REINFORCING STEEL BARS

REINFORCING STEEL BARS FOR CONCRETE

REINFORCING STEEL BARS ARE CRUCIAL AND COSTLY IN CONCRETE
CONSTRUCTION, REQUIRING ACCURATE ESTIMATING. HOOKS, SPLICES,
AND TIE WIRE MUST BE INCLUDED IN THE ESTIMATE. “HAPHAZARD”
ESTIMATE OR “GUESSTIMATE” CAN LEAD TO ERRORS OR FRAUD.

DEFORMED REINFORCING BARS ARE ROUND BARS AND SOMETIMES

CALLED “REBARS”. THEY ARE WIDELY USED FOR ORDINARY CONCRETE
CONSTRUCTION. THESE REINFORCING BARS ARE AVAILABLE IN A LARGE
RANGE OF DIAMETERS. THE MOST COMMON COMMERCIAL LENGTH IS 6m.
 PHILIPPINE STANDARD BARS
NOMINAL PHILIPPINE UNIT WEIGHT
DIAMETER (MM) STANDARD KG./METER
DESIGNATION
6 6 mm 0.222
8 8 mm 0.395
10 10 mm 0.616
12 12 mm 0.888
16 16 mm 1.579
20 20 mm 2.466
25 25 mm 3.854
28 28 mm 4.833
32 32 mm 6.313
36 36 mm 7.991
 DEVELOPMENT LENGTH OF MAIN BARS
DEVELOPMENT LENGTH IS DEFINED AS THE LENGTH OF EMBEDMENT
NECESSARY TO DEVELOP THE FULL TENSILE OR COMPRESSIVE
STRENGTH (FULL YIELD STRENGTH) OF THE STEEL BAR

DEVELOPMENT LENGTH IN TENSION IS USUALLY APPLIED TO BEAMS,

FOOTINGS, PILE CAPS, SHEAR WALLS AND RETAINING WALLS WHILE
DEVELOPMENT LENGTH IN COMPRESSION IS USED MAINLY IN COLUMNS
WHERE STEEL BARS TRANSFER COLUMN LOADS TO A SUPPORTING
FOOTING OR LOWER COLUMN.

IN THE CASE OF STRAIGHT BARS (BARS WITHOUT HOOKS), THE FORCE IS

TRANSFERRED FROM THE COLUMN TO THE FOOTING BY BOND ALONG THE
EMBEDMENT LENGTH, AND A PART OF IT BY END BEARING AT THE TIP OF
THE BARS ON THE CONCRETE FOR A COMPENSATION MEMBERS.
 ANCHORAGE LENGTH OF STRAIGHT BARS –
MAIN REINFORCEMENTS
A STRAIGHT BAR IS A STEEL BAR WITHOUT HOOK AT THE END. THERE ARE TWO
TYPES OF DEVELOPMENT LENGTHS, FOR MAIN REINFORCEMENT NAMELY:

a) DEVELOPMENT LENGTH IN TENSION

b) DEVELOPMENT LENGTH IN COMPRESSION

THE MOST BASIC FACTORS THAT INFLUENCE THE REQUIRED DEVELOPMENT

LENGTH OF REBARS ARE:

a) CONCRETE TENSILE STRENGTH

b) COVER DISTANCE
c) SPACING OF REINFORCING STEEL BARS
d) PRESENCE OF TRANSVERSE REINFORCEMENTS
 DEVELOPMENT LENGTH IN TENSION

*PROPER ANCHORAGE OF STEEL BARS SHALL BE REFERRED TO THE APPROVED STRUCTURAL PLANS
DEVELOPMENT LENGTH COMPUTATION (TENSION)
0.03 𝑑𝑏 𝑓𝑦
Ld= (English Units)
𝑓𝑐′
db = BAR DIAMETER (IN)
fy = YIELD STRENGTH (PSI)
fc’ = CONCRETE STRENGTH (PSI)

NOTE:

MINIMUM DEVELOPMENT LENGTH , 300 mm (12 INCHES)

 SAMPLE COMPUTATIONS
FOR 3/8 in (10 mm) , fc’ = 3,000 PSI , fy = 40,000 psi
3
0.03
8
(40000)
Ld= = 8.21 in (209 mm)
3000
NOTE:

MINIMUM DEVELOPMENT LENGTH , 300 mm (12 INCHES)

