Q21GATE 2009MCQ

During a CBR test, the load sustained by a remolded soil specimen at 5.00 mm penetration is 50 kg. The CBR value of the soil will be

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📖 Explanation

\begin{aligned} \text{CBR}(\%)&=\frac{\begin{array}{c}\text { Load sustained by }\\\text { specimend at 5.0 mm penetration} \end{array}}{\begin{array}{c}\text { Load sustained by standard }\\\text{aggregates at 5.0mm penetration} \end{array}} \times 100 \\\\ &=\frac{50}{2055} \times 100=2.4 \% \end{aligned}

Q22GATE 2009MCQ

Which of the flowing stress combinations are appropriate in identifying the critical condition for the design of concrete pavements ?

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Q23GATE 2008MCQ

It is proposed to widen and strength and existing 2-lane NH section as a divided highway. The existing traffic in one direction is 2500 commercial vehicles (CV) per day. The construction will take 1 year. The design CBR of soil subgrade is found to be 5 percent. Given : traffic growth rate for CV = 8 percent, vehicle damage factor =3.5 (standard axles per CV), design life = 10 years and traffic distribution factor =0.75. The cumulative standard axles (msa) computed are

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📖 Explanation

Number of commercial vehicles per day, A = existing traffic $\times$ traffic distribution factor $=2500 \times (1.08) \times 0.75=2025$ Annual growth rate of commercial vehicles =8 % Vehicle damage factor, F=3.5 Design life, n=10 years, Thus, the cumulative standard axles (msa) may be calculated as

\begin{aligned} N_{S}=& \frac{365 A\left[(1+r)^{n}-1\right]\$}{r} \times F \\ &=\frac{365 \times 2025 \times \left[\left(1+\frac{8}{100}\right)^{10}-1\right]\$}{\frac{8}{100} \times 3.5} \\ =& 37.47 \approx 37 \end{aligned}

Q24GATE 2007MCQ

The following data pertains of the number of commercial vehicles per day for the design of a flexible pavement for a national highway as per IRC:37-1984 : Assuming a traffic growth factor of 7.5% per annum for both the types of vehicles, the cumulative number of standard axle load repetitions (in million) for a design life of ten years is

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📖 Explanation

\begin{aligned} N_{9} &=\frac{365 A_{1}\left[(1+r)^{n}-1\right]\$}{r} \times F_{1} \\ &=\frac{365 \times 2000\left[\left(1+\frac{7.5}{100}\right)^{10}-1\right]\$}{\left(\frac{7.5}{100}\right) \times 10^{6}} \times 5 \\ &=51.6 \mathrm{msa} \\ N_{s 2} &=\frac{365 A_{2}\left[(1+r)^{n}-1\right]\$}{r} \times F_{2}\\ &= \frac{365 \times 200\left[\left(1+\frac{7.5}{100}\right)^{10}-1\right]\$}{\left(\frac{7.5}{100}\right) \times 10^{6}} \times 6 \\ &= 6.2 \\ N_{s}&= N_{9}+N_{s_{2}}=51.6+6.2 \\ &= 57.8 \mathrm{msa} \end{aligned}

Q25GATE 2007MCQ

The width of the expansion join is 20 mm in a cement concrete pavement. The paying temperature is $20^{\circ}C$ and the maximum slab temperature in summer is $60^{\circ}C$ . The coefficient of the thermal expansion of concrete is $10\times 10^{-6} mm/mm/^{\circ}C$ and the joint filler compresses up to 50% of the thickness. The spacing between expansion joints should be

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📖 Explanation

The spacing of expansion joint is given by,

\begin{aligned} L_{e} &=\frac{\delta^{\prime}}{100 C\left(T_{2}-T_{1}\right)} \\ \text { Given } \delta &=\frac{\text { Width of expansion joint }}{2} \\ &=\frac{20}{2}=10 \mathrm{mm} \\ L_{e} &=\frac{1}{100 \times 10 \times 10^{-6}(60-20)}=25 \mathrm{m} \end{aligned}

