Perancangan Struktur Beton Bab_3

1ENCV600101 - Perancangan Struktur Beton 1

Ch 3 Lentur: Balok dengan tulangan tekan

Sjahril A. RahimDepartemen Teknik Sipil FTUI

2014

Pengaruh tulangan tekan pada Kekuatan dan Perilaku

b

h

As

b

h d

As

As'd'

d

a1 c1 Cc1

T=Asfy

j1d=(d-a/2)

j2d

Cs

Cc2C

T=Asfy

a2 c1c2 s

cu

cu

a2 < a1Dgn tulangantekan

Tanpa tulangan tekan

s

s

Perbandingan

Tulangan tarik As Tulanga tekan As=0 Cc1=T T=Asfy Cc1 > Depth a1 > Lengan momen: j1d Mn=Asfy(j1d) j1 little >

Tulangan tarik As Tulangan tekan As C=Cc2+Cs=T T=Asfy Cc2 a2 Lengan momen: j2d Mn=Asfy (j2d) j2

Addition of comp. steel little effect on the usable ultimate moment

Peningkatan kapasitas momen akibat tulangan tekan

2Alasan untuk menyediakan tulangan tekan

Mengurangi defleksi beban yang berkelanjutan

Peningkatan daktilitas Perubahan kegagalan mode dari keruntukan

tekan menjadi keruntuhan tarik Memudahkan fabrikasi

Mengurangi defleksi beban yang berkelanjutan

Meningkatkan daktilitas Merubah moda keruntuhan dari tekan ke tarik

3Analysis of Beams with tension and Compression reinforcement

As

d=dt

d'

h

b

As'

As1

d=dt

d'As'

As2

d=dt(d-d') (d-a/2)

Cc

Cs

Cc

Cs

T1=As1fy T2=As2fy

0.003 0.85fc'

a/2fs'

fs=fyT

+a

0.85fc'

a/2

c a

sy

Analysis of Beams with tension and Compression reinforcement

The strain, stresses distribution and internal forces see figure b,c and d.The cross section was hypothetically divided in to two beams;Beam 1, consisting of the compression reinforcement at the top and sufficient at the bottom so that T1 = Cs, and beam 2, consisting of the concrete web and the remaining tensile reinforcement, as shown in Figure e and f.The strain and stress in the compression reinforcement:

003,0''

c

dcs sss Ef

If ys ' then ys ff '

s

yy E

fwhere

Case 1: Compression steel yields

The beam can be divided in to two imaginary beams, each with C=T:Beam 1: Consist of reinforcement in tension and compression and resists moment as steel force couple.

''1

'1

1'

1

ddfAMAA

fAfA

TC

yss

ss

ysys

s

As1

d=dt

d'As'

(d-d')

Cs=As'f's

T1=As1fy

Beam 1


Beam 2: Consist of the concrete plus the remaining steel:

bafCAA

ffAAA

cc

ss

ys

sss

'

'1

12

85,0

Since C=T for beam 2, where T=(As-As)fy, the depth of the compression stress block, a, is

bfcfAA

a yss'85,0

'

As2

d=dt (d-a/2)

Cc

T2=As2fy

a

0.85fc'

a/2

4Case 1: Compression steel yields

The nominal moment capacity of beam 2 is:

2

)( '2adfAAM yssn

The total moment capacity of a beam with compression steel is

2

)()'( '' adfAAddfAM yssysn


Correction due to concrete displaced by steel compression:

285,01'85,01 ''

'' adffAAddf

ffAM ycssy

y

csn

y

css f

fAA'

'1

85,01

and

bfc

ff

fAA

ay

y

css

'85,0

85,01'

'

where

Determination of Whether fs=fy in Compression Reinforcement

Tulangan tekan leleh, jika:

003,0''

c

dcs sss Ef

''

ys ff 'kemudian

MPa

MPafEf y

s

yys 000.200

'

Untuk mengkonfirmasi bahwa tulangan tekan leleh, tunjukkan

Determination of Whether fs=fy in Tension Reinforcement

Dengan kedalaman sumbu netral diketahui, asumsi kelelehan ketegangan baja tulangan dapat diperiksa. Dari segitiga serupa dalam distribusi regangan linear pada gambar, ekspresi berikut dapat diturunkan:

cuscus

ccdsehingga

ccd

Untuk mengkonfirmasi tulangan tarik leleh, tunjukkan

MPa

MPafEf y

s

yys 000.200

As

d=dt

d'

h

b

As'

