Month: June 2021

What is Railway Sleeper? Types of Railway Sleepers

What is Railway Sleeper?

A railway sleeper is a part of the railway track. Sleepers are placed below the rails and over the ballast. The main function of the sleepers is to distribute the loads coming on the rails over a large area of the ballast and support the rails to hold them firmly in place. It provides stability to the permanent way.

Different Types of Railway Sleepers

How many types of railway sleepers are there? According to the use of materials, railway sleepers are classified into the following categories:

Steel Sleepers.
Timber or Wooden Sleepers.
Cast Iron Sleepers.
Concrete Sleepers.

1. Steel Sleepers

Due to the growing scarcity of wooden sleepers, and their short life span, steel sleepers have now become more popular in every part of the world.

Advantages And Disadvantages of Metal Sleepers — Types of Railway Sleepers – #1. Steel Sleepers

Advantages of Steel Sleepers

The following are the advantages of Steel sleepers:

1. Steel sleepers are uniform in strength and durability.

2. The efficiency of fittings is better in Steel sleepers and therefore less creep develops.

3. They are economical because they are longer in life and easier to maintain.

4. In steel sleepers, the Gauge can be easily adjusted and also easily maintained.

5. For Steel sleepers, there is no need for frequent renewal.

6. They have a good scrap value.

7. They are not susceptible to fire and vermin attacks.

Disadvantages of Steel Sleepers

The following are the disadvantages of steel Sleepers:

1. More ballast is required than other types of sleepers.

2. These sleepers require a number of fittings which causes difficulties in maintenance.

3. These sleepers are liable to rusting/corrosion.

4. These sleepers are not suitable for track circuiting.

5. Steel sleepers are unsuitable for bridges, level crossings, points and crossings, etc.

6. These sleepers are only suitable for stone ballast.

Note: Steel sleepers and cast iron sleepers are the types of metal sleepers

2. Wooden Sleepers

Wooden sleepers are considered the best as they fulfill almost all the requirements of an ideal sleeper. In most cases, the most commonly used timbers for sleepers are oak, sal(Shorea robusta), and teak.

Wooden Sleepers - With Their Advantages and Disadvantages — Types of Railway Sleepers – #2. Wooden Sleepers

Advantages of wooden sleepers

Following are the advantages of wooden sleepers

1. The initial cost of a wooden sleeper is low.
2. Timber is easily available in all parts of the USA, UK, and India.

3. Fittings for wooden sleepers are few and simple in design.
4. These sleepers are able to resist shocks and vibrations due to heavy moving loads and also provide a noise-less track.

5. Wooden sleepers are easy to lay, relay, pack, lift, and maintain.
6. They are suitable for all types of ballast.
7. They are best for track-circuited operations.

Disadvantages of wooden sleepers

The following are the disadvantages of wooden sleepers
1. The service life of wooden sleepers is short(12 to 15 years)as compared to other types of sleepers.

2. These sleepers are subjected to wear, decay, and attack by white ants, spike killing, cracking, etc.

3. It is difficult to maintain the gauge in the case of wooden sleepers.

4. The track is easily disturbed. Hence alignment maintenance is difficult.

5. The maintenance cost of wooden sleepers is high as compared to other sleepers.

6. The scrap value of these sleepers is very low.

7. They are susceptible to fire.

3. Cast Iron Sleeper

Cast iron sleeper is another type of sleeper, which is extensively used in India. It is also used in South America, Burma, and Pakistan. Cast iron slippers have been used in India since 1870.

Advantages of cast-iron sleepers

Following are the pros or benefits or advantages of cast-iron sleepers

1. The probability of cracking is less.

2. They have less corrosion effects.

3. The scrap value is high in the case of cast iron sleepers.

4. The manufacturing process of cast iron sleepers is comparatively easy.

5. Maintaining work for this type of sleeper is minimum and easy. Therefore, less-skilled labor is required.

6. It provides good longitudinal as well as lateral resistance.

7. Cast iron sleepers are not affected due to the irregular falling of fire from the steam engines.

8. The efficiency of fittings is also better in cast iron sleepers; therefore, less creep develops. Read Also: Creep of Rails

Disadvantages of cast-iron sleepers

Following are the cons or drawbacks or disadvantages of cast-iron sleepers.

