Highway Engineering Test 2

Calculation:

i) Generally traffic volume in terms of vehicle per hour is considered during peak hour of a day

For 15 minute peak hour factor

i.e \(\frac{{15}}{{60}}\;hr\)

= ¼

ii) Flow of traffic stream is given by

\(q = \frac{{3600}}{{{H_t}}}\)

Where H_t is average time headway in sec

q is traffic volume in veh/hr

H_t = 6 sec

\(q = \frac{{3600}}{6}\)

q = 600 veh/hr

The bitumen adhesion test is carried out in order to decide the suitability of the road aggregate for use in road construction.

Concept:

The ruling gradient is the maximum gradient on a particular section, but if a curve lies on a ruling gradient, the resistance due to gradient is increased by that due to curvature and this further increases the resistance beyond the ruling gradient.

In order to avoid resistances beyond the allowable limits, the gradients are reduced on curves and this reduction in known as the grade compensation for the curve.

Calculation:

Given gradient = 1 in 200 = 1 / 200

Degree of curve = 2°

Grande compensation = -0.04% per degree of curve.

Grrade provided = Ruling rradient – grande compensation

\(= \frac{1}{{200}} - \frac{{0.04 \times 2}}{{100}}\)

Concept:

Lowest point on the valley curve:

The lowest point on the valley curve is to be located for providing the cross drainage facility. The lowest

point on the valley curve will be on the bisector of the angle between the grades, if the gradients on

either side are equal. When the gradients are not equal, the lowest point lies on the side of flatter

grade, and this point is from the tangent point of the first gradient, n₁ at a distance, X_o given by:

\({X_0} = L \times {\left( {\frac{{{n_1}}}{{2N}}} \right)^{1/2}}\)

Where X₀ = for distance from end of first tangent point t₀ the lowest point in metres.

n₁ = first gradient.

N = |n₁ – m₂|, L = length of valley curve

Calculation:

Descending gradient = 4%

\({n_1} = \frac{{ - 4}}{{100}}\)

Ascending gradient = 1 in 40

\({n_2} = \frac{1}{{40}}\)

\(V = \left[ {\frac{{ - 4}}{{100}} - \frac{1}{{40}}} \right] = 0.065\)

L = 200 M.

\(x = 200 \times {\left( {\frac{{0.04}}{{2 \times 0.065}}} \right)^{1/2}}\)

\(= 110.94\;m \sim 111\;m\)

Viscosity denotes the fluid property of bituminous material and it is a measure of resistance to flow. At the application temperature, this characteristic greatly influences the strength of resulting paving mixes. Low or high viscosity during compaction or mixing has been observed to result in lower stability values. At high viscosity, it resists the compactive effort and thereby resulting mix is heterogeneous, hence low stability values.

Important point:

Flow (q) = Density (k) × Velocity (V)

For Greenshields’s Model

\(V = {V_f} \times \left( {1 - \frac{k}{{{k_j}}}} \right),\)

Where, V = Velocity at any instant

                V_f = Free mean velocity

                k = Density of the flow

                k_j = Jam density of the flow

Maximum capacity (q_max) = V_max × K_max

As per Greenshields’s model

\(Vmax\; = \;\frac{{{V_f}}}{2}\;\& \;{k_{max}} = \frac{{{k_j}}}{2}\;\)

Maximum capacity ⇒ \(\left( {{q_{max}}} \right) = \frac{{{V_f}}}{2} \times \frac{{{k_j}}}{2} = \frac{{{V_f}{k_j}}}{4}\) 

Concept:

Burmister’s Method:

Assumptions

• Materials in each layers are isotropic, homogeneous and elastic.
• Pavement forms a stiffer layer having higher value of E than that of subgrade.
• Layers are in constant contact.
• Surface layer is infinite in horizontal direction but finite in vertical direction.

This method gives the following design deflection values for pavements with different layers –

1. For flexible pavement –

\(\Delta = 1.5\frac{{pa}}{{{E_s}}} \times {F_2}\)

1. For rigid pavement –

\(\Delta = 1.18\frac{{pa}}{{{E_s}}} \times {F_2}\)

Where, p = contact pressure at road surface due to wheel load (kg/m²)

E_s = modulus of elasticity of subgrade (kg/cm²), a = radius of contact area(cm) and

F₂ = deflection factor which depends upon E_s/E_P  and h/a,  for single layer system F₂ = 1.

