Friday, February 28, 2014

Critical Stress Combination for Rigid Pavements

Hi, How you doing?
As we know that the design of a pavement is governed by the analysis so, this is very important to know the various combinations of stresses that will be induced in the slab. There are two kinds of loading:

  1. Traffic Loading
  2. Temperature variation.
Traffic loading will induce the compressive or tensile stresses in the pavement and temperature variation results in to kinds of stresses. One is known as warping stress and another is frictional stress.

That makes total of three kinds of stresses that will be induced in the pavement slab. Now, we have to find out the critical combination of these three.

  • In Summer Season:
In summer season the average temperature is higher and it results in the overall expansion of the slab. This induces the frictional stresses which will be compressive so as to restrict the expansion. Remember friction will always act in the opposite direction of the movement of the bottom of the slab. That means when it tries to expand, friction will try to contract it or restrict the expansion. That makes it to induce the compressive stress.

In the after mid-noon hours in summer season, the temperature at the top layer will be higher as compare to the bottom layer. That will make the slab to tend to warp in downward direction. Under the self weight of the slab warping will be restricted and therefore will result in the tensile stress in the bottom layer.

 Downward warping results in contraction of the bottom layer which is restricted through the induction of the tensile warping stress. So, warping stress of tensile nature is induced.
Now, at the edge region the traffic load stress of tensile nature is also induced and that is where the critical load combination is formed.

Critical stress combination(At the edge)  = Traffic stress + Warping stress - frictional stress.

  • In Winter Season:
In the after mid-noon hours in winter season again, traffic load stress and the warping stresses will be tensile in the bottom layer. Also, the frictional stresses will be tensile because, there is overall contraction of the pavement due to lower average temperature and therefore friction will act along the tensile direction.
So, the Critical stress combination(At the edge)  = Traffic stress + Warping stress + frictional stress.

At the corners the critical load combination is that of the traffic load and the warping stress at the top layer at the mid-night hours.
See, in the night the temperature at the bottom layers will be higher as compare to the temperature at the top layers so, there will be upward warping. That will induce tensile warping stress at the top layer.
 So, Critical stress combination(At the corner)  = Traffic stress + Warping stress 

Thank you for your time!


Wednesday, February 26, 2014

Flash(Open cup) and Fire point Test - Highway Engineering Lab

Flash and fire point (open) test apparatus( courtesy: Aimil Ltd.)
Hi, I am going to explain flash(open) and Fire point test.

  • AIM: To find out the flash and fire point of a given bitumen sample using open cup.
  1. Flash and Fire point apparatus (Open Cup). This apparatus consist of the a test cup, heating plate, heater, test flame applicator and supports.
  2. Thermometer (range -6 to 400 degrees Celsius).

  • THEORY: Bitumen is a volatile material and when temperature is increased, vapors are formed.
In order to work safely with the bitumen at the site, one must know the temperature at which bitumen may become hazardous to work with.
Flash and Fire points are the lowest temperature at which, sufficient amount of vapors are formed to catch the flash and fire respectively when a flame is brought near to it. At flash point sufficient amount of vapors are formed so as to catch a momentarily flash.
Flash and fire point vary for different variety of Bitumen. 

  1. Heat up the bitumen sample to a pouring temperature and then pour it down into the cup up to the marking. Stir the bitumen properly to remove any foreign gases present.
  2. Put the thermometer in place and start raising the temperature and 5 to 6 degrees per minute.
  3. When the temperature is about 25 degrees Celsius near to the expected flash point, start the flame and bring it near to the vapors.
  4. Note down the temperature at which the vapors catch a momentarily flash.
  5. Now keep raising the temperature and note down the temperature at which fire is started.
  • RESULT: 
  1. Flash point of the given bitumen sample = ......
  2. Fire point of the given bitumen sample = .....
NOTE: For Indian standard specifications for flash and fire point of Bitumen, please refer to IS:1209-1978.

Wednesday, February 19, 2014

Factors affecting the Stopping sight distance


Factors effecting the Stopping sight distance

As per IRC(Indian Roads Congress), Stopping sight distance is such a distance along the center line of the road that if there is an obstruction of 0.15 m at the other end, it must become visible to the eye of a driver at a height of 1.2 m from the road surface .

