Reading Material: Forest Engineering
For B. Sc. Forestry Third Year First Semester
Compiled by: Vishwa Nath Khanal
Kathmandu Forestry College (KAFCOL)
Koteswor
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Forest Engineering - compiled by Vishwa Nath Khanal
Table of Contents
Unit 1. Forest Roads, Trails and Simple Bridge (18)
....... 2
1.1 Forest road alignment and surveys. ............................................................................... 2
1.2 Design and construction of forest roads ...................................................................... 14
1.3 Importance of Geometric Design................................................................................. 23
1.4 Factors Controlling Geometric Design of Roads......................................................... 24
1.5 Road construction: introduction, type of road construction ....................................... 25
1.6 Base Course Design (thickness design, CBR method) ................................................ 28
1.7 Road Subgrade Soil ..................................................................................................... 28
1.8 Special Considerations of Hill Roads .......................................................................... 30
1.9 Failure and Maintenance of Road ................................................................................ 31
1.10 Stabilization of Road ................................................................................................. 33
1.11 Bridges ....................................................................................................................... 35
Unit 2. Design and Construction Procedure of Road Side Drainage, Retaining Walls and
Breast Walls ........................................................................................................................... 39
2.1 Definitions of Road Drainage and its importance ....................................................... 39
2.2 Types of Road Side Drainage (rectangular, trapezoidal triangular and circular) ........ 40
2.3 Design of Longitudinal Drains for Peak Runoff.......................................................... 42
2.4 Retaining walls ........................................................................................................... 42
2.5 Design and Construction Procedure of Retaining Wall ............................................... 45
2.6 Stability Analysis of Retaining Wall ........................................................................... 46
Unit 3 Designs and Construction Procedure of Forest House Cottages ................................ 47
3.1 structural principles ..................................................................................................... 47
3.2 forces stress moments and reactions ............................................................................ 47
3.3 Analysis of simply supported structures with BMD and SF ....................................... 51
3.4 wood RCC construction :............................................................................................ 56
3.5 Site selection criteria for a building ............................................................................. 68
3.6 Orientation of a building .............................................................................................. 69
Unit 4. Estimation and costing (4) ......................................................................................... 72
4.1. Estimate the quantities of the following items of a two roomed forest guard house.. 73
4.2. Estimation and costing for retaining walls, breast wall and check dams. .................. 74
4.3. Forest Roads ............................................................................................................... 83
4.4 Factors affecting the cost of a project: ......................................................................... 84
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Unit 1. Alignment, Design and Construction Procedure of Forest Roads, Trails and
Simple Bridge
(18)
General
The forest officers have to design construct and maintain forest roads, trails, drainage,
bridges and culverts, forest houses and cottages and masonry structures.
So that forest officers do have some knowledge of civil engineering like engineering
drawing, surveying, estimation and costing, design, construction and maintenance of roads,
bridges, houses and masonry structures.
Forestry students gave to study or cover a broad field of civil engineering in a limited
amount of time. Forest engineering plays vital role for the successful management and
exploitation of forest.
Definition of forest road
Road is a way on which people, animals and wheeled vehicles move. Products of forest are
generally bulky. Transportation of forest product from place of origin to the place of sale or
place of distribution affects their value. Forest resources are not yet used due to lack of
roads.
Transportation is vital for the economic development of any region. Inadequate
transportation facilities retard the process of socio-economic development of the country.
Good roads in forest are national wealth as they help to exploite forest resources for the
development of the nation.
Purpose of Forest road.
1. To move of transport men and material from one place to another place.
2. To transport forest product from place of origin to place of distribution.
3. To safe guard the forest resource and animals
4. To plan and manage forest properly.
1.1 Forest road alignment and surveys.
Alignment requirement
The basic requirements of a good forest road alignment between two terminal stations are:
1. short
2. easy
3. safe, and
4. economical
Short: It is desirable to have a short (or shortest) alignment between two terminal points.
Safe: The alignment should be safe enough for construction and maintenance from the view
point of stability of natural hill slopes.
Easy: The alignment should be such that it is easy to construct and maintain the road with
minimum problems. Also the alignment should be easy for the operation of vehicles with
easy gradients and curves.
Economical: The road alignment can be said economical, if the total cost including initial
cost, maintenance cost and vehicle operation cost is lowest.
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The alignment should be such that is should offer maximum utility by reserving maximum
population and products. Initials cost can be decreased by avoiding high embankments and
deep cuttings, and the alignment should be chosen in such a manner to balance cutting and
filling.
Factors controlling alignment (in brief)
The position (or the layout of the centerline) of the road on the ground is called the
alignment. Various factors which control the highway alignment are
1. obligatory points
2. traffic
3. geometric design
4. economy
5. Other factors.
1. Obligatory points
The points which control the alignment of a highway are known as obligatory points, these
control points may be divided into two categories.
i) Points though which the alignment should pass.
ii) Points through which the alignment should not pass.
i) Points through which the alignment should pass may cause the alignment to deviate from
the shortest path. The obligatory points through which alignment should pass may be bridge
site, mountain pass, and intermediate town of a quarry.
ii) The obligatory points through which the alignment should not pass also may cause
necessary to deviate form the shortest alignment. The obligatory points though which
alignment should not pass may be religious places (like temple, mosques, church), very
costly structures, long stretches of very hard rock cutting.
2. Traffic
The alignment should suit the requirement.
3. Geometric design
Geometric design factors such as gradient, radius of curves, sight distance also influence the
alignment of the highway. As far as possible the gradient should be flat and less than ruling
of design gradient.
4. Economic
The alignment finalized based on the above factory should also be economical. In working
out the economy, the initial cost maintenance and operation cost should be taken into
account.
5. Other factors
Various other factors which may govern the alignment are drainage, hydrological factors,
and political considerations etc.
Step in new road project work.
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The steps in a new road project should consider the basic requirements of an ideal alignment
that are short, easy, safe and economical. The alignment should be such that it would offer
maximum utility by serving maximum population and products.
Once the road is aligned and constructed, it is not easy to change the alignment due to
increase in value of adjoining land and construction of costly structures by the road side.
A new road should be aligned very carefully as improper alignment should result in one or
more of the following disadvantages:
i) Increase in construction cost.
ii) Increase in maintenance cost.
iii) Increase in vehicle operation cost.
iv) Increase in accident rate.
While finalizing the alignment of a new road, the importance of careful considerations need
not be overemphasized.
Cant or super elevation
When a vehicle passes from a straight to a curved path, centrifugal force and weight of the
vehicle act on the vehicle.
Centrifugal force is a function of the speed of the moving vehicle. It always acts at the
center of gravity of the vehicle. Centrifugal force has two effects on the moving vehicle.
They are to overturn the vehicle and to skid the vehicle laterally.
To counter this tendency the outer edge of the road is raised. This rise of outer edge is called
super elevation or cant. The super elation 'e' is the ratio of the height of outer edge to the
horizontal width.
NL
 tan
e = ML
In practice the inclination  with the horizontal is very small. Therefore, the value of tan 
is practically equal to sin 
e  tan  sin   EB  E  eB (as cant is very small). The total rise in outer edge of the
pavement with respect to the inner edge, e = eB.
Analysis of super elevation of cant.
When a vehicle passes from a straight to a circular curve of radius R meter at a speed of V
m/sec, two force acts on the vehicle.
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WV
acting horizontally outwards through the CG.
gr
ii) The weight of the vehicle W acting vertically downwards through the CG.
iii) The frictional force developed between the wheels and the pavement counteractions (act
against) transversely along the pavement surface towards the center of the curve.
i) The centrifugal force, P =
P = centrifugal force
W = weight of the vehicle
RA = reaction at wheel A
RB = reaction at wheel B
FA = frictional force at wheel A
FB = frictional force at wheel B
f = co-efficient of lateral friction.
Resolving all forces horizontally, for equilibrium condition, Pcos  = Wsin  + FA + FB.
At limiting equilibrium condition, Pcos  = Wsin  + f(RA + RB)
or Pcos  = Wsin  + f (Wcos  + Psin  )
or (Pcos  - f sin  ) = Wsin  + f Wcos 
dividing both side by Wcos  , WP (1-f tan  ) = tan  + f
tan   f
or WP =
1  f tan 
For design purposes
1
1
f = 0.15 tan  = , : f tan  = 0.15*
= 0.01
15
15
1-f tan  = 1-0.01 = 0.99  1.0
tan   f
:. WP =
= f + tan  = f + e.............(1)
1.0
But P/W = V2/gR....................(2)
v2
From (1) and (2), e + f =
.................(3)
gR
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If f is neglected or assumed to zero, e =
v
gR
If speed is taken in kilometer per hour then, V =
1000V
 0.278V
3600
(0.278V ) 2
V2
:. e + f =

9.8R
127 R
2
2
v
V
:. e =

gR 127 R
v2
V2
If, e = 0, f =

gR 127 R
Where
e = rate of super elevation = tan Q
f = design value of co-efficient of lateral friction = 0.15
v = speed of the vehicle, m/sec
R = radius of horizontal curve, m
g = acceleration due to gravity = 9.8 m/sec2
Steps for super elevation design
Step 1: the super elevation for 75 % of design speed (v m/sec. or V kmph) is calculated
neglecting the friction.
e=
(0.75v) 2
(o.75V ) 2
V2
, or e =
,

127 R
225R
gR
e
V2
..........................(4)
225 R
Step 2: If the calculated value of 'e' is less than 7 % or 0.07 the values so obtained is
provided.
If the value of e exceed as per equation (4), then provide the maximum super elevation equal
to 0.07 and proceed with step 3 or 4.
Step 3: check the co-efficient of friction developed for the maximum value of e = 0.07 at the
full value of design speed.
v2
v2
V2
 f 
 0.07 
 0.07
e+f=
gR
gR
127 R
If the value of f thus calculated is less than 0.15, the super elevation of 0.07 is safe for the
design speed. If not, calculate the restricted speed as given in step 4.
Step 4: As an alternative to step 3, the allowable speed (Va m/sec or Va kmph) at the curve
is calculated by considering the design co-efficient of lateral friction and maximum super
elevation, i.e.,
va 2
Va 2

e + f = 0.07 + 0.15 = 0.22 =
gR 127 R
Calculate the safe or allowable speed,
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va  0.22 gR  2.156R m/sec. or va  2.156 R m/sec.
Va  27.94 R kmph..........................(5)
If the allowable speed is higher than design speed then the design is adequate and provides a
super elevation of 'e' equal to 0.07.
If the allowable speed is less than the design speed then the speed is limited to the allowable
speed Va kmph.
Appropriate warning sign and speed limit regulation sign are installed to restrict and regulate
the speed at such curves when the safe speed Va is less than the design speed V.
Example 4.8 : Khanna and Justo
The radius of a horizontal curve is 100m. . The design speed is 50 kmph and the design coefficient of lateral friction is 0.15
a) Calculate the super elevation required of full lateral friction is assumed to develop.
b) Calculate the co-efficient of friction needed if no super elevation is provided.
c) Calculate the equilibrium super elevation if the pressure on inner and outer wheels should
be equal.
Solution
a) Super elevation is given by the relation
2
v2
V
e+f=

gR 127 R
Here R = 100m, V =50kmph, f = 0. 15
50 2
1
e + 0.15 =
or e + 0.15 = 0.197 or e = 0.197- 0.15 or e = 0.047 =
21.28
127  100
i.e, super elevation is 1 in 21.3
b) If no sup revelation is provided , e = 0,
V2
50 2
e+f =
or 0+f =
or f = 0.197
127 R
127  100
c) For the pressure on inner and outer wheels to be equal or for equilibrium super elevation
counter acting centrifugal force fully, f = 0,
V2
50 2
1
or , e  0 
or , e  0.197 
e+f=
127 R
127  100
5.08
i.e., in equilibrium super elevation rate is 1in 5.1, however this rate of super elevation being
very high cannot be provided.
Examples 4.9: Khanna
A two lane road with design speed 80 kmph has horizontal curve of radius 480 m. Design
the rate of super elevation for mixed traffic. By how much should the outer edges of the
pavement be raised with suspect to the centerline, if the pavement is rotated with respect to
the center line and the width of the pavement at the horizontal curves is 7.5 m. Width of the
pavement at the horizontal curves is 7.5m.
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Solution
Here V = 80 kmph, R = 480m, B = 7.5m and e = ?, E = ?
For mix traffic condition the super elevation should fully counteract the centrifugal force for
75% of design speed. Superelevation (for 75 % of design speed) is calculated neglecting the
friction.
0.75v 2 = 0.75V 2 = V 2
e=
gR
127 R
225 R
80 2
 0.059 , Since this value is less than 0.07, the super elevation of 0.059
225  480
may be adopted.
0.059  7.5
e.B
Raising of outer edge with respect to centre, E =
or E =
 0.22m
2
2
or e =
Example 4.10: Khanna
Design the rate of super elevation for a horizontal highway curve of radius 500 m and speed
100 kmph
Solution
For mixed traffic condition, the rate of super elevation is given by:
V2
100 2
e=

 0.089
225R 225  500
As this value is greater than the maximum super elevation of 0.07, the actual super elevation
to be provided is restricted to 0.07.
Check: For co-efficient of lateral friction developed for full speed,
V2
100 2
e +f =
or 0.07  f 
or , f  0.157  0.07, orf  0.087
227 R
227  500
As this value is less than 0.15, the design is safe with a super elevation of 0.07
Example 4.11: Khanna
The design speed of a highway is 80 kmph. There is a horizontal curve of radius 200 m on a
certain locality. Calculate the super elevation needed to maintain this speed. If the maximum
super elevation of 0.07 is not to be exceeded calculate the maximum allowable speed on this
horizontal curve as it is not possible to increase the radius. Safe limit of transverse coefficient of friction is 0.15.
Solution
Super elevation for 75 % of design speed is calculated neglecting the friction.
V2
80 2