Q26GATE 2006MCQ

A subgrade soil sample was tested using standard CBR apparatus and the observations are given below. Assuming that the load-penetration curve is convex throughout, the CBR value(%) of the sample is

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📖 Explanation

\begin{aligned} \text{CBR}_{2.5}&=\frac{\begin{array}{c}\text { Load sustained by }\\\text { specimend at 2.5 mm penetration} \end{array}}{\begin{array}{c}\text { Load sustained by standard }\\\text{aggregates at 2.5mm penetration} \end{array}} \times 100 \\ &=\frac{60.5}{1370} \times 100=4.4 \%\\ \text{Also} \\ \text{CBR}_{5.0}&=\frac{\begin{array}{c}\text { Load sustained by }\\\text { specimend at 5.0 mm penetration} \end{array}}{\begin{array}{c}\text { Load sustained by standard }\\\text{aggregates at 5 mm penetration} \end{array}} \times 100 \\ &=\frac{80.5}{2055} \times 100=3.9 \%\\ \end{aligned}

Thus, the higher value of CBR which is obtained at 2.5 mm penetration is adopted i.e. 4.4%.

Q27GATE 2006MCQ

Using IRC:37-1984 "Guidelines for the Design of Flexible Pavements" and the following data, choose the total thickness of the pavement. NO. of commercial vehicles when construction is completed = 2723 veh/day Annual growth rate of the traffic =5% Design life of the pavement = 10 years Vehicle damage factor =2.4 CBR value of the subgrade soil=5%

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📖 Explanation

Number of standard axles in msa is given by,

\begin{aligned} N_{s} &=\frac{365 A\left[(1+r)^{n}-1\right]\$}{r \times 10^{6}}+F \\ &=\frac{365 \times 2723 \times \left[\left(1+\frac{5}{100}\right)^{10}-1\right]\$}{\frac{5}{100} \times 10^{6}} \times 2.4 \\ &=30 \mathrm{msa} \end{aligned}

Now, for 5 % CBR value, the total thickness of the pavement for 30 msa is given as 670 mm.

Q28GATE 2006MCQ

In case of governing equation for calculating wheel load stresses using Westergaard's approach, the following statements are made : I. Load stresses are inversely proportional to wheel load. II. Modulus of subgrade reaction is useful for load stress calculation

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Q29GATE 2005MCQ

The following observations were made of an axle-load survey on a road. The standard axle-load is 80 kN. Equivalent daily number of repetitions for the standard axle-load are

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📖 Explanation

From graph, $0 B^{\prime}=\frac{400 \times 5}{40}=50$ Equivalent daily number of Repetitions for the standard axle-load 80 are, 400+50=450

Q30GATE 2005MCQ

For a 25 cm thick cement concrete pavement, analysis of stresses gives the following values : Wheel load stress due to corner loading : 30 kg/c $m^{2}$ Wheel load stress due to edge loading : 32 kg/c $m^{2}$ Warping stress at corner region during summer : 9 kg/c $m^{2}$ Warping stress at corner region during winter : 7 kg/c $m^{2}$ Warping stress at edge region during summer : 8 kg/c $m^{2}$ Warping stress at edge region during summer : 6 kg/c $m^{2}$ Frictional stress during summer : 5 kg/c $m^{2}$ Frictional stress during winter : 4 kg/c $m^{2}$ The most critical stress value for this pavement is

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📖 Explanation

There are four cases which should be considered: (i) - During summer, critical combination of stresses at edges =wheel load stress due to edge loading +warping stress at edge region during summer - frictional stress during summer $= 32 + 8 - 5 = 35\text{ kg/cm}^{2}$ (ii) During winter, critical combination of stresses at edges. =wheel load stress due to edge loading +warping stress at edge region during winter + frictional stress during winter $=32 + 6 + 4 = 42 \text{kg/cm}^{2}$ (iii) During summer, critical combination of stresses at corner regions =wheel load stress at cor,ner loading + warping stress at corner region during summer $=30 +9=39 \text{kg/cm}^{2}$ (iv) During winter, critical combination of stresses at corner region =wheel load stress at corner loading + Warping stress at corner region during winter $=30+7=37 \mathrm{kg} / \mathrm{cm}^{2}$ Hence, the critical combination is $42 \mathrm{kg} / \mathrm{cm}^{2}$