Formula used:-

Design deflection for single-layered flexible pavement is given by,

\(\Delta = 1.5\frac{{pa}}{{{E_s}}}\)

Radius of contact area is given by,

\(a = \sqrt {\left( {\frac{P}{p}} \right) \times \frac{1}{\pi }} \)

Calculation:

Wheel Load, P = 40kN

Modulus of elasticity for subgrade and pavement, ES = EP = 20MPa

Tyre pressure, p = 0.50 MPa

\(a = \sqrt {\left( {\frac{P}{p}} \right) \times \frac{1}{\pi }} \Rightarrow a = \sqrt {\frac{{40 \times {{10}^3}}}{{0.5 \times {{10}^6}}} \times \frac{1}{{3.14}}} = 0.1596m\)

Design deflection for flexible pavement

\(\Delta \; = 1.5\frac{{pa}}{{{E_s}}}\)

\( \Rightarrow \Delta \; = 1.5 \times \frac{{0.5 \times 0.1596}}{{20}} = 0.00598m\; \approx 5.98mm\;\)

Concept:

The optimum signal cycle is given by as per Webster’s Method

\({C_o} = \frac{{1.5L + 5}}{{1 - Y}}\)

where

L = total lost time per cycle, in sec

L = 2n + R

n = number of phases

R = all red time

Y = Sum of Ratio of normal flow to the saturation flow

Calculation:

Total lost time = 2n + R (n = no. of phases, R = All red time)

Space mean speed

\(= \frac{{3.6dn}}{{\mathop \sum \nolimits_{i = 1}^n {t_i}}} = \frac{{3.6 \times 10 \times 3}}{{\frac{{10}}{{60}} + \frac{{10}}{{50}} + \frac{{10}}{{40}}}}\)

⇒ 175.13 kmph

\({V_t} = \frac{{{\rm{\Sigma }}{V_i}}}{n} =3.6\times \frac{{50 + 60 + 40}}{3} = 180\;kmph\)

Utility factor for 1000 tons of agricultural product = 1

Utility factor for 1000 tons of industrial product = 1 × 10 = 10

For Road A:

Total utility units served by the road:

10 × 0.5 + 8 × 1 + 3 × 2 + 15 × 1 + 10 × 1.2 = 46

Utility per unit length = 46/15 = 3.07

For Road B:

Utility per unit length \(= \frac{{16 \times 0.5 + 3 \times 1 + 1 \times 2 + 11 \times 1 + 0 \times 10}}{{10}}\) = 2.4

For Road C:

Utility per unit length \(= \frac{{20 \times 0.5 + 10 \times 1 + 2 \times 2 + 20 \times 1 + 10 \times 0.8}}{{18}}\) = 2.89

∴ Decreasing order of priority is A, C and B.

Concept:

CBR test is strength test conducted on the soil by introducing surcharge load at the compaction rate of 1.25 mm per minute on a completely soaked soil sample passing through 20 mm sieve size.

\({\rm{CB}}{{\rm{R}}_{\rm{\delta }}} = \frac{{{{\rm{P}}_{\rm{\delta }}}{\rm{\;of\;soil}}}}{{{{\rm{P}}_{\rm{\delta }}}{\rm{\;of\;standard\;crushed\;aggregate}}}} \times 100\)

δ = displacement in mm

P_δ = Load corresponding to ‘δ’ settlement

P_s = Load for standard crushed aggregate:

If CBR 2.5 > CBR 5 → Test Accepted

If CBR5 > CBR2.5 → Test Repeated

And higher value is recorded

Calculation:

CBR_2.5 = 68.5/1370 × 100 = 5%

CBR_5.5 = 88.5/2055 × 100 = 4.30%

Concept:

Practical capacity of a rotary road way is given by

\({Q_p} = \frac{{280 \times W \times \left( {1 + \frac{e}{W}} \right) \times \left( {1 - \frac{p}{3}} \right)}}{{\left( {1 + \frac{W}{L}} \right)}}\)

Where,

W is width of weaving section

e is entry width

L is length of weaving section

p is proportionality ratio

\(p = \frac{{weaving\;traffic}}{{Total\;traffic}}\)

Calculation:

Given,

W = 14 m, e = 8.5 m

L = 70 m,

Weaving traffic = 1000 PCU

Total traffic = 2000 PCU

\(p = \frac{{weaving\;traffic}}{{Total\;traffic}}\)

\(= \frac{{1000}}{{2000}}\)

= 0.5

\({Q_p} = \frac{{280 \times W \times \left( {1 + \frac{e}{W}} \right) \times \left( {1 - \frac{p}{3}} \right)}}{{\left( {1 + \frac{W}{L}} \right)}}\)

\({Q_p} = \frac{{280 \times 14 \times \left( {1 + \frac{{8.5}}{{14}}} \right) \times \left( {1 - \frac{{0.5}}{3}} \right)}}{{\left( {1 + \frac{{14}}{{70}}} \right)}}\)

= 4372.86 PCU/hr/lane

≈ 4373 PCU/hr/lane

Concept:

The design traffic in terms of the cumulative number of standard axles to be carried during the design life of the road as per IRC 37:2012 is:

\(N = \frac{{365 \times \left[ {{{\left( {1 + r} \right)}^n} - 1} \right] \times A \times D \times F}}{r}\)

Where,

N = Cumulative number of standard axles to be cater for in the design in (msa)

A = Initial traffic in the year of completion of construction in terms of number of commercial vehicles per day.