 0.142
e=
225R 225  200
As this value is greater than the maximum super elevation of 0.07, the actual super elevation
to be provided is restricted to 0.07.
Check: for co-efficient of lateral friction developed
V2
80 2
or 0.07  f 
, or , f  0.251  0.07, or , f  0.181
e+f=
127 R
127  200
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As this value is greater than the maximum allowable safe friction co-efficient of 0.15 and
also as the radius can not be increased, the speed has to be restricted.
Hence, the maximum allowable speed (Va kmph) on this curve is obtained by assuming the
Va 2
Va 2
Va 2
full value of co-efficient of friction of 0.15, e + f =
or.07  .15 
, or.22 
127 R
127 R
127 R
:. Va = 27.94 R = 27.94  200 = 74.75 kmph.
Hence, the speed may be restricted to less than 74 or say 70 kmph at this curve.
Widening of pavement of Horizontal curves
(Extra widening, Mechanical and Psychological widening).
On horizontal curves, it is common to widen the pavement when the radius of horizontal
curve is less than about 300m. At horizontal curve the width of road pavement is widened
due to the following reasons.
1. An automobile has a rigid wheel base and only the front wheels can be turned. When a
vehicle takes a turn to negotiate a horizontal curve, the rear wheel do not follow the
same path as that of front wheels.
2. While two vehicles cross or overtake at horizontal curve there is a psychological
tendency to maintain a greater clearance between the vehicles, than on straights for
increased safety.
3. More right distance on the sides of pavements is required for safe driving with
permissible speed.
4. To have a greater visibility at curve the drivers find a difficulty to steer the vehicles
along the line of the lane.
Mechanical widening
The mechanical widening calculated above is required for one vehicle negotiating a
horizontal curve along one traffic lane. Hence in a road having 'n' traffic lanes, as 'n' vehicles
can travel simultaneously, the total mechanical widening required is given by:
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Wm 
2
nl
where, Wm = mechanical widening in m.
2R
Psychological widening.
The widening of pavement is future widened due to psychological reasons such as to
provide for greater manoeuvrability (make a movement or series of moves skillfully and
carefully) of steering at higher speeds, to allow for the extra space requirements for the
overhangs of vehicles and to provide greater clearance for crossing and overtaking vehicles
on the curves.
Additional psychological widening 'Wps' is dependent on the design speed V of the vehicle
and the radius R of the curve.
An empirical formula for psychological widening is given by:
V
Wps 
9.5 R
Hence the extra widening, We = Wm + Wps
nl 2
V
:. We=

2 R 9.5 R
Where n = number of traffic lanes.
l = length of wheel base of longest vehicle, m.
(l = 6.1 or 6.0 m for commercial vehicle, if not known)
V = design speed, kmph
R = radius of horizontal curve, m.
Extra widening = mechanical widening + psychological widening
:. We = Wm + WPs
Example 4.14: Khanna
Calculate the extra widening required for a pavement of width 7m on a horizontal curve of
radius 250 m if the largest wheel base of vehicle expected on the road is 7.0m. Design speed
is 70 kmph. Compare the value obtained with Indian Road Congress recommendation.
Solution
nl 2
V

2 R 9.5 R
Here, n =2 (two lanes for pavement width of 7.0 m), R = 250 m, V =7.0 kmph, l = 7.8 m,
2 * 72
70

 0.196  0.466  0.662m
:.We =
2 * 250 9.5 250
IRC recommends extra widening of 0.6m when the radius of the curve is 101 to 300m.
Extra widening, We = Wm + Wps,
Example 4.15: Khanna
Find the total width of a pavement on a horizontal curve for a new national highway to be
aligned along a ruling terrain with a ruling minimum radius. Assume necessary data.
Solution
Assume the following data: V = 80 kmph, W = 7 m, n = 2, l = 6m, e = 0.07, f = 0.15.
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2
R ruling =
2
V
80

 229.06m, say230m
127(e  f ) 127(.07  .15)
nl 2
V
2  62
........
Extra widening =
= 0.157 + 0.555 = 0.712m



2 R 9.5 R 2  230 9.5 230
Total pavement width on curve = W+We = 7.0 + 0.71 = 7.71m
Sight Distance, Sight Distance consideration
Sight distance may be defined as the length of road visible ahead to the driver from a
specified height above the carriageway while driving a vehicle. Visibility is important for
safe vehicle operation. Sight distances to be considered in the design are:
 Stopping (or absolute minimum) sight distance.
 Safe overtaking (or passing) sight distance, and
 Safe sight distance for entering into uncontrolled intersections.
The standards for right distance should satisfy the following three conditions:
1. Driver traveling at the design speed has sufficient sight distance or length of road visible
ahead to stop the vehicle, in case of any obstruction on the road ahead, without collision.
2. Driver traveling at the design speed should be able to safely overtake, at reasonable
intervals, the sluggard to traffic of opposite direction.
3. Driver entering an uncontrolled intersections (particularly unsignalised intersection) has
sufficient visibility to enable him to take control of his vehicle and to avoid colliding
collision with another vehicle
Most of the accidents or highways take place because of the inability of the drivers to stop
the vehicles well before it may collide with an obstruction or obstacle. The failure to stop the
vehicle in time may be due to i) the break failure, and ii) inadequate sight distance.
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a) sight distance at horizontal curve
Figure a,b,c, sight distance consideration
Apart from the three situations mentioned above intermediate sight distance and Head light
sight distance are considered by the IRC in highway design.
Intermediate sight distance: Intermediate sight distance is defined as twice the stooping
sight distance. When overtaking sight distance can not be provided, intermediate sight
distance is provided to give limited overtaking opportunities to fast vehicles.
Head light sight distance: Head light sight distance is the distance visible to a driver during
night driving under the illumination of the vehicle head lights. This sight distance is critical
at up gradients and at the ascending stretch of the valley curves.
1. Stopping Sight Distance (SSD)
The minimum sight distance available on a road at any spot should be of sufficient length to
stop a vehicle traveling at design speed, safely without collision with any other obstruction.
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The minimum stopping sight distance should be equal to the TWICE the stopping distance
to enable both vehicle coming from opposite directions to stop.
The sight distance available on a road to a driver at any instance depends on:
 Feature of the road ahead
 Height of the driver's eye above the road surface
 Height of the object above the road surface
The distance within which a motor vehicle can be stopped depends upon the following
factors:
1. Total reaction time of the driver : perception time and break reaction time
2. Speed to the vehicle
3. Efficiency of brakes
4. Frictional resistance between the road and tyres
5. Gradient of the road, if any.
Stopping distance = lag distance + breaking distance.
Lag distance is the distance traveled by the vehicle during the total reaction time.
2. Overtaking Sight Distance (ODS)
The minimum overtaking sight distance (OSD) or the safe passing sight distance may be
defined as the minimum distance open to the view of the driver of a vehicle who is intending
to overtake a slower vehicle ahead with safety against the traffic coming from the opposite
direction.
For fast moving vehicle it is necessary to overtake the slow moving vehicles. It may not be
possible to provide the facility to overtake slow moving vehicles throughout the length of a
road. In such cases facilities for overtaking slow vehicles with adequate safety should be
made possible at frequent distance intervals.
Breaking distance is the distance traveled by the vehicle after the application of the brakes to
a dead stop position.
Stopping distance (SD) = lag distance + braking distance
SD =( vt + v2/2gf ), m.
Where, v = speed of vehicle in m/sec. t = reaction time = 2.5 sec., f =d esign co-efficient of
friction = 0.4 to 0.35, g = 9.8m/sec, If speed is kmph,
V2
SD = 0.278 V.t +
254 f m
A is the overtaking or fast moving vehicle at design speed V kmph or v m/sec.
B is the overtaken or slow moving vehicle with uniform speed vb m/sec or Vbkmph
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C is the vehicle coming from the opposite direction at design speed v m/sec or V kmph.
It is assumed that vehicle A is forced to reduce its speed to the speed Vb of the slow vehicle
B and moves behind it allowing a space S, till there is opportunity for safe overtaking
operation.
d1  vb  t ,
or , d1  2vb, m.
S  (0.7vb  6), m
Where, t = 2 second ie, reaction time of the driver,
'S' is given by empirical formula.:
d2 = ( b + 25), where, b = vb *T.
Now the time depends on speed of overtaken vehicle B and the acceleration of overtaking
vehicle A.
'T' may be calculated by equating the
distance
1
d 2 to (vb.T  aT 2 ),
2
 d 2  b  25  vb.T  25...........i,
1
d 2  vb.T  .aT 2 ..............ii
2
equation , vb.T  25  vb.T 
1
45
aT 2 T 
sec, wher , S  (0.7vb  6), m
2
a
d 2  (vb.T  25), m
d 3  (v  T ), m.
OSD  d 1  d 2 foronewayt raffic .
OSD  d 1  d 2  d 3
 vb.t   vb.t  25  v.T  foronewayt raffic .
inkmphuni, ODS  (0.28.Vb.t  0.28Vb.T  25  0.28V .T )
where, Vb  speedinkmph, t  2 sec, V  speedinkmph, T 
14.45
, A  accelerati oninkmph / sec
A
s  (0.2vb  6)
1.2 Design and construction of forest roads
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. Geometric design
The geometric design of road is the most important aspect of road construction. The
geometric design of a road should be such as to provide optimum efficiency in traffic
operations with maximum safety at reasonable cost. The geometric design of highways deals
with the following elements:
 Cross-section elements: pavement width, formation width, surface characteristics,
cross- slope etc.
 Sight distance
 Horizontal alignment details
 Vertical alignment details
 Intersection elements.
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Camber or Cross - Slope
Camber is the slope provided to the road surface in the transverse direction to drain off the
rain water from the road surface.
Cross slope is considered important to prevent the entry of surface water into the sub-grade
soil through pavement and to remove the rain water from the pavement surface as quickly as
possible.
Usually the camber is provided on the straight roads by raising the center of the
carriageway. The minimum camber needed to drain off surface water depends on the type of
pavement surface and the amount of rain fall.
The rate of camber or cross-slope is expressed as 1 in n (1 vertical to n horizontal) and as a
percentage.
Type of Camber
a) Parabolic Camber
In parabolic or elliptic camber profile is flat at the middle and steeper towards the edges,
which is preferred by fast moving vehicles as they have to frequently cross the crown line
during overtaking operation on a two lane highway.
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c) Composite Camber
Composite camber consists of two straight slopes with parabolic portion at the center.
S.N Type of road surface
.
1
Cement concrete or high type bituminous surface
Heavy to Light
to
1 in 60
2
Thin bituminous surface
1 in 40 to
1 in 50
3
Water bound macadam and gravel pavement
1 in 33
to
1 in 40
4
Earth
1 in 25
to
1 in 33
1 in 50
The cross slope for shoulders should be 0.5 % steeper than the cross slope of adjoining
pavement.
Providing Camber in the Field
For providing the desired amount and shape of camber, templates or camber boards are
prepared with the specified
40
camber.
v b  40kmph  v b 
 11.11m / sec .
3.6
In the case of parabolic
2
camber the general equation
a  0.99m / sec
x2
OSD  d 1  d 2  d 3 fortwo  waytraffic
may be adopted.
y
a
d 1  vb.t  11.11  2  22.22m
nw
Here, a 
for a pavement
45
2
d 2  vb.T  25  vb
 2(0.7vb  6)
a
of width W and cross- slope 1
in n.
S  90.7vb  6  0.7  11.11  6  13.78m
2x 2
45
4  13.78
Hence, y 
T

 7.46 sec .
nw
a
0.99
: .d 2  11.11  7.46  2  13.78  82.88  27.56  110.44m
d 3  vT  19.44  7.46  145.02m
: .OSD  d 1  d 2  d 3  22.22  110.44  145.02  277.68m
say278m
Overtaking Zone
The OSD and pavement width should be sufficient for safe overtaking operation. Signposts
should be installed at sufficient distance in advance to indicate the start of overtaking zones.
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Similarly the end of the overtaking zones should also be indicated by appropriate signpost
installed ahead at distances specified above.
OSD = d1  d 2 for one-way road.
OSD = d1  d 2  d 3 for two-way road.
The minimum length of overtaking zone should be THREE times the OSD.
It is desirable that the length of overtaking zones is kept FIVE times the OSD.
Example 4.6: Khanna
The speed of overtaking and overtaken vehicles are 70 and 40 kmph, respectively on a two
way traffic road. If the acceleration of overtaking vehicle is 0.99 m/sec,
a) Calculate safe overtaking sight distance
b) Mention the minimum length of overtaking zone and
c) Draw a neat-sketch of the overtaking zone and show the positions of the sign posts.
Solution
a) Here, V  70kmph  v 
70  1000 70

 19.44m / sec .
60  60
3.6
40
 11.11m / sec, a  0.99m / sec 2
3.6
OSD  d1  d 2  d 3 fortwowayt raffic
Vb  40kmph  vb 
d1  vb.t  11.11  2  22.22m
45
 2(0.7vb  6)
a
S  (0.7vb  6  0.7  11.11  6  13.78m
d 2  vb.T  25  vb.
45
4  13.78

 7.46 sec
a
0.99
 d 2  11.11  7.46  2  13.78  82.88  27.56  110.44m
T
d 3  vT  19.44  7.46  145.02m
OSD  d1  d 2  d 3  22.22  11044  145.02  277.68m  say278m
b) Minimum length of overtaking zone = 3*OSD = 3*278 = 834m
Desirable length of overtaking zone = 5* OSD =5 *278 = 1390m
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Example 4.7: Khanna
Calculate the safe overtaking sight distance for a design speed of 96 kmph. Assume all other
data suitably.
Solution
V = 96kmph, Assume Vb = 80kmph, A = 2.5kmph/sec., t = 2sec.
V  96kmph, assumeVb  80kmph, A  2.5kmph / sec, t  2 sec .
v
96 *1000
80 *1000
2.5 *1000
 26.76m/sec.v b 
 22.22m/sec., a 
 0.96m/sec 2
60 * 60
60 * 60
60 * 60
d1  0.28Vb.t  0.28 * 80 * 2  44.8m  45m.
vb * t  22 * 22 * 2  44.44m  45m
d 2  0.28Vb.T  25where, S  0.2Vb  6  0.2 * 80  6  22m
vb.T  25  .7vb  6  .7 * 22.22  6  21.55  22m
14.45
14.4 * 22
T

 11.26 sec .
A
2.5
45
4 * 22

 11.29 sec .
a
.69
: .d 2  0.28 * 80 * 11.26  2 * 22  296.22m  say297m
T
 22.22 * 11.29  2 * 22  294.86m  295m
d 3  0.28VT  0.28 * 96 *11.26  302.67m  303m
 v.T  26.67 * 11.29  301.10m  302m
: .OSDforone  waytraffic  d1  d 2  45  297  342m
 45  295  340m
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OSDfortwo  waytraffic  d1  d 2  d 3  45  297  303  645m
 45  295  302  642m
Gradient
Gradient is the slope (rate of rise or fall) along the length of the road with respect to the
horizontal.
V
 slope  gradient
H
Gradient is expressed as:
1. As an angle i.e, tan 
tan  
5
1