USE MINIMUM Ld = 300 mm

 SAMPLE COMPUTATIONS
FOR 5/8 in (16 mm) , fc’ = 3,000 PSI , fy = 40,000 psi

5
0.03
8
(40000)
Ld= = 13.69 in (348 mm)
3000

Ld = 348 mm

*THE DEVELOPMENT LENGTHS OF REBARS IN TENSION DECREASES AS THE STRENGTH OF CONCRETE

INCREASES
 DEVELOPMENT LENGTH IN COMPRESSION

*PROPER ANCHORAGE OF STEEL BARS SHALL BE REFERRED TO THE APPROVED STRUCTURAL PLANS
DEVELOPMENT LENGTH COMPUTATION (COMPRESSION)
0.02 𝑑𝑏 𝑓𝑦
Ld= ≥ 0.0003 x db x fy
𝑓𝑐′
(English Units)
db = BAR DIAMETER (IN)
fy = YIELD STRENGTH (PSI)
fc’ = CONCRETE STRENGTH (PSI)

NOTE:

MINIMUM DEVELOPMENT LENGTH , 200 mm (10 INCHES)

 SAMPLE COMPUTATIONS
FOR 3/8 in (10 mm) , fc’ = 3,000 PSI , fy = 40,000 psi
3
0.02
8
(40000)
Ld= = 5.48 in (140 mm)
3000
Vs.
0.0003 x 3/8 x 40000 = 4.5 in (114.3 mm)
NOTE:

MINIMUM DEVELOPMENT LENGTH , 200 mm (10 INCHES)

USE MINIMUM Ld = 200 mm

 SAMPLE COMPUTATIONS
FOR 5/8 in (16 mm) , fc’ = 3,000 PSI , fy = 40,000 psi

5
0.02
8
(40000)
Ld= = 9.12 in (232 mm)
3000
Vs.
0.0003 x 5/8 x 40000 = 4.5 in (190.5 mm)
Ld = 232 mm
*THE DEVELOPMENT LENGTHS OF COMPRESSION MEMBERS ARE SHORTER THAN TENSION MEMBERS
 ANCHORAGE OF TENSION BARS BY HOOKS – MAIN
REINFORCEMENTS
IN CASE WHEN THE DESIRED TENSILE STRESS OF A REINFORCING BAR CANNOT BE
DEVELOPED BY BOND ALONE ALONG THE DEVELOPMENT LENGTH, IT IS SPECIAL TO
PROVIDE ANCHORAGE AT THE ENDS OF THE BAR. THIS IS USUALLY DONE BY:

a) 90 DEGREE HOOKS
b) 180 DEGREE HOOKS

THE TOTAL LENGTH OF A STANDARD HOOK IS EQUAL TO THE LENGTH OF BEND

PLUS THE EXTENSION LENGTH AT THE FREE END OF THE BAR.
DIMENSIONS AND BEND RADII
TOTAL ANCHORAGE LENGTH

90 DEGREE HOOK

180 DEGREE HOOK

Ldh = Lds + Lh
 LENGTH OF EXTENSION (Lextn)
90 DEGREE HOOK

180 DEGREE HOOK

Lh = Lbend + Lextn
LENGTH OF BEND (Lbend) Lh = Lbend + Lextn

C= 2π x Radius
90 – Lbend = 0.25C= 0.25 (2 π ) (R + 0.5db)
90 – Lext = 12db

C= 2π x Radius
180 – Lbend = 0.5C= 0.50 (2π) (r + 0.5db)
90 – Lext = 4db or 65 mm(minimum)
 SAMPLE COMPUTATIONS
CALCULATE THE TOTAL ANCHORAGE LENGTH OF 20 MM DIAMETER
BAR WITH 90 DEGREE BEND IN A COLUMN HAVING A WIDTH OF
600 MM.

COMPUTE:
A. Lds
B. Lh (Lbend + Lextn)
C. Ldh (Lds + Lh)
S
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C
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 SAMPLE COMPUTATIONS
Note:
a. Lds r = 3 db for Ø10 to Ø25 mm
= 4 db for Ø28 to Ø36 mm

Lds = 600width – 50 – 20db – 3(20)3db

Lds = 600width – 50 – 20db – 3(20)3db

Lds = 470 mm
 SAMPLE COMPUTATIONS
Note:
b. Lh (Lbend + Lextn) r = 3 db for Ø10 to Ø25 mm
= 4 db for Ø28 to Ø36 mm

Lbend = 0.25(2π)(3db + 0.5db)

Lbend = 0.25(2π)[3(20) + 0.5(20)]
Lbend = 109.96 mm
Lextn = 12db
Lextn = 12(20) = 240 mm
Lh = 109.96 + 240 = 349.96 or 350 mm
 SAMPLE COMPUTATIONS
c. Ldh (Lds + Lh)

Ldh = 470 + 350 = 820 mm

 SAMPLE COMPUTATIONS
CALCULATE THE TOTAL ANCHORAGE LENGTH OF 25 MM DIAMETER
BAR WITH 180 DEGREE BEND IN A COLUMN HAVING A WIDTH OF
600 MM.

