How to Estimate Formworks of Columns, Beams, and Girders
Formworks for Construction
Formworks are frameworks of materials used to enclose concrete mixtures of different shapes and sizes. It is a temporary structure built in a way that makes it easy to assemble and disassemble. Formworks must be economical, stable, and reusable. The materials used for formworks are wood, metal, plastic, and composite materials, but wood and metal are the two most popular construction materials used for formworks.
In building formworks, you have to make sure that they can withstand pressure, tear and wear. There are five examples provided. The first few problems below explain how to estimate plywood and wood frames. Also, you will learn how to estimate metal black sheets, vertical supports, and circumferential ties for circular formworks.
Estimating formworks
John Ray Cuevas
Formworks for Square and Rectangular Columns
Formwork for square and rectangular columns consists of plywood forms surrounding the four long lateral faces of a column. It is wise to use plywood in making formwork for square and rectangular columns because it is economical and handy. There are a few things to consider in estimating plywood and wood frames. Consider first the size of the plywood.
There are few commercial sizes for plywood. Consider next the size of the lumber or wood frame. Lumber or a wood frame supports the plywood. Lastly, consider the type of framework to use. The continuous rib type and stud type are the two types of wood frameworks for formworks. Formworks for columns start from the boundary between the footing and the column. Learn the step-by-step procedure from the examples below.
| Thickness (mm) | Width (m/ft) | Length (m/ft) |
|---|---|---|
4 | 0.90 m. (3 ft.) | 1.80 m (6 ft.) |
4 | 1.20 m. (4 ft.) | 2.40 m. (8 ft.) |
6 | 0.90 m. (3 ft.) | 1.80 m (6 ft.) |
6 | 1.20 m. (4 ft.) | 2.40 m. (8 ft.) |
12 | 0.90 m. (3 ft.) | 1.80 m (6 ft.) |
12 | 1.20 m. (4 ft.) | 2.40 m. (8 ft.) |
20 | 0.90 m. (3 ft.) | 1.80 m (6 ft.) |
20 | 1.20 m. (4 ft.) | 2.40 m. (8 ft.) |
25 | 0.90 m. (3 ft.) | 1.80 m (6 ft.) |
25 | 1.20 m. (4 ft.) | 2.40 m. (8 ft.) |
| Area (Square Inches) | Lengths (ft) | Length (m) |
|---|---|---|
2" x 2" | 6 | 1.83 |
2" x 3" | 8 | 2.44 |
10 | 3.05 | |
12 | 3.666 | |
14 | 4.27 | |
16 | 4.88 | |
18 | 5.49 | |
20 | 6.10 | |
22 | 6.71 | |
24 | 7.32 |
| Size of Wood Frame/Lumber | 1/4" Thickness | 1/2" Thickness |
|---|---|---|
2" x 2" | 29.67 board foot | 20.33 board foot |
2" x 3" | 44.50 board foot | 30.50 board foot |
Problem 1: Continuous Rib-Type Form for Square Columns
A storage warehouse consists of eight concrete posts with dimensions of 0.20 m x 0.20 m x 3.00 m. If you are to create a continuous rib-type formwork for this warehouse, how much will you need for the following materials?
a. 1/4" x 4' x 8' Phenolic Plywood
b. 2" x 2" x 20' Wood Frame
Estimating formworks of square columns
John Ray Cuevas
A. Solving for the 1/4" x 4' x 8' Plywood Form
1. The formula for solving the required number of pieces of plywood for formworks of square columns is P = 2 ( a + b ) + 0.20. P is the perimeter of the column you want to estimate, a is the shorter side, and b is the longer side. The constant value 0.20 is the value to consider for the lapping of form joints.
Perimeter = 2 ( a + b ) + 0.20
Perimeter = 2 ( 0.20 + 0.20 ) + 0.20
Perimeter = 1.00 meter
2. Multiply the value of the perimeter obtained by the height of the column. The resulting value is the total surface area of the lateral faces of the column.
Area = Perimeter ( Length of a column )
Area = 1.00 ( 3.00 )
Area = 3.00 square meters
3. Solve for the total area of the columns in the storage warehouse. Multiply the obtained area by the number of columns in the storage warehouse. There are eight concrete posts, so multiply by 8.
Recommended
Total Area = Area ( Number of columns )
Total Area = 3.00 ( 8 )
Total Area = 24 square meters
4. Solve for the area of the plywood that you will use. In this case, the required size of plywood is 4' x 8'. Converting that to meters, it is equal to 1.20 meters x 2.40 meters.
