The document discusses forms, scaffolding, and staging used in concrete construction. It describes the different types and materials used for forms, including wood, metal, plastic, and composite. Forms are classified based on material, shape, construction method, and use. Proper form selection considers cost, strength, and durability. The document also provides guidance on estimating the materials needed for scaffolding and forms for columns, beams, slabs, and other concrete elements.

1. FORMS,
SCAFFOLDING and
STAGING

2. FORMS
 It is a temporary boarding, sheating or pan
used to produce
the desired shape and size of concrete.
 Forms must be simple and economically
designed in such manner that they are easily
removed and reassembled without damage to
themselves or to the concrete.

3. SELECTION OF FORMS ARE BASED ON
Cost of the Materials
Construction and assembling cost
The number of times it could be used
Strength and resistance to pressure and
tear and wear

4. CLASSIFICATION OF
FORMS

5. As to Materials
Wood
Metal
Plastic
Composite

6. As to Shape
Straight Circular, etc

7. Solid or Hollow cast
Single
Double
As to Methods of Construction
Ordinary
Unit

8. As to Uses
Foundation
Wall
Steps
Beam and Girders
Slab
Sidewalk, etc

9. Construction of forms consist of:
 Retaining Board
 Supporter or Studs
 Braces
 Spacer
 Tie wire
 Bolts and Nails

10. Types of Post and Wall Form
Continuos
Full unit
Layer unit
a) Continuos
b) Sectional

11. Greasing of Forms
 Forms are constantly greased before its use.
 Crude oil is the most economical and satisfactory materials for this
purpose
PURPOSE:
a) To make the wood waterproof
b) Prevent the adherence of concrete into the pores of the
wood

12. Plywood as Form has the following advantages
 It is economical in terms of labor cost.
 It is lightweight and handy
 It has smooth surface which may not require
plastering
 Less consumption of nails
 Ease of assembling and disassembling
 Available

13. Thickness
4, 6, 12, 20, 25
Standard Commercial Sizes
0.90 x 1.80 meters
1.20 x 2.40 meters

14. FORMS FOR SQUARE
AND RECTANGULAR
COLUMN

15. Consideration in determining the materials for square and
rectangular column forms
The thickness of the board to be
used.
The size of the frame.
Types of frameworks to be
adopted
a) Continuous rib type
b) Stud type

16. Example No.1
 Six concrete posts at 4.00
meters high with a
uniform cross sectional
dimensions of 0.30 X
0.30m. specify the use of
6mm (1/4”) marine
plywood on a 2”X2” wood
frame. List down the
materials required

17. A. Solving for the Plywood
1) Find the lateral perimeter of one column using the formula
P= 2(a+b) +0.20
P= 2(0.30+0.30) +0.20
P=1.40


18. 2) Multiply P by the column height and the number of
columns to find the total area of forms.
Area=1.40 X 4.00 X 6 columns
A= 33.6 sq. m.
3) Divide this area by 2.88, the area of one plywood to get
the number of plywood required.
No. of Plywood : (33.6/2.88) = 11.7 say 12 pcs.

19. B. Solve for the 2”X2” wood frame by direct counting
From Figure 5-2, by direct counting of the frame:
12 pcs. 2”X2”X16’ = 56 bd ft.
1 pcs. 2”X2” X10’= 3.3 bd. ft
________________________
Total = 356 bd ft

20. C. Solving the 2”X 2” frame with the Aid of Table 5-2
1) Refer to Table 5-1. For 2X2 frame under Post 6 mm (1/4”) thick,
multiply the number of plywood found by 29.67.
12 Plywood X 29.67= 356 board foot.
2) Order: 12 pcs.
1.20 X 2.40 (4’X8’) plywood
356 board feet
2”X2” lumber

21. FORMS FOR
CIRCULAR COLUMN

22. From Figure 5-4, determine the required metal black sheet
form for 8 circular columns 4.00 meters high each with a
uniform cross- sectional diameter of 60 centimetres.

23. Solution:
1)Solve for the circumference of one column
C= 3.1416 X 0.60m. = 1.88 meters
2) Multiply by column height to find the surface area
Area: 1.88 X 4.00 = 7.52 sq. m
3) Find the area of the 8 columns, multiply
Total surface area: 7.52 X 8 = 60.16 sq. m

24. 4) Find the number of sheet required. Refer to Table 5-2.
Using 1.20 X 2.40m. black sheet, multiply:
No. of sheet: 60.16 X 0.347= 21 pcs.

