#construction video tutorial

This construction video introduces detailed process for finding out the quantities of bricks as well as quantities of mortar in one cubic meter.

Usually, 500 standard size first class bricks are required to build one meter cube of brick work. But it is dependent on the size of brick you prefer.

The following two types of brickworks are found in a building.

Brickwork in Clay: This type of brickwork contains wet clay for assembling the bricks. The clay should be organized with acceptable amount of sand. The density of clay/earth mortar is 12 mm and the maximum height of building built up with clay is constrained to 2.5 m.

2. Brick work in Cement or Lime Mortar: This type of Brick work contains cement or lime mortar. Cement mortar comprises of Cement and sand having water in exact ratios and the lime mortar contains lime and Surkhi with water in proper ratios.

The density of the joint in this type of work is maintained not in excess of 10.0 mm. Other materials like stones and cement blocks are also utilized in brick work.

Sometimes corners are also made of bricks where it is required to keep the brick open (without plaster). This type of brick work is called ?facing brickwork?.

To get more details on How many bricks in one cubic meter process, go through the following exclusive civil engineering video tutorial.

Initially, you have to determine the area of the room where the tiles will be placed. Suppose, there is an L-shaped room and it is divided as part A and part B.

Length of room A = 3 meter Breadth of room A = 2 meter Length of room B = 6 meter Breadth of room B = 2 meter

At first, calculate the area of A as follow :- Area = Length x Breadth After putting the above values of room A, we get Area = 3 x 2 = 6 square meter

Then, calculate the area of B as follow :- Area = Length x Breadth After putting the above values of room A, we get Area = 6 x 2 = 12 square meter

Therefore, total area = area of A + area of B = 6 + 12 = 18 square meter Suppose the length of one tile = 600 mm = 0.6 m and breadth = 600 mm = 0.6 m So, area of one tile = length x breadth = 0.6 x 0.6 = 0.36 square meter There, no. of required tiles = Total Area / Area Of One Tile = 18 / 0.36 = 50 Nos. (approx)

But, you have to also consider the wastage that may occur in construction site. The wastage differs from site to site as well as according to room sizes. Assume, standard wastage is 5%.

So, 50 + 5% of 50 = 50 + (5/100 x 50) = 50 + 2.5 = 52.5 numbers = 53 numbers

To get more clear idea, go through the following video tutorial.

The following details of thumb rule is given :-

Minimum size of an RCC column should remain not under 9 x 9 inches containing 4 numbers of steel bars (12 mm) with standard fe 500 steel i.e. the strength of the steel is 500 N/mm2.

In RCC column, least grade of concrete should be M20. The proportion of M20 grade of concrete is 1: 1.5 : 3 (cement : sand : aggregate).

According to thumb rule, the span of the column should be 5m and it fully secured.

The longitudinal bars of the column should not be under 12 mm.

The stirrups with 8 mm diameter should be utilized by maintaining a gap of 150 mm c/c in column.

To view the graphical representation, watch the following video tutorial.

Suppose the size of the angle is taken as 20 x 20 x 3 mm i.e. length = 20 mm; width = 20 mm and thickness = 3 mm

Now you have to determine the unit weight of 1 meter.

Suppose, the angle is divided in two parts i.e. part A and Part B. Horizontal section is taken as A and vertical section is taken as B.

Volume of A = Length x Thickness x Height of the angle

= (0.02 x 0.003 x 1) [after converting from millimeter to meter)

Volume of B = (0.017 x 0.003 x x1) (after deducting the thickness from base i.e. 20-3 = 17)

Now, you can determine the weight of steel as follow :-

Weight = Volume of A & B x Density of steel (density of steel = 7850 Kg/m3) = 0.9 Kg

So, the unit weight of steel = 0.9 Kg/m

In this way, you can find the unit weight for different sizes of angles.

To get more clear ideas, watch the following construction video tutorial.

To resolve this issue, Sami Ullah, the renowned engineer, presents a useful video for civil engineering students. After finding the volume of wheelbarrow, the process for batching of concrete materials will be simplified.

To start the calculation, you have to select a box and find it’s volume.

