Question 1

2/-0.33

What will be the theoretical maximum capacity (to nearest 10 units) for a single lane of highway given that the speed of the traffic stream is 40 kmph?

SOLUTION

Concept:

Theoretical maximum capacity is given by,

\(c = \frac{{1000\;v}}{s}\)

Where,

c = is capacity is vehicle/hr

v = speed in km/hr

s = minimum clear distance between two vehicles

∴ s = 0.2 v + 6

Calculation:

\(c = \frac{{1000\;v}}{{\left( {0.2\;v + 6} \right)}}\)

\(\therefore c = \frac{{1000\times 40}}{{\left( {0.2\times 40 + 6} \right)}}\)

⇒ c = 2857.14 ≈ 2860 veh/h

Question 2

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In a abrasion test loss in weight is 27 gms, then coefficient of hardness will be

SOLUTION

Concept:

Coefficient of hardness is calculated from Dorry abrasion test. It is an indication of the hardness i.e. resistance against the external weathering of aggregates.

It is given by, Coefficient of Hardness ⇒ \(20 - \frac{{Loss\;in\;weight}}{3}\)

For a good aggregate, it should not be less than 17.

For a medium quality aggregate, it is between 14 to 17.

Aggregates having a coefficient of hardness less than 14 are termed to be too soft for road work.

Calculation:

Given loss in weight = 27 gm∴ \({\rm{Coefficient\;of\;hardness}} = 20 - \frac{{27}}{3} = 11\)

Question 3

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Consider the following two statements in connection to passenger car unit (PCU) in geometric design of a highway and chose the correct option.

Statement A: PCU of a bus is 3 implies that a bus has an average running speed three times that of a passenger car.

Statement B: In general, PCU increases with increase in vehicle length and with decrease in average running speed of vehicle.

SOLUTION

Concept:

Passenger Car Unit (PCU) or Passenger Car Equivalent (PCE):

  • In transportation engineering, a Passenger Car Unit is a measure of the impact (disturbance on traffic flow) that a class of vehicles has on traffic variables (such as headway, speed, density) compared to a single standard passenger car.
  • PCU of a bus is 3 indicates a bus, due its size and average running speed, will cause the same impact on traffic flow that is introduced by the movement of 3 passenger cars.
  • In general, as the size or length of a vehicle class increases the inconvenience and disturbance caused by them on a traffic flow increase, hence PCU also increases.
  • Also, if the average running speed of a vehicle class is less then it will bring inconvenience, slowness and disturbance on traffic flow, hence PCU increases. 

Question 4

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Spot speed survey was conducted at a particular location and following data is obtained

i) Maximum traffic volume of a week = 400000 PCU/week

ii) Average traffic volume of a week = 280000 PCU/week

iii) Maximum traffic volume of a day = 36000 PCU/day

iv) Average traffic volume of a day = 24000 PCU/day

v) Average traffic volume of an hour = 4000 PCU/hour

The daily expansion factor for the given data is

SOLUTION

Concept:

Periodic Volume Counts – are used to calculate expansion factors needed to estimate the annual traffic volume.

Hourly expansion factor \( = \frac{{Total\;vol.\;\;for\;24\;hr - period}}{{Vol.\;\;for\;particular\;hour}}\)

Daily expansion factor \( = \frac{{Av.\;\;total\;vol.\;\;for\;a\;weak}}{{Av.\;\;vol.\;\;for\;particular\;day}}\)

Monthly expansion factor \(= \frac{{AADT}}{{ADT\;for\;particular\;month}}\)

Thus if 24 hr count at a location is done and hourly volume is calculated, we can calculate the hourly expansion factor for each hour.

Calculation:

Daily expansion factor \(= \frac{{Average\;total\;volume\;for\;a\;week}}{{Average\;total\;volume\;for\;a\;day}} = \frac{{280000}}{{24000}}\) = 11.67

Question 5

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A man starts his journey from home to office every day. He travels 500 m on foot to reach the nearest bus stop. He then takes the bus to the nearest metro station and get off there. Then by metro he travels another 30 km and reaches home finally by rickshaw. The total number of trips is

SOLUTION

Concept:

Trip Generation:

  • A trip is defined as the one-way movement having a single purpose and mode of travel between a point of origin & a point of destination.


Calculation:

The man has one origin and one destination with one way movement.

Total mode of travel used = Walk + Bus + Metro + Rickshaw

⇒ Total number of trips = 4.