0.003

c s

5Case 1: Compression steel does not yields

If the compression reinforcement does not yield, fs is not known, and different solution is required. Assuming that tensile steel yields, the internal forces in the beam are:

bafC

fAT

cc

ys

'85,0

Ignoring the correction to the compressive stress in the steel:'' )( ssss AEC 003,0'1 1'

ad

s

where

From the equilibrium:

ysssc

sc

fAadAEbaf

TCC

003,0'185,0 1''

Case 1: Compression steel does not yields

This can be reduced to the quadratic equation in a, given by:

0)'003,0()003,0()85,0( 1''2' dAEafAAEabf ssysssc

Once the depth of the stress block, a, is known, the nominalmoment capacity of the section is

)'(2

ddCadCM scn

Determination of Whether fs=fy in Tension Reinforcement

Dengan kedalaman sumbu netral diketahui, asumsi kelelehan ketegangan baja tulangan dapat diperiksa. Dari segitiga serupa dalam distribusi regangan linear pada gambar, ekspresi berikut dapat diturunkan:

cuscus

ccdsehingga

ccd

Untuk mengkonfirmasi tulangan tarik leleh, tunjukkan

MPa

MPafEf y

s

yys 000.200

As

d=dt

d'

h

b

As'

0.003

c s

Pengecekan regangan tarik net, tb

h ddt

d'

As

As' c

Dengan kedalaman sumbu netral diketahui, strain regangan tarik net pada tulangan tarik terjauh dapat di ceck,. Dari segitiga serupa dalam distribusi regangan linear pada gambar, ekspresi berikut dapat diturunkan:

cut

tcu

t

t

ccdsehingga

ccd

t

Untuk menjamin potongan balok daktail, pastikan bahwa regangan tarik net:t 0.004

untuk memenuhi SNI 2847:2013, Pasal 10.3.5

6Upper Limit on Reinforcement in Beams

Under-reinforced beams fails in a ductile manner Over-reinforced beams in a brittle manner SNI section 10.3.5 attempts to prevent non-ductile failures by limiting the strain t at nominal flexural strength condition shall be greater than or equal to 0,004:

004,0t

new

As

d=dt

d'

h

b

As'Cs

Cc

0.003 0.85fc'

a/2fs'

fs=fyT

c a

s=0.004

Upper Limit on Double Reinforcement in Beams

Strain compatibility: tscu

cu dc

tcu

cu dc004.0

Equilibrium: C=T

ysy

s

ys

cu

cu

y

c

yycsc

ffifff

ffif

ff

bdbdfbdfbcfCabf

''

max

'max

'

1max

max'

1'

'

'

004.085.0

''85.085.0

Tulangan minimum pada komponen struktur lentur

ACI 318/SNI 2847:2013 Pasal 10.5.1 menentukan luas minimum tulangan longitudinal berikut untuk potongan balok yang di lentur positif,

y

ww

y

cs f

dbdbf

fA 4.1

25.0 'min,

fc dan fy dalam Mpa.bw = lebar badan balok, d = tinggi efektif balok

(10.3)

Ties for Compression Reinforcement

As the ultimate load is approach the compression steel in a beam may buckle, causing the surface layer of concrete to spall off, possibly leading to failure. For this reason it is necessary to enclose compression reinforcement with closed stirrup or ties. Please refer to SNI 9.11 for details.