Volume of box (v) = 1 x 1 x 0.5 (1 = length, 1 = width and 0.5 = height) = 0.5 cft

Now fill this box by pouring materials and transfer the total quantities of materials (numbers of boxes) to the wheelbarrow to find out the volume of wheelbarrow easily.

If the wheelbarrow is filled with the volume of 3 boxes, the volume of wheelbarrow will be 3 x 0.5 = 1.5 cft

Or, if the wheelbarrow is filled with the volume of 4 boxes, the volume of wheelbarrow will be 4 x 0.5 = 2 cft.

For more clarification, watch the following video tutorial.

As per reinforcement details, there exist two sizes of bars namely main bar & distribution bar in the slab.

FUNCTION OF MAIN BARS: Main bars are placed at the shorter span direction to transfer the bending moment (B.M) developed at the bottom of the slab to the beam.

Main bars (M.B) are arranged at the shorter extent in the slab to encounter high bending Moment called as sagging or positive bending moment.

The purpose of main bars is to combat all the tensile stresses, bending moment (Sagging), and superimposed load (Dead load) which occur at the shorter extent of the slabs.

Main bars (M.B) are employed at the rear end of the slab.

The high dimension of steel in the slab is chosen for main bars.

In one way slab, the support for slab is given at one direction or one side, so it becomes possible to set the main bar at the shorter extent.

Main reinforcement bars are arranged in one way slab in one side (At shorter extent), but in two way slab, the bars arranged in both ways.

In Flat plate slabs, the main bars are arranged in one direction at bottom of the distribution bar (in shorter extent).

Main bars in the slabs must not be under 8 mm when (H.Y.S.D) is applied or 10 mm when the Plain bars are applied.

FUNCTION OF DISTRIBUTION BARS: Distribution bars (D.B) in the slab, plays an important role as it can withstand the shear stress. As the cracks are formed to shear stress on the top of the slab, so, the distribution bars should have been arranged at the longer span of the slab.

Distribution bars (D.B) are arranged at the direction of longer extent in the slab.

Distribution bars (D.B) are applied at the Top of the main bars perpendicularly.

For Distribution bars (D.B), it is required to choose the smaller dimensions of steel, as it’s objective is to defy the shear stress in the slab.

The purpose  of distribution bar is to allocate the super imposed load uniformly or resist the Shrinkage stresses formed because of Temperature variation (Winter,Summer).

In Flat plate slab the, distribution bars are arranged in one direction at Top of main bar (in longer span).

The distribution bars should not be under 8 mm in a diameter or not be over 1/8 of the thickness of the R.C.C slabs.

The distribution bars are arranged to retain the mesh in perfect place and sustain the center to center c/c, bars spacing among main bars.

Tips: While wiping down any filthy woodwork with the use of a household cleaner, sand the trim, doors, and windows with 120-grit silicon-carbide sandpaper. Sand old paint with the purpose of sticking new paint.

Sand bare wood to take away raised grain, level off wood filler, and soften over sharp edges. Once the sanding is completed, dust the woodwork by applying an old, damaged brush or a shop vacuum. Then with the help of a tack cloth, wipe it clean.

Apply primer to dry during the night, then sand smoothly with 220-grit sandpaper. Purge the dust with vacuum and tack cloth prior to use final coats.

Prior to apply a brush, soak it with water (for latex paint) or paint thinner (for oil-based paint). Deform the bristles to enter the fluid into the brush's ferrule. Spin or tap the brush dry to make it to clean, in so doing expanding its life.

Watch the following video to learn the complete process.

The calculation is made on the basis of the following dimension :-

The size of brick exclusive of cement mortar is taken as 190 x 90 x 90 mm The density of cement mortar is taken as 10 mm Proportion of cement mortar is given as 1:6 The volume of brickwork is taken as 1M3 The calculation will be as follow :- Volume of 1 brick exclusive of cement mortar = 0.19 x 0.09 x 0.09 = 0.001539 M3 (after converting the value to meter)

Volume of 1 brick including cement mortar = 0.2 x 0.1 x 0.1 = 0.002 M3 (after adding density of mortar with brick)

Number of bricks having cement mortar in 1M3 = 1/0.002 = 500 nos. (volume of brickwork = 1 and volume of 1 brick with cement mortar = 0.002 M3)