Question 6

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A plate bearing test was carried out on a subgrade using an 80 cm diameter rigid plate. A deflection of 9.6 × 10-4 m was caused by a pressure of 0.69 kg/cm2. The modulus of a subgrade reaction is _______ × 105 kg/m3.

SOLUTION

Concept:

The modulus of subgrade reaction for 80 cm diameter rigid plate

\({K_1} = \frac{P}{{\rm{\Delta }}}\)

Where P = Pressure

Δ = Deflection

Calculation:

\({K_1} = \frac{{0.69\;kg/c{m^2}}}{{9.6 \times {{10}^{ - 2}}cm}}\)

K1 = 7.1875 kg/cm3

Modulus of subgrade for standard plate of diameter 75 cm

\(K = \frac{{{K_1}{d_1}}}{d}\)

\(= \frac{{7.1875 \times 80}}{{75}}\)

= 7.667 kg/cm3

= 7.662 × 106 kg/m3= 7.667 × 105 kg/m3

Question 7

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Match the following

 List I

 List II

 1. 

 Wrapping crack

 P.

 Settlement of fill and sliding of side slopes.

 2.

 Longitudinal cracking

 Q.

 Due to excessive stresses slab develops

 cracking at the edge in an irregular pattern

 3.

 Shrinkage crack

 R.

 Develop during curing operation of pavement

 immediately after construction.

 4.

 Reflection cracking

 S. 

 Observed in bituminous overlays provided over

 existing cement concrete pavements.

SOLUTION

Wrapping crack - Develop during curing operation of pavement immediately after construction.

Longitudinal cracking - Settlement of fill and sliding of side slopes.

Shrinkage crack - Due to excessive stresses slab develops cracking at the edge in an irregular pattern

Reflection cracking - Observed in bituminous overlays provided over existing cement concrete pavements.

Flexible pavement failures: -

  1. Alligator (map) cracking
  2. Consolidation of pavement layer
  3. Shear failure
  4. Longitudinal cracking
  5. frost Heaving
  6. Reflection cracking
  7. Lack of binding to the lower course
  8. Formation of waves and corruption.

Rigid Pavement Failures: -    

  1. scaling of cement concrete
  2. shrinkage crack
  3. spalling of Joints
  4. Wrapping cracks
  5. Mud Pumping
  6. Structural cracks

 

Question 8

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Which one of the following test is performed in the laboratory to detect overheated or cracked bitumen and are more sensitive than the other test in terms of detection of crack?

SOLUTION

  • Softening point denotes the temperature at which the bitumen attains a particular degree of softening under the specifications of the test. The test is conducted by using the Ring and Ball apparatus.
  • Normally the consistency or stiffness of bituminous material can be measured either by penetration test or viscosity test. But for a certain range of consistencies, these tests are not applicable and Float test is used.
  • The bitumen content of a bituminous material is measured by means of its solubility in carbon disulfide and hence can be considered as purity test for bitumen. In the standard test for bitumen content (ASTM D4), a small sample of about 2 g of the asphalt is dissolved in 100 ml of carbon disulfide and the solution is filtered through a filtering mat in a filtering crucible.
  • Spot test is used to find out cracking in bitumen. It is performed in the laboratory to detect overheated or cracked bitumen and are more sensitive than the Solubility test in terms of detection of cracking

Question 9

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A bitumen concrete mix has average specific gravity of material 2.525 and theoretical specific gravity of 2.620. The density of bitumen used is 1.03 g/cc and percent bitumen content by weight in the mix is 4.5. The percent of volume voids in mineral aggregate and percent volume filled by bitumen respectively are _________.

SOLUTION

Concept:

Percentage of air voids (Vr) \(= \frac{{{{\rm{G}}_{\rm{t}}} - {{\rm{G}}_{\rm{m}}}}}{{{{\rm{G}}_{\rm{t}}}}} \times 100\)

Percentage of volume bitumen (Vb) \( = {{\rm{w}}_{\rm{b}}} \times \frac{{{{\rm{G}}_{\rm{m}}}}}{{{{\rm{G}}_{\rm{b}}}}}\)

Percentage of voids in mineral aggregate (VMA) = Vb% + VV%

Calculation:

\({{\rm{V}}_{\rm{v}}} = \frac{{2.620 - 2.525}}{{2.620}} \times 100 = 3.623{\rm{\;\% }}\)

\({{\rm{V}}_{\rm{b}}} = 4.5 \times \frac{{2.525}}{{1.08}} = 11.03{\rm{\;\% }}\)

VMA = Vb + VV = 11.03 + 3.623 = 14.65%

Volume filled by bitumen (VFB) = \(\rm{\frac{V_b \times 100}{VMA}}\)

\(\rm{V_{FB}=\frac{11.03 \times 100}{14.65}=75.29\%}\)

Question 10

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The following data pertains to the number of commercial vehicles per day for the design of a flexible pavement for a national highway as per IRC: 37-1984.

