FORMS, SCAFFOLDING and STAGING
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.
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
CLASSIFICATION OF FORMS
As to Materials
Wood
Plastic
Metal
Composite
As to Shape
Straight
Circular, etc
Solid or Hollow cast Single
Double
As to Methods of Construction Ordinary
Unit
As to Uses Foundation Wall
Steps
Beam Slab
and Girders
Sidewalk,
etc
Construction of forms consist of: Retaining
Board
Supporter or
Studs
Braces Spacer
Tie Bolts
wire
and Nails
Types of Post and Wall Form Continuos Full
unit Layer unit a) Continuos b) Sectional
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
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
Thickness 4,
6, 12, 20, 25
Standard Commercial Sizes 0.90 x
1.80 meters 1.20 x 2.40 meters
FORMS FOR SQUARE AND RECTANGULAR COLUMN
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
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
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
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.
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
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
FORMS FOR CIRCULAR COLUMN
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.
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 Area: 1.88 X 4.00 = 7.52 sq. m
surface area
3) Find the area of the 8 columns, multiply Total surface area: 7.52 X 8 = 60.16 sq. m
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.
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
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.
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)
FORMS FOR BEAM AND GIRDER
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.
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
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
Scaffolding and Staging
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.
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.
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
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
ESTIMATING SCAFFOLDING AND STAGING
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.
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.
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
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, 51 X 4.70 = 240 bd. ft.
multiply
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,
72 X 9.10= 655 bd. ft.
multiply:
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
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
STEEL PIPE SCAFFOLDIN GS
Steel pipe scaffolding can be used freely to prefabricate height and width according to the places and forms to install.
