Practicing questions from test series forms an essential part of the preparation for any competitive exam. Test series help in evaluating one's performance in a particular subject. It could then be used to shift the focus of one's study plan and prepare accordingly.

Also, the full-length test series that contains questions in the pattern of the actual exam helps one to have a feel of the real exam. Therefore, it is clear that test series are inevitable when it comes to preparing for any competitive exam.

**| Now**

Now is the correct answer to this question. But many have a different opinion. Some aspirants start the test series only before a month of the exam or so. Practicing questions alongside preparation is important. It gives an idea about what questions are expected out of a subject and at what difficulty are they asked. Both of these aspects are required to fine-tune one's preparation strategy. Therefore, start practicing questions in general and practice test series in particular.

One can refer to Where to Practise Questions for GATE 2023 to find sources available for practicing questions.

APSEd's test series is one of the most sought-after test series for the GATE exam preparation. APSEd's test series are specially curated by experts in each subject. It is then reviewed and corrected to make the test series error-free. All of these make the APSEd's test series one of the finest test series out there.

We at APSEd have launched test series for GATE CE, ES, and GE. This blog will explain the details of the GATE GE Test Series in detail.

APSEd's GATE GE test series contains 12 tests in total. Out of the 12 tests, 10 tests are subject tests and the rest 5 are full-length mock tests. The 10 subject tests include the following.

Engineering math

Aptitude

Remote sensing

GIS

GNSS

Maps

Surveying

Photogrammetry and two mix tests

All the questions are similar to the GATE level of questions. And each question has been provided with a clear explanation. Apart from this, APSEd's test series gives access to get in touch with the faculty for doubt clarifications.

APSEd's test series are not time-bound. Once purchased all the test series are available till the GATE exam schedule. One can attempt it several times as required. Also, all the test series are easily accessible via online browsers, making it one of the easily accessible test series for the GATE exam.

**| Quality Above Quantity **

Quality above quantity has been the key principle of APSEd's test series. We are aimed at providing mistake-free test series with completely new GATE level questions each year. Though there are platforms that could offer more tests, it is the quality that matters more at APSEd. Therefore, one can trust APSEd's test series for quality.

**| Trust is Earned when Actions Meet Words**

We at APSEd understand the meaning of the above-said quote. Therefore, we are sharing information from the GATE 2022 Geomatics Engineering Paper Analysis.

From the image, it is clear that APSEd's full-length test series almost follows the GATE exam pattern. Therefore, it is clear that APSEd's test series best captures the actual GATE exam pattern.

APSEd provides you with a detailed test report after the end of each test. It contains 3 sections as explained below.

The 1st section contains the overall accuracy score. It also contains the number of correct and wrong answers.

The 2nd section contains the time analysis. It gives details on how much time is spent on correct answers and wrong answers.

The 3rd section contains the leaderboard. It shows how well you have performed compared to other aspirants.

All these could be helpful in fine-tuning your preparation to excel in the GATE exam. Therefore, make use of the test series as much as possible.

As an introductory offer for test series, all the test series are available at a **60% offer**. This offer makes the APSEd's test series even more **"value for money product"**. Therefore, don't miss out on this opportunity. For more details visit APSEd's Test Series Page.

For more insights check out the video below. (Will be updated)

Practicing questions from test series forms an essential part of the preparation for any competitive exam. Test series help in evaluating one's performance in a particular subject. It could then be used to shift the focus of one's study plan and prepare accordingly.

Also, the full-length test series that contains questions in the pattern of the actual exam helps one to have a feel of the real exam. Therefore, it is clear that test series are inevitable when it comes to preparing for any competitive exam.

**| Now**

Now is the correct answer to this question. But many have a different opinion. Some aspirants start the test series only before a month of the exam or so. Practicing questions alongside preparation is important. It gives an idea about what questions are expected out of a subject and at what difficulty are they asked. Both of these aspects are required to fine-tune one's preparation strategy. Therefore, start practicing questions in general and practice test series in particular.

One can refer to Where to Practise Questions for GATE 2023 to find sources available for practicing questions.

APSEd's test series is one of the most sought-after test series for the GATE exam preparation. APSEd's test series are specially curated by experts in each subject. It is then reviewed and corrected to make the test series error-free. All of these make the APSEd's test series one of the finest test series out there.

We at APSEd have launched test series for GATE CE, ES, and GE. This blog will explain the details of the GATE ES Test Series in detail.

APSEd's GATE ES test series contains 12 tests in total. Out of the 12 tests, 10 tests are subject tests and the rest 2 are full-length mock tests. The 10 subject tests include the following.

Engineering math

Aptitude

Water and wastewater treatment

Water resource and environmental hydraulics

Global and regional environmental issues

Environmental chemistry

Solid and hazardous waste management

Environmental management

Air pollution

Microbiology

All the questions are similar to the GATE level of questions. And each question has been provided with a clear explanation. Apart from this, APSEd's test series gives access to get in touch with the faculty for doubt clarifications.

APSEd's test series are not time-bound. Once purchased all the test series are available till the GATE exam schedule. One can attempt it several times as required. Also, all the test series are easily accessible via online browsers, making it one of the easily accessible test series for the GATE exam.

**| Quality Above Quantity **

Quality above quantity has been the key principle of APSEd's test series. We are aimed at providing mistake-free test series with completely new GATE level questions each year. Though there are platforms that could offer more tests, it is the quality that matters more at APSEd. Therefore, one can trust APSEd's test series for quality.

**| Trust is Earned when Actions Meet Words**

We at APSEd understand the meaning of the above-said quote. Therefore, we are sharing information from GATE 2022 Environmental Science and Engineering Paper Analysis.

From the image, it is clear that APSEd's full-length test series almost follows the GATE exam pattern. Therefore, it is clear that APSEd's test series best captures the actual GATE exam pattern.

APSEd provides you with a detailed test report after the end of each test. It contains 3 sections as explained below.

The 1st section contains the overall accuracy score. It also contains the number of correct and wrong answers.

The 2nd section contains the time analysis. It gives details on how much time is spent on correct answers and wrong answers.

The 3rd section contains the leaderboard. It shows how well you have performed compared to other aspirants.

All these could be helpful in fine-tuning your preparation to excel in the GATE exam. Therefore, make use of the test series as much as possible.

As an introductory offer for test series, all the test series are available at a **60% offer**. This offer makes the APSEd's test series even more **"value for money product"**. Therefore, don't miss out on this opportunity. For more details visit APSEd's Test Series Page.

For more insights check out the video below. (Will be updated)

Practicing questions from test series forms an essential part of the preparation for any competitive exam. Test series help in evaluating one's performance in a particular subject. It could then be used to shift the focus of one's study plan and prepare accordingly.

Also, the full-length test series that contains questions in the pattern of the actual exam helps one to have a feel of the real exam. Therefore, it is clear that test series are inevitable when it comes to preparing for any competitive exam.

**| Now**

Now is the correct answer to this question. But many have a different opinion. Some aspirants start the test series only before a month of the exam or so. Practicing questions alongside preparation is important. It gives an idea about what questions are expected out of a subject and at what difficulty are they asked. Both of these aspects are required to fine-tune one's preparation strategy. Therefore, start practicing questions in general and practice test series in particular.

One can refer to Where to Practise Questions for GATE 2023 to find sources available for practicing questions.

APSEd's test series is one of the most sought-after test series for the GATE exam preparation. APSEd's test series are specially curated by experts in each subject. It is then reviewed and corrected to make the test series error-free. All of these make the APSEd's test series one of the finest test series out there.

We at APSEd have launched test series for GATE CE, ES, and GE. This blog will explain the details of the GATE CE Test Series in detail.

APSEd's GATE CE test series contains 20 tests in total. Out of the 20 tests, 15 tests are subject tests and the rest 5 are full-length mock tests. The 15 subject tests include the following.

Geotechnical engineering

Engineering math

Aptitude

Environmental engineering

Fluid mechanics

Construction materials

Construction management

Transportation engineering

Surveying

Irrigation engineering

Hydrology

Strength of materials

Steel structures

RCC

Structural analysis

All the questions are similar to the GATE level of questions. And each question has been provided with a clear explanation. Apart from this, APSEd's test series gives access to get in touch with the faculty for doubt clarifications.

APSEd's test series are not time-bound. Once purchased all the test series are available till the GATE exam schedule. One can attempt it several times as required. Also, all the test series are easily accessible via online browsers, making it one of the easily accessible test series for the GATE exam.

**| Quality Above Quantity **

Quality above quantity has been the key principle of APSEd's test series. We are aimed at providing mistake-free test series with completely new GATE level questions each year. Though there are platforms that could offer more tests, it is the quality that matters more at APSEd. Therefore, one can trust APSEd's test series for quality.

**| Trust is Earned when Actions Meet Words**

We at APSEd understand the meaning of the above-said quote. Therefore, we are sharing information from GATE 2022 Civil Engineering Paper Analysis.

From the image, it is clear that APSEd's full-length test series almost follows the GATE exam pattern. Therefore, it is clear that APSEd's test series best captures the actual GATE exam pattern.

APSEd provides you with a detailed test report after the end of each test. It contains 3 sections as explained below.

The 1st section contains the overall accuracy score. It also contains the number of correct and wrong answers.

The 2nd section contains the time analysis. It gives details on how much time is spent on correct answers and wrong answers.

The 3rd section contains the leaderboard. It shows how well you have performed compared to other aspirants.

All these could be helpful in fine-tuning your preparation to excel in the GATE exam. Therefore, make use of the test series as much as possible.

As an introductory offer for test series, all the test series are available at a **60% offer**. This offer makes the APSEd's test series even more **"value for money product"**. Therefore, don't miss out on this opportunity. For more details visit APSEd's Test Series Page.

For more insights check out the video below.

Field tests are performed to ascertain the quality of cement delivered to the field. These tests will give quick information about the quality of the cement. Below listed are the tests that are to be performed at the site before using the cement for construction.

Visual test: The colour of the cement must be greenish grey

Feel test: Thrusting the hand into the cement should give a cool feeling. Also, it should give a smooth feeling when rubbed between fingers.

Lump formation: Check with your hands for any lumps in the cement. If there are lumps, break them with your fingers. If the lumps don't break then the cement is considered spoiled due to air setting.

Floating test: The cement should float for some time before sinking when thrown into the water.

The objective of this test is to check the grinding of the cement. It can be checked using two methods i.e.,

Sieve test

Air Permeability Method

The test contains the following steps.

100 grams of cement is taken as a sample

The cement is sieved using a standard IS sieve no. 9 (90 microns)

Ensure that there are no lumps in the sample

Sieve the sample manually or mechanically for 15 minutes

Weigh the residue left on the sieve

This weight shall not exceed 10% for ordinary cement and 5% for rapid hardening or low-heat cement.