 1in 20
100 20
3. As 1 vertical in n horizontal, ie, 1 in n
1
1 5
5
1 in 20 =

 
 5%
20 20 5 100
2. As a percentage, ie, 5% =
Type of Gradient
1. Ruling gradient
2. Limiting gradient
3. Exceptional gradient
4. Minimum gradient.
Ruling gradient
Ruling gradient is the maximum gradient within which the designer attempts to design the
vertical profile of a road. Hence ruling gradient is also known as design gradient.
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Limiting gradient
Limiting gradient is steeper than ruling gradient. When topography of a place compels
adopting steeper gradients than ruling gradients, limiting gradients are used to avoid excess
cost of cutting the steeper slopes.
Exceptional gradient
Exceptional gradients is even steeper than limiting gradient. Exceptional gradient is
provided in some extra-ordinary situation over a short stretch of road alignment. However
the exceptional gradient should be strictly limited only for neither short stretches nor
exceeding about 100m at a stretch.
Minimum gradient
Minimum gradient is provided for easy drainage of rainwater. The minimum gradient should
depend on rainfall run off, type of soil, topography etc.
Radius of Horizontal curve
For a certain seeped of vehicle the centrifugal force is dependent on the radius of the
horizontal curve. To keep the centrifugal ratio within a low limit, the radius of the curve
should be kept correspondingly high.
The centrifugal force which is counteracted by the super-elevation and lateral friction is
v2
v2
given by the relation, e + f =

gR 127 R
Where, e = rate of super elevation = 0.07
(Maximum allowable super elevation has been fixed as 7 % or 0.07)
f =design co-efficient of lateral friction = 0.15
If the design speed is decided for a highway, then the minimum radius to be adopted can be
found from the above relationship.
2
v2
V2
v1
V2
e+f=

, : .Ruling 
, Alsoruling 
gR 127 R
(e  f ) g
127e  f 
When, minimum design speed V’ kmph is adopted, the absolute minimum radius of
horizontal curve R min is given by:
2
V1
R min =
127e  f 
In the above equation,
v and V= ruling design speed in m/sec. and kmph respectively.
V 1  min imumdesignspeed , kmph.
Example 4.13: Khanna
Calculate the values of ruling minimum and absolute minimum radius of horizontal curve of
National Highway in plan terrain. Assume ruling design speed and minimum design speed
values as 100 and 80 kmph respectively.
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Solution
V  100kmph, V 1  80kmph, e  0.07, f  0.15
Here, : .Ruling 
V2
100 2

 357.91m, say360m
127(e  f ) 127(.07  .15
2
v1
80 2
R min 

 229.00m, say230m
127(e  f ) 127(.07  .15)
Therefore ruling minimum radius of horizontal curve = 360m
Absolute minimum radius of horizontal curve = 230m.
Vertical curve its type (in brief)
Whenever two grades meet along the proposed alignment of a highway, vertical curves are
provided for gradual change of grade. An abrupt change in the rate of grade could otherwise
subject a vehicle to an impact, which may be either injurious or dangerous.
The vertical curve, thus, contributes to the safety, comfort and appearance. Generally an arc
of parabola is provided in vertical curve. The parabolic curve produces the best riding
qualities as the rate of change of grade is uniform throughout the curve.
The equation of parabola with vertical axis is given by:
Y = ax 2  bx
dy
Slope of this curve,
 2ax  b
dx
d2y
 r  2a  cons tan t .
Rate of change of slope, or rate of change of grade ( r ) =
dx 2
Thus, the grade changes uniformly throughout the curve, which is a desired condition.
Types of Vertical Curve
Vertical curves on highways may be classified into two categories.
1. Summit curves or crest curves with convexity upwards.
2. Valley or sag curves with concavity upwards.
Summit curve
Summit curves with convexity upwards are formed in any one of the causes given below.
The deviation angle between the two intersecting gradient is equal to the algebraic
difference between them.
1. An upgrade followed by a downgrade.
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2. An upgrade followed by zero grade or level.
3. An upgrade followed by another upgrade.
4. A down grade followed by another down grade.
2. Valley curves
Valley curves or sag curves are formed in any one of the cases given below. The lowest
point in the valley curve may be located from considerations of cross drainage.
1. A down grade followed by another upgrade.
2. A down grade followed by zero grade or level.
3. An upgrade followed by another upgrade
4. A down grade followed by another down grade.
1.3 Importance of Geometric Design
The geometric design of road is the most important aspect of road construction. The
geometric design of a highway deals with the dimensions and layout of visible features of
the highway such as alignment, sight distances and intersections. The geometric design of a
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road should be such as to provide optimum efficiency in traffic operations with maximum
safety at reasonable cost.
It is possible to design and construct the pavement of a road in stages but it is very
expensive and rather difficult to improve the geometric elements of road in stages at a later
date. Therefore it is important to plan and design the geometric features of the road cutting
the initial alignment itself taking into consideration the future growth of traffic flow and
possibility of the road being upgraded to a higher category or to a higher design speed
standard at a later stage.
Geometric design of highways deals with the following elements:
1. Cross-section elements: pavement width, formation width surface characteristics, crossslope etc.
2. Sight distance
3. Horizontal alignment details
4. Vertical alignment details
5. Intersection details.
The sight distance visible ahead of a driver, at horizontal and vertical curves and at
intersections govern the safe movement of vehicles. The changes in the road directions are
made possible by introducing horizontal curves. Super elevation is provided by raising the
outer edge of pavement to counteract the centrifugal force developed on a vehicle passing
over a horizontal curve. Extra pavement width is also provided on horizontal curves.
Vertical curves are provided in the vertical alignment of a road for gradual change of grade.
The efficiency, safety, speed, cost of operation and capacity of road system depends on the
intersection design. Design of road intersections with facilities for safe and efficient traffic
movement needs adequate knowledge of traffic engineering.
1.4 Factors Controlling Geometric Design of Roads
The important factor which control the geometric design of roads are:
1. Design speed
2. Topography
3. Traffic factors
4. Design hourly volume and capacity
5. Environmental and other factors.
Design speed
Design speed is the most important factor controlling the geometric design elements of
highways. Design of almost every geometric design element of a road is dependent on the
design speed. The cross section elements, sight distance, radius of curve, super elevation,
gradient, summit and valley curve lengths, pavement surface characteristic etc. all depend
mainly on the design speed of the road.
Topography
The topography or the natural features across the alignment influence the geometric design
significantly. The design standards specified for different classes of roads are different
depending on the terrain classification such as plain terrain, rolling terrain and mountainous
terrain.
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Traffic factors
Traffic factors that affect geometric design of roads are the vehicular characteristics and
human characteristics of road users. The important human factors which affect traffic
behaviour include the physical, mental, and psychological characteristics of drivers and
pedestrian.
The different vehicle classes such as passenger cars, buses, trucks, motorcycles etc, have
different speed and different dimension. However it is often necessary to consider some
standard vehicle as the design vehicle.
Design hourly volume and capacity
The traffic flow or volume fluctuates with time, such as low value during off-peak hours and
highest value during peak hours.
It will be uneconomical to design the roadway facilities for the peak traffic flow. Therefore a
reasonable value of traffic volume is decided for the design, which is known as the design
hourly volume.
Environmental and other factors
The environmental factors such as aesthetics, landscaping, air pollution, noise pollution and
other local conditions should be given due consideration in the geometric design of roads.
Important main roads and express-ways are designed for higher speed standard and
uninterrupted flow of vehicles by providing grade separated intersections and controlled
access.
1.5 Road construction: introduction, type of road construction (earthen road, graveled
road, WBM road)
Introduction
The purpose of road construction is to provide a firm and even surface for the carriage way
or the pavement to withstand the subsequent loads . Road construction may be divided into
groups:
 earth work
 pavement construction
Earthwork
The earthwork mainly consists of preparing the subgrade to a suitable grade (either in
excavation or embankment) depending upon the topography. Preparation of subgrade
includes site clearance, designed grade (cutting and filling), camber and adequate
compaction.
Pavement construction
Broken stone or soil (for stabilized soil) and binding material is spread on the prepared
subgrade and rolled in layers of about 15cm by sprinkling water.
Binding materials may be soil slurry, bituminous materials and cement. Type of aggregates,
binders, aggregate gradations and the proportion of the binders are decided depending upon
the desired strength of the pavement.
Types of Road Construction
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1. Earth road
2. Graveled road
3. Water bound macadam (WBM) road
4. Soil stabilized road
5. Bituminous road or black-top road
6. Cement concrete road.
Earth Road
A majority of roads in the forest are earth roads. Earth roads are fair weather roads and the
traffic passing on them is relatively light. Earth road is an example of low cost roads. Earth
road is constructed from the soil available at site and near by borrow pits. The camber
provided to the earthen road is very steep. The steep cross slope or camber helps to drain off
the rain water from the road surface as quickly as possible and to allow the pavement to get
dry soon after the rain; otherwise the soil being pervious, the water would damage the
pavement section by softening it.
The maximum cross slope of 1 in 20 to 1 in 33 is recommended to avoid erosion due to rain
water and formation of cross ruts. The performance of earthen road mainly depends upon the
effective maintenance and drainage. Construction of earthen road includes operations like
site clearance, grading (cutting or filling) and compaction. The formation level of earthen
road may be on embankment or excavation or at the existing ground surface.
In all the cases, site should be cleared off and the trees, shrubs, grass roots, rubbish and other
organic matter including top soil are to be removed before excavating earth for construction.
After clearing and grabbing the route, next operation is grading so as to bring the vertical
profile of earthen road to designed grade and camber. After grading next operation is
compaction. Bulldozers and blade graders are useful equipment to speed up this work.
Compaction increases density and stability, reduces settlement and lowers the adverse
effects of moisture. Hence proper compaction is essential for earthen road. It is always
desirable to compact the soil at the optimum moisture content that would give maximum dry
density. Soil compaction is made either by rolling or vibration. A certain percentage (say 95
%) of the standard density is aimed at in the field compaction. Sand replacement method is
considered quite satisfactory to find dry density.
Gravel Road
Gravel roads are superior to earth roads. Screened or crushed gravel and a suitable binder is
used in carriageway of gravel road. The camber in gravel road ranges between 1 in 25 to 1 in
30. This type of road can cater for about 100 tones to pneumatic tired vehicle or 60 tonnes of
iron tired vehicles per day per lane. Demanding upon the construction method. Gravel roads
are classified as:
 The feather edge type
 The trench type
The feather edge type: The feather edge type gravel road is constructed over the sub grade
with varying thickness so as to obtain the desired cross slope for the pavement surface.
The trench type: In trench type gravel road, the sub grade is prepared by excavating a
shallow trench. The trench type is preferred as there is a better confinement for the gravel.
Round stones and river gravel are not preferable as there is poor interlocking.
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Hard variety of crushed stone or gravel is used in pavement. There are no specifications of
material. Rounded stones and river gravel are not preferable as there is poor interlocking.
Water Bound Macadam (WBM) Road
John Macadam introduced new concept of road construction which became known after his
name as WBM road by the year 1827. Macadam suggested that heavy foundation stones are
not at all necessary at the bottom layer of pavement. Construction steps are:
1. Subgrade is compacted and prepared with a cross slope of 1 in 36 up to a desired width
(about 9 meters).
2. Broken stones of a strong variety, all passing through 5 cm size sieve are compacted to a
uniform thickness of 10 cm in bottom layer.
3. Broken stones of size 3.75 cm are compacted to a thickness of 10 cm in second layer.
4. Stones of size less than 2 cm are compacted to a thickness of 5cm in top layer and cross
slope of pavement surface is maintained 1 in 36.
In WBM pavement, crushed or broken aggregates are mechanically interlocked by rolling
and the voids are filled with screening and binding material with water. Compaction is done
by a three wheeled power roller of capacity 6 to 10 tonnes or by a vibratory roller. After the
application of screenings, the surface is sprinkled with water, wet slurry swept with brooms
to fill the voids and compacted. After final compaction the WBM course is allowed to set
overnight.
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1.6 Base Course Design (thickness design, CBR method)
In 1928 California Division of Highways in the USA developed California Bearing Ratio
(CBR) test method for evaluating the stability of soil subgrade and other flexible pavement
materials. In this method the strength of subgrade soil and the construction materials is
determined from an adhoc penetration test and the required thickness of pavement is
obtained from design curves evolved from experience.
The CBR test is carried out either in the laboratory on prepared specimens or in the field on
existing pavement layers including subgrade, subbase and basecourse. The CBR value is
calculated using the relation:
load (orpressure) sustainedbythespecim enat 2.5or 5mmpenetration
CBR % 
 100
load (orpressure) substainedbys tan dardaggregatesatthec orrespondingpenetrat ionlevel
Studies have shown that there exists a relationship between pavement thickness, wheel load
tyre pressure and CBR value within a range of 10 to 12 %.
CBR design curve for various loading conditions is given by the expression:
t
 1.75
1 

P 

CBR
p
^


1
2
1
1.75P A  2
=
 
 CBR ^ 
Where,
t = Pavement thickness, cm.
P = Wheel load, kg.
CBR = California bearing ratio, %.
P = Tyre pressure, kg/cm 2
A = Area of contact cm 2
However, these expressions are applicable only when the CBR value of the subgrade soil is
less than 12 %.
Example 7.4: Khanna
The CBR value of sub grade soil is 5 %, calculate total thickness of a pavement using design
formula. Assume 4100kg wheel load or medium light traffic of 200 commercial vehicles per
day for design.
Tyre pressure = 6kg/ cm 2
Solution
1
t
 1.75
1 2

