Reinforcing Steel Bars

RE-BARS

•  commonly used as a tension device in reinforced concrete and reinforced masonry structure.

• to strengthen and hold the concrete in compression.

• It is usually in the form of carbon steel bars or wires, and the surfaces may be deformed for a better bond with the concrete

Specifications and Grading

ASTM vs PNS Codes

Popular Nomenclature

Typical Application

Grade 33 / PNS 230

Structural GradeLow-rise Buildings and Low Loading

Conditions

Grade 40 / PNS 275

Intermediate GradeMedium-rise Structures /

Infrastructure Work

Grade 60 / PNS 415

High-Tensile GradeMedium & High-rise Structures /

Infrastructure

Specifications and Grading

Table of Standards

STANDARD WEIGHT OF DEFORMED ROUND STEEL BARS

Bar Diametermm∅

Unit WeightKg/m

Bar Commercial Length

6.0 m 7.5 m 9.0 m 10.5 m 12.0 m

10 0.616 3.696 4.620 5.544 6.468 7.392

12 0.888 5.328 6.660 7.992 9.324 10.656

16 1.578 9.468 11.835 14.202 16.569 18.936

20 2.466 14.796 18.495 22.194 25.893 29.592

25 3.853 23.118 28.898 34.677 40.457 46.236

28 4.834 29.004 36.255 43.506 50.757 58.008

32 6.313 37.878 47.348 56.817 66.287 75.756

36 7.990 47.940 59.925 71.910 83.895 95.880

Bar Splice• How to determine

splicing length of steel bars

Type of reinforcement:

– Tension Bars (25 x Bar size + 150mm)

– Compression Bars (20 x Bar size + 160mm)

Hooks and BendsNominal size

of bard 10 12 16 20 25 32 40

Hook allowance

h 180 210 240 280 340 400 480

Bend allowance

n 130 130 150 170 190 220 250

Radius r 30 36 48 60 75 96 120

Bends that form end anchorage

Semi- circular hooks

Reinforcement of Concrete Hollow Blocks

cm per Block per Sq.m per Block per Sq.m40 0.235 2.93 2 0.264 3.360 0.171 2.13 3 0.172 2.1580 0.128 1.6 4 0.138 1.72

Spacing Vertical Reinforcement Spacing layer

Horizontal ReinforcementLength of Bars in meters Length of Bars in meters

By Area Method: Solve for the area Spacing is given at plan, then refer to table above Multiply the area by the factor for both horizontal and

vertical then add. Convert to commercial length

Example:

Determine the number of CHB from a 4x3m wall, its vertical reinforcement spaced at 80cm, Horizontal reinforcement at every 3 layers.

Area = 4x3mArea = 12 sqm

Vertical Reinforcement , since spaced at 80cm.12 sqm x 1.60(factor) = 19.20 m longHorizontal bars at every 3 layers 12 sqm x 2.15(factor) = 25.80 m long

Total = 45 m Say, we use 6m comm’l length45 / 6m = 7.5 or 8 – 10mm x 6m long

# CHB = 12 x 13 = 156 pcs

Post and Column Reinforcement

• Determine the total length of the main steel bars ( Bend, height of footing – floor extended, depth of beam that will pass thru, floor slab thickness to which it will be connected)

Lateral Ties

• The ACI code provides that:

“All non pre-stressed bars for tied column shall be enclosed by a lateral ties of at least no. 3 in size for longitudinal bars no.

10 or smaller and at least no.4 in size for no. 11 to 18 and bundled longitudinal bars.”

• It just simply means that lateral ties spacing shall NOT exceed the following:1. 16 x the longitudinal bar diameter

2. 48 x the lateral tie bar diameter

3. The least dimension of the column

Stirrups for Beams and Girders

• Stirrups – structural reinforcing member that holds or binds together reinforcement of a beam or girder to a designed position.

Two types of Stirrups (commonly used)

Example:

A concrete beam with a cross sectional dim of 25x40 cm required 10mm open stirrups spaced. Prepare the order of 10mm steel bars for stirrups of 8 beams with the same category.

a. By direct counting there are 24 stirrups at 98cm or let’s say 1m long

b. Total number of stirrups = 192 pcsc. Using 6m – 10mm; divide = (192/6) = 32 pcs x 6m-10mmd. Your order: 32 pcs -10mm x 6m steel bars

End Part 1

One way reinforced concrete slab

Typical slab reinforcement

Sample:

Given data : Slab = 2.20 x 1.20 mUse 12mm diameter

a. L’ = 2.20 – 2 (0.04) = 2.12mb. W’ = 1.20 – 2(0.04) = 1.12mc. L’ / 0.40m = 2.12m/0/40m = 5.30d. W’/0.40cm = 1.12 /0.40 = 2.80e. Multiply 5.30 x 1.12 x 2 = 11.872m- one sidef. Multiply 2.80 x 2.12 x 2 = 11.872m – other sideg. Add = 23.744h. Divide comml length; say 6m = 3.96 or 4pcs- 12mm – 6m