÷ 2 = 67.5 mm
3. Elongation threshold = 1.8 × 67.5 = 121.5 mm
So the particle is elongated if its length is not less than 121.5 mm. This matches option D perfectly.
1. Upper sieve = 75 mm, Lower sieve = 60 mm
2. Mean dimension = (75 + 60) ÷ 2 = 67.5 mm
3. Elongation threshold = 1.8 × 67.5 = 121.5 mm
So the particle is elongated if its length is not less than 121.5 mm. This matches option D perfectly.
Elongation Threshold Examples
| Sieve Range | Mean Dim (mm) | Elongation Limit = 1.8 × Mean (mm) |
|---|---|---|
| 25 mm / 20 mm | 22.5 | 40.5 |
| 40 mm / 31.5 mm | 35.75 | 64.35 |
| 75 mm / 60 mm | 67.5 | 121.5 |
Key Concepts for Students
- Elongation factor = 1.8 (or 9/5); Flakiness factor = 0.6 (or 3/5) — memorise both.
- The elongation index test uses a length gauge with slots of width equal to 1.8 × mean sieve size for each size fraction.
- Combined flakiness + elongation index must be ≤40% per IS 383 for concrete aggregate.