Q31GATE 2004MCQ

The data given below pertain to the design of a flexible pavement : Initial traffic = 1213 cvpd Traffic growth rate = 8% per annum Design life = 12 years Vehicle damage factor = 2.5 Distribution factor = 1.0 The design traffic in terms of million standard axles (msa) to be catered would be

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📖 Explanation

The cumulative number of standard axle load,

\begin{aligned} N_{s} &=\frac{365 A\left[(1+r)^{n}-1\right]\$}{r} \times F \\ &=\frac{365 \times 1213 \times \left[\left(1+\frac{8}{100}\right)^{12}-1\right]\$}{8 / 100} \times 2.5 \times 10^{-6} \\ &=21.00 \mathrm{msa} \end{aligned}

Q32GATE 2004MCQ

In the context of flexible pavement design, the ratio of contact pressure to tyre pressure is called the Rigidity Factor. This factor is less than unity when the tyre pressure is

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Q33GATE 2003MCQ

The plate load test conducted with a 75 cm diameter plat on soil subgrade yielded a deflection of 2.5 mm under a stress of 800 $N/cm^{2}$ . The modulus of elasticity of the subgrade soil, in $kN/cm^{2}$ is

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📖 Explanation

The modulus of elasticity of the subgrade soil is given by,

\begin{aligned} E &=\frac{1.18 \mathrm{pa}}{\Delta} \\ &=\frac{1.18 \times 800 \times 75 \times 10}{2.5 \times 2} \\ &=141600 \mathrm{N} / \mathrm{cm}^{2}=141.6 \mathrm{kN} / \mathrm{cm}^{2} \end{aligned}

Q34GATE 2003MCQ

Temperature stresses in concrete pavements may cause the slab to crack. If a slab cools uniformly then the crack will develop at the which of the following locations of the slab ?

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Q35GATE 2002MCQ

Dowel bars in concrete pavement are placed

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Q36GATE 2002MCQ

The average daily traffic on a stretch of road is 300 commercial vehicles per lane per day. Design traffic repetitions for 10 years when vehicle damage factor is 2.5 and traffic growth rate is 7%, is

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📖 Explanation

\begin{aligned} N_{\mathrm{s}}&=365 A\left[\frac{(1+r)^{n}-1}{r}\right]\$ \times F\\ &=365 \times 300\left[\frac{(1+0.07)^{10}-1}{0.07}\right]\$ \times 2.5 \\ &=3.8 \mathrm{msa} \end{aligned}

Q37GATE 2002MCQ

In a concrete pavement (A) Temperature stress is tensile at bottom during day time (B) Load stress is compressive at bottom

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Q38GATE 2001MCQ

The radius of relatives stiffness for a 20 cm thick slab with $E = 3 \times 10^{5} kg/cm^{2}$ and poisson's ratio = 0.15, resting on a subgrade having modulus of 5 kg/ $cm^{3}$ is

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📖 Explanation

\begin{array}{l} \text { Given, } h=20 \mathrm{cm}, E=3 \times 10^{5} \mathrm{kg} / \mathrm{cm}^{2} \\ \mu=0.15, k=5 \mathrm{kg} / \mathrm{cm}^{3} \\ \qquad \begin{aligned} L &=\left[\frac{E h^{3}}{12 k\left(1-\mu^{2}\right)}\right]^{1 / 4} \\ &=\left[\frac{3 \times 10^{5} \times 20^{3}}{12 \times 5\left[1-(0.15)^{2}\right]\$}\right]^{1 / 4}\\ &79.98cm\approx80cm \end{aligned} \end{array}