COMPUTE:
A. Lds
B. Lh (Lbend + Lextn)
C. Ldh (Lds + Lbend)
S
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C
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 SAMPLE COMPUTATIONS
Note:
a. Lds r = 3 db for Ø10 to Ø25 mm
= 4 db for Ø28 to Ø36 mm

Lds = 600width – 50 – 25db – 3(25)3db

Lds = 600width – 50 – 25db – 3(25)3db

Lds = 450 mm
 SAMPLE COMPUTATIONS
Note:
b. Lh (Lbend + Lextn) r = 3 db for Ø10 to Ø25 mm
= 4 db for Ø28 to Ø36 mm

Lbend = 0.5(2π)(3db + 0.5db)

Lbend = 0.5(2π)[3(25) + 0.5(25)]
Lbend = 274.89 mm
Lextn = 4db or 65 mm (min)
Lextn = 4(25) = 100 mm > 65 mm , use 100mm

Lh = 274.89 + 100 = 374.89 or 375 mm

 SAMPLE COMPUTATIONS
c. Ldh (Lds + Lh)

Ldh = 450 + 375 = 825 mm

 ANCHORAGE OF TRANSVERSE REINFORCEMENTS

SPECIAL MENTION MUST BE GIVEN TO PROPER ANCHORAGE OF STIRRUPS (FOR

BEAMS) AND TIES (FOR COLUMNS) IN ESTIMATING STEEL BARS. STIRRUPS AND TIES
ARE SOMETIMES CALLED WEB OR TRANSVERSE REINFORCEMENTS. IN A SHORT
WORD, THEY ARE KNOWN AS HOOPS. TO DEVELOP THE YIELD STRENGTH OF
STIRRUPS AND TIES, A 90-DEGREE OR 135 DEGREE HOOK MUST BE PROVIDED AT
BOTH ENDS.
STANDARD HOOKS
FOR STIRRUPS
AND TIES
 HOOP LENGTH FOR MAIN WEB BARS
MAIN WEB BAR OR REINFORCEMENT IS A STIRRUP OR A TIE LOCATED AT THE
PERIMETER OF THE BEAM OR COLUMN SECTION TIED TO THE MAIN LONGITUDINAL
REINFORCEMENTS. THE REQUIRED LENGTH OF HOOP FOR MAIN WEB
REINFORCEMENTS CAN BE CALCULATED FROM THE FORMULA,

P=Lms + 3Lb + 2Lh

WHERE:

p = perimeter length of hoop, mm

Lb = length of corner bend
Lh = length of hook at common end for anchorage
Lms = length of straight bar at the middle between corner bends
 MIDDLE STRAIGHT LENGTH, Lms
Lms = 2 (x + y)
WHERE:

x=b-k
y=h-k
k = 2 (dc + db + 4db)
db = bar diameter
dc = Concrete cover (40 mm minimum for ties, stirrups and spirals)
 Lms = 2 (x + y)

MIDDLE STRAIGHT LENGTH, Lms

 CORNER BEND, Lb
C = 2πr
Lb =(0.25)(2π)(4db +0.5db)
 LENGTH OF HOOK, Lh (Lb + Lextn)
C = 2πr
Lb =(0.2590deg or 0.375135deg)(2π)(4db +0.5db)

+
Lextn =
 SAMPLE COMPUTATIONS

ESTIMATE THE TOTAL LENGTH OF A STIRRUP FOR A 400 X 600 MM

BEAM USING 10 MM REBAR. USE MINIMUM COVER OF 40 MM AND
135 – DEGREE BEND FOR CLOSED STIRRUPS.
SAMPLE COMPUTATIONS
 SAMPLE COMPUTATIONS
1. Compute x and y

b – beam width = 400

h – beam height = 600
r – radius bend = 4db r= 4 x 10 = 40 mm
db – bar diameter (tie or stirrup) = 10 mm
dc – concrete cover = 40 mm (minimum)
k – constant = 2( dc + db + 4db )
 SAMPLE COMPUTATIONS
1. Compute x and y

k – constant = 2( dc + db + 4db )
= 2 (40 + 10 + 40 ) = 180 mm

x=b–k x = 400 – 180 = 220 mm

y=h–k y = 600 – 180 = 420 mm
 SAMPLE COMPUTATIONS
2. Compute total length of stirrup, p

P = Lms + 3Lb + 2Lh

Lms = 2 (x + y) = 2 (220 + 420) = 1280 mm
Lb = 0.25C = 0.25 (2π)(4db +0.5db)
Lb = 0.25C = 0.25 (2π)[4(10) +0.5(10)] = 70.69 mm
Lh = 0.375 (2π)[4db + 0.5db] + 6db
Lh = 0.375 (2π)[4(10) + 0.5(10)] + 6(10) = 166.03 mm
 SAMPLE COMPUTATIONS
2. Compute total length of stirrup, p

