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Question 1
ప్రశ్న 1
For a proposed flexible pavement, the latest two-way commercial-vehicle count is 800 vehicles/day. Construction will finish after three years, and traffic is assumed to grow at 5% annually. Using A = P(1+r)^x, what is the opening-year count?
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Explanation:
Opening-year traffic is projected from the most recent measured count. The adopted expression is A = P(1+r)^x. Substituting the given data gives A = 800(1.05)^3 = 926.1 vehicles/day. Therefore approximately 926 commercial vehicles per day are expected at opening. This value is the starting traffic input for a later cumulative-loading computation, not the cumulative design traffic itself.
వివరణ:
Question 2
ప్రశ్న 2
Commercial traffic at the completion of construction is estimated as 1210 vehicles/day following two years of 10% annual growth. Using A = P(1+r)^x, what was the commercial-vehicle count at the last survey?
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Explanation:
The opening-year count includes two years of compounding from the survey date. Rearranging the relation gives P = A/(1+r)^x. Substitution yields P = 1210/(1.10)^2. The denominator is 1.21, so the original count was 1000 commercial vehicles/day. Applying the forward growth calculation to 1000 reproduces the given 1210 and independently checks the answer.
వివరణ:
Question 3
ప్రశ్న 3
A traffic count of 1000 commercial vehicles/day increases to 1210 vehicles/day over two years. Assuming a constant compound annual growth rate and no change in the counting basis, what is that rate?
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Explanation:
The growth relation is A/P = (1+r)^x. The ratio of the final to initial count is 1210/1000 = 1.21. Since the elapsed period is two years, 1+r = sqrt(1.21) = 1.10. Consequently r = 0.10, or 10% per year. This is a compound annual rate and must not be confused with simply dividing the 21% two-year growth by two.
వివరణ:
Question 4
ప్రశ్న 4
Both a single-lane two-way road and an ordinary undivided two-lane, two-way road carry 1200 commercial vehicles/day in both directions combined. Using IRC:37-2018 default distribution factors of 1.00 and 0.50 respectively, what is the ratio of the design-lane daily commercial traffic for the single-lane road to that for the two-lane road?
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Explanation:
Single-lane roads experience strongly channelized wheel-path repetitions. With the stated factor of 1.00, the first design count is 1200 vehicles/day. For the undivided two-lane road, the factor of 0.50 gives 600 vehicles/day. The required ratio is 1200/600 = 2. This comparison illustrates why the carriageway arrangement matters even when the total two-way traffic counts are identical.
వివరణ:
Question 5
ప్రశ్న 5
An intermediate-width undivided road has a 5.50 m carriageway and carries 1600 commercial vehicles/day in both directions. Applying the 0.75 lateral distribution factor specified for this class in IRC:37-2018, what commercial traffic is considered for design?
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Explanation:
The question describes the 5.50 m intermediate-width road category. The relevant specified factor is 0.75 of the total two-way commercial traffic. Multiplying 1600 by 0.75 gives 1200 commercial vehicles/day. The ordinary undivided two-lane factor of 0.50 should not be substituted for this different road class. Cumulative standard-axle loading would additionally require the growth factor, VDF and design period.
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Question 6
ప్రశ్న 6
An undivided four-lane single-carriageway road carries 2600 commercial vehicles/day in both directions combined. IRC:37-2018 specifies a design distribution factor of 0.40 for this road arrangement. What is the applicable design traffic in commercial vehicles/day?
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Explanation:
The four lanes in this question belong to one undivided carriageway. IRC:37-2018 uses 40% of its total two-way commercial-vehicle count for the design loading calculation. The daily design count is 2600 times 0.40 = 1040 vehicles. This differs from the factor used for a four-lane divided highway with two lanes per direction. The geometric distinction must be recognized before applying the traffic factor.
వివరణ:
Question 7
ప్రశ్న 7
A divided highway carries 6000 commercial vehicles/day in both directions combined. The heavier direction carries 55%, and the carriageway has two lanes in that direction. Using the IRC:37-2018 factor of 0.75 of directional traffic, determine the design-lane commercial count.
