HOW TO ESTIMATE THE COST OF A HOUSE (page 1)
It is important to how much a could cost from the beginning before initiating the project, the reasons for some uncompleted projects are as a result of not taking time to calculate the possible amount of money that could go into before it could be completed to a point someone could rent it out or inhabit it.
Many people make the mistake of starting a project without proper calculations in regard to their monthly income and the specified duration the could be ready according to their monthly savings and the total cost for the .
The following are the steps needed for easy calculations of an amount a could cost in naira value:
HOW MANY WALLS
the design of the how many walls that the would have, walls means the places will be raised to make up the entire . The number of the walls will determine how many that will be needed to erect all the walls altogether. If the total walls for a three bedroom flat are 15 walls altogether. The standard height of every should be between 12feet to 15 feet height, and every is made with 1 feet height and 2 feet breadth, while its widths ranges from 5inches up to 12 inches i.e. (0.42 feet to 1 feet), so with this we can obtain the total number of needed to erect the entire walls; below is the procedure to take:
Take the measurement of each walls length and height and calculate it by the simple formula for AREA=length x breadth assuming the value for a particular wall is length 20feet and height 15feet then the number of needed to erect such wall will be (20 X 15)/(1 X 2)= 300/2=150 required for that wall, do this to all the walls in the design and sum up the total to get an accurate number of you will need for your project and then consider which size of the thickness you may need ranging from the 5inch to 12inch and obtain the price of each from , for example, if 5inch is sold for N100 per , then the price of needed for the wall calculated above is N100 x 150= N15, 000.
THICKNESS OF NEEDED
Just as earlier mentioned the thickness of the needed in a depends on the overall height of the . If a is to have multiple floors over one another such as upstairs of one or two stories, the initial which constitutes the foundation of the may have its walls laid by 12inch , while the second which is referred to as the decking of the upstairs may have its wall raised with 9inch , and so forth. So the keyword is that the taller the the more the sequential arrangement of according to the inches to attain the height, the raising of walls will always start from bigger inches to the smallest in respect to the height from the to the roof.
A situation where the is not an upstairs may be a bungalow or a simple duplex that the total height does not exceed 30 feet the walls of the may be raised by 9inch or 6inch depending on the money available with the builder and the quality of the individual wants to achieve.
THE HEIGHTS OF THE WALLS
The height of the walls of the is another thing to consider, the average height of most wall is from 11 feet to 13 feet depending on the land topography of the and the type of the .
A bungalow may have the height mentioned above, but many upstairs and duplex have a lesser height in their walls for each , because if they are to have such height on each it could make the very tall and incur more cost of to the builder. Therefore, some of them have their height limited to 11feet or less.
Your calculation should be based on the total height of the walls the will have after it has been completed.
A good layer can estimate accurately the number of that can be laid on the walls per a bag of , this number of are used to divide the total number of calculated for the entire walls to give you the total number of the bags of required to lay all the on the walls.
SANDS PER WALLS
Though most builders do not measure the quantity of sand required for raising the walls of the due to the fact that it is readily available and cost less than other materials, but for the purpose of economics management in it becomes necessary to know the actual quantity of sand needed for the raising of the entire walls of the . If 4 wheelbarrow of sand was mixed with a bag of to lay about 100 it means that the total number of for the would be divided into 100 and the result is multiplied by 4, to give the total number of wheelbarrow of sand needed to raise all the walls of the . So the wheelbarrow of sand is now used to make an order on the quantity of sand needed.
STEEL RODS NEEDED PER WALL
The steel rods are used for the supports in the , they are used for the making of and lintels both for the and windows and for the chaining of the .
Obtaining the total quantity of rods needed for a may be easy if a careful approach is employed during your calculation. If the beams were to be tied in angle design it means 3 rods per pillar or wall for the lintel , but if the beams are to be formed in rectangle design it means 4 rods for a pillar or lintel on the wall. The total length of all the walls are put together to measure it with the lengths of rods which are mostly 38 feet to 40 feet in length, the total number rods for the total lengths of the walls will be multiplied into 3 time or four times depending on the design the and lintels would carry.
Note, that sizes of steel rods used in vary according to the size of and the height of the , which also includes the type of the . For a bungalow and small duplex the steel rods may be from 10mm to 14mm rods, also note that the smaller the size the number of rods needs on the beams of the and lintel but larger sizes require either three or four rods per pillar or beam.
ESTIMATED NUMBER OF WOOD BOARDS
The wood boards will be needed for the making of the lintels and . Before the or lintels are made they are first shaped by the action of wood fixed together with nails and the inside of the wood board is confirmed to be in accordance with the actual shape required for the pillar or lintel.
Obtaining the total number of wood boards needed for a is simply dividing the total lengths of the walls of the with the length of a particular wood board and then multiply it by two. let’s assume that the total length of the walls is 5000feet, and a single wood board is 15feet, therefore the number of wood board required for such would be, 5000/15 =334 wood boards, therefore multiply it with two because the boards will be on the either sides of the walls, so it becomes 334 x 2 = 668 wood boards.
HEIGHT OF THE ROOF AND WOOD REQUIRED FOR IT
The height of the roof determines the number of wood required as well as the sizes of the woods for the of the .
The modern buildings do have from 9feet up to 15 feet heights depending on the design of the . The widths of the are what determine the that will be created during the skeleton of the so to obtain the quantity of wood the will need, simply calculate the lines that the wood would make on the skeleton of the roof by assuming the width of the sheet to be 4 feet, therefore at every 4feet gap there will a line of wood in it running from the up to the down of the design, the same is applied to every segment of the roof, and then sum up the lengths obtains, by 16 feet which is the market length for 2 by 2 plywoods currently sold at N300 for one, also take the length of all the walls to obtain the number of 2 by 4 plywood that will be laid on the top of the wall on the surface of the last and divide it by 30 feet the usual length of 2 by 4 ply woods to obtain the total for it.
The next step is to calculate the size of 2 by 4 ply woods that will be used in lifting the heights of the house, another calculation will be the 2 by 4 ply wood that will be running across the width of the entire at each line created by the 2 by 2 ply wood. The number will be determine by the height of the roof and the number of lines created by the 2by 2 plywood and this also will depend on the width of the entire .
Additional woods can approximately be added in the case of joining and unforeseen designs.
Note, that the above calculation is with the exception of the roof for facial boards, and the wood for of the roof.
(THIS IS JUST THE PAGE ONE THE FOLLOWING IN THE NEXT PAGES UNDER THIS CATEGORY)
NUMBER OF ALUMINUM FOR ROOF
NUMBER OF ALUMINUM FOR FACIAL BOARDS
NUMBER OF WOODS FOR FACIAL BOARDS
NUMBER OF WOODS FOR PVC SKELETON
COST OF PVC AND FIXING COST
COST OF ONE BUNDLE OF PVC AND QUANTITY PER ROOM
COST OF PLASTERING SAND AND PLASTERING
COST OF AND
COST OF ELECTRIFYING THE HOUSE
COST OF TOILET SEATERS
COST OF SAND FILLING
COST OF GERMAN
COST OF TILING A
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