This test is done to check the specific surface of the cement based on which the fineness of the cement could be ascertained. Lea and Nurse Air permeability apparatus is used for this test. The test is based on the relationship between the flow of air through the cement bed and the surface area of the particles comprising the cement bed. Blain air permeability apparatus is also used for this test.

Standard consistency** **of cement is conducted to determine the quantity of water required to produce a cement paste of standard or normal consistency for use in other tests. This test is performed** ***with the help of* the *Vicat apparatus. *

The standard or normal consistency of a cement paste can be defined as that consistency that will permit the Vicat plunger (10 mm in diameter & 40-50 mm in length) to penetrate to a point 5 to 7 mm from the bottom of the Vicat mould. The following procedure is used to perform this test.

400 gm sieved cement is taken & 100 gm of water is added and mixed thoroughly for about 3 minutes. T

The paste is then filled into the Vicat mould, making it level with the top of the mould.

The filled-up mould is placed centrally below the movable rod fitted with a plunger.

The bottom surface of the plunger is brought in contact with the surface of the cement paste and the reading of the scale is taken.

The rod is then quickly released and the penetration is noted.

If the rod penetrates 5 to 7 mm

Otherwise, the trial paste should be made with a varying quantity of water and the test is repeated as above till the desired penetration of 5 to 7mm is obtained.

The cement paste remains plastic for a short period of time. As the time lapses, the plasticity gradually disappears and the paste changes into a solid mass. This process is known as the setting of cement. The time to reach this stage is known as setting time. The setting time is divided into two parts namely initial setting time and final setting time.

The time at which the cement paste loses its plasticity is termed the initial setting time. The time taken to reach the stage when the paste becomes a hard mass is known as the final setting time.

The initial and final setting time test on cement is performed with the help of the Vicat apparatus. The initial setting time of cement shall be the time elapsing between the time when the water is added to the cement and the time at which the needle (1 mm square or 1.13 mm in dia and 40 to 50 mm in length) penetrate to a point 5 mm from the bottom of the Vicat mould. The following procedure is used to perform this test.

400 gm sieved cement is taken and water is added to it @ 0.85 P by weight of cement. Where P is the percentage of water required for normal consistency paste.

A stopwatch is started at the instant of adding water. The paste is mixed thoroughly for about 3 minutes. The paste is then filled into the Vicat mould, making it level with the top of the mould.

The filled-up mould is placed centrally below the movable rod fitted with a needle.

The bottom surface of the needle is brought in contact with the surface of the cement paste and the reading of the scale is taken.

The rod is then quickly released and the penetration is noted.

The procedure is repeated until the needle fails to penetrate the flock for about 5 mm measured from the bottom of the mould. The time from the stopwatch is recorded which gives the initial setting time.

The cement shall be considered finally set when the final setting needle makes only an impression on the top surface of the paste.

The soundness test of cement is performed to identify the presence of excess free lime and magnesia in the cement. Excess lime could produce an expansion in the volume of cement which is undesirable. Therefore, ascertaining the absence of such free lime becomes essential.

This test is performed with the help of the Le-Chatelier apparatus. A Le-Chtalier apparatus consists of a cylindrical brass mould and two indicator points. A glass plate is used along with Le-Chatlier's apparatus for this test. The following procedure is used to perform this test.

Place the Le-Chatlier's apparatus on a lightly oiled glass sheet and fill it with cement paste formed by gauging cement with 0.78 times the water required to give a paste of standard consistency.

Cover the mould with another piece of a lightly oiled glass sheet, place a small weight on this covering glass sheet and immediately submerge the whole assembly in water at a temperature of 27 ± 2°C for 24 hours.

Measure the distance separating the indicator points to the nearest 0.5 mm.

Submerge the mould again in water at the temperature prescribed above.

Bring the water to boiling, with the mould kept submerged, in 25 to 30 minutes, and keep it boiling for three hours.

Remove the mould from the water, allow it to cool and measure the distance between the indicator points.

The difference between these two measurements indicates the expansion of the cement. This must not exceed 10 mm for ordinary, rapid hardening and low heat portland cement.

If in case the expansion is more than 10 mm as tested above, the cement is said to be unsound.

Compressive testing forms an important part of understanding the grade of cement. The following procedure is used for this test.

200 grams of cement and 600 grams of sand are taken. The water of quantity (P/4 + 3) % of the total mass of cement and sand (where P is the standard consistency of cement) is taken.

Mix cement and sand in dry condition with a trowel for one minute and then add water and mix until a uniform colour is obtained. The time of mixing shall not be less than 3 minutes and not more than 5 minutes.

Immediately after the mixing, cement mortar is placed in the cube mould. Before placing the mortar, apply oil to the inner surface of the cube mould. For expelling the entrained air and avoiding honeycombing, the mortar shall be prodded 20 times in about 8 seconds and then compacted by vibration.

After compaction, finish the top surface of the cube in the mould by smoothing the surface with the blade of a trowel.

Keep the filled moulds in a moist closet or moist room for 24 hours.

At the end of that period, remove the mortar cube from the moulds and immediately submerge it in clean, fresh water and keep it there until taken out just prior to testing.

A compression testing machine or universal testing machine is used for testing. Test 3 cubes for compressive strength for each time period as per detailed specifications. Such as 3 cubes for 3 days test, 3 cubes for 7 days test and 3 cubes for 28 days test.

The cubes testing shall be conducted on their sides. Do not use any packing between the cube and the steel plates of the compression testing machine, when testing is conducted.

The load shall be steadily and uniformly applied and the rate of loading should be 35 N/mm2/min.

The pressure at which the cube fails gives the compressive strength of the cement.

With this, we hope we had covered all the important topics related to the test on cement. Do you want us to cover a topic of your interest in a simple manner? Let us know in the comments below.

Before getting into details of paper combinations, let us see some general information about the GATE 2023. The GATE exam is conducted by IITs. Each year different IITs take up the role of conducting the exam. Conducting the exam comprises everything from the initial announcement to the release of the GATE scorecard. GATE 2023 is being conducted by IIT Kanpur. For more details check out GATE 2023 IIT Kanpur: Exam Dates, Pattern, Eligibility, Exam Papers & Paper Combinations blog.

IIT Kanpur has announced important dates for the GATE 2023 exam. The exam dates are announced as 4th, 5th, 11th, & 12th February. For more details check out GATE 2023 Official Notification | Important Dates for GATE 2023 blog.

A candidate who is currently studying in the 3rd or higher years of any undergraduate degree program OR has already completed any government-approved degree program in

Engineering / Technology / Architecture / Science / Commerce / Arts is eligible for appearing in the GATE 2023 exam.

One important disclaimer here is, that **GATE is not an admission-ensuring examination.** Qualifying in the GATE examination does not guarantee admission/scholarship. Admission to any institute is fully dependent on the admitting institute's criteria for educational qualification. Similarly, a GATE qualification does not assure a Public Sector Undertaking (PSU) job, as it depends on the recruitment procedure of the concerned PSU.

GATE exam will be conducted as an ONLINE Computer Based Test (CBT) in specific exam centers only. The online examination papers will contain some questions,

(i) Multiple Choice Question (MCQ) type, where only one option out of four options is correct.

(ii) Multiple Select Question (MSQ) type, where one or more than one option out of four options is/are correct,

(iii) Numerical Answers Type (NAT) where the answer must be keyed in by the candidate using a virtual keypad.

One more thing to note is, that no calculators will be allowed, instead, the candidates can use an on-screen virtual calculator provided for the examination.

The GATE 2023 exam contains the following papers.

Aerospace Engineering (AE)

Agricultural Engineering (AG)

Architecture and Planning (AR)

Biomedical Engineering (BM)

Biotechnology (BT)

Civil Engineering (CE)

Chemical Engineering (CH)

Computer Science and Information Technology (CS)

Chemistry (CY)

Electronics and Communication Engineering (EC)

Electrical Engineering (EE)

Environmental Science and Engineering (ES)

Ecology and Evolution (EY)

Geomatics Engineering (GE)

Geology and Geophysics (GG)

Instrumentation Engineering (IN)

Mathematics (MA)

Mechanical Engineering (ME)

Mining Engineering (MN)

Metallurgical Engineering (MT)

Naval Architecture and Marine Engineering (NM)

Petroleum Engineering (PE)

Physics (PH)

Production and Industrial Engineering (PI)

Statistics (ST)

Textile Engineering and Fibre Science (TF)

Engineering Sciences (XE)

Humanities & Social Sciences (XH)

Life Sciences (XL)

Now that we know some basic knowledge about GATE 2023 and the papers available, let us now see about combinations of papers that are available for aspirants.

The paper combinations are grouped as primary papers and a list of secondary papers that could be taken as the second paper. Each primary paper can have one or more GATE papers as a secondary paper. The table given below shows the secondary papers that are available for primary papers.

For instance, if a candidate chooses civil engineering paper as the primary paper, he/she can choose any one paper from AE, AG, AR, ES, GE, NM, or XE papers. Similarly, for all other primary papers, any one of the secondary papers can be chosen.

Do note that due to unforeseen circumstances, any paper combinations might be removed from the above table. Also, candidates may be allocated to a different exam center but within the same city for taking the second paper. Now that we know bout paper combinations we will now see a bit about preparation for GATE 2023.

Self-study requires self-motivation and an appropriate strategy to prepare for the exam. Self-motivation is an internal factor and it solely depends on the individual. Preparation strategy on the other hand can be generalized for all. There are three key factors to be considered for self-study which are listed as follows.

Resources

Time

Plan

We had explained a lot about this in the GATE 2023 Self Study | A Complete Strategy for Self Preparation blog, do check that out. We also published blogs on preparation strategies and sources for GATE 2023, all of which are listed below.

With a great determination for preparation from your side comes great offers from our APSEd side. We are excited to announce that we are offering a massive 55% offer on our GATE courses. For more details check out the APSEd website.

Aggregate is used in layers of the flexible pavement. Therefore, testing of aggregates becomes important. Properties like strength, toughness, etc. are all checked as explained below. As such aggregate testing could be classified into two broad categories as mentioned below.

Test for Strength of Aggregate

Test for Shape of Aggregate

This test is done to find the strength of aggregates. The test procedure involves the following steps.

The aggregate specimen is weighed (W1).

The specimen is filled in a cylindrical mold of diameter 15.2cm in three layers with each layer being subjected to 25 blows.

A plunger of diameter 15cm is placed on the top of the mold.

The entire setup is placed in a compression testing machine and is subjected to a load rating of 4 tons/min until it reaches a load of 40ton.

After attaining the max load, the load is released and is allowed to settle down.

The specimen is taken and sieved in a 2.36mm sieve. The weight of material retained in the sieve is found (W2).