Type of commercial vehicle

Axle load (kN)

Number of (cv/day)

Single axle with single

wheel on either side

55

200

60

250

Single axle with dual

wheels on either side

65

150

75

200

Tandem axle with dual

wheels on either side

80

230


Calculate the equivalent vehicle damage factor for the given above data.

SOLUTION

Concept:

Vehicle damage factor (VDF) for different conditions is given by:

Single axle with single wheel on either side \(= {\left( {\frac{{{\rm{axle\;load\;in\;kN}}}}{{65}}} \right)^4}\) 

Single axle with dual wheels on either side \(= {\left( {\frac{{{\rm{axle\;load\;in\;kN}}}}{{80}}} \right)^4}\) 

Tandem axle with dual wheel on either side \(= {\left( {\frac{{{\rm{axle\;load\;in\;kN}}}}{{148}}} \right)^4}\) 

Tridem axle with dual wheel on either side \(= {\left( {\frac{{{\rm{axle\;load\;in\;kN}}}}{{224}}} \right)^4}\) 

Calculation:

\({\rm{VDF}} = \frac{{{{\rm{N}}_1}}}{{{\rm{\Sigma N}}}} \times {\left( {\frac{{{\rm{axle\;load}}}}{{{\rm{standard\;axle\;load}}}}} \right)^4}\)

\({\rm{VDF}} = \frac{{200}}{{1030}} \times {\left( {\frac{{55}}{{65}}} \right)^4} + \frac{{250}}{{1030}} \times {\left( {\frac{{60}}{{65}}} \right)^4} + \frac{{150}}{{1030}} \times {\left( {\frac{{65}}{{80}}} \right)^4} + \frac{{200}}{{1030}} \times {\left( {\frac{{75}}{{80}}} \right)^4} + \frac{{230}}{{1030}} \times {\left( {\frac{{80}}{{148}}} \right)^4}\)VDF = 0.508

Question 11

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An observer counts 360 vehicles/hour arriving at a spot on highway location. Assume that the arrival of vehicles at this location follows Poisson’s distribution, the probability of 3 or more vehicles arriving over 20 seconds interval is

SOLUTION

Concept:

Poisson’s Distribution model:

  • To find out the probability of having ‘n’ vehicles arriving in time ‘t’ Poisson’s distribution model is followed.

As per this model probability of ‘n’ vehicles arriving in time ‘t’ is given by, 
\(P\left( n \right) = \frac{{{{\left( {\lambda t} \right)}^n}\;{e^{ - \lambda t}}}}{{n!}}\)

Where P(n) = Probability of n-vehicles arriving in time ‘t’.

λ = Average vehicular flow is vehicle/sec

t = Duration of time over which vehicles are counted in sec.

Calculation:

\(\lambda = \frac{{360}}{{60 \times 60}} = \frac{1}{{60}}\;veh/sec\)

And, t = 20 sec,

Probability of 3 or more vehicles = 1 – Probability of 0, 1 or 2 vehicles

Now,

Probability of zero vehicles over 20 seconds

\(P\left( 0 \right) = \frac{{{{\left( {0.1 \times 20} \right)}^0} \times \left( {{e^{ - 0.1 \times 20}}} \right)}}{{0!}}\) = 0.135

Probability of 1 vehicle over 20 seconds

\(P\left( 1 \right) = \frac{{{{\left( {0.1 \times 20} \right)}^1} \times {e^{ - 0.1 \times 20}}}}{{1!}}\) = 0.27

Probability of two vehicles over 20 seconds

\(P\left( 2 \right) = \frac{{{{\left( {0.1 \times 20} \right)}^2} \times {e^{ - 0.1 \times 20}}}}{{2!}}\) = 0.27

∴ P(n ≥ 3) = 1- [P(0) + P(1) + P(2)]

= 1 – [0.135 + 0.27 + 0.27] = 0.325 = 32.5%

Question 12

2/-0.33

A spot speed study on a highway gives the following six spot speeds viz. 45 kmph, 40 kmph, 52 kmph, 51 kmph, 39 kmph and 42 kmph. What will be the time mean speed and space mean speed of the spot study?

