Forms, Scaffolding and Staging

7/22/2019 Forms, Scaffolding and Staging

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FORMS, SCAFFOLDING and

STAGING


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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 man

they are easily removed and reassembled without damage to th

or to the concrete.


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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


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CLASSIFICATION

OF FORMS


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As to Materials

Wood

Metal

Plastic

Composite


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As to Shape

Straight Circular, etc


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Solid or Hollow cast

Single

Double

As to Methods of Construction

Ordinary

Unit


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As to Uses

Foundation

Wall

Steps

Beam and Girders

Slab

Sidewalk, etc


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Construction of forms consist of:

Retaining Board

Supporter or Studs

Braces

Spacer

Tie wire

Bolts and Nails


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Types of Post and Wall Form

Continuos

Full unit

Layer unit

a) Continuos

b) Sectional


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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 waterproofb Prevent the adherence of concrete into the pores of the wood


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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


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Thickness

4, 6, 12, 20, 25

Standard Commercial Sizes

0.90 x 1.80 meters

1.20 x 2.40 meters


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FORMS FOR

SQUARE AND

RECTANGULARCOLUMN


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Consideration in determining the materials for square and rectangular colum

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


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Example No.1

Six concrete posts at 4

with a uniform cross

dimensions of 0.30 X

the use of 6mm (1/4

on a 2X2 wood fram

materials required


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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


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2) Multiply P by the column height and the number of columns to find

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

required.

No. of Plywood : (33.6/2.88) = 11.7 say 12 pcs.


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B. Solve for the 2X2 wood frame by direct counting

From Figure 5-2, by direct counting of the frame:

12 pcs. 2X2X16 = 56 bd ft.1 pcs. 2X2 X10= 3.3 bd. ft

________________________

Total = 356 bd ft


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C. Solving the 2X 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

by 29.67.

12 Plywood X 29.67= 356 board foot.

2) Order: 12 pcs.

1.20 X 2.40 (4X8) plyw

356 board

2X2 lum


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FORMS FOR

CIRCULARCOLUMN

From Figure 5 4 determine the required metal black sheet form for 8 circula


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From Figure 5-4, determine the required metal black sheet form for 8 circula

4.00 meters high each with a uniform cross- sectional diameter of 60 centime


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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

4) Fi d th b f h t i d R f t T bl 5 2


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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.


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5) Find the number of Vertical Support (ribs) at 15 cm spacing distance. Refer

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:


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6) Convert to commercial length of steel bars says 6.00 meters long. Divide:

1,504/6.00 =251 pcs. (consult the plan what

bars used)

7) Solve for the Circumferential Ties. Again, refer to Table 5-2.


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Multiply:

Ties: 60.16 X 9.52 = 572.7 say 573 meters

8) Convert to commercial length of steel bars say 6.00 meters D

573.00/6.00= 95.5 say 96 pcs( consult the p

kind of bars used)

FORMS FOR


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FORMS FORBEAM AND

GIRDER

Ten concrete beams with cross sectional dimensions of 0.30 by 0.60 meter has a uniform


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4.50 meters. Using 4X8 plywood form on 2X2 lumber frame. List down the mat

A. Finding the Plywood Form


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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

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 2X2 Wood Frame


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1) Refer to Table 5-1. Under column beam using 6mm thick plywood on 2 X2 fram

25 X 25.06=626 bd. ft.

2) Order : 25 pcs. X 4 X 8 plywood form

626 board ft. 2 X2 lumber


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Scaffolding andStaging


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Scaffolding

caffolding is a temporary structure of wooden polesand planks providing platform for workers to stand

on while erecting or repairing of building. It is

further defined as temporary framework for other

purposes.


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Staging

tagingis a more substantial frameworkprogressively 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.


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The different parts of scaffolding to consider are:

Vertical SupportBase of Vertical Support ( as needed)

Horizontal member

Diagonal Braces

Blocks and weighs

Nails or bolts


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Cost of forms refer to:

Initial cost of materials

Assembling cost

The number of times it could be

usedDurability to resist pressure, and

tear and wear


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ESTIMATING

SCAFFOLDINGAND STAGING

A reinforced concrete building has 9 columns with a clear height of 4 00 meters


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A reinforced concrete building has 9 columns with a clear height of 4.00 meters

figure 5-8. Determined the required scaffolding under the following specification

Vertical support: 2 X2 Horizontal and Diagonal braces.

A. Scaffolding for Columns


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1) Find the total length of the 9 columns.

4.00 X 9 columns= 36 meters

2) Refer to Table 5-3. Using 2X 3 vertical support, multiply:

36 X 7.00= 252 bd. ft 2X 3 X 14 ft.


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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 Beams1) Find the total length of 6 beams


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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.

C. Scaffolding for Concrete Slab


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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 2X 3 support, multiply:

72 X 9.10= 655 bd. ft.


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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


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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 3Floor Slab Form..25 4 X 8 plywood

STEEL PIPE


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STEEL PIPE

SCAFFOLDINGS

Steel pipe scaffolding can be used freely to prefabricate height and width

to the places and forms to install


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to the places and forms to install.


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Documents

Forms, Scaffolding and Staging