P 

CBR
p
^


1
=
1.75P A  2
4100 
  = 35.5cm.
 CBR ^ 
1.7 Road Subgrade Soil
(Significance of subgrade soil, Characteristics of soil, Desirable properties, Soil moisture,
Dry density and Soil strength).
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Based on grain size of soil, soil have been classified as gravel, sand, silt and clay. The limits
of the grain size are shown below:
Gravel
Sand
Silt
Clay
coarse medium fine course medium fine coarse medium fine
or
colloidal
Note: values are in mm.
Significance of Subgrade Soil
Subgrade soil is an integral part of a road pavement structure as it provides the support to the
pavement from beneath. The properties of the subgrade soil are important in deciding the
thickness requirement of pavement. A subgrade with lower stability requires thicker
pavement to protect it from traffic loads.
Soil is the therefore considered as one of the principal highway material. The main function
of the subgrade is to give adequate support to the pavement. The formation of waves,
corrugations, rutting and shoving in blacktop pavements, and the phenomena of pumping,
blowing and consequent cracking in cement concrete pavement are generally developed due
to the poor sub grade. In addition to stability, incompressibility is also important, as
differential settlement may cause failures.
Characteristics of Soil
Soil is formed by the disintegration of rocks, by the action of water, frost, temperature
variations, pressure and animal life. The characteristics of soil depend on the size, shape,
surface texture, and chemical composition of its grains.
Based on the individual grain size of soil particles, soils have been classified as gravel, sand,
silt and clay. Moisture and dry density influence the engineering behavior of a soil mass.
Desirable properties of soil
The desirable properties of soil as a highway material are:
1. Stability
2. Incompressibility
3. Permanency of strength
4. Minimum changes in volume
5. Good drainage
6. Ease in compaction
The soil should possess adequate stability or resistance to permanent deformation under
loads and resistance to weathering.
Minimum changes in volume will ensure minimum variation in differential expansion and
differential strength values.
Good drainage is essential to avoid excessive moisture retention and to reduce the potential
forest action.
Ease in compaction ensures higher dry density and strength.
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Soil Moisture
It is well known fact that maximum dry density would be obtained in optimum moisture
content for a particular type and amount of compaction. Hence it is always desirable to
compact the soil at the optimum moisture content. The moisture content of the soil may be
found before compaction.
Dry density
If the moisture is controlled at the optimum moisture content, then the next control needed is
dry density. Sand replacement method is considered quite satisfactory to obtain the dry
density.
A certain percentage (say 100 or 95%) of the standard density is generally aimed at in the
field compaction. Thus by field checks, it is possible to control the construction to achieve
adequate compaction.
Subgrade Soil Strength
The factors on which the strength of soil depends are:
1. Soil type
2. Moisture content
3. Dry density
4. Internal structure of the soil
5. The type and mode of stress application.
Stability of resistance to deformation of the soil is important in highway construction.
In a soil mass, the deformation is largely due to slippage between soil particles. The strength
property or the stability of soil is often determined from its stress-deformation
characteristics.
Soil strength are evaluation by i) shear tests, ii) bearing tests, and iii) penetration tests.
1.8 Special Considerations of Hill Roads (a typical example of hill road showing all
component of hill road)
Hill Road
A hill road may be defined as the road which passes through a terrain with a cross slope of
25 percent or more. A hilly or mountainous area is characterized by a highly broken relief
with widely differing elevations, steep slopes, deep gorges and a number of water courses.
The route length of hill road ineffectively increases due to complex topography. Selection of
a suitable alignment, construction and maintenance is a difficult job.
Construction of road in hilly region disturbs the natural stability condition. Prevention of
soil erosion and stabilization of hill slopes has been major problems in the maintenance of
hill roads. At many locations of the hill road, the land slides and slips occurs blocking traffic
during the heavy rainfall. The effects of heavy rainfall are serious in the construction and
maintenance of hill roads.
Geometric standards as of plains can not be adopted in hills. Massive and costly protective
works are required at many places in the hill roads resulting in heavy expenditure. Large
quantity of earthwork including blasting of hard rock may be needed in hill road
construction.
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Various steps in the alignment of a hill road include map study, reconnaissance, preliminary
survey and detailed survey. In order to divert the water from the hill slope, catch water
drains are provided, running parallel to the roadway. Water from the catch water drains is
diverted by sloping drains and carried across the road by means of culvert. Side drain is
provided only on the hill side of the roads and not on both sides. As far as possible cross
drainage should be taken under the road and at right angle to it.
Retaining wall is constructed to maintain stability of soil and to prevent land slide in hill
road. Main function of retaining wall is to retain the back filling. Breast walls are provided
on the inner side of the hill road to give support to cut hill or to prevent the cut surface of
hill form sliding.
1.9 Failure and Maintenance of Road
Need for maintenance
Road maintenance may be defined as to maintain and upkeep the pavement in excellent
condition.
Even if the roads are well designed and constructed, they may require maintenance. The
extent of maintenance will depend on several factors including pavement type. Roads are
damaged due to heavy mixed traffic and adverse climatic condition. Various types of
failures in pavement do takes place due to several causes ranging from minor failures to
major failure.
Road maintenance involve the assessment of road condition, diagnosis of the problem and
adopting the most appropriate maintenance steps. Road maintenance is an essential and
important component to upkeep the pavement in good condition.
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General causes of pavement failure
Some of the general causes of pavement failures that require maintenance are:
1. Defects in the quality of materials used.
2. Defects in construction methods and quality control during construction
3. Inadequate surface or sub-surface drainage resulting in the stagnation of water in the
subgrade.
4. Increase in traffic volume and magnitude of wheel loads.
5. Settlement of foundation of embankment of the fill material.
6. Factors like heavy rainfall, soil erosion, high water table, snow fall, frost action etc.
Classification of Maintenance Work
Road maintenance works may be broadly classified as:
1. Routine maintenance repairs: Routine maintenance include filling up of pot holes and
patch repairers, maintenance of shoulders and cross slope, upkeep of the road side drains
clearing choked culverts road signs arboriculture planting in a shady place etc.
2. Periodic maintenance: Periodic maintenance includes renewals of wearing course of
pavement surface and preventive maintenance of various items.
3. Special repairs: Special repairs include strengthening of pavement structure or overlay
construction re construction of pavement widening of roads repairs of damages caused
by floods providing additional safety measures like inlands sings etc.
Maintenance of Earth Roads
Usual damages in earth roads needing frequent maintenance are:
1. Formation of dust in dry weather.
2. Formation of longitudinal ruts along wheel path of vehicles.
3. Formation of cross ruts due to rain water.
4. Formation of gullies in side-slopes.
5. Growing of vegetation inside drains and silting in drains
Remedy of damages in earth roads are:
1. The dust nuisance may be remedied by frequent sprinkling of water or treating the
surface with chemicals (calcium chloride).
2. Formation of ruts need periodic maintenance by spreading moist soil along the ruts and
compacting in properly.
3. Frequent reshaping of the camber is necessary.
4. Cross-ruts should be repaired frequently during and after the monsoon or stabilized layer
may be provided on the top.
5. Formation of gullies in side slopes may be treated by dressing the side slopes.
6. Side drains may be maintained by silt clearance and removing vegetation form them.
Maintenance of WBM Roads
WBM roads are damaged rapidly due to heavy mixed traffic and adverse climatic
conditions.
In dry weather dust is formed and during rains mud is formed.
The steel tyred bullock carts cause severe wear and tear to the WBM surface.
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The fast moving vehicles raise dust in dry weather and churn-up (stir or move about
violently) mud in wet conditions.
Due to the combined effect of traffic and the rain water washing away the soil binder from
the surface, the stone aggregates get loose on the surface layer. In such situation, some stone
aggregates get displaced causing ruts and potholes on the road surface.
Remedies
1. To prevent the aggregate from getting loosened from the road surface, a thin layer of
moist soil should be spread over the surface and compacted periodically as a part of
periodic maintenance, particularly after the monsoons.
2. Dust nuisances can be effectively prevented by providing bituminous surface dressing
course over WBM pavement.
3. Potholes and ruts formed should be patched up. Wet soil binder is applied on the patched
surface and the surface is rammed.
1.10 Stabilization of Road
Introduction
The greatest problem in the construction of new road network and improvement of village
road is the inadequacy of the available fund. The road engineer have to cope with the
problems such as lowering the initial construction cost construction pavements and
strengthening existing pavements in stages to cater for the increasing demands of mixed
traffic and maintaining road for all weather conditions.
It has been shown by various investigators that soil stabilization methods have considerable
scope in decreasing the initial construction cost of pavements and for stage construction.
The term soil stabilization means the improvement of the stability of soil by the use of
controlled compaction or by the addition of suitable admixture or stabilizers.
Basic principles of soil stabilization are:
1. Evaluating the properties of given soil.
2. Deciding the defective and economic method of stabilization.
3. Designing the stabilized soil mix.
4. Compacting the stabilized layers.
Soil stabilization may result the following changes:
1. Increase in stability.
2. Change in physical characteristics.
3. Change in chemical properties.
4. Retaining minimum strength by water proofing.
Objective of Soil Stabilization
If the stability of the local soil is not adequate for supporting wheel loads, the properties are
improved by soil stabilization techniques.
Principle of soil stabilized road construction involves the effective utilization of local soils
and suitable admixture or stabilizers.
Soil stabilization methods are used in decreasing the initial construction cost of pavements
and for stage construction.
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Greatest problem in developing countries is limited fund available to build road network to
meet the growing needs of the road traffic. Low cost road can be upgraded such as sub-base
course to higher specification at a later date in stages.
Construction cost can be considerably decreased by selecting local materials including local
soils for the construction of lower layers of the pavements such as the sub-base course.
Soil stabilization improves the stability density and strength of lower layers of the
pavements.
Various Techniques of Soil Stabilization
Various techniques of soil stabilization are:
1.
Proportioning techniques : Various locally available soils and aggregates are mixed
in suitable proportions and compacted to achieve the desired objective. For example,
the stability of a fine grained soil can be improved by the addition of gravel and sand
in suitable proportions. Like-wise the stability of a cohesionless sand may be
improved by the addition of some cohesive soil.
2. Cementing agents: The strength of the stabilized soil can considerably be increased
by the addition of cementing agents like Portland cement, lime or lime fly-ash.
Bituminous materials also impart binding effect to non-cohesive soils.
3. Modifying agents: Undesirable properties of certain soils such as high plasticity,
swelling etc. can be modified by adding the stabilizer in small proportion making
them more useful as construction material. Such stabilizers may be called modifiers.
The most common modifier used in the case of highly plastic soils is lime. Port land
cement also acts as modifier in some cases.
4. Water proofing agents: A compacted soil mass which is normally stable enough
may become weaker by the ingress of water. If the absorption of water can be
stopped or retarded by means of some water proofing agent, it will be possible to
make use of such materials with advantage. Bituminous materials is the most
common water proofing agent.
5. Water repelling agents: Vinsol resin and other resinous materials perform almost
the same function as water proofing agents.
6. Water retaining agents: Some non-cohesive soils have sufficient stability when the
compacted layer possesses slight moisture content, but the soil may become loose
and less stable when completely dried. Calcium chloride is likely to be useful to
retain some moisture to impart some apparent cohesion and thus retain the stability.
Calcium chloride reduces the dust nuisance in unsurfaced roads.
7. Heat treatment: Thermal stabilization has useful aspects as regards clayey soils.
Heat treated soil may be used as a soft aggregate in mechanical soil stabilization that
results reduction in swelling properties of soil.
8. Chemical stabilization: chemicals have also been successfully used as additives in
soil cement and soil line stabilization.
Adequate compaction of the stabilized layers is the most essential requirements in all the
methods or techniques of soil stabilization.
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1.11 Bridges
Definition of Bridge
"A bridge is a structure which provides passage over an obstacle or opening without closing
the way underneath."
As ideal bridge should be economical, serve the intended functions with safety and
convenience, and give aesthetic elegant (quality of being attractive, stylish) look.
Importance of Bridge
For the first time the primitive (of the earliest time) man used an accidentally fallen tree
across a stream as a bridge to cross the stream in search of food and shelter.
In about 200 BC Romans built stone arch bridges in around 600 AD. Chinese built a single
span stone arch bridge which is still fit for vehicular traffic. This bridge is situated at about
350 km south of Peking.
The oldest pedestrian bridge about 2500 years old is still standing across the river Meles in
Smyrna, Asia Minor. It is a stone slab bridge.
With the progress of civilization, the science of bridge engineering also developed. World's
first modern cantilever bridge was built across river Main in Germany (span 129m) in 1867
AD.
The importance of bridges has been felt from the very primitive age. Great battles have been
fought for capturing the cities and bridges.
The mobility of an army in war times is greatly affected by the availability of bridges to
cross the river. That is why military training puts social emphasis on imparting how to build
a new bridge quickly while advancing and how to destroy existing bridges while retreating
(going back). Thus the inventors which have abridged the distance have done the great
service to man kind.
Selection of Bridge Site
Selection of bridge site is very important from the point of view of economy and safety.
Selection of suitable bridge site depends upon the factors like engineering, economy,
demand of traffic, condition of stream, aesthetic and elegant look.
If there are conflicting claims for the bridge site selection, it is necessary to justify in the
basis of sound technical and economical grounds. Social benefits and economy should
govern the new bridge site selection.
Following points should be considered in selecting bridge site. They are:
1. Good foundation: It is desirable to have good foundation for abutments and piers.
2. Firm and well defined banks: Both banks should be firm and high enough to avoid
river training works.
3. Straight reach of the stream: Upstream as well as down stream should be straight for
sufficient length as river may change its course on a loop or curve by eroding bank.
4. Stream line flow: The flow of the river should be stream line flow as turbulent flow
will cause scouring of the bed near the piers foundation which will damage the bridge.
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5. Minimum length of bridge: It is desirable to have narrow width of the stream so that
minimum length of bridge will result reduction in the cost of the bridge.
6. Right angle crossing of river: As far as possible the centre line of the bridge should be
at right angle to the flow of water.
7. Straight approaches: Approaches on both ends of the bridge must be straight at least
for a distance of 15 meters to provide a minimum sight distance for the design speed.
Other factors like free from fault zone, and reasonable proximity to a direct alignment of the
road to be commented should consider in selecting the bridge site.
Loads on the Bridges
Various types of loads to be considered in designing bridges are:
1. Dead load
2. Live load
3. Impact load or dynamic effect of the live load
4. Wind load
5. Horizontal forces due to current of water
6. Longitudinal forces due to braking of vehicles or tractive effort at the time of starting
7. Centrifugal force
8. Buoyancy pressure
9. Earth pressure
10. Temperature stresses
11. Deformation stresses
12. Secondary Stresses
13. Erection stresses
14. Seismic force
15. Lateral load
1. Dead load: Dead load is the self weight of the structural members of the bridge.
2. Live Load: Live load is the load due to the road traffic.
3. Impact load: Fast moving loads on uneven surface of the bridge have jumping action
which produces shocks and vibrations on the structure. Impact load is considered about
10 to 25 % of live load.
4. Wind load: All bridge structures should be designed considering lateral wind forces.
Wind load act horizontally and should be considered 300 to 450 kg/m.
5. Horizontal forces due to current of water: Any part of bridge submerged in running
water should be able to withstand the horizontal forces due to current of water.
6. Longitudinal forces: Longitudinal forces are caused due to braking vehicles or tractive
effort at the time of starting.
7. Centrifugal force: Centrifugal force is the force due to the movement of the vehicle on
the curves.
8. Buoyancy force: Buoyancy force is considered for submersible bridges.
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9. Earth pressure: Earth pressure should be considered in the design of abutment.
10. Temperature stresses: Temperature stresses due to expansion and contraction of
structural component should be considered in the design of bridge.
11. Deformation stresses: Deformation stresses are considered for steel bridges only.
12. Secondary stresses: Secondary stresses develop due to deformation of certain members,
shrinkage of concrete, movement of supports etc.
13. Erection stresses: The component of the bridge fabricated in the workshop need to be
erected. Provision for erection stresses should be made in the design.
14. Seismic force: Seismic forces are due to the earthquake.
15. Lateral load: Lateral loads are due to lateral movement of roadway traffic. Lateral load
is taken 600 kg/m.
Wooden Beam and Girder Bridge
Figure: Cross-section of Girder Bridge
Wooden girder bridges are constructed where good quality timber is available in sufficient
quality at cheaper rates.
Girder bridges are constructed under 20' span with simple wooden beams. For span larger
than 20' (between 20' to 30') longitudinal beams are provided with truss.
Size of beams and members of trusses depend upon number of beams or trusses used, kind
of traffic and strength of material used.
Flooring consists of wooden planks laid transverse on longitudinal beams (road bearer).
Timber Bridges have the following advantages and disadvantages.
Advantages
 can be constructed quickly in less time
 initial costs is less
 no skilled labor is needed
Disadvantages
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



