### How to calculate the materials required for 4" brickworks?/ Brick, cement, & sand calculations for single brick wall.

Let us consider a single brick wall of size 10ft × 30ft.× 0.333 ft. (4")  as shown in the drawing.

Now, we will build this wall with a conventional brick of size 9 inches × 4 inches × 3 inches as shown in the image below.

Given data:

Length of the brick wall = 30 ft.

Height of the brick wall = 10 ft.

The thickness of the brick wall = 0.333 ft.

Brick size = 9 inches × 4 inches × 3 inches

Mortar thickness = 10mm = 0.39 inches

To calculate the number of bricks required to build this wall, first, we will find out the volume of the wall and the volume of a single brick with or without mortar.

The volume of the brick wall

= Length × Height × Thickness

= 10ft × 30ft × 0.333 ft.

= 100cu ft.

Now, the volume of the brick

= l × h × t

= 9 inches × 3 inches × 4 inches

( By converting them into ft.)

= 0.75ft × 0.25 ft × 0.333 ft.

=0.0625 cu ft.

From the drawing given below, you can observe that the addition of the mortar thickness comes on the 2 sides of every brick.

So, the volume of the brick with mortar

= 9.39 inches  ×3.39 inches  × 4 inches

=127.3284 cubic inch.

=0.0737cu ft.

1.  The number of bricks required

= volume of the wall ÷ volume of the brick with mortar

= 100 cu ft. ÷ 0.0737 cu ft.

=1357 nos.

The volume of mortar required

= volume of the brick wall - the volume of the total no. of bricks

= 100 cu.ft. - (1357 nos. × 0.0625 cu.ft.)

=   100 cu ft. - 84.812 cu ft.

=15.187 cu ft.

Let us prepare this 15.187 cu ft. of mortar in a 1: 6 ratio to build this 100 cu ft. of the brick wall.

Dry volume of the cement mortar

= 1.33 × 15.187

= 20.198 cu ft.

2.  Volume of the sand

= (6  ÷ ( 6 + 1)) × 20.198

= 0.8571 × 20.198

= 17.312 cu ft.

3. Volume of the cement

= total dry mortar vol. - volume of sand

= 20.198 cu ft - 17.312 cu ft.

=2.886 cu ft.

As you know, one bag of cement = 1.226 cu ft.

Number of cement bags

= 2.886 cu ft. ÷ 1.226 cu ft.

=   2.354bags.

Let us now find out the materials required for 1 cu ft. & 1 CUM. of  4" brick wall.

1. Bricks:

Bricks required for 1 cu ft. of brick wall

= 1357 nos. ÷ 100

= 13.57 nos.

Bricks required for 1 CUM of brick wall

= 13.57nos. × 35.315

= 479 nos.

2. Cement:

Cement required for 1 cu ft. of brick wall

= 2.354 bags. ÷ 100

= 0.0235 bags.

Cement required for 1 CUM of brick wall

= 0.0235 bags × 35.315

= 0.83 bags.

3. Sand:

Sand required for 1 cu ft. of brick wall

= 17.312 cu ft. ÷ 100

= 0.1731 cu ft.

Sand required for 1 CUM of brick wall

= 0.1731 cu ft. × 35.315

= 6.11 cu ft.

### What will be the cost of block wall masonry?/Rate analysis of concrete block masonry.

Let us consider solid concrete block masonry as shown in the drawing for calculation purposes.

Given data:

Length of the block masonry = 16ft.

Height of the block masonry = 10ft.

The thickness of the block masonry = 0.5ft.

The total volume of the block masonry

= length × height × thickness

= 16ft. × 10ft. × 0.5ft.

= 80 cu ft.

The total area of block masonry

= length × height

=16ft. × 10ft.

= 160 sq ft.

A. Material calculation:

Read the article "Calculating the quantity of materials in a 100 cubic ft. block wall." before proceeding further.

No. of blocks required/100 cu ft. of wall = 210 nos.

No. of Cement bags required/100 cu ft of wall = 1.038 bags

The volume of sand required/100 cu ft. of wall = 7.634 cu ft.

Note: The quantity of blocks, cement & sand/ 100 cu ft. is taken from the above-mentioned article.

1. No. of  concrete blocks required

= [(vol. of block wall ÷ 100 cu ft.) × blocks/100 cu ft.]

=  [(80 cu ft. ÷ 100 cu ft.) × 210 nos.]

= 168 nos.

2. No. of  cement bags required

= [(vol. of block wall ÷ 100 cu ft.) × bags/100 cu ft.]

=  [(80 cu ft. ÷ 100 cu ft.) × 1.038 bags]

= 0.83 bags.