\(A = P{\left( {1 + r} \right)^x}\)

x = number of years between the last count and the year of completion of construction

P = Number of commercial vehicles as per last count

r = annual growth rate of commercial vehicles in decimal

n = design life of pavement in years

D = Lane distribution factor

F = Vehicle damage factor

Calculation:

As all other factors except “r” is constant, so

\({N_1} = \frac{{365 \times \left[ {{{\left( {1 + 0.07} \right)}^n} - 1} \right] \times A \times D \times F}}{{0.07}}\)

\({N_1} = 6529.85\;ADF\)

\({N_2} = \frac{{365 \times \left[ {{{\left( {1 + 0.1} \right)}^n} - 1} \right] \times A \times D \times F}}{{0.1}}\)

\(\begin{array}{l} {N_2} = 7803.7\;ADF\\ \% \;increase = \frac{{{N_2} - {N_1}}}{{{N_1}}} \times 100 \end{array}\)

\(\% \;increase = \frac{{7803.7 - 6529.85}}{{6529.85}} \times 100\)

% increase = 19.51 %

Concept:

Voids in mix aggregate (VMA) is given by

\(VMA = {V_b} + {V_a}\)

Where V_b is percentage bitumen by volume

\({V_b} = {G_m} \times \frac{{{w_b}}}{{{G_b}}}\)

Where,

G_m is Bulk specific gravity of mix

W_b is percentage weight of bitumen

G_b is specific gravity of bitumen

V_a is percentage air voids given by

\({V_a} = \frac{{{G_t} - {G_m}}}{{{G_t}}} \times 100\)

Where,

G_t is theoretical specific gravity

G_m is Bulk specific gravity

Voids filled by bitumen (VFB) is given by

\(VFB = \frac{{{V_b}}}{{VMA}} \times 100\)

Calculation:

Given,

G_m = 2.324, G_b = 1.10

G_t = 2.546, W_b = 5 %

\({V_a} = \frac{{{G_t} - {G_m}}}{{{G_t}}} \times 100\)

\({V_a} = \frac{{2.546 - 2.324}}{{2.546}} \times 100\) = 8.72 %

\({V_b} = {G_m} \times \frac{{{w_b}}}{{{G_b}}}\)

\({V_b} = 2.324 \times \frac{5}{{1.1}}\) = 10.56 %

i) \(\;VMA = {V_b} + {V_a}\)

VMA = 10.56 + 8.72 = 19.28 %

ii) \(VFB = \frac{{{V_b}}}{{VMA}} \times 100\)

\( = \frac{{10.56}}{{19.28}} \times 100\) = 54.78 %

Concept:

Length of the transition curve:

a) Based on rate of change of centrifugal acceleration

\({{\rm{L}}_{\rm{s}}} = \frac{{{{\rm{v}}^3}}}{{{\rm{CR}}}}\)

b) Based on super elevation criteria

i) If rotated about outer edge:

\({\rm{Ls\;}} = {\rm{\;eN}}\left( {{\rm{W\;}} + {\rm{\;We}}} \right)\)

ii) If rotated about center edge

\({\rm{Ls\;}} = \frac{{{\rm{eN}}\left( {{\rm{W\;}} + {\rm{\;We}}} \right)}}{2}\)

c) Based on IRC recommendations

\({{\rm{L}}_{\rm{s}}} = \frac{{2.7{{\rm{V}}^2}}}{{\rm{R}}}\)

Where

L_s = Length of transition curve

v = speed of vehicle in m/s

c = rate of change of centrifugal acceleration

R = radius of the curve

e = super elevation provided

N = rate of change of super elevation

W = width of pavement

W_e­ = Width of extra widening

Given:

Horizontal curve radius (R) = 600 m

Design speed (v) = 80 × 5/18 m/s = 22.22 m/s

Total pavement width (W) = 7.5 m

The number of lanes (n) = 2

Wheelbase = l = 6 m

Allowable rate of super elevation = 1 in 150

Change of centrifugal acceleration = 0.507

Calculation:

(i) Length of transition curve (based centrifugal acceleration)

\({{\rm{L}}_{\rm{s}}} = \frac{{{{\rm{v}}^3}}}{{{\rm{CR}}}} = \frac{{{{\left( {80 \times \frac{5}{{18}}} \right)}^3}}}{{0.507 \times 600}} = 36.07{\rm{\;m}}\)

(ii) Based on superelevation criteria

\({\rm{Ls\;}} = {\rm{\;eN}}\left( {{\rm{W\;}} + {\rm{\;We}}} \right)\)

The pavement rotating about an outer edge

\({\rm{e}} = \frac{{{{\left( {0.75{\rm{V}}} \right)}^2}}}{{{\rm{gR}}}} = \frac{{{{\left( {0.75 \times 22.22} \right)}^2}}}{{9.81 \times 600}} = 0.0471\)

e = 4.71% < 7% ⇒ Hence Okay

∴ e = 0.047

For radus greater than 300 m, extra widening is not required. ⇒ W_e = 0

∴ Total width = 7.5 = 7.50 m

\({\rm{Ls\;}} = {\rm{\;eN}}\left( {{\rm{W\;}} + {\rm{\;We}}} \right) = {\rm{\;}}150 \times 0.047{\rm{\;}} \times 7.50 = 52.86{\rm{\;m}}\)

(iii) By using empirical formula

\({{\rm{L}}_{\rm{s}}} = \frac{{2.7{{\rm{V}}^2}}}{{\rm{R}}} = \frac{{2.7 \times {{80}^2}}}{{600}} = 28.8{\rm{\;m}}\)

∴ L_s = (Maximum of above three condition)

Various tests conducted on bitumen for testing its various properties are as follows:

Concept:

Moving observer method:

Moving car or moving observer method of traffic stream measurement has been developed to provide simultaneous measurement of traffic stream variables. It has the advantage of obtaining the complete state with just three observers and a vehicle.

The mean speed by moving the observer method is given as follows

\({{\rm{V}}_{\rm{s}}} = \frac{{\rm{L}}}{{{{\rm{t}}_{\rm{w}}} - \frac{{{{\rm{m}}_{\rm{w}}}}}{{\rm{q}}}}}\)

t_w = Observation time when the observer is moving with the stream

m_w = Net vehicles overtake the observer when it is moving with the stream

q = Flow density when the observer is at rest in vehicles/min

Calculation:

Total time taken by the student to reach home = 40 min

Student’s stop time = 15 min

∴ t_w = 40 – 15 = 25 min = 0.416 hr

L = 5 km

\({\rm{q}} = \frac{{45{\rm{\;vehicles}}}}{{15{\rm{\;min}}}} = 3\frac{{{\rm{veh}}}}{{{\rm{minute}}}} = 180{\rm{\;veh}}/{\rm{hr}}\)

m_w = 60 vehicles

\({{\rm{V}}_{\rm{s}}} = \frac{{\rm{L}}}{{{{\rm{t}}_{\rm{w}}} - \frac{{{{\rm{m}}_{\rm{w}}}}}{{\rm{q}}}}}\)

\({{\rm{V}}_{\rm{s}}} = \frac{5}{{0.416 - \frac{{60}}{{180}}}} = 60.48{\rm{\;km}}/{\rm{hr}}\)

Hence the speed of vehicle stream = 60 km/hr

Concept:

Thickness of pavement by single layer elastic theory is given by,

\(T = \sqrt {{{\left( {\frac{{3P}}{{2\pi {E_s}{\rm{\Delta }}}}} \right)}^2} - {a^2}}\) 

Contact pressure, \(\left( p \right) = \frac{P}{{\pi {a^2}}}\)

Where, P = Design wheel load (in kg)

E_s = Elastic modulus (in kg/km²)

a = radius of contact area (in cm)

Δ = permissible deflection (in cm)

Calculation:

Design wheel load = 4200 kg

Tyre pressure = 6.0 kg/cm²

Elastic modulus = 150 kg/cm²

Permissible deflection = 0.25 cm

\(p = \frac{P}{{\pi {a^2}}} \Rightarrow 6 = \frac{{4200}}{{\pi \times {a^2}}} \Rightarrow a = {\left( {\frac{{4200}}{{6 \times \pi }}} \right)^{1/2}} = 14.92\;cm\) 

\(T = \sqrt {{{\left( {\frac{{3 \times 4200}}{{2\pi \times 150 \times 0.25}}} \right)}^2} - {{\left( {14.92} \right)}^2}} = 51.352\)