1. Ballast requirement is more compared to the other types of sleepers.

2. In these sleepers, more fittings are required, and frequent Inspection is essential.

3. Cast iron sleepers are not usable for track-circuited areas as they are good conductors of electricity.

4. They are not suitable for level crossings and bridges.

4. Concrete Sleepers

The first idea about concrete sleepers was conceived by Monier of France in 1884. However, it takes several years to become popular in different countries. After world war II it was rapidly developed due to the scarcity of timber. Especially in countries like the United Kingdom, Germany, France, Russia, and Japan. During world war II, many of the railway lines were destroyed in these countries, and the shortage of wood forces them to find a suitable alternative sleepers material of timber sleepers. Finally, the concrete sleeper became one of the most popular sleeper types.

Types of Railway Sleepers – #3. Cast Iron Sleeper

Advantages of Concrete Sleepers

Following are the Pros or Benefits or advantages of concrete sleepers

1. Concrete sleepers give more strength and stability to the track.

2. They have great resistance to buckling of the track.

3. A concrete sleeper is a poor conductor of electricity. For this reason, they are preferred for track circuited areas.

4. It has a very long lifespan. It can last 30 to 40 years.

5. It has economic benefits due to its long lifespan and minimal joints.

6. Concrete sleepers are free from natural decay and the effect of chemical action.

7. They are not attacked by insects or vermin.

8. They are not affected by the adverse environmental effects.

9. They can be produced in a large volume using local resources.

Disadvantages of Concrete Sleepers

Following are the cons or drawbacks or disadvantages of concrete sleepers

1. It has no scrap value

2. The maintenance and replacement process is difficult.

3. Transportation is difficult as they are heavy in weight.

4. They are heavily damaged at the time of replacement.

5. Concrete sleepers are not suitable for better packing.

Read Also:

Railway Sleepers – MCQ

Reinforcement Details In Column – Building Construction

Reinforcement Details In Column

The column is a very important part of buildings. Generally, they are used to transfer the load of the beams and slabs to the footing of the buildings. Therefore, it is very important to make it in a proper way, so that it can easily bear all loads which act over it. In this article, I will discuss the reinforcement details of a column.

Based on shape & size, columns may be of different types, such as rectangular, square, circular, etc.

Minimum reinforcement in column

Now, 1st thing which comes to our mind is that what should be the minimum reinforcement in a column.

i) In the case of a square column, the minimum longitudinal reinforcement should be 4 nos. As per IS code, the minimum dia of the longitudinal bar should not be less than 12 mm, and the dia of a stirrup should not be less than 8 mm. Stirrup should be placed at a distance of 150 mm (center-to-center distance).

Note: In my experience, you can also use a dia of 10 mm bar in the column if the structure is very simple and only for a height of 10 to 12 ft from the ground level. And, also the distance between the two columns should not be more than 8 ft.

ii) In the case of a rectangular column, the minimum longitudinal reinforcement should be 4 nos. The minimum dia of bars should be the same as the above-mentioned.

iii) In the case of a circular column, the minimum longitudinal reinforcement should be 6 nos. Dia of bars should be the same as the above-mentioned.

Coloumn Type	Minimum Numbers of Longitudinal bar	Minimum dia of longitudinal bar	Dia of stirrup
Square column	4	12 mm	8 mm (150 mm center to center distance)
Rectangular column	4	12 mm	8 mm
circular column	6	12 mm	8 mm

What is the Minimum Size of a Column?