**Crushing value = ((W1 - W2 ) / W1) *100**

For a good quality aggregate, the crushing value should be as low as possible. IRC recommends a crushing strength of 30% for the wearing course and 45% for the base course.

This test is done to find the toughness characteristics of aggregates. The test procedure involves the following steps.

The aggregate specimen is weighed (W1).

The specimen is filled in a cylindrical mold in three layers with each layer being subjected to 25 blows.

A hammer weighing 13.5 to 14kg is dropped on the specimen from a height of 38cm 15 times.

The specimen is taken and sieved in a 2.36mm sieve. The weight of material passing the sieve is found (W2).

**Impact value = (W2 / W1) *100**

For a good quality aggregate, the impact value should be as low as possible. IRC recommends an impact value of 30% for the wearing course and 45% for the base course.

Abrasion test is done to find the wearing and tearing characteristics of the aggregate. Los Angels abrasion test is usually performed. The test procedure involves the following steps.

A specimen of aggregate (5 - 10kgs) is taken and weighed (W1)

Los Angel's apparatus consists of a hollow drum with a diameter of 70cm and length of 50cm. The cylinder is placed in a horizontal position.

The specimen is loaded up in the cylinder and the abrasion charges are placed. (Abrasion charges are spheres made of cast iron having a diameter of 48mm each and a weight of 390-450gm each). The drum is rotated at 30-33 rpm for a total revolution of 500-1000.

The specimen is taken and sieved in a 1.7mm sieve and passing material weight is obtained (W2).

**Abrasion Value = (W2/W1) * 100**

The abrasion value shall not exceed 30% for the wearing course and 45-50% for the base course.

This test is done to check the durability characteristics of the aggregate. The test procedure involves the following steps.

A sample of aggregate is taken (W1)

It is immersed in a solution of sodium sulfate or magnesium sulfate for 16 to 18 hours

After immersion, the sample is taken out from the solution and it is dried at a temperature of 105 to 110°C

The dried aggregate is immersed once again in the magnesium sulfate or sodium sulfate solution and the process is repeated 10 times

After 10 cycles the weight of the specimen of aggregate is obtained (W2)

**Soundness Value = ((W1-W2) / W1) * 100**

The soundness value shall not exceed 12% in the case of sodium sulfate solution and 18% in the case of magnesium sulfate solution.

Three types of tests are available to determine the shape of aggregates. Below mentioned are the tests available.

Flakiness Test

Elongation Test

Angularity Test

A flakiness test is done to find the thickness of the aggregate i.e., the least lateral dimension. The thickness of the aggregate or least lateral dimension should not be less than 0.6 times the mean size of the aggregate i.e., 0.6(x+y/2), where, x smaller size of the aggregate in the sample, y is the larger size of the aggregate.

The flakiness index is the ratio of the weight of particles/aggregates smaller than 0.6 times the mean size to the weight of the total sample of aggregate. In other words, % by weight of particles smaller than 0.6 times the mean size.

**Flakiness index = (W2/W1) * 100**

where,

W2 - the weight of aggregate passing through a slot having size 0.6 times the mean size of the aggregate

W1 - Total weight of the aggregate specimen

The flakiness index shall not exceed 15%

Elongation represents the greatest dimension of aggregate which shall not exceed 1.8 times the mean size of the aggregate.

The elongation index is defined as the % of the weight of aggregate having size more than 1.8 times the mean size.

**ELongation index = (W2/W1) * 100**

where,

W2 - the weight of aggregate retained on slot having size 1.8 times the mean size of aggregate

W1 - Total weight of the aggregate sample

The elongation index shall not exceed 15%. The combined value of flakiness and elongation index shall not exceed 30%

Angularity represents the opposite of roundness. If the aggregate is angular beyond a certain limit more voids will be present which is not preferable. According to Indian Standard, the maximum percentage of solids in aggregate is 67%. This means the standard voids in the aggregate is 33%. If the void percentage is greater than 33% then it means that the aggregate is angular.

The angularity number is defined as the percentage of void in excess of 33% or the percentage of solid lower than 67%.

**Angularity Number = 67% - ((Vs/V)*100)**

where,

Vs - the volume of solids

V - total volume of aggregates

**Angularity Number = 67% - ((Ws/Gs*V)*100)**

where,

Ws - the weight of solids

Gs - specific gravity of solids

The angularity number shall be in a range of 0 to 11%.

With this, we hope we had covered all the important topics related to the test on pavement materials. The test on bitumen will be covered in our next blog in this series. Or do you want us to cover a topic of your interest in a simple manner? Let us know in the comments below.

As said, flow through pipes is the flow of a fluid through a closed conduit without or with a free surface of the flow. Flow through pipes having a free surface is treated as open channel flow and is dealt with separately. Here we will see about full flowing fluids in pipes i.e., under pressure flows.

Also, there are two types of flows i.e., laminar flow and turbulent flow. If Reynold's number is less than 2000 then the flow is called laminar flow. And if Reynol's number is greater than 4000 then it is called turbulent flow.

The fluid flowing through a pipe experience some resistance to its flow due to which some of the energy of the flow is lost. Based on the magnitude of energy lost it is classified into two types.

It is the energy lost due to frictional resistance offered by the pipe material. It depends on the length of the pipe, the velocity of flow, the diameter of the pipe, and the frictional coefficient. Methods for calculating major energy losses are listed and explained below.

Darcy-Weisbach Formula

Chezy's Formula

Loss of head due to friction is calculated as,

**hf = (4*f*L*V^2) / (2*d*g)**

where

hf - head loss due to friction

f - coefficient of friction which is calculated as,

**f = 16/Re,**

for Re<2000 (Laminar flow)

**f = 0.079 / (Re^(1/4)),**

for Re varying from 4000 to 10^6 (Turbulent flow)

L - length of the pipe

V - mean velocity of flow

d - diameter of the pipe

Chezy's formula for headloss s given as,

**V = C * (mi)^(1/2)**

where,

C - Chezy's constant

m - hydraulic mean depth or hydraulic radius = d/4 for pipes

i - headloss per unit length

Losses that happen because of the below-mentioned reasons are grouped as minor losses.

Sudden expansion of the pipe

Sudden contraction of the pipe

Bend in pipe

Pipe fittings

Obstruction in pipe

Formulas for calculating the losses that happen because of the above-mentioned reasons are explained further.

**he = ((V1 - V2)^2) / (2*g)**

where

V1 - velocity of flow before the expansion of the pipe

V2 - velocity of flow after the expansion of the pipe

**he = (0.5*V2^2) / 2*g**

where

V2 - velocity of low after contraction of the pipe

This loss occurs when a liquid enters the pipe from a reservoir or a tank.

**hi = (0.5*V^2) / 2*g**

where

V - velocity of flow

This loss happens at the exit where the velocity of flow is dissipated as a free jet or is lost in the reservoir.

**ho = (V^2) / 2*g**

where

V - velocity at the outlet of the pipe

Obstruction in a pipe causes a reduction in the cross-sectional area of the pipe, which then leads to losses.

**Hob = ((V^2) / 2*g) * ( (A / (Cc (A -a)) - 1)**

where

V - velocity of floe

A - area of the pipe

a - maximum area of obstruction

**hb = (k*V^2) / 2*g**

where

k - coefficient of bend/pipe fitting

V - velocity of flow

When pipes of different lengths and different diameters are connected from end to end it is called a series connection. Here, the rate of flow remains the same in a series connection. But the total head loss is obtained as the sum of head losses. In general, minor losses are neglected to obtain the equivalent diameter of pipes connected in series. Applying this, we get the formula for equivalent diameter as,

When single pipe branches out into two or more pipes, the branched pipes are called parallel pipes. Here, the head loss remains the same in all the pipes. But the total discharge is given as the sum of discharges through each pipe. In general, minor losses are neglected to obtain the equivalent diameter of pipes connected in parallel. Applying this, we get the formula for equivalent diameter as,

With this, we hope we had covered all the important topics related to the flow through pipes. Want us to cover a topic of your interest in a simple manner? Let us know in the comments below.

GATE exam is conducted by IITs. Each year different IITs take up the role of conducting the exam. Conducting the exam comprises everything from the initial announcement to the release of the GATE scorecard. GATE 2023 is being conducted by IIT Kanpur. For more details check out GATE 2023 IIT Kanpur: Exam Dates, Pattern, Eligibility, Exam Papers & Paper Combinations blog.

IIT Kanpur has announced important dates for the GATE 2023 exam. The exam dates are announced as 4th, 5th, 11th, & 12th February. For more details check out GATE 2023 Official Notification | Important Dates for GATE 2023 blog.

Self-study requires self-motivation and an appropriate strategy to prepare for the exam. Self-motivation is an internal factor and it solely depends on the individual. Preparation strategy on the other hand can be generalized for all. There are three key factors to be considered for self-study which are listed as follows.

Resources

Time

Plan

We had explained a lot about this in the GATE 2023 Self Study | A Complete Strategy for Self Preparation blog, do check that out. We also published blogs on preparation strategies all of which are listed below.

What we didn't cover in detail in all these blogs are the places where one can find practice questions for the GATE exam. Therefore, we will cover this in detail now.

Previous year question papers are important for preparation. It is required not only for GATE exam preparation but for preparation for any other competitive exam as well. Also, previous years' question papers are the best sources to find problems for practice. For instance, practicing questions from the previous 10-year question paper means practicing 650 questions, which is more than enough to get a good GATE score.

Other than practicing questions one can also understand the exam pattern, difficulty of the exam, distribution of questions across different topics, and so on, from solving the previous years' question papers. The importance of previous papers cannot be overstated than this. Therefore, one should try and solve as many previous year questions as possible.

But where to find these previous years' question papers? Well, the answer is simple. All the previous years' question papers are available on IIT Kanpur's GATE 2023 website. One can just enter the question paper year and can find the question paper of the required subject that was asked in that particular year. After locating the question paper one can just download the paper and can start solving it. The link for previous years' question papers available on the IIT Kanpur's GATE website is provided below.

Next comes the question banks for the GATE exams. Previous years' question papers can be used only for practicing a topic. No questions will be asked from previous question papers. This is the reason why one should also consider question banks as a source while practicing GATE questions.

Question banks contain a huge number of questions that are regularly updated. These banks contain new questions as well and as such these questions have a higher chance of appearing in the current GATE exam, though the exact values might differ. Because of this, it is important to practice questions from the question bank as well.

But where to find these question banks? There are a number of websites to download these question banks. Some of them are available for free and some are paid. We at APSEd offer question banks for three papers i.e., Civil, Environmental, and Geomatics engineering. Question banks for other papers could be found on the **made easy** website. Links for question banks for civil, environmental, and geotechnical, that are available with APSEd are provided below.