In order to analyse the stopping sight distance which is required to design a given highway we have to consider the following factors which, effect the sight distance.

  1. The total reaction time of the driver
  2. Speed of the vehicle
  3. Friction between the Tyre and the pavement surface
  4. Break efficiency
  5. Gradient of the road 
  • The total reaction time of the driver: Total reaction time of a driver is the time from the instance the obstruction is visible to the driver to the instance when he effectively applies the break. 
 In the total reaction time vehicle moves at the speed at which the driver is moving or taken as the design speed. So, if the total reaction time of the driver is more, more will be the distance traveled and more will be the stopping sight distance.
Stopping sight distance can be phased into two parts: 1) Perception time 2) Brake reaction time

Perception time is the time taken by the driver to realize that there is an obstruction on the road.
Brake reaction time is the time taken by the driver to effectively apply the brakes. This depends upon the skills of the driver, friction of the tyre and road surface and the brake efficiency.

PIEVE theory is used to analyse the total reaction time. According to this theory total reaction time of a drive is phased into four classes:

  1. Perception time
  2. Intellect time
  3. Emotion time
  4. Volition time
  • Speed of the vehicle: Speed of the vehicle affects the distance traveled by the driver in the total reaction time, more the speed more will be the distance.  This is known as lag distance.
Similarly,the distance traveled by the driver after the application of the brakes. More the speed more will the braking distance.

  • Friction between the Tyre and the road surface: The friction between the Tyre and the road surface depends upon the type of road surface and the condition of the tyre. 
Also, it depends upon the speed of the vehicle. More the friction, less will the stopping sight distance required but, if less is the friction, more it will be.
  • Brake Efficiency: 100% brake efficiency means the rotation of the Tyre is completely locked, but it will surely result in the skidding of the vehicle. Efficiency of the brakes are considered by reducing the original value of the friction in a range of 0.35 to 0.40.
  • Gradient of the road: Gradient may be positive or negative and accordingly the required stopping sight distance will be less and more respectively. In case of upward(positive) gradient, a component of the gravity force will help in stopping the vehicle.
In case of the downward gradient the component of the gravity will be in opposite direction to the direction of the friction force so this will increase the stopping sight distance.

Thanks for your visit!

Saturday, February 15, 2014

Tri-axial method of flexible pavement design


  • In this method of pavement design we use the Tri-axial shear test to determine the Elastic modulus of the sub-grade material or those of the other layers. Boussinesq's equation is used to determine the thickness of the pavement.
  • In order to include the factors, amount of traffic and rain-fall 'X' and 'Y' are multiplied to the equation which gives us the pavement thickness. More the traffic and amount of rainfall higher will be the values of 'X' and 'Y'.
  • A stiffness factor is introduced when the Elastic modulii are different for the sub-grade and pavement layers. This factor (Es/Ep)^(1/3) is multiplied to the former value of thickness in which, the moduli are considered equal.
  • In order to find out the thickness for the different layers we simply use the stiffness factor according to which, thickness is inversely proportional to the Elastic modulus raised to a power of 1/3.

Thursday, February 13, 2014

Road Margins - Highway Engineering

Road margins are the various cross sectional elements of the road except the carriageway or pavement width. 
Various road margins are shoulders, foot-paths, cycle-tracks, frontage paths, driveways, lay byes, side slopes and guide rails. I will discuss each, one by one.

