HIGHWAY AND TRANSPORTATION ENGINEERING

2 Lane & 2 Way Highway Construction

Estimated period of construction = 3+6= 9 years

(CVPD) ADT of commercial vehicles = 5500nos

VDF from chart

Design life(n)= 15+9=24years

Traffic growth rate ()= (5-6)%

Lane distribution factor from code CBR of soil below 500 mm subgrade =5%

CBR of the 500mm of the subgrade from borrow pits = 10%

Ans

From IRC 37:2012 ,We have VDF for 5500 no of vehicles/day=4.5 for (Rolling /plain terrain)

Let us assume the traffic frowth rate = 5.5%

As per IRC37:2012 the distribution factor for 2 lane 2 way roads(D) = 75%

=>Computation of design traffic :-

365[(1+)n 1
N= A D VDF

Where N = The cumulative number of standard axles to be catered for in the design in terms of msa

A = Initial traffic in the year of completion of construction in terms on the no.of commercialvehicles/day.

= (1 + )

= 5500(1+0.055)9

= 8905 no./day

So the initial traffic = 8905

= 5.21 *108 msa

1.75
t=
%

t= Thickness of Pavement ,cm

P=Wheel load

=Tyre pressure,kg/cm2

A= Area of contact

= 2 =

Standard load , 2.5cm = 1370kg

5cm = 2055kg

For 5% below 500mm of subgrade We contact pressure P=

For heavy traffic where load = 5443kg

From chart,for CBR=5%

Minm thickness will be= 45cm

Pavement thickness(T)=54cm

Thickness of sub grade course ,

(tsb)=T-Tsb

Tb=Tsb-ts
For heavy traffic ,wheel load =5443kg
From the chart for CBR=10%

The depth of construction /pavement thickness(T) 35cm

(a) Thickness = (8+3+6) = 17 cm

Radius of relative stiffness = 90cm

Design wheel load = 4000 kg

Load capacity of dowel system = 40%

Joint width ,Z = 3cm

Permissible stress = 1000 kg/cm3

Bending stress = 1500 kg /cm3

Bearing stress ,fb = 100 kg/cm3

(b) Thickness = (12+9)=21 cm

Width of road = 7 cm

Unit weight of conc. () = 2400 kg/cm2

Co-efficient of friction = 1.5

Allowable working tensile stress in steel = 1750kg/cm2

Bond stress of deformed bars = 24.6 kg/cm2

Answer of (a) Permissible bearing stress in Conc. Is calculated as under :-

(10.16)
Fb = [ fck =400kg/cm2]
9.525

(10.163.2)400
Fb = = or,b= 7.78
9.525

8
Assuming ,spacing b/w the dowel bars = 30cm

First dowel bars is placed at a distance = 15cm( from the pavement edge)

Length of bar = 50cm

No.of dowel bars participating in load transfer when wheel load in just over dowel bar close to
the edge of the slab

= 1+

90
= 1+
30

= 1+3= 4 dowels

The total load transferred to the system ,

(
= 1+
9030
90
+
9060
90
+
9090
90
)pt
2 1
= (1+ + ) pt = 2pt
3 3

40000.4
Load carried by outer dowel bar =
2

pt = 800 kg

4 (8)4
=>Check for bearing stress = I = = = 201.06 cm4
64 64

Relative thickness of dowel bar in conc. = = /4

415008
= [ ]1/4
42106 201

= 0.12
2+
Bearing stress in dowel bar = (Pt* K) *
4 3
 {2+(0.123)}
= (800*41500) * ( )
40.123 2106 20

= 28.198 kg/cm2 < fb (safe).

Answer of (b) spacing and length of plain bar

71.524000.21
Area of steel ,(As) = = = 3.024 cm2/m
1750

Assuming , a diameter of tie bar = 16mm

1.62
The c/s area ,(A) = = 2.012 2 2
4

Perimeter of tie bar ,(P) = d = 1.6 = 5.03cm

2
Spacing of tie bars = (A/As) = 3.024 100 = 66

Assume 70

2 217502
Length of bars, L= = = 56.57
24.65.03

As per IRC 58:2002,including length 10cm for loss of bond due to painting and another 5 cm for
tolerance in placement

L = (56.57 + 10 + 5) = 71.57cm

Say = 72cm