Finally, the questions are also available in books for each topic. This is not as important as the above-mentioned sources and one can practice questions from books only after practicing questions from previous years' question papers and question banks. Still, if you intend to practice questions from books, refer to some standard books that are available for a subject and practice questions from them. We had already shared a blog on Best Books for GATE Civil Engineering, do check that.

This is a special type of source as it is available only just before the exam. It contains questions in the same GATE exam pattern and is useful for quick revisions. Sometimes it is also as offered as subject-wise tests by various platforms.

We at APSEd offer full-length tests before the GATE exam. It is available in a GATE-like environment. After completion of the exam, it also provides insights and analysis of your performance which could be used for focussing your preparation. Do check out the question paper analysis blogs that are shared below to understand the resemblance between APSEd full-length tests and the actual GATE 2022 paper.

With these, we come to the end of this blog. For any queries do reach us out at support@apsed.in.

Each vehicle occupies certain width while traveling in a straight path. But in a curved path, the width requirement for a vehicle increases. This is because of the fact that both the front and back wheels follow different paths while traversing a curved section. Because of this, the carrying capacity of the road would decrease. Therefore, extra widening becomes indispensable in curved sections to maintain the carrying capacity throughout the curved section.

Apart from this, the extra widening is also required to meet the following.

The first reason, as said, is due to the rigid wheelbase of the vehicle. This leads to from and back wheels following different paths at the curved section. This is called off-tracking.

Next, due to a speed higher than the design speed, the existing superelevation will not be sufficient to counteract the centrifugal force. This could lead to lateral skidding.

To have greater visibility on the cured sections, the driver could take the vehicle to the outer edge of the pavement.

To maintain a sufficient gap with vehicles in the opposite direction.

IRC recommends providing extra widening if the radius of the horizontal curve is less than or equal to **300m**. IRC also recommends the minimum value of extra widening based on the radius of horizontal curves as follows.

The width of the extra widening is the sum of two components i.e.,

Psychological Widening

Mechanical Widening

As per IRC, psychological widening is totally dependent on the design speed and the radius of the horizontal curve. This is totally due to the psychological factor and doesn't depend on the wheelbase of the vehicle.

**Wpsy = v / (9.5 * R^(1/2))**

where

v - design speed in km/hr

R - radius of the horizontal curve in m

Note that, psychological widening need not be considered for single-lane roads.

This is dependent on the length of the wheelbase of the vehicle. This is given by the expression,

**Wmech = (n * l^2) / (2*R)**

where

n - is the number of lanes

l - length of the vehicle

R - radius of the horizontal curve

The total widening required is the sum of psychological widening and mechanical widening.

**Wextra = Wpsy + Wmech**

**Wextra = {v / (9.5 * R^(1/2))} + {(n * l^2) / (2*R)}**

For greater than two lanes, the extra width of widening is equal to the product of 1/2 of the total width of widening for two lanes and the number of lanes.

This blog forms a small part of the topic of **"Geometric Design of Highways"**. The geometric design of highways is an important topic from the GATE exam perspective and each year at least one question pops from the topic. Almost all topics within the geometric design of highways are already covered in our previous blogs. Links for the same are provided below.

Before getting into rigid pavement design it is important to understand certain characteristics of rigid pavement. As said, rigid pavements are made of concrete, i.e., plain cement concrete, reinforced cement concrete, or prestressed concrete.

The concrete slab can be used as both wearing surface and base course. But in general, a rigid pavement consists of a concrete slab below which a granular base or sub-base is provided. The provision of a base course improves the life of the rigid pavement.

Unlike flexible pavement where the load is transferred by grain to grain action, the load in the rigid pavement is transferred by slab action.

The main difference between rigid pavement and flexible pavement is that the critical condition of stress in the rigid pavement is the maximum flexural stress occurring in the slab due to wheel load and temperature changes whereas in flexible pavement it is the distribution of compressive stresses.

Westergard is considered the pioneer in the analysis of rigid pavements. IRC has also adopted the Westergard method for the design of rigid pavements with slight modifications and recommendations.

Before getting into details of Westergard's design it is important to understand certain terms related to concrete in pavement design.

The concrete grade to be used for rigid pavement should be

**M40**According to IS456:2000, the flexural strength of concrete is given by the relation

where fck is the characteristic compressive strength of concrete.**fcr = 0.7*(fck^0.5),**It is also called the modulus of rupture or bending tension.According to IRC, the modulus of elasticity of concrete of rigid pavement shall be,

(i.e., 5000*(fck^0.5)) for M40 concrete.**Ec = 3 * 10^4 MPa**The poison ratio of concrete shall be

**0.15**

Now let us discuss some important concepts that Westergard had theorized for rigid pavement design.

The design of rigid pavement is dependent on the modulus of subgrade reaction. It is defined as the ratio of pressure applied on a circular plate to the settlement of the plate. It is represented by the letter k. Its unit is kgf/cm^3.

**k = p/Δ**

where,

k - modulus of subgrade reaction

p - pressure applied

Δ - settlement of the plate

According to IRC, the settlement of a 75 cm circular plate shall be 1.25 mm. Therefore,

**k = p/0.125**

The maximum value of subgrade reaction for rigid pavement is 5.5 kgf/cm^3.

The subgrade offers a certain amount of resistance to the slab deflection. This is dependent on the properties of both the subgrade material and slab.

The resultant deflection of the slab which is also the deformation of the subgrade is a direct measure of the magnitude of subgrade pressure. The pressure deformation characteristic of rigid pavement is thus a function of the relative stiffness of the slab to that of the subgrade.

Westergard coined this term "Radius of relative stiffness (l)"

**l = ((Eh^3/12k(1-(µ^2)))^(1/4)**

where,

l - radius of relative stiffness (cm)

E - modulus of elasticity of cement concrete (kg/cm^2)

µ - Poisson's ratio of concrete

h - slab thickness (cm)

k - modulus of subgrade reaction (kg/cm^3)

As the name suggests, this gives the equivalent area of the slab that is effective in resisting the bending moment applied to it. Westergard approximates this by the radius of wheel load distribution and slab thickness.

**b = (1.6a^2 + h^2)^(1/2) - 0.675h**

where,

b - equivalent radius of resisting section (cm)

a - radius of wheel load distribution (cm)

h - slab thickness (cm)

For a better understanding both the formulas are given in a pictorial format below.

There are three locations namely the interior, edge, and corner where the differing conditions of slab continuity exist.

**Interior Loading:** Load is applied in the interior of the slab surface at any place remote from the edges

**Edge Loading: **Load is applied at the edges of the slab at any place remote from a corner

**Corner Loading:** Load is applied at the corner that is formed by the intersection of two edges.

In most cases, critical stresses caused by these loads will be given directly. We should just find the most critical combinations out of the given loads and use that load as the design load for calculation. But still, it is important to understand how these stresses are calculated from the given load. Westergard provided a simple method to find the stresses for all conditions as given below.

According to Mr. Westergard, the flexural stress developed in concrete due to wheel load is inversely proportional to the square of the thickness of the rigid pavement.

**fload = (P/h^2) * Q**

where,

P - wheel load

h - thickness of the pavement

Q - stress coefficient, **Q = l/b**

l - radius of the relative stiffness of concrete

b - equivalent radius of contact of the wheel

According to IRC the wheel load on a rigid pavement is 5100 kgf i.e., 51 kN. And the thickness of the rigid pavement should not be less than 15cm.

Note that this formula cannot be directly used to compute stresses for interior, edge, and corner loading and requires some more components to calculate the stresses. But still, if the stress coefficient for a loading type, load, and thickness are given one can use this formula to calculate the critical stress.

Apart from wheel load stresses a rigid pavement is also subjected to stresses due to temperature change. The variation in temperature across the depth of the slab is caused by daily variation in temperature whereas the seasonal variation in temperature causes an overall increase or decrease in slab temperature. Accordingly, there are two types of stresses namely,

Warping stresses

Frictional stresses

**Warping Stress: **Whenever the top and bottom surfaces of a concrete pavement simultaneously possess different temperatures, the slab tends to warp upward or downward. This is caused due to daily variation in the temperature.

According to IRC, the warping stress shall be determined by Mr. Bradbury's formula which is based on Hooke's law.

**f = α * Δt * E**

where

α - coefficient of thermal expansion (10 * 10^-6 /°C)

Δt - change in temperature

E - young's modulus of concrete

**Frictional Stress: **This is caused due to uniform rise and fall of temperature in the slab which results in the overall expansion and contraction of the slab. As the slab is in contact with the subgrade, the subgrade tries to restrain this expansion/contraction by friction thereby inducing the frictional stresses.

**Sf = WLf / (2*10^4)**

where

Sf - unit stress developed (kg/cm^2)

W - the unit weight of concrete, kg/cm^3

f - coefficient of subgrade restraint

L - slab length (m)

B - slab width (m)

Combinations of the above-mentioned stresses need to be considered while calculating the critical combination of stresses. The most critical combination occurs during daytime only as the wheel load stress and warping stress cause bending tension in the bottom fiber. Whereas, during night thought the bending tension due to wheel load remains the same, warping stress causes compression in the bottom-most fiber.

The following conditions are considered to provide the critical combinations.

**During Summer: **Critical combination of stresses = wheel load stress + warping stress - frictional stress, at the edge region

**During Winter: **Critical combination of stresses = wheel load stress + warping stress + frictional stress, at the edge region

**At Corner: **Critical combination of stresses = wheel load stress + warping stress

The greater value of resultant stress determined at the edge and corner shall not exceed the modulus of rupture of concrete.

**f(wheel) + f(warping) + f(friction) <= fcr = 0.7*(fck^(1/2))**

Apart from designing the thickness of the slab expansion joints and contraction joints also need to be designed.

The expansion joint is an assembly designed to hold the structure together while safely absorbing temperature-induced expansion. According to IRC, the expansion gap in concrete should be 25mm, and the net expansion gap is 25/2 = 12.5mm.

Spacing between expansion joints (L)

Δl = L * α * Δt

25/2 = L * 10 * 10^-6 * 28

**L = 44.64 m**

Dowel bars are large diameter bars placed along the traffic direction i.e., longitudinal direction. The design strength of dowel bars is taken as 40% of wheel load. The design of dowel bars is dependent on the following parameters,

Shearing strength of dowel bars

Bending strength of dowel bars

Bearing strength of the concrete near dowel bars

Contraction joints are provided to form a weak plane in the pavement surface such that cracks will occur along the plane rather than any other undesired locations. The length of the contraction joint is given by two different expressions based on the presence of dowel bars.

**Case 1: Dowel bars are not present**

**Lc = (2*fct) / (γcc*f)**

where,

fct - allowable tension in concrete = 0.8 kgf/cm^2

γcc - unit weight of the concrete = 2400 kgf/m^3

f - coefficient of friction = 1.5

Substituting all the values in the above equation, we get, the length of the contraction joint as 4.5m.