The recommended minimum size of a column is 9 inc by 9 inc (9″ × 9″). But, sometimes 9 inc by 5 inc column is also used( However, it is not recommended to use)

Arrangement of Lateral Ties and Links in Column

Typically arrangement of lateral ties and links in columns are as follows:

i) Single Tie

ii) Two Tie

iii) Single Tie and Link

iv) Two Ties

v) Single Tie and Two Link

vi) Three Ties

vii) Single Tie and Three Links

viii) Three Ties

ix) Single Tie and Four Links

Reinforced Brickwork and Their Advantages

Ultimate Compressive Load for an Axially Loaded Short Column

Materials Required Per Unit (Km) Length of Track

Materials Required Per Unit(km) Length of Track

It is necessary to work out exactly the quantities of various materials required to lay track. Otherwise, the excess materials will have to be taken back and work will be delayed. In this article, we will discuss the number of rails, sleepers, fish plates, fish bolts, dog spikes, and bearing plates required for each kilometer of track length.

1. Number of rails per km of track

= [latex] \frac{100}{length\; of \; rail }\times 2[/latex]

In case of 12.80 m length of the rail

The number of rails will be required for each kilometer(per km) of track length

= [latex] \frac{100}{12.80}\times 2[/latex] =156.25. say 157 nos.

2. Number of sleepers per km of track length

= [latex] \frac{no.\; of \; rails\; per \; km}{2}\times (n+x) [/latex]

Where (n+x) is the expression for sleeper density.

if x = 3, then the approximate number of sleepers required in the above case of 12.80 m rail would be equal to

[latex] \frac{157}{2}\times \left ( 12.80+ 3 \right ) [/latex] = 1240.3 say 1241 nos.

3. Number of fish plates per km of track length

= 2 × number of rails per km(kilometer) of track

= 2 × 157 = 314 nos.

4. Number of fish bolts per km of track

= 4 × number of rails per km of track length

Then, the number of fish bolts will be required for each kilometer(per km) of track length

= 4 × 157 = 628

5. Number of bearing plates per km of track

= 2 × number of sleepers per km of track

The number of rails will be required for each kilometer(per km) of track length

= 2 × 1241 = 2482 nos.

6. Number of dog spikes per km of track

= 4 × number of sleepers per km of track length

Then, the number of dog spikes will be required for each kilometer(per km) of track length

= 4 × 1241 = 4964 nos.

Read Also:

Types of Railway Sleepers

TYPES OF RAILS

5 Major Cross Sectional Elements of a Railway Track

How to Determine the Frequency of Irrigation – Problem & Solution

How to Determine the Frequency of Irrigation

Mathematical Problem

The following data are given to determine the frequency of irrigation

Field capacity of soil = 45 %
Permanent wilting point = 19 %
Density of soil = 1.5 g/cm³
Depth of root zone = 65 cm
Daily consumptive use of water = 18 mm

Solution:

Available moisture = Field Capacity – Permanent wilting point = 45 – 19 = 26

Let readily available moisture = is 75 % of the available moisture,

So, readily available moisture is = 26 × 0.75 = 19.5 %

Optimum moisture content = 45 – 19.5 = 25.5 %

Now, by applying irrigation water the moisture content is to be raised from 25.5 % to 45 %.

Calculate the depth of water to be applied in each watering, by using the following equation:

[latex]D_{w} = \frac{W_{s}\times d}{W_{w}}\times [F_{c}-M_{0}][/latex]

Where,

W_s = Unit Weight of soil or density of soil = 1.5 g/cm³
W_w = Density of water = 1g/cm³
d = Depth of root zone = 65 cm
F_c = Field capacity = 45 %
M₀ = Optimum moisture content = 25.5 %

After putting the all value,

[latex]D_{w} = \frac{1.5\times 65}{1}\times\left [ 0.45 – 0.255 \right ][/latex]

[latex]Or, D_{w} = 97.5 \times 0.195 = 19.01 [/latex] cm

Now, we can calculate the frequency of irrigation by using the following equation

[latex]f_{w} = \frac{D_{w}}{C_{u}}[/latex]

Where,

f_w = Frequency of watering or frequency of irrigation
C_u = Daily consumptive use of water =
D_w = Depth of water to be applied in each watering

After putting all values, the frequency of irrigation

[latex]f_{w} = \frac{19.01}{1.8} = 10.56 = 11 days (say)[/latex]

Read Also:

Find the Intensity of Rainfall

Determine Runoff by using Rational Method

Types Of Irrigation

Find the Intensity of Rainfall When the duration of Rainfall is Given

Find the intensity of Rainfall When the duration of Rainfall is Given

First of all, we need to know the formula used to determine rainfall intensity. There are three types of empirical formulas –

a) For storm duration of 5 to 20 minutes

[latex] i = \frac{672}{t+10} [/latex]

b) For storm duration of 20 to 100 minutes

[latex] i = \frac{1020}{t+10} [/latex]

c) Where rainfall is frequent

[latex] i = \frac{3430}{t+18} [/latex]

Here, i = Intensity of rainfall in mm/hr, t = duration of the storm in minutes.

Following are the two mathematical examples for the determination of rainfall intensity

Mathematical Problem: 1

Find the intensity of rainfall when the storm continues for a period of 75 minutes.

Solution:

As the duration of rainfall is in between 20 to 100 minutes, so, the formula will be
[latex] i = \frac{1020}{t+10} [/latex]
The intensity of rainfall,

[latex] i = \frac{1020}{75+10} [/latex]

[latex] Or, i = \frac{1020}{85} = 12 mm/hr [/latex]

Mathematical Problem: 2

Find the intensity of rainfall when the rainfall occurs frequently and the total duration of rainfall is 115 minutes:

Solution:

As the rainfall occurs frequently, we will use 3rd formula, [latex] i = \frac{3430}{t+18} [/latex]
So, the intensity of rainfall is

[latex] i = \frac{3430}{115+18} [/latex]

[latex] Or, i = \frac{3430}{133} = 25.78 mm/hr [/latex]

Read Also:

Determine the Frequency of Irrigation

Determine Runoff Coefficient

Types of Rainfall

How to Determine Runoff by using Rational Method – Mathematical Problem & Solution

How to Determine Runoff by using Rational Method

First of all, we need to understand which formula or expression is used for the determination of runoff. In the rational method, the runoff is given by the following expression:

[latex]Q = \frac{KiA}{36}[/latex]

Where,

Q = Runoff in cumec
K = Coefficient of runoff
i = Intensity of rainfall in cm/hr
A = Catchment area in hectares

The above formula may also be expressed as follows,

[latex]Q = \frac{KiA}{360}[/latex]

This formula applies when the intensity of rainfall is given in mm/hr. Otherwise, all units are the same as above.

Mathematical Problem

A catchment area of 35 sq km consists of two-thirds rural area and one-third urban area. The rainfall in the whole catchment area is recorded as 30 mm/hr. Find the total runoff from the catchment area.

Solution:

We know the total are = 35 sq km = 3500 hectares ( 1 sq km = 100 hectares)

Now,

Rural area = [latex]3500 \times \frac{2}{3}[/latex] = 2333.33 hectares

Urban area = [latex]3500 \times \frac{1}{3}[/latex] = 1166.66 hectares.

Rainfall intensity i = 30 mm/hr.

Assuming, runoff coefficient for urban area = 0.50 and for rural area = 0.30

In rational method, the runoff is given by,

[latex]Q = \frac{KiA}{360}[/latex]

Or, [latex]\large Q = \frac{(0.30\times 25\times 2333.33) + (0.50\times 25\times 1166.66)}{ 360 }[/latex]

Or, [latex]\large Q = \frac{17499.97+14583.25}{360}[/latex]

Or, Q = 89.12 cumec.

How to Calculate Annual Runoff

Burkli Ziegler Formula – To Determine the Peak Runoff Rate

How to Determine Runoff Coefficient

How to Determine Runoff Coefficient – Mathematical Problem & Solution

How to Determine Runoff Coefficient

When the nature of the surface of different areas and the impermeability factors of the areas are known or given, then the value of the runoff coefficient(K) may be determined by the following expression:

[latex]K= \frac{a_{1}p_{1}+a_{2}p_{2}+\cdot \cdot \cdot \cdot \cdot }{A}[/latex]

Where,

a₁, a₂ = Area of different surfaces.
p₁, p₂ = Impermeability factors of those surfaces.
A = Total surface area.