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life is less about 15 to 20 years.
are liable to catch fire easily
deteriorate much faster than other materials
not suitable for large span.
need constant maintenance
Suspension Bridge
In suspension bridge, two parallel cables are stretched between the two banks and a platform
is hung to cross the river. Suspension bridge is best suited for very long span.
Suspension bridge are constructed in places where it is not possible to construct intermediate
support due to deep gorge (narrow opening with steep sides), huge velocity of flowing water
and unstable foundation.
Suspension bridge of forest requirement carry general laden men and mule traffic. Two main
cables are carried over top of towers (piers) at the ends. Free end of these cables are fixed to
firm anchorages buried beneath ground.
The road way is supported by transverse cross-pieces or wooden planks. These cross-pieces
or planks are connected at ends by longitudinal beams. Small ropes are called suspender
hanged from longitudinal ropes. Cables and ropes must e prevented from corrosion.
Timber Cantilever Bridge
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Timber cantilever bridge is suitable foe hilly areas where good quality of timber is available
in abundance and other construction materials are not easily available. Timber cantilever
bridge is suitable for light loads.
Cantilever bridge consists of two systems of counter balanced beam. Number of separate
beams is placed on top of the other. End of each beam projects beyond the end of the beam
on which it rests. This process is continued till the width can be spanned over a single beam.
End of beams resting on abutments are counter-weighted to make the construction stable.
Counter balanced beams are built out from either side of the banks.
Unit 2. Design and Construction Procedure of Road Side Drainage, Retaining Walls
and Breast Walls
2.1 Definitions of Road Drainage and its importance
Drain = pipe channel or trench etc carrying away water sewage, wastewater etc.
Drainage = draining or being drained.
Definition of Road Drainage
Drainage is the process of removal of gravity water (free water) from a soil mass in order to
keep it in a stable condition. Road drainage may be defined as the process of removing and
controlling excess surface water and sub-soil water from the road way.
Drainage may be classified as: i) surface drainage and ii) sub-surface drainage
Surface drainage is the removal and diversions of surface water from the roadway and
adjoining land. Sub-surface drainage is the removal and diversion of excess soil water from
the subgrade. Sub-surface drainage helps in improving the properties of soil which improves
the stabilization of soil.
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Importance of Road Drainage
Drainage is an essential and important part of road construction as excess water on shoulders
and pavement edge causes decreases in stability of soil mass that may causes the failure of
earth slopes and pavement.
Road drainage is important because of following reasons:
1. Excess moisture in sub grade soil decreases stability of soil mass.
2. Increase in moisture causes reduction in strength of pavement materials like stabilized
soil and water bound macadam.
3. Poor drainage causes pavement failure by the formation of waves and corrugations in
flexible pavements.
4. In some clayey soils, variations in moisture content causes variation in volume of sub
grade which may result pavement failure.
5. Sustained contact of water with bituminous pavement causes failures due to stripping of
bitumen from aggregates and formation of potholes.
6. Prime causes of failure in rigid pavements by mud pumping are due to the presence of
water in fine subgrade soil.
7. Excess water on shoulders and pavement edge causes considerable damage.
8. Excess moisture causes increase in weight and thus increases in stress which results
reduction in strength of the soil mass. This is one of the main reasons of failure of earth
slopes and embankment foundations.
9. In places where breezing temperature are prevalent in winter, the presence of water in
the sub grade can cause considerable damage to the pavement due to frost action.
10. Surface water may cause erosion of soil from top of unsurfaced roads and slopes of
embankment.
2.2 Types of Road Side Drainage (rectangular, trapezoidal triangular and circular)
Depending upon the shape, types of road side drainage is rectangular, trapezoidal, triangular
and circular. Theoretically, a semicircular type of cross-section is the most economical
section as it gives a maximum hydraulic radius, maximum discharge and a minimum wetted
surface.
Rectangular Section
Rectangular channels are commonly used for road side drainage. Most economical section is
that which has the least wetted perimeter for a given area of flow.
Wetted perimeter P = b + 2d, A = b*d, b = A/d
2
A
A
dp
dp
A
A
 p   2d , forP min,
 0
  2  2  0, or 2  2, A  2d
d
d
dd
dd
d
d
2
A 2d
b 
 2d  b  2d , p  b  2d  2d  4d , p  4d
d
d
A 2d 2 d
d
hydroulicr adius , R  
 , A  2d 2 , P  4d , b  2d , andR 
p
4d
2
2
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Trapezoidal Section
Trapezoid is the commonest shape for channels. From practical consideration of avoiding
very big depths, trapezoidal section is adopted for higher discharge. Most economical
trapezoidal section with slope varying is that having angle of sides Q = 30º with respect to
vertical. Most economical trapezoidal section with depth and slope varying for maximum
discharge, the necessary conditions are:
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2.3 Design of Longitudinal Drains for Peak Runoff
Run off means the draining or flowing off of precipitation from a catchment area through a
surface channel. The excess precipitation moves over the land surfaces to reach smaller
channels. Flows from several small channels join bigger channels and flows from these in
turn combine to form a larger stream, and so on, till the flow reaches the catchments outlet.
The flow in this mode where it travels over the surface and reaches the catchments outlet is
called surface runoff.
Consider a rainfall of uniform intensity and very long duration occurring over a basin. The
run off rate gradually increases from zero to a constant value. The runoff increases as more
and more flow from remote areas of the catchment reach the outlet.
The peak value of the runoff is given by Qp = C i A
Where,
Qp = peak discharge, m3 /s
C = co-efficient of run off <1
I = intensity of rainfall, mm/h
A = area of catchment in km2
The co-efficient of runoff 'C' depends upon the nature of the surface, surface slope and
rainfall intensity.
Value of the co-efficient 'C' in agricultural area:
Area
Tight clay, cultivated
Wood land
Sandy loam, cultivated
Flat
0.50
0.40
0.30
Hilly
0.70
0.60
0.40
Empirical formula
The empirical formulae used for the estimation of the flood peak are essentially regional
formula based on statistical correlation of the observed peak and important catchments
properties.
Empirical formula is applicable only in the region from which they were developed and
when applied to other areas they can at best give approximate values.
Dickens formula (1865)
3
Qp = CD A 4 , where Qp = maximum flood discharge, m 3 /s
A = catchment area in km2
CD = Dickens constant with value between 6 to 30
Dickens formula is used in the central and northen part of the India.
Ryves formula (1884)
Qp = CR
A = catchment area in km2
CR = Ryves constant with value between 6.8 to 10.2
Ryves formula is in use in Ttamil Nadu, Karnataka and Andrapradesh.
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English formula (1930)
124 A
Qp=
, Where Qp=maximum flood discharge, m 2 / S
A  10.4
A= catchment area in km2 , Inglish formula is in use in naharastra.
Fullers formula (1914)
Qtp=Ct A0.8(1+0.8 log T)
Where Qtp= max.24h flood with a frequency f types in m 3 /s
A=catchments area in km2
Ct= a constant value between 0.18 to 1.88
Fuller's formula derived for catchments in USA
2.4 Retaining Wall: Definition, Type
Definition Retaining of Wall
Retaining wall may be defined as relatively rigid wall used for supporting the soil mass
laterally so that the soil can be retained.
Retaining wall is important structure in hill road to provide stability to the road way and to
prevent landslide.
The main function of retaining wall is to retain the back filling.
Hill road is generally constructed by the elevation and the excavated material is used for
back filling behind the retaining wall which forms part of the road.
Types of Retaining of Wall
Gravity retaining wall: Gravity retaining walls depend upon their weight for stability.
Gravity retaining walls are usually constructed of plain concrete or masonry. Such walls are
not economical for large heights.
Semi-gravity retaining wall
The size of the section of a gravity retaining wall may be reduced if a small amount of
reinforcement is provided near the back face. Such walls are known as semi-gravity walls.
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Cantilever retaining wall
Cantilever retaining walls are made of RCC. The wall consists of a thin stem and a base slab
cast monolithically. This type of cantilever retaining walls is found to be economical up to 6
to 8 m height
Counterfort retaining wall
Counterfort retaining walls have thin vertical slabs known as counterforts spaced across the
vertical stem at regular intervals. The counterforts tie the vertical stem with the base slab.
Counter forts are on the side of the backfill.
The counterfort retaining walls are economical for a height more than 6 to 8 m.
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2.5 Design and Construction Procedure of Retaining Wall
Before the actual design, the lateral earth pressure and bearing capacity of soil must be
evaluated.
Pressure on retaining wall varies from zero at top of wall to maximum at base of wall.
Weight of wall passes through C.G of wall and acts vertically downwards. The resultant of
these two forces (earth pressure and weight of wall) should be within the middle third of the
base of wall.
Dry stone masonry is preferred in retaining wall as it permits easy drainage of seeping
water. If the height is more, stone masonry is laid in cement mortar and weep holes are
provided at 1m vertical height and 1.2m apart horizontally to drainage seeping water.
Some practical rules in the design of Retaining Walls:
Rule 1: Stability of retaining wall is preserved when the retaining walls have the following
dimensions:
Material
Dry stone masonry
Dressed stone masonry
masonry in mortar
Thickness
Base
Top
0.5H
0.25H
Width of
foundation
0.75H
Remarks
up to 6' height
0.4H
0.2H
0.6H
6'to12' height
0.35H
0.175H
0.52H
over 12'height
or Stone
Mass concrete 1:2:4
H
 1'.
10
Keep top width of wall 1'6"to 2'0"minimum
Keep earth face vertical.
Give front face a batter of 1 in 4 (1 horizontal in 4 vertical).
Wall up to 6' height may be constructed dry stone masonry.
Walls greater than 6'height may be constructed to 6' in dry stone masonry and rest of the
height in cement mortar.
For extreme strength whole wall to be constructed in cement mortar with stone masonry.
1. Minimum depth of foundation =
2.
3.
4.
5.
6.
7.
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Rule 2:
8. Remember base width of retaining wall.
1
9. Top width = of base width
2
1
10. Width of foundation =1 of base width
2
2.6 Stability Analysis of Retaining Wall
Any material will be stable at a certain slope angle. If the slope angle is greater than its safe
value, the slope will be unstable and soils located at higher elevation tend to move
downwards and outwards. In such cases, a retaining structure is to be provided to maintain
the stability of soil.
In the absence of a retaining wall the soil mass located at higher elevation will have a
tendency to move towards the lower elevation and the soil mass will be unstable. The
stability of retaining wall is due to the self weight of the wall and the passive resistance
developed in front of the wall.
Pa = active earth pressure acting at a height H/3 from the base.
Pp = passive earth pressure in front of the wall.
Fr = base sliding resistance.
W = weight of wall.
Stability analysis: Consider a cantilever wall carrying a sloping backfill as shown in the
figure.
For stability, a retaining wall should satisfy the following conditions:
1. Check against sliding: The retaining wall should be stable against sliding the factor of
 Fr  1.5
safety against sliding shall be a minimum of 1.5 Fs 
Pa * h
Where, Fs = Factor of safety against sliding
Fr = Base sliding resistance
Pa = Active earth pressure acting at a height H/3 from the base.
2. Check against overturning
The wall should be stable against overturning the factor of safety against overturning shall
be a minimum of 1.5 for granular backfill and 2 for cohesive soil.
Mr  2
Fo=
Mo
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Where,
Fo = Factor of safety against overturning
Mr = Resisting moment
Mo = Overturning moment.
3. Check against bearing capacity failure
The base of the wall should be stable against bering capacity failure.
The factor of safety against bearing capacity failure shall be a minimum of 2 for granular
backfill and 3 for cohesive backfill.
Unit 3 Designs and Construction Procedure of Forest House Cottages
3.1 structural principles
Design of any structure should fulfill its functional requirement and structural requirement.
Functional requirement: functional requirement includes:
1. Fit for the purpose
2. Aesthetic, ceiling height acoustic system.
3. No obstacle inside the building.
4. Lighting ventilation and air-conditioning.
Structural requirement: structural requirement includes:
1. System of framework and material used.
2. Economical design of building component to cater the functional requirement for
reasonable period of time without failure.
3. Safe.
Four step of structural design are:
1. Determination of the loads, force that framework require to support.
2. Selection of suitable structural arrangement and material.
3. Analysis of internal stresses in the adopted framework.
4. Economical design of framework member to safely resist the internal stresses produced.
3.2 forces stress moments and reactions
Force or load or weight - external forces acting on a body or member are termed as loads.
Forces in structure or load are the weight that structure has to bear.
Force=mass*acceleration. the unit of force is the Newton (N), which is the force required to
give 1 kg. Mass 1m/sec2 acceleration. Thus 1 N is 1kg.m/ sec2. The Newton is a derived unit
that is in depend dent of the acceleration of gravity.
Approximating the acceleration of gravity 9.81m/ sec2, 1 kg of mass exerts a force of 9.81N
on its support point. Conversion factors: MKS to SI units, i kgf(kg)=9.81N, i tone(t) =9.81
Kn.
While designing building structure loads are calculated according to the standard code (eg.
building code of Nepal is BS code etc).
Types of load
i) Dead load: dead load is the weight of the structure itself.
ii) Live load: live load is the weight that can be removed from structure like people,
furniture etc in building and vehicle in bridge.
iii) Wind load: wind load is the lateral force produced by wind.
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iv) sesmic load: sesmic load is the force due to the earth quick
Stress
Stress may be defined as the load per unit area. When an external load is applied to a
member, internal resistance is set up by the material. Thus resistance is called stress.
load W
N
Stress=