**Case 2: Dowel bars are present at contraction joint**

**Lc = (2*****σst*Ast) / (f*****γcc*B*h)**

where,

σst - allowable tensile stress in steel = 140 N/mm^2 for Fe250

Ast - area of tensile reinforcement (mm^2)

f - coefficient of friction = 1.5

γcc - unit weight of the concrete = 2400 kgf/m^3

B - width of the pavement (m)

h - depth of the pavement (m)

Tie bars are smaller diameter bars placed in the perpendicular direction to the traffic flow. It is used to connect two panels. The design is based on friction and not the wheel load. The length of the tie bars is found as the sum of the development lengths of bars in both panels.

With this, we hope we had covered all the important topics related to the design of rigid pavements. Want us to cover a topic of your interest in a simple manner? Let us know in the comments below.

A summit curve is a type of vertical curve with an upward gradient i.e., convexity in an upward direction.

It is used to join two gradients that fall under the following types.

Ascending and descending

Ascending and ascending

Ascending and horizontal

If you notice you will find that the summit curve is not used when a descending gradient meets an ascending gradient. The curve used in such a situation is called a valley curve. We had already covered valley curves in the length of the valley curve blog.

The summit curve is provided in the shape of a simple parabola. It is chosen because it has a constant slope.

Summit curves are designed based on a single criterion. This criterion is nothing but the sight distance required at each section of the highway.

If you recall from the valley curve design criteria there were two, i.e., sight distance and comfort criteria. But why is the comfort criterion not used here?

Let us assume a fast-moving vehicle traversing from ascending to horizontal gradient over a summit curve. As the convexity is upward, the centrifugal force acts in the upward direction i.e., against gravity. Due to this upward centrifugal force, no pressure is exerted on the suspension system of the vehicle and therefore causes no discomfort for the passengers within. This is the reason why comfort is not considered a criterion in summit curve design.

The length of a summit curve depends on the following parameters.

Deviation angle,

**N = |(****±n1) - (±n2)|**Stopping sight distance (SSD)

Height of an obstacle from the road surface (h) ( i.e., 15 cm as per IRC)

Height of the eye level of the driver from the road surface (H) (i.e., 1.2m as per IRC)

There are four cases using which the length of the summit curve is found as discussed below.

L = (N(S^2)) / (2((h^0.5)+(H^0.5))^2)

Substituting h = 0.15m and H = 1.2m we get,

**L = (NS^2) / 4.4**

where,

N = |(±n1) - (±n2)|

S - Stopping sight distance

L = 2S - (2((h^0.5)+(H^0.5))^2)/N

Substituting h = 0.15m and H = 1.2m we get,

**L = 2S - (4.4 / N)**

where,

N = |(±n1) - (±n2)|

S - Stopping sight distance

L = (N(S^2)) / (2((H^0.5)+(H^0.5))^2)

Substituting H = 1.2m we get,

**L = (NS^2) / 9.6**

where,

N = |(±n1) - (±n2)|

S - Stopping sight distance

L = 2S - (2((H^0.5)+(H^0.5))^2)/N

Substituting h = 0.15m and H = 1.2m we get,

**L = 2S - (9.6/ N)**

where,

N = |(±n1) - (±n2)|

S - Stopping sight distance

For a better representation of the formulas please refer to the image below.

With this, we hope we had covered all the important topics related to the summit curve. Want us to cover a topic of your interest in a simple manner? Let us know in the comments below.

Before we get into flexible pavement design, it is important to get familiarised with various aspects of flexible pavement. As said, the flexible pavement has little to no flexural strength, and they showcase flexible structural action under the action of loads. If the lower layer of the flexible pavement is undulated, it is reflected back to the surface layer as deformations.

Typically, flexible pavement consists of four layers as listed below.

Soil subgrade

Sub-base course

Base course

Surface course

The flexible pavement transfers the vertical compressive vehicular road by grain to grain transfer through the points of contact to its lower layers.

The compressive stress is maximum on the surface and is equal to the contact pressure under the wheel. As we go down to the next layers, the stress is spread out in the shape of a truncated cone and hence the lower layers have to withstand only lesser stresses.

Because of this ability to spread out and lower the stresses, only the surface course should be made of high-quality material to withstand high compressive stress and wear & tear by vehicular traffic. The lower layers can be made using lower quality materials as they need to withstand lesser stresses only and there is no vehicular motion.

Bituminous concrete is one of the best flexible pavement materials. Other materials like crushed aggregates, gravel, soil-aggregate mixture, etc. are used in other layers of flexible pavement.

Below given is the list of methods available for designing a flexible pavement.

Empirical method

California bearing ration method (CBR)

Mcleod method

Group index method

Semi-empirical method

Triaxial test method

Theoretical method

Mr. Burmester method

Out of these, empirical methods are the most commonly used methods for designing a flexible pavement.

Group Index Method

A group index is an arbitrary value assigned to soil types based on percent fines, liquid limit, and plasticity index. The GI value is given as,

GI = 0.2a + 0.005ac + 0.01bd

where,

a - that portion of material passing 0.074mm sieve, greater than 35 and not exceeding 75 percent (expressed as whole number 0 to 40)

b - that portion of material passing 0.074mm sieve greater than 15 and not exceeding 35 percent (expressed as whole number 0 to 40)

c - that value of liquid limit in excess of 40 and less than 60 (expressed as whole number 0 to 20)

d - that value of plasticity index exceeding 10 and not more than 30 (expressed as a whole number from 0 to 20)

GI values of soil vary from 0 to 20. The higher the GI value, the weaker is the soil, and for a given traffic load, the greater the pavement thickness required.

Design steps

GI value of the soil is found

Vehicular traffic is g=found and grouped as shown below

The total thickness of pavement (surface, base, and sub-base) is found using the group index chart corresponding to the GI value

Before getting into the details of CBR design it is important to know about the CBR test. The procedure followed in a CBR test is shown below.

The diameter of the CBR mould is 150 mm with a height of 130mm

The specimen of soil subgrade is placed in the mould and a plunger of 50mm diameter is kept on the top surface. It is allowed to penetrate into the specimen

The rate of penetration is 1.25mm per minute

Load is applied for the plunger to penetrate and the load at which penetration is 2.5mm is noted

Further load is applied to reach a penetration of 5mm and the load causing this penetration is noted

The specimen is removed and a standard broken aggregate is filled in the mould

Procedure 4 to 5 is repeated for this standard aggregate and corresponding values are noted.

The standard pressure required for 2.5mm & 5mm penetration for standard aggregate is 70 kg/cm^2 and 105 kg/cm^2 respectively.

The ratio between load or pressure sustained by the specimen at 2.5 mm penetration and load or pressure sustained by the standard aggregate at 2.5 mm penetration is found.

Similarly, the ratio between load or pressure sustained by the specimen at 5 mm penetration and load or pressure sustained by the standard aggregate at 5 mm penetration is found.

The greater of these two ratios is taken as the CBR value of the soil subgrade specimen. Generally, the ratio based on 2.5 mm penetration is higher and is taken as the CBR value.

This method is based on design charts that are prepared by the California state highway department after performing extensive CBR tests on existing highways. The basis of the design chart is that a material with a given CBR required a certain thickness of pavement layer as cover for a given vehicular load.

First, the soaked CBR value of the soil subgrade is found

An appropriate design curve is chosen by considering the design wheel load

From the chart, for the chosen design curve, the total thickness of pavement required to cover the soil subgrade can be found

To know the thickness of the sub-base course, CBR of the sub-base course is found and by using the same design curve, the total thickness of the pavement required above the sub-base is found. The thickness of the sub-base could be found as the difference between the total pavement thickness and thickness of pavement required above the sub-base.

Similarly, the thickness of all the layers of flexible pavement could be found

There are two methods recommended by the Indian Road Congress for the design of flexible pavements using the CBR method as listed below.

Based on the number of vehicles

Based on the cumulative standard axle

But the recommendations remain the same for both of these methods as mentioned below. IRC has recommended a CBR design chart for use in India. IRC has also recommended some points on using the CBR method of design which are listed as follows.

The CBR test should be performed on remoulded soils in the lab only

The subgrade soil should be compacted according to Proctor's method at optimum moisture content

For the construction of flexible pavement in heavy rainfall zone, the soil specimen must be soaked in water before testing

The top 50 cm of the subgrade should be compacted to at least 95% of proctor density

At least three samples of the same material and density should be tested. The maximum permissible variation for various CBR values is given below. If the variation is more for the same sample, then six samples must be tested.

The thickness of the wearing course should not be counted in total pavement thickness as they don't serve any structural functionality.

An estimate of the traffic must be found at the end of the expected life using existing traffic conditions and probable growth. It is found as,

**A = P * [(1+r)^(n+10)]**

where,

A - design number of heavy vehicles per day (laden weight > 3 tons)

P - present number of vehicles

r - the rate of growth of vehicles (generally 7.5%)

n - number of years between the last count of traffic and estimated construction completion

10 - represents the design life of the highway

**Note:**

The design is based on the number of heavy vehicular movements per day. Based on this, the design chart has seven design curves from A to G. Number of vehicles associated with each curve is given in the table below.

This is the revised method suggested by IRC and is most commonly used nowadays for flexible pavement design. Instead of the number of vehicles per day, cumulative standard axle load is found. The revised design charts now contain seven curves categorized based on CBR value with the cumulative standard axle on the x-axis and total pavement thickness on the y-axis.

Also called the million standard axles (MSA), it is the total number of standard axles ruing on the flexible pavement during the life of the flexible pavement. It is given as,

**MSA = (A * 365 * [(((1+r)^n) - 1) / r] * VDF * LDF) / (10^6)**

where,

A - present number of commercial vehicles per day

r - the rate of growth of commercial vehicles (7.5%)

n - the life of flexible pavement (10 to 20 years)

VDF - vehicle damage factor

LDF - lane distribution factor

For single lane, LDF = 1

For double lane, LDF = 0.75

For four lane, LDF = 0.4

**VDF = ((V1 * (Wa1/Ws)) + (V2 * (Wa2/Ws)) + (V3 * (Wa3/Ws)) + ...) / (V1 + V2 + V3 + ...)**

where,

Vi - the percentage of 'I type' of vehicle

Wai - the weight of that 'I type' vehicle

Ws - 8.2 tons (standard)

We have already covered various types of fluid flow such as steady flow, unsteady flow, uniform flow, and nonuniform flow in our types of open channel flow blog. Here, we will discuss combinations of these types of flow along with examples.

At a given section, fluid characteristics like velocity and density don't change with time. Flow through a constant diameter pipe is a good example.