P = Lms + 3Lb + 2Lh

P = 1280 + 3(70.69) + 2(166.03)
P= 1824.13 mm
 SAMPLE COMPUTATIONS

ESTIMATE THE TOTAL LENGTH OF A STIRRUP FOR A 500 X 800 MM

BEAM USING 20 MM REBAR. USE MINIMUM COVER OF 40 MM AND
90 – DEGREE BEND FOR CLOSED STIRRUPS.
SAMPLE COMPUTATIONS
 SAMPLE COMPUTATIONS
1. Compute x and y

b – beam width = 500

h – beam height = 800
r – radius bend = 4db r= 4 x 20 = 80 mm
db – bar diameter (tie or stirrup) = 20 mm
dc – concrete cover = 40 mm (minimum)
k – constant = 2( dc + db + 4db )
 SAMPLE COMPUTATIONS
1. Compute x and y

k – constant = 2( dc + db + 4db )
= 2 (40 + 20 + 80 ) = 280 mm

x=b–k x = 500 – 280 = 220 mm

y=h–k y = 800 – 280 = 520 mm
 SAMPLE COMPUTATIONS
2. Compute total length of stirrup, p

P = Lms + 3Lb + 2Lh

Lms = 2 (x + y) = 2 (220 + 520) = 1480 mm
Lb = 0.25C = 0.25 (2π)(4db +0.5db)
Lb = 0.25C = 0.25 (2π)[4(20) +0.5(20)] = 141.37 mm
Lh = 0.25 (2π)[4db + 0.5db] + 12db
Lh = 0.25 (2π)[4(20) + 0.5(20)] + 12(20) = 381.37 mm
 SAMPLE COMPUTATIONS
2. Compute total length of stirrup, p

P = Lms + 3Lb + 2Lh

P = 1480 + 3(141.37) + 2(381.37)
P= 2666.85 mm
 CROSS TIES AND SUPPLEMENTARY TIES
FOR LARGE HEAVILY LOADED COLUMNS, CROSSTIES (OR CROSS BARS) AND
SUPPLEMENTARY TIES ARE USED TO RESIST SHEAR STRESS. FOR ESTIMATE
PURPOSES, CROSS TIE IS DEFINED AS SINGLE TIE BAR WITH 135 DEGREES HOOK AT
ONE END AND 90 DEGREE AT THE OTHER END.

SUPPLEMENTARY TIE IS ADDITIONAL CLOSED TIE PLACED AT THE MIDDLE PART OF

THE COLUMN CROSS SECTION ON TOP OF THE MAIN RECTANGULAR HOOP ALONG
THE PERIMETER OF THE COLUMN.

THE EXACT LENGTH AND WEIGHT OF CROSS TIES AND SUPPLEMENTARY TIES ARE
IMPORTANT IN ESTIMATE OF TRANSVERSE REINFORCEMENTS FOR COLUMNS
BECAUSE OF THEIR CONSIDERABLE QUANTITY.
CROSS TIES AND SUPPLEMENTARY TIES
 CROSS TIES
A CROSS TIE IS A CONTINUOUS REINFORCING BAR WITH A SEISMIC HOOK ( WITH A
BEND NOT LESS THAN 135 DEGREES AND A 6 BAR DIAMETER EXTENSION (6DB) AT
ONE END AND A HOOK WITH NOT LESS THAN 90 DEGREES AND AT LEAST 6 BAR
DIAMETER EXTENSION AT THE OTHER END. THE LENGTH OF A CROSS TIE CAN BE
COMPUTED FROM,

Lcr = L135 + Lm + L90

Where:
LCR – length of cross tie
L135 – length of 135 – degree hook,mm
Lm – length of middle straight bar, mm
L90 – length of 90 degree hook, mm
 CROSS TIES

Lm = h – [2dc + db + 2(4db)]

Where:
h – column dimension under consideration
dc – concrete cover
db – bar diameter of crosstie
4db – radius of bend
 CROSS TIES

Lm = h – [2dc + db + 2(4db)]
 SAMPLE COMPUTATIONS
DETERMINE THE TOTAL LENGTH OF 10MM CROSS TIE FOR 300 X 400
COLUMNS PARALLEL TO THE LONG DIMENSION.
 SAMPLE COMPUTATIONS
a. L135= Lbend + Lextn
Lb = (135/360)C = 0.375(2π)(4db +0.5db)

Lb = (135/360)C = 0.375(2π)[4(10) +0.5(10)] = 106.03 mm

Lextn = 6db = 6(10) = 60 mm

a. L135= 106.03 + 60 = 166.03 mm

 SAMPLE COMPUTATIONS
b. L90= Lbend + Lextn
Lb = (90/360)C = 0.25(2π)(4db +0.5db)

