# How to Estimate Formworks of Columns, Beams, and Girders

*Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.*

## 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 it is 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 it can withstand the pressure and tear and wear. There are five examples provided. The first problems below explain how to estimate plywoods and wood frames. Also, you will learn how to estimate metal black sheets, vertical supports, and circumferential ties for circular formworks.

## 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 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 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 a wood framework for formworks. Formworks for columns start from the boundary between footing and 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 of dimensions 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

**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 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 it by 8.

Total Area = Area ( Number of columns )

Total Area = 3.00 ( 8 )

Total Area = 24 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 = 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 foot )

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 is 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

**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 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 it 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 foot )

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 are 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 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.

**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.

Length = 94.20 ( 25 )

Length = 2,355.00 meters

2. The commercial length of vertical ribs is 6.00 meters. Divide the total length obtained by 6.00 to get the number of pieces.

Number of pieces = Total Length / 6.00 meters

Number of pieces = 2,355.00 / 6.00

Number of pieces = __393 pieces of vertical support ribs__

**C. Solving for Circumferential Ties for Circular Columns**

1. Multiply the total area obtained by the multiplier in Table 4. The multiplier is 9.52 meters.

Length = 94.20 ( 9.52 )

Length = 897 meters

2. The commercial length of steel bars is 6.00 meters. Divide the total length obtained by the commercial length of steel bars.

Number of pieces = Total Length / 6.00 meters

Number of pieces = 897.00 / 6.00

Number of pieces = __150 pieces of circumferential ties__

## Formworks for Beams and Girders

Just like in square columns, formworks for beams and girders use plywood as well. But you still have the option to use metal sheets. There is a slight difference in the formula for the solving the formworks of beams. Since beams are horizontal structural elements, only three faces of the beam use plywood forms. These faces are the side faces and the bottom face. Covering of top faces are not allowed since it will serve as the passage of concrete.

## Problem 4: Formworks for Concrete Beams

A residential house has eight concrete beams having dimensions of 0. 30 m x 0.40 m. x 3.00 m. Estimate the following materials needed:

a. 1/4" x 4' x 8' Phenolic Plywood

b. 2" x 2" Wood Frame

**A. Solving for 1/4" x 4' x 8' Phenolic Plywood**

1. The formula for solving the required number of pieces of plywood for formworks of beams is P = 2 ( d ) + b + 0.10. 'P' is the perimeter of the three sides you want to estimate, 'd' is the length of the vertical side and 'b' is the bottom form. The constant value 0.10 is the value to consider for the lapping of form joints.

Perimeter = 2 ( d ) + b + 0.10.

Perimeter = 2 ( 0.40 ) + 0.30 + 0.10

Perimeter = 1.20 meters

2. Multiply the value of perimeter obtained by the length of the beam. The resulting value is the total surface area of the three faces of the beam.

Area = Perimeter ( Length of a beam )

Area = 1.20 ( 3.00 )

Area = 3.60 square meters

3. Solve for the total area of the beams. Multiply the obtained area by the number of beams. There are eight concrete beams so multiply it by 8.

Total Area = Area ( Number of beams )

Total Area = 3.60 ( 8 )

Total Area = 28.80 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 = 28.80 square meters / 2.88 square meters

Number of pieces =__ 10 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 = 10 pieces ( 29.67 board foot )

Total Board Foot = 296.7 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 = 297 / ( (2) (2) (20/12) )

Number of pieces = __45 pieces of 2" x 2" x 20' wood frame / lumber__

## Problem 5: Formworks for Concrete Girders

There are four concrete girders in a residential house. The girder has a general dimension of 0.40 m x 0.60 m x 6.00 m. Estimate the following materials:

a. 1/2" x 4' x 8' Marine Plywood

b. 2" x 3" Lumber Frame

**A. Solving for 1/2" x 4' x 8' Marine Plywood**

1. The formula for solving the required number of pieces of plywood for formworks of girders is P = 2 ( d ) + b + 0.10. 'P' is the perimeter of the three sides you want to estimate, 'd' is the length of the vertical side and 'b' is the bottom form. The constant value 0.10 is the value to consider for the lapping of form joints.

Perimeter = 2 ( d ) + b + 0.10.

Perimeter = 2 ( 0.60 ) + 0.40 + 0.10

Perimeter = 1.70 meters

2. Multiply the value of perimeter obtained by the length of the girder. The resulting value is the total surface area of the three faces of the girder.

Area = Perimeter ( Length of a girder )

Area = 1.70 ( 6.00 )

Area = 10.20 square meters

3. Solve for the total area of the girders. Multiply the obtained area by the number of girders. There are four concrete girders so multiply it by 4.

Total Area = Area ( Number of girders )

Total Area = 10.20 ( 4 )

Total Area = 40.80 square 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 = 40.80 square meters / 2.88 square meters

Number of pieces = __15 pieces of 1/2" x 4' x 8' Marine Plywood__

**B. Solving for the 2" x 3" Wood Frame**

1. Multiply the number of marine plywood obtained by the multiplier from Table 3.

Total Board Foot = 15 pieces ( 44.50 board foot )

Total Board Foot = 668 board feet of 2" x 3" x 20' wood frame / lumber

Number of pieces = Total Board foot / Volume of Lumber in cubic inches

Number of pieces = 668 / ( (2) (3) (20/12) )

Number of pieces = __67 pieces of 2" x 3" x 20' wood frame / lumber__

**© 2018 Raycaptiosus**

## Comments

**Raycaptiosus (author)** from Philippines on August 20, 2020:

Hi Dominic, you can allot about 20% - 35% of the total cost for labor cost for residential houses. But, it is still up to you if you want to go higher or lower the given range. Hope it helps.

**Raycaptiosus (author)** from Philippines on August 11, 2020:

Hi, Maykel! Phenolic boards can be used 2-5 times depending on the brand. There are also brands that can be used for not less than 5 times.

**Maykel** on August 10, 2020:

How many times can you use the phenolic boards as forms po (for large construction projects)??

**Raycaptiosus (author)** from Philippines on July 31, 2020:

Thank you, Reynaldo!

**Al Trell** on July 22, 2020:

Can you help me estimate some structural drawings? just the bill of materials.

**Blizzflickz** on July 02, 2020:

formworks for slab?

**Joana** on July 01, 2020:

Hi I just want to ask how did you get multiplier for wood frame lumber?

**Raycaptiosus (author)** from Philippines on June 22, 2020:

Noted! Just in case you would like me to send you estimation materials/resources, feel free to e-mail me and I'll happily help you.

**GENIEL** on June 22, 2020:

Estimate of Rebars, Painting, Ceiling, Nails, Scaffolding and Staging. Thanks

**Raycaptiosus (author)** from Philippines on March 08, 2020:

What other topics you would like me to cover regarding estimation in construction? Please leave your comments here!

**shay** on February 25, 2020:

Can the Multiplier for Wood Frames or Lumber table used for 10ft. lumber as well?

**Raycaptiosus (author)** from Philippines on January 06, 2020:

You're very welcome. If you have any questions regarding the article just write it in the Questions and Answers capsule.

**ren** on December 06, 2019:

very informative. thanks