**∂V/∂t = 0, ∂Q/∂t = 0**

At a given time, fluid characteristics like velocity and density don't change with space.

**∂y/∂S = 0, ∂V/∂S = 0**

**Example:** Flow through a constant diameter pipe with constant discharge

**Reason:** As the discharge is constant, the velocity remains the same at a given section over time i.e., steady flow. Also, as the area of flow is constant, at a given time, the velocity between any two sections will remain constant i.e., uniform flow.

**Example:** Flow through a tapering diameter pipe at a constant rate.

**Reason:** As the discharge is constant, the velocity remains the same at a given section over time i.e., steady flow. But, as the pipe is tapered, the cross-sectional area decreases thereby increasing the velocity over distance. Therefore, at a given time, the velocity between two sections changes ie., non-uniform flow.

**Example:** Fluid flow through a constant diameter pipe with varying discharge.

**Reason:** As the discharge changes, the velocity at a given section changes over time i.e., unsteady flow. But, as the area of flow is constant, at a given time, the velocity between any two sections will remain constant i.e., uniform flow.

**Example:** Fluid flow through tampering diameter pipe with varying discharge.

**Reason:** As the discharge changes, the velocity at a given section changes over time i.e., unsteady flow. Also, as the pipe is tapered, the cross-sectional area decreases thereby increasing the velocity over distance. Therefore, at a given time, the velocity between two sections changes ie., non-uniform flow.

The flow of fluids can be expressed mathematically using the continuity equation. It is based on the law of conservation of mass. The general form of the continuity equation for a three-dimensional flow having steady, unsteady, uniform, non-uniform, compressible, and incompressible flow, is given as,

**∂ρ/∂t + ∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z = 0**

where,

ρ - density of fluid i.e., constant

u, v, w - velocity component in x, y, and z direction

For steady flow, **∂ρ/∂t = 0**

For incompressible flow, **ρ = constant**

For uniform flow, **∂u/∂x = 0**

Therefore, the continuity equation for a steady, uniform, and incompressible flow is given as,

**∂u/∂x + ∂v/∂y + ∂w/∂z = 0**

In rotational flow, the fluid particles rotate about their mass center due to tangential stress caused by the viscosity of the fluid.

In such cases, the angular velocity of rotation is given by,

**wz = (1/2) * ((∂v/∂x) - (∂u/∂y))**

In the case of irrotational flow, the fluid particles don't rotate, and hence wz = 0.

**∂v/∂x = ∂u/∂y**

Acceleration in general is given as,

a = dV/dt = (∂V/∂s) * (ds/dt) = V * (∂V/∂s)

In the case of fluid flow, each velocity component is a function of x, y, z, and time (t). Therefore, the acceleration component for each velocity component should be found.

**ax = (u * (∂u/∂x)) + (v * (∂u/∂y)) + (w * (∂u/∂z)) + (∂u/∂t)**

**ay = (u * (∂v/∂x)) + (v * (∂v/∂y)) + (w * (∂v/∂z)) + (∂v/∂t)**

**az = (u * (∂w/∂x)) + (v * (∂w/∂y)) + (w * (∂w/∂z)) + (∂w/∂t)**

a = axi + ayj + azk, in vector format.

The terms (∂u/∂t), (∂v/∂t), and (∂w/∂t), are called **temporal acceleration** and they become zero in the case of steady flows.

The terms (u * (∂u/∂x)) + (v * (∂u/∂y)) + (w * (∂u/∂z)), (u * (∂v/∂x)) + (v * (∂v/∂y)) + (w * (∂v/∂z)), and (u * (∂w/∂x)) + (v * (∂w/∂y)) + (w * (∂w/∂z)), are called **convective acceleration** and they become zero in the case of uniform flow.

Therefore, in the case of a steady uniform flow, the acceleration is always zero.

The velocity potential function is a scalar function such that its negative derivative along any direction will give the velocity component in that direction.

**u = -∂φ/∂x**

**v = -∂φ/∂y**

The continuity equation in terms of the velocity potential function is called the Laplace function. For a steady, uniform, and incompressible flow, the Laplace equation should be satisfied.

Irrotational flow in terms of velocity potential function

The stream function is a scalar function such that its derivative along any direction gives the velocity component in the perpendicular direction, in the clockwise or anti-clockwise direction.

**u = ∂ψ/∂y, v = -∂ψ/∂x**

Irrotational flow in terms of stream function

**Note: **

Discharge between two points can be found as the difference in stream function between the two points.

Vorticity for a two-dimensional fluid flow is given as,

**∂v/∂x - ∂u/∂y**Equation of streamline is given as,

**dx/u = dy/v****∂ψ/∂y = ∂φ/∂x**

Streamline - curve line obtained during the flow of fluid particles such that a tangential drawn represents the resultant velocity of flow

Streamtube - formed by a number of streamlines

Pathline - line traced by fluid particles during a period of tie

Streak line - Locus of all fluid particles at any time instant which passes through a fixed point

We hope we have covered all the important details related to fluid kinematics. Test your knowledge by solving a GATE:2005 problem given below.

As said, hydrostatic force is the resultant force acting on a submerged surface due to the pressure loading of the fluid into which the surface is submerged.

If the liquid is at rest, then there is no tangential force acting on the submerged surface and hence the total pressure will act normal to the surface in contact. Therefore, hydrostatic force is the resultant force acting on a submerged surface when the liquid is at rest.

The location of the hydrostatic force is the center of pressure. The center of pressure is always below the center of gravity of the surface in contact.

In this case, the submerged surface is parallel to the surface of the water. The hydrostatic force can be simply calculated as the product of pressure exerted by the fluid and area of the surface.

Hydrostatic force on a horizontal plane,

**Fh = (γ*h) * A**

where,

γ - the unit weight of the fluid

h - depth at which the submerged plane is located

A - surface area of the submerged plane

In this case, the submerged surface is vertical i.e., perpendicular to the surface of the water. Integration is required to derive the hydrostatic force on a vertical plane formula.

Hydrostatic force on a vertical plane,

**Fv = γ*A*h'**

where,

γ - the unit weight of the fluid

A - total surface area

h' - the center of gravity of the surface

The center of pressure is given as,

**hcp = h' + (Ig/Ah')**

where,

Ig - the moment of inertia of the surface about its center of gravity

A - total surface area

h' - the center of gravity of the surface

The center of pressure (hcp) is always greater than the center of gravity (h') i.e., hcp > h'. Therefore, the center of pressure always lies below the center of gravity.

In this case, the submerged surface has a curved profile, unlike the plain surfaces that we saw earlier. An example of such a surface is a sluice gate. In these cases, two forces are considered i.e., horizontal and vertical, and then the resultant force is found.

**Horizontal force** is determined by considering the vertical surface i.e., the vertical projection of the curved profile.

The horizontal force is given as,

**Fh = γ * A * h'**

where,

γ - the unit weight of the fluid

A - area of the vertical projection of the profile

h' - the center of gravity of the surface

**Vertical force** is the weight of liquid acting on the curved surface in contact.

The vertical force is given as,

**Fv = γ * A contact * l = γ * Volume**

where,

γ - the unit weight of the liquid

A contact - contact area of the profile

l - length of the profile

**Resultant force, R = ((Fh^2) + (Fv^2))^(1/2)**

Let us understand this through an example problem.

**Question: **A spillway is shown below. Find the magnitude of the resultant force that is acting on the curved profile of the spillway.

**Solution:**

Horizontal force, **Fh = γ * A * h'**

Fh = 9.81 * (2*6) * (2/2)

Fh = 117.72 kN

Vertical force, **Fv = γ * A contact * l **

Fv = 9.81 * (((兀/4)*(d^2))/2) * 6

Fv = 92.457 kN

**Resultant force, R = ((Fh^2) + (Fv^2))^(1/2)**

R = 149.687 kN

We hope we had covered all the important aspects of hydrostatic forces on submerged bodies. This is an important topic preparing which could be really helpful in the GATE exam. We cover more such important topics on three topics per week basis at APSEd blog page. Do check them out!

GATE exam is conducted by IITs. Each year different IITs take up the role of conducting the exam. Conducting the exam comprises everything from the initial announcement to the release of the GATE scorecard. GATE 2023 is being conducted by IIT Kanpur.

Below shared are the important dates as announced by the IIT Kanpur.

The most important dates out of the above-given dates are:

30th August 2022 - Opening date of online application

30th September 2022 - Closing date of online application

3rd January 2023 - Admit cards

4th, 5th, 11th & 12th February - Exam dates

Below given are some of the other useful information on GATE 2023. We had already posted a blog on GATE 2023, do check that out for more detailed insights.

GATE exam will be conducted as an ONLINE Computer Based Test (CBT) in specific exam centers only. The online examination papers will contain some questions,

(i) Multiple Choice Question (MCQ) type, where only one option out of four options is correct.

(ii) Multiple Select Question (MSQ) type, where one or more than one option out of four options is/are correct,

(iii) Numerical Answers Type (NAT) where the answer must be keyed in by the candidate using a virtual keypad.

One more thing to note is, that no calculators will be allowed, instead, the candidates can use an on-screen virtual calculator provided for the examination.

Every year, the GATE exam may or may not modify the exam papers. But, there are some standard sets of papers that will for sure appear in GATE 2023. The same is listed below.

Aerospace Engineering

Agricultural Engineering

Architecture and Planning

Biomedical Engineering

Biotechnology

Civil Engineering

Chemical Engineering

Computer Science and Information Technology

Chemistry

Electronics and Communication Engineering

Electrical Engineering

Environmental Science and Engineering

Ecology and Evolution

Geomatics Engineering

Geology and Geophysics

Instrumentation Engineering

Mathematics

Mechanical Engineering

Mining Engineering

Metallurgical Engineering

Naval Architecture and Marine Engineering

Petroleum Engineering

Physics

Production and Industrial Engineering

Statistics

Textile Engineering and Fibre Science

Engineering Sciences

Humanities & Social Sciences

Life Sciences

Let us now clarify some of the frequently asked questions (tentative as per previous year notifications)

**1. Is there any age limit to appear for the GATE exam?**

NO. There is NO age limit to appear for GATE 2022.

**2. Are there any restrictions on the number of times one can appear for GATE?**

NO. One can appear for the GATE examination any number of times.

**3. Can I appear in any of the GATE 2023 papers?**

Although the candidate is free to choose any of the papers (up to two from the given combinations), the candidate should select a paper appropriate to the discipline of the qualifying degree.

**4. If I have not uploaded the photograph as per the specifications mentioned, will my application be rejected?**

YES, without a proper photograph, the application will be rejected.

**5. If I have not uploaded a signature as per the specifications mentioned, will my application be rejected?**

YES, without a proper signature, the application will be rejected.