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Explanation:
Design traffic for a divided carriageway must first be assigned to the appropriate direction. The heavier direction carries 6000 times 0.55 = 3300 commercial vehicles/day. The stated two-lane directional distribution factor is 0.75. Multiplying 3300 by 0.75 gives 2475 vehicles/day for structural design. It would be incorrect to use 0.75 of the undivided two-way total or apply the 55% share a second time.
వివరణ:
Question 8
ప్రశ్న 8
A dual three-lane carriageway has three lanes in each direction. A particular direction carries 2500 commercial vehicles/day at opening. With the IRC:37-2018 directional lane-distribution factor of 0.60, what daily traffic is assigned for pavement design?
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Explanation:
The question supplies traffic for one direction, not the combined two-way road. Under the stated guidance, the factor for a dual three-lane carriageway is 0.60 of the relevant directional count. Multiplication gives 2500 times 0.60 = 1500 vehicles/day. No additional half-of-two-way factor is appropriate because the input is already directional. The result is a commercial-vehicle count that must still be converted into equivalent axle repetitions for design.
వివరణ:
Question 9
ప్రశ్న 9
A divided highway has four lanes in each direction, with 3600 commercial vehicles/day travelling in the design direction. If the applicable IRC:37-2018 lateral distribution factor is 0.45, what is the design-lane traffic?
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Explanation:
The stated 3600 commercial vehicles/day is the traffic for the design direction. For a dual four-lane carriageway, the relevant specified distribution factor is 0.45. Multiplying gives 3600 times 0.45 = 1620 commercial vehicles/day. It is not appropriate to apply the 0.60 factor that belongs to a dual three-lane carriageway. Directional data and the correct number of lanes must both be identified before computing this input.
వివరణ:
Question 10
ప్రశ్న 10
In an axle-load survey, 60% of commercial vehicles have an assumed VDF of 2.0 and the remaining 40% have an assumed VDF of 5.0. The proportions are based on vehicle counts. What is the weighted average VDF?
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Explanation:
A weighted VDF must reflect the relative number of vehicles in each surveyed group. The first contribution is 0.60 times 2.0 = 1.2. The second contribution is 0.40 times 5.0 = 2.0. Their sum is 3.2 equivalent standard axles per commercial vehicle. Using a simple unweighted average would be incorrect because the two categories contain unequal shares of the traffic.
వివరణ:
Question 11
ప్రశ్న 11
An undivided two-lane, two-way road carries 1200 commercial vehicles/day in both directions. Surveys yield directional VDF values of 3.0 and 5.0, and the design adopts the higher VDF of 5.0 for the common carriageway. With D = 0.50, how many equivalent standard axles/day are represented for the design calculation?
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Explanation:
The total two-way traffic for the undivided road is 1200 commercial vehicles/day. The adopted lateral distribution factor is 0.50, giving 600 design commercial vehicles/day. The question explicitly requires the higher directional VDF of 5.0 to be used for the shared pavement. Therefore the daily equivalent loading is 600 times 5.0 = 3000 standard-axle repetitions. The result is a conservative computation under the stated VDF-selection assumption and is not a physical count of individual axles.
వివరణ:
Question 12
ప్రశ్న 12
One tyre carries a vertical load of 20 kN on a flexible pavement at an assumed uniform contact pressure of 0.50 MPa. What equivalent tyre-contact area is obtained from A = P/p?
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Explanation:
Contact area equals the applied vertical load divided by the assumed pressure. The tyre load of 20 kN equals 20,000 N. The specified pressure of 0.50 MPa equals 500,000 N/m². Division gives 20,000/500,000 = 0.040 m². The result is an equivalent uniformly loaded contact area rather than a description of the exact shape of a real tyre footprint.
వివరణ:
Question 13
ప్రశ్న 13
An 80 kN standard single axle is fitted with dual tyres at both ends, giving four tyre contact patches. Under the simplified assumption that all four tyres share the axle load equally, what load acts through each tyre?
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Explanation:
The specified single axle supports a total vertical load of 80 kN. Dual tyres at each of its two ends give four tyre contact patches. The problem assumes uniform sharing of the load among these four tyres. Accordingly each tyre carries 80/4 = 20 kN. Real load distribution can differ, but the calculation is exact under the stated equal-sharing assumption.
వివరణ:
Question 14
ప్రశ్న 14
An undivided two-lane, two-way road has an opening-year traffic count of 950 commercial vehicles/day. Using a design factor D = 0.50 and average VDF = 3.6, what is the corresponding opening-year equivalent standard-axle loading per day?