  • Shoulders: Shoulders are the parts of the formation width except carriage ways. In other words, shoulder is the part of the pavement which is non-surfaced. 
They are used by the vehicular traffic as the emergency lanes or sometimes as the service lanes for the repairing of the non-expected problems. IRC recommends a minimum value 2.5 m for the shoulders for two lane rural roads. A width of 4.6 m is recommended so that a truck can be accommodated without interfering with the adjacent lane. 
  • Footpaths: Foot paths or pedestrian paths are the smoothly paved paths used by the pedestrians to walk parallel to the pavements. Footpaths are smoothly surfaced in order to attract the pedestrians to walk over them. Footpaths are necessary where the pedestrian traffic is considerable.
  • Cycle-tracks: Cycle tracks are provided to carry the cyclists. They are generally provided in the urban roads where the design speed is very high and it becomes necessary to separate the high speed traffic from the low speed traffic. Minimum of 2 m width is recommended for single lane and 1 m is added for constructing the extra lanes of the cycle tracks.
  • Frontage Paths: They are provided in front of the commercial buildings where it becomes a necessity. In urban areas they are provided with the dividers to separate from the vehicular lanes.
  • Drive-ways: Drive ways are provided to reach the buildings like fuel pumps or service centers. Drive ways should have a less width as much as possible because they are hindrance to the pedestrian traffic travelling along the foot-paths.
  • Lay Byes: Lay Byes are the paved areas provided at some places on the sides of the lanes for providing a stoppage for the vehicles. 
  • Parking Lanes: They are provided for parking the vehicular traffic generally needed in the market centers or the urban areas. Parallel parking is recommended in order to avoid the hindrance to the traffic and a minimum of 3 m parking lane is recommended.
  • Side Slopes: A proper, flatter/gentle side slopes are necessary in case of filling or cutting for the stability of the pavements. If possible they must be provided with the landscaping so making them a pleasant and aesthetic appearance and making more stable.

  • Guide Rails: Guide rails are provided on the roads on filling generally on the hills, on the outer edges, for the psychological as well as physical protection to the driver. 

Thursday, February 6, 2014

Solution- GATE 2014 -Civil Engineering- set -I

I hope GATE 2014 was a good experience for you. I don't remember values and figures but language of the questions can make you remind the question so, I will post here the manner in which the question was to be solved.This post contains the questions asked in set I(9:00AM to 12:00PM).

 I will keep updating this post. I will give you the possible manner in which question was to be solved. Questions are not in exact serial number because it is difficult to remember.


Q.1 Match group I with Group II.

GroupI                                                        Group II
P. Alidade                                               A. Chaining
Q. Arrow                                                B. Plane Table

Solution:  P-B, Q-A,

Q.2 Following staff readings were taken using a level. Find the R.L. of station S.

Station           B.S.     F.S.        R.L.(m)
P                                           100
Solution: In this question first you have to add the B.S. to the R.L. of bench mark(P), to get the Height of instrument and then minus the F.S. at Q, then add the B.S. at Q, then minus the F.S. at R, then add up the B.S. at R and minus the for sight at S and you will get the R.L. of the S.
Remember, the Fore sights were given with the negative signs at some stations, so automatically two negative signs will produce a positive sign.

Structural Engineering:

Q.1 The influence line diagram for the shear force at point P at a distance of L/4 from end A in a simply supported beam AB will be.

Solution: Remember, there were four diagrams. The diagram which has a negative linear variation of shear force on the left of the section from a value of 0 at A to 0.25 units just on left of P and then 0.75 when just on right of P and then varying linearly to a value of 0 at B, is the right diagram for the influence line of shear force at P.

Q.2 Central singly concentrated collapse load for the given propped cantilever beam of length L will be:
(A) 2Mp/L  (B) 4Mp/L   (C) 6Mp/L  (D) 8Mp/L

Solution:  (C) 6.Mp/L
Given beam is a propped cantilever so, there will be two hinges, one at the fixed end and one at the center, where the load is acting.
Equating the external work done to the internal work done,
Wp.L/2.O= Mp.O + 2.Mp.O
=> Wp = 6.Mp/L

Q.3 Find out the static indeterminacy:

(A) Two (B) One  (C) Zero   (D) Three.

Solution: (C) Zero
You were given a rigid frame with two pinned end supports and one internal hinge. Two pinned supports produce 4 external support reactions and also with one internal hinge our number of equations of static equilibrium gets increased by one so, total is four.
So, static indeterminacy = 4-4 = 0.

Q.4 A given reinforced concrete beam of M25, has a neutral axis at a distance of y  from the top face. If the maximum stress developed in the top fiber is x, find out the curvature of the beam. (If you remember the exact values, please post in comment box).

Solution: Here you have to use the flexure formula,
f/y = E/R    => 1/R = f/(E.y)

where, f= flexural stress in the top fiber, y = distance of the neutral axis from the top fiber
E is Thomas Young's modulus of elasticity = 5000(fck)^(1/2)  here, fck = 25 Mpa

Q.5 Find out the value of force P and its angle of application, in order to lift the given body of weight 'W' without any swing.