25. 5) Find the number of Vertical Support (ribs) at 15 cm
spacing distance. Refer again to Table 5-2.
Multiply:
Vert. support: 60.16 X 25 = 1, 504 meters

26. 6) Convert to commercial length of steel bars says 6.00
meters long. Divide:
1,504/6.00 =251 pcs. (consult the plan what
kind of steel bars used)
7) Solve for the Circumferential Ties. Again, refer to Table 5-
2.

27. Multiply:
Ties: 60.16 X 9.52 = 572.7 say 573 meters
8) Convert to commercial length of steel bars say
6.00 meters Divide:
573.00/6.00= 95.5 say 96 pcs( consult the
plan what kind of bars used)

28. FORMS FOR BEAM
AND GIRDER

29. Ten concrete beams with cross sectional dimensions of 0.30 by 0.60
meter has a uniform clear span of 4.50 meters. Using ¼” 4’X8’
plywood form on 2”X2” lumber frame. List down the materials
required.

30. A. Finding the Plywood Form
1) Find the lateral perimeter of the beam
P=2(d) + b + 0.10
2) Substitute data in the formula:
P=2(0.60) + 0.30 + 0.10=1.60
3) Multiply P by the length and number of beams to get the
area of the forms.
Area: 1.60 X 4.50m. X 10 columns
A=72 sq. m.
4) Divide by 2.88 to get the number of plywood required.
No. of Plywood : 72/2.88 = 25 pcs

31. B. Solving for 2”X2” Wood Frame
1) Refer to Table 5-1. Under column beam using 6mm ¼ “ thick plywood
on 2” X2” frame, multiply:
25 X 25.06=626 bd. ft.
2) Order : 25 pcs. ¼ “ X 4’ X 8’ plywood form
626 board ft. 2” X2” lumber

32. Scaffolding and
Staging

33. Scaffolding
Scaffolding is a temporary
structure of wooden poles and
planks providing platform for
workers to stand on while erecting
or repairing of building. It is
further defined as temporary
framework for other purposes.

34. Staging
Staging is a more substantial
framework progressively built
up as a tall building rises up.
The term staging is applied
because it is built up in stages
one story at a time.

35. The different parts of scaffolding to consider
are:
Vertical Support
Base of Vertical Support ( as needed)
Horizontal member
Diagonal Braces
Blocks and weighs
Nails or bolts

36. Cost of forms refer to:
Initial cost of materials
Assembling cost
The number of times it could be used
Durability to resist pressure, and tear
and wear

37. ESTIMATING
SCAFFOLDING AND
STAGING

38. A reinforced concrete building has 9 columns with a clear
height of 4.00 meters as shown on figure 5-8. Determined the
required scaffolding under the following specifications: 2” X 3”
Vertical support: 2” X2” Horizontal and Diagonal braces.

39. A. Scaffolding for Columns
1) Find the total length of the 9 columns.
4.00 X 9 columns= 36 meters
2) Refer to Table 5-3. Using 2”X 3” vertical support,
multiply:
36 X 7.00= 252 bd. ft 2”X 3” X 14 ft.

40. 3) Find the horizontal supports. Refer to Table 5-3, using 2”
X 2” lumber, multiply:
36 X 21.00= 756 bd. ft. 2” X 2” lumber
4) Find the diagonal braces. From Table 5-3, multiply:
36 X 11.7= 421 bd. ft. 2” X 2” lumber

41. B. Scaffolding for Beams
1) Find the total length of 6 beams
Length: ( 4.50 X 6) + (4.00 X 6)= 51 meters
2) Refer again to Table 5-3
a) For vertical support using 2” X 3” lumber, multiply:
51 X 6.00 = 306 bd. ft.
b) For horizontal support using 2” X 2” lumber,
multiply
51 X 4.70 = 240 bd. ft.

42. C. Scaffolding for Concrete Slab
1) Find the area of the concrete floor slab
Area= 4.50 X 4.00 X 4 units = 72 sq. m
2) Refer to Table 5-3. Using 2”X 3” support,
multiply:
72 X 9.10= 655 bd. ft.

43. D. Floor Slab Forms
1) Find the floor area:
Area =( 4.50 X 4.00 X 4 units) = 72 sq. m.
2) Divide by 2.88 effective covering of one plywood
72/ 2.88 = 25 pcs. 4’ X 8’ marine plywood

44. Summary of the Materials:
For Columns.................. 252 bd. ft. 2” X 3”
1,177 bd. ft. 2” X 2”
For Beams…………………..306 bd. ft. 2” X 3”
240 bd. ft. 2” X 2”
For Slab……………………….655 bd. ft. 2” X 3”
Floor Slab Form…………..25 4’ X 8’ plywood

45. STEEL PIPE
SCAFFOLDINGS

46. Steel pipe scaffolding can be used freely to
prefabricate height and width according to the places
and forms to install.