SOLUTION

Concept:

Spot Speed: Spot speed is the instantaneous speed of a vehicle at a specified section or location. In traffic engineering generally there are two type of spot speed ae used which are time mean speed and space mean speed.

Time Mean Speed: It represents the speed distribution of vehicles at a specific point for a particular time frame. It is the athematic mean of spot speeds of observed vehicles at the spot. Hence, \({{\rm{V}}_{\rm{t}}} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{V}}_{\rm{i}}}}}{{\rm{n}}}\)  where \({{\rm{V}}_{\rm{t}}}\) is time mean speed, \({{\rm{V}}_{\rm{i}}}\) is the spot speed of i-th vehicle and n is number of vehicles observed. If spot speeds are given in the form frequency then, \({{\rm{V}}_{\rm{t}}} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{V}}_{\rm{i}}}{{\rm{f}}_{\rm{i}}}}}{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{f}}_{\rm{i}}}}}\) where \({{\rm{f}}_{\rm{i}}}\) is the frequency of i-th vehicle.

Space Mean Speed: It represents the speed distribution of vehicles at a specific time for a particular stretch of road. It is the harmonic mean of spot speeds of observed vehicles at the spot. Hence, \({{\rm{V}}_{\rm{s}}} = \frac{n}{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\frac{1}{{\rm{V}}}}_{\rm{i}}}}}\)  where \({{\rm{V}}_{\rm{s}}}\) is space mean speed, \({{\rm{V}}_{\rm{i}}}\) is the spot speed of i-th vehicle and n is number of vehicles observed. If spot speeds are given in the form frequency then, \({{\rm{V}}_{\rm{s}}} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{f}}_{\rm{i}}}}}{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} \frac{{{{\rm{f}}_{\rm{i}}}}}{{{{\rm{V}}_{\rm{i}}}}}}}\)  where  is the frequency of i-th vehicle.

Calculations:

Given, the spot speeds are 45 kmph, 40 kmph, 52 kmph, 51 kmph, 39 kmph and 42 kmph respectively.

\(\therefore {\rm{\;}}{{\rm{V}}_{\rm{t}}} = \frac{{45 + 40 + 52 + 51 + 39 + 42}}{6} = 44.833{\rm{\;kmph}}\) 

\(\therefore {{\rm{V}}_{\rm{s}}} = \frac{6}{{\frac{1}{{45}} + \frac{1}{{40}} + \frac{1}{{52}} + \frac{1}{{51}} + \frac{1}{{39}} + \frac{1}{{42}}}} = 44.276{\rm{\;kmph}}\) 

Question 13

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For a 25 cm thick cement concrete pavement analysis of stresses gives the following values'

Wheel load stress due to corner loading                                30 kg/cm2

Wheel load stress due to edge loading                                  32 kg/cm2

Warping stress at corner region during summer                    9 kg/cm2

Warping stress at corner region during winter                        7 kg/cm2

Warping stress at edge region during summer                      8 kg/cm2

Warping stress at edge region during winter                          6 kg/cm2             

Frictional stress during summer                                             5 kg/cm2

Frictional stress during winter                                                 4 kg/cm2

The most critical stress value for this pavement is

SOLUTION

Critical combination of stresses:

1. Summer mid day- the critical stress is for edge region 

Critical stress = (Wheel load stress + Warping stress - Frictional stress)at edge region

2. Winter mid day- the critical stress is for edge region 

Critical stress = (Wheel load stress + Warping stress + Frictional stress)at edge region

2. Winter mid-night- the critical stress is for corner region

Critical stress = (Wheel load stress + Warping stress )at corner region

Critical Edge stress at summer mid day = 32 + 8 - 5 =  35 kg/cm2

Critical Edge stress at winter mid day = 32 + 6 + 4 = 42 Kg/cm2

Critical corner stress at winter mid night = 30 + 7 = 37 Kg/cm2

So the critical stress is = 42 Kg/cm2

Question 14

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When the design vehicle with a design speed of 90 kmph on a two-lane road is negotiating a horizontal curve of radius 245 m off-tracking is measured as 0.088 m. The required widening of carriage way of the two lane road on this curve will be ______ m.

SOLUTION

Off-tracking for n-number of lanes is given by.