3. The volume of sand required

=  [(vol. of block wall ÷ 100 cu ft.) × sand/100 cu ft.]

=  [(80 cu ft. ÷ 100 cu ft.) × 7.634 cu ft.]

6.11 cu ft.

B.  Labor calculation:

The labor rate per sq ft. of the block masonry ranges from INR 10/- to 15/- depending upon the region.

Let us consider an average rate of INR 12/- per sq ft.

The total cost of labor for block masonry

= total area of block masonry in sq ft.  × labor rate/ sq ft.

= 160 sq ft. × 12/-

= INR 1920/-

Go through the article👇

👀. What will be the cost of 9" brick wall masonry?/ Rate analysis of 9"  thick brick wall work.

Now, let us calculate the total cost of block masonry in table format.

 Sl. no. Item. Qty. Rate. in INR. Unit. Cost in INR. 1. Solid block 168 28/- Nos. 4704.00 2. Cement 0.83 400/- bag 332.00 3. Sand 6.11 60/- Cu ft 366.60 4. Curing Lump-sum 200.00 5. Miscellaneous Lump-sum 150.00 6. The material cost = 5752.60 7. Add 5% wastage = 287.63 8. Total material cost = 6040.23 9. Labour cost = 1920.00 10. The total cost of block masonry = 7960.23

Note: To get accurate results, insert your regional market rate of materials & labor, in the above table.

The cost of block masonry/sq ft.

= [total block masonry cost ÷ total block wall area]

= 7960.23 ÷ 160

= INR 49.75/sq ft.

The cost of block masonry/ cu ft.

= [total block masonry cost ÷ total block wall volume]

= 7960.23 ÷ 80

= INR 99.50/cu ft.

If you have any quarries, you can ask me in the comment box👇.

### What will be the cost of plastering in a building?/ How to calculate the cost of internal plastering?

Now, let us calculate the interior plastering cost of one room of a building as shown below.

Given data:

Width of the room = 14ft.

Length of the room = 16ft.

Height of the room = 10 ft.

One door of size = 2.5ft × 7.0 ft.

One window of size = 3ft × 4ft.

Area of wall to be plastered

= [{(2nos × area of  front & back wall) + (2nos × area of sidewalls)} - {(door size) +( window size)}]

= [{(2nos × 14ft. × 10ft.) + (2nos.× 16ft. × 10ft.)} - {(2.5ft. × 7.0ft.) +( 3ft. × 4ft.)}]

=  [{(280sq ft.) + (320 sq ft.)} - {(17.5 sq ft.) +( 12 sq ft)}]

= [ { 600sq ft.} - {29.5 sq ft.}]

= 570.5 sq ft.

Area of the ceiling to be plastered

= [14ft × 16ft.]

= 224 sq ft.

The total area of plastering

= wall area +ceiling area

= 570.5 sq ft. + 224 sq ft.

= 794.5 sq ft.

Usually, the thickness of wall plaster ranges from 10mm. to 15mm. & the thickness of the ceiling plaster range is between 6mm to 10mm.

For the calculation purpose, let us consider wall plaster of thickness 12mm. & the ceiling plaster of 8mm. thickness.

Data considered:

Wall plaster thickness = 12mm =0.0394 ft.

Ceiling plaster thickness= 8mm = 0.0262 ft.

Let us make plaster of  1:4 ratio.

The total volume of the plaster

= [vol. of wall plaster + vol. of ceiling plaster]

=[ (area of wall plaster × thickness) + (area of ceiling plaster × thickness)]

= [( 570.5 sq ft.× 0.0394 ft.) + ( 224 sq ft. × 0.0262 ft.)]

= 22.477 cu ft. +5.869 cu ft.

= 28.346 cu ft.

A. Material calculation:

Cement bag required per cu ft of plaster mortar = 0.217 bags

The volume of sand required/cu ft. of plaster mortar = 1.064 cu ft.

Note: The quantity of cement & sand/ cu ft. is taken from the above-mentioned article.

1. No. of  cement bags required

= vol. of plaster × bags/ cu ft.

= 28.346 cu ft × 0.217

= 6.15 bags.

2. The volume of sand required

= vol. of plaster × sand / cu ft.

= 28.346 cu ft. × 1.064

= 30.16 cu ft.

B.  Labor calculation:

The labor rate per sq ft. of the internal plastering ranges from INR 10/- to 15/- depending upon the region.

Let us consider an average rate of INR 12/- per sq ft.

The total cost of labor for plastering

= total area of plaster in sq ft.  × labor rate/ sq ft.

