# Pavement Design Questions and Answers – Highway Materials – Stone Aggregate Tests- 2

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This set of Pavement Design Questions and Answers for Experienced people focuses on “Highway Materials – Stone Aggregate Tests- 2”.

1. What type of pavement can aggregate with an aggregate impact value of 26% not be used in?
a) Cement concrete base course
b) Bitumen bound macadam base course
c) Bituminous concrete surface course
d) Water bound macadam base course

Explanation: For the bituminous concrete surface course, the impact value should not exceed 24%, so the sample in the question cannot be used. For cement concrete base course, it must not exceed 45%, bitumen bound macadam base course 35% and for water bound macadam base course 40%.

2. In a length gauge used for determination of elongation index, what is the length of gauge size for sieve sets 25 mm and 20 mm?
a) 25.65
b) 40.5
c) 81
d) 32.4

Explanation: The length of gauge size is obtained by multiplying the average of sieve sets with 1.8. So, for particles passing 25 mm sieve and retained on 20 mm sieve, the length of the gauge size would be =$$\frac{25+20}{2}$$×1.8=40.5 mm.

3. What is the weight of the charge used in the Los Angeles abrasion test?
a) 416.67 mg
b) 450.55 mg
c) 450.55 g
d) 416.67 g

Explanation: As per IS 2386 part 4, the steel balls are used as the charge for the test. They are of 48 mm diameter and each ball weighs around 416.67 g. Depending on the grading used for the test, the number of balls used changes.

4. The following particulars are obtained after conducting the test for angularity number:
Volume of cylinder = 3000 g
Specific gravity = 2.68

Sample number Weight of aggregate (g)
1 5350
2 5350
3 5400

What is the angularity number for the above sample?
a) 0
b) 0.5
c) 0.4
d) 1

Explanation: Angularity number is obtained using the formula, AN=$$67-\frac{100W}{CG_a}$$ Where W represents the mean weight of aggregate (g),
C represents the volume of the cylinder (g) and,
Ga represents the specific gravity
From the readings, mean value, W = $$\frac{5350+5300+5400}{3}$$=5350 g
AN=$$67-\frac{100W}{CG_a}=67-\frac{100×5350}{3000×2.68}=0.457\approx1$$
The angularity number is represented as a whole number and is rounded off to nearest one.

5. There are six types of tests to conduct the adhesion test on aggregates.
a) True
b) False

Explanation: The tests to determine the stripping value can be classified into six types. They are static immersion test, dynamic immersion test, chemical immersion test, immersion mechanical test, immersion trafficking test and coating test.

6. The result for soundness test is expressed in terms of ______
a) Percentage density gain
b) Percentage density loss
c) Percentage weight loss
d) Percentage weight gain

Explanation: To conduct the soundness test, the aggregates are subjected to solutions of sodium sulphate or magnesium sulphate as per IS 2386. After cycles of immersion and drying, the loss in weight of aggregate is noted down in percentage and that gives the soundness result.

7. The combined index is obtained by ¬______ the flakiness and elongation index.
a) Subtracting
c) Taking average
d) Multiplying

Explanation: Combined index is obtained as the sum of both flakiness and elongation index. The sample is first subjected to flakiness test, then the non-flaky particles are subjected to the elongation test. Their indices are found out and added together.

8. What is used as a charge in the Dorry abrasion test?
a) Sand
b) Steel balls
c) Steel shavings
d) Crushed gravel

Explanation: Dorry abrasion test was used way back and is not used these days. The charge used for the test was sand. It was let into the machine to cause abrasive action. Crushed gravel cannot be used as it is the same material and it would cause attrition, not abrasion.

9. After the abrasion test, the sample is passed through which sieve?
a) 2.36 mm
b) 1.45 mm
c) 1.7 mm
d) 2.5 mm

Explanation: The sample after removing from the abrasion machine is passed through 1.7 mm sieve and the weight is noted down. 2.36 mm is used in the impact and crushing test on aggregates.

10. The aggregate impact test was conducted on a sample and the following readings were taken:
Weight of sample taken () = 350 g
Weight of sample passing 2.36 mm sieve () = 92 g
What will be the aggregate impact value for the sample?
a) 26%
b) 26 g
c) 26.29 g
d) 26.29%

Explanation: The aggregate impact value is obtained as the ratio of the weight of sample passing 2.36 mm sieve to the weight of the sample taken for the test. For the readings in the question, the aggregate impact value can be found out as =$$\frac{92}{350}=26.29\approx26$$ % and is expressed as a whole number in percentage.

11. Which of the below test – procedure pairs are matched correctly?
a) Impact test – 2 layers tamped 25 times each
b) Angularity number – 3 layers tamped 100 times each
c) Bulk density – 2 layers tamped 25 times each
d) Crushing value – 3 layers tamped 50 times each

Explanation: As per IS 2386 part 3, in the impact test, the sample is filled in three layers and given 25 tamping each. As per IS 2386 part 1, the sample is filled in three layers and tamped 100 times each for the angularity number. As per IS 2386 part 3, in the bulk density test, the sample is filled in three layers and given 25 tamping each. As per IS 2386 part4, the sample for crushing test is filled in three layers and tamped 25 times each.

12. The polished stone value of an aggregate is reported as the ______ of the two values of ______
a) Mean, abrasion number
b) Mean, skid number
c) Sum, skid number
d) Sum, abrasion number

Explanation: There are two stages conducted for this test. The sample is polished in the first stage and subjected to pendulum type friction tester in the second stage. The result obtained is the skid number and the mean of two values of the same is taken as the polished stone value.

13. What would be the width of a slot in the thickness gauge if the sieve sets are 20 mm and 16 mm?
a) 20.2 mm
b) 32.4 mm
c) 10.8 mm
d) 13.5 mm

Explanation: The width of the slot is obtained by multiplying the mean of the sieve sets with 0.6. So, for the sieve sets 20 mm and 16 mm, the width of the slot would be =$$\frac{20+16}{2}$$×0.6=10.8 mm.

14. The details of the Los Angeles test conducted on a sample is given below:
Weight of sample taken = 5 kg
Weight of sample retained = 3.16 kg
What is the Los Angeles abrasion value for the sample?
a) 36.81%
b) 37%
c) 36.8%
d) 36%

Explanation: The value is found as the ratio of loss in mass/wear to the original weight and it is expressed in percentage. It is not rounded off to the nearest whole number. The value is obtained as =$$\frac{5-3.16}{5}$$×100 %=36.8 %

15. The friction coefficient of coarse aggregate should be more than 55.
a) True
b) False