 2
area A m
All externally applied loads deform an elastic material as the material undergoes
deformation; it sets up internal resistance to the deforming forces. The quantum of internal
resisting forces correspondingly increases with the increase in externally applied loads only
up to a certain limit beyoned which any increase in applied loads will continue the process
of deformation to stage of failure.
The deformation is known as strain and the resisting forces are called stresses.
Stress may be either tensile or compressed according to whether member is being stretched
or compressed.
Strain
Strain is a measure of deformation produced in a member by the load. Strain (for bodies
subjected to normal tensile or compressive stresses) may be defined as change in length per
unit length. Tensile stresses increase the lengths whereas compressive stresses decrease the
l
length e=
l
Where,
e= deformation per unit length or strain due to applied load
l = change in length caused by the applied load.
l=original length.
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Reaction
Condition for equilibrium of a beam is:
1. Algebraic sum of all forces is zero i.e, total downward force equals the total upward force
(usually the sum of support reactions).
2. algebraic sum of the moments of all forces including the support reactions is zero about
any point i.e. the sum of all the clockwise moments equals the sum of all the anticlockwise
moments about any point.
Equation of equilibrium: V  0,  H  0,  M  0
Calculation of reactions: taking moments about B,
1
wl
wl
R A  l  wl   R A 
, totalload  wl , R A  RB 
2
2
2
To determine the support reactions take moments about A. Rb*5=10*1+20*3 or
70
Rb*5=10+60 or Rb=
 14t.
5
Taking movements about B,
RA*5=10*4+20*2 or Rb=
80
 16
5
Ra+Rb=10+20, Ra+14t=30t, Ra=16t.
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Ans : R A  4.3t
RB  6.7t
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Hooke's law
hooke's law states that for materials subjected to simple tension or compression the stress is
proportional to the strain within elastic limit.
stress
stress
Stress and strain,
 cons tan t ,
 young's modulus of elasticity
strain
strain
Since the unit of stress f is N/ m 2 , the strain e being only a ratio the unit of youn's modulus
of elasticity shall be N/ m 2 , or kg/c m 2 ,
Moment:
moment is the turning effect produced by a force on the body on which it acts.
Moment=force*distance (i.e  distance), moment, M=p*l.
Where, P=force acting on the body,
l= perpendicular distance of the point and the line of action of the force.
Unit of moment = N-m ort-m or kg-m
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3.3 Analysis of simply supported structures with BMD and SF
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3.4 wood RCC construction :
allowable stress, wooden beam design, wooden, column, wooden joints basic requirements
for RCC design, theory of RCC foundation and forting with simple numerical example.
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Allowable stress
When an external load is applied to a member internal resistance is setup by the material.
load
This resistance is called stress, stress=
external forces applied to a body have the
area
tendency to deform the body which develops an internal resistance against the deforming
forces. This resistance increases with the increase in deforming force but only up to a certain
limit, beyond which the deforming force will cause the failure of that body. A material if
stressed beyond elastic limit does not return back to its original position when the load is
removed. A material fails when the stress in the material reaches the elastic limit stress in
simple tension/compression. Allowable stress is lower than the maximum stress that the
material is capable of taking. Structure is designed using the working stress method and
limit state method. Working stress is obtained diving to ultimate stress by the factor safety
as the available materials are victims of poor workmanship. Limit state is the acceptable
limit for the safety and serviceability of the structure before failure occurs. Limit state
method of design in RC structures is an entirely new concept.
table 3.1 structural properties and safe working stresses for some timbers are:
timber average modulus
of bending
horizontal bending stress in
density elasticity(N/mm2*103 stressing
share
compression(N/mm2)
2
@1.2%
tension(N/mm ) stress
MC
(N/mm2)
2
kg/m
babul 785
10.6
17.9
1.5
11
deodar 545
9.3
10
0.7
7.7
Sal
865
12.5
16.5
0.9
10.4
sisam 787
8.7
14.9
1.2
9.2
Wooden Beam Design:
Timber structures are used widely where thimber is available cheaply. A well seasoned
timber without any defects like knot, crack, warping etc is suitable for structural works.
Bending Equation:
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But, MR is the same as BM  M 
From equation (1), we have ,
EI
M E
or ,
 ...........2
R
I
R
f E

y R
f M E


y
I
R
This relation is known as the bending equation and is he general case of the beam element
subjected to pure bending.
E
Under a given condition of loading for a given beam
is a constant.
R
E=Young's modulus of elasticity kg/c m 2 ,
R= radius of curvature of beam under bending
I= moment of inertia of the section
I= b * d
3
12
forrec tan gular sec tion (cm 4 ormn4 )
I= d 64 forcircula r sec tion
f= bending stress (compressive or
from the neutral axis, kg/ cm 2
4
tensile) calculated at a point at a distance y
y = distance from the natural axis to edge of section of beam 9cm, mm).
y  d forrec tan gular sec tion (cm, ormm), y  d forcircula r sec tion (cm, ormm)
2
2
M= moment of resistance= Max. bending moment in the beam.
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1
i
M  f *  M  f * z.wher , Z  isknownas sec tion mod uluscm 3 ormm3
y
y
wooden beam design steps
1. The beam is designed for safe fiber stresses, horizontal share stresses and deflection
2. Generally timber beam is designed in rectangular section.
3. Rectangular beam with large depth is recognizable.
3
2
I
bd 12
bd
bd 2
 f
 f
M  f 
..............1 Width of beam shall not be
d
y
6
6
2
less than 50mm.
4. With the assumed ration of beam section, the dimension of beam can be worked out from
equation1.
5. Check the size of the beam with reference to the following criteria: a) Check for
horizontal share stress
b) Check for deflection
6. Check for horizontal share stress. Maximum horizontal shear stress  safe working stress,
maximum
horizontal
share
stress,
3 v
f s  * , where, f s  max imum.sheastress, v  Maximumshearforce
2 bd
7. Maximum deflection in simply supported beam
wl 3
whenbeamis actedbypo int loadatmids pan.
i) 
48EI
5wl 4
whenbeamis actedbyud 2( w .
ii) 
m
384 EI
M  f
8. Max SF in case of UDL.
l

V= w   d wherel  spanoftheb eam.
2

-when a beam carries UDL, the SF at a distance 'd' from the edge of the beam is critical.
9. Check for deflection
 Maximum deflection allowabled eflection
l
 allowable deflection=
incaseofsi mply sup portedbeam.
240
Design procedure of timber beam
i) A beam section is usually chosen which can resist maximum bending moment
occurring over its span.
ii) Check the share stress for the chosen beam section to be within the permissible limits.
iii) Check the deflection for the chosen beam section to be within the permissible limits.
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design for bending : M=f.z
note: calculated f=  thepermissiblebendin gstress.
M
M
1.5V
 f ,Z 
, designforshare : max 
Z
f
bd
Numerical:
Ex 3.4.1. A 4 meter long beam wirh rectangular section of 10 cm width and 20cm depth is
simply supported at the ends. If it is soaded with a uniformly distributed load of 400 kg/m.
through out the span determine the maximum stress in the beam.
Solution:
Given: l=4m, b=10m, d=20cm, UDL=400kg/m
Max stress=?
l

4

We know Max. SF, v= w  d   400  0.2   400(2  0.2)  400  1.80orv  720kg.
2

2

3
V
3
720
Maximum shear stress, f f s  
 
 5.40kg / cm 2  f s  5.40kg / cm 2
2 b  d 2 10  20
ex.
3.4.2. Design a 5m. Long sal timber beam simply supported at its end to carry a uniform
load of 6kn/m and concentrated load of 10kn at centre of beam.
Allowable bending stress =16.5N/ mm 2
Allowable horizontal shear stress =0.9N/ mm 2
Unit weight of sal =365kg/ m 3
l
5000
Allowable deflection =

 20.83mm
240 240
Modulus of elasticity =12.5* 10 3 N/ mm 2
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d
 1.5
b
2.5
Max bending moment max= 20*2.5-6*2.5*
orn / max  50  18.75  31.25KN .m , with
2
the
bending
equation,
we
have
2
2
f
f
b(1.5b)
M
M
bd
 or 3

 M  f.
orM  f 
,
I
y bd / 12 d
6
6
2
2.25b 3
or 31.25 1000 1000  16.5 
 b  171.57mm
6
assumeb  175mm  d  1.5 171.57  257.36mm  275mm
selfweight ofthebeam  865 10 * .175  .275  416.28M / m  0.42 KN / m
 totalUDL  6kn / m  0.42kn / m  6.42kn / m
Let the depth to width ratio of beam as 1.5 ie.
Maximum
bending
orM  52.62  20.06  32.56kn / m,
max imumshearforce, v  21.05K
Check
i) for allowable bending stress:
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moment= 21.05  2.5  6.42  2.5 
2.5
2
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M
32.56  1000  1000 275
f 
y

 14.76 N / mm 2 16.5 N / mm 2 safe
3
I
2
275
175 
12
ii ) forallowab ledeflecti on
3 v
3 21.05  1000
f  .
 
 0.656 N / mm 2  0.9 N / mm 2 safe
2 bd 2
175  175
wl 3
5wl 4
175  275 3
3
iii ) forallowab ledeflecti on :

, EI  12.5  10 
 3.79  1012 mm 2
48EI 384 EI
12
3
4
10  1000  5000
5  6.42  5000


 6.87  13.78  20.65mm20.83mm 2 safe
12
48  3.79  10
384  3.79  1012
ex.3.4.3. A timber beam 150mm wide, 300mmdeep is simply supported over a span of 4m.
Find the maximum UDL that the beam can carry, if the bending stress is not to exceed 8N/
mm2 ans : 9KN / m.
Solution:
With the bending equation, we have
M
f
I
bd 3 / 12
bd 2
 , orM  f   f 
, orM  f 
...............1
d
I
y
Y
6
2
Maximum bending moment at mid span is
wl 2 l
l l wl 2 wl 2 wl 2
wl 2
  w  


M 
...................2
M=
8
2
2 4
4
8
8
8
Equating equation 1and 2 we get,
wl 2
bd 2
w  4000 2 8  150  300 2
 f
, or

,
8
6
8
6
8  150  300  300  8 8  15  3  3  8
or , w 

 9 N / mm
4000  4000  6
40  4  6
1000 N
9 KN
w  9 N / mm  9 

 9 KN / m
1000mm
m
ex:3.4.4 . A timber beam of rectangular section, simply supported over a span of 4m, is
carrying a UDL of 2000kg. /m over the entire span. Calculate the width and depth of the
beam if the stress is not to exceed 80 kg./cm2. Take the ratio of depth to width as 1.5 (ans:
24cm*36cm)
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Solution:
Max bending moment at midspan, M=4000*2-2000*2*
2
or M= 8000-4000=4000kg.m
2
:. M=4000kgm=4000*100kg/cm
with bending equation, we have
f
M
I
bd 3 / 12
bd 2
 orM  f   f 
 f
,
d
I
y
y
6
2
2
b1.5b 
2.25b 3
orM  f 
or ,4000  100  80 
6
6
400  100  6
orb 3 
 b  23.71cm  24cm
80  2.25
 d  1.5, b  1.5  24cm  36cm
 b  24cm,
d  36cm
wooden column design
Column is compression member normally rectangle, square, circular, section are used in
wooden columns.
Design data for solid rectangle columns
type
limit of slenderness ratio
a) short
l
11
d
l
11  k g
d
l
k g
d
b) intermediate
c) long
permissible
stress
Fc
compressive
 1 l 4
Fc 1  (
) 
 3 k g d 
0.329 E
l d 
2
Where,
Fc = permissible compressive stress for short column.
l= unsupported length of column.
d= lasts dimension of column.
E
k g = 0.702
Fc
E= modules of elasticity.
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Note: In any case l/d ration should not exceed for solid timber columns.
ex 3.4.5
Find the allowable Arial load on a deodar column 200mm*150mm in cross section and
a) 3.0m, b) 4.5m in length. Use the basic values. F f c  7.8N / mm2 , E  9500 N / mm2
Solution:
a)l  3.0m,
3 *1000mm
 20,
d
150mm
9500
K g  .702
 24.5
7.8
11 l K g ,
d
There fore it is an intermediate column. For intermediated column permissible compressive
stress is given by:
 1 l 4
3000 4 
 1
Fc 1  (
)   7.81  (
) 
 3 24.5 *150 
 3 k g d 
 0.871  .444 / 3  78 * .852  6.65 N / mm 2
Permissible axial load =permissible compressive stress *area
=6.65N/mm2*200mm*150mm
=199500N=199.5KN
4.5 *1000
b)l  4.5m, l 
 30, K g  24.5, l  K g , therefore it is a long column for long
d
d
150
column,
permissible
compressive
stress
is
given
by
2
0.39 E 0.329 * 9500 N / mm

 3.47 N / mm 2 , permissible axial load= permissible
2
2
30
l
d
compressive*area
= 3.47 N / mm2 *150mm*200mm=104100N=104.1KN.
l
 3*
 
ex. 3.4.6
design a section of wood column made up of sisham to carry an axial load of 200 KN. the
length of the column is 40 m. the column may be subjected to seasonal wetting and drying,
given the following values for outside location:
Fc  8.4 N / mm2 , E  8900 N / mm2
Solution
E
8900
K g  .702
 .702
 .702 * 32.55  22.85
Fc
8.4
arearequired 
load
200 KN
200 *1000


mm 2  23809.52mm 2
2
stress 8.4 N / mm
8.4
Side of a square column = 23809.52mm2  154.30mm  155mm
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l  4 *1000  25.81, l  K Therefore it is a long column. For long column, permissible
d
d g,
155
compressive stress is given by:
0.329 E 0.329 * 8900 N / mm 2