= 794.5sq ft. × 12/-

= INR 9,534/-

Now, let us calculate the total cost of plastering in table format.

 Sl. no. Item. Qty. Rate. in INR. Unit. Cost in INR. 1. Cement 6.15 400/- Bags 2460.00 2. Sand 30.16 60/- Cu ft 1809.60 3. Chicken mesh Lump-sum 200.00 4. Miscellaneous Lump-sum 150.00 5. The material cost = 4619.60 6. Add 5% wastage = 230.98 7. Total material cost = 4850.58 8. Labour cost = 9534.00 9. The total cost of plastering = 14,384.58

The cost of internal plaster/sq ft.

= [total plastering cost ÷ total plastered area]

= 14384.58 ÷ 794.5

= INR 18.10/sq ft.

If you have any quarries, you can ask me in the comment box👇.

### How to calculate the concrete volume of slump cone?

The volume of the slump cone is calculated by the formula

= [ (h ÷ 3 ) ×{ A1+A2 + √A1× A2 }]

Where

h = height of the slump cone.

A1 = area of the top circular surface.

A2 = area of the bottom circular surface.

Given data:

Height of the slump cone h = 300mm.=0.3m

Diameter of the top surface d1 = 100mm.=0.1m.

Diameter of the bottom surface d2 = 200mm.=0.2m

First, we will calculate the area A1 & A2

A1 = [( π ×  d12) ÷ 4]

=  [( 3.142 ×  0.12) ÷ 4]

=  [0.03412 ÷ 4]

= 0.007855 sqm.

A2=  [( π ×  d22) ÷ 4]

=  [( 3.142 ×  0.22) ÷ 4]

=  [0.12568 ÷ 4]

= 0.03142 sqm.

The volume of slump cone

= [ (h ÷ 3 ) ×{ A1+A2 + √A1× A2 }]

= [ (0.3 ÷ 3 ) ×{ 0.007855+0.03142 + √0.007855× 0.03142 }]

= [ (0.1) ×{ 0.039275 + √0.0002468 }]

= [ (0.1) ×{ 0.039275 + 0.01571}]

=[ 0.1  × 0.054985]

=0.00549 cu m.

= 0.1939 cu ft.

Alternate method:

Volume of the slump cone

= [ ( π × h ÷ 3) × { r12 +r1 × r2  +  r22 }]

Where, r1 = radius of top circular surface.

r2 = radius of bottom circular surface.

h = height of the slump cone.

Here,

r1 = d1 ÷ 2

= 0.1÷ 2 =0.05m.

r2 = d2 ÷ 2

= 0.2÷ 2 =0.1m.

The volume of slump cone

= [ ( π × h ÷ 3) × { r12 +r1 × r2  +  r22 }]

= [ ( 3.142× 0.3 ÷ 3) × { 0.052 + 0.05 × 0.1  +  0.12 }]

= [(0.3142) × { 0.0025 + 0.005 + 0.01}]

= [ 0.3142 × 0.0175]

= 0.00549 cu m.

= 0.1939 cu ft.

### How to do a layout of the building plan with a 3 - 4 - 5 rule?/ Using 3-4-5 method in construction work.

#### 1. What is the 3-4-5 rule or the 3-4-5 method?

According to the Pythagoras theorem, in a right angle triangle,

c2=a2+ b2

Where a,b, & c are the sides of the triangle.

Suppose, if the side AB = 3ft. & side BC = 4ft. then,

Side AC2=AB2+ BC2

i.e. AC = √AB2+ BC2

AC = √32+ 42

AC = √9 + 16

AC = √25 = 5ft.

Now, let us observe how we can use this method to do a layout in construction work.

#### 2. How to do a layout of the building plan with a 3-4-5 rule?

Let us consider a building ABCD, having four walls, i.e. AB, BC, CD, & DA.

To make the junction point B of the walls at a right angle, let us measure & mark 3ft. on side AB & 4ft. on side BC before constructing these walls.

Now, we have to adjust the alignment of these sides until we get the length of EF as 5ft.

This procedure makes the wall corners of the building at a right angle to each other.

A similar step has to be followed to all 4 walls of the building, to make their corners at a right angle to each other.

#### 3. Where we can use the 3-4-5 rule in the construction works?

This 3-4-5 rule can be applied to do the layout of all types of structures like residential buildings, underground tanks, outhouses, etc. If the corner of the site or plot is perpendicular to each other, you can use this method to complete the layout work. All you need is a measuring tape & a lineout string.

#### 4. What should be the unit of the 3-4-5 method?

You can use any unit of measurement in this 3-4-5 rule. Suppose, if you measure the sides in 3m & 4m length, the hypotenuse should be measured for 5m. length. Similarly for 3ft, & 4ft, measurement, the hypotenuse becomes 5ft.