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Explanation:
The daily design count is obtained by applying the lateral distribution factor to the given total commercial traffic. That count is 950 times 0.50 = 475 vehicles/day. Each design vehicle represents an average of 3.6 standard axles under the stated VDF. Multiplying 475 by 3.6 produces 1710 equivalent standard axles/day. This is a daily loading figure and has not yet been accumulated over the pavement design life.
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Question 15
ప్రశ్న 15
For a constant-traffic illustrative design, an undivided road has 1400 opening-year commercial vehicles/day in both directions, D = 0.50, VDF = 4.0, and a design period of 15 years. With zero annual growth, what is the cumulative loading in million standard axles?
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Explanation:
At zero growth the cumulative standard-axle count is 365 times A times D times F times n. Here the initial design-lane equivalent loading is 1400 times 0.50 times 4.0 = 2800 standard axles/day. Multiplying by 365 and 15 years gives 15,330,000 repetitions. Expressing this quantity in millions gives 15.33 msa. The no-growth expression is the zero-growth limit of the usual geometric-growth calculation.
వివరణ:
Question 16
ప్రశ్న 16
Using the IRC:37-2018 cumulative-traffic expression N = 365 A D F [(1+r)^n - 1]/(r × 10^6), calculate N for opening-year two-way traffic A = 1200 CVPD, D = 0.50, F = 4.0, r = 0.05 and n = 10 years.
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Explanation:
The traffic growth multiplier is [(1.05)^10 - 1]/0.05 = approximately 12.5779. The initial design-lane equivalent loading is 1200 times 0.50 times 4.0 = 2400 standard axles/day. Multiplying 2400 by 365 and the growth multiplier gives approximately 11,018,234 standard axles. Dividing by one million yields 11.018 msa to three decimals. The count is explicitly opening-year traffic, so no additional preconstruction growth factor is used.
వివరణ:
Question 17
ప్రశ్న 17
A traffic survey records 1000 two-way commercial vehicles/day two years before pavement opening. Assume 5% annual growth both before opening and throughout the following 10-year design period, D = 0.50 and VDF = 3.5. Applying A = P(1+r)^x followed by the IRC cumulative-traffic expression, what is N to three decimals?
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Explanation:
Traffic must first be advanced from the survey year to the opening year. The opening count is A = 1000(1.05)^2 = 1102.5 commercial vehicles/day. For ten years at 5% growth, the accumulation factor is [(1.05)^10 - 1]/0.05 = approximately 12.5779. Therefore N = 365 times 1102.5 times 0.50 times 3.5 times 12.5779 divided by one million. This gives approximately 8.858 msa, with preconstruction growth included only in the opening-year count.
వివరణ:
Question 18
ప్రశ్న 18
A flexible pavement was initially calculated to carry 12 msa using VDF = 3.0. If the traffic count, growth, distribution factor and design period remain unchanged but the adopted VDF increases to 4.0, what is the revised cumulative loading?
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Explanation:
The cumulative design-traffic expression is directly proportional to VDF when its other inputs are fixed. The old calculation used a VDF of 3.0. The new VDF is 4.0, so the loading multiplier is 4/3. Applying this to 12 msa gives 12 times 4/3 = 16 msa. This comparison highlights why an axle-load survey can materially alter pavement design loading even without any change in commercial-vehicle counts.
వివరణ:
Question 19
ప్రశ్న 19
A zero-growth design has N = 7.30 msa, opening-year two-way traffic A = 1000 CVPD, a distribution factor D = 0.50 and a design period of 10 years. What average VDF must have been used?
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Explanation:
For zero growth the cumulative loading in standard axles is 365 A D F n. Rearranging gives F = N times 10^6 divided by (365 A D n). Substituting yields F = 7,300,000/(365 times 1000 times 0.50 times 10). The denominator is 1,825,000, so F = 4.0. This inverse calculation recovers the assumed average equivalent axle damage per design commercial vehicle.
వివరణ:
Question 20
ప్రశ్న 20
An undivided two-lane road carries 1000 opening-year CVPD in both directions, with D = 0.50 and VDF = 4.0. If zero traffic growth is assumed and the required cumulative loading is 14.60 msa, what design period satisfies the calculation?