**6. Do I need to fill out two forms to appear in two papers?**

NO. The two papers option may be selected based on the given set of combinations in a SINGLE form only.

**7. When will I receive my Admit Card?**

Admit Card can ONLY be downloaded from the GATE 2022 website from the first week of January 2023. Admit Cards will NOT be sent by post or as an e-mail attachment.

**8. Will I be provided with any white paper for rough work and calculations during the examination?**

Only one scribble pad at any point of time will be provided to the candidate that can be used to do the rough work. To obtain a second scribble pad, the candidate MUST return the first scribble pad. The candidates have to return any scribble pad in their possession at the end of the examination. The virtual scientific calculator will be available on the computer screen.

**9. Will there be any arrangement at the examination centre for the safe keeping of my personal items such as my mobile phone?**

NO. Such arrangements will not be made at the examination centre. If a candidate brings personal belongings including mobile phones, they have to be kept outside the examination hall at the candidate’s own risk. The GATE officials are not responsible for the safe keeping of your personal items.

**10. Is there a relaxation in qualifying marks and fees for the EWS (Economically weaker section) candidates?**

The application fee for EWS candidates will be the same as that of General candidates, however, the qualifying mark is 90% of that of the General candidates.

**11. What will happen if I am not able to produce an EWS certificate at the time of admission/counselling etc.?**

If you are not having a valid EWS certificate at the time of admission/counselling/recruitment, you will be considered a general category candidate. However, GATE has nothing to do with subsequent admission or the recruitment process.

Do go through the attached video for more insights and ideas.

Hope we had covered all the necessary aspects of important dates as announced by the IIT Kanpur. Until the detailed notification arrives, one can check out the syllabus of civil, environmental, and geomatics papers posted in our APSEd blogs.

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Oil and Natural Gas Corporation is a MahaRatna public sector company. It is the largest domestic producer and supplier of crude oil and natural gas products. It is at the forefront of meeting the Nation's crude oil demand and is therefore one of the best PSUs to work in.

ONGC is also consistent in recruiting students from various disciplines through GATE scores. This year too there is no exception as the ONGC has released the notification for the recruitment. The official notification is attached below.

Below tabulated are the roles offered based on the GATE paper.

**Total Vacancy**: will be updated

Start date of online registration of application:

Last date of online registration of application:

Let us see the educational qualification and GATE Paper qualifications for the vacancies.

Essential Qualification: Graduate Degree in Civil Engineering with a minimum of 60% marks

Eligible GATE Paper: GATE CE

Essential Qualification: Engineering Graduate in Environment Engineering/ Environment Science with minimum 60% marks OR Graduate in an Engineering discipline with minimum 60% marks with M.Tech./ME in Environment Engineering/ EnvironmentScience

Eligible GATE Paper: GATE ES

Essential Qualification: Graduate Degree in any Engineering with a minimum of 60% marks

Eligible GATE Papers: GATE ES/CE/ME/EE/IN/PE/CH

Note: The abovementioned qualification is based on the previous year's notification. It may or may not vary for this year. Till the detailed notification arrives one could follow this for guidance.

It will be updated soon

It will be updated soon

The candidate should clear all the eligibility criteria discussed above including the age criteria.

The GATE 2021 Score carries a weight of 60% to the final selection. The qualification carries 25% weightage (20% marks for the essential qualification and 5% for in-line Ph.D.)

Then a personal interview will be held for the selected candidates, which carries 15%.

Candidates who qualify for the interview shall be empaneled on the Merit List.

**Note:** The selection procedure mentioned above is based on the previous year's notification. It may or may not vary for this year. Till the detailed notification arrives one could follow this for guidance.

As said, valley curves are a type of vertical curves used in the vertical alignment of highways. It is also called a sag curve.

A valley curve has upward concavity. Such a situation is possible only when a descending gradient meets the ascending gradient of a highway. Therefore, valley curves are located at places where a descending portion of the highway meets the ascending section of the highway.

Generally, the cubic parabola is adopted for a valley curve. But, for small angles of deviations, all types of transition curves like a spiral, lemniscate, and cubic parabola are suitable.

While traversing a valley curve, the centrifugal force acts downwards and exerts pressure on tyre. The tyre is now subjected to both the weight of the car and centrifugal force due to the valley curve. This causes discomfort to the passengers. Therefore, the length of a valley curve is designed so as to minimize the discomfort to passengers.

The comfort criteria can be satisfied only by transition curves which help in gradually increasing and decreasing the centrifugal force as the car traverses the valley curve. Therefore, the length of the valley curve as per comfort criteria is calculated as the length of two transition curves for descending and ascending sections of the highway.

The headlight sight distance is the amount of highway that should be visible at any time in night conditions in order for the vehicle to stop before the obstacle. Numerically it is the same as the stopping sight distance.

In a valley, during the daytime, there is no problem in satisfying the sight distance. But during the nighttime, it becomes a problem. Therefore headlight sight distance is also considered a criterion.

As said, the comfort criteria can be satisfied only by transition curves. Therefore, the length of the valley curve is calculated as the length of two transition curves for descending and ascending sections of the highway. The formula is given as,

**L = 2 * Lt1 = 2 * ((V^3)*N/C)^(1/2)**

where,

C = rate of change of centrifugal acceleration. Generally fixed at 0.6 m/s^3

**N = |(****±n1) - (±n2)|**

n1 - descending gradient which is in -ve

n2 - ascending gradient which is in +ve

V - design speed of the highway in m/s

Length of valley curve as per headlight sight distance is derived using basic trigonometry. The formula is given as,

For L > SSD

**L = (N*S^2) / (1.5 + 0.035*S)**

For L < SSD

**L = 2S - ((1.5 + 0.035*S) / N)**

where,

S - headlight sight distance (stopping sight distance)

**N = |(****±n1) - (±n2)|**

n1 - descending gradient which is in -ve

n2 - ascending gradient which is in +ve

V - design speed of the highway in m/s

Initially, we don't know whether the length of the curve is greater or lesser than the sight distance. Therefore, it is calculated on a trial and error basis. The formulas are better illustrated below.

**Question: **For a portion of the highway, descending gradient 1 in 25 meets ascending gradient 1 in 20. The design speed is given as 90 kmph. Calculate the length of the valley curve (L) based on comfort and headlight distance criteria. Take S value as 153.66 m. Assume L is greater than SSD.

**Solution: **

N = |(±n1) - (±n2)|

N = |-1/25 - 1/20| = 0.09

**Comfort Criteria:**

**L = 2 * ((V^3)*N/C)^(1/2)**

L = 2 * (((0.278*90)^3) * 0.09 / 0.5)^(1/2)

L = 106.066 m

**Headlight Distance Criteria:**

For L > SSD

**L = (N*S^2) / (1.5 + 0.035*S)**

L = (0.09*153.66^2) / (1.5 + 0.035*153.66)

L = 308.96 m

Therefore, the length of the transition curve as per the **comfort criterion is 106.66m** and as per **headlight sight distance is 308.96m. **Usually, a higher value is adopted for design.

That's all for this blog. Want us to cover a topic of your interest? Fill out the form below and let us know.

Soil is a three-phase system in general. It contains air, water, and solids as a part of the system. Soil can also exist as a two-phase system depending on the field conditions. The two-phase systems that could exist are listed below.

Dry soil system - solids and air

Saturated soil system - solids and water

Before getting into various terminologies it is important to understand certain terms related to weights and volumes of phase as listed below.

Volumes of phases

Va - Volume of air

Vw - Volume of water

Vv - Volume of voids

Vs - Volume of solids

V - Total volume of the soil mass

Weight of phases

Wa - Weight of air (zero)

Ww - Weight of water

Wv - Weight of material occupying voids (neglected)

Ws - Weight of solids

W - Total weight of the soil mass

There are six basic relations using weights and volumes of phases as discussed below.

The porosity of a soil mass is the ratio of the volume of voids to the total volume of the soil mass. It is commonly expressed as a percentage and ranges from 0 to 100%.

**n = (Vv/V) * 100 **

where,

Vv = Vw + Va

V = Vw + Va + Vs

The void ratio of a soil mass is the ratio of the volume of voids to the volume of solids in the soil mass. Its value is always greater than zero.

**e = Vv/Vs**

Void ratio is more commonly used than porosity as the volume of solids (Vs) remains constant upon application of pressure.

The degree of saturation of a soil mass is the ratio of the volume of water in the voids to the volume of voids. It is commonly expressed as a percentage and ranges from 0 to 100%.

**S = (Vw/Vv) * 100**

For fully saturated soil mass Vv = Vw and the degree of saturation becomes 100%. For a dry soil mass, Vw = 0, and hence the degree of saturation becomes 0.

The percentage of air voids of a soil mass is the ratio of the volume of air to the total volume of soil mass. It is commonly expressed as a percentage and ranges from 0 to 100%.

**na = (Va/V) * 100**

The air content of a soil mass is the ratio of the volume of air to the total volume of voids. It is commonly expressed as a percentage and ranges from 0 to 100%.

**ac = (Va/Vv) * 100**

The water content of a soil mass is the ratio of the weight of water to the weight of solids of the soil mass. It is commonly expressed as a percentage and ranges from 0 to 100%.

**w = (Ww/Ws) * 100**

There are six unit weights that are required to be understood all of which are discussed further.

The bulk unit weight of a soil mass is the weight per unit volume of the soil mass. It is also called the 'mass unit weight'.

**γ = W/V, **

W = Ww + Ws

V = Va + Vw + Vs

The unit weight of solids is the weight of soil solids per unit volume of solids alone. It is also called the 'absolute unit weight'.

**γs = Ws/Vs**

The unit weight of water is the weight per unit volume of water. The unit weight of water is 9.81 kN/m^3 at 4°C which is commonly used as the standard.

**γw = Ww/Vw**

The saturated unit weight is nothing but the bulk unit weight in a saturated condition.

Submerged unit weight is its unit weight in submerged condition.

**γ' = (Ws)sub/V**

**γ' = γsat - γw**

The dry unit weight of a soil mass is the soil solids per unit of total volume.

**γd = Ws/V**

There are two specific gravity terms relating to the soil as discussed further.

The mass specific gravity of a soil is the ratio of mass unit weight of the soil to the unit weight of water. This is also called the 'bulk specific gravity or 'apparent specific gravity.

**Gm = ****γ/γw**

The specific gravity of solids is the ratio of the unit weight of solids to the unit weight of water. It is also called the 'absolute specific gravity or 'grain specific gravity. This term is relatively constant as it is based on the unit weight of solids (γs) and hence used in almost all relations.