Area of plywood = 1.20 (2.40)
Area of plywood = 2.88 square meters
5. Divide the total area by the area of one plywood.
Number of pieces = Total Area / Area of plywood
Number of pieces = 24 square meters / 2.88 square meters
Number of pieces = 8.333 = 9 pieces of 1/4" x 4' x 8' Phenolic Plywood
B. Solving for the 2" x 2" Wood Frame
1. Multiply the number of phenolic plywood obtained by the multiplier from Table 3.
Total Board Foot = 9 pieces ( 29.67 board feet )
Total Board Foot = 267 board feet of 2" x 2" x 20' wood frame / lumber
2. Get the number of pieces of wood lumber by dividing the total board foot by the volume of the lumber in cubic inches.
Number of pieces = Total Board foot / Volume of Lumber in cubic inches
Number of pieces = 267 / ( (2) (2) (20/12) )
Number of pieces = 41 pieces of 2" x 2" x 20' wood frame / lumber
Problem 2: Continuous Rib Type Form for Rectangular Columns
Ten concrete posts' dimensions are 0.40 x 0.50 x 6.00 m. Estimate the required amount of the following materials:
a.1/4" x 4' x 8' Phenolic Plywood
b. 2" x 3" Wood Frame
Estimating formworks of rectangular columns
John Ray Cuevas
A. Solving for the 1/4" x 4' x 8' Phenolic Plywood Form
1. The formula for solving the required number of pieces of plywood for formworks of rectangular columns is P = 2 ( a + b ) + 0.20. P is the perimeter of the column you want to estimate, a is the shorter side, and b is the longer side. The constant value 0.20 is the value to consider for the lapping of form joints.
Perimeter = 2 ( a + b ) + 0.20
Perimeter = 2 ( 0.40 + 0.50 ) + 0.20
Perimeter = 2.00 meters
2. Multiply the value of the perimeter obtained by the height of the column. The resulting value is the total surface area of the lateral faces of the column.
Area = Perimeter ( Length of a column )
Area = 2.00 ( 6.00 )
Area = 12.00 square meters
3. Solve for the total area of the columns. Multiply the obtained area by the number of columns. There are ten concrete posts, so multiply by 10.
Total Area = Area ( Number of columns )
Total Area = 12.00 ( 10 )
Total Area = 120 square meters
4. Solve for the area of the plywood you will use. In this case, the required size of plywood is 4' x 8'. Converting that to meters, it is equal to 1.20 meters x 2.40 meters.
Area of plywood = 1.20 (2.40)
Area of plywood = 2.88 square meters
5. Divide the total area by the area of one plywood.
Number of pieces = Total Area / Area of plywood
Number of pieces = 120 square meters / 2.88 square meters
Number of pieces = 41.67 = 42 pieces of 1/4" x 4' x 8' Phenolic Plywood
B. Solving for the 2" x 3" Wood Frame
1. Multiply the number of phenolic plywood obtained by the multiplier from Table 3.
Total Board Foot = 42 pieces ( 44.50 board feet )
Total Board Foot = 1869 board feet of 2" x 3" x 20' wood frame / lumber
2. Get the number of pieces of wood lumber by dividing the total board foot by the volume of the lumber in cubic inches.
Number of pieces = Total Board foot / Volume of Lumber in cubic inches
Number of pieces = 1869 / ( (2) (3) (20/12) )
Number of pieces = 187 pieces of 2" x 3" x 20' wood frame / lumber
Formworks for Circular Columns
You can't use plywood for formworks of circular columns. Plywood is not bendable. Instead, you use metal sheets for formworks of circular columns. You can either use plain galvanized iron sheets or black metal sheets.
The primary use of these sheets is formworks for circular, elliptical, and all other shapes with irregularity. They are perfect materials for formworks of circular columns because they conform to the shape of concrete.
| Size of Black Metal Sheets | Number of Black Metal Sheets Per Square Meter | Length of 15 cm. Spacing Vertical Ribs (m) | Length of 20 cm. Spacing Vertical Ribs (m) | Length of Circular Ties (m) |
|---|---|---|---|---|
0.90 m. x 2.40 m. | 0.462 | 25 | 18 | 9.52 |
1.20 m. x 2.40 m. | 0.347 | 25 | 18 | 9.52 |
Problem 3: Metal Black Sheets for Circular Columns
A two-story office building has ten circular concrete columns with a diameter of 50 centimeters and a height of 6.00 meters. Determine the required number of 0.90 m. x 2.40 m. metal black sheets, 20 cm. vertical supports, and circumferential ties for the circular columns.
Estimating formworks for circular columns
John Ray Cuevas
A. Solving for Metal Black Sheet for Circular Columns
1. Solve for the circumference of a circular column. The formula for the circumference of a circle is C = πD or C = 2πr. C is the circumference of the circle, D is the diameter of the circle, and r is the radius of the circle. The circular column has a diameter of 50 centimeters. Convert this measurement to meters:
Circumference = π ( 0.50 meters )
Circumference = 1.57 meters
2. Solve for the area of a circular column. Multiply the circumference obtained by the total height of a circular column. The height of the column given is 6.00 meters.
Area = Circumference ( Height of a column )
Area = 1.57 meters ( 6.00 meters )
Area = 9.42 square meters
3. Solve for the total area of the columns of the two-story office building. There are ten circular concrete columns in the building. Multiply the area of one column by the number of columns.
Total Area = Column Area ( Number of Columns )
Total Area = 9.42 ( 10 )
Total Area = 94.20 square meters
4. Solve for the number of black metal sheets required. Table 4 shows the number of black metal sheets per square meter. Multiply the total area by 0.462.
Number of sheets = Total Area ( 0.462 )
Number of sheets = 94.20 ( 0.462 )
Number of sheets = 44 sheets of 0.90 m. x 2.40 m. metal black sheets
B. Solving for Vertical Support Ribs for Circular Columns
1. Given the spacing distance of vertical supports is 20 cm, solve for the number of pieces of vertical ribs. Multiply the total area obtained by the multiplier in Table 4. The multiplier is 25.00 meters.