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. Single Double Ordinary Unit Solid or Hollow cast
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. 4, 6, 12, 20, 25 Thickness Standard Commercial Sizes
0.90 x 1.80 meters
1.20 x 2.40 meters

14. FORMS FOR SQUARE AND RECTANGULAR COLUMN

15. The thickness of the board to be used. The size of the frame.
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. P=1.40 A. Solving for the Plywood
1) Find the lateral perimeter of one column using the formula
P= 2(a+b) +0.20
P= 2( ) +0.20
P=1.40

18. 2) Multiply P by the column height and the number of columns to
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
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= 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 = 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: 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: X 25 = 1, 504 meters

26. 6) Convert to commercial length of steel bars says 6. 00 meters long
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: X 9.52 = 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
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
2) Substitute data in the formula:
P=2(0.60) =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
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
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= 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 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.