Road Aggregates – Minimum Coefficient of Hardness for Wearing Courses For stones intended for use in road work (e.g., wearing course metal), the coefficient of hardness should be greater than which minimum value to ensure adequate resistance to abrasion and attrition under traffic?

Difficulty: Easy

Correct Answer: 17

Explanation:


Introduction:
Aggregates for road surfaces experience repeated abrasion and impact from traffic loads. The coefficient of hardness is a practical index indicating a stone’s ability to resist wear. This question checks the minimum acceptable value typically cited for road metal in standard texts and practice questions.


Given Data / Assumptions:

  • Application: wearing course aggregates in pavements.
  • Index: coefficient of hardness derived from standardized abrasion/attrition tests.
  • Objective: ensure adequate field performance and service life.


Concept / Approach:

A higher coefficient indicates better resistance to abrasion and attrition, which is vital to maintaining surface texture, skid resistance, and thickness under traffic. Traditional examination references prescribe a threshold value beyond which the material is considered fit for road use.


Step-by-Step Solution:

1) Recognize the role of hardness in resisting polishing and wear.2) Relate hardness indices to field durability in macadam and bituminous mixes.3) The accepted benchmark in many syllabi for road metal is a coefficient of hardness greater than 17.4) Therefore, choose 17 as the minimum threshold value.


Verification / Alternative check:

Historic laboratory manuals for highway materials list minimum hardness requirements around this value for stone aggregates used in wearing courses.


Why Other Options Are Wrong:

Values 10–15 are too low for durable wearing surfaces; 20 is more stringent than the commonly cited minimum and may reject otherwise serviceable aggregates.


Common Pitfalls:

Confusing coefficient of hardness with other indices like impact value or Los Angeles abrasion value; memorizing the wrong threshold due to similar-looking numbers.


Final Answer:

17

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