Lb = (90/360)C = 0.25(2π)[4(10) +0.5(10)] = 70.69 mm

Lextn = 6db = 6(10) = 60 mm

b. L90= 70.69 + 60 = 130.69 mm

 SAMPLE COMPUTATIONS
c. Lm = h – [2dc + db + 2(4db)]

Lm = 400 – [2dc + db + 2(4db)]

Lm = 400 – [2(40) + 10 + 2(4)(10)]

Lm = 400 – [2(40) + 10 + 2(4)(10)]

Lm = 230 mm
 SAMPLE COMPUTATIONS

Lcr = L135 + Lm + L90

Lcr = 166.03 + 230 + 130.69
Lcr = 526.72 mm
 SUPPLEMENTARY TIES
SUPPLEMENTARY TIE IS AN ADDITIONAL WEB REINFORCEMENT USED TO RESIST
SHEAR STRESSES IN A STRUCTURAL COLUMN. THEY ARE TYPICALLY BARS BENT
INTO RECTANGULAR SHAPES AND LOCATED PERPENDICULAR TO LONGITUDINAL OR
MAIN REINFORCEMENTS. A LENGTH OF SUPPLEMENTARY TIE IS DETERMINED FROM,

Lst = Lms + 3Lb + 2Lh

 SUPPLEMENTARY TIES
Lst = Lms + 3Lb + 2Lh
Where:
Lst – length of supplementary tie, mm
Lms – middle straight length = 2(xc + y)
xc – center to center distance between outer longitudinal bars to be
enclosed by the supplementary tie
xc = [h-(2 x concrete cover + db)] / n
n = number of bar spaces
y = b – k (b is the shorter dimension of the column)
k = 2 (dc + db + 4db)
Lb , length of corner bends = 0.25C= 0.25(2π)(4db + 0.5db)
Lh = length of two hooks at common end of anchorage
db = diameter of supplementary ties
h – longer column dimension
 SAMPLE COMPUTATIONS
A. WITH 90 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

Lst = Lms + 3Lb + 2Lh

 SAMPLE COMPUTATIONS
A. WITH 90 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

 SAMPLE COMPUTATIONS
A. WITH 90 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

Lms = 2(xc + y) y=b–k dc = 40 mm

db = 10 mm
Xc = [400 – (2x40 +10)]/3 = 103.3 mm
k = 2(dc + db +4db) = 2[40 + 10 +4(10)] = 180 mm
y = 200 – 180 = 20
Lms = 2(103.3 + 20) = 246.6mm
Lb = 0.25(2π)(4db + 0.5db) = 0.25(2π)[4(10) + 0.5(10)] =70.69mm
Lh for 90 degree hook = 0.25(2π)[4db + 0.5db] + 6db
Lh for 90 degree hook = 0.25(2π)[4(10) + 0.5(10)] + 6(10) = 130.69 mm
 SAMPLE COMPUTATIONS
A. WITH 90 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

Lst = 246.6 + 3(70.69) + 2(130.69)

Lst =720.0 mm
 SAMPLE COMPUTATIONS
B. WITH 135 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

Lst = Lms + 3Lb + 2Lh

 SAMPLE COMPUTATIONS
B. WITH 135 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

 SAMPLE COMPUTATIONS
B. WITH 135 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

Lms = 2(xc + y) y=b–k dc = 40 mm

db = 10 mm
Xc = [400 – (2x40 +10)]/3 = 103.3 mm
k = 2(dc + db +4db)
y = 200 – 180 = 20
Lms = 2(103.3 + 20) = 246.6mm
Lb = 0.25(2π)(4db + 0.5db) = 0.25(2π)[4(10) + 0.5(10)] =70.69mm
Lh for 135 degree hook = 0.375(2π)[4db + 0.5db] + 6db
Lh for 135 degree hook = 0.375(2π)[4(10) + 0.5(10)] + 6(10) = 166.03
mm
 SAMPLE COMPUTATIONS
A. WITH 90 – DEGREE HOOKS

FOR 200 X 400 MM WITH 10MM DIAMETER SUPPLEMENTARY TIE

Lst = 246.6 + 3(70.69) + 2(166.03)

Lst =790.73 mm
 SPIRALS
CIRCULAR SPIRALLY REINFORCED COLUMNS ARE NEEDED IN BUILDINGS, BRIDGES
AND IN BORED PILES. MAIN LONGITUDINAL STEEL BARS IS CIRCULAR COLUMNS
ARE HELD TOGETHER BY CONTINUOUS TRANSVERSE REINFORCEMENTS CALLED
SPIRALS. THE LENGTH OF SPIRAL, Ls PER TURN OR ONE ROUND CAN BE COMPUTED
BY THE FORMULA,