 4.40 N / mm 2 , to carry axial load of 200KN, area
2
25.812
l
d
axialload
200 KN

mm 2  45454.55mm 2
= permissiblecompressivestress 4.40 N /
 
 sideofasquarecolumn  45454.55mm 2  213.20mm  215mm
Permissible axial load=permissible compressive stress* area
 4.40 * 215 * 215  203,390 N  203.39KN  200KNsafe
:. Section of 215mm*215mm required sisham wood column is safe to carry an axial load of
200 Kn.
Wooden joints
Wood is widely used in construction field. Timber is used for building component like
column, beam floor, roof truss, door, window, stair etc. during the construction of any
structure timber is used as form work (centering and shuttering) scaffolding, shoring etc.
joints play an important role in timber construction in several ways, such as increasing
structural stability improving aesthetic values etc. but the joints are construed to be the
weakest part f any structure.
The fastenings should be designed and placed in such a manner as to avoid failure of joint
by share.
Type of wooden joints
i) Farmers joints
ii) Lap joint
iii) Bolted joints
iv) Nail joint
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RCC foundation and footing
Foundation is the lowest part of any structure that transfers the load of the structure to the
sub-soil underneath. Foundation provides rigidity to the structure and prevents the
differential settlement by evenly loading the substrate.
The direct bearing foundation which is constructed as a s[read under the base of a wall is
called footing. spread foundation is provided for structures of moderate height built on
sufficiently firm dry ground.
Design aspect
1. Loads
Dead load live load self weight of the booting itself (assumed @10% for wall foundation
and @20% column and piers)
2. safe bearing capacity of soil
the maximum pressure which the soil can bear safety without any risks of shear failure and
settlement is called safe bearing capacity of soil safe bearing capacity of soil de[ends upon
soil type, grain, size, moisture content, compaction and water table.
s.n type of soil
safe
bearing
capacity of soil
KN/m2
1.
Hard rock
3240
2.
laminated rock
1620
3.
gravel sand and gravel
440
4.
coarse sand compact and dry
440
5.
medium sand, compact and dry
245
6.
find sand, silt
150
7.
soft clay
100
8.
very soft clay
50
3. Depth foundation
According to rankings formula the depth of foundation may be determined by the following
formula.
P (1  sin  ) 2
D s
, where ,
(1  sin  ) 2
d=depth of foundation
Ps = safe bearing capacity of soil in Kn/m2
 = unit weight of soil in KN/m3
 = angle of repose or angle of internal friction of the soil.
4. Width of foundation (incase of wall foundation)
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P
Where
Ps
B= width of foundation perimeter in m.
P= Total load of foundation perimeter in KN.
Ps = safe bearing capacity of soil in Kn/m2
B
5. Area of foundation (in case of column foundation)
P
Where,
Ps
A= area of foundation in square meter
P= total load to foundation in Kn
Ps = safe bearing capacity of soil KN/m2
A
Basic requirements for RCC design
RCC means reinforced cement concrete. RCC =PCC+ steel semi for cement. The ingredient
of PCC or concrete is cement, fine aggregate (sand), coarse aggregate and water. Concrete is
very strong in compression but tensile strength of concrete is just 10%of its compressive
strength. Steel rod is equally strong in tension and compression but steel rod cannot develop
full compressive strength because of buckling effect, the combination of steel and converter
is economical because compressive forces are borne by concrete and tensile corves are borne
buy steel. Steel and concreter corms concrete forms perfect bind or grip. RCC is widely used
in construction as RCC structures can be constructed in any desired shape. Uses of RCC in
construction are.
i) Building construction - foundation, basement, column, beam, wall, roof, floor, staircase
etc.
ii) Other engineering project- bridge culvert, road, pavement, retaining wall, dam etc.
Advantages of RCC.
1. Combination of steel and concrete is economical.
2. RCC can be molded to any desired shape.
Ingredients of RCC are easily available.
4. RCC structures are durable and fire resisting.
5. RCC structures required less maintenance cost.
6. RCC structures are impermeable to moisture and easily with stand effects of atmospheric
agencies.
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ex. 3.4.8. A wall carries a load of 100KN/m including its own weight, find a suitable width
and depth of its foundation. Take Ps  90Kn / m 2 ,   16KN / m 3 and  30 0.
Solution:
P
We know, B  , where , B= width of foundation, P= total load of foundation per meter,
Ps
Ps= safe bearing capacity of soil.
100 Kn / m 10m

 1.11, B  1.11
90 Kn / m2
9
1
1
B=
90(1  ) 2 90 *
Ps (1  sin  ) 2
2 
4  90  0.625m, D  0.625m
D

2
1
9
16 * 9
 (1  sin  )
16(1  ) 2 16 *
2
4
ex: determine the area and minimum depth of the foundation of a column carrying a load of
700 KN including its own weight. the safe bearing capacity of soil is 110Kn/m2, angle of
repose is 300 and the unit weight of soil is 16KN/m3
Solution:
Given A  ?, D  ?, P  700KN , Ps  110KN / m 2 ,   30 0 ,   16KN / m3
Totalloado ffondation ( p)
we know, area of foundation A=
safebearin gcapacttyo fsoil ( p s )
P
700 KN
or , A 

 6.36m 2 ,
Ps 110 KN / m 2
depthoffou ndation , D 
p s (1  sin  ) 2
 (1  sin  ) 2
1
110(1  ) 2 100 * 1
110 KN / m
(1  sin 30 )
2 
4  110  0.76m
D