\(\frac{{{l^2}}}{{2R}} = offtracking = 0.088\)

\(\frac{{{l^2}}}{{2 \times 245}} = 0.088\)

L = 6.56 m

Where L = Length of wheel base

R = Radius of the curve

Total widening required on two lane road

\({E_w} = n \times \frac{{{l^2}}}{{2R}} + \frac{V}{{9.5\sqrt R }}\)

N = no. of lanes = 2

\({E_w} = \frac{{2 \times {{\left( {6.56} \right)}^2}}}{{2 \times 245}} + \frac{{90}}{{9.5\sqrt {245} }}\)

Ew = 0.1756 + 0.605Ew = 0.781 m

Question 15

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For predicting the flow on a highway, a theoretical speed-density model is established and assumed as \({\rm{V}} = ({{\rm{V}}_{\rm{f}}} - 10){{\rm{e}}^{0.45 - \frac{{\rm{k}}}{{{\rm{\;}}{{\rm{k}}_0}}}}}\) where V is speed in km/hr and k is density in vehicles per km. Statistical analysis of vehicular data shows that the free stream speed, Vf = 80 km/hr and optimum density, k0 = 150 vehicles/ km. What is the maximum flow (in vehicles per hour corrected to next integer) obtained as per this model? 

SOLUTION

Concept:

Flow-density-speed relationship:

Speed (v) is defined as the rate of motion in distance per unit of time. Generally, it is expressed in km/ hr or m/s unit.

Flow or volume (q) is defined as the number of vehicles that pass a point on a highway or a given lane or direction of a highway during a specific time interval. It is expressed in vehicles/hour.

Density (k) is defined as the number of vehicles occupying a given length of highway or lane and is generally expressed as vehicles/km.

The relationship between Speed (v), flow (q) and density (k) is given by, q = k × V.

Calculation:

Given, \({\rm{V}} = ({{\rm{V}}_{\rm{f}}} - 10){{\rm{e}}^{0.45 - \frac{{\rm{k}}}{{{{\rm{k}}_0}}}}}\)  and \({\rm{q\;}} = {\rm{\;k\;}} \times {\rm{V}} = {\rm{k}}({{\rm{V}}_{\rm{f}}} - 10){{\rm{e}}^{0.45 - \frac{{\rm{k}}}{{{{\rm{k}}_0}}}}}\) 

When flow to be maximum, then \(\frac{{{\rm{dq}}}}{{{\rm{dk}}}} = 0{\rm{\;\;\;}}\therefore \left( {{{\rm{V}}_{\rm{f}}} - 10} \right)\left[ {{\rm{k}} \times {{\rm{e}}^{0.45 - \frac{{\rm{k}}}{{{{\rm{k}}_0}}}}} \times \left( { - \frac{1}{{{{\rm{k}}_0}}}} \right) + {{\rm{e}}^{0.45 - \frac{{\rm{k}}}{{{{\rm{k}}_0}}}}}} \right] = 0{\rm{\;}}\)

\(\therefore - \frac{{\rm{k}}}{{{{\rm{k}}_0}}} + 1 = 0{\rm{\;\;}}\therefore {\rm{k}} = {{\rm{k}}_0}\) 

Hence, maximum capacity, \({{\rm{q}}_{{\rm{max}}}} = {{\rm{k}}_0} \times {\rm{\;}}({{\rm{V}}_{\rm{f}}} - 10){{\rm{e}}^{0.45 - \frac{{{{\rm{k}}_0}}}{{{{\rm{k}}_0}}}}} = \frac{{{{\rm{k}}_0}\left( {{{\rm{V}}_{\rm{f}}} - 10} \right)}}{{{{\rm{e}}^{0.55}}}}{\rm{\;}}\)

Now given, Vf = 80 km/hr and k0 = 150 vehicles/ km.

Hence, \({{\rm{q}}_{{\rm{max}}}} = {\rm{\;}}\frac{{{{\rm{k}}_0}\left( {{{\rm{V}}_{\rm{f}}} - 10} \right)}}{{{{\rm{e}}^{0.55}}}} = \frac{{150 \times \left( {80 - 10} \right)}}{{{{\rm{e}}^{0.55}}}} = 6057.97 \approx 6058{\rm{\;vehicles}}/{\rm{\;hour}}\)

Trick/Tip:

Optimum Density is defined as the density at which the traffic flow is maximum for any model. So, in question if optimum density is explicitly mentioned and given, then calculate flow at optimum density and it will be the maximum flow as per the model.

