Highway Engineering Test 3

As per IRC 37:2012, clause 6.3.2, the model considers the vertical strain in subgrade as the only variable for rutting, which, is a measure of bearing capacity of the subgrade. Rutting in granular layer also is lower when the vertical subgrade elastic strains are given by equation:

\(N = 4.1656 \times {10^{ - 8}} \times {\left[ {1 / {{\rm{\varepsilon }}_V}} \right]^{4.5337}} \)

Ns = Number of cumulative standard axle.

= Vertical strain in the subgrade which is the only variable for rutting.

Concept:

Cutback Bitumen

The viscosity of bitumen is reduced by volatile diluents. Cut back bitumens are available in three types.

1. Rapid Curing (RC)

2. Medium Curing (MC)

3. Slow Curing (SC)

• The cutbacks are designated by numerals representing progressively thicker or viscous cutbacks.
  For example, RC-2 is thicker than RC-1 but RC-2, MC-2, and SC-2 have the same viscosity.

Bituminous Emulsion:

• The emulsion is a two-phase system consisting of two immiscible liquids.
• The bitumen/tar content in emulsion range from 40 to 60% and the remaining portion is water. 
• The emulsion is used especially in maintenance and patch repair works. The main advantage of the emulsion is that is can be used in wet weather even when it is raining. Emulsions can be used for soil stabilization in deserts.

Tar and Bitumen:

• Bitumen is a petroleum product whereas tar is produced by the destructive distillation of coal or wood.
• Bitumen is soluble in ‘carbon disulfide’ and ‘carbon tetrachloride’ but tar is soluble in only toluene.
• Tar is more temperature susceptible resulting in great variation in viscosity with temperature.

Hence the correct answer is P → False, Q → True, R → False

Raveling

It is the on-going separation of aggregate particles in a pavement from the surface downward or from the edges inward. Usually, the fine aggregate wears away first and then leaves little "pock marks" on the pavement surface. As the erosion continues, larger and larger particles are broken free and the pavement soon has the rough and jagged appearance typical of surface erosion. 

Bird baths (Depressions)

Depressions are localized pavement surface areas with slightly lower elevations than the surrounding pavement. This is also known as bird baths, the localized depression results in water accumulation. Thus, depressions are very noticeable after a rain when they fill with water

Potholes

Small, bowl-shaped depressions in the pavement surface that penetrate all the way through the asphalt layer down to the base course. They generally have sharp edges and vertical sides near the top of the hole. Potholes are the result of moisture infiltration and usually the end result of untreated alligator cracking. As alligator cracking becomes severe, the interconnected cracks create small chunks of pavement, which can be dislodged as vehicles drive over them. The remaining hole after the pavement chunk is dislodged is called a pothole.

 Subsidence

Concept:

Group Index (GI): It is used to give a rating to the quality of soil within its group.
GI = 0.2a + 0.005ac + 0.01 bd

Where,

a = % passing 75 μ seive – 35 ; a ≯ 40

b = % passing 75 μ seive – 15 ; b ≯ 40

c = w - 40 ; c ≯ 20

d = I_p – 10 ; d ≯ 20

Calculation:

w_L = 45%, w_P = 25%, P = 60%

∴ I_P = 45 – 25 = 20%

a = P – 35 = 60 – 35 = 25

b = P – 15 = 60 – 15 = 45 > 40 ⇒ Adopt b = 40

c = w_L – 40 = 45 – 40 = 5

d = I_p – 10 = 20 – 10 = 10

∴ G. I = 0.2 × 25 + 0.005 × 25 × 5 + 0.01 × 40 × 10 = 9.625

By using linear interpolation:

Concept:

Plate bearing test is used to evaluate the support capability of sub-grades and bases.

Modulus of subgrade reaction is given as

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

Where, P = Pressure corresponding to a settlement of 1.25 mm

Δ = Design Deflection  considered as 1.25 mm

Modulus of subgrade reaction (k1) for any other plate of a radius (a­1) is given by:

\({{\rm{k}}_1} = \frac{{{\rm{k}} \times {\rm{a}}}}{{{{\rm{a}}_1}}}\)

Calculation:

Load (corresponding to 1.25 mm deflection) = 1260 kg

Pressure \( = \frac{{Load}}{{Area}} = \frac{{1260}}{{\left( {\frac{\pi }{4} \times {{0.3}^2}} \right)}} = 17825.35\;kg/{m^2}\) 

\(\therefore {K_{30}} = \frac{P}{{\rm{\Delta }}} = \frac{{17825.35}}{{\left( {\frac{{1.25}}{{1000}}} \right)}} = 14260.28 \times {10^3}\;kg/{m^2}/m\)

Also,

K₃₀ × 30 = K₇₅ × 75

⇒ 14260.28 × 10³ × 30 = K₇₅ × 75

Concept:

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

N = Number of vehicles after ‘n’ years, r = traffic growth rate, A = Initial traffic, VDF = Vehicle damage factor, and LDF = Lane distribution factor

Calculation:

Since it is a divided highway, thus the traffic in one direction is considered as design traffic.

