Highway Engineering Test 1

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

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

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. 

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

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.

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

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

• 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

Concept:

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

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

Percentage of voids in mineral aggregate (V_MA) = V_b% + V_V%

Calculation:

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

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

V_MA = V_b + V_V = 11.03 + 3.623 = 14.65%

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

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

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:

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

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

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%

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.

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

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

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

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

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

E_w = 0.1756 + 0.605

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, V_f = 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.

Concept:

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

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

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%