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 MaterialsConstruction and assembling costThe number of times it could be

usedStrength and resistance to

pressure and tear and wear

CLASSIFICATION OF FORMS

As to Materials

Wood

Metal

Plastic

Composite

As to Shape

Straight Circular, etc

Solid or Hollow cast

SingleDouble

As to Methods of Construction

OrdinaryUnit

As to Uses

FoundationWallStepsBeam and GirdersSlabSidewalk, etc

Construction of forms consist of:

Retaining BoardSupporter or Studs

BracesSpacerTie wire

Bolts and Nails

Types of Post and Wall Form

ContinuosFull unitLayer unit

a) Continuosb) 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 meters1.20 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 columnsA= 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 surface area

Area: 1.88 X 4.00 = 7.52 sq. m

3) Find the area of the 8 columns, multiplyTotal 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 Form1) Find the lateral perimeter of the beam

P=2(d) + b + 0.102) 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 columnsA=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 SupportBase of Vertical Support ( as needed)Horizontal memberDiagonal BracesBlocks and weighsNails or bolts

Cost of forms refer to:

Initial cost of materialsAssembling costThe number of times it could be

usedDurability 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 Columns1) Find the total length of the 9 columns.

4.00 X 9 columns= 36 meters2) 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” lumber4) Find the diagonal braces. From Table 5-3, multiply:

36 X 11.7= 421 bd. ft. 2” X 2” lumber

B. Scaffolding for Beams1) Find the total length of 6 beams

Length: ( 4.50 X 6) + (4.00 X 6)= 51 meters2) 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.

C. Scaffolding for Concrete Slab1) Find the area of the concrete floor slab

Area= 4.50 X 4.00 X 4 units = 72 sq. m2) Refer to Table 5-3. Using 2”X 3” support,multiply:

72 X 9.10= 655 bd. ft.

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.