Solution: There were numbers of forces acting on the body with their respective angles mentioned. In order to lift up the body without any swing we had to use the following two equations:
Algebraic sum of all the horizontal forces = 0
Algebraic sum of all the vertical forces = W
We have two unknowns, P and its angle and also have two equations, so can be easily found.

Transportation Engineering:

Q.1 The design speed of a road is V km, and the co-efficient of longitudinal friction for the road and tyre is f, if the stopping sight distance of the vehicle is 's' find out the total reaction time of the driver.

Solution: In this question we need to use the equation  V.t + V^2/ (2gf) = s
here, all the values are given except t, which can be easily found.

Q.2 Equation of the velocity(Km/Hr) for a given traffic flow is u= 70+0.7k, where k(Veh/Km) is the density of the flow, the capacity of the flow is:
(A)..   (B)   .. (C) ..   (D)..

Hydraulics and water resource Engineering

Q.1 Two parallel pipes diverges from a single pipe and after same length again converge at the another end. If the diameter of the first pipe is four times the diameter of the second pipe, the velocity ratio in first first pipe to that in second pipe will be
(A) 1  (B) 2  (C) 4  (D) 3

Solution: (B)
As the head loss in two pipes will be same,
we can equate the head loss equations of the two. given d1=4.d2.
use Hl= f.l.v^2/(2gd).

Q.2 A1 and V1 are area of cross section and velocity at section 1, and A2 is the area of the section at section two of a section of a continuous flow. What is the velocity at section 2?

Solution: Continuity equation can be applied to get the velocity at section 2.
Note: Values were given in the question and it was a subjective question)

Q.3 Diameter of the Venturimeter and of a section are given, velocity head difference at the throat of Venturimeter and section is given also, velocity at the section is given. Find out the co-efficient of discharge for the Venturimeter.

Solution:  Here first you have to calculate the velocity at the Venturimeter throat which, can be calculated from the given velocity head difference between the section and the throat.
The discharge at the section is calculated by the given data of the area and velocity,
finally equate the discharge to the formula of the discharge for the Venturimeter.
Final equation will give the value of co-efficient of discharge.

To be continued...

Monday, February 3, 2014

Group Index Method of Pavement Design

Group Index method of flexible pavement design is an empirical method which is based on the physical properties of the soil sub-grade.  To design the thickness of the pavement you have to go through the following steps:

  1. Find out the Group Index Value(GI) of the soil.
  2. Use the design charts to find out the thickness of the pavement and layers.
Finding GI Value : Now let's discuss Group Index(GI) first. Group Index is a number assigned to the soil based on its physical properties like particle size, Liquid limit and plastic limit. It varies from a value of 0 to 20, lower the value higher is the quality of the sub-grade and greater the value, poor is the sub-grade.
To find out the value of GI we, can either use the following equation, or we can use the charts

GI = 0.2a + 0.005 a.c. + 0.01b.d

a= percentage of soil passing 0.074 mm sieve in excess of 35 percent, not exceeding 75.
b= percentage of soil passing 0.074 mm sieve in excess of 15 percent, not exceeding 55.
c= Liquid limit in percent in excess of 40.
d= Plasticity index in excess of 10.

The second method to find out the GI value is to use two charts whose numerical values added together to get the GI value.

Using design charts to find the thickness: 
Once the GI value is found next we can use the design charts given by IRC(Indian Roads Congress) to find out the thickness of the pavement and its thickness of various layers. Charts are prepared based on the GI value and one can find the thickness of the pavement to be designed for different traffic volume. 

Traffic volume is classified in three categories as Light, Medium and Heavy. When nos. of vehicles per day is less than 50, this is light traffic, greater then 50 and less than 300 is medium and greater than 300 is classified as heavy traffic. 

You have follow the corresponding curve for the given traffic and for the one among the three different curves you can find the total value of the thickness of the pavement. Then there are two other curves, one to find out the thickness of the sub-base and one to find the thickness of the base course.

Limitation of Group Index method: Limitation of this method is that this is based only on the physical properties of the soil and does not consider the strength parameters.

Thank you!

Solved- Traffic Flow Distribution among Routes at Equilibrium

Hi, Prob: There are two routes, such that the time consumed is given by the following.  t1 = 15 + 0.005q1 ,  t2 = 12 + 0.003q2,  in p...