Where:
DS = diameter of turn of the spiral in horizontal projection
DS = D – 2 (dc + 0.5db )
dc = concrete cover πDs
db = diameter of spiral LS= 𝐶𝑂𝑆β
β = angle of inclination of the spiral
− 𝟏 0.50 Ss
β = 𝑡𝑎𝑛 Ds
Ss = pitch of the spiral, mm (center to center
distance between two spirals)
 SPIRALS

THE TOTAL NUMBER OF TURNS OF SPIRALS CAN BE COMPUTED

FROM

H
Nr = Ss
WHERE H IS HEIGHT OF CIRCULAR COLUMNS
 SPIRALS
THE TOTAL LENGTH OF SPIRAL, Lst EXCLUDING SPLICES CAN BE
OBTAINED FROM THE EXPRESSION,

LST = Nr x Ls

WHEN THE TOTAL SPIRAL LENGTH, LST EXCEEDS THE AVAILABLE

STANDARD COMMERCIAL LENGTH OF REINFORCING STEEL BARS,
SPLICES MUST BE CONSIDERED. LENGTH OF SPIRAL SPLICES WILL
BE DISCUSSED IN THE NEXT SECTION.
 SAMPLE COMPUTATIONS
ESTIMATE THE TOTAL LENGTH OF 12 MM DIAMETER SPIRAL FOR A
600 MM CIRCULAR REINFORCED WITH 8-25 MM DIAMETER
LONGITUDINAL BARS. THE HEIGHT OF THE COLUMN IS 3.00 m AND
THE SIZE OF THE SPIRAL IS 12 MM WITH 75 MM PITCH. ASSUME A
40 MM CONCRETE COVER AND A SPLICE LENGTH OF 300 MM FOR
SPIRAL.
 SAMPLE COMPUTATIONS
a. Length of Spiral, LS per turn
Ds = D – 2 (dc + 0.5db)
Ds = 600 – 2 (40 + 0.50 x 12)
Ds = 508 mm

− 𝟏 0.50
Ss
β = 𝑡𝑎𝑛 Ds
− 𝟏 0.50x75
β = 𝑡𝑎𝑛 508

β = 4.22 degree
 SAMPLE COMPUTATIONS
a. Length of Spiral, LS per turn
Ds = D – 2 (dc + 0.5db)
Ds = 600 – 2 (40 + 0.50 x 12)
Ds = 508 mm

− 𝟏 0.50
Ss
β = 𝑡𝑎𝑛 Ds
− 𝟏 0.50x75
β = 𝑡𝑎𝑛 508

β = 4.22 degree
 SAMPLE COMPUTATIONS
β = 4.22 degrees
a. Length of Spiral, LS per turn Ds = 508 mm

πDs
LS=
COSβ
π x 508
LS=
COS4.22

LS= 1600 mm
 SAMPLE COMPUTATIONS
b. TOTAL NUMBER OF TURNS
H = 3 X 1000
Nr = = 40
Ss 75
 SAMPLE COMPUTATIONS
C. TOTAL NUMBER LENGTH OF SPIRALS, LST

Lst = NR X Ls = 40 x 1.60 = 64 mm
 SAMPLE COMPUTATIONS
ASSUMING 6m LENGTH STANDARD REBAR AND SPLICE LENGTH OF
0.30 M, EFFECTIVE LENGTH IS 6.0 – 0.30 (SPLICE ASSUMED) = 5.70 m.
HENCE, THE TOTAL NUMBER OF 6m REBAR IS:

n = L/Bar Effective Length

n = 64/5.7 = 11.22 pcs
= 11 pcs - 12mm x 6.0m
= 1 pc - 12mm x 1.25m (0.22 pc x 5.7m)
 BAR SPLICES

REINFORCING STEEL BARS ARE SOLD BY SUPPLIERS IN

VARIOUS COMMERCIAL LENGTHS FROM 6 TO 13.5 METERS.
USUALLY, 6m LENGTH BAR IS THE MOST WIDELY AVAILABLE
WHICH REQUIRES SHORT VEHICLES TO TRANSPORT. 6M
LENGTH BARS ARE MORE CONVENIENT FOR STRUCTURAL
MEMBERS NOT MORE THAN 6 METERS.
 BAR SPLICES
LOCATION OF SPLICES IN REINFORCEMENT ARE USUALLY
INDICATED IN THE STRUCTURAL DETAILED PLANS. SPLICES ARE
USUALLY MADE BY LAPPING THE BARS WITH SUFFICIENT LENGTH
T TO TRANSFER STRESS BY BOND FROM ONE BAR TO THE OTHER.
THE LAPPED BAS ARE HELD TOGETHER BY TIE WIRES SO THAT
THEY STAY IN POSITION AND IN CONTACT DURING CONCRETE
POURING. THERE ARE THREE CASES IN BAR SPLICING:

a. LAP SPLICES IN TENSION

b. LAP SPLICES IN COMPRESSION
c. LAP SPLICES IN COMPRESSION AND BENDING (BIAXIAL)
DEVELOPMENT LENGTH COMPUTATION (TENSION)