7Analysis procedure of a Beam with Compression Reinforcement: Compression

reinforcement yields Assuming that fs=fy and fs=fy and divide the beam

into two components Compute a for beam 2 Check if compression reinforcement steel yields Check if fs=fy and whether the section is tension-

controlled Check if As As,min Compute Mn

Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields

Assume fs=fyfs=fy

'1 ss AA 12 sss AAA

bffAA

ac

yss'

'

85,0)(

1

Beam 1 Beam 2Y

cus cdc

''

1ac


1

ys ff '

Y2

Check yielding ofcompression reinforcement ys

'

s

yy E

f

cus ccd

ys


2

yy

cs f

bdbdff

A 4,14

'

min,

min,ss AA Increase AsN

YBeam 1 Beam 2

Check for minimum reinforcement

''1 ddfAM ysn

2

)( '2adfAAM yssn

21 nnn MMM 3

8Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields

cut

t ccd

3

t0.005

0.005

9Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement Does

Not Yields3

''

'

'85,0

ssss

cus

cc

AECc

cdbafC

)'(2

ddCadCM scn

4

Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement Does

Not Yields

cut

t ccd

4

t0.005

0.005

10

Latihan 2 Analisis

d'

b

dh

2D28 mm

6D28 mm

Figure 1

Case 2: d=80 mm

Rectangular beam with compression reinforcementCompute Mn for the beam shown in Figure 1, where fc=20 MPa, fy=400 MPa, b=300 mm, d=810 mm, dt= 935 mm, d=80 mm and h=900 mm.

Design procedure of a Beam with Compression Reinforcement:

1. Calculate the maximum moment that can be resisted by the undereinforced section with =max, or 0.005 to ensure =0.90. The corresponding tensile area is As=bd, and Mn=Asfy(d-a/2), with a=(Asfy)/(0.85fcb)

2. Find the excess moment, if any, that must be resisted, and set M2=Mn, as calculated in step 1, M1=Mu/-M2. Now As from step 1 is defined as As2, ie., that part of the tension steel area in the double reinforced beam that works with the compression force in the concrete, As-As=As2.

3. Tentatively assume that fs=fy. Then As=M1/(fy(d-d). Alternatively, if the compression reinforcement is known not to yield, go to step 6.

4. Add an additional amount of tensile steel As1=As. Thus, the total tensile steel reinforcement area As is As2 from step 2 plus As1.

5. Analyze the doubley reinforced beam to see if fs=fy; that is, check the tensile reinforcement ratio against cy.


6. If

11

M2=Mn

21 MMM u

ss AA 2

sy

6

1

c=a/1

cus cdc

''

f s=fy

)'(1'

ddfMA

ys

As1=As

As=As1+As2

2


yn


Assume fs=fyfs=fy

'1 ss AA 12 sss AAA

bffAA

ac

yss'

'

85,0)(

3

Beam 1 Beam 2Y

cus cdc

''

1ac

2


3

ys ff '

Y4

Check yielding ofcompression reinforcement ys

'

s

yy E

f

cus ccd

ys


4

yy

cs f

bdbdff

A 4,14

'

min,


YBeam 1 Beam 2


''1 ddfAM ysn

2

)( '2adfAAM yssn

21 nnn MMM 5

12


cut

t ccd

5

t0.005

0.005

13


Not Yields

9

Y

10

y

ww

y

cs f

dbdbff

A 4,14

'

min,


Y


Check for tension control

cus ccd

ys

1ac


Not Yields10

''

'

'85,0

ssss

cus

cc

AECc

cdbafC

)'(2

ddCadCM scn

11


Not Yields

cut

t ccd

11

t0.005

0.005

14

Daftar referensi

1. James K Wight, James McGregor : Reinforced Concrete, Mechanics and Design, Sixth Edition, Pearson, 2012

2. ________________, Persyaratan beton struktural untuk bangunan gedung, SNI 2847:2013, Badan Standarisai nasional

3. ________________, Building Code requirements for Structural Concrete, ACI 318-2011, American Concrete Institute

Alternative determination of Whether fs=fy in Tension Reinforcement

The tension steel will yield if a tension or balanced failure occurs. If fs=fy, the balance conditions:

yc

yysss

fbdffffAA

600600

'85,0/

1

''

Substituting =As/bd and =As/bd gives

yy

c

by

s

fff

ff

60060085,0' '1

'

ys

bysys

ffffff

)/'()/'( '' If then

Old

Upper Limit on Tension Reinforcement in Beams with Compression Steel

b)'(75,0)'(

where

yy

cb ff

f600

60085,0''

1

yy

c

by

s

fff

ff

60060085,0' '1

'

or

Old

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Perancangan Struktur Beton Bab_3