Question 16

2/-0.33

The length of summit curve required for a stopping sight distance of 180 m at the junction of an up-gradient of 1 in 200 and a downgradient of 1 in 200 is -

SOLUTION

Concept:

 

Length of L vertical curve (L) is given by:

Condition

Summit curve

Valley curve

S < L

\(\frac{{{\rm{N}}{{\rm{S}}^2}}}{({\sqrt {2{\rm{H}}} + \sqrt {2{\rm{h}}})^2}}\)

\(\frac{{{\rm{N}}{{\rm{S}}^2}}}{{2{{\rm{h}}_1} + 2{\rm{S}}\tan {\rm{\alpha }}}}\)

S > L

\(\frac{{2{\rm{S}} - {{\left( {\sqrt {2{\rm{H}}} - \sqrt {2{\rm{h}}} } \right)}^2}}}{{\rm{N}}}\)

\(\frac{{2{\rm{S}} - 2{{\rm{h}}_1} + 2{\rm{S}}\tan {\rm{\alpha }}}}{{\rm{N}}}\)

Where,

N = deflection angle, H = height of driver cycle, h = height of object on road surface,

S = stopping sight distance, and α = beam light angle

Calculation:

SSD = 180 m,

\({n_1} = \frac{1}{{200}}\),

\({n_2} = \left( {\frac{{ - 1}}{{200}}} \right)\)

\(N = \left( {{n_1} - {n_2}} \right) = \left( {\frac{1}{{200}} - \left( {\frac{{ - 1}}{{200}}} \right)} \right) = \frac{2}{{200}} = \frac{1}{{100}}\)

H = 1.2 m, h = 0.15 m (As per IRC)

Assume the length of curve > stopping sight distance

\( \Rightarrow L = \frac{{N{S^2}}}{{{{\left( {\sqrt {2H} + \sqrt 2 \;h} \right)}^2}}}\)

\(\Rightarrow L = \frac{{\frac{1}{{100}} \times {{180}^2}}}{{{{\left( {\sqrt {2 + 1.2} + \sqrt {2 \times 0.15} } \right)}^2}}} = 73.68\;m\)

 

∵ L < SSD ⇒ Our assumption is wrong.

Applying the another formula,

\(L = 2S - \frac{{4.4}}{N} = 2 \times 180 - \frac{{4.4}}{{\left( {\frac{1}{{100}}} \right)}}\) = -80 m

⇒ The gradient change does not need any summit curve, hence the minimum curve length is provided for aesthetic purpose.

Minimum curve length = 60 m

Question 17

2/-0.33

A 18cm thick cement concrete slab is resting on Subgrade having modulus of subgrade as 4.5 kg/cm3. If the thickness of cement concrete slab and modulus of subgrade is increased by 40% and 50% respectively, then the radius of relative stiffness is increased by?

Take Modulus of elasticity of cement Concrete = 210000 kg/cm2

Poisson’s ratio of concrete = 0.15

SOLUTION

Concept:

Radius of relative stiffness is given by:

\(l = \;{\left( {\frac{{E{h^3}}}{{12k\left( {1 - {\mu ^2}} \right)}}} \right)^{\frac{1}{4}}}\)

where,

l = radius of relative stiffness (cm)

h = slab thickness (cm)

μ = Poisson’s ratio

k = Modulus of subgrade (kg/cm3)

E = Modulus of elasticity of cement Concrete (kg/cm2)

Calculation:

\(l = \;{\left( {\frac{{E{h^3}}}{{12k\left( {1 - {\mu ^2}} \right)}}} \right)^{\frac{1}{4}}}\)

\(l = \;{\left( {\frac{{210000 \times {{\left( {18} \right)}^3}}}{{12 \times 4.5\left( {1 - {{\left( {0.15} \right)}^2}} \right)}}} \right)^{\frac{1}{4}}}\)

l = 69.40cm

Now, h1 = 1.40 × 18 = 25.2cm

k1 = 1.50 × 4.5 = 6.75 kg/cm3

\({l_1} = \;{\left( {\frac{{210000 \times {{\left( {25.2} \right)}^3}}}{{12 \times 6.75\left( {1 - {{\left( {0.15} \right)}^2}} \right)}}} \right)^{\frac{1}{4}}}\)

l1 = 80.70cm

\(\% \;increase = \left( {\frac{{80.70 - 69.40}}{{69.40}}} \right) \times 100\)

= 16.29 ≅ 16.3%

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