0.03 𝑑𝑏 𝑓𝑦
Ld= (English Units)
𝑓𝑐′

NOTE:

MINIMUM LAP SPLICE LENGTH IS 300 mm (12 INCHES)

 LAP SPLICES IN TENSION
THE REQUIRED LENGTH OF LAP FOR TENSION SPLICES IS
DETERMINED IN TERMS OF THE DEVELOPMENT LENGTH, Ld IN
TENSION. A BEAM WITH SPLICES SHOULD BE MADE AS DUCTILE
AS THE ONE WITHOUT SPLICES. SPECIAL PROVISIONS ARE
INTENDED TO ASSURE THAT NO SPLICE FAILUR WILL OCCUR
WHEN THE FULL NOMINAL STENGTH IN FLEXURE IS REACHED AT
THE SPLICE LOCATION.
FOR PURPOSES OF ESTIMATE THE SPLICE LENGTH IN TENSION:

Lspt= 1.3Ld ≥ 300 mm

Lap splice is not applicable for bars larger than 36mm diameter
(NSCP 2010).
LAP SPLICES IN TENSION

RC BEAM BAR SPLICES

 LAP SPLICES IN TENSION
FOR SINGLE BAR Lspt= 1.3Ld

FOR BUNDLED BAR

Lspt= 1.3Ld x 1.20

Lspt= 1.56Ld For 3-bar bundle &

2-bar bundle

Lspt= 1.3Ld x 1.33

Lspt= 1.73Ld For 4-bar bundle

BUNDLED BARS (TENSION)
 LAP SPLICES IN TENSION

WHEN BARS OF DIFFERENT SIZE ARE LAP SPLICED IN

TENSION, SPLICE LENGTH SHALL BE THE LARGER OF:

a. Ld OF THE LARGER BAR

b. TENSION LAP SPLICE LENGTH OF THE SMALLER BAR.
 SAMPLE COMPUTATIONS

CALCULATE THE LAP LENGTH FOR A 36 MM

DIAMETER BAR IN A BEAM WITH FY = 410 MPA
(60000PSI) AND FC’ = 27.5 MPA (4000 PSI).
 SAMPLE COMPUTATIONS
a. CALCULATE DEVELOPMENT, Ld
0.03 𝑑𝑏 𝑓𝑦
Ld= (English Units)
𝑓𝑐′

36
0.03 x ( in) x 60000 psi
Ld= 25.4
4000 𝑝𝑠𝑖

=40.34 in (1.024 m)
 SAMPLE COMPUTATIONS
b. COMPUTE LAP LENGTH, LSPT

LSPT= 1.3Ld

LSPT= 1.3 x 1024 = 1.331 m

 SAMPLE COMPUTATIONS

A BEAM WITH FY = 410 MPA (60000PSI) AND

FC’ = 27.5 MPA (4000 PSI) HAS A BAR
DIAMETER OF 25 MM AT SUPPORT AND 20
MM AND MIDSPAN. WHAT IS THE SPLICE
LENGTH?
 SAMPLE COMPUTATIONS

WHEN BARS OF DIFFERENT SIZE ARE LAP SPLICED IN TENSION,

SPLICE LENGTH SHALL BE THE LARGER OF:

a. Ld OF THE LARGER BAR

b. TENSION LAP SPLICE LENGTH OF THE SMALLER BAR.
 SAMPLE COMPUTATIONS

A. LD OF THE LARGER BAR

25
0.03 𝑥 60000
Ld= 25.4
= 28.01 in (711.51 mm)
4000
 SAMPLE COMPUTATIONS

B. SPLICE LENGTH OF SMALLER BAR, LSPT = 1.3 LD

20
0.03 𝑥 60000
1.3Ld= 1.3 𝑥 25.4
4000
= 29.13 in (739.97 mm)
 SAMPLE COMPUTATIONS

C. Ldbig VS Lsptsmall

711.51 mm vs 739.97 mm
711.51 mm < 739.97 mm

We choose the bigger value so, Lspt = 739.97 mm

 LAP SPLICES IN COMPRESSION
REINFORCING STEEL BARS IN COMPRESSION ARE SPLICED
MAINLY IN COLUMNS. EMBEDMENT LENGTH LENGTH FOR
COMPRESSION STEEL IS SHORTER THAN THOSE REQUIRED FOR
TENSION BARS BECAUSE OF THE PRESENCE OF BEARING AT THE
ENDS OF THE BARS WHICH HELPS DEVELOP THE LOAD.
 LAP SPLICES IN COMPRESSION