#### 5. Can we measure the length in multiples of the 3-4-5 method?

Yes. As you can observe in the drawing, if you measure 6ft. & 8ft. along the sides, the hypotenuse should be measured for 10ft. length.

Similarly, by multiplying 3-4-5 by 3, we can use the measurement 9 -12-15 to carry the layout work.

#### 6. Can we take any other measurements to make the corners at right angles?

Yes. As you can observe in the drawing, the two sides of a building are 10ft. & 13ft. in length.

Now, AC2=AB2+ BC2

AC = √AB2+ BC2

AC = √132+ 102

AC = √169+ 100

AC = √269 =16' 5"

Here, you have to calculate the hypotenuse for any convenient length you have taken. By calculating the diagonal of building ABCD, you can fix all the corners at right angles easily.

Here, AC=BD=16'5"

By fixing the intersection points of the building sides & diagonals at A, B, C, &D, the building can be set in a truly rectangular shape.

### What is a parapet wall? - The purpose & design of a parapet wall.

#### 1.  What is a parapet wall?

The boundary i.e. built over the edges of terrace, balcony, or roof, is called a parapet wall. The material used in constructing the wall may be brick, glass, RCC grills, mild steel, stainless steel, or a combination of all these materials.

#### 2. What is the purpose of building a parapet wall?

The main purposes of constructing a parapet wall are,

1. Protection:

The parapet wall provides protection for people, pets, & especially children. It prevents falling from the rooftop, terrace, or balcony by acting as a barrier.

2. Appearance:

The parapet wall gives an aesthetic look to the building. A creative combination of different materials helps to improve the architectural appearance of a building.

3. Privacy:

A parapet wall provides the needed privacy to you & your belongings on the balcony or over the terrace.

4. Space:

The wall creates a working space to utilize for any purpose. for eg. terrace gardening, exercise & gym., solar paneling, etc.

#### 3.  What should be the height of a parapet wall?

The height of the parapet wall should be as shown in the below table.

 Height of parapet wall. Recommended height In feet In metre In centimetre Minimum height 2.5 ft. 0.761m. 761cm. Maximum height 5.0 ft. 1.52m. 1520cm. Common height 3.0 ft 0.91m 910cm.

#### 4.  What should be the thickness of the parapet wall?

From a safety point of view, the minimum thickness of the parapet wall should be 9 inches. A gradient should be provided for the top plaster surface so that the rainwater drains off easily.

### Minimum space required for car parking in a residential building.

When you build a house, you will keep aside some space for the parking of car. The dimensions of individual cars are different & in the future, parking becomes difficult when you change the model of your four-wheeler.

I have made a drawing of the car parking space with boundaries, where we can park our car comfortably without any congestion.

Nearly, all models of Indian cars fit within the dimensions of 6.5ft width × 14ft. length.

Space required to open the car door  👉  2.5ft. on both sides.

Front clearance           👉  1.5ft.

Back clearance          👉  1.5ft.

Masonry wall thickness  👉  0.5ft.

Total built-up area required = 17.5ft × 13 ft.

= 227.5 sq ft.

= 21.14 sqm.

Total area required for parking the vehicle without boundaries

= 17ft. × 12 ft.

= 204 sq ft.

= 18.96 sqm.

### Why there is an air vent in the overhead tank?/ Why do we provide T- pipe in the overhead tank?

#### 1. Why do we provide an air vent in the overhead tank?

When we utilize the water from the overhead tank, the air occupies the empty space created by outflow water. When we pump water into the tank, the air needs an immediate outlet to move out.

The air vent helps in the easy circulation of air in the overhead tank. The air bubble separates out of water due to their lighter wt. & passes through the air vent pipes.

#### 2. What are the different locations of providing an air vent?

As you can observe in the above drawing, the air vents are provided in three different locations.

1. The air vent can be provided directly over the top surface of the tank.

2. The air vent can be installed perpendicular to the outlet or delivery pipe.

3. In the smaller capacity overhead tank, the overflow pipe acts as an air vent for circulation. Adding an extra T - pipe can be avoided as overflow pipe does this job.

#### 3. What happens if an air vent is not provided in the overhead tank?

If we don't provide an air vent in the overhead tank, the following 3 things will happen.

1. The air creates pressure over the pump as the air does not have an easy outlet.

2. The air flows in the outlet pipe and the laminar flow of water gets disturbed.

3. If the tank is in an empty state,  the air flows into the plumbing pipe when you pump water. Initially, these air bubbles come out of your plumbing tap creating a nuisance.