**Gs = γs/γw**

Below mentioned are some of the important relations between the above-discussed terminologies. Note that G in the below relations refers to grain specific gravity i.e., Gs.

e = n/(1 - n)

n = e/(1 + e)

e = wG/S

ac = 1 - S

na = ac * n

γ = ((G + Se)*γw) / (1 + e)

γsat = ((G + e)*γw) / (1 + e)

γd = (G*γw) / (1+e)

γsub = (G - 1)*γw / (1 + e)

γd = γbulk / (1+w)

γd = (1 - na) * ((G*γw) / (1+e))

For your easy reference, all the relations are attached as an image below which could be downloaded and used for preparation.

**Question: **If the porosity of the soil sample is 20%, the void ratio is? (GATE: 1997)

**Solution:**

n = 0.2

e = n/(1 - n)

e = 0.2/0.8 = 0.25

**Therefore, the void ratio of the soil sample is 0.25**

Hope you enjoyed reading through the blog. Want to stay updated on our blog feed? Get subscribed to APSEd blogs by filling out the form below.

The flow in an open channel is classified into the following types.

Steady and unsteady flow

Uniform and non-uniform flow

Laminar and turbulent flow

Sub-critical, critical, and super-critical flow

In a steady flow, the flow characteristics such as flow depth, the velocity of flow, etc. at any point do not change with respect to time. Mathematically it is expressed as,

**∂V/∂t = 0, ∂Q/∂t = 0**

On the other hand, if the flow characteristics like the depth of the flow, the velocity of the flow, etc. at any point in an open channel change with respect to time then it is called an unsteady flow. Mathematically it is expressed as,

**∂V/∂t ≠ 0, ∂Q/∂t ≠ 0**

For a given length of the channel if the depth of the flow, the velocity of the flow, the slope of the channel, etc. remains constant then it is called a uniform flow. In other words, the flow characteristics remain constant with space. Mathematically it is expressed as,

**∂y/∂S = 0, ∂V/∂S = 0**

On the other hand, if the depth of flow, the velocity of the flow, etc. changes with respect to space i.e., along a section of the channel, then it is a non-uniform flow. Mathematically it is expressed as,

**∂y/∂S ≠ 0, ∂V/∂S ≠ 0**

In reality, combinations of the above-mentioned flows such as steady uniform flow, steady non-uniform flow, unsteady uniform flow, and unsteady non-uniform flow, exist.

Non-uniform flow is also called a varying flow and is further classified as,

Gradually varied flow

Rapidly varied flow

**Gradually Varied Flow: **If the depth of flow in a channel changes gradually over a long length of the channel then it is called a gradually varied flow.

**Rapidly Varied Flow: **If the depth of the channel changes abruptly over a very small length of the channel, then it is called a rapidly varied flow.

The flow in an open channel flow is said to be laminar if Reynold's number is less than 500 to 600. Reynold's number is given as,

Re = ρVR/μ,

where,

V - mean velocity of flow

R - Hydraulic radius or hydraulic mean depth = cross-section of flow/wetted perimeter

ρ - density of the liquid

μ - viscosity of the liquid

If Reynold's number is greater than 2000, then it is called a turbulent flow. If Reynold's number remains between 500 to 2000 then the flow is said to be in a transition state.

This classification is based on the Froude number. The Froude number is expressed as,

Fe = V/(gD)^(1/2)

where,

V - mean velocity of flow

D - hydraulic depth = wetted area/top width of the channel

If the Froude number is less than one then it is called a sub-critical flow. It is also called a tranquil flow.

The flow is critical if the Froude number is equal to one. If the Froude number is greater than one then the flow is super-critical.

We know that the total energy of a flowing fluid per unit weight is given as,

Total energy = z + h + V^2/2g

If the channel bottom in which the fluid is flowing is taken as datum then the above equation will become,

Total energy = h + V^2/2g

The above equation is known as the specific energy of the flowing liquid. Therefore, specific energy is defined as the total energy per unit weight of the liquid with respect to the bottom of the channel.

The specific energy curve is a curve that shows the variation of specific energy with respect to depth. It is obtained as follows.

**E = h + (V^2)/2g = Ep + Ek**

Ep - the potential energy of the flow

Ek - kinetic energy of the flow

Let us consider a rectangular channel in which a steady non-uniform flow is taking place at a discharge of Q. Let b and h be the width and depth of the channel respectively. Discharge per unit width is given as,

q = Q/b = which is a constant

Velocity can be rewritten as, V = Q/A = Q/(b*h) = q/h

Substituting 'V' in the specific energy equation,

**E = h + (q^2)/(2gh^2)**

where,

Ep = h

Ek = q^2/(2gh^2)

The above equation shows the variation of specific energy variation with depth. Therefore, with the help of this equation, the specific energy curve can be plotted by directly finding the specific energy values for a given discharge at different depths or by plotting the potential energy curve (which is a straight line) and kinetic energy curve (which is a parabola) separately and then combining them. The blue curve below represents the specific energy.

Critical depth of flow is that of flow at which the specific energy is minimum. From the above graph, it is the depth of flow at point 'C'. Mathematically the critical depth can be obtained by differentiating the specific energy equation with respect to depth and equating it to zero.

dE/dh = o

d/dh[h + (q^2)/(2gh^2)] = 0 (q^2/2g = constant)

Differentiating and rearranging the terms, we get,

**hc = (q^2/g)^(1/3)**

Critical velocity is the velocity of flow at critical depth. It is found from the critical depth formula as follows.

hc = (q^2/g)^(1/3)

Taking cube on both sides, we get,

hc^3 = q^2/g

we know that q = Q/b = b*h*V/b = hc*Vc

Substituting 'q' in the above equation and rearranging the terms, we get,

**Vc = (g*hc)^(1/2)**

It is obtained by substituting the critical depth value (hc) in the specific energy equation. It is expressed as,

**Emin = 3hc/2**

Except for the point at the minimum specific energy, there are two possible values of depth possible for any given value of specific energy. These two values are called alternate depths. One of the depths is greater than the critical depth while the other is less than the critical depth.

**Question: **The conjugate depths at a location in a horizontal rectangular channel, 4m wide, are 0.2m and 1m. The discharge in the channel is? (GATE 1991)

**Solution: **

Conjugate depths are nothing but alternate depths. It is given that,

h1 = 0.2 m

h2 = 1 m

**E = h + (q^2)/(2gh^2)**

E1 based on h1 = 0.2 + (q^2/0.785)

E2 based on h2 = 1 + (q^2/19.62)

Equating the above two equations (as at conjugate depths the specific energy is same) and solving for q, we get,

q = 1.0849

Q = q * b = 4*1.0849 = 4.339 cum/s

**Therefore, discharge = 4.339 cum/s**

Hope you found all the required information on open channel flow. There is one more interesting sub-topi called "Hydraulic jump" in the open channel flow topic. Want us to cover hydraulic jumps? Comment "Hydraulic Jumps" or any other topic of your interest in the form below.

Before getting into the Webster method, it is important to understand some of the technical terms related to traffic signals as discussed further.

Traffic signal works on the time-sharing principle. Cycle length is the time taken to complete one full cycle of the signal at an intersection. For instance, it is the time taken for a signal to go from red, yellow, green, and then come back to the red signal.

The green and red interval is the amount of time for the green and red signals respectively.

Change interval is the amount of time for the yellow signal. The yellow time is also called the amber time. Amber time can be calculated using simple formula as explained below.

In the above case, as the vehicle can see the amber signal from its stopping sight distance, so, the vehicle can slow down and come to halt before the signal turns red.

But in the above case, the signal is amber only after the vehicle crosses its stopping sight distance. Therefore, sufficient time must be provided so that the vehicle can pass the signal without any disturbance.

The distance needed to be traveled by the vehicle is the summation of the stopping sight distance (SSD) of the vehicle, length of the carriageway (W), and length of the vehicle (L). Now, the amber time is found as,

**Amber time = (SSD + W + L) / v,**

where,

v is the velocity of the vehicle (or design speed)

Clearance interval is the amount of time for pedestrians to cross and extra time for vehicles to clear the intersection.

Phase is the number of paths crossing at an intersection. For example, in a four-armed intersection, the number of phases is also four. It is also given as the summation of the green interval, change, and clearance interval.

In a traffic signal, once the signal is green, the vehicle that is first in queue will take some time to react to the signal and start moving. The second vehicle will take slightly lesser time than the first vehicle and so on. This time will decrease and will eventually reach a constant time called the headway.

The extra time in excess of the headway taken by the vehicles upfront the queue is called lost time. Each phase will have the lost time and needs to be factored in to calculate the optimum cycle length.

Saturation flow is the highest amount of vehicular flow that is possible. It is given as the inverse of headway. If the headway is in seconds, then the saturation flow is given as,

Saturation flow = 3600/headway in vehicles per hour.

Observed volume is the actual observed volume of traffic flow that is happening at the intersection. It is also represented as vehicles per unit time.

The critical flow ratio at a phase is the ratio between the observed volume of flow to the saturation flow occurring at all the phases of an intersection. It is given as,

**Critical flow ratio at ith phase = observed volume / saturation flow = v/s at ith phase**

Using the above parameters, Webster created a simple formula to calculate the optimum cycle length for an intersection. The optimum cycle length is also taken as the total cycle time for a signal system. Webster's formula is given as,

**Optimum cycle length (Co) = (1.5*L + 5) / (1 - y), **

where,

L - total lost time including all red time,

**L = (n * Lost time at a phase) + All red time**

n - number of phases

All red time is usually taken as zero

Lost time at a phase is usually taken as 2 seconds

y - is the summation of the critical flow ratio at all the phases

Green time for a road 'a' by Webster's method is given as,

**Ga = (ya/y) * (Co - L),**

where,

ya - critical flow ratio for road 'a'

y - summation of all critical flow ratio

Co - Optimum cycle length

L - lost time including all red time

**Question: **The normal flow of traffic on crossroads A and B are 400 and 250 vehicles per hour respectively. The saturation of flow for roads A and B are estimated as 1250 and 1000 vehicles per hour respectively. The all-red time for pedestrians to cross is 12 seconds. Design a two-phase traffic system by Webster's method.

**Solution: **

By Webster method, Optimum cycle length (Co) = (1.5*L + 5) / (1 - y),

n = 2

Lost time = 2 s (usually taken)

All red = 12 s (given)

L = (n * Lost time at a phase) + All red time

L = (2*2) + 12 = 16

Road A:

Observed volume (va) = 400

Saturation flow (sa) = 1250

Road B:

Observed volume (vb) = 250

Saturation flow (sb) = 1000

Critical flow ratio on road A (ya) = va/sa = 400/1250 = 0.32

Critical flow ratio on road B (yb)= vb/sb = 200/1000 = 0.25

y = ya + yb = 0.32 + 0.25 = 0.57

Now, Co = (1.5*16 + 5) / (1 - 0.57) = 67.44

**Therefore, the optimum cycle length is 67.44 s**

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