COLUMN MIDHEIGHT BAR SPLICES BEAM – COLUMN BAR SPLICES

 LAP SPLICES IN COMPRESSION
0.02 𝑑𝑏 𝑓𝑦
Ld= ≥ 0.0003 x db x fy
𝑓𝑐′
(English Units)
NOTES:
MINIMUM LAP LENGTH FOR ESTIMATE PURPOSES ONLY

Lsc= 1.3Ld ≥ 300 mm

 LAP SPLICES IN COMPRESSION

WHEN BARS OF DIFFERENT SIZE ARE LAP SPLICED IN

TENSION, SPLICE LENGTH SHALL BE THE LARGER OF:

a. Ld OF THE LARGER BAR

b. TENSION LAP SPLICE LENGTH OF THE SMALLER BAR.
 LAP SPLICES IN COMPRESSION
THE NSCP SPECIFIES THAT FOR CONCRETE STRENGTHS LESS
THAN 20.70 MPA (3000 PSI), THE COMPUTED LAP SPLICES
SHALL BE INCREASED BY 1/3 (33%). SINCE IT’S DIFFICULT TO
DETERMINE THE TENSILE STRESSES IN COLUMNS (BIAXIAL
BENDING) AND BEYOND THE SCOPE OF WORK OF THE
ESTIMATOR, THE LAP SPLICE LENGTHS LSC IN COMPRESSION
MEMBERS ARE ASSUMED EQUAL TO 1.3 Ld FOR ESTIMATE
PURPOSES AND SAFETY.
 REINFORCING STEEL BARS FOR FOOTINGS
FOR CONCRETE SURFACES TO BE IN CONTACT WITH GROUND,
A PROTECTIVE COVERING OF AT LEAST 50MM (2 INCHES) IS
REQUIRED WHEN THE USE OF FORMS IS NECESSARY AS IN
THE SIDES OF FOOTINGS. IF THE CONCRETE IS POURED IN
DIRECT CONTACT WITH THE GROUND WITHOUT THE USE OF
FORMS, AS IN THE BOTTOM SIDES OF FOOTINGS, A CONCRETE
COVERING OF AT LEAST 75 mm (3 INCHES) MUST BE
PROVIDED. BAR SPLICING IN ORDINARY FOOTINS IS SELDOM
ENCOUNTERED . BUT LAP SPLICE LENGTHS FOR MAT AND
GRID FOOTINGS CAN BE COMPUTED BASED ON THE PREVIOUS
SECTIONS REGARDING BEAM SPLICES.
REINFORCING STEEL BARS FOR FOOTINGS
 SAMPLE COMPUTATIONS
A 1.50M X 1.50M FOOTING IS REINFORCED WITH 25MM BARS SPACED
AT 0.20 M IN BOTH DIRECTIONS. ESTIMATE THE TOTAL AMOUNT OF
REINFORCING STEEL BARS. A 180 DEGREE BEND IS SPECIFIED.
 SAMPLE COMPUTATIONS

A. DETERMINE THE HALF LENGTH (FOR SIMPLIFIED CALCULATION

Where:
Lx=Ly = 0.50L – (Dpc + db + 3db) Dpc = protective cover
3db = radius of bend for 25 mm diameter
Lx=Ly = 0.50(1500) – [50 + 25 + 3(25)] db = bar diameter

Lx=Ly = 750 – 150 = 600 mm

 SAMPLE COMPUTATIONS

B. Length of hook Lh

Lh = Lb + L extn
Lb = 0.5(2π)(3.5db)
Lb = 0.5(2π)[(3.5(25)]= 274.89mm
Lextn = 4db
Lextn = 4 x 25 = 100 mm

Lh = 274.89 + 100 = 154.98 = 374.89 = 375 mm

 SAMPLE COMPUTATIONS

C. TOTAL LENGTH OF FOOTING BAR

Lf = 2(Lx/y + Lh)
Lf = 2(600 + 375) = 1950 mm or 1.95 m

USE LF = 2 METERS TO INCLUDE WASTE

 SAMPLE COMPUTATIONS

D. TOTAL NUMBER OF FOOTING BAR

ALONG X AND Y DIRECTION (SINCE IT IS A SQUARE FOOTING

Nx = Ny = [(1.50 – 0.10)/20] + 1
Nx = Ny = 8 pcs – 25mm x 2.0 m

Ntotal = Nx + Ny = 16 pcs – 25mm x 2.0 m