∴ Number of commercial vehicles per day = 2500 

Growth rate = 10%

∴ Number of commercial vehicles at the end of construction  = A = 2500 × (1 + 0.1)² = 3025

Cumulative standard Axles (N_s) \( = \frac{{365\;A\;\left[ {{{\left( {1 + r} \right)}^n} - 1} \right]}}{r} \times LDF \times VDF\)

Concept:

Vehicle Damage Factor (VDF): Different commercial vehicles have different weights and the conversion of the design number of commercial vehicles during the design life of the road into a number of standard axle repetitions. This is done using VDF.

The vehicle damage factor (VDF) is a multiplier for converting the number of commercial vehicles of different axle loads and axle configurations to the number of standard axle-load repetitions.

Mathematically,

\({\rm{VDF}} = \frac{{{\rm{Total\;number\;of\;standard\;axles}}}}{{{\rm{Total\;number\;of\;vehicles\;surveyed}}}}\)

This is used in the estimation of cumulative standard axles during the design life of the road which in turn is used in CBR method of design of flexible pavements.

Calculation:

Hence 450 vehicles surveyed is equivalent to 2453.33 standard axles

\({\rm{VDF}} = \frac{{{\rm{Total\;number\;of\;standard\;axles}}}}{{{\rm{Total\;number\;of\;vehicles\;surveyed}}}} = \frac{{2453.33}}{{450}} = 5.452\)

Concept:

The total force developed in the cross section of concrete pavement due to movement of half the length of the slab (L_c/2) is equal to the frictional resistance due to the restraint at the interface in half the slab length.

\({{\rm{S}}_{\rm{f}}} \times {\rm{h}} \times {\rm{b}} \times 100{\rm{\;}}\left( {{\rm{kg}}} \right) = {\rm{b}} \times \left( {\frac{{{{\rm{L}}_{\rm{c}}}}}{2}} \right) \times \frac{{\rm{h}}}{{100}} \times {\rm{W}} \times {\rm{f\;}}\left( {{\rm{kg}}} \right)\)

∴ S_f = (W L_c f)/(2 × 10⁴)

Where

S_r = stress developed due to inter-face friction in cement concrete pavement, W = unit weight of concrete, kg/m³ (about 2400 kg/m³), f = coefficient of friction at the interface (maximum value is about 1.5), L_c = spacing between the contraction joint = slab length (in meters) and b = slab width (in meters)

Calculation:

\({{\rm{L}}_{\rm{c}}} = \frac{{{{\rm{S}}_{\rm{f}}} \times 2 \times {{10}^4}}}{{{\rm{W}} \times {\rm{f}}}} = \frac{{0.5 \times 2 \times {{10}^4}}}{{2400 \times 1.1}}\)

L_c = 3.79 meters

Concept:

The equivalent C-value of the three layered pavement section is calculated by converting individual thickness of each layer into any one of the layer section by using the relation below:

\(\frac{{{t_x}}}{t} = {\left( {\frac{c}{{{c_x}}}} \right)^{\frac{1}{5}}}\)

where,

t_x = thickness of concerned pavement

t = thickness of individual pavement layer

C = C-value of individual pavement layer

C_X = C-value of concerned pavement

Calculation:

The individual thickness of each layer is converted to their respective Bituminous concrete layer.

\(\frac{{{t_b}}}{t} = {\left( {\frac{c}{{{c_b}}}} \right)^{\frac{1}{5}}}\)

\({t_b} = {\left( {\frac{c}{{{c_b}}}} \right)^{\frac{1}{5}}} \times t\)

\({t_b} = {\left( {\frac{c}{{70}}} \right)^{\frac{1}{5}}} \times t\)

For Base course (cement treated Base):

\({t_b} = {\left( {\frac{{260}}{{70}}} \right)^{\frac{1}{5}}} \times 23\)

t_b = 29.90cm

For sub-base course (Gravel):

\({t_b} = {\left( {\frac{c}{{70}}} \right)^{\frac{1}{5}}} \times t\)

\({t_b} = {\left( {\frac{{30}}{{70}}} \right)^{\frac{1}{5}}} \times 18 = 15.19cm\)

For Bituminous Concrete,

t_b = 11.5cm

Equivalent Total thickness of Bituminous conerete = 11.5 + 29.90 + 15.19

t_b = 56.59 cm

Actual pavement thickness = 11.5 + 23 + 18

T = 52.50 cm

Equivalent c-value

\(\frac{{{t_b}}}{T} = {\left( {\frac{c}{{{c_b}}}} \right)^{\frac{1}{5}}}\)

\(\frac{{56.59}}{{52.50}} = {\left( {\frac{c}{{70}}} \right)^{\frac{1}{5}}}\)

C = 101.86