3
0 2
1
9
144
16 KN / m
(1  sin 30 )
16 *
16(1  ) 2
4
2
2
0 2
3.5 Site selection criteria for a building
Besides food and clothing, man's next basic need is shelter for protecting himself against
natural calamities, wild animals, beasts and other fellow beings. There is universal tendency
in man to own a comfortable house suitably located.
Site selection for a building has important role on planning and designing of buildings.
Therefore, an architect has either to make a choice of suitable site or to plan building
structure to suit the available site.
A good planning and design of a building should fulfill the following aspects
I. Proper site selection
II. Proper Orientation
III. Building bye-laws
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IV. functional design of building
A properly selected site enhances the comfort and beauty of the building without any extra
expenditure. Natural defects of a site will involve considerable expenditure on construction
and maintenance of the bu8ilding. Site selection criteria varies with the purpose of the
building
General principles or factors to be considered in the selection of a site are:
1. The site should be selected keeping in view of the general purpose of the building.
2. The site should be situated in a locality to secure happy living conditions where the
neighbors belong to an equal status, social and friendly.
3. The plot should be in a locality where the various facilities like means of transport,
water supply, gas, electricity, telephone, schools, hospitals etc are available.
4. The site should not be irregular in shape or having any sharp corners and would
meet the requirements of the owner, preferably with possibilities of future
extensions.
5. The site should be situated on an elevated place so as to provide good and quick
drainage of rain water.
6. The soil surface of the site should be good enough to provide economical
foundations for the intended building sithout causing any problems.
7. The site should have a natural beauty and good landscape.
8. The site should be such as to ensure unobstructed natural light and air.
9. A site should be abandoned under adverse circumstances such as;
a. unhealthy, noisy or crowded localities
b. immediate neighborhood of rivers carrying heavy floods, badly maintained
drains
c. Water-logged areas
d. Industrial vicinity having smoke and obnoxious (very unpleasant) odours
(smell).
10. The site should be level (not much undulating)
11. The site should not be made up type as it may results cracks and differential
settlement.
3.6 Orientation of a building
Orientation is defined as a method of setting or fixing the direction of the plan of the
building in such a way that it derives functional comfort inside the building from the natural
gift such as sun, wind and rain. The knowledge of orientation is the first pre-requisite for
good planning.
Good orientation means proper placement of plan units of the building in relation to the sun,
wind, rain, topography, outlook and providing convenient access to the street. It should be
remembered that poor orientation of the buildings results in discomfort conditions inside the
building.
In poorly oriented buildings, operational cost is very high to create comfort conditions
through the mechanical devices.
Factors affecting orientation
The various factors affecting orientation are as follows:
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i.
ii.
iii.
iv.
Solar heat gain - The building should receive maximum solar heat in winter and
minimum in summer.
Prevalent wind - The direction of prevailing wind is welcomed in summer and
avoided in winter
Rainfall - The direction and intensity of rain.
Site conditions - Location of site either rural, urban, or sub-urba, neighbourhood and
surroundings.
In view of above data, particularly solar and climatic data, the orientation is made based on
the need of summer or winter comfort. For places where summer causes greater thermal
discomfort, the building as a whole should be oriented to intercept minimum solar radiation
in summer and vice-versa.
At hill stations, the winter season causes more discomfort. Therefore, the orientation of the
buildings should obtain maximum solar energy on the building in winter. The orientation of
the building in hill stations should be such that the living rooms are open on the south and
west sides of the sun.
3.7 Building components: definition and types (Walls, doors, windows, lintels, roofs, water
supply(plumbing) and septic systems)
1. Wall
The primary function of wall is to enclose or divide space. Secondary function of wall are
such as supporting the weight of the upper floors and roofs providing privacy, security
giving protection against heat and cold.
Types of wall
Walls are of several types depending upon their positions, functions and types of
construction.
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
Load bearing walls - Those walls which are designed to carry any superimposed load
are termed as load bearing walls.
Non-lead bearing walls - Those walls which support no vertical load are termed as
non-load bearing walls.
Curtain walls - In framed structures, external non-load bearing wall built between
columns is termed as curtain wall, panel wall or screen wall.
Partition wall - It is generally an interior non-load bearing wall dividing the space
within the building.
Shear Wall - Any wall designed to carry horizontal forces is termed as shear wall eg.
Basement wall.
Cavity or Hollow Wall - Cavity wall comprises of two leaves (i.e. thin walls) of
equal or unequal thickness separated by cavity (maximum 7.5 cm).
Blank wall - A wall without any opening is termed as blank wall.
Dwarf wall - Any wall which does not extend up to ceiling is termed as dwarf wall.
Parapet wall - The part of wall constructed entirely above the roof is termed as
parapet wall.
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x.
xi.
xii.
Masonary walls - Walls built of materials such as bricks, stones, clay or concrete,
usually in horizontal courses bonded with some form of mortar is termed as
masonary wall.
Retaining Walls - The main function of retaining walls is to resist the lateral pressure
of earth filling sand or granular material deposited behind it after its construction.
Breast walls - Breast wall is built to protect the freshly
cut surface of natural ground.
2. Doors
The main function of door is to serve as a access to the user or
connecting link between internal parts and also to allow the free
movement outside the building. Width of door varies depending
upon the purpose. Doors are of different types such as paneled
door battened door, flush door, glazed door, swing door, sliding
door, rolling shutter and collapsible steel door.
3. Windows:
Windows are generally provided for the proper ventilation and
lighting of a building. Number and size of window is determined
depending upon the requirements. Windows are openable and
fixed as per the requirement.
Fig. paneled door
4. Lintels
Lintels are provided to support the weight of the wall above the
door and window openings as the actual frame of door and
window is not strong enough to support the weight of the wall.
Lintels are like a beam and transfers the load vertically to
supporting walls. Lintels are made of various materials such as
wood, stone, brick, reinforced brick, and reinforced cement
concrete.
Fig. window
5. Roofs
Lintel
Roof is the uppermost part of a building whose main
function is to cover the building or structure to protect it
from the rain, sun, wind, heat, show etc. Durability of
building or structure depends upon the quality of roof.
Types of roof depend upon the climate and availability of
materials. There are main two types of roof namely
sloping roof and flat roof.
6. Water Supply and septic system in building
Fig: Lintel above door frame
Plumbing services include services like water supply, sanitation, drainage and gas supply
required in buildings, Water supply connection is making to kitchens, bathroom, toilets etc.
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The water used in building changes into waste water. This waste water is then conveyed
through drain pipes to join the municipal sewer laid underground under the street. The
conveyance system is also provided with underground chambers like manholes etc.
In rural areas where municipal sewers are not available, waste from toilets are safely
disposed to septic tank. Soap water, kitchen water, surface water, and paper should not be
allowed to drain into the septic tank.
The effluent from the septic tank is disposed by absorption in soil through soakpit. The soak
pit should be raised above ground level and no surface water should be allowed to enter into
it.
Unit 4. Estimation and costing (4)
4.1
Estimation and Costing for two roomed forest guard house.
4.2
Estimation and costing for retaining walls, breast walls and check dams.
4.3
Estimation and costing for forest roads Earthwork) trails and road side drains
4.4
Factors affecting the cost of a project.
Practical
1. Drawing building plans, elevations and sections and its cost estimate (6).
2. Design and drawing plans, elevations and sections of retaining wall and brest wall
and its cost estimate (4).
3. Design and drawing of cross sections of forest road in the plain and hilly area and
calculation of earth work volume.
4. Field observation of road components(4)
Estimation
Estimate is a judgment or calculation, not necessarily detailed or accurate. Estimate is made
to calculate roughly the cost, size, value etc of something.
An estimator should picture the object (building, structure, etc) in his mind from the study of
drawings and specifications. One who understands and can read drawing may calculate the
quantities. In preparing and estimate, the quantities of different items of work are calculated
by splitting a complete project into different items of work.
Costing
Cost is the price to be charged for or paid for something. In another word cost is the amount
of money needed for something. Costing is an estimate of how much money will be required
for something.
A complete project is splitted in to various activities and different item of works are
calculated from the drawings and rate of each item of works is obtained or competed
by analysis of rate of different items. Multiplying the quantities of different items by their
respective rates, we can compute the cost estimate of various activities and of a complete
project.
For all engineering works it is required to know beforehand the probable cost of
construction, which is known as the estimated cost.
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The main purpose of estimating is to know beforehand:
i.
The reasonably accurate cost of the work that involves materials, manpower,
tools and equipments, cost of establishment and supervision, overhead or
reasonable profit, taxes etc.
ii.
For plain, shallow and thin items where the thickness is less such as plastering,
formworks, painting etc shall be taken in square meter or square feet. To
compute the area, length and breadth or height is measured.
iii.
For those items, when one dimension is extremely large than other two
dimension like rope, cable, pipe etc shall be taken in running meter or running
feet.
iv.
Piece work or job works like wash basin, tube light, fan etc shall be taken in
number.
Methods of taking out quantities
For the calculation of quantities the dimensions (length, breath and height or depth) are
taken from the drawing -plan, elevations and sections. There is no hard and fast rule for
finding out the dimensions but they should be taken out accurately.
1. Long wall and short wall method: In this method, the length of the long wall (wall
running in longitudinal direction) measured out-to out and the length of the short
wall (wall running in transverse direction ) measured in to in. Then the quantity is
calculated by multiplying the length of the corresponding wall, breadth and height or
depth of wall.
2. Centre Line method: In this method, the sum total length of center line of walls (long
and short) has to be found out. this method is quick requires special attention and
consideration at the junctions, meeting points of partition or cross walls etc. for each
junction half breadth of the respective footing has to be deducted from the total
center line length. For different types of walls, each set of walls have to be dealt
separately.
Drawing: Drawing is a picture, plan or diagram. Engineering drawing includes a) Plans, b)
Elevations and c) Sections.
Plan: It is the view presented when looking vertically downwards at an object.
Elevation: It is the view obtained when looking in a horizontal direction towards the object.
Section: It is the true shape of an object presented after it has been cut through (or assumed
to have been divided) by a plane.
Specification: Specify or state something clearly and definitely about materials,
workmanship of an item of works. Specification is a description that indicates exactly what
is required.
4.1. Estimate the quantities of the following items of a two roomed forest guard house
1. Earth work in excavation in foundation
2. P.C.C in foundation.
3. Brick masonry in C.M (1:6) in foundation and plinth.
4. 2.5 cm cement concrete damp proof course (DPC)
5. Brick masonry in superstructure
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6. RCC in roof slab.
7. Wood work door and window frame.
8. Door window shutter.
9. Cement plaster.
10.Coloring/painting.
4.2. Estimation and costing for retaining walls, breast wall and check dams.
Practical: Design and drawing plans, elevations and sections of retaining wall and its cost
estimate.
Retaining wall: Retaining walls are of various types such as reinforced cement concrete,
stone masonry, brick masonry etc. Retaining wall is most important structure in hill road to
provide stability to the road way and to the scope.
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Retaining wall is constructed to maintain stability of soil and to prevent landslide. The main
function of retaining wall is to retain the back filling. It is provided where the hill section is
partly in cutting and partly in filling. In the absence of retaining wall the soil mass located at
higher elevation will have a tendency to move towards the lower elevation and the soil mass
will be unstable.
Dry stone masonry is preferred in retaining wall as it permits easy drainage of seeping
water. If the height is more, stone masonry is laid in cement mortar and weep holes are
provided at 1 m vertical height and 1.2 m apart horizontally to drainage seeping water.
Pressure on retaining wall varies from zero at to of wall to maximum at base wall. Weight of
wall passes through C.G of wall and acts vertically downwards. The resultant of these two
force (earth pressure and weight of wall) should pass within the middle third of the base of
wall. The wall is likely to be push bodily outwards due to earth pressure and may slide along
its base. Due to earth pressure any portion of the wall may be sheared off across the section.
Design of Retaining walls/Breast walls: Some Practical Rules:
Rule:
1. Stability of retaining wall is preserved when the retaining walls and breast walls have the
following dimensions.
Material
thickness
Base
Top
width of foundation
Remarks
Dry stone masonry
0.5 H
0.25 H
0.75 H
0.2 H
0.6
0.17 H
0.52 H
Up to 6'
height
6' to 12'
height
Over
12'
height
Dressed stone masonry or Stone 0.4 H
masonry in mortar
Mass concrete 1:2:4
0.35 H
i.
ii.
iii.
iv.
v.
Min depth of foundation =H/10 +1
Keep top width of wall 1'6" to 2'0" minimum.
Keep earth face vertical
Give front face a batter of 1 in 4 (here 1 is horizontal & 4 vertical)
Walls unto 6' height may be constructed dry stonemasonry.
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vi.
vii.
Walls greater than 6'height may be constructed top 6' in dry stone masonry and
rest of the height in cement masonry.
For extreme strength whole wall to be constructed in cement mortar with stone
masonry.
Rule 2:
a. Remember base width of R. wall.
b. Top width = 1/2 of Base
c. Width of foundation = 1 1/2 of Base width.
Breast wall:
The purpose of breast wall is to prevent the cut surface of hill from sliding. Breast walls
are provided on the inner side of the road to give support to cut hill. Breast walls prevent
land slides, which may block the road. Weep holes are provided to relieve (or reduce)
the pressure in the back of the wall.
Brest walls have a front batter of 1 in 2 and back batter of 1 in 3. If the height is above
6'0"; the top 6'0" alone should be in random rubble masonry and the remaining portion in
(1:6) cement mortar. Breast walls are not meant for protecting the slopes of cutting in
bad soil where there is likelihood of slipping and thus avoid blockages.
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Prepare a detailed estimate of a Retaining wall of 120'0" in length whose cross-section is
given. Walls top 6' height in dry stone masonry and rest of the height in stone masonry
in cement mortar (1:6). 1 1/4 HDP pipe at 3 layers vertically and @ 3'6" c/c horizontally
in weep holes.
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4.3. Forest road:
Road is a way on which people, animals or wheeled vehicles moves. Good forest roads are
national wealth as they help to exploite resources for the development of the nation.
The purpose of forest roads are to move or transport men and materials, transport forest
products, safeguard the forest resource and animals, and manage forest properly.
The requirements of a good road aligment are short, easy, safe and economical.
Basic principles of design of road:
a) Alignment and survey
b) Geometric design
c) Drainage
d) Retaining wall/ Breast wall
a) Alignment and survey: Determine the obligatory points through which the road
alignment should pass & should not pass. Points through which alignment should pass are
bridge site, town, hill pass, tourist spot etc and points through which alignment should not
pass are marshy place, waterlogged area, unstable area, long stretches of very hard rock
cutting etc. Engineering survey for road alignment consists of four stages:
i.
Map study
ii.
Reconnaissance
iii.
Preliminary survey, and
iv.
Detailed survey.
In first or initial stage, probable alternate alignment of road can be located on the map. In
second stage, reconnaissance survey is made to decide the most feasible route. During
reconnaissance, geological information, sources of materials etc are noted down. From the
details collected during reconnaissance, the alignment proposed after map study may be
altered or even changed completely. (Equipment- Abney level, Barometer).
After reconnaissance, preliminary survey of probable alternate route is made that help to
compare different alignment, to work out the cost of alternate route and to finalize the best
alignment from all considerations. In this survey, centerline of road is determined, centerline
profits and typical x-sections are taken, soil survey, material survey topographical features
are recorded.
In final and detailed survey, the alignment finalized after the preliminary survey is to be
located on the field by establishing the centre line. Next detailed survey should be carried
out for collecting the information's necessary of the preparation of plans and construction
details for the road project.
b) Geometric design: Geometric design of a road consists of gradient, width of carriage
way, design speed, camber, super lavation sight distance, overtaking right distance, radius of
horizontal and vertical curves etc.
c) Drainage: Side drains on either sides of road are provided to drain off the surface rain
water as well as the run off rain water. When side drain is unable to cope with the quantity
of water coming to the drain, cross-drains are provided at suitalbel intervals to discharge the
excess water down the hill. Provision of weep holes are made in retaining and breast walls
to relieve the seeping water. When the catchment area of the water shed above the road is
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large and the intensity of rainfall is heavy cathchwater drain is provided running paralled to
the road. Cathch water drains is diverted by sloping drains and carried across the road by
means of culverts.
d) Retaining wall, Breast wall: The function of retaining wall is to retain the back filling
and the purpose of breast wall is to prevent the cut hill from sliding. The ideals in road
design are: Good alignment, easy gradient, sufficient width, properly designed curves,
proper drainage, correct camber and superrelavation. Economical construction in material,
labour and transport.
4.4 Factors affecting the cost of a project:
General: A comparative statement of tenders received for a work often shows considerable
differences among the tendered runs, and between tendered runs and the estimated rest of
the work.
Lack of thorough knowledge of the various factors affecting the price of an item of work
must be considered as the chief cause for such differences.
Many other factors such as cut-throat competition to keep staff and machinery going, rushed
tenders, firms of different standards are also responsible to some extent that affects the cost
of a project.
The various factors affecting the unit price of an item or work may be grouped under the
five main component parts. These are:
1. Material
2. Labour
3. Plant
4. Overhead
5. Profit.
1. Materials: As a generalization it may be stated that the cost of materials in a unit price of
an item of work is nearly fifty percent of the total price. Various factors which affect the cost
of materials are: i) Quantity of material -The estimate of the amount of materials required
must be accurate. If materials purchased and transported on site are found to be short of the
requirement, it may result: a) Stoppage of work, b) purchase and transportation of materials
of ten at a higher cost than the cost involved in bulk purchase and transportation.
ii) Quality of materials
iii) Large quantities
2. Labor is the most uncertain factor in ascertaining the estimated cost of any item of work.
Cost of labour is determined assuming hourly or daily output of a workman and find out the
total number of man hours or days required for and item of work.
Output of different artisan various according to their skill, day to day weather, efficient sit e
organization and management. Higher wages, clean and healthy camp, less working hours
improve the output. Rate of wages should in lude the provision for the insurance, holidays
with pay, workmen's compensation and welfare obligation.
3. Plant: Owing to the shortage of labour the use of plant is increasing and is likely to
continue to do so. On large extensive work use of plant will be economical. If there is no
sufficient space for the movement of plant or if the amount of work is small the extensive
use of plant may prove impracticable or wasteful. Cost of work will depend upon the
decision of the estimator whether plant can be used. Purchase of plant would be economic
only it there is possibility of sufficient work ahead where it can be used. If pant can be used,
whether it is cheaper to purchase or to hire the plant.
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4. Overhead charges: Costs of office expenses, erecting office sheds, rents, stationery,
electricity, telephone, salaries of staff, postage etc are included in estimating overhead
charges. Every firm maintains the account of expensed under these headings. 10 percent
overhead charge is taken while analysing the rates of an item of work.
5. Profit: The margin of profit that a contractor may expect depends upon many factors. If a
firm has contracts to the limit of its capacity they will be prepared to take on new work at a
comparatively high figure. Another firm, in need of work to keep his workmen and
equipment fully occupied may expect only a nominal profit margin. The cost of work is
another factor on which the profit margin will depend. More the cost of a work, smaller is
the % of profit.
Calculation of Earthwork Volume:
Cross-section of earthwork of road in banking or in cutting is usually in the form of
trapezium, and the quantity of Earthwork may be calculated by the following methods:
Figure : Banking and Cutting
Sectional Area = Area of central rectangular portion + Area of two side triangular portion
=Bd+2(1/2xsdxd)
=Bd+ Sd2
S: 1 is the ratio of side slopes as Horizontal: Vertical
For 1 vertical, horizontal is S
.
. . for d vertical, horizontal is Sd
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Quantity = (Bd+ Sd2) x L
When the ground is in slope ,the height of bank or the depth of cutting will be different at
the two ends of the section, and mean height or depth may be taken for 'd'.
d  d2
Mean height or depth, dm= 1
(
2
d  d2
1. Mid-sectional Area Method Q=(Bdm+ Sd2m) x L where dm= 1
2
Station
or
chain age
depth
or
Height
Mean depth
or delight
dm
Area
central
portion
Ddm
of
Area of sides
sd2m
Total
sectional Area
Bdm+Sd2m
length
between
station
L
Quantity
Q=(Bdm+ Sd2m) x l
Embankm Cutting
ent
2. Mean-sectional area method
Quantity =Mean sectional area*length
A  A2
2
2
Q 1
 L, wher , A1  Bd 1  Sd1 , A2  Bd 2  Sd 2 , whereD1and , d 2 are tge height or
2
depth at two ends.
Station or
Chain age
Height
or depth
Area of
central
portion Bd
Area of
sides Sd2
Total
Sectional
Area
Bd+Sd2
Mean
sectional
Area Am
Length
between
Station L
Quantity
Q=(Bd+Sd2) x L
Embank Cutting
ment
3. Prismoidal Formula
L
Quantit , Q  ( A1  A2  4 Am )
6
2
  d1  d 2  2
L
 d1  d 2  
2
2
2
  Bd 1  Sd1  Bd 2  Sd 2  4 B
  S
 
6
2
2




 



 

Area of turfing of side slopes - to prevent damage by erosion and by rain water.
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sd 
2
 d  ,
2
mean sloping width of the embankment  s 2 d 2  d 2
 d s2 1
Station
or Depth
Chainage
height
or Mean depth Breadth or Length Total area of both
or height
sides slope between side slopes
Stations 2*L* d s 2  1
d s2 1
L
Ex.1. Gurucharan singh/Jagadish singh
Calculate the volume of earthwork for 100 meter length of road for a portion of a road in an
uniform groud.
Height of banks at the two ends is 0.75 m and 1.25m. Formation width is 10 meter and side
slopes of embankment are 2:1 Assume that there is no transverse slope.
Solution:
1. mid sectional area method:
d  d 2 0.75  1.25
weknowQ  ( Bd m  5d 2 m)  L, whereB  10m, Dm  1

 1.00m
2
2
s2


Q  10  1 / 00  2  (1.00) 2  100  (10  2)  100  1200cmm
2. Mean sectional area method:
We know,
A  A2
2
2
Q 1
 L, wher , A1  Bd 1  Sd1 , A2  Bd 2  Sd 2
2
A1  10  0.75  2  (0.75) 2  7.5  1.125  8.625Sq.m
A2  10  11.25  2  (1.25) 2  12.5  3.125  15.625sq.m.
A1  A2
 8.625  15.625 
L 
  100  1212.50sq.m
2
2


3. prismoidal formula
we know,
Q
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L
2
Q  A1  A2  4 Am , whereAm  Bd m  sd m
6
100

8.625  15.625  4(10  1  2  12 )
6
100
8.625  15.625  48

6
100

 72.25  12.4.167 sq.m
6
Ex. 3 Punmia
RL of ground along the center line of a proposed road from chain age 10 to20 are given
below:
The formation level at the 10th chain age is 107 and the road is downward gradient of 1 in
100 downward. Formation width of road is 10 meter and side slopes of banking are 2:1
(horizontal) . Length of chain is 30 meter.
Draw longitudinal section of the road and a typical cross-section, and prepare an estimate of
earthwork at the rate of Rs 275.00%


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Ex. 2 Punmia
a) Calculate the area of the side slopes of portion of a bank for a length of 200 meter.
The ratio of side slope is 2:1.
b) If the side slopes are to be provided with 15 cm thick stone pitching, calculate the
cost of pitching at the rate of Rs 1500 per cum.
Solution:
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We know, area of the two side slopes = 2 L * d s 2  1
d  d 2 2.5  3.5
Mean height, d m  1

 3m
2
2
Sloping breadth at the mid section= d m s 2  1  3 * 2 2  1,  6.71m
Area of the two side slopes=2*200*6.71=2684 sq.m
Quantity of stone pitching= Area*thickness
=2684*0.15=402.60sq.m
Cost of stone pitching=quantity *rate
402.60sqm*Rs1500/sq.m
=Rs603.900.00
Ex.4 B.N Dutta
Estimate the cost of earthwork for a portion of road for 400 meter length from the following
data: Formation width of the road is 10 meter side slopes 2:1 in banking 1 ½ : 1 in cutting
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The road passes from banking to cutting in between the station 30 and 31. The distance
where it passes through zero i.e, ground level, may be determined as follows:
The two triangles on either side of zero point are symmetrical.
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Home Task
Exe5. B N Dutta
Prepare a detailed estimate for earthwork for a portion of a road from the following data
Formatin width of road is 10meter, side slope 2: 1 in Banking and 1 1/2: 1 in cutting.
Draw longitudinal section of the road and a typical cross-section. Find the area of the side
slopes and cost of turfing the side slopes at the rate of Rs 600% sq.m.
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