= 51.352 cm

Concept

eactual(main) = - eactual(branch)

 \(\therefore {{\rm{e}}_{{\rm{main}}}} = \frac{{{\rm{G}}{{\rm{V}}^2}}}{{127{\rm{R}}}}\)

Calculation

 Given , 

 Cant Deficiency = 50 mm, Degree of main curve = 9° 

 Degree of branch curve = 8° 

Radius of branch curve \(\left( {{{\rm{R}}_{\rm{b}}}} \right) = \frac{{1719}}{9} = 191{\rm{\;metres}}\)

Radius of main curve \(\left( {{{\rm{R}}_{\rm{m}}}} \right) = \frac{{1719}}{8} = 214.88{\rm{\;metres}}\)

(E_d) Cant deficiency = 50 mm

\(\therefore {{\rm{e}}_{{\rm{main}}}} = \frac{{{\rm{G}}{{\rm{V}}^2}}}{{127{\rm{R}}}}\)

\(\therefore {{\rm{e}}_{{\rm{main}}}} = \frac{{1000 \times {{40}^2}}}{{127 \times 214.88}}\)

∴ e_main = 58.63 mm

∴ Theoretical super elevation on main branch

(e_{theorotical(main)}) = 58.63 mm

Now

e_{theorotical(main)} = e_actual(main) + e_d

∴ e_actual(main) = 58.63 – 50 = 8.63 mm

e_actual(main) = - e_{actual(branch)}

∴ e_o(branch) = - 8.63 mm

Also,

e_{actual(branch)} + ed  = etheorotical(branch)

∴ e_{theorotical(branch)} = 50 – 8.63

∴ e_{theorotical (branch)} = 41.37 mm

\(\therefore 41.37 = \frac{{1000 \times {{\rm{V}}^2}}}{{127 \times 191}}\)

Concept:

Correction for elevation

ICAO recommends that basic runway length should be increased at the rate of 7% per 300 m rise in elevation above mean sea level.

∴ Correction for elevation \( = \frac{7}{{100}} \times runway\;length \times \frac{{Airport\;elevation}}{{300}}\)

Correction for temperature

Rise in temperature = ART - SAT  

Where,

ART is Atmospheric reference temperature

SAT Standard atmospheric Temperature

As per ICAO, Basic runway length after correction for elevation should be further increased at the rate of 1% for every 1° C rise of airport reference temperature

correction for temperature \(= Correctedlength \times \frac{1}{{100}} \times Rise\;in\;temperature\)

Calculation

Correction for elevation

7% increase for 300 m rise

So for 600 m rise = 7 × 2 = 14% increase

∴ Runway length after elevation correction.

= 1800 + 1800 × 0.14

= 2052 m

Temperature Correction

Airport Reference temperature

\({\rm{ART}} = {\rm{Ta}} + \frac{{{\rm{Tm}} - {\rm{Ta}}}}{3}\)

\(= 15 + \frac{{30 - 15}}{3} = 20^\circ {\rm{c}}\)

Standard temperature = 15 – 0.0065 × 600

= 11.10c [6.5°c fore 1000m]

DT = 20 – 11.1 = 8.9°c

∴ Corrected = 1% for 1⁰c difference \(= \frac{{8.9}}{{100}} \times 2052 = 182.6{\rm{\;m}}\) 

Corrected value = 2052 + 182.6

= 2234.6 m = 2235 m

∴ % increase in length \(= \frac{{2235 - 1800}}{{1800}} \times 100\% \;\)

Concept:

Total cycle length (C_o) as per Webster’s approach:

\({{\rm{C}}_{\rm{o}}} = \frac{{1.5{\rm{\;L}} + 5}}{{1 - {{\rm{Y}}_0}}}\)

L = Total time lost per cycle in seconds

Y₀ = y₁ + y₁ + y₃ +…. y_n

Where,

n = Number of phases

y_n = Maximum value of ratio of design flow to saturation flow for different phase in n number of phases

For phase 1: (North-South phase)

North direction: \(\frac{{\rm{q}}}{{\rm{s}}} = \frac{{1500}}{{3000}} = 0.5\)

South direction: \(\frac{{\rm{q}}}{{\rm{s}}} = \frac{{900}}{{3000}} = 0.3\)

∴ For phase 1: y₁ = 0.5

For phase 2: (East-West phase)

East direction: \(\frac{{\rm{q}}}{{\rm{s}}} = \frac{{1000}}{{2500}} = 0.4\)

West direction: \(\frac{{\rm{q}}}{{\rm{s}}} = \frac{{750}}{{2500}} = 0.3\)

∴ For phase 2: y₂ = 0.48