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Explanation:
With constant traffic, cumulative standard-axle loading equals 365 A D F n. The daily equivalent loading is 1000 times 0.50 times 4.0 = 2000 standard axles. The annual equivalent loading is therefore 730,000 standard axles. Dividing the target 14,600,000 by 730,000 gives 20 years. This is an arithmetic design-period inference and does not establish the acceptability of any particular pavement layer arrangement.
వివరణ:
Question 21
ప్రశ్న 21
In a laboratory CBR test, the measured penetration load at 2.5 mm is 2.648 kN and the corresponding reference load is 13.24 kN. What is the CBR at that penetration?
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Explanation:
The CBR at a specified penetration is the measured load divided by the standard reference load at that penetration. Both values are given in kN, so the units cancel. The load ratio is 2.648/13.24 = 0.20. Multiplying by 100 gives a CBR of 20%. This calculation refers to the 2.5 mm result and should not be confused with the final design CBR selection process.
వివరణ:
Question 22
ప్రశ్న 22
Three laboratory specimens prepared using the same subgrade soil under comparable prescribed conditions give CBR values of 5%, 6% and 7%. What is their arithmetic mean CBR for the material characterization step?
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Explanation:
The arithmetic mean is the total of the three specimen values divided by the number of specimens. The sum is 5 + 6 + 7 = 18 percentage points. There are three specimens, so the mean is 18/3 = 6%. This is the mean for the specified soil specimens only. Selecting an overall design CBR for a road stretch may require additional sampling and application of the relevant IRC design-value procedure.
వివరణ:
Question 23
ప్రశ్న 23
For a subgrade CBR of 4.5%, estimate resilient modulus using the IRC:37-2018 relationship MR = 10 × CBR (MPa), applicable when CBR does not exceed 5%.
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Explanation:
The specified CBR of 4.5% falls within the range at or below 5%. The relevant IRC estimate is MR = 10 times the numerical CBR percentage in MPa. Substituting 4.5 gives 10 times 4.5 = 45 MPa. The result is a resilient modulus estimate rather than a CBR expressed in pressure units. A different power-law correlation is used for CBR values above 5% in this guidance.
వివరణ:
Question 24
ప్రశ్న 24
Using the IRC:37-2018 empirical relation MR = 17.6(CBR)^0.64 MPa for CBR greater than 5%, estimate subgrade resilient modulus for CBR = 10%, to one decimal place.
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Explanation:
A CBR of 10% lies above the 5% threshold for the stated correlation. The appropriate expression is MR = 17.6 times (10)^0.64 MPa. The value of (10)^0.64 is approximately 4.3652. Multiplication by 17.6 gives approximately 76.83 MPa. The requested one-decimal result is 76.8 MPa and should not be obtained using the lower-CBR linear relation.
వివరణ:
Question 25
ప్రశ్న 25
An illustrative flexible-pavement trial section has an assumed total layer thickness of 610 mm above the prepared subgrade. The proposed wearing course is 40 mm, bituminous binder/base course is 100 mm and granular base is 250 mm. What remaining thickness is available for the granular sub-base, if all four layer thicknesses must sum to 610 mm?
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Explanation:
The total thickness is defined as the sum of the four proposed layers above the subgrade. The three specified thicknesses add to 40 + 100 + 250 = 390 mm. Subtracting 390 mm from the total of 610 mm gives 220 mm. That is the remaining granular sub-base thickness for the stated section. This arithmetic balance does not establish structural adequacy, which still requires the applicable material, fatigue, rutting and drainage checks.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. Approximately 926 vehicles/day. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. 1000 vehicles/day. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. 10%. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. 2:1. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: B. 1200 vehicles/day. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. 1040. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. 2475 vehicles/day. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 1500 vehicles/day. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. 1620 vehicles/day. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: D. 3.2. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: B. 3000. సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 0.040 m². సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: D. 20 kN. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: A. 1710 standard axles/day. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: C. 15.33 msa. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. 11.018 msa. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 8.858 msa. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. 16 msa. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 4.0. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: B. 20 years. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: B. 20%. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 18%. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 45 MPa. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. 76.8 MPa. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: C. 220 mm. సమాధానం: C.